Algebra I_Solutions Manual

January 26, 2017 | Author: Hector I. Areizaga Martinez | Category: N/A
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Solutions Manual

Copyright © by The McGraw-Hill Companies, Inc. All rights reserved. Printed in the United States of America. Permission is granted to reproduce the material contained herein on the condition that such material be reproduced only for classroom use; be provided to students, teachers, and families without charge; and be used solely in conjunction with Glencoe Algebra 1. Any other reproduction, for use or sale, is prohibited without prior written permission of the publisher. Send all inquiries to: Glencoe/McGraw-Hill 8787 Orion Place Columbus, OH 43240-4027 ISBN: 0-07-827752-3 1 2 3 4 5 6 7 8 9 10

009

07 06 05 04 03 02

Algebra 1 Solutions Manual

Contents Chapter 1

The Language of Algebra . . . . . . . . . . . . . . . . . . . . . . . 1

Chapter 2

Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Chapter 3

Solving Linear Equations . . . . . . . . . . . . . . . . . . . . . . 62

Chapter 4

Graphing Relations and Functions . . . . . . . . . . . . . . . . . 122

Chapter 5

Analyzing Linear Equations . . . . . . . . . . . . . . . . . . . . 191

Chapter 6

Solving Linear Inequalities . . . . . . . . . . . . . . . . . . . . . 244

Chapter 7

Solving Systems of Linear Equations and Inequalities . . . . . . 305

Chapter 8

Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364

Chapter 9

Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400

Chapter 10

Quadratic and Exponential Functions . . . . . . . . . . . . . . . 438

Chapter 11

Radical Expressions and Triangles . . . . . . . . . . . . . . . . . 491

Chapter 12

Rational Expressions and Equations . . . . . . . . . . . . . . . . 533

Chapter 13

Statistics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 578

Chapter 14

Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 601

Prerequisite Skills . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633 Extra Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 643 Mixed Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766

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Chapter 1 Page 5

The Language of Algebra 1-1

Getting Started

1. 8  8  64 4. 23  6  138 6. 68  4 

68 4

2. 4  16  64 3. 18  9  162 57 5. 57  3  3  19  17 7.

72 3

 24

8.

90 6

Page 8

 15

1 32 11  41 8 2  4 18



7. 44  4  4  4  4 4  256 8. the product of 4 and m to the fourth power 9. one half of n cubed 10. Lorenzo will receive the difference of $20 and p the cost of the peanuts. The word difference implies subtract, so the expression can be written as 20  p.

11 2 1 52

1

The perimeter is 5 2 feet. 12. P  2l  2w

1

5

2

1

1

 2 42 8  2 25 4    

10 2 84 8  50 4 2 2 85 8  50 4 1 2 85 4  50 4 3 135 4

2

Pages 8–9 3

1

14.

1.2 6 7.2

1.8 15. 1.83.2  4 1 8 14 4 1 4 4 0 17.

3 4



4

3.9  0.5 1.95

7.6 16. 1.4  10.6 4 9 8 84 8 4 0

3

12 1

3 4



 12  4 

2

3

19.



5

3

1

12 1

 34 1

9

5 16

3

18. 1 3  4  3  4 3

1

5

5

1

4

9

 14

9 12



1

5 16

12  9

20.

5 6



2 3



5 6 2

1

18. One-half implies multiply by 2, and cube implies raise to the third power. So the expression can be 1 written as 2 n3.

1

3 2

3

5

 16  4

15

 36 

12 9

Practice and Apply

11. The word sum implies add, so the expression can be written as 35  z. 12. Sample answer: Let x be the number. The word sum implies add, so the expression can be written as x  7. 13. The word product implies multiply, so the expression can be written as 16p. 14. Sample answer: Let y be the number. The word product implies multiply, so the expression can be written as 5y. 15. Sample answer: Let x be the number. Increased by implies add, and twice implies multiply by 2. So the expression can be written as 49  2x. 16. And implies add, and times implies multiply. So the expression can be written as 18  3d. 17. Sample answer: Let x be the number. Two-thirds 2 implies multiply by 3, and square implies raise to the second power. So the expression can be 2 written as 3 x 2.

The perimeter is 135 4 feet. 13.

Check for Understanding

1. Algebraic expressions include variables and numbers, while verbal expressions contain words. 2. Sample answer: The perimeter is the sum of the four sides of the rectangle. Thus, the perimeter is the sum of two l’s and two w’s, or 2l  2w. 3. Sample answer: Let a be the variable. Then, a to the fifth power is a5. 4. The word sum implies add, so the expression can be written as j  13. 5. Sample answer: Let x be the number. Less than implies subtract in reverse order, and times implies multiply. So the expression can be written as 3x  24. 6. 92  9  9  81

9. P  2(l  w)  2(5.6  2.7)  2(8.3)  16.6 The perimeter is 16.6 meters. 10. P  2(l  w)  2(6.5  3.05)  2(9.55)  19.1 The perimeter is 19.1 centimeters. 11. P  4s



Variables and Expressions

5

4

19. The word for implies multiply. So the amount of money Kendra will have is represented by the expression s  12d.

1

 14

5 12

1

Chapter 1

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47. You can use the expression 4s to find the perimeter of a baseball diamond. Answers should include the following. • four times the length of the sides and the sum of the four sides • ssss 48. D; More than implies add, and times implies multiply. So the expression can be written as 6  2x or 2x  6. 49. B; 4  4  4  43 and c  c  c  c  c4

20. The word square implies raise to the second power. So the area of the circle is represented by the expression r2.2 21. 62  6  6 2  36 22. 82  8  8  64 23. 34  3  3  3  3  81 24. 63  6  6  6  216 25. 35  3  3  3  3  3  243 26. 153  15  15  15  33753 27. 106  10  10  10  10  10  10  1,000,000

Page 9 50.

28. 1003  100  100  100  1,000,000 29. The word for implies multiply. So the cost of buying b dozen bagels for $8.50 a dozen can be represented by the expression 8.5b. Also the cost of buying d dozen donuts for $3.99 a dozen can be represented by the expression 3.99d. The word and implies add. So the cost of buying b dozen bagels and d dozen donuts can be represented by the expression 8.5b  3.99d. 30. The word for implies multiply. So the amount of miles driven for two weeks can be represented by the expression 14m. The mileage on Sari’s odometer after her trip is the sum of 23,500 and 14m or the expression 23,500  14m. 31. 7 times p 32. 15 times r 33. three cubed 34. five to the fourth power 35. three times x squared plus four 36. 2 times n cubed plus 12 37. a to the fourth power times b squared 38. n cubed times p to the fifth power 39. Sample answer: one-fifth 12 times z squared 40. Sample answer: one-fourth 8 times g cubed 41. 3 times x squared minus 2 times x 42. 4 times f to the fifth power minus 9 times k cubed

54.

Maintain Your Skills 1

14.3  1.8 16.1

1 3

2

51.

5

09 91

10.00 52. 3.24 6.76

6

 5  15  15

55.

3.2 11 15.3 1.04 53. 4.8  6 14 4  4.3 96 312 9 6 416 0 4.472 3 4

1

11

3 8

4

2

7

 15 56.

9

 6  12  12  12

1

1

3

4

2

3

989 1

6 57.

7 10

3

7

5

7

5

 5  10  3 1

 10  3 2

7

1

 6 or 16

Page 10

Reading Mathematics

c; 9  2  n 2. b; 4  (n  6) 3. f; n  52 h; 3(8  n) 5. g; 9  (2  n) 6. d; 3(8)  n a; (n  5)2 8. e; 4  n  6 Sample answer: one more than five times x Sample answer: five times the quantity x plus one Sample answer: three plus the product of seven and x 12. Sample answer: the sum of three and x multiplied by seven 13. Sample answer: the sum of six and b divided by y 14. Sample answer: six plus the quotient of b and y

1. 4. 7. 9. 10. 11.

43. Sum implies add, and product implies multiply. 1 So the expression can be written as x  11x. 44. Sum implies add, product implies multiply, and twice implies multiply by 2. So the expression can be written as 2lw  2lh  2wh. 45. The word by implies multiply, so the expression can be written as 3.5m.

1-2 Page 13

Check for Understanding

1. Sample answer: First add the numbers in parentheses, (2  5). Next square 6. Then multiply 7 by 3. Subtract inside the brackets. Multiply that by 8. Divide, then add 3. 2. Sample answer: (2  4)  3 3. Chase; Laurie raised the incorrect quantity to the second power.

46. The area of the square can be represented by the expression a2 or a  a . The perimeter can be represented as a  a  a  a or 4a. If a  4, then a  a  4  4  16 and 4  a  4  4  16. Thus, the value of a is 4.

Chapter 1

Order of Operations

2

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4. (4  6)7  (10)7 5. 50  (15  9)  50  24  70  26 6. 29  3(9  4)  29  3(5)  29  15  14 7. [7(2)  4]  [9  8(4)]  [14  4]  [9  8(4)]  10  [9  8(4)]  10  [9  32]  10  41  51 8.

(4  3) 2  5 9  3

   

17. 15  3  2  15  6 18. 22  3  7  22  21  21  43 19. 4(11  7)  9  8  4(18)  9  8  72  9  8  72  72 0 20. 12(9  5)  6  3  12(14)  6  3  168  6  3  168  18  150 21. 12  3  5  42  12  3  5  16  4  5  16  20  16 4 22. 15  3  5  42  15  3  5  16  5  5  16  25  16 9 23. 288  [3(9  3)]  288  [3(12)]  288  [36] 8 24. 390  [5(7  6)]  390  [5(13)]  390  [65] 6

(12) 2  5 9  3 144  5 9  3 720 9  3 720 12

 60 9.

3  23 52 (4)

 

3  8 52 (4) 11 52 (4) 11

 25 (4) 11

 100 10. hk  gj  6  12  4  8  72  4  8  72  32  40 11. 2k  gh2  j  2  12  4  62  8  2  12  4  36  8  24  4  36  8  24  144  8  168  8  160 12.

2g(h  g) gh  j

25.

   

2  64  4  8 2  8 128  32 2  8 96 2  8 96 16

6 26.

4  62  42  6 4  6

 

2  4(6  4)  468 2  4(2) 468 8(2) 468 16 468 16  24  8 16  16

 

4  36  16  6 4  6 144  96 4  6 48 4  6 48 24

2 27.

1 13. 3 packages 2 additional of software plus packages 1442443 123 1442443  20.00 2  9.95  or 20.00 2  9.95 14. Evaluate 20.00  2  9.95. 20.00  2  9.95  20.00  19.90  39.90 The cost of 5 software packages is $39.90.

Pages 14–15

2  82  22  8 2  8

[ (8  5) (6  2) 2 ]  (4  17  2) [ (24  2)  3 ]

28. 6 

Practice and Apply



[ (13) (4) 2 ]  (4  17  2) [ (24  2)  3 ]



[ (13) (16) ]  (4  17  2) [ (24  2)  3 ]



208  (68  2) [ (24  2)  3 ]



208  34 [ (24  2)  3 ]



174 [ (24  2)  3 ]



174 [ 12  3 ]



174 4



87 2

1

or 43 2

3 2 3 7  (2  3  5) 4  6  3 2 3 7  (6  5) 4 2  7  6  3 3  14 9  6  33  14  6  [ 3  1] 62 4

15. (12  6)  2  6  2  12 16. (16  3)  4  13  4  52

3

Chapter 1

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29. Evaluate n(2n  3) for n  4. n(2n  3)  4(2  4  3)  4(8  3)  4(11)  44 The area of the rectangle is 44 cm2.2. Derrick 30. Samantha sells 60 floor seats

and

70 balcony seats

and

60(7.5)



70(5)



or 60(7.5)



70(5)



sells 50 floor seats

38.

1xy 2

2



3y  z (x  y) 2

 90 balcony seats.



90(5)



90(5)

  

31. Evaluate 60(7.5)  70(5)  50(7.5)  90(5). 60(7.5)  70(5)  50(7.5)  90(5)  450  350  375  450  800  375  450  1175  450  1625 Samantha and Derrick have collected $1625. 32. x  y2  z2  12  82  32  12  64  32  12  64  9  76  9  85 33. x3  y  z3  123  8  33  1728  8  33  1728  8  27  1736  27  1763 34. 3xy  z  3  12  8  3  36  8  3  288  3  285 35. 4x  yz  4  12  8  3  48  8  3  48  24  24 36.

2xy  z3 z

    

  39.

37.

    

x  z y  x



2y  x y2  2

  

9 3  8  3  (12  8) 2 4 9 24  3  (12  8) 2 4 9 21  (12  8) 2 4 9 21  42 4 9 21  16 4 36 21  16 16 15 16

12  32 2  8  12  82  2 8  12 12  9 16  12  64  2 8  12 3 4  64  2 8  12 3 4  2 32 3

3

4

 3  2  32 9

4

 2  32 9  16

4

 2  16  32   

144 4  32 32 148 32 37 5 or 4 8 8

40. 100 cells after 20 minutes

250 cells after 20 minutes plus 144424443 123 144424443 100.2 250.2  or 100  2  250  2 Evaluate 100  2  250  2. 100  2  250  2  200  250  2  200  500  700 The total number in both dishes is 700 bacteria cells. 41. the sum of salary, commission, and 4 bonuses 4 bonuses salary plus commission plus 1442443 42. 14243 123 1442443 123

2  12  8  33 3 2  12  8  27 3 24  8  27 3 192  27 3 165 3

4b s  c  or his earnings e is s  c  4b. 43. Evaluate s  c  4b for s  42,000, c  12  825 or 9900, and b  750. s  c  4b  42,000  9900  4  750  42,000  9900  3000  51,900  3000  54,900 Mr. Martinez earns $54,900 in a year.

12  82  3  3 3 12  64  3  3 3 768  3  3 3 768  9 3 759 3

 253

Chapter 1

2



 55 xy2  3z 3

2

2



and

2

2

1442443 123 1442443 123 1442443 123 1442443

50(7.5) 50(7.5)

3  8  3 1128 2  (12  8) 3 3  8  3  1 2 2  (12  8)



4

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44. Sample answer: Using 1, 2, 3: 1  2  3  6; 1  2  3  7; 1  2  3  5; 3  2  1  1; (2  1)  3  3 45. Use the order of operations to determine how many extra hours were used and then how much the extra hours cost. Then find the total cost. Answers should include the following. • 6 [4.95  0.99(n)]  25.00 • You can use an expression to calculate a specific value without calculating all possible values. 46. A; Evaluate a  b  c for a  10, b  12, and c  17. a  b  c  10  12  17  22  17  39 47. B; (5  1)3  (11  2)2  (7  4)3  43  92  33 3  64  81  27  145  27  172 48. Enter:

49. Enter:

4 9 91

0.5000 0.0075 0.4925

2.884 66. 5.214.9968   10 4 4 59 4 16 436 416 208 208 0 1

1

68. 4 8  1 2    

( .25 ⫻ .75 2 ) ⫼ ( 7 ⫻ .75 3 ) ENTER 0.0476190476 ( 2 ⫻ 27.89 2

50. Enter:

64.

( 12.75 ( 12.75

2 )

)



(

27.89

3

ENTER 2.074377092 12.75 2 ) ⫼

3

) 2

27.89

12.75

 70.

5 6

4

2

5

4

3

1

3 2 3  4 2  4 12 8

5.600  1.612 7.212

67.

6.42 5  2.3 1 9 27 5 12 8 50 14.7 77 5

69.

3 5

1 11

5

3

19 7 21 95  35 35 116 35 11 3 35

 27  5    

5  6 5 

71. 8 

2 9

2 3 8

2

19 8

9

 12

ENTER 1.170212766

Page 15

1

33  8 33  8 33  8 21 8 5 28

1

65.

4

Maintain Your SkillsMaintain Your Skills



8 1



36 1

9

2

1

 36

51. Product implies multiply, so the expression can be written as a3  b4. 52. Less than implies subtract in reverse order, and times implies multiply. So the expression can be written as 3y2  6. 53. Sum implies add, increased by implies add, and quotient implies divide. So the expression can be b written as a  b  a. 54. Times implies multiply, sum implies add, increased by implies add, twice implies multiply by 2, and difference implies subtract. So the expression can be written as 4(r  s)  2(r  s). 55. Triple implies multiply by 3, and difference implies subtract. So the expression can be written as 3(55  w3).3). 56. 24  2  2  2  2  16 57. 121  121 58. 82  8  8 59. 44  4  4  4  4  64  256 60. Five times n plus n divided by 2 61. 12 less than q squared 62. the sum of x and three divided by the square of the quantity x minus two 63. x cubed divided by nine

1-3 Page 18

Open Sentences Check for Understanding

1. Sample answer: An open sentence contains an equals sign or inequality sign. 2. Sample answer: x  7 3. Sample answer: An open sentence has at least one variable because it is neither true nor false until specific values are used for the variable. 4. Replace x in 3x  7  29 with each value in the replacement set. x

3x  7  29

True or False?

?

10 31102  7  29 S 23 29 ?

11 31112  7  29 S 26 29 ?

12 31122  7  29 S 29  29 ?

13 31132  7  29 S 32 29 ?

14 31142  7  29 S 35 29 ?

15 31152  7  29 S 38 29

false false true ✓ false false false

The solution of 3x  7  29 is 12.

5

Chapter 1

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11. Replace x in 3(12  x)  2 28 with each value in the replacement set.

5. Replace x in 12(x  8)  84 with each value in the replacement set. 12(x  8)  84

x

True or False?

?

10 12110  82  84 S 24 84

false

?

11 12111  82  84 S 36 84

?

1.5 3(12  1.5)  2 28 S 29.5 28

false

?

12 12112  82  84 S 48 84

2

false

?

13 12113  82  84 S 60 84 ?

3

false

?

15 12115  82  84 S 84  84 3

3

2

x  5  1 20 2 ?

or C 

13. Solve C  C

3 1 20

false

C

 5  1 20 S 1 20  1 20

 5  1 20 S 20 1 20

2 ?

3

3

2 ?

3

2

2 ?

3

13

3

3

false

3

1 4  5  1 20 S 1 20 1 20 2

3

Pages 19–20

3

b

True or False?

x

7.2(x  2)  25.92

1.2

7.211.2  22  25.92 S 23.04 25.92

1.4

7.211.4  22  25.92 S 24.48 25.92

1.6

7.211.6  22  25.92 S 25.92  25.92

1.8

7.211.8  22  25.92 S 27.36 25.92

? ? ? ?

12 17

false

18 false 21

true ✓

25

false

6

w2

b

2 3 4 5 6

true ✓

?

true ✓

?

true ✓

?

true ✓

?

true ✓

?

true ✓

24  2122 13 S 20 13 24  2132 13 S 18 13 24  2142 13 S 16 13 24  2152 13 S 14 13 ?

24  2162 13 S 12 13

12 17 18 21

True or False?

?

24  2112 13 S 22 13

25

14

3500  4 . 14 3500  4 14 14,000 14

Practice and Apply

b  12  9

True or False?

?

12  12  9 S 0 9

false

?

17  12  9 S 5 9

false

?

18  12  9 S 6 9

false

?

21  12  9 S 9  9

true ✓

?

25  12  9 S 13 9

false

34  b  22

True or False?

?

34  12  22 S 22  22 ?

34  17  22 S 17 22 ?

34  18  22 S 16 22 ?

34  21  22 S 13 22 ?

34  25  22 S 9 22

The solution of 34  b  22 is 12.

false

The solution set for 24  2x 13 is {0, 1, 2, 3, 4, 5}. Chapter 1



4

15. Replace b in 34  b  22 with each value in the replacement set.

27  x w3 The solution is 27. The solution is 3. 10. Replace x in 24  2x 13 with each value in the replacement set.

1

.

The solution of b  12  9 is 21.

The solution of 7.2(x  2)  25.92 is 1.6. 14  8 8. 4(6)  3  x 9. w  2

24  2102 13 S 24 13

3500

14. Replace b in b  12  9 with each value in the replacement set.

7. Replace x in 7.2(x  2)  25.92 with each value in the replacement set.

24  2x  13

the number of days.

false

The solution of x  5  1 20 is 4.

24  3  x

divided by

C  1000 The number of Calories a person would have to burn each day is 1000 Calories.

true ✓

1  5  1 20 S 1 5 1 20

the number of Calories the number perpound times of pounds

3500  4 14

9 S 10

3 4

0



3 1 20

3 4

x

C

false



1

is

3

1 2

1

The number of Calories

13

1 2

14

true ✓

3(12  3)  2 28 S 25 28

3

1 4

1

true ✓

?

1442443 123 1442443 123 1442443 14243 1442443

True or False?

1 4

2 ?  5

true ✓

?

12.

6. Replace x in x  5  1 20 with each value in the replacement set. x

3(12  2)  2 28 S 28 28

The solution set for 3(12  x)  2 28 is {2, 2.5, 3}.

true ✓

The solution of 12(x  8)  84 is 15. 2

false

?

2.5 3(12  2.5)  2 28 S 26.5 28

false

14 12114  82  84 S 72 84

True or False?

3(12  x)  2  28

x

6

true ✓ false false false false

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16. Replace a in 3a  7  31 with each value in the replacement set. 3a  7  31

a

True or False?

?

3(0)  7  31 S 7 31

0

?

3

3(3)  7  31 S 16 31

5

3(5)  7  31 S 22 31

8

3(8)  7  31 S 31  31

? ? ?

10 3(10)  7  31 S 37 31

4a  5  17

false

0

4(0)  5  17 S 5 17

3

4(3)  5  17 S 17  17

? ?

5

4(5)  5  17 S 25 17

8

4(8)  5  17 S 37 17

? ?

10 4(10)  5  17 S 45 17

false

3

40

5

40 5

 4  0S4 0

8

40 8

10

40 10

?

1 S 93

0

 4  0S0  0



17 8

false

7 ? 17 8

S

21 8



17 8

false

The solution of x  4 

17 8

is 8.

7

7

false true ✓

1 2

false

12 3

 2  4S2 4

17

17 3

 2  4 S 33 4

18

18 3

 2  4S4  4

21

21 3

 2  4S5 4

25

25 3

7

1

false

25

1 2



1

True or false?

7 ? 25  12 12

S 12 12

7

? 25 12

S 12 12

7

? 25 12

S 12  12

7

? 25 12

S 12 12

1  12 

1

false

12

1 2  12 

2

2  12 

7

13

25

false

19

25

false

25

25

true ✓

31

25

false

25

1

The solution of x  12  12 is 1 2. 8

2

22. Replace x in 5 (x  1)  15 with each value in the replacement set.

1 6 1 3 1 2 2 3

2 1x 5

8

 12  15

1  12  158 S 157 158 ? 8 8 8 2 1  1 2  15 S 15  15 5 13 ? 8 3 8 2 1  1 2  15 S 5 15 5 12 ? 8 2 2 2 8  1 2  15 S 3 15 5 13

True or False?

?

2 1 5 6

2

8

false true ✓ false false 1

The solution of 5 (x  1)  15 is 3. 23. Replace x in 2.7(x  5)  17.28 with each value in the replacement set.

True or False?

12

25

x  12  12

x

 4  0 is 10.

?

3

21. Replace x in x  12  12 with each value in the replacement set.

true ✓

24

x

false

2.7(x  5)  17.28 ?

?

2

? ?

24 b 3

19 8

4

b

The solution of

S

7 8

19. Replace b in 3  2  4 with each value in the replacement set.

?

7 ? 17 8

7 8

false

?

b

true ✓

false

false

 4  0S1 0

b 3

17 8

4

false

?

40 a



x

?

The solution of

17 8

5 8

false

40 3

S

5 8

 4  0 S undefined 0

?

false

7 ? 17  8 4

true ✓

True or False?

40 0

17 8



18. Replace a in  4  0 with each value in the replacement set.

0



3 8

40

True or False?

15 8

4

3 8

40 a

40 a

17 8

S

1 8

false

The solution of 4a  5  17 is 3.

a

7

with each value in the

7 ? 17 8

1 8

True or False?

?

17 8

x4

x

The solution of 3a  7  31 is 8. 17. Replace a in 4a  5  17 with each value in the replacement set. a

7

20. Replace x in x  4  replacement set.

1 S 63

4

1.2 2.7(1.2  5)  17.28 S 16.74 17.28

false

?

1.3 2.7(1.3  5)  17.28 S 17.01 17.28

true ✓

?

1.4 2.7(1.4  5)  17.28 S 17.28  17.28

false

?

1.5 2.7(1.5  5)  17.28 S 17.55 17.28

True or False? false false true ✓ false

The solution of 2.7(x  5)  17.28 is 1.4.

false

 2  4 is 18.

7

Chapter 1

PQ249-6481F-01[001-009] 26/9/02 4:06 PM Page 8 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

24. Replace x in 16(x  2)  70.4 with each value in the replacement set.

1

p  4 [7(8)  4(25)  12]

True or False?

x

16(x  2)  70.4

2.2

16 (2.2  2)  70.4 S 67.2 70.4

2.4

16 (2.4  2)  70.4 S 70.4  70.4

2.6

16 (2.6  2)  70.4 S 73.6 70.4

? ? ? ?

16 (2.8  2)  70.4 S 76.8 70.4

2.8

1

35. p  4 [7(23 )  4(52 )  6(2) ] 1

p  4 [56  100  12] 1

false

p  4 [156  12]

true ✓

p  4 [144]

false

p  36 The solution is 36.

1

false

1

36. n  8 [6(32 )  2(43 )  2(7) ]

The solution of 16(x  2)  70.4 is 2.4. 25. Replace x in 21(x  5)  216.3 with each value in the replacement set.

1

n  8 [54  128  14] 1

True or

n  8 [182  14]

False?

n  8 [164]

false

n  21 The solution is 21.

21(x  5)  216.3

x

1

n  8 [6(9)  2(64)  14]

?

3.1 21 (3.1  5)  216.3 S 170.1 216.3 ?

4.2 21 (4.2  5)  216.3 S 193.2 216.3

1

false

?

37. Replace a in a  2  6 with each value in the replacement set.

true ✓

5.3 21 (5.3  5)  216.3 S 216.3  216.3 ?

6.4 21 (6.4  5)  216.3 S 239.4 216.3

false

a

The solution of 21(x  5)  216.3 is 5.3.

6

26.

7

The adult the student is no number price number price more $30. of adults times 1442443 per ticket plus 1442443 of students times 1442443 per ticket 123 than 123 1442443 123

123



2

a



8

123



3



s

$30

9

or 2a  3s 30 27. Evaluate 2a  3s for a  4.50 and s  4.50. 2a  3s  2(4.50)  3(4.50)  9  13.50  22.50 The cost for the family to see a matinee is $22.50. 28. Evaluate 2a  3s for a  7.50 and s  4.50. 2a  3s  2(7.50)  3(4.50)  15  13.50  28.50 The cost for the family to see an evening show is $28.50. 29. 14.8  3.75  t 30. a  32.4  18.95 11.05  t a  13.45 The solution is 11.05. The solution is 13.45. 12  5  3 60 12

31. y  15 y

g

7(3)  3

33. d  4(3  1)  6 21  3 4(2)

d

24 8

11

a 13 14 15 16 17

15  6  7 90 9

6

6

d36 d9 The solution is 9.

a

4(14  1) 3(6)  5 4(13)  18  5

a

52 13

true ✓

?

true ✓

6  2 6 6S4 6 6 7  2 6 6S5 6 6 ?

8  2 6 6S6  6 ?

9  2 6 6S7  6 ?

10  2 6 6 S 8  6 ?

11  2 6 6 S 9  6

a  7  22

false false false

True or False?

?

true ✓

?

true ✓

13  7 6 22 S 20 6 22 14  7 6 22 S 21 6 22 ?

15  7 6 22 S 22  22 ?

16  7 6 22 S 23  22 ?

17  7 6 22 S 24  22

false false false

a

39. Replace a in 5 2 with each value in the replacement set. a 5

 2

True or False?

7

a

7

5

5 ?

5

2S1 2

false

10

10 ?

5

2S2 2

true ✓

15

15 ?

5

2S3 2

true ✓

20

20 ?

5

2S4 2

true ✓

25

25 ?

5

2S5 2

true ✓

7

a47 a  11 The solution is 11.

The solution set for Chapter 1

false

The solution set for a  7  22 is {13, 14}.

g  10 The solution is 10. 34. a 

True or False?

?

The solution set for a  2  6 is {6, 7}. 38. Replace a in a  7  22 with each value in the replacement set.

32. g  16

y5 The solution is 5. d

10

a26

8

a 5

2 is {10, 15, 20, 25}.

PQ249-6481F-01[001-009] 26/9/02 4:06 PM Page 9 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

44. Replace b in 2b  5 with each value in the replacement set.

2a

40. Replace a in 4 8 with each value in the replacement set. 2a 4

a 12 14 16 18 20 22

8

21122 ? 4

8S6 8

4

8S7 8 8S8 8

21182 ? 4

8S9 8

4

8 S 10 8

21222 ? 4

8 S 11 8

The solution set for

2a 4

45.

3.4

4(3.4)  3 10.6 S 10.6 10.6

3.6 3.8

?

false

?

true ✓

?

true ✓

?

true ✓

413.62  3 10.6 S 11.4 10.6 413.82  3 10.6 S 12.2 10.6 4(4)  3 10.6 S 13 10.6

4

4.2

614.22  5 23.8 S 20.2 23.8

4.5

614.52  5 23.8 S 22 23.8

?

?

$39

4.8

614.82  5 23.8 S 23.8 23.8

5.1

615.12  5 23.8 S 25.6 23.8

5.4

? ?

615.42  5 23.8 S 27.4 23.8

3a  4 3102 4 S 0 4

0 1 3 2 3

1 1 13

?

1 12

18,995



n



$10.95



true ✓

2 39122  10.95 102.50 S88.95 102.50

true ✓ true ✓ 1

6

39n  10.95  102.50 ?

$102.50

Reasonable? too low

?

3 39132  10.95 102.50 S127.95 102.50 too high

true ✓

?

1 2



n

Examine The solution set is {0, 1, 2}. You can buy 2 sweaters and stay within your budget. 49. The solution set includes all numbers less than or 1 equal to 3. 50. You can use equations to determine how much money you have to spend and how you can spend your money. Answers should include the following. • Sample answers: calculating gasoline required for a trip: cooking time

true ✓

5

6220

true ✓

?

3 13 4 S 4 4



123 14243 123 14243 123 14243

?

true ✓

3112 4 S 3 4

15,579

glasses of soda.

39  10.95 102.50 49.95 102.50 The estimate is too low. Increase the value of n.

false

true ✓

113 2 4 S 1 4 2 ? 313 2 4 S 2 4

3



plus

?

True or False?

?

6

39n  10.95 102.50

The solution set for 6a  5 23.8 is {4.8, 5.1, 5.4}. 43. Replace a in 3a 4 with each value in the replacement set. a

plus

glasses of juice

39(1)  10.95 102.50

false

?

is

glasses of milk

or 39n  10.95 102.50 48. Explore You can spend no more than $102.50. So the situation can be represented by the inequality 39n  10.95 102.50. Plan Since no replacement set is given, estimate to find reasonable values for the replacement set. Solve Start by letting n  1 and then adjust values up or down as needed.

True or False?

6a  5  23.8

Total number of glasses

1442443 123 1442443 123 1442443 14243

The solution set for 4a  3 10.6 is {3.4, 3.6, 3.8, 4}. 42. Replace a in 6a  5 23.8 with each value in the replacement set. a

1

$102.50. each sweater times of sweaters plus for shipping more than 14243

true ✓

?

5

false

or g  15,579  6220  18,995 46. Evaluate 15,579  6220  18,995. 15,579  6220  18,995  21,799  18,995  40,794 The average American drinks 40,794 glasses. 47. The cost of the number the charge is not

True or False?

4(3.2)  3 10.6 S 9.8 10.6

false

?

?

g

41. Replace a in 4a  3 10.6 with each value in the replacement set.

3.2

1 2 6 5S5  5

true ✓

1 22

2132 6 5 S 6  5

1442443

8 is {12, 14, 16}.

4a  3  10.6

true ✓

?

The solution set for 2b  5 is 1, 12, 2 .

false

a

2

3

false

1 2 6 5S3 6 5

true ✓

?

1 12

2122 6 5 S 4 6 5

2 12

false

21202 ?

2

2

true ✓

True or False?

?

2112 6 5 S 2 6 5

1 12

true ✓

21162 ? 4

1

true ✓

21142 ?

2b  5

b

True or False?

The solution set for 3a 4 is 0, 3, 3, 1, 1 3 .

9

Chapter 1

51. B; Replace n in

15  n2 2  5

57. 53  3(42)  53  3(16)  125  3(16)  125  48  173

6 28 with each value in

19  32 2  n

the replacement set. 15  n2 2  5

n

2

 28

?

11

19  3 2  n

5

15  52 2  5

7

15  72 2  5

9

15  92 2  5

True or False?

19  33 2  7

6 28 S 16 37 6 28

15  112 2  5

13

15  132 2  5

19  3 2  11 2

19  3 2  13 2

?

23

?

7

6 28 S 28 37  28 ?

6 28

7 S 62 34

The solution set for

 26 1 59. [5(2  1)]4  3  [5(3)]4  3  [15]4  3  50,625  3  50,628

true ✓ false false

60.

1 6



2 5

1

6 3

 28

15  n2 2  5

19  32  n 2

false

6 28 is {5, 7}.

5 6



15 16

4

9 3

3 7

2

15 16

63.

6 14



12 18

25 32

1

6

6

12 18

 14  7 2



6 21 7

2

7 8 13



2 11

66.

3 11



7 16

8  2  11 16 143

 13 

65.



4 9

4  4  9 16 63

7

8

3  7  16

Page 20

4 7



 11 

67.

21 176

2 9



24 25

2

9 3



24 25

16 75

Practice Quiz 1

1. 2. 3. 4. 5.

twenty less than x five times n plus two a cubed n to the fourth power minus one 6(9)  2(8  5)  6(9)  2(13)  54  2(13)  54  26  28 6. 4[2  (18  9)3]  4[2  (2)3]  4[2  8]  4[10]  40 7. 9(3)  42  62  2  9(3)  16  62  2  9(3)  16  36  2  27  16  36  2  27  16  18  11  18  29

1

t 14s  r2  2 14  5  22 1

 2 120  22 1

 2 1222 1

11 55. Sum implies add, and times implies multiply. So the expression can be written as (r  s)t2. 1

Evaluate (r  s)t2 for r  2, s  5, and t  2.

7

4

8.

56. Decreased by implies subtract, so the expression can be written as r5  t. 1 Evaluate r5  t for r  2 and t  2.

15  22 2

314  2  72

 314  

1



1



r5  t  25  2  32  2

10

132 2

 2  72 9 314  2  72 9 318  72 9 3112 9 3

3

1

 31 2 Chapter 1

3 7

4

5



Evaluate t(4s  r) for r  2, s  5, and t  2.

112 22 1  172 1 4 2



 21

6

Maintain Your Skills

 172

4 9

5

62.

53. Increased by implies add, and times implies multiply. So the expression can be written as r 2  3s. Evaluate r2  3s for r  2 and s  5. r2  3s  22  3  5 43 5  4  15  19 54. Times implies multiply, and sum implies add. So the expression can be written as t(4s  r).

112 22

61.

1

64.

1r  s2t2  12  52

2 5

 15

52. C; 27  3  (12  4)  27  3  8 98  17

Page 20

1

1

2

6 28 S 43 7  28 ?

26

 2  13 26

6 28 S 8 38 6 28

11

38  12 2  13

true ✓

19  32 2  5

19  32 2  9

58.

3

9.

5  42  10  2 6  5

   

11. From Exercises 9 and 10 we determine that four score and seven years is 87 years.

5  16  10  2 6  5 80  10  2 6  5 90  2 6  5 88 11

Pages 23–25

8 10. Replace n in 2n2  3  75 with each value in the replacement set. 2n2  3  75

n 4 5 6 7 8 9

True or False?

?

2  42  3  75 S 35  75 2

?

2

?

2

?

2  5  3  75 S 53  75 2  6  3  75 S 75  75 2  7  3  75 S 101 75 ?

2  82  3  75 S 131 75 2

92

?

 3  75 S 165 75

true ✓ true ✓ true ✓ false false

18. Multiplicative Inverse Property; 1 1 n  2, since 1  2  2.

false

The solution set for 2n2  3  75 is {4, 5, 6}.

1-4

19. Multiplicative Inverse Property; 1 n  1, since 4  4  1. 20. Substitution Property; n  5, since (9  7)(5)  2(n). 21. Substitution Property; n  3, since 3  (2  8)  n  10. 22. Multiplicative Identity Property;

Identity and Equality Properties

Page 23

Check for Understanding

1

1. No; for 1 to be an additive identity, the sum of any number a and 1 must be a. However, 3  1  4  3. 2. Sample answer: 5  3  2, and 3  2  4  1, so 5  4  1; 5  7  8  4, and 8  4  12, so 5  7  12. 3. Sample answer: You cannot divide by zero. 4. Multiplicative Property of Zero; n  0, since 13  0  0. 5. Additive Identity Property; n  17, since 17  0  17.

n  3, since 3 52 

n  2, since 6 24.

1

2  3 125  251 2

3 4

112  22  6  1  6.

[ 4  (7  4) ] 3

 4 [4  3] Substitution; 7  4  3

34

3 4 3

4

1 25.

7. 6(12  48  4)  6(12  12) Substitution; 48  4  12  6(0) Substitution; 12  12  0 0 Multiplicative Property of Zero; 6  0  0 8.

1 25

3 1  3. 23. Multiplicative Identity Property;

6. Multiplicative Inverse Property; 1 n  6, since 6  6  1.

1 15  15

Practice and Apply

12. Multiplicative Identity Property; n  1, since 12  1  12. 13. Multiplicative Identity Property; n  5, since 5  1  5. 14. Reflexive Property; n  5, since 8  5  8  5. 15. Reflexive Property; n  0.25, since 0.25  1.5  0.25  1.5 16. Additive Identity Property; n  0, since 8  0  8. 17. Additive Identity Property; 1 1 1 n  3, since 3  0  3.

2 3

4

Substitution; 4  3  3 3 4

Multiplicative Inverse;



4 3

1

[ 3  (2  1) ] 2

 3 [3  2] Multiplicative Identity; 2  1  2

34

2 3 2

3

1

2

 8  0  12

3

Substitution; 3  2  2 Multiplicative Inverse;

26. 2 (3  2  5) 

 (1 + 8  0)  12 Multiplicative Inverse; 1 15  15  1  (1  0)  12 Multiplicative Property of Zero; 8 00  1  12 Additive Identity; 1  0  1  12 Multiplicative Identity; 1  12  12

2 3



3 2

1

1 33 1

 2(6  5)  3  3 Substitution; 3  2  6  2 112  3  1

 2  33

1 3

Substitution; 6  5  1 Multiplicative Identity; 2 12

9. four score and seven 1 442443 123 1 424 3

 4(20) 7 or 4(20)  7 10. 4(20)  7  80  7 Substitution; 4  20  80  87 Substitution; 80  7  87

11

21

Multiplicative Inverse; 1 331

3

Substitution; 2  1  3

Chapter 1

1

35.

27. 6  6  5 (12  4  3)

Eight players times 14243 123

1

 6  6  5(3  3) Substitution; 12  4  3 6

1 6

 5102

8

Substitution; 3  3  0

 1  5(0)

Three players times 3

4

36.

37.

38.

39. 40. 41.

1169.10

or 1169.10  y  1169.10 y  0, since 1169.10  0  1169.10. 33.

increase Pay for from grade yr to 10 yr E-4 at 6 yr times 6 1442443 1442443 123



1653

is

y



1653

123

pay for grade E-4 at 10 yr. 144424443

42.

Base salary

is the bonus for plus 12 touchdowns same as

base salary



$50,000



$350,000

bonus for plus 76 points. 

$50,000

or 350,000  50,000  350,000  50,000

Chapter 1

($400,000





$50,000

base salary

plus



($400,000



1601 yards

14243

$1,000,000

$50,000)

$50,000)

12

4.5 yards plus per carry

123 14243



$50,000)

or 4(400,000  1,000,000  50,000) So the expression can be written as 8(400,000  100,000  50,000)  3(400,000  50,000  50,000)  4(400,000  1,000,000  50,000). 8(400,000  100,000  50,000)  3(400,000  50,000  50,000)  4(400,000  50,000  50,000)  8(550,000)  3(400,000  50,000  50,000)  4(400,000  50,000  50,000) Substitution; 400,000  100,000  50,000  550,000  8(550,000)  3(500,000)  4(500,000) Substitution; 400,000  1,000,000  50,000  1,450,000  4,400,000  1,500,000  2,000,000 Substitution; 4  1,450,000  5,800,000  7,900,000 Substitution; 4,400,000  1,500,000  5,800,000  11,700,000 Sometimes; Sample answer: true: x  2, y  1, z  4, w  3, 2 1 and 4 3 and 2  4  8 3  1  3; false: x  1, y  1, z  2, w  3, 1 1 and 2 3 however, 1(2)  2 3  (1)(3). You can use the identity and equality properties to see if data is the same. Answers should include the following: • Reflexive: r  r, or Symmetric: a  b so b  a. • Oklahoma, week 1  a, week 2  b, week 3  c. a  b and b  c so a  c. A; In words: If one quantity equals a second quantity, then the second quantity equals the first. B; Substitution Property, since 10  8  2. False, since 4  5  1 and 1 is not a whole number. True, for any whole numbers a and b the product ab is also a whole number. 1

14243 123 144424443 14243 14243 123 14243

$350,000



76 points 12 plus touchdowns plus scored

43. False, since 1  2  2 and number.

or 1653y  1653 y  1, since 1653  1  1653. 34.



Four players times

grade E-2 at 12 yr. 144424443



base salary

14243 123 14243 123

 80(2.5  1)  40(10  6) 25(5  3) or 25(5  3) + 80(2.5 1) + 40(10  6) 31. 25(5  3)  80(2.5  1)  40(10  6)  25(2)  80(2.5  1)  40(10  6) Substitution; 532  25(2)  80(1.5)  40(10  6) Substitution; 2.5  1  1.5  25(2)  80(1.5)  40(4) Substitution; 10  6  4  50  120  160 Substitution; 40  4  160  330 Substitution; 170  160  330 increase pay for 32. Pay for 123

$100,000

or 3(400,000  50,000  50,000)

Additive Identity; 1  0  1 1 2 28. 3  5(4  2 )  1  3  5(4  4)  1 Substitution; 22  4  3  5(0)  1 Substitution; 4  4  0 301 Multiplicative Property of Zero; 5  0  0 31 Additive Identity; 3  0  3 2 Substitution; 3  1  2 29. 7  8(9  32)  7  8(9  9) Substitution; 32  9  7  8(0) Substitution; 9  9  0 70 Multiplicative Property of Zero; 8 00 7 Additive Identity; 7  0  7 30. Profit for profit for profit for and 80 buttons and 40 caps 25 peanuts 123 1442443 1442443 123 14243

is



14243 123 14243 123 1442443 123 14243

Multiplicative Property of Zero; 5  0  0

from grade plus yr to 12 yr E-2 at 3 yr 123 3 1442443 1442443  y 1169.10

($400,000

or 8(400,000  100,000  50,000)

Multiplicative Inverse; 1 661

10



35% of offensive base weight salary plus below 240 lb plus plays 14243 123 1442443 123 14243

1 2

is not a whole

Page 25

Maintain Your Skills

x

10  x  6

3

10  3 7 6 S 7 7 6

?

?

10  5 7 6 S 5  6

5

?

10  6 7 6 S 4  6

6

?

10  8 7 6 S 2  6

8

x

True or False?

1 2

true ✓

1 3

false

1 4

false false

1 5

The solution set for 10  x 6 is {3}. 45. Replace x in 4x  2 58 with each value in the replacement set. 4x  2  58

x 11

4  11  2 6 58 S 46 6 58

12

4  12  2 6 58 S 50 6 58

13 14

? ?

4  13  2 6 58 S 54 6 58 ?

4  14  2 6 58 S 58  58 ?

4  15  2 6 58 S 62  58

15

1 6

false

false

5.9

5.9 ? 2

3 S 2 20 3

false

3S3 3

true ✓

1 S 3 20

6.1

6.1 ? 2

3

3

true ✓

6.2

6.2 ? 2

3 S 3 10 3

true ✓

6.3

6.3 ? 2

3

3 S 3 20

The solution set for

x 2

3

3 1 32

3 3.25 3.5 3.75 4

?

8  3  32 S 24  32 ?

8  3.25  32 S 26  32 ?

8  3.5  32 S 28  32 ?

8  3.75  32 S 30  32 ?

8  4  32 S 32  32

6

3 10

S 30 6

1

?

 24 6

3 10

S5 6

1 5

?

3 10

S 10  10

3

3

false

3 10

S 30  10

11

3

false

2 2

1 6

6 ?

6

1

1

7 10

3 10

true ✓

3 10

true ✓

3 10

 2x 6

true ✓

3 10

2x  1  2

is

512, 13, 14 6.

?

1

True or False? 1

true ✓

?

false

?

2  3  1  2S5 2 1 2  32

false

?

 1  2S6 2

50. (3  6)  32  (9)  32 99 1 51. 6(12  7.5)  7  6(4.5)  7  27  7  20 52. 20  4  8  10  5  8  10  40  10 4 53.

true ✓

16  22 2 16

 3192  

3 is {6, 6.1, 6.2, 6.3}.

8x  32

1 3

?

3

5 16

false

The solution set for 2x  1  2 is 1 4 .

47. Replace x in 8x  32 with each value in the replacement set. x

S 10 6

2  2  1  2S3 2

3

1

3 10

2

1

false

3

6 ? 2

?

2  14  1  2 S 12  2

5.8

6

7 10

1

2

7 10 7 10

 22 6

14

True or False?

19

7 10

True or False?

true ✓

5.8 ? 2

9 S 2 10

7 10

3

x

x

3

 2x  10

true ✓

46. Replace x in 2 3 with each value in the replacement set. x

7 10

with each value in the

49. Replace x in 2x  1  2 with each value in the replacement set.

true ✓

The solution set for 4x  2 58 is {11, 12, 13}.

x 2

3 10

The solution set for

True or False?

?

7

48. Replace x in 10  2x 6 replacement set.

44. Replace x in 10  x 6 with each value in the replacement set.

54.

True or False? true ✓ true ✓ true ✓

55.

true ✓ true ✓

The solution set for 8x  32 is {3, 3.25, 3.5, 3.75, 4}.

56.

57.

13

182 2

 3192

16 64  16

3192

 4  3(9)  4  27  31 [62  (2  4)2] 3  [62  (6)2]3  [36  (6)2]3  [36  12]3  [24]3  72 2 2 9(3)  4  6  2  9(3)  16  36  2  27  16  36  2  27  16  18  11  18  29 Sample answer: Let x be the number. Sum implies add, and twice implies multiply by 2. So the expression can be written as 2x2  7. 10(6)  10(2)  60  10(2)  60  20  80

Chapter 1

58. (15  6)  8  9  8  72 59. 12(4)  5(4)  48  5(4)  48  20  28 60. 3(4  2)  3(6) 61. 5(6  4)  5(2)  18  10 62. 8(14  2)  8(16)  128

4. 3(x  3)  3x  3 The rectangle has 3 x-tiles and 9 1-tiles. The area of the rectangle is x111x111x111 or 3x  9. Therefore, 3(x  3)  3x  3 is false. x3

3

The Distributive Property

1-5 Page 28

Algebra Activity

x 1 1

5

6. Rachel; Sample answer: 3(x  4)  3(x)  3(4)  3x  12 The rectangle has 3 x-tiles. The area of the rectangle is 3x  12. Therefore, Rachel is correct.

x4

3

1 1 1 1 1

 5x  10

Pages 29–30

2x  1

Chapter 1

x x

x x

1 1

x x x

1 1 1

1 1 1 1 1 1

1 1 1

 3x  12

Check for Understanding

1. Sample answer: The numbers inside the parentheses are each multiplied by the number outside the parentheses then the products are added. 2. Sample answer: 4ab  3b  a  2ab  7ab 3. Courtney; Ben forgot that w4 is really 1  w4. 4. 6(12  2)  6  12  6  2  72  12  60 5. 2(4  t)  2  4  2  t  8  2t 6. (g  9)5  g  5  9  5  5g  45 7. 16(102)  16(100  2)  16(100)  16(2)  1600  32  1632

3. 2(2x  1) The rectangle has 4 x-tiles and 2 1-tiles. The area of the rectangle is x  x  1  x  x  1 or 4x  2. Therefore, 2(2x  1)  4x  2.

2

 5x

x x x x x

x2 1 1 1 1 1

 3x  9

 2x  2

2. 5 (x  2) The rectangle has 5 x-tiles and 10 1-tiles. The area of the rectangle is x11x11x11x11 x  1  1 or 5x  10. Therefore, 5(x  2)  5x  10.

x x x x x

1 1 1 1 1 1

32

x1

x x

1 1 1

5. x(3  2)  3x  2x The rectangle has 5 x-tiles. The area of the rectangle is x  x  x  x  x or 3x  2x or 5x. Therefore, x(3  2)  3x  2x is true.

1. 2(x  1) The rectangle has 2 x-tiles and 2 1-tiles. The area of the rectangle is x  1  x  1 or 2x  2. Therefore, 2(x  1)  2x  2.

2

x x x

 4x  2

14

1 12

1

2

8. 3 17 1172  3  17 17

9. 10.

11. 12. 13.

1

28. 4(8p  4q  7r)  4(8p)  4(4q)  4(7r)  32p  16q  28r 29. number of number of people at people at number Olympic aquatics of days times Stadium plus center 14243 123 14243 123 14243

 3 1172  17 1172 1

 51  1  52 13m  m  (13  1)m  14m 3(x  2x)  3(x)  3(2x)  3x  6x  (3  6)x  9x There are no like terms. 14a2  13b2  27 is simplified. 4(3g  2)  4(3g)  4(2)  12g  8 average number of price of plus tip haircuts times each haircut 14243 123 1442443 123 14243  ($19.95  $2) 12

30.

31.

32.

or 12(19.95  2) 14. 12(19.95  2)  12(19.95)  12(2)  239.40  24  263.4 Ms. Curry earned $263.40.

Pages 30–31

33.

Practice and Apply

34.

15. 8(5  7)  8  5  8  7  40  56  96 16. 7(13  12)  7  13  7  12  91  84  175 17. 12(9  5)  12  9  12  5  108  60  48 18. 13(10  7)  13  10  13  7  130  91  39 19. 3(2x  6)  3  2x  3  6  6x  18 20. 8(3m  4)  8  3m  8  4  24m  32 21. (4  x)2  4  2  x  2  8  2x 22. (5  n)3  5  3  n  3  15  3n

1

1

2

23. 28 y  7  281y2  28

1

24. 27 2b 

1 3

 28y  4

4  (110,000  17,500) or 4(110,000  17,500) 4(110,000  17,500)  4(110,000)  4(17,500)  440,000  70,000  510,000 The attendance for the 4-day period was 510,000 people. 5  97  5(100  3)  5(100)  5(3)  500  15  485 8  990  8(1000  10)  8(1000)  8(10)  8000  80  7920 17  6  (20  3)6  20(6)  3(6)  120  18  102 24  7  (20  4)7  20(7)  4(7)  140  28  168

1 12

1

1

35. 18 2 9  18 2  9

2

 18(2)  18  36  2  38

37.

12

1 12

1

1

36. 48 3 6  48 3  6

1 9

2

 48132  48  144  8  152

116 2

hours hours hours on number using meeting telephone of weeks times e-mail plus in person 123 plus 1 14243 123 14243 123 1 4424 43 4424 43 12



(5





12

18)

or 12(5  12  18)

38. 12(5  12  18)  12(5)  12(12)  12(18)  60  144  216  420 She should plan 420 h for contacting people. monthly monthly monthly 39. charge number charge charge of months times for medical plus for dental plus for vision

1442443 123 1442443 123 1442443 123 1442443

12

6

1 7



(78



20



12)

or 6(78  20  12)

40. 6(78  20  12)  6(78)  6(20)  6(12)  468  120  72  660 The cost to the employee is $660.

2  2712b2  27 1 2 1 3

 54b  9 25. a(b  6)  a(b)  a(6)  ab  6a 26. x(z  3)  x(z)  x(3)  xz  3x 27. 2(a  3b  2c)  2(a)  2(3b)  2(2c)  2a  6b  4c

15

Chapter 1

41.

number monthly charge monthly charge of months times for medical for dental plus

1

l  w  13 2

1442443 123 144424443 123 144424443

42. 44. 45. 46. 47. 48. 49. 50.

51.

(78  50



12



(20  15)

3 w 2

or 12[(78  50)  (20  15)] 12[(78  50)  (20  15)]  12 [128  35]  12(128)  12(35)  1536  420  1956 The cost to the employee is $1956. 2x  9x  (2  9)x 43. 4b  5b  (4  5)b  11x  9b There are no like terms. 5n2  7n is simplified. 3a2  14a2  (3  14)a2  17a2 12(3c  4)  12(3c)  12(4)  36c  48 15(3x  5)  15(3x)  15(5)  45x  75 2 6x  14x  9x  6x2  (14  9)x  6x2  5x 3 3 4 4y  3y  y  (4  3)y3  y4  7y3  y4 6(5a  3b  2b)  6(5a)  6(3b)  6(2b)  30a  18b  12b  30a  (18  12)b  30a  6b 5(6m  4n  3n)  5(6m)  5(4n)  5(3n)  30m  20n  15n  30m  (20  15)n  30m  5n

1

3 2 w 2

 13 2

3 2 w 2



2 3 2 w 3 2

1

27 2

2  23 1272 2

w2  9 w3 Substitute w  3 into the area formula. 1

l  w  13 2 1

l  3  13 2 1

l  42 1

Therefore, the rectangle has a length of 4 2 units and a width of 3 units. 55. You can use the Distributive Property to calculate quickly by expressing any number as a sum or difference of more convenient numbers. Answer should include the following. • Both methods result in the correct answer. In one method you multiply then add, and in the other you add then multiply. 56. D; 3(x  y)  2(x  y)  4x  3x  3y  2x  2y  4x  3x  2x  4x  3y  2y  (3  2  4)x  (3  2)y  1x  5y  x  5y 57. C; c  7(2  2.8  3  4.2)  7(5.6  12.6)  7(5.6)  7(12.6)  39.2  88.2  127.4

7 x 7 1 52. x2  8 x  8  x2  8 x  8x

178  18 2x

 x2 

1

 w  13 2

6

 x2  8 x 3

 x2  4 x 53. a 

a 5

5

2

1

Page 31

2

 5a  5a  5a  5a 

155  15  25 2 a

58. 59. 60. 61. 62. 63. 64.

8

 5a

1

54. Area  l  w  13 2

w  5  perimeter  5  121l  w2 2 1

1

 5  12l  2w2 1 2

Solve w  w w

2  5w 3 w 5 5 3 w 2 5 3 w 2

12

  

2 w 5 2 w 5

2

is

d



number of number of feet per second times seconds 1129



2

or d  1129(2) 65. 1129(2)  (1000  100  30  1)2  1000(2)  100(2)  30(2)  1(2)  2000  200  60  2  2258 Sound travels 2258 feet in 2 seconds. 66. 3ab  c2  3  4  6  32 3469  12  6  9  72  9  63

for l.

12

l

3

Substitute 2 w for l into the Area formula.

Chapter 1

total distance

14243 123 144424443 123 1442443

 5l  5w 2 l 5 2 l 5 2 l 5 2 l 5 5 2 l 2 5

Maintain Your Skills

Symmetric Property Substitution Property Multiplicative Identity Property Multiplicative Inverse Property Multiplicative Inverse Property Reflexive Property

16

67. 8(a  c)2  3      68.

6ab c 1a  22

7.

6  4  6

 314  22    

69. 1a  c2

6. 5  3  6  4  5  4  3  6  (5  4)  (3  6)  20  18  360

8(4  3)2  3 8(1)2  3 813 83 11

1a 2 b 2  14  32 14 2 6 2 4  6  172 1 2 2 10  172 1 2 2

 12 1142  168 Area is 168 cm2.2. 72. A  s2  8.52  72.25 Area is 72.25 m2.a i

Commutative and Associative Properties Check for Understanding

1. Sample answer: The Associative Property says the way you can group numbers together when adding or multiplying does not change the result. 2. Sample answer: Division is not Commutative. For example, 10  2  2  10. 3. Sample answer: 1  5  8  8  1  5; (1  5)8  1(5  8) 4. 14  18  26  14  26  18  (14  26)  18  40  18  58



64  10

1

2

114 q  234 q2  2q 3 1  1 4  24 2 q  2q

1442443 1 (p 2

1 1p 2



 2q)

 2q2  4q  2 1 p2  2 12q2  3

1

1

3

1

 2 p  q  4q

1 2 3 1  2 p  11  4 2q 

1 1 32  22  4 1 1 32  22  4

1

5 6

5 6

 3q  2q  13  22q  5q 11. 3(4x  2)  2x  3(4x)  3(2)  2x  12x  6  2x  12x  2x  6  (12x  2x)  6  (12  2)x  6  14x  6 12. 7(ac  2b)  2ac  7(ac)  7(2b)  2ac  7ac  14b  2ac  7ac  2ac  14b  (7ac  2ac)  14b  (7  2)ac  14b  9ac  14b 13. 3(x  2y)  4(3x  y)  3(x)  3(2y)  4(3x)  4(y)  3x  6y  12x  4y  3x  12x  6y  4y  (3  12)x  (6  4)y  15x  10y 14. half the sum by three-fourths q of p and 2q increased 1442443 144424443

1



2



 2 1242 1142

4

3

1

1

5.

3

10. 4 q  2q  24 q  4 q  24 q  2q

71. A  2 bh

1 22

1

5 6

 130 8. 4x  5y  6x  4x  6x  5y  (4x  6x)  5y  (4  6)x  5y  10x  5y 9. 5a  3b  2a  7b  5a  2a  3b  7b  (5a  2a)  (3b  7b)  (5  2)a  (3  7)b  7a  10b

70. A  l  w 95  45 Area is 45 in2.2.

1 32

3

 156 

8

Page 34

3

 16  94  16  94 

 16  94 

24  6 314  22 144 314  22 144 3162 144 18

 172 152  35

1-6

5 6

2

1 p 2

1

 q

3

 2 p  14 q

17

3 q 4

3 q 4 3 q 4

Distributive Property Multiply. Associative () Distributive Property Substitution

Chapter 1

15. number of area of triangles times each triangle 1442443 123 1444244 43 1  4 bh 2

11 2

3

1

1

3

2

1

2

2

26. 3 7  14  14  3 7  14  14 1

 48  14

112  5.2  7.862 1  1 4  2 2  15.2  7.862

 60

4 2bh  4

27.

5 28

2  24  6 3

1

5

 2 8  24  6 3 2

 63  6 3

 2  40.872  81.744 The area of the large triangle is 81.744 cm2.

28.

 420 cost cost cost Friday Saturday Sunday

cost Monday

14243 14243 14243 14243

Pages 34–36

$72  $63  $63  $72 72  63  63  72  72  63  72  63  (72  63)  (72  63)  135  135  270 The total cost is $270.

Practice and Apply

16. 17  6  13  24  17  13  6  24  (17  13)  (6  24)  30  30  60 17. 8  14  22  9  8  22  14  9  (8  22)  (14  9)  30  23  53 18. 4.25  3.50  8.25  4.25  8.25  3.50  (4.25  8.25)  3.50  12.50  3.50  16 19. 6.2  4.2  4.3  5.8  6.2  4.3  4.2  5.8  (6.2  4.3)  (4.2  5.8)  10.5  10  20.5 1

1

1

1

29. cost tax cost tax cost tax cost tax for for for for for for for for Friday 123 Saturday 1424 3 1424 3 Monday 14243 Saturday 123 Friday 14243 Sunday 1424 3 Sunday 1424 3 Monday

$72 $5.40 $63  $5.10  $63 $5.10 $72  $5.40

72  5.40  63  5.10  63  5.10  72  5.40  72  63  72  63  5.40  5.10  5.40  5.10  (72  63)  (72  63)  (5.40  5.10)  (5.40  5.10)  135  135  10.50  10.50  (135  135)  (10.50  10.50)  270  21  291 The total cost including tax is $291. 30. Sample answer: Sales from Sales from Sales from Sales from 3 new 5 used 2 older 2 DVDs videos releases videos 14243

20. 6 2  3  2  2  6 2  2  3  2

1

2

 62  2  13  22 1

1

75  12 3

3

3

3

14243

21. 2 8  4  3 8  28  3 8  4

1

3

3

2

 28  3 8  4   

6 58 6 98 3 94

4

31.

22. 5  11  4  2  5  2  11  4  (5  2)  (11  4)  10  44  440 23. 3  10  6  3  3  3  10  6  (3  3)  (10  6)  9  60  540 24. 0.5  2.4  4  0.5  4  2.4  (0.5  4)  2.4  2  2.4  4.8 25. 8  1.6  2.5  8  2.5  1.6  (8  2.5)  1.6  20  1.6  32

Chapter 1

32.

33.

34.

18

14243

14243

2(3.99)  3(4.49)  2(2.99)  5(9.99) 2(3.99)  3(4.49)  2(2.99)  5(9.99)  2(3.99)  2(2.99)  3(4.49)  5(9.99)  2(3.99  2.99)  3(4.49)  5(9.99) Two expressions to represent total sales can be 2(3.99)  3(4.49)  2(2.99)  5(9.99) and 2(3.99  2.99)  3(4.49)  5(9.99). 2(3.99  2.99)  3(4.49)  5(9.99)  2(6.98)  3(4.49)  5(9.99)  13.96  13.47  49.95  27.43  49.95  77.38 The total sales of the clerk are $77.38. 4a  2b  a  4a  a  2b  (4a  a)  2b  (4  1)a  2b  5a  2b 2y  2x  8y  2y  8y  2x  (2y  8y)  2x  (2  8)y  2x  10y  2x  2x  10y x2  3x  2x  5x2  x2  5x2  3x  2x  (x2  5x2)  (3x  2x)  (1  5)x2  (3  2)x  6x2  5x

44. twice the sum of s and t decreased by 1442443

35. 4a3  6a  3a3  8a  4a3  3a3  6a  8a  (4a3  3a3)  (6a  8a)  (4  3)a3  (6  8)a  7a3  14a 36. 6x  2(2x  7)  6x  2(2x)  2(7)  6x  4x  14  (6x  4x)  14  (6  4)x  14  10x  14 37. 5n  4(3n  9)  5n  4(3n)  4(9)  5n  12n  36  (5n  12n)  36  (5  12)n  36  17n  36 38. 3(x  2y)  4(3x  y)  3(x)  3(2y)  4(3x)  4(y)  3x  6y  12x  4y  3x  12x  6y  4y  (3x  12x)  (6y  4y)  (3  12)x  (6  4)y  15x  10y 39. 3.2(x  y)  2.3(x  y)  4x  3.2(x)  3.2(y)  2.3(x)  2.3(y)  4x  3.2x  3.2y  2.3x  2.3y  4x  3.2x  2.3x  4x  3.2y  2.3y  (3.2x  2.3x  4x)  (3.2y  2.3y)  (3.2  2.3  4)x  (3.2  2.3)y  9.5x  5.5y 40. 3(4m  n)  2m  3(4m)  3(n)  2m  12m  3n  2m  12m  2m  3n  (12m  2m)  3n  (12  2)m  3n  14m  3n 41. 6(0.4f  0.2g)  0.5f  6(0.4f )  6(0.2g)  0.5f  2.4f  1.2g  0.5f  2.4f  0.5f  1.2g  (2.4f  0.5f )  1.2g  (2.4  0.5)f  1.2g  2.9f  1.2g 42.

3 4

2(s  t)  s 2(s  t)  s  2(s)  2(t)  s Distributive Property  2t  2s  s Comm. ()  2t  s(2  1) Dist.  2t  s(1) Sub.()  2t  s Mult. Id.  s  2t Comm. () 45. five times the by 3xy product of x and y increased 1442443 123 14444244443

 3xy 5(x  y) 5(xy)  3xy  5(xy)  3(xy) Associative ()  xy(5  3) Distributive Property  xy(8) Substitution  8xy Commutative () 46. the product of 6 and the the sum of square of z increased by seven, z2, and 6 1442443 1442443 144424443

6  z2  (7  z2  6) 2 2 6z  (7  z  6)  6z2  (z2  7  6) Commutative ()  (6z2  z2)  (7  6) Associative ()  z2(6  1)  (7  6) Distributive Property  z2(7)  13 Substitution  7z2  13 Commutative () 47. three times the six times sum of x and half the sum of x144424443 and y squared decreased by 144424443 of y squared 1442443 6(x  y2)

6 1x 

3

2

3

2

3

2

4

1

 3x  42y2

4

123s  s2  43t 3 2 4  4  1 3  1 2 s  3t 3

1

2

4

 4  3s  3t

1 2

3 1

2

3

1 2

2

43. 2p  5 2 p  2q  3  2p  5 2 p  5 2q  3 3

6

2

 2p  10 p  5q  3

3 1 2 6 2 3 6 2  1 2  10 2 p  5 q  3

 2p  10 p  5 q  3

23

6

2

 10 p  5 q  3 

2 3

23

2

 3 1x2  3 3

1

3

1

2

112 y22

2

Distributive Property Commutative () Distributive Property Commutative ()

48. Sometimes; Sample answer: 4  3  3  4, but 4  4  4  4. 49. You can use the Commutative and Associative Properties to rearrange and group numbers for easier calculations. Answers should include the following. • d  (0.4  1.1)  (1.5  1.5)  (1.9  1.8  0.8) 50. C; 6(ac  2b)  2ac  6(ac)  6(2b)  2ac  6ac  12b  2ac  6ac  2ac  12b  (6ac  2ac)  12b  (6  2)ac  12b  8ac  12b

4

3 1

6 1 y2 2

 x(6  3)  y2 6  2

2

 4  3s  s  3t

5

1

3 x

1 2 y 2

1

3 x  2y2



 6x  3x  6y2  2 y2

 4  3s  3t  s

3

y2 2

 6 1x2 

 3 1s  2t2  s  4  3 1s2  3 12t2  s 2

s

1442443 123

6

 10 p  5 q

19

Chapter 1

66. 4a  3b  4  2  3  5 835  8  15  23

51. B; the Commutative Property implies that 6  5  5  6.

Page 36

Maintain Your Skills

52. 5(2  x)  7x  5(2)  5(x)  7x  10  5x  7x  10  (5  7)x  10  12x  12x  10 53. 3(5  2p)  3(5)  3(2p)  15  6p 54. 3(a  2b)  3a  3(a)  3(2b)  3a  3a  6b  3a  3a  3a  6b  (3a  3a)  6b  (3  3)a  6b  0a  6b  6b 55. 7m  6(n  m)  7m  6(n)  6(m)  7m  6n  6m  7m  6m  6n  (7m  6m)  6n  (7  6)m  6n  13m  6n 56. (d  5)f  2f  (d)(f )  5( f )  2f  df  5f  2f  df  (5  2)f  df  7f 57. t2  2t2  4t  (1  2)t2  4t  3t2  4t 58. 3(10  5  2)  21  7  3(10  10)  21  7 Substitution; 5  2  10  3(0)  21  7 Substitution; 10  10  0  0  21  7 Multiplicative Property of Zero; 3  0  0 03 Substitution; 21  7  3 3 Additive Identity; 033 59. 12(5)  6(4)  60  6(4)  60  24  36 60. 7(0.2  0.5)  0.6  7(0.7)  0.6  4.9  0.6  4.3 61. 8[62  3(2  5)]  8  3  8[62  3(7)]  8  3  8[36  3(7)]  8  3  8[36  21]  8  3  8[15]  8  3  120  8  3  15  3  18 62. 2x  7  2  4  7 63. 6x  12  6  8  12 87  48  12  15  60 64. 5n  14  5  6  14 65. 3n  8  3  7  8  30  14  21  8  16  13

Chapter 1

Page 36

Practice Quiz 2

1. j; Additive Identity Property, since a  0  a 2. c; Substitution Property of Equality, since 18  7  11 3. i; Commutative Property, since a  b  b  a 4. f; Reflexive Property of Equality, since a  a 5. g; Associative Property, since (a  b)  c  a  (b  c) 6. d; Multiplicative Identity Property, since 1aa 7. b; Multiplicative Property of 0, since a  0  0 8. a; Distributive Property, since a(b  c)  ab  ac 9. h; Symmetric Property of Equality, since if a  b, then b  a 10. e; Multiplicative Inverse Property, since a b  1 b a

1-7 Page 39

Logical Reasoning Check for Understanding

1. Sample answer: If it rains, then you get wet. Hypothesis: it rains Conclusion: you get wet 2. Sample answer: Counterexamples are used to disprove a statement. 3. Sample answer: You can use deductive reasoning to determine whether a hypothesis and its conclusion are both true or whether one or both are false. 4. Hypothesis: it is January Conclusion: it might snow 5. Hypothesis: you play tennis Conclusion: you run fast 6. Hypothesis: 34  3x  16 Conclusion: x  6 7. Hypothesis: Lance does not have homework Conclusion: he watches television If Lance does not have homework, then he watches television. 8. Hypothesis: a number is divisible by 10 Conclusion: it is divisible by 5 If a number is divisible by 10, then it is divisible by 5. 9. Hypothesis: a quadrilateral has four right angles Conclusion: it is a rectangle If a quadrilateral has four right angles, then it is a rectangle. 10. The last digit of 10,452 is 2, so the hypothesis is true. Conclusion: the number is divisible by 2. Check: 10,452  2  5226 ✓ 2 divides 10,452.

20

33. The hypothesis is false. If the VCR cost less than $150, we know Ian will buy one. However, the hypothesis does not say Ian won’t buy a VCR if it costs $150 or more. Therefore, there is no valid conclusion. 34. The conditional statement does not mention DVD players. There is no way to determine if the hypothesis is true. Therefore, there is no valid conclusion. 35. The conclusion is true. If the cost of the VCRs is less than $150, the hypothesis is true, also. However, if the cost of the VCRs is $150 or more, the hypothesis is false. There is no way to determine the cost of the VCRs. Therefore, there is no valid conclusion. 36. People move to other states. 37. There is a professional team in Canada. 38. Girls can wear blue clothes. 39. Left-handed people can have right-handed parents. 40. x  2, y  3; Is 2  3 even? 2  3  6 and 6 is even, but 3 is not even. ? 41. 8 is greater than 7. 42. n  15; 4  15  8 52 2(8)  16 52 52 ✓ but 15  15 16  16

11. The conclusion is true. If the last digit of the number is 2, the hypothesis is true also. However, if the last digit is an even number other than 2, the hypothesis is false. There is no way to determine if the last digit is 2. Therefore, there is no valid conclusion. 12. The conclusion is true. 946 is divisible by 2. However, since the last digit is 6, the hypothesis is false. Therefore, there is no valid conclusion. 13. Anna could have a schedule without a science class. 14. a book that has more than 384 pages ?

12 7 1 11 ? 16. x  15; 3  15  7 52 52 52 ✓ but 15  15 15. x  1;

17. A;

? x  1, 12 7 1 11

Pages 40–42

Practice and Apply

18. Hypothesis: both parents have red hair Conclusion: their children have red hair 19. Hypothesis: you are in Hawaii Conclusion: you are in the tropics 20. Hypothesis: 2n  7 25 Conclusion: n 16 21. Hypothesis: 4(b  9)  68 Conclusion: b  8 22. Hypothesis: a  b Conclusion: b  a 23. Hypothesis: a  b, and b  c Conclusion: a  c 24. Hypothesis: it is Monday Conclusion: the trash is picked up If it is Monday, then the trash is picked up. 25. Hypothesis: it is after school Conclusion: Greg will call If it is after school, then Greg will call. 26. Hypothesis: a triangle has all sides congruent Conclusion: it is an equilateral triangle If a triangle has all sides congruent, then it is an equilateral triangle. 27. Hypothesis: a number is divisible by 9 Conclusion: the sum of its digits is a multiple of 9 If a number is divisible by 9, then the sum of its digits is a multiple of 9. 28. Hypothesis: x  8 Conclusion: x 2  3x  40 If x  8, then x 2  3x  40. 29. Hypothesis: s 9 Conclusion: 4s  6 42 If s 9, then 4s  6 42. 30. $139 is less than $150, so the hypothesis is true. Conclusion: Ian will buy a VCR. 31. $99 is less than $150, so the hypothesis is true. Conclusion: Ian will buy a VCR. 32. The conclusion is false. Ian did not buy a VCR, and he would have bought one if the cost were less than $150. Therefore, the VCR cost $150 or more.

6

1

43. x  3, y  2;

6 3



1 2

1

11 but

6 3

 1 and

1 2

1

44. Sample answer: P

Q

R

45. Sample answer: R

P

Q

46. See students’ work. There will probably be both examples and counterexamples. 47. Numbers that end in 0, 2, 4, 6, or 8 are in the “divisible by 2” circle. Numbers whose digits have a sum divisible by 3 are in the “divisible by 3” circle. Numbers that end in 0 or 5 are in the “divisible by 5” circle. 48. Sample answer: If a number is divisible by 2 and 3, then it must be a multiple of 6. 49. There are no counterexamples to the conclusions obtained in Exercises 47 and 48. 50. No; Sample answer: Let a  1 and b  2; then 1 * 2  1  2(2) or 5 and 2 * 1  2  2(1) or 4. 51. You can use if-then statements to help determine when food is finished cooking. Answers should include the following. • Hypothesis: you have small, underpopped kernels Conclusion: you have not used enough oil in your pan • If the gelatin is firm and rubbery, then it is ready to eat. If the water is boiling, lower the temperature.

21

Chapter 1

52. 8, since 14  8  12  112  12  100 53. C; # 4  

66. Multiplicative Property of Zero n  0, since 36  0  0. 67. 5(7)  6  x 35  6  x 41  x The solution is 41. 68. 7(42)  62  m 7(16)  36  m 112  36  m 76  m The solution is 76.

43 2 64 2

 32

Page 42

Maintain Your Skills

54. 2x  5y  9x  2x  9x  5y  (2x  9x)  5y  (2  9)x  5y  11x  5y 55. a  9b  6b  a  (9b  6b)  a  (9  6)b  a  15b 3

2

5

3

5

69. p   

2

56. 4g  5f  8g  4g  8g  5f

134 g  58 g2  25f 3 5 2  1 4  8 2 g  5f 





70. 71.

11 2 g  5f 8 3 2 1 8g  5f

72.

57. 4(5mn  6)  3mn  4(5mn)  4(6)  3mn  20mn  24  3mn  20mn  3mn  24  (20mn  3mn)  24  (20  3)mn  24  23mn  24 58. 2(3a  b)  3b  4  2(3a)  2(b)  3b  4  6a  2b  3b  4  6a  (2b  3b)  4  6a  (2  3)b  4  6a  5b  4 59. 6x2  5x  3(2x2)  7x  6x2  5x  6x2  7x  6x2  6x2  5x  7x  (6  6)x2  (5  7)x  12x2  12x 60. gallons gallons used used gallons number flushing showering used of days toilet and bathing in sink times 123 14243

73. 74. 75. 76. 77.

78.

14243 1442443 14243

61. 62. 63. 64. 65.

Chapter 1

2

28  2 22  8 28  4 14 7

2 The solution is 2. The word product implies multiply, so the expression can be written as 8x4. Times implies multiply, and decreased by implies subtract. So the expression can be written as 3n  10. More than implies add, and quotient implies divide. So the expression can be written as 12  (a  5). 40%  90  0.4(90)  36 23%  2500  0.23(2500)  575 18%  950  0.18(950)  171 38%  345  0.38(345)  131.1 42.7%  528  0.427(528)  225.456  225.5 67.4%  388  0.674(388)  261.512  261.5

1-8

(100  80  8)  d (100  80  8)d  (100)(d)  (80)(d)  (8)(d)  100d  80d  8d Two expressions to represent the amount of water used in d days can be (100  80  8)d and 100d  80d  8d. Multiplicative Identity Property n  64, since 1(64)  64. Reflexive Property n  7, since 12  7  12  7. Substitution Property n  5, since (9  7)5  (2)5. Multiplicative Inverse Property 1 n  4, since 4  4  1. Additive Identity Property n  0, since 0  18  18.

22  113  52

Page 46

Graphs and Functions Check for Understanding

1. The numbers represent different values. The first number represents the number on the horizontal axis and the second represents the number on the vertical axis. 2. Sample answer: A dependent variable is determined by the independent variable for a given function. 3. See students’ work. 4. Sample answer: Alexi’s speed decreases as he rides uphill, then increases as he rides downhill.

22

Height (cm)

5. Just before jumping from a plane, the skydiver’s height is constant. After she jumps, her height decreases until she lands. When she lands the skydiver’s height above the ground is zero. Graph B shows this situation. 6. Time is the independent variable, as it is unaffected by the height of the object above the ground. Height is the dependent quantity, as it is affected by time. 7. The ordered pairs can be determined from the table. Time is the independent variable, and the height above the ground is the dependent variable. So, the ordered pairs are (0, 500), (0.2, 480), (0.4, 422), (0.6, 324), (0.8, 186), and (1, 10). 500 8.

40 35 30 Cost

450 400 350 300 250 200 150 100 50 0

14. The ordered pairs can be determined from the table. The time parked is the independent variable, and the cost is the dependent variable. So, the ordered pairs are (0, 0), (1, 1), (2, 2), (3, 2), (4, 4), (5, 4), (6, 4) (7, 5), (8, 5), (9, 5), (10, 5), (11, 5), (12, 30), (13, 30), (14, 30), (15, 30), (16, 30), (17, 30), (18, 30), (19, 30), (20, 30), (21, 30), (22, 30), (23, 30), (24, 30), (25, 45), (26, 45), (27, 45), (28, 45), (29, 45), (30, 45), (31, 45), (32, 45), (33, 45), (34, 45), (35, 45), and (36, 45). 15. 45

25 20 15 10 5 0

0.2

0.4 0.6 Time (s)

0.8

4

8 12 16 20 24 28 32 36 Time

16. From 7:00 A.M. Monday to 7:00 A.M. Tuesday is 24 hours. From 7:00 A.M. Tuesday to 7:00 P.M. Tuesday is 12 hours. From 7:00 P.M. Tuesday to 9:00 P.M. Tuesday is 2 hours. Therefore, from 7:00 A.M. Monday to 9:00 P.M. Tuesday is 24  12  2  38 hours. The cost for the first 24 hours is $30. The cost for the next 14 hours is $15. Thus, the cost of parking from 7:00 A.M. Monday to 9:00 P.M. Tuesday is 30  15 or $45. 17. The number of sides of the polygon is the independent variable as it is unaffected by the sum of the interior angles, and the sum of the interior angles is the dependent variable as it is affected by the number of sides of the polygon. 18.

1.0

Height

9. Paul releases the ball above the ground. The height of the ball then decreases as it approaches the height at which the catcher catches the ball.

Time

900

Practice and Apply

Sums

Pages 46–48

10. Michelle gets a fever and takes some medicine. After a while her temperature comes down, then slowly begins to go up again. 11. Rashaad’s account is increasing as he makes deposits and earns interest. Then he pays some bills. He then makes some deposits and earns interest, and so on. 12. As it moves along, a radio-controlled car has a constant speed. When it hits the wall, its speed is zero. Graph C shows this situation. 13. A person’s income starts at a certain level, then, in general, their income increases. This is represented by starting at a certain height, then gradually increasing. Graph B shows this situation.

720 540 360 180 0

1

2

3

4 5 Sides

6

7

19. If we look for a pattern in the numbers 180, 360, 540, 720, and 900, we see that the numbers are all multiples of 180. In fact: 180  1  180 360  2  180 540  3  180 720  4  180 900  5  180 Thus, the next three numbers would be 6  180  1080, 7  180  1260, and 8  180  1440. Therefore, we would predict that the sum of the interior angles for an octagon is 1080, for a nonagon is 1260, and for a decagon is 1440.

23

Chapter 1

20. A car’s value decreases as it gets older, but if it is taken care of, the value of the car again increases.

24. B; The value 4 on the horizontal axis corresponds to the highest point on the curve. 25. A; cost for charge for minus profit is each CD each CD 123 123 14243

123 14243

 (2.50 0.35)  number cost of times minus equipment of CDs 123 14243 14243 1442443

Value

p

Years

  n or p  2.15n  875

21. When the block of ice is removed from the freezer the temperature starts to increase. The temperature increases until it approaches the room temperature, at which time the temperature gradually becomes constant.

Page 48

Temperature (˚F )

22a. At each value of the independent variable the dependent variable is 23 units greater. The following table shows this relationship. Lisa’s Age 5 10 15 20 25 30 35 40 Mallory’s 28 33 38 43 48 53 58 63 Age The ordered pairs for this data are (5, 28), (10, 33), (15, 38), (20, 43), (25, 48), (30, 53), (35, 58), and (40, 63). Graph the ordered pairs. Then draw a line through the points. 90

Mallory’s Age

80 70

U.S. Commercial Radio Stations by Format, 2000

60 50 40 30

Number of Stations

20 10 5 10 15 20 25 30 35 40 45 Lisa’s Age

2400 2200 2000 1800 1600 1400 1200 1000 800 600 0

24

ck

Format

Ro

Co

s ie ld O k al s/T w y Ne ar t or ul mp Ad te n Co ry t un

22b. We know Mallory is 23 years older than Lisa. So, for Mallory to be twice as old as Lisa, Lisa must be 23 years old. The point on the graph that corresponds to this situation is (23, 46). 23. Real-world data can be recorded and visualized in a graph and by expressing an event as a function of another event. Answers should include the following. • A graph gives you a visual representation of the situation, which is easier to analyze and evaluate. • During the first 24 hours, blood flow to the brain decreases to 50% at the moment of the injury and gradually increases to about 60%. • Significant improvement occurs during the first two days.

Chapter 1

Maintain Your Skills

26. Hypothesis: you use a computer Conclusion: you can send e-mail 27. Hypothesis: a shopper has 9 or fewer items Conclusion: the shopper can use the express lane 28. ab(a  b)  (ab)a  (ab)b Distributive Prop.  a(ab)  (ab)b Commutative ()  (a  a)b  a(b  b) Associative ()  a2b  ab2 Substitution 29. Substitution Property n  3, since (12  9)(4)  (3)(4). 30. Multiplicative Property of Zero n  0, since 7(0)  0. 31. Multiplicative Identity Property n  1, since (1)(87)  87. 32. Step 1 Draw a horizontal axis and a vertical axis. Label the axes. Add a title. Step 2 Draw a bar to represent each category. The vertical scale is the number of radio stations using each format. The horizontal scale identifies the formats used.

Time

0

875

Page 49

Algebra Activity (Follow-Up of Lesson 1-8)

1-9

1. Sample answer: In 1900 there were 15,503,000 students, and in 1920 there were 21,578,000 students. Therefore, we could estimate that there

Pages 53–54

were  18,540,500, or about 18,540,000 students in 1910. In 1970 there were 45,550,000 students, and in 1980 there were 41,651,000 students. Therefore, we could estimate that there were 45,550,000  41,651,000 2

4.

5.

 43,600,500 or about 43,600,000 students in 1975. Sample answer: 55,000,000. By looking at general trends of the graph, we see that it is increasing, and the amount of students will be around 55,000,000 in 2020. Sample answer: For Exercise 1 we averaged the enrollments for 1900 and 1920 and then for 1970 and 1980. For Exercise 2 we calculated the increase in students per year from 1900 to 1980, then added 40 times that amount to the 1980 enrollment. If the U.S. population does not increase as quickly as in the past, then the number of students may be too high. The year is the independent variable since it is unaffected by the number of students per computer, and the number of students per computer is the dependent variable since it is affected by the year. The ordered pairs can be determined from the table with x representing the number of years since 1984. The orderd pairs are (0, 125), (1, 75), (2, 50), (3, 37), (4, 32), (5, 25), (6, 22), (7, 20), (8, 18), (9, 16), (10, 14), (11, 10.5), (12, 10), (13, 7.8), (14, 6.1), (15, 5.7). Graph the ordered pairs.

Students per Computer

3.

14243

123

8940   8940 of the students were from Germany. 7. The percentage representing the number of students from Canada is 0.15%. The percentage representing the number of students from the United Kingdom is 0.05%. So, find 0.15  0.05 or 0.1% of 14.9 million. 0.1% of 14,900,00 equals 14,900 123 1442443 14243 0.0006

14,900,000

14243

123

8. 9.

130 120 110 100 90 80 70 60 50 40 30 20 10 0

Check for Understanding

1. A circle graph compares parts to the whole. A bar graph compares different catagories of data. A line graph shows changes in data over time. 2. See students’ work. 3. Sample answer: The percentages of the data do not total 100. 4. The bar for the number of schools participating in basketball shows 321 and the bar for the number of schools participating in golf shows 283. So, there were 321 – 283 or 38 more schools participating in basketball than in golf. 5. The bar showing the least amount is for tennis. Therefore, of the sports listed, tennis is offered at the fewest schools. 6. The percentage representing the number of students from Germany is 0.06%. So, find 0.06% of 14.9 million. 0.06% 14,900,00 equals 8940. of 1442443 14243 123

15,503,000  21,578,000 2

2.

Statistics: Analyzing Data by Using Tables and Graphs

10. 11.

0.001  14,900,000 14,900  14,900 more students were from Canada than from the United Kingdom. No; the data do not represent a whole set. A bar graph would be more appropriate, since a bar graph is used to compare similar data in the same category. Sample answer: The vertical axis scale shows only partial intervals. The vertical axis needs to begin at 0.

Pages 54–55

Practice and Apply

film stock plus processing plus prep for telecine plus 12. 1 442443 123 1442443 123 144424443 123 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Years Since 1984

3110.40





621.00



60.00

telecine plus tape stock equals total cost.

14243 123 1442443 14 424 43 1442443

A prediction is 1 student per computer because it does not seem likely that schools would have more computers than students.

1000.00



73.20



4864.60

The total cost is $4864.60. 13. original

backup total cost of tape stock plus tape stock equals digital video . 1442443 123 1442443 14 424 43 144424443 10.00 10.00 20.00  

The total cost of using digital video is $20.00, and from Exercise 12 the total cost of using 35-millimeter film is $4864.60. 4864.60  20.00  243.23 The cost of 35-millimeter film is about 250 times as great.

25

Chapter 1

Page 55

14. The section of the graph representing books purchased in the spring is 19%, so find 19% of 25 million. 19% 25,000,000 equals 4,750,000 1 23 of { 14243 123 14243 0.19  25,000,000  4,750,000

Heart rate

About 5 million books were purchased in the spring. 15. The section of the graph representing books purchased in the summer is 15%, so find 15% of 15,000. 15% 15,000 equals 2250. 1 23 of { 14243 123 1 23 0.15  15,000  2250

Distance (mi)

23. 4x  5  42 4x  47

She should expect to sell about 2250 books. 16. The vertical axis is extended and does not begin at 0. It gives the impression that the number of cable television systems is decreasing rapidly. 17. Yes, the graph is misleading because the sum of the percentages is not 100. To fix the graph, each section must be drawn accurately and another section that represents “other” toppings should be added. 18a. To show little increase, bunch the values on the vertical scale closer together.

3

x  11 4 Thus, x  12 is a sample counterexample. 24. Sample answer: 3

x  2;

but

Percentage of U.S. Households

3 ? 7 1 2 3 7 1 2 1 3  3 2 2



25. a rectangle with length 6 inches and width 2 inches 6  6  2  2  16 16  16 ✓ However, each side is not 4 inches long. 26. 7a  5b  3b  3a  7a  3a  5b  3b  (7a  3a)  (5b  3b)  (7  3)a  (5  3)b  10a  8b 27. 4x2  9x  2x2  x  4x2  2x2  9x  x  (4x2  2x2)  (9x  x)  (4  2)x2  (9  1)x  6x2  10x

Color Television Ownership, 1980–2000 100 90 80 70 60 ’80 ’85 ’90 ’95 ’00 Year

18b. To show rapid increase spread the values on the vertical scale further apart. Color Television Ownership, 1980–2000 Percentage of U.S. Households

Maintain Your Skills

22. Pedro’s heart rate increases as he exercises, and continues to increase until he is done sprinting. His heart rate decreases during his last walk until it returns to a normal rate.

1

2

1

1

1

1

5

7

2

100

112 n  13 n2  123 m  12 m2 1 1 2 1  12  3 2n  13  2 2m



95 90 85

 6n  6m

80

5

1

 6n  1 6m

’80 ’85 ’90 ’95 ’00 Year

18c. See students’ graphs and explanations. 19. Tables and graphs provide an organized and quick way to examine data. Answers should include the following. • Examine the existing pattern and use it to continue a graph to the future. • Make sure the scale begins at zero and is consistent. Circle graphs should have all percents total 100%. The right kind of graph should be used for the given data. 20. C; From the second to the third day the temperature increase was about 8F. 21. C; A line graph shows changes in data over time.

Page 56

Spreadsheet Investigation (Follow-Up of Lesson 1-9)

1. Enter the data in a spreadsheet. Use Column A for the years and Column B for the sales. Select the data to be included in the graph. Then use the graph tool to create a line graph. Snowmobile Sales Sales (millions)

1200 1000 800 600 400 200 0 1990

Chapter 1

1

28. 2n  3m  2m  3n  2n  3n  3m  2m

26

1992

1994 Year

1996

1998

21. 3  2  4  3  8  11

2. Enter the data in a spreadsheet as in Exercise 1. Select the data to be included in your graph. Then use the graph tool to create a bar graph.

22.

Snowmobile Sales

110  62 8

1

2

Sales (millions)

1200 1000

23. 18  42  7  18  16  7 27 9 24. 8(2  5)  6  8(7)  6  56  6  50 25. 4(11  7)  9  8  4(18)  9  8  72  9  8  72  72 0 26. 288 [3(9  3)]  288 [3(12)]  288 36 8 27. 16 2  5  3 6  8  5  3 6  40  3 6  120 6  20 28. 6(43  22)  6(64  22)  6(64  4)  6(68)  408

800 600 400 200 0 1990

1992

1994 Year

1996

1998

3. Yes; you can change the scales to begin at values other than zero, or change the intervals on the scale to be misleading.

Chapter 1 Study Guide and Review Page 57 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

4

8

Vocabulary and Concept Check

a; Additive Identity Property e; Multiplicative Identity Property g; Multiplicative Property of Zero f; Multiplicative Inverse Property h; Reflexive Property j; Symmetric Property i; Substitution Property k; Transitive Property b; Distributive Property d; Associative Property

29. 13  12 3 

14  62 15  22

 33    

Pages 57–62

Lesson-by-Lesson Review

11. a number x to the fifth power

30. t2  3y    

14444444244444443 x5

The algebraic expression is x5. 12. five times a number x squared 123 1444 4424 44443 123

ty x



14  62

15  22 10  15  22 10 33  10 10 27  10

33

 27  1  26 42  3  2 31. xty3  3  4  23 16  3  2 3 4 8 16  6  12  8 22  96

4  2 3 8 3 2 23

 x2 5 The algebraic expression is 5x2. twenty-one 13. a number x sum of 1442443

32.

21 x  The algebraic expression is x  21. 14. twice a number x difference of eight

33. x  t2  y2  3  42  22  3  16  22  3  16  4  19  4  23 34. 3ty  x2  3  4  2  32 3 4 29  12  2  9  24  9  15 35. 8(x  y)2  2t  8(3  2)2  2  4  8(1)2  2  4 8 12 4 82 4 88  16

 

1442443 14243

14444244443 144 424 443 14243

2x 8  The algebraic expression is 2x  8. 15. 33  3  3  3 16. 25  2  2  2  2  2  27  32 17. 54  5  5  5  5  625 18. the product of two and a number p squared 19. the product of three and a number m to the fifth power 20. the sum of one half and two

27

Chapter 1

36. x  22  13 9 The solution is 9. 38. m  



14  28 4  3 42 7



6 The solution is 6. 42. b   

4 2  4  5 15 S 13  15

714  32

5 2  5  5 15 S 15 15

18 9



6 84 6



6 2  6  5 15 S 17 15

4 5 6 7 8

x27 ?

4  2 7 7S6 7 ?

5  2 7 7S7 7 ?

6  2 7 7S8 7 7 ?

7  2 7 7S9 7 7 ?

8  2 7 7 S 10 7 7

7 2  7  5 15 S 19 15

96 6 8 2 16 4

4

10  x  7 ?

10  4 6 7 S 6 6 7 ?

5

10  5 6 7 S 5 6 7

6

10  6 6 7 S 4 6 7

7

10  7 6 7 S 3 6 7

8

10  8 6 7 S 2 6 7

? ? ?

true ✓

?

8 2  8  5 15 S 21 15

true ✓

The solution set for 2x  5 15 is {5, 6, 7, 8}. 48. 2[3 (19  42)]  2[3 (19  16)] Substitution; 42  16  2[3 3] Substitution; 19  16  3 2 1 Substitution; 3 3  1 2 Multiplicative Identity; 2 12

6172  2132 2

4  6122 42  6 16  12 36 4

49.

1 2

 2  2[2  3  1] 1

 2  2  2[6  1] Substitution; 2  3  6 1

 22  25 125

Substitution; 6  1  5 Multiplicative Inverse; 21

1 2

Substitution; 2  5  10

 1  10

true ✓

 11Substitution; 1 + 10  11 50. 42  22  (4  2)  42  22  2 Substitution; 4  2  2  16  22  2 Substitution; 42  16  16  4  2 Substitution; 22  4  12  2 Substitution; 16  4  12  10 Substitution; 12  2  10 51. 1.2  0.05  23  1.2  0.05  8 Substitution; 23  8  1.15  8 Substitution; 1.2  0.05  1.15  9.15 Substitution; 1.15  8  9.15 52. (7  2)(5)  52  5(5)  52 Substitution; 7  2  5  5(5)  25 Substitution; 52  25  25  25 Substitution; 5  5  25 0 Substitution ()

true ✓

53. 3(4 4) 2  4 (8)

True or False? false false true ✓ true ✓ true ✓

The solution set for x  2 7 is {6, 7, 8}. 46. Replace x in 10  x 7 with each value in the replacement set. x

true ✓

?

 14 9 The solution is 14. The solution is 9. 44. y  5[2(4)  13]  5[8  1]  5[7]  35 The solution is 35. 45. Replace x in x  2 7 with each value in the replacement set. x

true ✓

?

4 The solution is 4. 43. x 

18 3 71122

false

?

2 The solution is 2. 41. n 

True or False?

?

39. x  12  3 

2x  5  15

x

21  3

64  4 17 68 17

4 The solution is 4. 40. a 

47. Replace x in 2x  5 15 with each value in the replacement set.

37. y  4  32 49  13 The solution is 13.

True or False?

1

1

true ✓

 3(1) 2  4 (8) Substitution; 4 4  1

true ✓

31 3

true ✓

The solution set for 10  x 7 is {4, 5, 6, 7, 8}.

1 182 4

1 (8) 4

32

Substitution; 12  1 Multiplicative Identity; 313 Substitution;

1 4

82

1 Substitution; 3  2  1 54. 2(4  7)  2(4)  2(7) 55. 8(15  6)  8(15)  8(6)  8  14  120  48  22  72

Chapter 1

28

56. 4(x  1)  4(x)  4(1)  4x  4 57. 3

1

1 3

2

p 3

72. five times the decreased by 2x sum of x and y 1 44424443 123 144424443

1 2  31 p2

 2x 5(x  y) 5(x  y)  2x  5(x)  5(y)  2x Distributive Property  5x  2x  5y Commutative ()  3x  5y Substitution 73. twice the product the product increased by of p and q of p and q 144424443

1 3

 1  3p 58. 6(a  b)  6(a)  6(b)  6a  6b 59. 8(3x  7y)  8(3x)  8(7y)  24x  56y 60. 4a  9a  (4  9)a  13a 61. There are no like terms. 4np  7mp is simplified. 62. 3w  w  4v  3v  (3w  w)  (4v  3v)  (3  1)w  (4  3)v  2w  1v  2w  v 63. 3m  5m  12n  4n  (3m  5m)  (12n  4n)  (3  5)m  (12  4)n  8m  8n 64. 2p(1  16r)  2p(1)  2p(16r)  2p  32pr 65. 9y  3y  5x  (9y  3y)  5x  (9  3)y  5x  12y  5x 66. 3x  4y  2x  3x  2x  4y  (3x  2x)  4y  (3  2)x  4y  5x  4y 2 67. 7w  w  2w2  7w2  2w2  w  (7w2  2w2)  w  (7  2)w2  w  9w2  w 1

1

14444244443

1442443

 5( p  q) pq 2pq  pq  (2  1)pq Distributive Property  3pq Substitution 74. the sum of eight six times a plus times b and twice a 1442443 123 1444442444443 6a  (8b  2a) 6a  8b  2a  6a  2a  8b Commutative ()  (6a  2a)  8b Associative ()  (6  2)a  8b Distributive Property  8a  8b Substitution 75. three times the sum of x squared the square of x and seven times x plus 144424443 123 1444442444443 3  x2 (x2  7  x)  3x2  x2  7x  (3x2  x2)  7x Associative ()  (3  1)x2  7x Distributive Property  4x2  7x Substitution 76. Hypothesis: it is 7:30 A.M. Conclusion: school begins If it is 7:30 A.M., then school begins. 77. Hypothesis: a figure is a triangle Conclusion: it has three sides If a figure is a triangle, then it has three sides.

1 1 1 2 1 1  13 2  2 2m  n

68. 3 2m  2m  n  3 2m  2m  n

78. x  13, y  12;

?

13 7 12 13 7 12 ✓ but 21132 31122

 4m  n 69. 6a  5b  2c  8b  6a  5b  8b  2c  6a  (5b  8b)  2c  6a  (5  8)b  2c  6a  13b  2c 70. 3(2  3x)  21x  3(2)  3(3x)  21x  6  9x  21x  6  (9x  21x)  6  (9  21)x  6  30x  30x  6 71. 6(2n  4)  5n  6(2n)  6(4)  5n  12n  24  5n  12n  5n  24  (12n  5n)  24  (12  5)n  24  17n  24

79. a  15, b  1, c  12;

?

?

15 7 1 and 15 7 12 15 7 1 and 15 7 12 ✓ but 1 12 80. When taking off, the airplane has zero altitude. Then the plane gains altitude. The plane’s altitude remains constant as it flies around. Then the altitude decreases until the plane lands. Graph C shows this situation.

29

Chapter 1

7 difference of a number x squared 5. { 44443 144424443 14444424 7 x2  The algebraic expression is 7  x2. 6. 5(9  3)  3  4  5(12)  3  4  60  3  4  60  12  48 7. 12  6 3  2 8  72 3  2 8  24  2 8  48 8 6 8. a2b  c  22  5  3 4 53  20  3  23 9. (cd)3  (3  1)3 10. (a  d )c  (2  1)3  33  (3)3  27 9 11. y  (4.5  0.8)  3.2 12. 42  3(4  2)  x y  5.3  3.2 42  3(2)  x y  2.1 16  3(2)  x The solution is 2.1. 16  6  x 10  x The solution is 10.

Helium (cm3 )

81. The amount of helium in a balloon increases from zero at a steady rate until it bursts. When the balloon bursts, the amount of helium in the balloon is again zero.

Time

82. Construct a table with 2 rows. The first row is labeled Earth Years and the second row is labeled Mars Years. Since one year on Earth is about 0.54 years on Mars, pick some value for Earth Years and multiply by 0.54 to get values for Mars Years. Earth Years

5

10

15

20

25

Mars Years

2.7

5.4

8.1

10.8

13.5

83.

50

Mars Years

40 30 20

13.

10

0

10

20 30 40 Earth Years

50

84. The section of the graph representing students who chose the amusement park is 45% of the circle, so find 45% of 120. 45% 120 equal 54. { 1 23 123 { 1 23 of

n n n 7 3

1

or 2 3.

14. 32  2  (2  2)  9  2  (2  2) Substitution; 32  9 920 Substitution; 2  2  0 70 Substitution; 9  2  7 7 Additive Identity; 7  0  7 15. (2  2  3)  22  32  (4  3)  22  32 Substitution; 2  2  4  1  22  32 Substitution; 4  3  1  1  4  32 Substitution; 22  4 149 Substitution; 32  9 59 Substitution; 1  4  5  14 Substitution; 5  9  14 16. 2m  3m  (2m  3m)  (2  3)m  5m 17. 4x  2y  2x  y  4x  2x  2y  y  (4x  2x)  (2y  y)  (4  2)x  (2  1)y  2x  3y 18. 3(2a  b)  5a  4b  3(2a)  3(b)  5a  4b  6a  3b  5a  4b  6a  5a  3b  4b  (6a  5a)  (3b  4b)  (6  5)a  (3  4)b  1a  7b  a  7b

Chapter 1 Practice Test Page 63 d; Reflexive Property of Equality a; Substitution Property of Equality c; Transitive Property of Equality a number x sum of 13 1442443 14243 123

13 x  The algebraic expression is x  13.

Chapter 1

n

The solution is

0.45  120  54 54 students chose the amusement park. 85. The section of the graph representing students who chose the amusement park is 45% of the circle. The section of the graph representing students who chose the water park is 25%. So, find 45 – 25 or 20% of 180. 20% 180 equal 36. 1 23 of { 1 23 123 { 0.2%  180  36 There were 36 more students who chose the amusement park than the water park.

1. 2. 3. 4.

23  13 2  1 8  1 3 7 3 1 23

30

2. C; 2    r  2  3.14  4  6.28  4  25.12 3. B; There are about 50 weeks in a year. 80  50  4000

19. Running for 15 minutes does not mean you can run for a few hours. 20. 2x  3 9 2x 12 x 6 Thus, x  6 is a counterexample. 21. When a basketball is shot, its height starts above the ground. Then it increases until the basketball reaches its maximum height. The height of the ball decreases until it gets to the basket, at which time it falls back to the ground.

4 1250 48 2 5. D; charge number cost of replaced plus per hour times of hours parts 14243 123 14243 123 144424443 4. A;

Height

  ($36 h) $85 or (36  h)  85 6. D; 3(2x  3)  2(x  1)  3(2x)  3(3)  2(x)  2(1)  6x  9  2x  2  6x  2x  9  2  (6x  2x)  (9  2)  (6  2)x  11  8x  11 7. B; 3 is a positive integer, but 9 is not divisible by 2. 8. B; The section of the graph representing foreignborn people from Asia is 25.2% of the circle. The section of the graph representing foreignborn people from Central America is 34.5% of the circle. 25.2  34.5

Time

Height

22. The nickel’s height is at a maximum when it is dropped. It’s height decreases until it hits the stack of pennies. The height of the nickel increases as it bounces off the pennies.

5

9. Five eighths of the students are girls, so 1  8 or 3 3 of the students are boys. Find 8 of 32. 8 3 of 32 8 { { { 3  32 8

Time

23. The section of the graph representing students who chose chocolate is 62% of the circle. The section of the graph representing students who chose vanilla is 32% of the circle. So, find 62 – 32 or 30% of 200. 30% 200 equal 60. 1 23 of { 12 3 123 { 0.3  200  60 There were 60 more students who chose chocolate than vanilla. 24. The section of the graph representing students who chose chocolate is 62% of the circle. The section of the graph representing students who chose vanilla is 32% of the circle. So, 62  32 or 94% of the students chose either chocolate or vanilla. 25. D; 2 is a prime number, but is not odd.

There are 12 boys in the class. 10.

1

3

2

cost of the books

plus

tax

[ (8.99  13.99)



0.06(8.99  13.99) ]

$25 minus

123 14243 144424443 123 14444244443

25



25  [(8.99  13.99)  0.06 (8.99  13.99)] 25  [(8.99  13.99)  0.06 (8.99  13.99)]  25  [22.98  0.06(22.98)]  25  [22.98  1.38]  25  24.36  0.64 She should receive $0.64 in change. 11. The bars with the least difference in height are representing 1999. Therefore, 1999 was the year with the least difference in home runs. 12. 15% of 80 equals 1 12. B;123 23 123 123

1 424 3

12   80 12.5 of 50 equals 123 123 123 1 424 3 123 12.5 0.25  50  12 12.5 0.15 25%

Chapter 1 Standardized Test Practice Pages 64–65

equal 12. 123 {  12

1

1. B; 1  4  800  200  4  800  200  200  200  40,000

31

Chapter 1

13. A;

10 2 3

2

15. C;

 10 3

1 1a 4

3

 10  2 

10 3  1 2



10  3 12



53 11 15 1

 15 15 7 15 14. B; Replace x in 2x  1 2x  1 with each value in {1, 2, 3, 4, 5}. 2x  1  2x  1

1

2  1  1 6 2  1  1S1 6 3

? ?

2  2  1 6 2  2  1S3 6 5 ?

2  3  1 6 2  3  1S5 6 7 ?

4

2  4  1 6 2  4  1S7 6 9

5

2  5  1 6 2  5  1 S 9 6 11

?

ac  bc 4

True or False? true ✓ true ✓ true ✓

Hours

true ✓

17c. This graph shows that only the old pump was pumping at first, and then after some hours the new pump started pumping also.

true ✓

Notice that the expression 2x  1 is always 2 units greater than the expression 2x  1.

Chapter 1

1

Gallons

x

3

 4 1ac  bc2

16. C; (26  39)  (39  13)  (26  39)  (13  39)  (26  13)  39  39  39  392 17a. Sample answer: The new pump pumps many more gallons per hour than the old pump. The new pump pumps about twice as many gallons per hour as the old pump. 17b. The gallons pumped per hour by both pumps at the same time is the sum of the number of gallons pumped per hour by each pump individually.

1

2

1



5



 b2c  4 3a 1c2  b 1c2 4

32

Chapter 2 Real Numbers Page 67 1.

5.

Getting Started

2.2 0.16 2.36

1 4

2

2. 0 12 1 1 3 .4 4.5 8.9

3.

8

6.

3

 3  12  12

19. mean  4.

33

6.4 8.8 512 51 2 56.3 2 1 2

1

3

11

3

2

1 1

5

3

 10  4  10

8.

2



8 There are six numbers in the set. When ordered from least to greatest, the third number is 7 and 7  7 the fourth is 7. So, the median is 2 or 7. 7 appears two times, and the other numbers of the set appear only once. So, the mode is 7.

6

1

5 4



366

 12 7.

18 4.25 76 . 50   4 25 34 00 34 00 0

4 9

1

4

3

 3  91

3 8

3



4 3

or



14. b  3  

36 4

1

1

 0.3(8)  2.4 17. mean   

2

15. xy  7  0.3  2.1

2  4  7  9  12  15 6 49 6 1 86

 

145 22  45  45 16

 25

2-1

Rational Numbers on the Number Line

Page 70

Check for Understanding

5

6.

6

5432 1 0 1 2 3 4 5 6 7 8

7. 4 3 2 1

0

1

2

3

4

8.

There are six numbers in the set. When ordered from least to greatest, the third number is 7 and 7  9 the fourth is 9. So, the median is 2 or 8. Every number in the set appears only once. So, there is no mode. 18. mean 

23.

123 22  23  23

1. Let a be any integer. Since a can always be a written in the form 1 , a is also a rational number. The statement is always true. 2. Sample answer: Absolute value is how far from zero a number is. 3. Sample answer: describing distances in a given direction such as north versus south, or left versus right 4. The dots indicate each point on the graph. The coordinates are {2, 1, 2, 5}. 5. The bold arrow on the left means that the graph continues indefinitely in that direction. The 11 9 7 5 3 coordinates are . . .,  2 , 2, 2, 2, 2 .

1

16. y(a  b)  0.3 2  4

22.

4

 12

3

1 34

21. 0.92  0.9  0.9  0.81

9

9 1 4

20. 112  11  11  121 1 13

9. 3a  2  3  2  2 10. 2x  5  2  7  5 62  14  5 4  19 11. 8( y  2.4)  8(0.3  2.4)  8(2.7)  21.6 1 1 1 12. 4(b  2)  4 4  2 13. a   2  2 2

1 2 1 8  414  4 2 9  414 2

7  19  2  7  4  9 6 48 6

1

1 2

0 1 2 4 5

1

5 3

9. 9 8 76 5 4 3 2 1 0 1

10. 2 is two units from zero in the negative direction. 02 0  2 11. 18 is eighteen units from zero in the positive direction. 018 0  18 12. 2.5 is two and one half units from zero in the positive direction. 02.5 0  2.5

23  23  23  12  12  14 6 107 6 5 176

There are six numbers in the set. When ordered from least to greatest, the third number is 14 and 14  23 1 the fourth is 23. So, the median is or 182. 2 The number 23 appears three times, and the other numbers of the set appear only once. So, the mode is 23.

5

13. 6 is five-sixths unit from zero in the negative direction.

0 56 0  56

33

Chapter 2

14. 57  0x  34 0  57  018  34 0  57  052 0  57  52 5 15. 19  021  y 0  19  021  4 0  19  017 0  19  17  36 16. 0z 0  0.26  00.76 0  0.26  0.76  0.26  0.50 17. Rational Numbers Integers

1 2

Whole Numbers 0

40 4 2 53 3 13

Pages 71–72

34. 38 is thirty-eight units from zero in the negative direction. 038 0  38 35. 10 is ten units from zero in the positive direction. 010 0  10 36. 97 is ninety-seven units from zero in the positive direction. 097 0  97 37. 61 is sixty-one units from zero in the negative direction. 061 0  61 38. 3.9 is three and nine tenths units from zero in the positive direction. 03.9 0  3.9 39. 6.8 is six and eight tenths units from zero in the negative direction. 06.8 0  6.8

1.25 2 1 5   3 5 2 0.33 2.98 49.98

23

40. 56 is twenty-three fifty-sixths unit from zero in the negative direction.

Practice and Apply

41.

18. The dots indicate each point on the graph. The coordinates are {4, 2, 0, 2, 4}. 19. The dots indicate each point on the graph. The coordinates are {7, 6, 5, 3, 2}. 20. The bold arrow on the right means that the graph continues indefinitely in that direction. The coordinates are {2, 3, 4, 5, 6, . . .}. 21. The bold arrow on the left means that the graph continues indefinitely in that direction. The coordinates are {. . ., 0, 0.2, 0.4, 0.6, 0.8}. 22. The dots indicate each point on the graph. The 5 2 coordinates are 2, 3, 1, 3, 1 .

5 5

42.

43. 44. 45.

23. The dots indicate each point on the graph. The 1 4 7 8 coordinates are 5, 5, 5, 5, 2 . 24. 5 4 3 2 1

0

1

2

3

4

46.

25. 1 0 1 2 3 4 5 6 7 8 9 10

26. 7

6

5

4

3

2

47.

27. 2 1

0

1

2

3

4

5

6

28. 9 8 7 6 5 4 3 2

48.

29. 7 6 5 4 3 2 1

0

30. 1  2  1 3

3

1 3

0

2 3

1 11 12 2 3

3

49.

31. 4 3 2 1

0

1

2

3

32. 9 8 7 6 5 4 3 2 1 0 1

50.

33. 6 4 2

Chapter 2

0

2

4

6

8

35 80

is thirty-five eightieths unit from zero in the positive direction.

0 3580 0  3580

108 6 4 2 0 2 4 6 8 10

6

6

0 2356 0  2356

10

34

Therefore, the percents of change from least to greatest are 10.6, 2.9, 1.4, 0.2, 1.7, 4.3, 4.7, 5.3, 8.5, and 10.0. Philadelphia, PA; Sample answer: It had the greatest absolute value. Wayne, MI; Sample answer: It had the least absolute value. 48  0x  5 0  48  012  5 0  48  07 0  48  7  55 25  017  x 0  25  017  12 0  25  029 0  25  29  54 017  a 0  23  017  6 0  23  011 0  23  11  23  34 043  4a 0  51  043  4  6 0  51  043  24 0  51  019 0  51  19  51  70 0z 0  13  4  05 0  13  4  5  13  4  18  4  14 28  13  0z 0  28  13  05 0  28  13  5  15  5  20

51. 6.5  08.4  y 0  6.5  08.4  3.2 0  6.5  05.2 0  6.5  5.2  1.3 52. 7.4  0y  2.6 0  7.4  03.2  2.6 0  7.4  00.6 0  7.4  0.6 8 53.

1 6

0

7

0

60. Sample answer: You can plot the data on a number line to visualize their relationship. Answers should include the following. • Determine the least and greatest values of the data, and use those as the endpoints of the line. • Find the absolute value of each number. 61. D; 08 0  2  8  2 6 62. C; Zero is a whole number, but zero is not a natural number.

0 23  127 0 1 1  6  0 12 0 1

 b  12  6  1

1

 6  12

Page 72

1

4

1

1

2 0 5 0 123  12 2  0 56 0 2 1 5  13  2 2  6

54. b  2  6 

7

5

66 1

2

3

0 54  1 0  25 1 2  040  5

55. 0 c  1 0  5 

1

2

45 13

 20

1

2 0 5 0 1 12 5 1  4  12  2 2

56. 0c 0  2  2  4  2  2 1

5

5

15 4

or 3 4

42 

3

57.  0x 0  1  0x 0 Therefore, 0x 0   0x 0 can be thought of as what number when multiplied by 1 is equal to itself? The number 0 is the only number that satisfies that characteristic. i.e. 1  0  0. So, 0x 0  0. What number is zero units from zero? x  0, or the solution is 0. 58. 10

0

10

Maintain Your Skills

63. The graph shows that Mr. Michaels’s sales were less than 10 cars for each month except December. In December Mr. Michaels sold 12 cars. Therefore, his greatest sales happened in December. 64. The graph shows that Mr. Michaels’s change in sales between any two consecutive months was always less than 7 cars except between November and December when it was 8 cars. Therefore, the greatest change in sales occurred between November and December. 65. The graph shows that Mr. Michaels sold 3 cars in February, July, and October. 66. The volume starts high, decreases when she turns it down, remains constant while she is on the phone, then increases to its starting volume when she gets off the phone.

20

Volume

Time

67. 8x  2y  x  8x  x  2y  (8x  x)  2y  (8  1)x  2y  9x  2y 68. 7(5a  3b)  4a  7(5a)  7(3b)  4a  35a  21b  4a  35a  4a  21b  (35a  4a)  21b  (35  4)a  21b  31a  21b 69. 4[1  4(5x  2y)]  4[1  4(5x)  4(2y)]  4[1  20x  8y]  4[1]  4[20x]  4[8y]  4  80x  32y

30

59. 11 is eleven units from zero in the negative direction. For Bismarck, 011 0  11. 5 is five units from zero in the negative direction. For Caribou, 05 0  5. 4 is four units from zero in the negative direction. For Chicago, 04 0  4. 9 is nine units from zero in the negative direction. For Fairbanks, 09 0  9. 13 is thirteen units from zero in the negative direction. For International Falls, 013 0  13. 7 is seven units from zero in the positive direction. For Kansas City, 07 0  7. 34 is thirty-four units from zero in the positive direction. For Sacramento, 034 0  34. 33 is thirty-three units from zero in the positive direction. For Shreveport, 033 0  33.

70.

3 8

1

4

88

71.

7 12

3

1

1

2 72.

7 10

1

4

 12  12 3

7

2

 5  10  10 9

 10

73.

3 8

2

9

16

 3  24  24 25

 24 1

 1 24

35

Chapter 2

74.

5 6

1

5

3

266

75.

3 4

1

9

4

2

8

3

5

6 9 15

1

1

18

15

 2  30  30

77.

3

7

14

7

 18  18  18 7

 30 

7 9

 18

1

1 10

Page 76

7

2 1

Check for Understanding 3

1

13

16. 134  (80)  134  (80)  134  80  214 The difference is 214.

Pages 76–78

1 72 8 7   1 0 14 0  0 14 0 2 8 7   1 14  14 2 8

1

 14 8

25

32

9. 12  15  60  60

1 0 3260 0  0 2560 0 2 32 25   1 60  60 2 

7

 60

10. 18  23  18  (23)  ( 023 0  018 0 )  (23  18)  5 11. 12.7  (18.4)  12.7  (18.4)  12.7  18.4  31.1 12. (3.86)  1.75  (3.86)  (1.75)  ( 03.86 0  01.75 0 )  (3.86  1.75)  5.61 13. 32.25  (42.5)  32.25  (42.5)  32.25  42.5  ( 042.5 0  032.25 0 )  (42.5  32.25)  10.25 Chapter 2

Practice and Apply

17. 8  13  ( 013 0  08 0 )  (13  8) 5 18. 11  19  ( 019 0  011 0 )  (19  11) 8 19. 41  (63)  ( 063 0  041 0 )  (63  41)  22 20. 80  (102)  ( 0102 0  080 0 )  (102  80)  22 21. 77  (46)  ( 077 0  046 0 )  (77  46)  123 22. 92  (64)  ( 092 0  064 0 )  (92  64)  156 23. 1.6  (3.8)  ( 01.6 0  03.8 0 )  (1.6  3.8)  5.4 24. 32.4  (4.5)  ( 032.4 0  04.5 0 )  (32.4  4.5)  36.9 25. 38.9  24.2  ( 038.9 0  024.2 0 )  (38.9  24.2)  14.7 26. 7.007  4.8  ( 07.007 0  04.8 0 )  (7.007  4.8)  2.207 27. 43.2  (57.9)  ( 057.9 0  043.2 0 )  (57.9  43.2)  14.7

 2  14  14

5

2 1 42 2 1 55 2 42 55  1 60 2  1 60 2 42 55  1 60 2  60 55 42   1 0 60 0  0 60 0 2 55 42   1 60  60 2  60

4. 15  (12)  ( 015 0  012 0 )  (15  12)  27 5. 24  (45)  ( 024 0  045 0 )  (24  45)  69 6. 38.7  (52.6)  ( 052.6 0  038.7 0 )  (52.6  38.7)  13.9 7. 4.62  (12.81)  ( 04.62 0  012.81 0 )  (4.62  12.81)  17.43

1 12

11

15. 10  12  60  60

1. Sample answer: 5  5 2. Sample answer: To subtract a real number, add its opposite. 6 6 3. Gabriella; subtracting 9 is the same as adding 9.

4 7

47

 90

Adding and Subtracting Rational Numbers

2-2

8.

1 27 2 20 27   1 0 90 0  0 90 0 2 20 27   1 90  90 2  90  90

 3 or 1 3 76.

27

20

 12

4

20

14. 9  10  90  90

 3  12  12

36

28. 38.7  (61.1)  ( 061.1 0  038.7 0 )  (61.1  38.7)  22.4 29.

6 7

2

18

37. To find the total score, find the sum of her scores. 100  200  500  (300)  400  (500)  300  500  (300)  400  (500)  800  (300)  400  (500)  ( 0800 0  0300 0 )  400  (500)  (800  300)  400  (500)  500  400  (500)  900  (500)  ( 0900 0  0500 0)  (900  500)  400 Her total score was 400 points. 38. First find the net yards gained by finding the sum of 6, 8, and 3. 6  (8)  3  ( 08 0  06 0 )  3  (8  6)  3  2  3  ( 03 0  02 0)  (3  2) 1 Over the three plays, they gained 1 yard from their 20-yard line. Therefore, they were on the 21-yard line. 39. 19  8  19  (8)  ( 019 0  08 0 )  (19  8)  27 40. 16  (23)  16  (23)  16  23  39 41. 9  (24)  9  (24)  9  24  33 42. 12  34  12  (34)  ( 034 0  012 0 )  (34  12)  22 43. 22  41  22  (41)  ( 041 0  022 0)  (41  22)  19 44. 9  (33)  9  (33)  9  33  ( 033 0  09 0 )  (33  9)  24 45. 58  (42)  58  (42)  58  42  ( 058 0  042 0)  (58  42)  16 46. 79.3  (14.1)  79.3  (14.1)  79.3  14.1  93.4 47. 1.34  (0.458)  1.34  (0.458)  1.34  0.458  1.798 48. 9.16  10.17  9.16  (10.17)  ( 09.16 0  010.17 0 )  (9.16  10.17)  19.33

14

 3  21  21 32

11

 21 or 1 21 30.

3 18

6

51

108

 17  306  306 159

 306 53

 102 4

3

20

33

31. 11  5  55  55

1 0 3355 0  0 2055 0 2 33 20   1 55  55 2 

13

 55 2

17

8

17

32. 5  20  20  20

1 0 1720 0  0 208 0 2 17 8   1 20  20 2 

1

4

9

 20 9

2

1 135 2 64 135   1 0 240 0  0 240 0 2 64 135   1 240  240 2 64

33. 15  16  240  240

1

16

13

2

199

 240

1 26 2 16 26   1 0 40 0  0 40 0 2 16 26   1 40  40 2 16

34. 40  20  40  40

42

 40 1

1

1

2

21

1

 20 or 1 20

2 33 12  1 0 8 0  0 8 0 2 33 12  1 8  8 2

35. 4 8  1 2 

33 8

1

12

 8

21 8 5  28 17 67 3 25  50



17

36. 1 50 

1

2

1

184

  50

2

1 0 0  0 6750 0 2 184 67   1 50  50 2 

184  50

117

  50

17

 2 50

37

Chapter 2

58. 2  (6)  (4)  (4)  ( 02 0  06 0 )  (4)  (4)  (2  6)  (4)  (4)  8  (4)  (4)  ( 08 0  04 0 )  (4)  (8  4)  (4)  12  (4)  ( 012 0  04 0 )  (12  4)  16 59. Under; yes; it is better than par 72. 60. Subtract to find the change in value. week 8 week 1 value 123 minus 123 value 123

49. 67.1  (38.2)  67.1  (38.2)  67.1  38.2  105.3 50. 72.5  (81.3)  72.5  (81.3)  72.5  81.3  153.8 1

2

1

4

51. 6  3  6  6

1 42 1 4   1 0 6 0  0 6 0 2 1 4  16  6 2 1

 6  6

5

 6 52.

1 2

4

5

8

 5  10  10 5 10



1



8 10

11,257.24  9791.09  1466.15. The value changed by 1466.15.

2

1 0 0  0 105 0 2 8 5   1 10  10 2 8 10



3

61. From week 1 to week 2 the change was: 10,126.94  9791.09  335.85. From week 2 to week 3 the change was: 10,579.85  10,126.94  452.91. From week 3 to week 4 the change was: 10,810.05  10,579.85  230.20. From week 4 to week 5 the change was: 10,951.24  10,810.05  141.19. From week 5 to week 6 the change was: 10,821.31  10,951.24  10,821.31  (10,951.24)  ( 010,951.24 0  010,821.31 0 )  (10,951.24  10,821.31)  129.93. From week 6 to week 7 the change was: 11,301.74  10,821.31  480.43. From week 7 to week 8 the change was: 11,257.24  11,301.74.  11,257.24  (11,301.74)  ( 011,301.74 0  011,257.24 0 )  (11,301.74  11,257.24)  44.50. Therefore, week 7 had the greatest change from the previous week. 62. Week 8 had the least change from the previous week. 63. Sometimes; the equation is false for positive values of x, but true for all other values of x. 64. Sample answer: If a team gains yards, move right on the number line. If a team loses yards, move left on the number line. Answers should include the following. • Move right or left, depending on whether the Giants gained or lost yards on each play. Where you end will tell you how many yards the Giants lost or gained. • Instead of using a number line, you can use the rules for adding and subtracting rational numbers. 65. C; 57  87  57  (87)  ( 057 0  087 0 )  (57  87)  144

 10 7

1

3

2

1 32 14 3  16  1 16 2 14

53. 8  16  16  16

14

3

 16  16

1 0 14 0 0 163 0 2 14 3   1 16  16 2   16 

1

1 32

11

 16

1 92 1 9  12  1 12 2 1

54. 12  4  12  12

1

9

 12  12

1 0 129 0  0 121 0 2 9 1   1 12  12 2 

8

 12 2

3 1

1

9

55. 2 4  6 3  4  27

19 3

1

76

 12  12

2

1 0 0  0 2712 0 2 76 27   1 12  12 2 

76 12

49

1

 12 or 4 12

3

31

53

81

56. 5 10  1 50  10  50  

265 81  50 50 92 17 or 325 25

57. A score of 70, or 2 under par, is written as 2. A score of 66, or 6 under par, is written as 6. A score of 68, or 4 under par, is written as 4. During the four rounds, he shot 2, 6, 4, and 4.

Chapter 2

38

73. Replace a in 3a  5 7 with each value in the replacement set A.

66. B; 5  (8)  5  (8) 58 85

3a  5  7

a

32  5 7 7S1 7

2

Page 78

Maintain Your Skills

33  5 7 7S4 7 34  5 7 7S7 7 3  5  5 7 7 S 10 7 7 3  6  5 7 7 S 13 7 7

6

c  12  214

C

True or False?

1 ?

1

1 ?

1

1 ?

1

1

1

true ✓

1 ?

1

1

1

true ✓

1 ?

1

3

1

true ✓

3

1 4

1 4

 2 6 24 S 4 6 24

1 2

1 2

 2 6 24 S 1 6 24

3 4

3 4

 2 6 24 S 14 6 24

1

true ✓

1

true ✓

1

1  2 6 24 S 12 6 24

114

14  2 6 24 S 14 6 24

1

1

1

The solution set for c  2 6 24 is

514, 12, 34, 1, 114 6.

75. Less than implies subtract in reverse order, and square implies raised to the second power. So the expression can be written as q2  8. 76. Less than implies subtract in reverse order, and times implies multiply. So the expression can be written as 2k  37. 1 2

2

1

2

3  23 1

1

Other 8%

3

b

b  1.3  1.8

0.3

0.3  1.3  1.8 S 1.6 1.8

? ?

0.4  1.3  1.8 S 1.7 1.8 ?

0.5  1.3  1.8 S 1.8  1.8 ?

0.6  1.3  1.8 S 1.9  1.8 0.7  1.3  1.8 S 2.0  1.8

4

 

True or False?

1 4

2

1

2

5  45

79.

2

3 4

3

12 5 2 25

5

1 3

5

6  46

2

5

 10

80. 4  5  1  5

72. Replace b in b  1.3  1.8 with each value in the replacement set B.

1

78.

1

3

0.7

true ✓

The solution set for 3a  5 7 is {5, 6}. 1 1 74. Replace c in c  2 6 24 with each value in the replacement set C.

77.

?

true ✓

?

1

0.6

false

?

5

Leave It 25%

0.5

false

?

4

Drink It 67%

0.4

false

?

3

67. 12.2  08  x 0  12.2  08  4.8 0  12.2  03.2 0  12.2  3.2  15.4 68. 0y 0  9.4  3  07.4 0  9.4  3  7.4  9.4  3  16.8  3  13.8 69. 24.2  018.3  z 0  24.2  018.3  10 0  24.2  08.3 0  24.2  8.3  15.9 70. Sample answer: A category labeled “other” representing 8% would have to be added so that the data would sum to 100%. 71. Draw a circle with three sections. Make one section 67% of the circle and label it with the phrase ‘drink it.’ Make another section 25% of the circle and label it with the phrase ‘leave it.’ Make the third section 8% of the circle and label it with the word ‘other.’ Cereal Milk

True or False?

?

8

1

81. 8  58  18  58

82.

1

5

1 5

7 9

7

4

 12  9  12 1  

3 28 3 1 93

false

2-3

false

Multiplying Rational Numbers

true ✓

Page 81

true ✓

Check for Understanding

1. ab will be negative if either one factor is negative, and the other is positive. Let a  2 and b  3: 2(3)  6. Let a  2 and b  3: 2(3)  6. 2. Sample answer: calculating a $2.00 monthly banking fee for the entire year: 12  ($2)  $24 3. Since multiplication is repeated addition, multiplying a negative number by another negative number is the same as adding repeatedly in the opposite, or positive, direction.

true ✓

The solution set for b  1.3  1.8 is {0.5, 0.6, 0.7}.

39

Chapter 2

4. (6)(3)  18 6. (4.5)(2.3)  10.35 8.

1 21 2  5 3

10 21

2 7

3

5. 5(8)  40 7. (8.7)(10.4)  90.48 9.

1 21 2  4 7 9 15

32. 5(5)(2)  (3)(2) 6

28 135

33.

10. 5s(6t)  5(6)st  30st 11. 6x(7y)  (15xy)  6(7)xy  15xy  42xy  15xy  (42  15)xy  57xy

1 22 12

 3  4 1

13. np  2 3 4

15

7

  8 or 18

112 22123  22 1 4  12 2213 2 1 4  1 4 21 3 2

14. n2(m  2) 

4

 12 1

3 15. To find how much honey 675 honeybees make, multiply the number of bees by the amount an average bee makes. 1

675  12   

675 12 225 4 1 56 4 1

675 bees make 56 4 teaspoons of honey.

Pages 81–83 Practice and Apply 16. 5(18)  90 18. 12(15)  180 20. 47(29)  1363

17. 8(22)  176 19. 24(8)  192 21. 81(48)  3888

22.

23.

145 2138 2  1240 3

 27

1 3 2156 2  1530

24. 5

1

1

1

4 15

21

1

2 1

16

21

15

26. 35 72   5  2

27.

21



1 2 2167 2  1235

25. 5

 2 1

20 1125 2149 2  108 5

 10

240 10

2

 24

1 22

2  195 2 152 2 45

 10 9

2 1

 42 28. 7.2(0.2)  1.44 30. (5.8) (2.3)  13.34 Chapter 2

 (2)(4)

8 34. 6(2x)  14x  6(2)x 14x  12x  14x  (12  14)x  26x 35. 5(4n)  25n  5(4)n25n  20n  25n  (20  25)n  45n 36. 5(2x  x)  5(2  1)x  5(1)x  5x 37. 7(3d  d )  7(3  1)d  7(4)d  28d 38. 2a(3c)  (6y)(6r)  2(3)ac  (6)(6)yr  6ac  (36)yr  6ac  36yr  6ac  36ry 39. 7m(3n)  3s(4t)  7(3)mn  3(4)st  21mn  (12)st  21mn  12st 40. To find the change in price of 35 shares, multiply the number of shares by the change in price of one share. 35  (61.66  63.66)  35  (2)  70 The change in price of 35 shares was $70. 41. To find how much money you gained or lost, multiply the number of shares by how much you gained or lost on one share. 50  (61.69  64.38)  50  (2.69)  134.5 You lost $134.50 or $134.50. 42. 5c2  5(4.5)2 43. 2b2  2(3.9)2  5(20.25)  2(15.21)  101.25  30.42 44. 4ab  4(2.7)(3.9)  10.8 (3.9)  42.12 45. 5cd  5(4.5)(0.2)  22.5(0.2)  4.5 46. ad  8  (2.7)(0.2)  8  0.54  8  7.46 47. ab  3  (2.7)(3.9)  3  10.53  3  13.53 2 48. d (b  2a)  (0.2)2 [3.9  2(2.7)]  (0.2)2 [3.9  (5.4)]  (0.2)2 [3.9  5.4]  (0.2)2 [9.3]  (0.04)[9.3]  0.372

12. 6m  6 3

1 32 1 15  2 1 4 2

2 (11)(4) 11

29. 6.5(0.13)  0.845 31. (0.075)(6.4)  0.48

40

49. b2 (d  3c)  (3.9)2 (0.2  3  4.5)  (3.9)2 (0.2  13.5)  (3.9)2 (13.7)  (15.21)(13.7)  208.377

60.

4 5

1 32

1

 20

2

50. To find the length of the union, multiply 5 by the fly. 2 5

6  

61. 42  (14)  42  (14)  42  14  56 62. 14.2  6.7  14.2  (6.7)  ( 014.2 0  06.7 0 )  (14.2  6.7)  20.9 63.

12 5 2 25 2

The union is 2 5 feet long. 51. To find the price after 7 months, subtract the number of months times the drop in price per month from the starting price. 1450  7(34.95)  1450  244.65  1205.35 The price of a computer was $1205.35. 52. To find the degrees difference, multiply the number of 530-foot rises in altitude by the temperature drop in one 530-foot rise. 64,997 530

53.

54.

55.

56.

57. 58.

1 15 2 16 15   1 0 20 0  0 20 0 2 16 15   1 20  20 2 16

 4  20  20

4 32 1 0 1 2 3 4 5 6

64. 2 1

0

1

2

3

4

65. 1

1  3 0

2 1 3

2

66. Graph c; Before Brandon fills the balloon with air, the balloon has no air in it. The amount of air in the balloon increases as he fills it. The amount of air decreases after Brandon lets it go until the balloon has a minimal amount of air left in it. Graph c shows this situation. 67. Sample answer:

129,994 530

 (2)  

 245 The amount of degrees difference was about 245F. To find how many plastic bottles are used in one day, multiply the number of hours in a day by the number of plastic bottles used every hour. 24  2,500,000  60,000,000 About 60 million plastic bottles are used in one day. To find how many plastic bottles are used in one week, multiply the number of hours in a week by the number of plastic bottles used every hour. 168  2,500,000  420,000,000 About 420 million plastic bottles are used in one week. Positive; the product of two negative numbers is positive and the even number of negative factors can be divided into groups of two. Sample answer: Multiplying lets consumers calculate quickly the total of several similar items. Answers should include the following. • Coupons are negative values because adding a negative number is the same as subtracting a positive number. • Multiply 13.99 and 1.50 by three, then, add the products. B; 2x(4y)  2(4)xy  8xy B; 2ab  2(4)(6)  (8)(6)  48

?

x  5;

25  4  6 66✓ but 5 5 68. Sample answer: ? a  4; 04 0 7 3 4 7 3✓ but 4 3 69.

5 8

5

1

 2  82

70.

2 3

1

5

1

 16 3

6 4

71. 5  4  5  3 20 3 2  63 4 1 8  23 1 4 3 1  13 2 1 4 5 56 1 3 2 3

2

75.

1 2

4 5

3

8

6

5

5

72. 1  5  1  2 5

2



73.

1

2  4  32

1

 22 74.

7 9

5

2 6

7

 6  95 3 14

 15

76.

7 8

2

7

3

 3  82 21

 16 5

 116

Page 83

Maintain Your Skills

59. 6.5  (5.6)  ( 06.5 0  05.6 0 )  (6.5  5.6)  12.1

41

Chapter 2

Page 83

Practice Quiz 1

10.

1. The dots indicate each point on the graph. The coordinates are {4, 1, 1, 6}. 2. 32  0x  8 0  32  015  8 0  32  023 0  32  23 9 3. 15  7  ( 015 0  07 0 )  (15  7)  8 4. 27  (12)  27  (12)  27  12  39 5. 6.05  (2.1)  ( 06.05 0  02.1 0 )  (6.05  2.1)  8.15

1 22

3

28

 4

12.

 650a

13.

2ab ac

  

1 82 15 8  20  1 20 2

 (6b  18)  (2)

1 12 1 1  6b 1 2 2  18 1 2 2

2(3)(4.5) (3)(7.5) (6) (4.5) (22.5) 27 22.5

14.

a

15. b 

a c

3

3

 (4.5)  7.5 7.5 3

22.5

8

 (13.5)  (1.67)  1.67 16. To find the number of visitors the site had in 1999, divide the number of visitors in 2000 by eight. 419,000  8  52,375 There were 52,375 visitors in 1999.

7

7. 9(12)  108 8. (3.8)(4.1)  15.58 9. (8x)(2y)  (3y)(z)  (8)(2)xy  (3)(1)yz  16xy  (3)yz  16xy  3yz 10. mn  5  (2.5)(3.2)  5  8  5  ( 08 0  05 0 )  (8  5)  3

Pages 86–87 17. 19. 21. 23. 24. 25. 26.

Dividing Rational Numbers

Practice and Apply

18. 78  (4)  19.5 64  (8)  8 20. 108  (0.9)  120 78  (1.3)  60 42.3  (6)  7.05 22. 68.4  (12)  5.7 23.94  10.5  2.28 60.97  13.4  4.55 32.25  (2.5)  12.9 98.44  (4.6)  21.4 1

1

1

3

1

1. Sample answer: Dividing and multiplying numbers with the same signs both result in a positive answer while dividing or multiplying numbers with different signs results in a negative answer. However, when you divide rational numbers in fractional form, you must multiply by a reciprocal. 1 1 1 2. Sample answer: 2, since 1  2 and 2 7 2.

1

3

 

Chapter 2



35



2

  3 or 11 3 2

7

30. 5  7  5  2 35

1

  2 or 17 2 31.

16 36

24

16

60

24

960

33.

5 4 10 12 5 6

14 32

1

 12

10 9

2

42

63

1512

1

14

1

25

 25  32  12 

24

 1736

or 1 9



31

32. 56  63  56  31

 60  36  24  864

2 3



5

29. 7  5  7  3

3. To divide by a rational number, multiply by its reciprocal. 4. 96  (6)  16 5. 36  4  9 6. 64  5  12.8 7. 64.4  2.5  25.76 

 48  16

2

9.

1

3

 12

Check for Understanding

4 5

3

28. 4  12  4  12

27. 3  4  3  4

 12 

(7.5)(4.5) 4(3) 33.75 12

 2.81

3

 20

8.

 

 (4.5) 

2 3

cb 4a

 1.2

15

2 1 3  12 2 36 1 18

1101 2

 3b  (9)  3b  9

1 0 15 0 0 208 0 2 15 8   1 20  20 2

2 3

 650a  10

 (6b  18) 2

  20 

Page 86

650a 10

 65a 6b  18 2

 20  20

2-4

11.

 7

6. 4  5  20  20 15

25  3 4

350 384 175 192

2

27

 31

34.

80 25

1 22

1 32

80

45. To find the number of pillows, divide the total amount of fabric by the amount of fabric needed for each pillow.

 3  25  2 240

  50 24

1

4

1 52

3

1 32

1 32

222 5

9

36

1 82

1248 3

 

81c 9

 81c  9  81c  9c

39.

8r  24 8

12

38.

105g 5

1 9

 105g  5  105g  21g

115 2

 (8r  24)  (8)

1 12 1 1  8r 1 8 2  24 1 8 2

 (8r  24) 8

47.

7h  35 7

 (7h  35)  (7)

1 12 1 1  7h 1 7 2  35 1 7 2

 1h  (5)  h  5 40a  50b 2



(8) (6.5) (3.2) 52 3.2

51.

n  p m

 

 (40a  50b)  2

112 2 1 1  40a 1 2 2  50b 1 2 2

53.

m  2n n  q

54.

m  p q

 

8  3.2 5.4 4.8 5.4

 0.89

8  2(6.5)  (5.4) 8  13 6.5  5.4

 6.5

p  3q q  m

3.2  3(5.4)

 (5.4)  (8) 3.2  (16.2)

 (5.4)  (8) 

 [8f  (16g)]  8

118 2 1 1  8f 1 8 2  (16g) 1 8 2



 [8f  (16g)]

3.2  16.2 5.4  8 19.4 13.4

 1.45 55. To find the average loss per month, divide the total loss by the number of months.

 1f  (2g)  f  2g 44.

52.

 1.76

 14c  6d

5x  (10y) 5



(6.5) (3.2) 8 20.8 8

21

113 2 1 1  42c 1 3 2  18d 1 3 2

8f  (16g) 8



 11.9

 (42c  18d)  3  (42c  18d)

43.

6.5  3.2 8 9.7 8



 20a  25b 42c  18d 3

np m

 1.21

 (40a  50b)

42.

48.

 16.25  2.6 49. mq  np  (8) (5.4)  (6.5) (3.2)  43.2  20.8  2.08 50. pq  mn  (3.2) (5.4)  (8) (6.5)  17.28  (52)  0.33

 (7h  35) 7

41.

mn p



 1r  (3)  r  3 40.

18 7 4 27

She can make 2 pillows. 46. To find the average change in revenue, divide the change in revenue by the number of years. (2,764,000,000  2,800,000,000)  8  36,000,000  8  4,500,000 The average change in revenue for each of the 8 years was $4,500,000.

 416 37.

4

 14

2

or 44 5

36. 156  8  156  3 

7

 27

35. 74  3  74  5 

9

42  14  2  4

  5 or 4 5

23,985  12  1998.75 The average loss per month was $1998.75. 56. To find the fraction that is pure gold, divide the karat value by the karat value of pure gold.

 [5x  (10y)]  5

115 2 1 1  5x 1 5 2  (10y) 1 5 2

 [5x  (10y)] 

10

10  24  24

 1x  (2y)  x  2y

5

 12 Therefore, 10-karat gold is 5 7 1  12 or 12 not gold.

43

5 12

pure gold, and

Chapter 2

70. Substitution Property; 1.2  3.8  5 71. 8b  12(b  2)  8b  12(b)  12(2)  8b  12b  24  (8  12)b  24  20b  24 72. 6(5a  3b  2b)  6(5a)  6(3b)  6(2b)  30a  18b  12b  30a  (18  12)b  30a  6b 73. 3(x  2y)  2y  3(x)  3(2y)  2y  3x  6y  2y  3x  (6  2)y  3x  4y

57. Build up the fraction so that the denominator is 24. The numerator will be the karat value of the gold. 2 3

2.8

 3.8 16

 24 Therefore, jewelry that is

2 3

gold is 16-karat gold.

58. Since 4 is even, 4 is divisible by 2. Therefore, if a number is divisible by 4, then it is divisible by 2. Note that 3 . 4 . 6  72 and 8 . 9  72. Therefore, if a number is divisible by 72, then it is divisible by 3, 4, 6, 8, and 9. Since 8 and 9 have no common factor, no number smaller than their product will be divisible by both 8 and 9. Now, note that 72 . 5 . 7  2520, so 2520 is divisible by 1, 2, 3, 4, 5, 6, 7, 8, and 9. Since 72, 5, and 7 have no common factor, no number smaller than their product will be divisible by 72, 5, and 7. Thus, 2520 is the least positive integer that is divisible by all whole numbers from 1 to 9. 59. Sample answer: You use division to find the mean of a set of data. Answers should include the following. • You could track the mean number of turtles stranded each year and note if the value increases or decreases. • Weather or pollution could affect the turtles. 60. D; 6.25  10  12  0.625  12  7.5 61. E; 6x  1  6 

17 3

74. mean  

 36 There are five numbers in the set. When ordered from least to greatest, the third number is 38. So, the median is 38. The number 40 appears twice, and the other numbers of the set appear only once. So, the mode is 40. 75. mean  

1

76. mean  

64.

1 (5) 4



5 4



1 14

63. 2.5(1.2)  3 65. 1.6(0.3)  0.48

66. 8  (6)  8  (6) 86  14 67. 15  21  15  (21)  ( 021 0  015 0 )  (21  15)  6 68. 7.5  4.8  7.5  (4.8)  ( 07.5 0  04.8 0 )  (7.5  4.8)  12.3 5

1 12

77. mean  

1 42 15 4  24  1 24 2 15

4

 24  24

1 0 15 0 0 244 0 2 15 4   1 24  24 2   24  11

 24

Chapter 2

79  84  81  84  75  73  80  78 8 634 8

 79.25 There are eight numbers in the set. When ordered from least to greatest, the fourth number is 79 and 79  80 the fifth is 80. So, the median is or 79.5. The 2 number 84 appears twice, and the other numbers of the set appear only once. So, the mode is 84.

69. 8  6  24  24 15

1.2  1.7  1.9  1.8  1.2  1.0  1.5 7 10.3 7

 1.47 There are seven numbers in the set. When ordered from least to greatest, the fourth number is 1.5. So, the median is 1.5. The number 1.2 appears twice, and the other numbers of the set appear only once. So, the mode is 1.2.

Maintain Your Skills

62. 4(11)  44

3  9  0  2  11  8  14  3 8 50 8

 6.25 There are eight numbers in the set. When ordered from least to greatest, the fourth number is 3 and 3  8 1 the fifth is 8. So, the median is 2 or 52  5.5. The number 3 appears twice, and the other numbers of the set appear only once. So, the mode is 3.

 34  1  35

Page 87

40  34  40  28  38 5 180 5

44

2-5

Statistics: Displaying and Analyzing Data

Page 91

Check for Understanding

11. The greatest common place value is tens, so the digits in the tens place are the stems. Stem Leaf 5 4556 6 0149 7 03578 8 0035888 9 0 10 0 2 5 11 0 5 0 4  54 12. Occurring three times, the most frequent value is 88. 13. The mode is not the best measure as it is higher than most of the values.

1. They describe the data as a whole. 2. Sample answer: Line Plot:   1



  

 



2

3

4

5

Line Graph:

Pages 92–94

Practice and Apply

14. The lowest value is 35, and the highest value is 54, so use a scale that includes those values. Place an  above each value for each occurrence.   1

2

3

4

5

6

15. The lowest value is 2.0, and the highest value is 2.5, so use a scale that includes those values. Place an  above each value for each occurrence.     



2

1

                  

10 12 14 16 18 20 22 24

5. The lowest value is 0, and the highest value is 14, so use a scale that includes those values. Place an  above each value for each occurrence.           

1 0

2

4

6

8

 

0

1

 

 

2

3

16. The lowest value is 1, and the highest value is 8, so use a scale that includes those values. Place an  above each value for each occurrence.

      

  

 

34 36 38 40 42 44 46 48 50 52 54

7

3. A set of data with one extremely high value would be better represented by the median, not by the mean. Sample answer: 13, 14, 14, 28 4. The lowest value is 10, and the highest value is 22, so use a scale that includes those values. Place an  above each value for each occurrence.



    

10 12 14

    

       

3

5

  7

9

17. 18. 19. 20.

23 of the 44 teams were not number 1 seeds. 29 of the 44 teams were seeded higher than third. Sample answer: Median; most of the data are near 2. The greatest common place value is ones, so the digits in the ones place are the stems. Stem Leaf 5 89 6 03569 7 01123 5 08  5.8 21. The greatest common place value is tens, so the digits in the tens place are the stems. Stem Leaf 1 88 2 2366689 3 011234 4 7 1 08  18

6. Occurring four times, the most frequent value is 0. 7. The mean, 6.75, and the median, 6.5, both represent the data accurately as they are fairly central. 8. The greatest common place value is tens, so the digits in the tens place are the stems. Stem Leaf 6 46888 7 1236 8 01688 9 3 6 0 4  64 9. The greatest value is 12.9, and the least value is 9.3. Therefore, the difference is 12.9  9.3 or 3.6. 10. Median; most of the data clusters higher, near the median, 11.8.

45

Chapter 2

38. Sample answer: 4, 4, 4, 5, 5, 5, 7, 8, 8, 8, 8, 18

22. The greatest common place value is hundreds, but the hundreds digit in every number is a one. Thus we use the digits in the tens place as the stems. Stem Leaf 10 0 0 4 5 5 6 6 7 8 9 9 11 0 0 0 0 1 1 2 2 2 2 3 3 4 4 4 4 4 5 67778888889 12 0 0 0 0 1 1 2 5 8 13 4 10 00  100 23. Occuring six times, the most frequent temperature is 118. 24. No; the mode is higher than most of the data. 25. See students’ work. 26. Occuring three times, the most frequent magnitude was 7.5. 27. Mean or median; both are centrally located and the mode is too high. 28. The lowest value is 8, and the highest value is 40, so use a scale that includes those values. Place an  above each value for each occurrence.     

8





mean  

39.

40.

41.

42.



12 16 20 24 28 32 36 40

29. 7 of the 10 countries won fewer than 25 gold medals. 30. When ordered from least to greatest, the fifth value is 13 and the sixth is 14. So, the median is 13  14 or 13.5. 2 31. Sample answer: Yes; most of the data are near the median. 32. The greatest common place value is tens, so the digits in the tens place are the stems. Stem Leaf 1 688999 2 00000112334568889 9999 3 00001334467 4 37 1 06  16 33. 22 of the 40 vehicles get more than 25 miles per gallon. 34. Sample answer: Mean; the median is too high, and the modes are either too high or too low. 35. The greatest common place value is tens, so the digits in the tens place are the stems. Stem Leaf 3 047 4 5 29 6 27 7 7 8 4 5 3 00  30 36. The interval that corresponds to the stem value 3 has the most values. The interval is from 30 to 39. 37. Every number in the set appears only once. So, there is no mode.

Chapter 2

43.

4  4  4  5  5  5  7  8  8  8  8  18 12 84 12

7 There are twelve numbers in the set. When ordered from least to greatest, the sixth number 5  7 is 5 and the seventh is 7. So, the median is 2 or 6. The number 8 appears four times, and the other numbers of the set appear three times or less. So, the mode is 8. High school: $33,184  $23,061  $10,123; some college: $39,221  $27,757  $11,464; Bachelor’s degree: $60,201  $41,747  $18,454; Doctoral degree: $81,687  $60,079  $21,608 Sample answer: The highest median salaries are earned by people whose highest level of education is a bachelor’s degree. Sample answer: because the range in salaries is often very great with extreme values on both the high end and low end Sample answer: They can be used in marketing or sales to sell the most products to a specific group. Answers should include the following. • a line plot showing the number of males with the names from the beginning of the lesson • By finding out the most popular names, you can use the popular names on more of your items. C;

mean  

7.5  7.5  7.5  7.5  7.9  7.9  7.9  8.3  9.1  11 10 82.1 10

 8.21 44. C; 7 of the 10 wingspans are less than 8 inches.

Page 94

Maintain Your Skills

45. 56  (14)  4 46. 72  (12)  6 47. 40.5  3  13.5 48. 102  6.8  15 49. 2(6x)  5x  2(6)x  5x  12x  5x  (12  5)x  17x 50. 3x(7y)  4x(5y)  3(7)xy  4(5)xy  21xy  20xy  (21  20)xy  41xy 51. 5(3t  2t)  2(4t)  5(3t)  5(2t)  2(4t)  5(3)t  5(2)t  2(4)t  15t  10t  8t  (15  10  8)t  3t 52. x dollars d dollars plus per week for 12 weeks 14243 123 14243 123 123 d  x  12 or d  12x 53. y  3x  16  3.5 54. xz  3  5.9  3  16  15  45  3 1  15

46

55. 2x  x  (y  4)  2  5  5  (16  4) 2554  10  5  4 54 9 56.

58.

60.

62.

64.

2

x  z 2y

2

5  9 2  16 25  9  2  16 16  2  16 16  32 1 2 54 54  6  60  6 60 9  10 42 42  6  48  6 48 7 8 28 28  4  52  4 52 7  13 84 84  6  90  6 90 14  15

Page 95



57.

12 18

4. There are four 5s and 52 total cards. 4

P(5)  52 1

 13 1

The probability of selecting a 5 is 13 or about 8%. 5. There are two red 10s and 52 total cards.

12  6

 18  6

2

P(red 10)  52

2

3

1

 26 1

The probability of selecting a red 10 is 26 or about 4%. 6. There are 16 odd numbered cards and 52 total cards. 59.

21 30

21  3

16

 30  3

P(odd number)  52

7

4

 10 61.

32 64

 13

32  32

 64  32

The probability of selecting an odd numbered card 4 is 13 or about 31%. 7. There is 1 way to pick the queen of hearts and 1 way to pick the jack of diamonds. So there are 1  1 or 2 ways to pick the queen of hearts or jack of diamonds

1

2 63.

16 36

16  4

 36  4 4

9

2

P(queen of hearts or jack of diamonds)  52 1

 26 The probability of selecting the queen of hearts or 1 jack of diamonds is 26 or about 4%. 8. There are three numbers on the spinner that are multiples of 3, and there are 10 – 3 or 7 numbers that are not multiples of 3.

Reading Mathematics

1. Sample answer: The data show the acreage and number of visitors in thousands for selected state parks and recreation areas in 1999. 2. Sample answer: The footnote indicates that the number of visitors includes those staying overnight. 3. The data is for 1999. 4. The unit indicator is thousands. 5. The table has the value 1016 in the row labeled New York and the column labeled Acreage. Therefore, New York has 1,016,000 acres. 6. The largest number in the column labeled Visitors is 76,736, which is in the row labeled California. Therefore, California had the greatest number of visitors with 76,736,000 visitors.

3

odds of a multiple of 3  7 The odds of getting a multiple of 3 on the spinner are 3:7. 9. There are three even numbers less than 8 on the spinner, and there are 10  3 or 7 numbers that are not even or not less than 8. 3

odds of an even number less than 8  7 The odds of getting an even number less than 8 on the spinner are 3:7. 10. There are seven numbers on the spinner that are odd or blue, and there are 10  7 or 3 numbers that are even and not blue. 7

2-6 Page 98

odds of an odd number or blue  3

Probability: Simple Probability and Odds

The odds of getting an odd number or blue on the spinner are 7:3. 11. There are six numbers on the spinner that are red or yellow, and there are 10  6 or 4 numbers that are blue.

Check for Understanding

1. Sample answers: impossible event: rolling a number greater than 6; certain event: rolling a number from 1 to 6; equally likely event: rolling an even number. 3 2. The probability is 5 which means there are 3 favorable outcomes for every 5  3 or 2 unfavorable outcomes. Thus the odds are 3:2. 3. Doug; Mark determined the odds in favor of picking a red card.

6

odds of red or yellow  4 The odds of getting a red or yellow number on the spinner are 6:4.

47

Chapter 2

18. There are 70  100  80 or 250 coins with value less than $1.00 and 300 total coins.

12. The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48, of which 4, 8, 12, 16, 24, and 48 are multiples of 4. There are 4 factors that are not a multiple of 4 and 10 total factors.

250

P(value less than $1.00)  300 5

6

4

P(not a multiple of 4)  10

 0.83 The probability of selecting a coin with value less than $1.00 is 65 or about 83%. 19. There are 80  50 or 130 coins with value greater than $0.10 and 300 total coins.

2

5 The probability of selecting a factor that is not a 2 multiple of 4 is 5 or 40%. 13. The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48, of which 12, 24, and 48 have both 4 and 6 as factors. There are 3 factors of 48 that have 4 and 6 as two of their factors and 10 total factors of 48.

130

P(value greater than $0.10)  300 13

 30  0.43 The probability selecting a coin with value greater 13 than $0.10 is 30 or about 43%. 20. There are 80  50 or 130 coins with value at least $0.25 and 300 total coins.

3

P(has 4 and 6 as factors)  10 or 0.3 The probability of selecting a factor that has 4 3 and 6 as two of its factors is 10 or 30%.

130

Pages 99–101

P(value at least $0.25)  300

Practice and Apply

13

 30

14. There are 80 quarters and 300 total coins.

 0.43 The probability of selecting a coin with value at least $0.25 is 13 or about 43%. 30 21. There are 70  100  80  50 or 300 coins with value at most $1.00 and 300 total coins.

80

P(quarter)  300 

4 15

 0.27 4 The probability of selecting a quarter is 15 or about 27%. 15. There are 100 dimes and 300 total coins.

300

P(value at most $1.00)  300 1 The probability of selecting a coin with value at most $1.00 is 1 or 100%. 22. There are 15 ways for the sum of two dice to be less than 7 and 36 total outcomes.

100

P(dime)  300 1

3  0.33 1

The probability of selecting a dime is 3 or about 33%. 16. There are 70 ways to pick a nickel and 50 ways to pick a dollar. So there are 70  50 or 120 ways to pick a nickel or a dollar and 300 total coins.

15

P(sum less than 7)  36 5

 12  0.42 5 The probability of the sum being less than 7 is 12 or about 42%. 23. There are 21 ways for the sum of two dice to be less than 8 and 36 total outcomes.

120

P(nickel or dollar)  300 2

5  0.40 The probability of selecting a nickel or a dollar is 2 or 40%. 5 17. There are 80 ways to pick a quarter and 70 ways to pick a nickel. So there are 80  70 or 150 ways to pick a quarter or a nickel and 300 total coins.

21

P(sum less than 8)  36 7

 12  0.58 The probability of the sum being less than 8 is or about 58%. 24. There are 0 ways for the sum of two dice to be greater than 12 and 36 total outcomes.

150

P(quarter or nickel)  300 1

 2 or 0.5 The probability of selecting a quarter or a nickel 1 is 2 or 50%.

P(sum is greater than 12) 

7 12

0 36

0 The probability of the sum being greater than 12 is 0 or 0%.

Chapter 2

48

33. There are 3 polygons that have more than one right angle and 6 total polygons.

25. There are 36 ways for the sum of two dice to be greater than 1 and 36 total outcomes. 36

3

P(sum is greater than 1)  36

P(more than one right angle)  6

1 The probability of the sum being greater than 1 is 1 or 100%. 26. There are 20 ways for the sum of two dice to be between 5 and 10 and 36 total outcomes.

1

 2 or 0.5 The probability of selecting a polygon that has more than one right angle is 1 or 50%. 2 34. There is one day in April that is the 29th and 30 total days in April. The probability that the 1 person’s birthday is the 29th is 30 or about 3%. 35. There are 15 days in July after the 16th and 31 total days in July. The probability that the 15 person’s birthday is after the 16th is 31 or about 48%. 36. There are three as in the name, and there are 24  3 or 21 letters that are not an a.

20

P(sum is between 5 and 10)  36 5

9  0.56 The probability of the sum being between 5 and 5 10 is 9 or about 56%. 27. There are 25 ways for the sum of two dice to be between 2 and 9 and 36 total outcomes. P(sum is between 2 and 9) 

3

odds of an a  21

25 36

1

7

 0.69 The probability of the sum being between 2 and 9 25 is 36 or about 69%. 28. There are 3 triangles and 6 total polygons.

The odds of selecting an a from the name are 1:7. 37. There are four ts in the name and there are 24  4 or 20 letters that are not a t. 4

odds of a t  20 or

3

P(triangle)  6

1 5

The odds of selecting a t from the name are 1:5. 38. There are eleven vowels in the name and there are 24  11 or 13 consonants.

1

 2 or 0.5 1

The probability of selecting a triangle is 2 or 50%. 29. There is 1 pentagon and 6 total polygons.

11

odds of a vowel  13

1

P(pentagon)  6

The odds of selecting a vowel from the name are 11:13. 39. There are thirteen consonants in the name and there are 24  13 or 11 vowels.

 0.17 1 The probability of selecting a pentagon is 6 or about 17%. 30. There are 3 polygons that are not triangles and 6 total polygons.

13

odds of a consonant  11 The odds of selecting a consonant from the name are 13:11. 40. There are four uppercase letters in the name and there are 24  4 or 20 lowercase letters.

3

P(not a triangle)  6 1

 2 or 0.5 The probability of selecting a polygon that is not a 1 triangle is 2 or 50%. 31. There are 4 polygons that are not quadrilaterals and 6 total polygons.

4

odds of an uppercase letter  20 or

1 5

The odds of selecting an uppercase letter from the name are 1:5. 41. There are nine lowercase vowels in the name and there are 24  9 or 15 letters that are not lowercase vowels.

4

P(not a quadrilateral)  6 2

3

9

odds of a lowercase vowel  15 or

 0.67 The probability of selecting a polygon that is not a 2 quadrilateral is 3 or about 67%. 32. There are 3 polygons that have more than three sides and 6 total polygons.

3 5

The odds of selecting a lowercase vowel from the name are 3:5. 42. There are five stamps from Canada and there are 32  5 or 27 stamps that are not from Canada.

3

5

P(more than three sides)  6

odds of being from Canada  27

1

 2 or 0.5

The odds the stamp is from Canada are 5:27. 43. There are twelve stamps from Mexico and there are 32  12 or 20 stamps that are not from Mexico.

The probability of selecting a polygon that has 1 more than three sides is 2 or 50%.

12

odds of being from Mexico  20 or

3 5

The odds the stamp is from Mexico are 3:5.

49

Chapter 2

52. 100  40  25(100  2  25)  100  40  25(100  50)  100  40  25(50)  4000  1250  5250 2  252  100  35  2  625  100  35  1250  3500  4750 The area of the region not shaded is 5250 cm2, and the area of the shaded region is 4750 cm2.

44. There are 32  3 or 29 stamps not from France and there are 3 stamps from France. odds of not being from France 

29 3

The odds the stamp is not from France are 29:3. 45. There are 3  8  1  3 or 15 stamps not from a North American country and there are 32 – 15 or 17 stamps from a North American Country. odds of not being from a North American 15 country  17

5250

odds against shaded region  4750

The odds the stamp is not from a North American country are 15:17. 46. There are 3  1 or 4 stamps from Germany or Russia and there are 32  4 or 28 stamps not from Germany or Russia. 4 1 odds of being from Germany or Russia  28 or 7

21

 19 The odds against placing a piece on a shaded region are 21:19. 53. 100  35  3500 100(40  25)  100  65  6500 The area of the green rectangle is 3500 cm2, and the area outside the green rectangle is 6500 cm2.

The odds the stamp is from Germany or Russia are 1:7. 47. There are 5  8 or 13 stamps from Canada or Great Britain and there are 32  13 or 19 stamps not from Canada or Great Britain. 13 odds of being from Canada or Great Britain  19

3500

odds of green rectangle  6500 7

 13

The odds the stamp is from Canada or Great Britain are 13:19. 3 48. The probability that it will occur is 7, so the

The odds of a piece being placed within the green rectangle are 7:13. 54. There are 24 players who hit more than 35 home runs and 46 total players.

probability that it will not occur is 74. 3

24

P(more than 35)  46

4

odds it occurs  7 : 7 or 3:4

12

 23

The odds that the event will occur are 3:4.

 0.52 The probability of selecting a player who hit more 12 than 35 home runs is 23 or about 52%. 55. There are forty-two players who hit less than 45 home runs and there are 46  42 or 4 players who hit 45 or more.

2

49. The probability that it will occur is 3, so the 1

probability that it will not occur is 3. 1

2

odds against it occuring  3 : 3 or 1:2. The odds that the event will not occur are 1:2. 50. There are four cards from the coworkers in the bowl and there are 80  4 or 76 cards not from the coworkers.

odds of less than 45  

4

odds a coworker wins  76

The odds of selecting a player who hit less than 45 home runs are 21:2. 56. The number of home runs is 38 and the total number of bats is 439.

1

 19 51.

The odds one of the coworkers will win are 1:19. 2  252  100  35  2  625  100  35  1250  3500  4750 100(35  40  25)  100(100)  10,000 The area of the shaded region is 4750 cm2, and the total area is 10,000 cm2.

38

P(home run)  439  0.09 The probability the player hits a home run the 38 next time the player bats is 439 or about 9%. 57. There is one winning game card and 1,000,000 nonwinning game cards, so there is one winning card and a total of 1,000,001 cards.

4750

P(shaded region)  10,000

1

19

P(grand prize)  1,000,001

 40 or 0.475

 0.000001 The probability of selecting the winning game 1 card is 1,000,001 or about 0.0001%. 58. No; even with 100 game cards the odds of winning are only 100:999,901. It would require several hundred thousand cards to significantly increase the odds of winning.

The probability the piece is placed on a shaded 19 region is 40 or 47.5%.

Chapter 2

42 4 21 2

50

68. 12.2  7.8  ( 012.2 0  07.8 0 )  (12.2  7.8)  4.4

59. There are 6 ways for a head to appear on at least one of them and 7 total possible outcomes. 6

P(at least one head)  7

13

70.

12

 

2 5 4 5

74.

   

1

66.

ab c

    

1 0 10 0 0 127 0 2 10 7   1 12  12 2   12 

0 23 0  23

1 6

is one-sixth unit from zero in the positive direction.

0 16 0  16

75. 62  6  6  36 77. (8)2  (8)(8)  64

76. 172  17  17  289 78. (11.5)2  (11.5)(11.5)  132.25

79. 1.62  1.6  1.6  2.56

80.

81. 49 2  49 49

2  16 16 82. 16 15 15 15

1 2

2

65. 2a  b  2 3  5

2 1

2

2

or 80%

1 12

10

73. 3 is two-thirds unit from zero in the negative direction.

63. The greatest common place value is tens, so the digits in the tens place are the stems. Stem Leaf 5 8.3 6 4.3 5.1 5.5 6.7 7.0 8.7 9.3 7 0.0 2.8 3.2 5.8 7.4 7.4 5 0 8.3  58.3 1

1

71. 4.25 is four and twenty-five hundredths units from zero in the positive direction. 04.25 0  4.25 72. 8.4 is eight and four tenths units from zero in the negative direction. 08.4 0  8.4

Maintain Your Skills

2

7

1

12

64. b  c  5  2

1 52

 6  12  12

 4

P(satisfy inequality)  15

Page 101

7 12

5

 8

3

 13:12 62. D; There are 12 numbers that satisfy the inequality and 15 total numbers. 

1 32 2 3   1 0 8 0  0 8 0 2 2 3  18  8 2 2

 12

odds against occurring  25 : 25

4 5

1 32

1

69. 4  8  8  8

 0.86 6 The probability of at least one head is 7 or about 86%. 60. Sample answer: Probabilities are often used for strategy like placing a certain pitcher against a batter who has a low probability of hitting a pitch from that pitcher. Answers should include the following. • baseball: using the probability that a team can get base runners out; basketball: the probability that a player can make a basket from a certain place on the court; auto racing: the probability that a set of tires will hold out for the remainder of a race • Odds in favor of an event and odds against an event are frequently used. 13 61. B; The probability the event will not occur is 25.

2 2 3  5 2 5 3  2 10 6 5 2 3 or 13

Page 101



1 21 2

1125 22  125  125 1

16 81

2

Practice Quiz 2

1. 136  (8)  17

2

3. (46.8)  4  11.7



1

256 225

1 32

21

or

2

31 1225

1 82

2. 15  8  15  3

3  5 1 2 1 2 1 3  5  2 2 1 15  2 2 2 15  1 4 15

25

 144

4.

3a  9 3

120

  3   40

 (3a  9)  3

113 2 1 1  3a 1 3 2  9 1 3 2

 (3a  9)  1a  3 a3

67. 4.3  (8.2)  ( 08.2 0  04.3 0 )  (8.2  4.3)  3.9

51

Chapter 2

5.

4x  32 4

2.

 (4x  32)  4

114 2 1 1  4x 1 4 2  32 1 4 2

4 boys BBBB

 (4x  32)

 1x  8 x8 6.

15n  20 5

 (15n  20)  (5)

1 12 1 1  15n 1 5 2  20 1 5 2

 3n  (4)  3n  4

7. The lowest value is 0.4, and the highest value is 5.0, so use a scale that includes those values. Sample answer: scale from 0 to 5 Place an  above each value for each occurrence.         

0

1

2

   



 



3



5

4

1 boy, 3 girls BGGG GBGG GGBG GGGB

4 girls GGGG

3. There are 2 possibilities for a one-child family, 4 possibilities for a two-child family, 8 possibilities for a three-child family, and 16 possibilities for a four-child family. Following the pattern, there should be 32 possibilities for a five-child family and 64 possibilities for a six-child family. The pattern represents powers of 2. 4. There are 3 ways to have 2 boys and 1 girl and 8 total outcomes. Therefore, the probability of having 2 boys and 1 girl in a three-child family is 3 or 37.5%. 8 5. There are 6 ways to have 2 boys and 2 girls and 16 total outcomes. Therefore, the probability of having 2 boys and 2 girls in a four-child family is 6 3  8 or 37.5%. 16 6. Sample answer: Each number in each row shows the number of ways to have boys and girls for a given number of children. 7. The fifth row labeled Row 4 represents a fourchild family. The second number from the right is 4 and represents how many ways to have 3 girls and 1 boy. There are 1  4  6  4  1 or 16 total outcomes. Therefore, the probability of having 1 boy in a four-child family is 4  1 or 25%.

 (15n  20) 5



3 boys, 2 boys, 1 girl 2 girls BBBG BBGG BBGB BGBG BGBB BGGB GBBB GBBG GBGB GGBB

8. There are 3 ways for the sum of two dice to be 10 and 36 total outcomes. 3

P(sum of 10)  36 1

 12  0.08 1 The probability of the sum being 10 is 12 or about 8%. 9. There are 26 ways for the sum of two dice to be greater than or equal to 6 and 36 total outcomes. 26

P(sum  6)  36

16

4

13

 18

2-7

 0.72 The probability of the sum being greater than or equal to 6 is 13 or about 72%. 18 10. There are 30 ways for the sum of two dice to be less than 10 and 36 total outcomes.

Page 107

30

5

6

Page 102

5 6

Rational numbers: 0.5, 0.125, 0.3; Irrational numbers: 22, 23, 3. There is no real number that can be multiplied by itself to result in a negative product.

Algebra Activity (Follow-Up of Lesson 2–6)

4. 225 represents the negative square root of 25.

1. There are 2 more possibilities for 2 boys, 1 girl and 2 more possibilities for 1 boy, 2 girls. 3 boys BBB

Chapter 2

2 boys, 1 girl BBG BGB GBB

1 boy, 2 girls BGG GBG GGB

Check for Understanding

1. Sometimes; the square root of a number can be negative, such as 4 and 4, which are both square roots of 16. 2. Rational numbers are numbers that, when written as decimals, terminate or repeat. Irrational numbers do not terminate nor do they repeat. Sample answer:

P(sum 10)  36  0.83 The probability of the sum being less than 10 is or about 83%.

Square Roots and Real Numbers

25  52 S 225  5 5. 21.44 represents the positive square root of 1.44.

3 girls GGG

1.44  1.22 S 21.44  1.2

52

18. Write each number as a decimal.

16

6. ; 3 49 represents the positive and negative square roots of 16 49

147 22 and 1649  147 22



;3

230  5.477225575 . . . or about 5.48

16 . 49

16 49

;

4

59  5.444444444 . . . or about 5.44 13  13.0

4 7

1 230

7. 232 represents the positive square root of 32. 32 

5.662

0.18 6 5.44 6 5.48 6 13.0 The numbers arranged in order from least to 1 4 greatest are 30, 59, 230, 13.

S 232  5.66

8. Because 264  8, this number is an integer and a rational number. 9. Because 8 and 3 are integers and 8  3  2.66666 . . . is a repeating decimal, this number is a rational number. 10. Because 228  5.29150262 . . . , which is not a repeating or terminating decimal, this number is irrational. 56 11. Because 7  8, this number is a natural number, a whole number, an integer, and a rational number. 12. 6

5

4

3

2

19. C; Replace a in 2a 6  ?

22 6 

21. 281 represents the positive square root of 81. 81  92 S 281  9 22. 25.29 represents the positive square root of 5.29. 5.29  2.32 S 25.29  2.3

1

23. 26.25 represents the positive square root of 6.25. 6.25  2.52 S 26.25  2.5 24. 278 represents the negative square root of 78. 78  8.832 S 278  8.83 25. 294 represents the negative square root of 94.

 0.333333 . . .

94  9.702 S 294  9.70

1 . 3

36

26. ; 3 81 represents the positive and negative square

15. You can use a calculator to find an approximation 2 for 9. 2 9

roots of 36 81

 0.222222 . . .

0.2  0.222222 . . . Therefore,

2 9

2

100

27. ; 3 196 represents the positive and negative square roots of 100 196

6 26.

3



11014 22 and

100

10

; 3 196  ; 14

17. Write each number as a decimal. 1 8 1 8

6

;3

26  2.449489743 . . . Therefore,

169 22 and 8136  169 22

36

 0.166666 . . . 1 6



36 . 81

; 3 81  ; 9

 0.2.

16. You can use a calculator to find approximations 1 for 6 and 26. 1 6

Practice and Apply

49  72 S 249  7

The heavy arrow indicates that all points to the right of 7 are included in the graph. The dot at 7 indicates that 7 is included in the graph. 14. You can use a calculator to find an approximation 1 for 3. Therefore, 0.3 6

1

20. 249 represents the positive square root of 49.

13.

1 3

with 2.

22

Pages 107–109

2

0

1 2a

1.41 6 0.71

The heavy arrow indicates that all numbers to the left of 3.5 are included in the graph. The circle at 3.5 indicates 3.5 is not included in the graph. 8 7 6 5 4 3 2 1

 0.1825741858 . . . or about 0.18

100 . 196 100  196

11014 22

5

;7

 0.125  0.3535533906 . . . or about 0.35

0.15  0.15151515 . . . or about 0.15 15  15.0 15.0 6 0.125 6 0.15 6 0.35 The numbers arranged in order from least to 1 1 greatest are 15, 8, 0.15, 3 8.

53

9

9

25

25

28.

3 14 represents the positive square root of 14. 9 9  0.802 S 3 14  0.80 14

29.

3 42 represents the positive square root of 42. 25 25  0.772 S 3 42  0.77 42

Chapter 2

49. Because  3.141592654 . . . , which is not a repeating or terminating decimal, this number is irrational. d 50. Replace d in 3 16 with 28.

30. ; 2820 represents the positive and negative square roots of 820. 820  28.642 and 820  (28.64)2 ; 2820  ; 28.64 31. ; 2513 represents the positive and negative square roots of 513. 513  22.652 and 513  (22.65)2 ; 2513  ; 22.65 32. Because 222  4.69041576 . . . , which is not a repeating or terminating decimal, this number is irrational. 36 33. Because 6  6, this number is a natural number, a whole number, an integer, and a rational number. 34. Because 1 and 3 are integers and 1  3  0.333333 . . . is a repeating decimal, this number is a rational number. 35. Because 5 and 12 are integers and (5  12)  0.416666 . . . is a repeating decimal, this number is a rational number.

28

3 16  1.32 The ball would take 1.32 s to reach the ground. 3 51. Replace d in 224d with 434.

3 24 1 43 4 2  21050  32.40 3

Jerome should not get a ticket. He was traveling at about 32.4 mph. 52. 14 13121110 9 8 7 6

The heavy arrow indicates that all numbers to the right of 12 are included in the graph. The circle at 12 indicates that 12 is not included in the graph. 53. 1

82 20

36. Because 3  2.024845673 . . . , which is not a repeating or terminating decimal, this number is irrational. 37. Because 246  6.782329983 . . . , which is not a repeating or terminating decimal, this number is irrational. 38. Because 210.24  3.2, which is a terminating decimal, this number is a rational number. 39. Because 54 and 19 are integers and 54  19  2.8421052631579 is a terminating decimal, this number is a rational number. 40. Because 3 and 4 are integers and (3  4)  0.75 is a terminating decimal, this number is a rational number. 41. Because 220.25  4.5, which is a terminating decimal, this number is a rational number. 18 42. Because 3  6, this number is a natural number, a whole number, an integer, and a rational number. 43. Because 22.4025  1.55, which is a terminating decimal, this number is a rational number. 44. Because 68 and 35 are integers and 68  35  1.9428571428571 is a terminating decimal, this number is a rational number. 45. Because 6 and 11 are integers and 6  11  0.54545454 . . . is a repeating decimal, this number is a rational number. 46. Because 25.5696  2.36, which is a terminating decimal, this number is a rational number.

3

4

5

6

7

8

9

54. 10

9

The heavy arrow indicates that all points to the right of 10.2 are included in the graph. The dot at 10.2 indicates that 10.2 is included in the graph. 55. 2

1

0

The heavy arrow indicates that all numbers to the left of 0.25 are included in the graph. The circle at 0.25 indicates 0.25 is not included in the graph. 56. 5 4 3 2 1

0

1

2

3

4

The heavy arrows indicate that all numbers to the right and left of 2 are included in the graph. The circle at 2 indicates that 2 is not included in the graph. 57. 8 6 4 2

0

2

4

6

8

The heavy line and arrows indicate that all numbers to the right and left of 6 and 6 are included in the graph. The circles at 6 and 6 indicate that 6 and 6 are not included in the graph. 58. You can use a calculator to find an approximation for 15. 25  2.23606797 . . . 5.72  5.727272 . . . Therefore, 5.72 7 25. 59. You can use a calculator to find an approximation for 18.

78

47. Because 3 42  1.36277028 . . . , which is not a repeating or terminating decimal, this number is irrational. 48. Because 29.16  3.02654919 . . . , which is not a repeating or terminating decimal, this number is irrational.

Chapter 2

2

The heavy arrow indicates that all points to the left of 8 are included in the graph. The dot at 8 indicates that 8 is included in the graph.

18  2.82842712 . . . 2.63  2.636363 . . . Therefore, 2.63 6 28.

54

67. Write each number as a decimal. 4.83  4.83838383 . . . or about 4.84 0.4  2.82842712 . . . or about 2.83

60. You can use a calculator to find approximations for 1 1 and . 7 1 7 1 27

27

 0.142857142 . . .

28 3

8  0.375 4.84 6 0.375 6 0.4 6 2.83 The numbers arranged in order from least to 3 greatest are 4.83, 8, 0.4, 18.

 0.37796447 . . .

Therefore,

1 7

6

1 27

.

61. You can use a calculator to find approximations for 2 2 and . 3 2 3 2 23

23

68. Write each number as a decimal.

 0.666666 . . .

165  8.06225774 . . . or about 8.06 2 65  6.4

 1.15470053 . . .

Therefore,

2 3

6

2 23

127  5.19615242 . . . or about 5.20 8.06 6 6.4 6 5.20 The numbers arranged in order from least to 2 greatest are 165, 65, 127. 69. Write each number as a decimal. 2122  11.0453610 . . . or about 11.05 4 79  7.444444 . . . or about 7.44

.

62. You can use a calculator to find approximations 1 for and 131 . 31 1

231

231 231 31

 0.17960530 . . .  0.17960530 . . .

Therefore,

1 131



131 . 31

2200  14.1421356 . . . or about 14.14 7.44 6 11.05 6 14.14 The numbers arranged in order from least to 4 greatest are 79, 1122, 1200.

63. You can use a calculator to find an approximation 12 for 2 . 22 2 1 2

 0.707106781 . . .

70. Replace h in 1.41h with 1500. 1.421500  1.4  38.73  54.222 The tourists can see about 54.2 miles. 71. Replace h in 1.41h with 135. 1.4 2135  1.4  11.62  16.268 Replace h in 1.41h with 85.

 0.5

Therefore,

22 2

1

7 2.

64. Write each number as a decimal. 20.42  0.648074069 . . . or about 0.648 0.63  0.63636363 . . . or about 0.636 14  0.66666666 . . . or about 0.667 3 0.636 6 0.648 6 0.667 The numbers arranged in order from least to 14 greatest are 0.63, 10.42, 3 .

1.4185  1.4  9.22  12.908 Marissa can see about 16.268  12.908  3.36 or about 3.4 miles farther than Dillan. 72. Replace h in 1.41h with 120.

65. Write each number as a decimal. 10.06  0.244948974 . . . or about 0.245 0.24  0.24242424 . . . or about 0.242 19 12

1.41120  1.4  10.95  15.33 The lighthouse keeper cannot see the boat. The lighthouse keeper can only see about 15.3 miles. 73. They are true if q and r are positive and q 7 r.

 0.25

0.242 6 0.245 6 0.25 The numbers arranged in order from least to 19 greatest are 0.24, 10.06, 12 . 66. Write each number as a decimal. 1.46  1.46464646 . . . or about 1.46 0.2  1.41421356 . . . or about 1.41 12 1

6  0.1666666 . . . or about 0.17 1.46 6 0.17 6 0.2 6 1.41 The numbers arranged in order from least to 1 greatest are 1.46, 6, 0.2, 12.

55

Chapter 2

74. The area of a square is given by A  s2 where s is the side length. If A  s2 then 1A  s. To find the side lengths, find the square root of the area. 1  12 S 21  1 4  22 S 24  2 9  32 S 29  3 16  42 S 216  4 25  52 S 225  5

81. There are 20 even numbers in a deck of cards, and there are 52  20 or 32 cards that are not even numbers. 20

odds of an even number  32 5

8 The odds of selecting an even number from a deck of cards are 5:8. 82. There are 12 face cards in a deck of cards, and 52  12 or 40 cards that are not face cards.

The perimeter of a square is 4 times the side length. To find the perimeter of each square, multiply each of the side lengths by 4. 414 428 4  3  12 4  4  16 4  5  20

40

odds against a face card  12 

The odds against selecting a face card from a deck of cards are 10:3. 83. There are four aces in a deck of cards, and 52  4 or 48 cards that are not aces.

Squares Area (Units2 ) 1 4 9 16 25

Side Length 1 2 3 4 5

odds against an ace  Perimeter 4 8 12 16 20



84. 85.

75. The length of the side is the square root of the area. 76. The perimeter of a square is 4 times the side length. The side length of a square is the square root of the area. Therefore, the perimeter of a square is 4 times the square root of the area. For a square whose area is a units2, the perimeter is 41a units. 77. Sample answer: By using the formula Surface

86.

87.

height  weight

Area  3 , you need to use square 3600 roots to calculate the quantity. Answers should include the following. • You must multiply height by weight first. Divide that product by 3600. Then determine the square root of that result. • Sample answers: exposure to radiation or chemicals; heat loss; scuba suits • Sample answers: determining height, distance

88.

78. B; 27 79. B; Write each number as a decimal. 3

6  0.5 6

3  2 0.5 7 2 ✓

Page 109

48 4 12 1

The odds against selecting an ace from a deck of cards are 12:1. Sample answer: Mean; the median and mode are too low. 4(7)  3 (11)  28  33  28  (33)  ( 028 0  033 0 )  (28  33)  61 3(4)  2(7)  12  (14)  ( 012 0  014 0 )  (12  14)  26 1.2(4x  5y)  0.2(1.5x  8y)  1.2(4x)  1.2(5y)  0.2(1.5x)  0.2(8y)  1.2(4)x  1.2(5)y  0.2(1.5)x  0.2(8)y  4.8x  6y  (0.3)x  1.6y  4.8x  6y  0.3x  1.6y  4.8x  0.3x  6y  1.6y  (4.8  0.3)x  (6  1.6)y  5.1x  7.6y 4x(y  2z)  x(6z  3y)  4x(y)  4x(2z)  x(6z)  x(3y)  4(1)xy  4(2)xz  1(6)xz  1(3)xy  4xy  (8)xz  6xz  (3)xy  4xy  8xz  6xz  3xy  4xy  3xy  8xz  6xz  (4  3)xy  (8  6)xz  7xy  14xz

Chapter 2 Study Guide and Review

Maintain Your Skills

80. There are two red 4s in a deck of cards, and there are 52  2 or 50 cards that are not a red 4.

Page 110

Vocabulary and Concept Check

1. true; 26 is twenty-six units from zero in the negative direction. 2. true; definition of a rational number 3. true; 2144 represents the principal square root of 144. 144  122 S 2144  12

2

odds of a red 4  50 1

 25 The odds of selecting a red 4 from a deck of cards are 1:25.

Chapter 2

10 3

56

22. 2  10  2  (10)  ( 02 0  010 0 )  (2  10)  12 23. 9  (7)  9  (7) 97  16 24. 1.25  0.18  1.07 25. 7.7  (5.2)  7.7  (5.2)   7.7  5.2  ( 07.7 0  05.2 0 )  (7.7  5.2)   2.5

4. false; Because 2576  24, this number is an integer and a rational number. 5. true; 152  225 6. false; sample answer: 3 7. false; sample answer: 0.6 or 0.666 . . . 8. true; key concept of multiplication

Pages 110–114

Lesson-by-Lesson Review

9. 4 32 1 0 1 2 3 4 5 6

10. 3 2 1

0

1

2

3

4

5

6

6 5 4 3 2 1

0

1

2

3

26.

11. 12. 32  0y  3 0  32  08  3 0  32  05 0  32  5  27 13. 3 0x 0  7  3 04 0  7  34  7  12  7 5 14. 4  0z 0  4  09 0 49  13 15. 46  y 0x 0  46  8 04 0  46  8 # 4  46  32  14 16. 4  (4)  ( 04 0  04 0 )  (4  4) 0 17. 2  (7)  ( 07 0  02 0 )  (7  2)  5 18. 0.8  (1.2)  ( 00.8 0  01.2 0 )  (0.8  1.2)  2 19. 3.9  2.5  ( 03.9 0  02.5 0 )  (3.9  2.5)  1.4 1

1 12

5 6

1 12

1 12

9

1 12

 2  2  2 9

1

22 

10 2

5

27.

1 8



1 2  243  11624 2 3 16  24  1 24 2 2 3

3

16

 24  24

1 0 1624 0  0 243 0 2 16 3   1 24  24 2



13

 24

28. (11)(9)  99 30. 8.2(4.5)  36.9 32.

3 4

7

29. 12(3)  36 31. 2.4(3.6)  8.64

1 1 21

9

2

9

 12  48

21

33. 3 10  30

7

 10

 16

3

34. 8(3x)  12x  8(3)x  12x  24x  12x  (24  12)x  12x 35. 5(2n)  9n  5(2)n  9n  10n  9n  (10  9)n  1n n 36. 4(6a)  (3)(7a)  4(6)a  (3)(7)a  24a  21a  (24  21)a  45a

1 12 2 1   1 0 8 0  0 8 0 2 2 1  18  8 2 2

20. 4  8  8  8

21.

9 2

37.

54 6

38.

74 8

 54  6  9

3

 8

1 22 5 2   1 0 6 0  0 6 0 2 5 2   16  6 2 5

 3  6  6

 (74  8)

 9.25 39. 21.8  (2)  10.9 40. 7.8  (6)  1.3 41. 15 

3

6

134 2  15  43 

1

2

60 3

 20

57

Chapter 2

42.

21 24

1

21

3

52. There is 1 S and 12 total letters.

 3  24  1

1

P(S)  12

63

 24  43.

14  28x 7

21 8

 0.08 1 The probability of selecting an S is 12 or about 8%.

5

or 28

 (14  28x)  (7)

1 12 1 1  14 1 7 2  28x 1 7 2

53. There are 3 Es and 12 total letters.

 (14  28x) 7

3

P(E)  12 1

 4 or 0.25

 2  (4x)  2  4x 44.

5  25x 5

The probability of selecting an E is

115 2 1 1  5 1 5 2  25x 1 5 2

10

P(not N)  12 5

6  0.83 The probability of selecting a letter that is not N is 5 or about 83%. 6 55. There are 2 Rs and 1 P. So there are 2 1 or 3 letters that are R or P and 12 total letters.

 1  5x 4x  24y 4

 (4x  24y)  4

114 2 1 1  4x 1 4 2  24y 1 4 2

 (4x  24y)

3

P(R or P)  1

 1x  6y  x  6y 46. xz  2y  (4)(3)  2 (2.4)  12  4.8  16.8 47. 2

1

 4 or 0.25 The probability of selecting an R or P is

12yz 2  212 324 2 4.8  2 1 3 2

2x  z 4

 3y  

50

10

 39 The odds of selecting a dime are 10:39. 57. There are 90 pennies, and there are 75  50  30 or 155 coins that are not a penny. 90

odds of a penny  155 18

 31

  2.75  7.2  4.45 49. The lowest value is 12, and the highest value is 30, so use a scale that includes those values. Place an  above each value for each occurance.

5

10

15

20

The odds of selecting a penny are 18:31. 58. There are 90  50  30 or 170 coins that are not a nickel, and there are 75 nickels. odds of not a nickel  

25

170 75 34 15

The odds of selecting a coin that is not a nickel are 34:15. 59. There are 75  50 or 125 coins that are a nickel or a dime, and there are 90  30 or 120 coins that are not a nickel or a dime.

   

30

The greatest common place value is tens, so the digits in the tens place are the stems. Stem Leaf 1 222344555566 77778899999 2 01112668 3 0 1 0 2 = 12 50. Sample answer: Mean; the median and mode are too low. 51. Sample answer: Median; it is closest in value to most of the data.

Chapter 2

or 25%.

odds of a dime  195

2(4)  3  3(2.4) 4 8  3  7.2 4 11  7.2 4

            

1 4

56. There are 50 dimes, and there are 90  75  30 or 195 coins that are not a dime.

 2(1.6)  3.2

48.

or 25%.

54. There are 10 letters that are not N and 12 total letters.

 (5  25x)  5  (5  25x)

45.

1 4

125

odds of a nickel or a dime  120 25

 24 The odds of selecting a coin that is a nickel or a dime are 25:24. 60. 2196 represents the positive square root of 196. 196  142 S 2196  14 61. ;21.21 represents the positive and negative square roots of 1.21. 1.21  1.12 and 1.21  (1.1)2 ;21.21  ;1.1

58

4.  0x 0  38   02 0  38  2  38  40 5. 34  0x  21 0  34  07  21 0  34  014 0  34  14  20 6. 12  0x  8 0  12  01.5  8 0  12  06.5 0  12  6.5  5.5 7. 19  12  ( 019 0  012 0 )  (19  12)  7 8. 21  (34)  21  (34)  21  34  ( 034 0  021 0 )  (34  21)  13 9. 16.4  (23.7)  ( 023.7 0  016.4 0 )  (23.7  16.4)  7.3 10. 6.32  (7.41)  6.32  (7.41)  6.32  7.41  13.73

62. 2160 represents the negative square root of 160. 160  12.652 S 2160  12.65 4

63. ; 3 225 represents the positive and negative 4 square roots of 225. 4 225

 4

1152 22 and 2254  1152 22 2

; 3 225  ; 15

64. Because 16 and 25 are integers and 16  25  0.64 is a terminating decimal, this number is a rational number. 264

65. Because 2  4, this number is a natural number, a whole number, an integer, and a rational number. 66. Because 248.5  6.96419413 . . ., which is not a repeating or terminating decimal, this number is irrational. 67. You can use a calculator to find an approximation 1 for . 249 1  0.125 8

1 249

 0.142857142 . . .

Therefore,

1 8

6

1 249

.

7

2

2

33

3

7

70. Replace d in 93

3 216

7

1

 36

1

3 3.

13. 5(19)  95 15. 96  (0.8)  120

d3

729

3 216



38

20

1 0 21 0 0 2036 0 2 20 21   1 36  36 2

3 216 with 9.



1 20 2 20 21  36  1 36 2 21

  36 

1

3

1 52

21

3 3  0.577350269 p 34

1

 36  36

 0.866025403 p

Therefore,

1 0 7 0 0 166 0 2 7 6   1 16  16 2

12. 12  9  36  36

4

7 9.

69. You can use a calculator to find approximations 3 1 for 3 4 and 3 3. 3 4

6

 16

 0.4444444 . . .

Therefore,

7

  16 

3 3  0.816496580 . . . 4 9

3

11. 16  8  16  16

68. You can use a calculator to find approximations 2 4 for 3 3 and 9.

1

1

1 12

17. 8  (5)  8  5 1

 40

27

14. 56  (7)  8 16. (7.8)(5.6)  43.68 15

3

15

4

18. 32  4  32  3 60

 96 5

 1.84 The worst part of the hurricane will last about 1.8 hours.

 8 19. 5(3x)  12x  5(3)x  12x  15x  12x  (15  12)x  27x 20. 7(6h  h)  7(6h)  7(h)  7(6)h  7h  42h  7h  (42  7)h  35h 21. 4m(7n)  (3d)(4c)  4(7)mn  3(4)dc  28mn  (12)dc  28mn  12cd

Chapter 2 Practice Test Page 115 1. absolute value 2. rational 3. sample space

59

Chapter 2

22.

36k 4

31. The lowest value is 58, and the highest value is 74, so use a scale that includes these values. Place an  above each value for each occurance.

 36k  4 1

 36k  4  36  9k 23.

9a  27 3

114 2k

           

 (9a  27)  (3)

1 12 1 1  9a 1 3 2  27 1 3 2

58 60 62 64 66 68 70 72 74

 (9a  27) 3

32. Sample answer: The median and mode can be used to best represent the data. The mean is too high. 33. B; There are 7 hard-rock songs and 8  7  5 or 20 total songs.

 3a  (9)  3a  9 24.

70x  30y 5

 (70x  30y)  (5)

1 2 1  70x 1 2  30y 1 5 2  (70x  30y)

7

P(hard-rock)  20 or 0.35

1 5

The probability a hard-rock song will be playing 7 is 20 or 35%.

1 5

 14x  (6y)  14x  (6y)  14x  6y

Chapter 2 Standardized Test Practice

25. 264 represents the negative square root of 64.

Pages 116–117

64  82 S 264  8

1. B; 518.4  (9  8)  518.4  72  7.2 2. C; 20 in 10 minutes, 40 in 20 minutes, 60 in 30 minutes, 80 in 40 minutes. Extending the pattern would give 100 in 50 minutes and 120 in 60 minutes. 3. C; The points on the number line show 1, 0, 1, 2, and 3. 4. D; 0 4 0  4 04 0  4 07 0  7 09 0  9 4 6 7 6 9 5. A; 3.8  4.7  ( 04.7 0  03.8 0 )  (4.7  3.8)  0.9 6. D; 3(2m)  7m  3(2)m  7m  6m  7m  (6  7)m  13m 7. D; The least value is 3|1  31. 8. A; 1 box has a prize and 4  1 or 3 boxes do not have a prize.

26. 23.61 represents the positive square root of 3.61. 3.61  1.92 S 23.61  1.9 16

27. ; 3 81 represents the positive and negative square 16 roots of 81. 16 81



149 22 and 8116  149 22

16

4

; 3 81  ; 9

28. You can use a calculator to find an approximation 1 for . 23

1 23 1 3

 0.577350269 p  0.33333333 p

Therefore,

1

1

23

7 3.

29. You can use a calculator to find an approximation for

1

3 2.

1

3 2  0.7071067812 8 11

p

 0.727272727 p

Therefore,

1

32

6

8 . 11

1

odds of choosing a prize  3

30. You can use a calculator to find approximations 23 for 20.56 and 2 .

The odds of choosing a prize are 1:3. 9. B; 210  3.16 10. The length of the label will be the circumference of the jar. 2 r  2(3.14)(4)  6.28(4)  25.12 The length of the label is 25.12 cm.

20.56  0.748331477 p 23 2

 0.866025403 p

Therefore, 20.56 6

Chapter 2

23 . 2

60

11.

5  1 4  12  3  2

4  12  3  2 4 4  4  2 4 4  8 4 12 1 3

18. D; 49  72 and 49  (7)2 a could be 7 or 7. 19a. cost per number monthly Total cost 1is minute 123 times 14243 of minutes 123 plus 14243 fee. 14243 23 14243

4    

C  m  y  x or C  my  x 19b. For plan A, evaluate my  x with m  0.3, y  150, and x  5.95. my  x  0.3  150  5.95  45  5.95  50.95 For plan B, evaluate my  x with m  0.1, y  150, and x  12.95. my  x  0.1  150  12.95  15  12.95  27.95 For plan C, evaluate my  x with m  0.08, y  150, and x  19.99. my  x  0.08  150  19.99  12  19.99  31.99 Plan B is the least expensive. 20a. Without a key, you cannot determine what the values are. 20b. If the key is 3 02  3.2, then the data are ten times as great as they would be if the key is 3 02  0.32.

12. Replace m in 4m  3  9 with each value in the replacement set. 4m  3  9 4  0  3  9 S 3  9 42  3 9S5  9 43  3 9S9  9 4  5  3  9 S 17  9

m 0 2 3 5

True or False? false false true ✓ false

The solution of 4m  3  9 is 3. three times the 2p plus difference of m and n 123 123 1444424443 2p  3  (m  n) or 2p  3(m  n) 14. Hypothesis; 3x  3 7 24 15. C; x is x units from zero, and x is x units away from zero. 0x 0  0x 0 16. B; y 6 x 1 1  y 6 x  xy xy 13.

1 x

6

1 y

1

17. A; n 7 1 1 nn 7 1n 1 7 n

61

Chapter 2

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Chapter 3 Page 119

Solving Linear Equations 16. The base is 300, and the part is 21. Let p represent the percent.

Getting Started

1. Greater than implies add, and half of implies 1 multiply by 2. So, the expression can be written as 1 t  5. 2

a b 21 300

2. Product implies multiply, and divided by implies divide. So, the expression can be written as 7s  8y. 3. Sum implies add, times implies multiply, and square implies raise to the second power. So, the expression can be written as 3a  b2. 4. Decreased by implies subtract, so the expression can be written as w5  37. 5. Times implies multiply, and subtracted from implies subtract. So, the expression can be written as 95  9y. 6. Sum implies add, and divided by implies divide. So, the expression can be written as (r  6)  12. 12 4

7. 3  6 

 18 

2  6  4 2

 

2100 300

19  5 7

a b 15 5

14 7

1500 5

12 4

a b 12 60

1200 60

a b 16 10

1600 10

p

 100 p

 100



60p 60

p

 100 p

 100



10p 10

160  p Sixteen is 160% of 10. 20. The base is 50, and the part is 37.5. Let p represent the percent. a b 37.5 50

p

 100

p

 100 p

 100

37.5(100)  50p 3750  50p

p

 100

5(100)  20p 500  20p

3750 50



50p 50

75  p Thirty-seven and one half is 75% of 50.

20p 20

25  p Five is 25% of 20.

Chapter 3

5p 5

16(100)  10p 1600  10p

3

0 15. The base is 20, and the part is 5. Let p represent the percent.





20  p Twelve is 20% of 60. 19. The base is 10, and the part is 16. Let p represent the percent.

1

500 20

p

 100

12(100)  60p 1200  60p

14. 4 (24)  2 (12)  6  6

a b 5 20

p

 100

300  p Fifteen is 300% of 5. 18. The base is 60, and the part is 12. Let p represent the percent.

23 5 1

300p 300

15(100)  5p 1500  5p

12  4 2 8 2

3



7p Twenty-one is 7% of 300. 17. The base is 5, and the part is 15. Let p represent the percent.

4 11. (25  4)  (22  1)  21  (4  1)  21  3 7 12. 36  4  2  3  9  2  3 73  10 13.

p

 100

21(100)  300p 2100  300p

 18  3  15 8. 5(13  7)  22  5(6)  22  30  22 8 9. 5(7  2)  32  5(5)  32  5(5)  9  25  9  16 10.

p

 100

62

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5.

the sum of is the Five times m and n same as seven times n. 123 123 14243 1 424 3 123 123 { 5  (m  n)  7  n The equation is 5(m  n)  7n. 6. Words: Area equals one half times the base times the height. Variables: Let A  area, b  base, and h  height.

Writing Equations

3-1 Page 122

Algebra Activity

1. The area of the front is the product of the length and the height or lh. 2. The area of the back is the product of the length and the height or lh. 3. The area of one side is the product of the width and the height or wh. 4. The area of the other side is the product of the width and the height or wh. 5. The area of the top is the product of the length and the width or lw. 6. The area of the bottom is the product of the length and the width or lw. surface area is 7. The the sum of the areas of the faces. 1 4444244 443 { 1444444442444444443 

S

one the the Area half base height. 1 23 equals 123 1 23 times 123 1 23 times 123 123

Formula:

Circumference equals the product of two, pi, and the radius. Variables: Let C  circumference and r  radius. Circumference 1442443

Fourteen 9.

8



1 b 3 {

plus

d

equals

3 4 {



{

six times d.



{

2a {

One-third of b minus three-fourths equals two times a.

10. Sample answer: The original cost of a suit is c. After a $25 discount, the suit costs $150. What is the original cost of the suit? Darius’ number of pounds 11. current weight 14 442444 3 155

plus 1 23 

he wants to gain 144 442444 43 g

equals 1 4243 

160. 123 160

The equation is 155  g  160. 12. 155  g  160 Find g mentally by asking, “What number added to 155 equals 160?” g5 Darius needs to gain 5 pounds to reach his goal.

Pages 124–126 13.

Two hundred 1 4 44 244 43 200

Practice and Apply minus 123 

three 123 3

times 123 

x

{

x

is equal to 14 424 43 

nine. 123 9

The equation is 200  3x  9. 14. Rewrite the sentence so it is easier to translate. Twice plus three times s 1 is44244 identical to 1 424 3r 12 3 1 23 1 23 { 3 thirteen. 1 424 3 2r  3  s  13

The equation is 2r  3s  13. 15. Rewrite the sentence so it is easier to translate. One-third q 1 plus is44244 as much as q. 1 4424 43 23 25 { 1 3 twice 1 424 3

30  w or 30w 2d. No; 1900  52(30)  3460, which is less than 3500. 3. Sample answer: After sixteen people joined the drama club, there were 30 members. How many members did the club have before the new members? 4. 1 Two t decreased 23 times 123 { 14 4244by 3 eight 123 equals 123 seventy. 14243 

the product of two, pi, and the radius. 144424443

equals 123

Formula: C  2r The formula for the circumference of a circle is C  2r. 8. 14  d  6d { { { { {

lh  lh  wh  wh  lw  lw

30  10 30 . 10  300 money number per week times of weeks 14243 123 14 243 30  20 30 . 20  600 She will add $300 in 10 weeks and $600 in 20 weeks. 2c. money number per week times of weeks 14243 123 14 243

t

h

7. Words:

Check for Understanding





b

1

1. Explore the problem, plan the solution, solve the problem, and examine the solution. 2a. Misae has $1900 in her account at the beginning. 2b. money number per week times of weeks 14243 123 14 243

2



The formula for the area of a triangle is A  2 bh.

or S  2lh  2wh  2lw 8. If s represents the length of the side of a cube, then the surface area is given by S  2lh  2wh  2lw with l  s, h  s, and w  s. S  2lh  2wh  2lw  2s  s  2s  s  2s  s  2s2  2s2  2s2  6s2 The surface area is given by S  6s2.

Page 123

1 2



A

1 q 3



The equation is

1 q 3

25



2q

 25  2q.

16. 14 The44244 square of m 123 minus the cube 43 of n is sixteen. { 1 4243 43 14 424 2 3  n  16 m The equation is m2  n3  16.

70

The equation is 2t  8  70.

63

Chapter 3

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Volume equals the product of , the square of the radius of the base, and the height. Variables: Let V  volume, r  radius of base, and h  height.

17.

the sum of is equal Two times v and w 1 to43 1 two z. 123 123 14243 42 23 times 123 { 2  (v  w)  2  z The equation is 2(v  w)  2z. 18. the sum of is the Half of nine and p same as p minus three. 123 { 14243 1 424 3 { 123 123 1  (9  p)  p  3 2

26. Words:

the square of the the Volume times radius of the base times height. 1 424 3 equals 123  { 123 14 4424443 123 123 2

Formula: V    r  h The formula for the volume of a cylinder is V  r2h. 27. Words: The square of the measure of the hypotenuse is the square of the measure of one leg plus the square of the measure of the other leg. Variables: Let c  measure of hypotenuse, a  measure of one leg, and b  measure of other leg. Formula:

1

The equation is 2(9  p)  p  3. 19. Rewrite the sentence so it is easier to translate. The divided the is the twice the number g 1 by 3 number h same as sum of g and h plus seven. 14243 424 14243 1 424 3 1442443 123 123



g

h



2(g  h)



7

The equation is g  h  2(g  h)  7. 20. Rewrite the sentence so it is easier to translate. the square of the the square the square Five-ninths times sum of a, b, and c equals of a plus of c. 14243 123 144424443 123 14243 123 14243 5 9



(a  b  c)2

a2





c2

The square of the the square of the the square of the measure of the hypotenuse measure of one43 leg 1 plus of the other leg. 144 44424444 43 is {1 444244 23 measure 14444244443

5

The equation is 9(a  b  c)2  a2  c2.

c2

21. Rewrite the sentence so it is easier to translate. The Pacific Ocean 46% Earth. { 1 23 of { 123 144 42444 3 is

This formula for a right triangle is c  a2  b2. 28. Words: Temperature in degrees Fahrenheit is nine-fifths of the degrees Celsius plus thirty-two. Variables: Let F  degrees Fahrenheit and C  degrees Celsius. Temperature in the degrees degrees Fahrenheit is nine-fifths of Celsius plus thirty-two. 144424443 { 14243 { 14243 123 14243

Formula:



d 30.



C



32.

minus 2f { k2

{

fourteen 6 {

equals  {

five. 19 {

equals

nineteen.

2a { 3 p 4 {

six

17

{

Two times a 33.

{

plus



k squared plus

32.

 {

{

34.

2 w 5{



5

{



53

{

{

j

{

{

seventeen equals fifty-three minus j.



7a {

{

equals 

{

Three-fourths of p plus

(B  h)



14 {

{

Two times f 31.

The formula for the volume of a pyramid is 1 V  3Bh.



{

seven times a 1 2 {

one-half 1 w 2{



{

b

{

minus b. 

p

{

{

equals p. 

{

3

{

Two-fifths times w equals one-half times w plus three. 35. 7(m  n)  10n  17 1 424 3 { 1 23 { { Seven times equals ten plus seventeen. the sum of m and n times n 36. 4(t  s)  5s  12 { { { { 123 Four times the equals five times s plus twelve. quantity t minus s

25. Words:

Perimeter is twice the sum of the lengths of the two adjacent sides. Variables: Let P  perimeter, a  one side, and b  an adjacent side. the sum of the lengths Perimeter is two times of the two adjacent sides. 44424444 43 14243 { 123 123 144

Formula: P  2  (a  b) The formula for the perimeter of a parallelogram is P  2(a  b).

Chapter 3



29. { d

the product of the area Volume is one-third the44424444 base and its height. 14243 times 123 of 144 43 123 { 1 3

9 5



F

The formula for temperature in degrees Fahrenheit 9 is F  5C  32.

Formula: A  b  h The formula for the area of a parallelogram is A  bh. 24. Words: Volume is one-third times the product of the area of the base and its height. Variables: Let V  volume, B  area of base, and h  height.

V 

b2



2

P  0.46  E The equation is P  0.46E. 22. Rewrite the sentence so it is easier to translate. The cost number per yard times of yards plus 3.50 equals 73.50. 14243 123 1 424 31 23 1 23 123 123 1.75  f  3.50  73.50 The equation is 1.75f  3.50  73.50. 23. Words: Area is the base times the height. Variables: Let A  area, b  base, and h  height. Area the base times the height. 1 23 is { 1 424 3 1 23 14243

Formula:

a2



64

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1

Plan:

37. A  2h (a  b)

Write the formula for the volume of a sphere.

The area A of a trapezoid equals one-half times the product of the height h and the sum of the bases, a and b. 38. rt  d {

{

The fourVolume is thirds times 1 424 3 { 123 123

{

V

Rate times time equals distance. 39. Sample answer: Lindsey is 7 inches taller than Yolanda. If 2 times Yolanda’s height plus Lindsey’s height equals 193 inches, find Yolanda’s height. 40. Sample answer: The price of a new backpack is p dollars and the tax on the backpack is 0.055p. If the total cost of the backpack and the tax is $31.65, what is the price of the backpack? 41. Words: Volume equals one-third times the product of , the square of the radius of the base, and the height. Variables: Let V  volume, r  radius of base, and h  height.

Solve:

   

1 3







r2



h 1

Solve:

1 3

1912

1

V  3r2h 1

 3 (10)2(30) 1

 3 (100)(30)

11003 2(30)

 1000  3141.6 Examine: The volume of a cone with r  10 and h  30 is about 3142 cm3. 43. Words: Volume is four-thirds times  times the radius of the sphere cubed. Variables: Let V  volume and r  radius of sphere.

Formula:

fourVolume is thirds times 1 424 3 { 123 123 4 V  3 

r3



y



1928

a  (4a  15)  60 The equation is a  (4a  15)  60. 50. Find the solution for a  (4a  15)  60. Start by letting a  10 and then adjust values up or down as needed. a

a  (4a  15)  60 ?

10 10  (4  10  15)  60 S 65  60

 {

times 123

the radius of the sphere cubed. 14243





r3

8

4

9

The formula for the volume of a sphere is V  3r3. 44. Explore:



4 r3 3 4  (4)3 3 4  (64) 3 4 (64) 3 256  3

4  a  15 The expression is 4a  15. 49. Rewrite the sentence so it is easier to translate. The advertising 15 more than 4 times portion plus advertising portion { is { 60. 1442443 1 23 the 14444244443

1

 3 (100)(30) 

times 123

The equation is 1912  y  1928. 47. 1912  y  1928 Ask yourself, “What number added to 1912 equals 1928?” y  16 There were 16 years between the first Tarzan story and the naming of the town. 48. Rewrite the phrase so it is easier to translate. the advertising Four times portion plus 1 23 1 23 1442443 1 23 15 {

(r2h)



{

The year Tarzan the number of years before the year the was published plus the town was renamed equals town was renamed. 1 4442444 3 123 1444442444443 123 144424443

42. Explore: You know the formula for the volume of a cone. You need to evaluate the formula with r  10 and h  30. Plan: Write the formula for the volume of a cone. the product of , the square The volume equals 1 one-third times 14 of 424 r, and43 h. 14 4244 3 123 4243 123 

 

1912  y The expression is 1912  y. 46. Rewrite the sentence so it is easier to translate.

The formula for the volume of a cone is V  3 r2h.

V



 268.1 The volume of a sphere with r  4 is about 268 in3. 45. Rewrite the phrase so it is easier to translate. the number of years The year Tarzan before the town was was first published plus named Tarzana 144424443 123 144424443

the square of the the Volume equals one-third times  times radius of the base times height. 1 424 3 123 14243 123 { 123 14444244443 123 123 

4 3

Examine:

Formula:

V

V



the radius of the sphere cubed. 14243

?

5

5  (4  5  15)  60 S 40  60

7

7  (4  7  15)  60 S 50  60

? ?

8  (4  8  15)  60 S 55  60 ?

9  (4  9  15)  60 S 60  60

Reasonable? too high too low too low almost true ✓

The solution of a  (4a  15)  60 is 9. Therefore, 9 minutes were spent on advertising. 51. See students’ work.

You know the formula for the volume of a sphere. You need to evaluate the formula with r  4.

65

Chapter 3

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63. 12d  3  4d  12d  4d  3  (12d  4d)  3  (12  4)d  3  8d  3 2 64. 7t  t  8t  7t2  (t  8t)  7t2  (1  8)t  7t2  9t 65. 3(a  2b)  5a  3(a)  3(2b)  5a  3a  6b  5a  3a  5a  6b  (3a  5a)  6b  (3  5)a  6b  8a 6b 66. 5(8  3)  7  2  5(5)  7  2  25  14  39 67. 6(43  22)  6(64  4)  6(68)  408 68. 7(0.2  0.5)  0.6  7(0.7)  0.6  4.9  0.6  4.3 69. 1 70. 1 0.57 5.67 2.80 3.70 9.37 3.37 71. 4 72. 8 9 5.128 9.010 3. 4 0 7.3 5 1. 8 8 1.6 5

52. Words:

Surface area is the sum of the area of the faces. Variables: Let S  surface area, a  length of the sides of an equilateral triangle, and h  height of a prism. The height of the equilateral triangle is a 23 given by 2 .

Surface area of area of area is three times a side plus two times a base. 1 4243 { 123 123 123 123 123 123 123 Formula: 1 a 23 S  3  ah  2  2a 2 The formula for the surface area of this a2 23

triangular prism is S  3ah  2 . 53. Equations can be used to describe the relationships of the heights of various parts of a structure. Answers should include the following. • The equation representing the Sears Tower is 1454  a  1707. 54. B; a the One-fourth of number plus five equals number minus 123 seven. 14243 { 1 424 3 1 23 1 231 424 31 424 3 123 1 4



n

The equation is

1 n 4

 {

55. D;{ 7 Seven

times

Page 126





5

n



7

 5  n  7. (x  y) 123 the sum of x and y

{



35 {

equals

35.

Maintain Your Skills

56. 28100 represents the positive square root of 8100. 8100  902 S 28100  90 25

57.  3 36 represents the negative square root of 25 36



156 22 S  3 2536  56

73.

2 3

1

10

3

 5  15  15

74.

1 6

2

13

75.

7 9

2

7

4

5

 15 25 . 36

1

366 6

6

399

76.

3 4

1

9

2

 6  12  12 7

1

 12

9

58. 290 represents the positive square root of 90. 90  9.492 S 290  9.49

Page 127

Algebra Activity (Preview of Lesson 3–2)

59. 255 represents the negative square root of 55. 1. Step 1

55  7.422 S 155  7.42 60. There is 1 six and 6 total outcomes.

Model the equation.

x

1

P(6)  6  0.17 The probability of rolling a 6 is

1 1 6

1

or about 17%.

1

61. There are 3 even numbers and 6 total outcomes. P(an even number)  

3 6 1 2

1 1

or 0.5

1

1

1

1



1

1

1

1

1

1

1

1 1 1 1 1

x57 x  5  (5)  7  (5) Place 1 x tile and 5 positive 1 tiles on one side of the mat. Place 7 positive 1 tiles on the other side of the mat. Then add 5 negative 1 tiles to each side.

1

The probability of rolling an even number is 2 or 50%. 62. There are 4 numbers greater than two and 6 total outcomes. 4

P(a number greater than 2)  6 2

3 The probability of rolling a number greater than 2 2 is 3.

Chapter 3

1

66

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Step 2

Place 1 x tile and 4 positive 1 tiles on one side of the mat. Place 27 positive 1 tiles on the other side of the mat. Then add 4 negative 1 tiles to each side. Step 2 Isolate the x term.

Isolate the x term.

x 1

1

1

1

1

1

1

1

1



1

1

1

1

1

1

1

1

1

1

x

1

1

x2

1

1

 1

1

–1

1

1

Group the tiles to form zero pairs. Then remove all the zero pairs. The resulting equation is x  2. 2. Step 1 Model the equation.

x

–1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

–1

1

–1

1

1

1

1



1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

–1

1

1

1

–1

1

1

1

–1

1

1

1

–1

1

x 1 1 1

1



x  (2)  28 x  (2)  2  28  2 Place 1 x tile and 2 negative 1 tiles on one side of the mat. Place 28 positive 1 tiles on the other side of the mat. Then add 2 positive 1 tiles to each side. Step 2 Isolate the x term. 1

1

1

x  23 Group the tiles to form zero pairs. Then remove all the zero pairs. The resulting equation is x  23. 4. Step 1 Model the equation.

1

x



1

1

1

1

1

1

1

1

1

1

1

x  (3)  4 x  (3)  3  4  3 Place 1 x tile and 3 negative 1 tiles on one side of the mat. Place 4 positive 1 tiles on the other side of the mat. Then add 3 positive 1 tiles to each side. Step 2 Isolate the x term. \ x

–1

1

–1

1

–1

1



x  30 Group the tiles to form zero pairs. Then remove all the zero pairs. The resulting equation is x  30. 3. Step 1 Model the equation.

1

1

1

1

1

1

1

x7 x 1

1 1 1

1

1 1



Group the tiles to form zero pairs. Then remove all the zero pairs. The resulting equation is x  7.

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1 1

x  4  27 x  4  (4)  27 (4)

67

Chapter 3

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5. Step 1

Group the tiles to form zero pairs. Then remove all the zero pairs. The resulting equation is x  5. 7. You can model the equation a  b by putting an a tile on one side of the mat, and putting a b tile on the other side of the mat. You can add a c tile to each side of the mat. The resulting equation would be a  c  b  c. 8. You can model the equation a  b by putting an a tile on one side of the mat, and putting a b tile on the other side of the mat. You can add a negative c tile to each side of the mat. The resulting equation would be a  (c)  b  (c) or a  c  b  c.

Model the equation.

x 1

1

1



1

1

1

1

1

1

1

1

1

1

x  3  4 x  3  (3)  4  (3) Place 1 x tile and 3 positive 1 tiles on one side of the mat. Place 4 negative 1 tiles on the other side of the mat. Then add 3 negative 1 tiles to each side. Step 2 Isolate the x term.

3-2 Page 131

x

1

1

1

1

1

1



1

1

1

1

1

1

1

Group the tiles to form zero pairs. Then remove all the zero pairs. The resulting equation is x  7. 6. Step 1 Model the equation.

x

1

1

1

1

1

1

1

1

1

1 1

1



1

1

1

1 1 1

1

1

1

1

x72 x  7  (7)  2  (7) Place 1 x tile and 7 positive 1 tiles on one side of the mat. Place 2 positive 1 tiles on the other side of the mat. Then add 7 negative 1 tiles to each side. Step 2 Isolate the x term.

x 1 1 1 1 1 1 1

1

1

1

1

1

1

1 1



1 1 1

1 1

1

1

1

x  –5 Chapter 3

Check for Understanding

1. Sample answers: n  13, n  16  29, n  12  25 2. The Addition Property of Equality and the Subtraction Property of Equality can both be used to solve equations. The Addition Property of Equality says you can add the same number to each side of an equation. The Subtraction Property of Equality says you can subtract the same number from each side of an equation. Since subtracting a number is the same as adding its inverse, either property can be used to solve any addition equation or subtraction equation. 3. (1) Add 94 to each side. g  94  (94)  75  (94) g  19 (2) Subtract 94 from each side. g  94  94  75  94 g  19 4. t  4  7 t  4  4  7  4 t  3 t  4  7 Check: ? 3  4  7 7  7 ✓ The solution is 3. 5. p  19  6 p  19  19  6  19 p  13 p  19  6 Check: ? 13  19  6 66✓ The solution is 13. 6. 15  r  71 15  r  15  71  15 r  56 15  r  71 Check: ? 15  56  71 71  71 ✓ The solution is 56.

x  –7

1

Solving Equations by Using Addition and Subtraction

68

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104  y  67 104  67  y  67  67 171  y Check: 104  y  67 ? 104  171  67 104  104 ✓ The solution is 171. 8. h  0.78  2.65 h  0.78  0.78  2.65  0.78 h  3.43 h  0.78  2.65 Check: ? 3.43  0.78  2.65 2.65  2.65 ✓ The solution is 3.43.

14. The average time in Europe 1442443

the average minus time in Japan 123 14424 43 35.5  16.8 35.5  16.8  18.7 The difference is 18.7 hours.

7.

2 3

9. 2 3

Pages 132–134

1

 w  12 2

1

2

 w  3  12  3 5

w6 2 w 3 2 5 ? 6 3 1 12  5 solution is 6.

Check:

The

1

12 1

12 1

12 ✓

10. 1 A4 number minus twenty-one 42443 1 424 3 1 442443 n  21 n  21  8 n  21  21  8  21 n  13 n  21  8 Check: ? 13  21  8 8  8 ✓ The solution is 13. 11. 1 A4 number increased by 37 42443 1 4442444 3 1 23

is 123 

8. 1 23 8

is 91. { 123 n  37  91 n  (37)  91 n  (37)  37  91  37 n  54 Check: n  (37)  91 ? 54  (37)  91 91  91 ✓ The solution is 54. 12. Words: The average time in Japan plus 8.1 equals the average time in the United States. Variable: Let /  the average time in Japan. The average the average time in Japan plus 8.1 equals time in the U.S. 1442443 123123 123 1442443 Equation: /  8.1  13. /  8.1  24.9 /  8.1  8.1  24.9  8.1 /  16.8 The average time is 16.8 hours.

Practice and Apply

15.  9  14  9  9  14  9  23  9  14 Check: ? 23  9  14 14  14 ✓ The solution is 23. 16. s  19  34 s  19  19  34  19 s  15 s  19  34 Check: ? 15  19  34 34  34 ✓ The solution is 15. 17. g  5  33 g  5  5  33  5 g  28 Check: g  5  33 ? 28  5  33 33  33 ✓ The solution is 28. 18. 18  z  44 18  z  18  44  18 z  26 18  z  44 Check: ? 18  26  44 44  44 ✓ The solution is 26. 19. a  55  17 a  55  55  17  55 a  38 a  55  17 Check: ? 38  55  17 17  17 ✓ The solution is 38. 20. t  72  44 t  72  72  44  72 t  28 Check: t  72  44 ? 28  72  44 44  44 ✓ The solution is 28. 21. 18 61  d 18  61 61  d 61 43  d 18  61  d Check: ? 18  61  43 18  18 ✓ The solution is 43.

24.9

69

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22.

23.

24.

25.

26.

27.

28.

6  m  (3.42) 6  3.42  m  (3.42)  3.42 2.58  m 6  m  (3.42) Check: ? 6  2.58  (3.42) 6  6 ✓ The solution is 2.58. 30. 6.2  4.83  y 6.2  4.83  4.83  y  4.83 11.03  y 6.2  4.83  y Check: ? 6.2  4.83  11.03 6.2  6.2 ✓ The solution is 11.03. 31. t  8.5  7.15 t  8.5  8.5  7.15  8.5 t  15.65 Check: t  8.5  7.15 ? 15.65  8.5  7.15 7.15  7.15 ✓ The solution is 15.65. 32. q  2.78  4.2 q  2.78  2.78  4.2  2.78 q  6.98 q  2.78  4.2 Check: ? 6.98  2.78  4.2 4.2  4.2 ✓ The solution is 6.98.

25  150  q 25  150  150  q  150 125  q 25  150  q Check: ? 25  150  125 25  25 ✓ The solution is 125. r  (19)  77 r  19  77 r  19  19  77  19 r  96 r  (19)  77 Check: ? 96  (19)  77 77  77 ✓ The solution is 96. b  (65)  15 b  65  15 b  65  65  15  65 b 50 Check: b  (65)  15 ? 50  (65)  15 15  15 ✓ The solution is 50. 18  (f )  91 18  f  91 18  f  18  91  18 f  73 18  (f )  91 Check: ? 18  (73)  91 91  91 ✓ The solution is 73. 125  (u)  88 125  u  88 125  u  125  88  125 u  37 125  (u)  88 Check: ? 125  [(37) ]  88 88  88 ✓ The solution is 37. 2.56  c  0.89 2.56  c  2.56  0.89  2.56 c  3.45 Check: 2.56  c  0.89 ? 2.56  3.45  0.89 0.89  0.89 ✓ The solution is 3.45. k  0.6  3.84 k  0.6  0.6  3.84  0.6 k  4.44 k  0.6  3.84 Check: ? 4.44  0.6  3.84 3.84  3.84 ✓ The solution is 4.44.

Chapter 3

29.

33.

3

5

3

5

x46 3

3

x4464 7

x  112 Check:

3

5

x46

3 ? 5 6 5 5  ✓ 6 6 7 The solution is 112. 3 7 a  5  10 3 3 7 3 a  5  5  10  5 1 a  10 3 7 Check: a  5  10 1 3 ? 7 10  5  10 7 7 10  10 1 The solution is 10. 1 5 2  p  8 1 1 5 1 2  p  2  8  2 1 p  18 1 5 Check: 2  p  8 1 1 ? 5 2  18  8 5 5 8✓ 8 1 The solution is 18. 7

112  4 

34.

35.

70



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2 3

36. 2 3

4

43.

A number eighteen equals 31. 1424 3 minus 1231 424 3 123 { Equation: n  18  31 n  18  31 n  18  18  31  18 n  49 n  18  31 Check: ? 49  18  31 31  31 ✓ The solution is 49. 44. What number equals 18? 14442 4443 decreased 14 44244 4by 3 77 { 1 424 31 23

 r  9 2

4

2

 r  3  9  3 1

r  19 2 3

Check: 2 3

1

4

 r  9 1

2

?

4

 19  9 4

4

9  9 ✓ 1

The solution is 19. 37.

2 3 2 4 5 3 2 15

4

v5 4

38. 4

v55 v

2 4 Check: v5 3 2 ? 2 4  15  5 3 2 2 3✓ 3 2 The solution is 15.

2 5 2 3 4 5 7 20

3

w4 3

 77  18 n 77  18 n  77  77  18  77 n  59 n  77  18 Check: ? 59  77  18 18  18 ✓ The solution is 59. 45. A number increased by 16 21. 1424 3 14 4244 31 23 is { 1 23 Equation: n  (16)  21 n  (16)  21 n  (16)  16  21  16 n  5 Check: n  (16)  21 ? 5  (16)  21 21  21 ✓ The solution is 5. 46. A number plus 43 is 1 102. 23 14 243 1 23 1 23 { Equation: n

3

w44 w

2 3 w4 5 2 ? 7 3  20  4 5 2 2 5✓ 5 7 solution is 20.

Check:

The

39. First, solve x  7  14. x  7  14 x  7  7  14  7 x  21 Now, replace x in x  2 with 21. x  2  21  2  19 The value is 19. 40. First, solve t  8  12. t  8  12 t  8  8  12  8 t  20 Now, replace t in t  1 with 20. t  1  20  1  19 The value is 19. 41. The length of the rectangle 1444 4424444 43 is 78. {

n  (43)  102 n  (43)  102 n  (43)  43  102  43 n  145 Check: n  (43)  102 ? 145  (43)  102 102  102 ✓ The solution is 145. Equation:

{

x  55  78 x  55  78 x  55  55  78  55 x  23 x  55  78 Check: ? 23  55  78 78  78 ✓ The solution is 23. 42. The width of the rectangle 144444244444 3 is 24. Equation:

y  17 y  17  24 y  17 17  24  17 y  41 y  17  24 Check: ? 41  17  24 24  24 ✓ The solution is 41.

Equation:

47.

What number minus 1 424 3 123

Equation:



n

1

3

1

3

one- is equal half to 3 1 23 1424 1 2



negative three-fourth? 1 442443 3

4

n  2  4 1

1

n  2  2  4  2 1

n  4 Check:

{ {

1

3

1 ?

3

3

3

n  2  4 1

4  2  4

 24

4  4 ✓ 1

The solution is 4.

71

Chapter 3

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58. Words:

The number of volumes at Harvard University plus the difference equals the number of volumes at the Library of Congress. Variable: Let x  the difference in volumes between the two libraries.

48.

is equal 19 plus 42 plus a number to 87. { 1 23 { 123 14 4244 3 14 4244 3 { Equation:19  42  n  87 19  42  n  87 61  n  87 61  n  61  87  61 n  26 Check: 19  42  n  87 ? 19  42  26  87 87  87 ✓ The solution is 26. 49. Sometimes; if x  0, x  x  x is true. 50. Always; any number plus 0 is always the number. 51. Words: The miles per gallon of a luxury car plus 10 equals the miles per gallon of a midsize car. Variable: Let   the miles per gallon of a luxury car. The miles per gallon of a luxury car plus equals 144424443 1 23 10 { 123

The number of volumes the the number of volumes at at Harvard University plus difference equals the Library of Congress. 14444244443 123 14243 123 1444442444443

Equation: 13.6  x  24.0 13.6  x  24.0 13.6  x  13.6  24.0  13.6 x  10.4 The Library of Congress has 10.4 million volumes more. 59. Words: The number of volumes at New York Public plus the difference equals the number of volumes at Harvard University. Variable: Let x  the difference in volumes between the two libraries.

the miles per gallon of a midsize car. 144424443

Equation:   10  34 The equation is   10  34. /  10  34 52. /  10  10  34  10 /  24 A luxury car travels 24 miles on a gallon of gasoline. the miles per gallon 53. The miles per gallon of a subcompact car 144424443 x

is { 

of a luxury car 144424443 24

plus 123 

13. { 13

37  34 37  34  3 A subcompact car travels 3 miles more. 55. Sample answer: 29 miles; 29 is the average of 24 (for the 8-cylinder engine) and 34 (for the 4-cylinder engine). 56. Words: The height of the Great Pyramid plus the amount of decrease equals the original height. Variable: Let d  the amount of decrease. the amount plus of4decrease 1 23 1 42443

Equation: 450  d The equation is 450  d  481 57. 450  d  481 450  d  450  481  450 d  31 The decrease in height is 31 ft.

Chapter 3



equals 123

The number of volumes at the New York Public 1 plus difference 144424443 23 14243

the number of volumes at the equals Library of Congress. 123 144424443

Equation: 11.4  x  24.0 11.4  x  24.0 11.4  x  11.4  24.0  11.4 x  12.6 The Library of Congress has 12.6 million volumes more. 61. Words: The total number of volumes is equal to the sum of the volumes at the Library of Congress, Harvard University, and New York Public. Variable: Let x  the total number of volumes.

the original equals 123

the number of volumes at Harvard University. 144424443

Equation: 11.4  x  13.6 11.4  x  13.6 11.4  x  11.4  13.6  11.4 x  2.2 Harvard University has 2.2 million volumes more. 60. Words: The number of volumes at New York Public plus the difference equals the number of volumes at the Library of Congress. Variable: Let x  the difference in volumes between the two libraries.

x  24  13 x  37 A subcompact car travels 37 miles on a gallon of gasoline. 54. The miles per gallon the miles per gallon of a subcompact car minus of a midsize car 144424443 123 144424443

The height of the Great Pyramid 144424443

The number of volumes at the New York Public 1 plus difference 144424443 23 14243

height. 14 42443

The total volumes at volumes at volumes at number is Lib. of C. plus Harvard plus NY Public. 14243 { 14243 123 14243 123 14243

481

Equation: x  24.0  13.6  11.4 x  24.0  13.6  11.4 x  49 The total number of volumes at the three largest U.S. libraries is 49 million.

72

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68. A; x 167  52 x 167  167  52  167 x  115

62. Words:

The number of one-year olds plus the difference equals the number of newborns. Variable: Let x  the difference between the number of one-year olds and newborns. The number of the one-year olds 1 plus 1442443 23 difference 14243 equals 123

Page 134

Maintain Your Skills

69. Words: Area is  times the radius squared. Variables: Let A  area and r  radius.

the number of newborns. 1442443

Equation: (679  634) x  (1379  1286) 679  634  x  1379  1286 1313  x  2665 1313  x  1313  2665  1313 x  1352 There are 1352 more newborns. 63. Words: The number of males plus the difference equals the number of females. Variable: Let x  the difference between the number of males and females.

Area 1 23

is

{



{

times 123

the radius squared. 144424443

Formula: A    r2 The formula for the area of a circle is A  r2. 70. Replace r in r2 with 16. r2  (16)2  (256) ≈ 804.2 The area is about 804 in2. 71. You can use a calculator to find an approximation for 22. 1 2

The number the the number of4males plus difference equals 1 of4females. 14 2443 1 23 1 442443 123 42443

 0.5

22  1.41421356 p

Equation: (1379  679  x  (1286  1707) 634  3714) 1379  679  1707  x  1286  634  3714 3765  x  5634 3765  x  3765  5634  3765 x  1869 There are 1869 more females. 64. Words: The total population is equal to the sum of the number of newborns, one-year olds, and adults. Variable: Let t  the total population.

Therefore,

1 2

6 22.

72. You can use a calculator to find an approximation 2 for 3. 3 4 2 3

 0.75  0.66666666 p

Therefore,

3 4

2

7 3. 3

73. You can use a calculator to find the value of 8. 3 8

The total the number of the number of the number population is newborns plus one-year olds plus of adults. 14243 { 14424431 23 1442443 123 14243

 0.375 3

Therefore, 0.375  8.

Equation: t  (1379  1286)  (679  634)  (1707  3714) t  1379  1286  679  634  1707  3714 t  9399 The total deer population is 9399. 65. If a  b  x, then a  b  x. If a  x  b  x, then by the transitive property a  a  x. If a  a  x, then x  0. If x  0, then a  x  b  x implies a  b. Therefore, if a  b  x, then a  b and x  0 would make a  x  b  x true. 66. Equations can be used to describe the relationships of growth and decline in job opportunities. Answer should include the following. • To solve the equation, add 66 to each side. The solution is m  71. • An example such as “The percent increase in growth for paralegals is 16 more than the percent increase in growth for detectives. If the growth rate for paralegals is 86%, what is the growth rate for detectives? d  16  86; 70%” 67. C; b 15  32 b  15  2  32  2 b 13  34

74. The greatest common place value is tens, so the digits in the tens place are the stems. Stem Leaf 3 12456 4 0123 5 246 3 01  31 75. The greatest common place value is ones, so the digits in the ones place are the stems. Stem Leaf 0 58 1 1247 2 3689 3 4 15 0 05  0.5 76. Hypothesis: y  2 Conclusion: 4y  6  2 77. Hypothesis: it is Friday Conclusion: there will be a science quiz 78. 4(16  42)  4(16  16) Substitution; 42  16  4(1) Substitution; 16  16  1 4 Multiplicative Identity; 4  1 4

73

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4. 2g  84

79. (25  52)  (42  24)  (32  25)  (16  16) Substitution  7  0 Substitution 7 Additive Identity 80. Replace x in 3x  2 2 with each value in the replacement set. 3x  2  2

x

30  2 7 2S2 2

true ✓

?

true ✓

1

31  2 7 2S5 7 2

2

32  2 7 2S8 7 2

?

2y2  1  0

True or False?

?

true ✓

1

2(1)2  1 7 0 S 1 7 0

3

2(3)2  1 7 0 S 17 7 0

5

2(5)2  1 7 0 S 49 7 0

?

true ✓

?

true ✓

7

14

84.

t  35

2 3

85.

5

2

5

8 38

87.

4

5 9

6. 36

3

5

88.



2 5

 

1 2 5 4

8 9



4 15

 

1

 14

 

7.

4

9

1 2  36149 2 a 4 9 36 16 ? 4 9 36 4 4 9 9



1

1

5

3

3

2

Solving Equations by Using Multiplication and Division

Page 138

Check for Understanding

12

8

9 4

12

1 2

8. 3.15  1.5y 3.15 1.5

8 15  9 4 120 36 10 3 1 33

3-3

4 k 5 5 4 k 4 5

5 8 9 40 10 k  36 or 9 1 k  19 4 8 Check: k9 5 4 1 ? 8 1 9 5 9 8 8 9✓ 9 1 The solution is 19.

 1.5y 1.5

2.1  y Check:

3.15  1.5y ? 3.15  1.5(2.1) 3.15  3.15 ✓ The solution is 2.1. 9.

1314 2p  212 1134 2p  52 4 5 4 13 p  13 1 2 2 13 1 4 2 20

10

p  26 or 13

1. Sample answer: 4x  12 2. Dividing each side of an equation by a number is the same as multiplying each side of the equation by the number’s reciprocal. 3. Juanita; to find an equivalent equation with 1n on one side of the equation, you must divide 1 each side by 8 or multiply each side by 8.

Chapter 3

5

The solution is 16.

1

89.

a 36 a 36

Check:

6

5 2

5

a  16

 10  9  10

 12 1 2

t  7 35 ?  7

5  5 ✓ The solution is 35.

7 0.3  0.1 5 35 1 5 70 3 10.5 4 5 0.22 1.50.3 30   30 30 30 0

1

86.

17t 2  7(5)

1

83.

6 .5  2 .8 52 0 13 0 18.2 0 7.12 2.5 17.8 00   17 5 30 25 50 50 0

 5

Check:

The solution set for 2y2  1 0 is {1, 3, 5}. 82.

t 7

5.

The solution set for 3x  2 2 is {1, 2}. 81. Replace y in 2y2  1 0 with each value in the replacement set. y

84 2

2(42)  84 84  84 ✓ The solution is 42.

false

?



g  42 Check: 2g  84

True or False?

?

0

2g 2

Check:

1314 2p  212 ? 1314 211013 2  212 1

1

22  22 ✓

The solution is

74

10 . 13

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16. 1634  86s

120. Five times a number is 1 10. 123 123 14243 { 23 5 n  120  5n  120 5n 5



1634 86

17.

5

12

7

 24

The discharge the the of a river is width times depth times 144424443 { 123 123 123 123

Equation: 3198  533  d 3198  533 d (0.6) 3198  533(0.6)d 3198  319.8 d 3198 319.8





11

v

1 v2

v  225 v

5  45

Check:

225 ?  5



319.8d 319.8

45

45  45 ✓ The solution is 225. 19.

Practice and Apply

2 n 3 3 2 n 2 3

12

 14 3

 2 (14)

n  21 2 n 3

Check: 2 3

55

 14

1212  14 ?

14  14 ✓ The solution is 21.

r  11 5r  55 Check: ?

20.

5(11)  55 55  55 ✓ The solution is 11. 14. 8d  48

2 g 5 5 2 g 2 5

12

 14 5

g  35

2

2 g 5

2 5

d6 Check:

1

 2 14

Check:

48 8

 14

1352  14 ?

14  14 ✓ The solution is 35.

8d  48 ? 8(6)  48 48  48 ✓ The solution is 6. 15. 910  26a 

11

5 5  5(45)

0.6

 5

910 26

b  7 77 ?  7

5  45

18.

the speed. 123

13. 5r  55



1 2  7(11)

11  11 ✓ The solution is 77.

10  d The Mississippi River is 10 m deep at this location.

8d 8

 11

Check:

n  60 The solution is 60. 12. Words: The discharge of a river is the width times the depth times the speed. Variable: Let d  the depth of the river.

5r 5

b 7 b 7

b  77

5

 2 (24)

Pages 138–140

86s 86

19  s Check: 1634  86s ? 1634  86(19) 1634  1634 ✓ The solution is 19.

120 5

n  24 The solution is 24. equals 24. 11. Two fifths of a number 1 1442443 { 14243 4243 123 2  n 24  2 n 5 5 2 n 2 5



21. 24

g 24 g 24

5

 12

1 2  241125 2 g  10

26a 26

Check:

35  a Check: 910  26a ? 910  26(35) 910  910 ✓ The solution is 35.

5 g  12 24 10 ? 5  12 24 5 5  12 12



The solution is 10.

75

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22. 45

z 45 z 45

2

5

28.

1 2  45125 2 z  18

z 2 5 45 18 ? 2 5 45 2 2 5 5

Check:

33

x  22 or



1

Check:

11.78 1.9

1

1

2

?

29. 5h  33 2

5h 5

6.272 0.49

2

11 . 15

1

30. 3p  45 1

3p 3

45



3 21 1  5 3 21 7 or 5 15 2 15

p p p

97.41 5.73

Check:

1

3p  45

1 22

?

1

3 15  45 1

1

45  45 ✓ 2

The solution is 15.

1235 2t  22 1135 2t  22 5 13 5 13 1  5 2 t  13(22)

31. First, solve 4m  10. 4m  10 4m 4



m m

110  13 6  813 3 25

10 4 5 2 1 22 1

Now, replace m in 12m with 22.

1 12 5  12 1 2 2

1

2t  22 ? 1235 218136 2  22

12m  12 22

 30 The value is 30.

22  22 ✓ 6

The solution is 813.

Chapter 3

?

The solution is

q  17 5.73q  97.41 Check: ? 5.73(17)  97.41 97.41  97.41 ✓ The solution is 17.

Check:

11115 2  323 2

2.8(3.5)  9.8 9.8  9.8 ✓ The solution is 3.5. 26. 5.73q  97.41

t

2

5h  33

33  33 ✓

9.8

t

1 2

5

?

27.

11 15

Check:

 2.8





h

m  3.5 2.8m  9.8 Check:

5.73q 5.73

33

5 11 1 h   3 5

k  12.8 Check: 0.49k  6.272 ? 0.49(12.8)  6.272 6.272  6.272 ✓ The solution is 12.8. 25. 2.8m  9.8 2.8m 2.8

1

The solution is 12.

1.9(6.2)  11.78 11.78  11.78 ✓ The solution is 6.2. 24. 0.49k  6.272 

1323 2x  512 ? 1323 21112 2  512

52  52 ✓

f  6.2 Check: 1.9f  11.78

0.49k 0.49

3 2

x  12



The solution is 18. 23. 1.9f  11.78 1.9f 1.9

1323 2x  512 1113 2x  112 3 11 3 11 x  11 1  2 2 11 1 3 2

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32. First, solve 15b  55. 15b  55 15b 15



b b

38.

12

1 2 11  31 3 2 2 33

39.

1 p 7

1

41. Replace  in /  7p with 65 and solve. 1

/  7p 1

65  7p 7(65)  7

50 



3



n

12

126t 126

3

8n  12

n  32 The solution is 32. Two and one half times a number equals one and one fifth.

14444244443 123 1 44244 3 14243 14444244443 1

1212 2n  115 12

6

5

n



1

15

132t 132

12

The solution is



60.5 126



60.5 132

t  0.46 It takes about 0.46 second. time for a time for a 45. two-seam fastball minus four-seam fastball 14444244443 1 424 3 14444244443 0.48  0.46 0.48  0.46  0.02 The difference is about 0.02 second.

2 6 5 12 25

5

n



d 

t  0.48 It takes about 0.48 second. 44. Replace r with 132 and d with 60.5 in rt  d and solve. rt  d 132t  60.5

8

3 8 n  3(12)

22



15.9  d The diameter was about 16 feet. 43. Replace r with 126 and d with 60.5 in rt  d and solve. rt  d 126t  60.5

Negative three eighths times a number equals 12. 1444442444443 123 1 44244 3 14243 123 8

117 2p

455  p There are 455 people in the group. 42. Replace C in C  d with 50 and solve. C  d 50  d

n  60 The solution is 60.

5 n 2 2 5 n 5 2

1

 7(350)

 50 There are 50 left-handed people.

12

37.

The number of people in left-handed people is one seventh of the world. { 1442443 { 14 42443 1   p /

1

117 9

1 32

3

 4(4.82)

40. Replace p in 7p with 350 and solve.

5 1 n  12 5 5 1 5 n  1(12) 1 5

8

 4.82

7

n  13 The solution is 13. 35. One fifth of a number is 12. 14243 { 1 44244 3 { 123 1 n  12 

36.

 4.82

1

84 7



4.82

The equation is /  7p.

n  12 The solution is 12. Negative nine times a number is 117. 34. 1 44424443 123 1 44244 3 { 123 n 9  117  9n  117 9n 9



14444244443

 11 The value is 11. Seven equals 84. 33. 1 times a number 1 424 3 123 1 44244 3 4243 123 7 n 84   7n  84 

n

n  3.615 The solution is 3.615.

Now, replace b in 3b with 33.

7n 7

13

1 13 n 4 n 3 3 4 n 4 3

2



1

1 2

55 15 11 3 2 33

3b  3

One and one third times a number is 4.82. 14444244443 123 1 44244 3 { 14243

12 . 25

77

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46. eight 123

times 123

The side length of each square is 4 inches. Area is side length squared. 123 { 144424443 A  42 2 A4 A  16 The area of the square is 16 in2. 53. A; 4t  20

the number of grams of hydrogen 14444244443

8  x The expression for oxygen is 8x. 47.

number of grams number of grams Number of grams of oxygen of water. plus equals of hydrogen 14444244443 123 1 424 3 

x

14444244443

8x



14444244443

477

The equation is x  8x  477. 48. Solve x  8x  477 x  8x  477 (1  8)x  477 9x  477 9x 9



4t 2

2t  10

Page 140

477 9



y

11 6 5 16 5

Now, replace y in 18y  21 with 16.

1 52 11  18 1 6 2  21

18y  21  18 16  21  33  21  12 The value is 12. 51. You can use the distance formula and the speed of light to find the time it takes light from the stars to reach Earth. Answers should include the following. • Solve the equation by dividing each side of the equation by 5,870,000,000,000. The answer is 53 years. • The equation 5,870,000,000,000t  821,800,000,000,000 describes the situation for the star in the Big Dipper farthest from Earth. 52. C; 2 times 2 times Perimeter is length plus width. 1442443 { 14243 123 14243 48  48  2(5x)  2x 48  10x  2x 48  (10  2)x 48  12x 48 12



2(5x)



Ten 1 23 times 123 10 

a4 number a 1 42443 a

to43 { 5 142  5

times 1 23 

The equation is 10a  5(b  c). 58. (5)(12)  60 59. (2.93)(0.003)  0.00879 60. (4)(0)(2)(3)  (0)(2)(3)  (0)(3)  (0) 61. 4 3 2 1

0

1

2

3

4

62. 654321 0 1 2 3 4 5 6 7 8 9 10

2x

63. 7 6 5 4 3 2 1

0

1

64. 7 6 5 4 3 2 1

0

1

65. Commutative Property of Addition 66. Associative Property of Multiplication 67. 2  8  9  16  9  25 68. 24  3  8  8  8 0

12x 12

4x

Chapter 3

Maintain Your Skills

m  14  81 m  14  14  81  14 m  67 Check: m  14  81 ? 67  14  81 81  81 ✓ The solution is 67. 55. d  27  14 d  27  27  14  27 d  13 Check: d  27  14 ? 13  27  14 14  14 ✓ The solution is 13. 56. 17  (w)  55 17  w  55 17  w  17  55  17 w  72 Check: 17  1w2  55 ? 17  31722 4  55 55  55 ✓ The solution is 72. 57. is equal

54.

x  53 There are 53 grams of hydrogen. 49. Replace x in 8x with 53. 8x  8(53)  424 There are 424 grams of oxygen. 50. First, solve 6y  7  4. 6y  7  4 6y  7  7  4  7 6y  11 6y 6

20

 2

78

the sum of b2 and c. 14 443 (b  c)

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69.

3 (17 8

3

 7)  8 (24) 

7.

72 8

9 70.

15  9 26  12

  

Page 140 1.

Surface area 14 243 S

2.

3.

4.

5.

6.

6 26  12 6 38 3 19

12

 18 3

 2 (18)

p  27 2 p 3

Check:

 18

? 2 (27)  3

18 18  18 ✓ The solution is 27. 8. 17y  391

Practice Quiz 1 equals 123 

2 p 3 3 2 p 2 3

four 1 23 4

times 123 



{



times 123 

17y 17

the square of the radius. 1 4424 43

391

 17

y  23 17y  391 Check:

r2

?

The formula for the surface area of a sphere is S  4r2. Replace r in 4r 2 with 7. 4r2  4 (7)2  4 (49)  4(49)  196  615.8 The surface area is about 615.8 cm2. d  18  27 d  18  18  27  18 d  45 Check: d  18  27 ? 45  18  27 27  27 ✓ The solution is 45. m  77  61 m  77  77  61  77 m  16 Check: m  77  61 ? 16  77  61 61  61 ✓ The solution is 16. 12  a  36 12  a  12  36  12 a  24 Check: 12  a  36 ? 12  1242  36 36  36 ✓ The solution is 24. t  (16)  9 t  16  9 t  16  16  9  16 t  7 Check: t  (16)  9 ? 7  (16)  9 99✓ The solution is 7.

17(23)  391 391  391 ✓ The solution is 23. 9. 5x  45 5x 5



45 5

x  9 Check:

5x  45 ?

5 (9)  45 45  45 ✓ The solution is 9. 2

5 d  10

10. 5

1 22

5

2 5 d  2 (10) d  25 2

5 d  10

Check:

?

2

5 (25)  10 10  10 ✓ The solution is 25.

Page 141 1. Step 1

Algebra Activity Model the equation.

x 1 1 1

x

 1 1 1

1 1 1 1 1 1

2x  3  –9

Place 2 x tiles and 3 negative 1 tiles on one side of the mat. Place 9 negative 1 tiles on the other side of the mat.

79

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Step 2

Place 3 x tiles and 5 positive 1 tiles on one side of the mat. Place 14 positive 1 tiles on the other side of the mat. Step 2 Isolate the x term.

Isolate the x term.

x 1 1 1

x



1 1 1 1

1

1 1 1

x

1

1

1

1

1 1 1

x

1

1

1

1

x

1

1

1

1

1

1

1

1

1

1

2x  3  3  –9  3

Since there are 3 negative 1 tiles with the x tiles, add 3 positive 1 tiles to each side to form zero pairs. Step 3 Remove zero pairs.

x

1 1 1

x

1 1 1



1

1

1

1

1

1

1 1 1

1 1 1

1 1

1 1

3x  5  5  14  5

Since there are 5 positive 1 tiles with the x tiles, add 5 negative 1 tiles to each side to form zero pairs. Step 3 Remove zero pairs.

1

1

1

1

1

1

1

1

1

1

2x  –6



1 1 1

x



1 1 1

x

 1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1 1

1 1

1 1

1 1



1 1

1 1

1 1

1 1

1 1

x



1

1

1

x



1

1

1

x



1

1

1

9 3x 3 3

Separate the tiles into 3 equal groups to match the 3 x tiles. Each x tile is paired with 3 positive 1 tiles. Thus, x  3.

3x  5  14

Chapter 3

1

Group the tiles to form zero pairs and remove the zero pairs. Step 4 Group the tiles.

x 1

1

3x  9

Separate the tiles into 2 equal groups to match the 2 x tiles. Each x tile is paired with 3 negative 1 tiles. Thus, x  3. 2. Step 1 Model the equation.

1

1

x

1 1

–6 2x  2 2

x

x

x

Group the tiles to form zero pairs and remove the zero pairs. Step 4 Group the tiles.

x



1

80

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3. Step 1

Separate the tiles into 3 equal groups to match the 3 x tiles. Each x tile is paired with 4 positive 1 tiles. Thus, x  4. 4. Step 1 Model the equation.

Model the equation.

x x x

 1 1

1

1

1

1

1

1

1

1

1

x x 1 1 1

1

1 1 1



1 1 3x  2  10

Place 3 x tiles and 2 negative 1 tiles on one side of the mat. Place 10 positive 1 tiles on the other side of the mat. Step 2 Isolate the x term.

x

1

1

1

1

1

1

1

1

1

1

1

1

1



1 1

1

Since there are 2 negative 1 tiles with the x tiles, add 2 positive 1 tiles to each side to form zero pairs. Step 3 Remove zero pairs.

x

x

1

1

1

x

1

1

1



1

1

1



1

1

1

1

1 1

1 1

1 1

1 1

8  4  2x  4  4

Since there are 4 positive 1 tiles with the x tiles, add 4 negative 1 tiles to each side to form zero pairs. Step 3 Remove zero pairs.

x

1

x

1 1 1

3x  2  2  10  2

1

1

1 1 1 1

1

1

Place 2 x tiles and 4 positive 1 tiles on one side of the mat. Place 8 negative 1 tiles on the other side of the mat. Step 2 Isolate the x term.

1

1

1

8  2x  4

x x

1

x

1 1 1

1 1

x

1 1 1

1

1

1

1

1

1 1

1

1

1 1

1

1

1 1

3x  12

Group the tiles to form zero pairs and remove the zero pairs. Step 4 Group the tiles.



12  2x

Group the tiles to form zero pairs and remove the zero pairs.

x



1

1

1

1

x



1

1

1

1

x



1

1

1

1

3x  12 3 3

81

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Step 4

Step 3

Group the tiles.

1 1 1 1 1 1

1 1 1 1 1 1



Remove zero pairs.

x x x x

x



x

1

1

1 1



1

1

1

1

1 1

1

1

1

1

1

1

1

x x x x

1

1

Group the tiles to form zero pairs and remove the zero pairs. Step 4 Group the tiles.

Separate the tiles into 2 equal groups to match the 2 x tiles. Each x tile is paired with 6 negative 1 tiles. Thus, x  6. 5. Step 1 Model the equation.

1

1

4x  8

–12  22x 2

x

1

 1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

x

1

1

x

1

1



x

8 4x 4 4

3  4x  11

Separate the tiles into 4 equal groups to match the 4 x tiles. Each x tile is paired with 2 positive 1 tiles. Thus, x  2. 6. Step 1 Model the equation.

Place 4 x tiles and 3 positive 1 tiles on one side of the mat. Place 11 positive 1 tiles on the other side of the mat. Step 2 Isolate the x term.

x x x x x 1

1

 1

1 1 1

x

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1



1

1

1 1 1 2x  7  1

3  4x  3  11  3

Place 2 x tiles and 7 positive 1 tiles on one side of the mat. Place 1 positive 1 tile on the other side of the mat. Step 2 Isolate the x term.

Since there are 3 positive 1 tiles with the x tiles, add 3 negative 1 tiles to each side to form zero pairs.

x x 1

1

1

1

1

1

1



1

1 1 1 1

1 1 1 1

1 1 1

1 1 1

2x  7  7  1  7

Chapter 3

82

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Step 2

Since there are 7 positive 1 tiles with the x tiles, add 7 negative 1 tiles to each side to form zero pairs. Step 3 Remove zero pairs.

x

1 1 1

x

1 1 1

1 1

1 1

1 1

1 1

1 1

1 1

1

1

1

1

1

1

1

1

1



x x x x 1 1 1 1 1 1 1



1

1

1

1

1

1

Group the tiles to form zero pairs and remove the zero pairs. Step 4 Group the tiles.

1

1

1

1

1

1

1

1

1

x

x

1

1

1

x

x

1

1

1



1 1 1 1

1

1

1

1

1

1

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1 1

x

1

Since there are 7 negative 1 tiles with the x tiles, add 7 positive 1 tiles to each side to form zero pairs. Step 3 Remove zero pairs.

2x  6



1

9  7  4x  7  7

1 1

1 1

x

Isolate the x term.

16  4x

Group the tiles to form zero pairs and remove the zero pairs. Step 4 Group the tiles.

2x  –6 2 2

Separate the tiles into 2 equal groups to match the 2 x tiles. Each x tile is paired with 3 negative 1 tiles. Thus, x  3. 7. Step 1 Model the equation.

x

1

1

1

1

1

1

1

1

1

1

1

1

x

1

1

1

1

x



x

x x 1

1

1

1

1

1

1

1

1

x x



1 1 1

16  44x 4

1 1 1

Separate the tiles into 4 equal groups to match the 4 x tiles. Each x tile is paired with 4 positive 1 tiles. Thus, x  4. 8. Step 1 Model the equation.

1 9  4x  7

Place 4 x tiles and 7 negative 1 tiles on one side of the mat. Place 9 positive 1 tiles on the other side of the mat.

x x 1 1 1

x 1

1

1

1

1

1



1 1 1 1 1

1 7  3x  8

83

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Place 3 x tiles and 7 positive 1 tiles on one side of the mat. Place 8 negative 1 tiles on the other side of the mat. Step 2 Isolate the x term.

x x x

Page 145

1 1 1

1

1

1

1

1



1

1 1

1 1 1 1

1 1 1 1

1 1 1

1 1 1

7  3x  7  8  7

Since there are 7 positive 1 tiles with the x tiles, add 7 negative 1 tiles to each side to form zero pairs. Step 3 Remove zero pairs.

x x x

1 1 1 1 1 1

1 1

1 1

1 1

1 1

1 1

1 1



Statement The result is 55. The product is added to 13. A number is multiplied by seven.

1 1

1 1 1 1 1 1 1

1 1

3x  15

x

Statement There are 2,187,000 bacteria on the seventh day. A bacteria population triples in number each day.

1 1 1 1 1



1 1 1

7.

1 1 1 1 1

x

42  7  6

Undo the Statement 2,187,000

2,187,000  36  3000

There were 3000 bacteria. 4g  2  6 4g  2  2  6  2 4g  4 4g 4

1 1



4 4

g  1 4g  2  6 ? 4(1)  2  6 ? 4  2  6 6  6 ✓ The solution is 1. 8. 18  5p  3 18  3  5p  3  3 15  5p Check:

3x  –15 3 3

Separate the tiles into 3 equal groups to match the 3 x tiles. Each x tile is paired with 5 negative 1 tiles. Thus, x  5. 9. First add 12 to each side, and then divide each side by 7.

15 5



5p 5

3p

Chapter 3

Undo the Statement 55 55  13  42

The number is 6. 6. Start at the end of the problem and undo each step.

Group the tiles to form zero pairs and remove the zero pairs. Step 4 Group the tiles.

x

Check for Understanding

1. Sample answers: 2x  3  1, 3x  1  7 2. (1) Add 4 to each side. (2) Multiply each side by 5. (3) Subtract 3 from each side. 3. Odd integers are two units apart. If we want to signify the odd integer before odd integer n, we need to subtract 2 units. Therefore, the expression can be written as n 2. 4a. Subtract 3 from each side. 4b. Simplify. 4c. Multiply each side by 5. 4d. Simplify. 4e. Subtract 4 from each side. 4f. Simplify. 4g. Divide each side by 2. 4h. Simplify. 5.Start at the end of the problem and undo each step.

1 1 1

1

Solving Multi-Step Equations

3-4

84

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18  5p  3 ? 18  5(3)  3 ? 18  15  3 ? 18  18 ✓ The solution is 3.

13. Twelve a number equals 34. 123 decreased 144244by 3 twice 1442443 123 1 23 2n  34 12  12  2n  34 12  2n  12  34  12 2n  46

Check:

3 a 2

9. 3 a 2

2n 2

 8  11

12

 19 2

 3(19)

a

38 3 2

a  123 3 a 2 3 2 123 2

Check:

n  (n  1)  (n  2) n  (n  1)  (n  2)  42 3n  3  42 3n  3  3  42  3 3n  39

 8  11

1 2  8  11 ? ?

19  8  11 11  11 ✓

3n 3

2

The solution is 123. 10. 2

1

b  4 2 b  4 2

2  2(17) b  4  2 30  4 ?  2 34 ?  2

17

26  2  2n 26  2  2  2n  2 24  2n

17 17

24 2

2n 2

Pages 145–148

5.6

Practice and Apply

16. Start at the end of the problem and undo each step.

n  28 0.2n  3  8.6 ? 0.2(28)  3  8.6 ? 5.6  3  8.6 8.6  8.6 ✓ The solution is 28. 12. 3.1y  1.5  5.32 3.1y  1.5  1.5  5.32  1.5 3.1y  6.82

Statement The result is 25. The quotient is added to 17. A number is divided by 4.

Check:





12  n The Hawaiian alphabet has 12 letters.

 0.2

3.1y 3.1

39 3

The English alphabet 123 equals 2 more than twice the Hawaiian alphabet. 14444244443 { 14243 1444442444443 26  2 + 2n

17  17 ✓ The solution is 30. 0.2n  3  8.6 11. 0.2n  3  3  8.6  3 0.2n  5.6 0.2n 0.2



 42

n  13 n  1  13  1 or 14 n  2  13  2 or 15 The consecutive integers are 13, 14, and 15. 15. Let n  the number of letters in the Hawaiian alphabet.

 17

b  4  34 b  4  4  34  4 b  30 Check:

46 2

n  23 The solution is 23. 14. Let n  the least integer. Then n  1  the next greater integer, and n  2  the greatest of the three integers. The sum of three consecutive integers is 42. { { 1444444442444444443

 8  8  11  8 3 a 2 2 3 a 3 2



Undo the Statement 25 25  17  8 8  4  32

The number is 32. 17. Start at the end of the problem and undo each step.

6.82 3.1

Statement The result is 75. The difference is multiplied by 5. Nine is subtracted from a number.

y  2.2 3.1y  1.5  5.32 Check: ? 3.1(2.2)  1.5  5.32 ? 6.82  1.5  5.32 5.32  5.32 ✓ The solution is 2.2.

Undo the Statement 75 75  5  15 15  9  24

The number is 24.

85

Chapter 3

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18. Start at the end of the problem and undo each step.

21. Start at the end of the problem and undo each step. Statement He had $13.45 left. He bought lunch for $6.55. He spent half for a haircut. He spent one third for gasoline.

Statement Undo the Statement The fourth question is worth $6000. $6000 The fourth question is worth twice as much $6000  2  $3000 as the third question. The third question is worth twice as much $3000  2  $1500 as the second question. The second question is worth twice as much $1500  2  $750 as the first question.

1 2

The sculpture lost its weight the 8th hour.

2

25. 24

1

5 4

22

1

The sculpture lost 2 its weight the 5th hour.

5 2

5

1

10  2  20

The sculpture lost 2 its weight the 3rd hour. 1 2

The sculpture lost its weight the 2nd hour. 1

The sculpture lost 2 its weight the 1st hour.



10 5

3c 3



18 3

c  6

4a 4



7g 7

7  g c 3

26. c 3

57 c 3 c 3

y 5

27. y 5

5575 3

96

9969

2

1 2  3(2)

5

c  6

3

28.

20  2  40

3

a 7

40  2  80 7

a 7

 2

9 

p 4

30. t 8

Statement Undo the Statement She went into the building. 0 She climbed the 044 remaining 4 rungs. 4  7  11 She went up 7 rungs. She backed down 11  5  6 5 rungs. 639 She moved up 3 rungs. She stood on the 2(9)  1  19 middle rung.

 6  12

8

4

1

p  56

1 2  8(6)  10

17  s  40 17  s  17  40  17 s  57 s 1



57 1

s  57

The ladder has 19 rungs.

86

m 5

31.

t 8

2  4(10)

959

1 p2

t  48

32.

5

4 4  4(14)

m 5

 6  31

 6  6  31  6

 6

17  s 4 17  s 4

p 4

p

 6  6  12  6 t 8

1 2  5(3)

4  14

 5

1 2  7(5)

t 8

 3

y  15

 3  2  3 a 7 a 7

y 5 y 5

9 

29.

a  35

The sculpture weighed 80 lb. 20. Start at the end of the problem and undo each step. Count the rungs from the top.

Chapter 3



49 7

25

5  2  10

The sculpture lost its weight the 4th hour.

3

$40  2  $60

5a 63  7g  14 63  14  7g  14  14 49  7g

5

1 2

$20  2  $40

n  2 15  4a  5 15  5  4a  5  5 20  4a 20 4

5 8

5 8

The sculpture lost 2 its weight the 6th hour.

24.

5 16

1

The sculpture lost 2 its weight the 7th hour.

5n 5

Undo the Statement

5 16

$13.45  $6.55  $20

He withdrew $60. Exercises 22–39 For checks, see students’ work. 22. 23. 5n  6  4 7  3c  11 5n  6  6  4  6 7  3c  7  11  7 5n  10 3c  18

The first question is worth $750. 19. Start at the end of the problem and undo each step. Statement After 8 hours, it 5 weighs 16 of a pound.

Undo the Statement $13.45

5

m 5 m 5

 25

1 2  5(25) m  125

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33. 6

3

3j  (4) 6 3j  (4) 6

Now, replace a in 5a  2 with 5. 5a  2  5(5)  2  25  2  27 The value is 27. 41. First, solve 2x  1  5. 2x  1  5 2x  1  1  5  1 2x  4

 12

4  6 (12)

3j  (4)  72 3j  (4)  (4)  72  (4) 3j  76 3j 3



j j 34.

76 3 76 3 1 253

2x 2

3d  1.2  0.9 3d  1.2  1.2  0.9  1.2 3d  2.1 3d 3



x2 Now, replace x in 3x  4 with 2. 3x  4  3(2)  4 64 2 The value is 2. 42. Two-thirds a number minus six is negative ten. 14243 of { 14243 123 { { 1442443 2  n  6  10 3

2.1 3

d  0.7 35. 2.5r  32.7  74.1 2.5r  32.7  32.7  74.1  32.7 2.5r  106.8 2.5r 2.5



106.8 2.5

2 n 3

r  42.72 36. 0.6  (4a)  1.4 0.6  (4a)  0.6  1.4  0.6 4a  0.8 4a 4



2 n 3

p 7 p 7

 1.8

1 2  7(1.8)

p  12.6 38. 3.5x  5  1.5x  8 (3.5  1.5)x  5  8 2x  5  8 2x  5  5  8  5 2x  3 2x 2

16 4

3

2 1

9z  4 8 5 9z  4 88 5 9z  4 5 9z  4 5 5

1

 5.4  8  13.4

3n 3

9z  4  67 9z  4  4  67  4 9z  63 

4n 4

n  (n  2)  (n  4) n  (n  2)  (n  4)  51 3n  6  51 3n  6  6  51  6 3n  45

 5.4

2  5(13.4)

9z 9



4n The solution is 4. 44. Let n  the least odd integer. Then n  2  the next greater odd integer, and n  4  the greatest of the three odd integers. The sum of three consecutive odd integers { is 51. { 144444444424444444443

x  12 or 1.5 39.

 4

n  6 The solution is 6. 43. Twenty-nine is thirteen added to four times a number. 1442443 { 14243 14243 123 123 14243 29  13  4  n 29  13  4n 29  13  13  4n  13 16  4n

 0.5  0.5  1.3  0.5 7

 6  6  10  6

1 2  32(4)

0.8 4

 0.5  1.3 p 7 p 7

 6  10

2 n 3 3 2 n 2 3

a  0.2 37.

4

2



 51

45 3

n  15 n  2  15  2 or 17 n  4  15  4 or 19 The consecutive odd integers are 15, 17, and 19.

63 9

z7 40. First, solve 3a  9  6. 3a  9  6 3a  9  9  6  9 3a  15 3a 3



15 3

a5

87

Chapter 3

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49. Let x  the length of the shortest side. Then x  2  the length of the next longer side, and x  4  the length of the longest side. The sum of the three sides is 54. 1444442444443

45. Let n  the least even integer. Then n  2  the next greater even integer, and n  4  the greatest of the three even integers. The sum of three consecutive even integers is 30. 1444444442444444443 { 123  30 n  (n  2)  (n  4) n  (n  2)  (n  4)  30 3n  6  30 3n  6  6  30  6 3n  36 3n 3



{ {

x  (x  2)  (x  4)  54 x  (x  2)  (x  4)  54 3x  6  54 3x  6  6  54  6 3x  48

36 3

3x 3

9

50.

92 7 2000(7) 

{ {

n  (n  1)  (n  2)  (n  3)  94 n  (n  1)  (n  2)  (n  3)  94 4n  6  94 4n  6  6  94  6 4n  88 

88 4

n  1  22  1 or 23 n  3  22  3 or 25 The consecutive integers are 22, 23, 24, and 25. 47. Let n  the least odd integer. Then n  2  the next greater odd integer, n  4  the next greater odd integer, and n  6  the greatest of the four odd integers. The sum of four consecutive odd integers is 8. 1444444442444444443

20 2

{ {

4 4

0.10

number times of4miles 123 1 243  x

is

{







 2000

1500 0.02

The area the area of of the square minus the rectangle is 1442443 123 1442443 { 4/ 10  10  

cost for one day. 14243 60

4 of 5 { { 4  5

4

10  10  4/  5 (10  10) 4

100  4/  5(100) 100  4/  80 100  4/  100  80  100 4/  20

45.05 0.1

4/ 4

x  450.5 Ms. Jones can drive 450.5 mi.

Chapter 3

2/ 2

x  75,000 Mr. Goetz must have $75,000 in sales. 54. Let   the length of the rectangle.

14.95  0.1x  60 14.95  0.1x  14.95  60  14.95 0.1x  45.05 0.1x 0.1



0.02x 0.02

n  2  1  2 or 1 n  6  1  6 or 5 The consecutive odd integers are 1, 1, 3, and 5. 48. Let x  the number of miles driven in one day. cost per mile 14 243

2

500  0.02  x 500  0.02x  2000 500  0.02x  500  2000  500 0.02x  1500

n  1 n  4  1  4 or 3

Price of car per day plus 14243 1 23 14.95 

1

2

10  / The person’s foot is about 10 in. long. 52. See students’ work. 53. Let x  the amount of sales. The monthly the amount salary plus of of sales is $2000. 1442443 1 23 2% { { 1442443 { 123

 8 n  (n  2)  (n  4)  (n  6) n  (n  2)  (n  4)  (n  6)  8 4n  12  8 4n  12  12  8  12 4n  4 

a  7000 2 2000 a  7000 2 2000 a  7000 2000 a  7000 2000 2000

14,000  a  7000 14,000  7000  a  7000  7000 21,000  a They can climb to about 21,000 ft. 51. 8  2/  12 8  12  2/  12  12 20  2/

n  22 n  2  22  2 or 24

4n 4

48 3

x  16 x  2  16  2 or 18 x  4  16  4 or 20 The lengths of the sides are 16 cm, 18 cm, and 20 cm.

n  12 n  2  12  2 or 10 n  4  12  4 or 8 The consecutive even integers are 12, 10, and 8. 46. Let n  the least integer. Then n  1  the next greater integer, n  2  the next greater integer, and n  3  the greatest of the four integers. The sum of four consecutive integers is 94. 1444444442444444443

4n 4





20 4

/5 The length of the rectangle is 5 in.

88

the area of the square. 1442443 (10  10)

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55. Never; Let n, n  2 be the even numbers, and m, m  2 be the odd numbers. The sum of two the sum of two consecutive consecutive even numbers equals odd numbers. 1442443 123 1442443

Step 2

n  (n  2)  m  (m  2) n  (n  2)  m  (m  2) 2n  2  2m  2 2n  2  2  2m  2  2 2n  2m 2n 2



58. D; 3 a 5

3



n



Step 2

22

Step 2 Step 3

 3

1 2  53 (3)

a  5 0  11y  33 Press

Step 3 Enter (

and choose 0, for solve.

7t 7

0  18  Step 2

Press

Press

and choose 0, for solve. ALPHA R ⫼ 0.8 )

Press

and choose 0, for solve.

h 7 h 7 h 7

Maintain Your Skills 91

 7

?

66. 15

4. Step 4 Press ALPHA [SOLVE] to reveal the solution. W  18. 6  6  12 

 0.7

7(13)  91 91  91 ✓ The solution is 13.

and choose 0, for solve. ALPHA W  2 ) ⫼

6  12 

0

t  13 7t  91 Check:

5

61. Step 1

 0.7

r  0.8 6

65. 7t  91

Enter 11 ALPHA Y  33. Step 4 Press ALPHA [SOLVE] to reveal the solution. Y  3. w  2 40 5 60. Step 1 Press

0.7  0.7 

r  0.8 6 r  0.8 6

Page 148

Step 3

Step 2

6.

Enter 7.2 ALPHA T 33.84. ALPHA Step 4 Press [SOLVE] to reveal the solution. T  4.7.

5 3 a 3 5

Step 2

⫼ ( ) 2

( ) 5

0.7. Step 4 Press ALPHA [SOLVE] to reveal the solution. R  5. 4.91  7.2t  38.75 64. Step 1 4.91  7.2t  38.75  38.75  38.75 7.2t  33.84  0

 19  19  16  19

59. Step 1

(

6

 19  16 3 a 5

and choose 0, for solve.

Step 3 Enter (

15  3n  22 3 a 5

0

ALPHA P

0.7 

63. Step 1

2



66

Step 4 Press ALPHA [SOLVE] to reveal the solution. P  17.

2m 2

• The alligator is about 93 or 10 years old. 57. B; 1 Fifteen minus three times a number equals negative twenty-two. 4243 123 123 123 14243 123 14444244443 15

6

Press

Step 3 Enter ( ) )

nm Thus, n must equal m, so n is not even or m is not odd. 56. By using the length at birth, the amount of growth each year, and the current length, you can write and solve an equation to find the age of the animal. Answers should include the following. • To solve the equation, subtract 8 from each side and then divide each side by 12.

or

p  (5) 2 p  (5) 6 2 p  (5) 6 2

62. Step 1

r 15 r 15

 8

1 2  15(8) r  120

Check:

r  15 120 ?  15

8 8

8  8 ✓ The solution is 120.

6

and choose 0, for solve.

ALPHA H ⫼ ( ) Step 3 Enter ( ) 18 7. Step 4 Press ALPHA [SOLVE] to reveal the solution. H  126.

89

Chapter 3

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2

1

3b  1 2

67.

2 3b 3 2 2 3 b

75.

3  2 3 3  2 2 9 b4 1 b  24 2 1 Check: 3b  1 2 2 1 ? 1 3 2 4  1 2 1 1 1 2  1 2 1 The solution is 2 4.

1 2

3

76.



Equation: m  18  47 The equation is m  18  47. 69. Solve m  18  47. m  18  47 m  18  18  47  18 m  29 There were 29 models in 1990. 70. There are two numbers divisible by 3, and there are 8  2 or 6 numbers not divisible by 3.

1 12 1 12

The odds of spinning a number divisible by 3 are 1:3. 71. There are four numbers equal to or greater than 5, and there are 8  4 or 4 numbers less than 5. 

 16(1)  16  16  4  20

1

1

2

 18(2)  18

114 2 119 2

n

82. the quantity 3 plus b 14243 divided by {y 14444244443 (3  b)  y Thus, the algebraic expression is (3  b)  y. 83. { 3 times a 1 plus square b 123 { 23 the 144 4244of 43

6

odds of less than 7  2

84.

3

1 86.

The odds of spinning a number less than 7 are 3:1.

88.

1

73. 7  3  7  3 6

 21

89.

2 7

2

38 1

 38 2

 24 1

 12

Chapter 3

2

Thus, the algebraic expression is 5m  2 .

4 4 1 1

The odds of spinning a number equal to or greater than 5 are 1:1. 72. There are six numbers less than 7, and there are 8  6 or 2 numbers not less than 7.

2

1

 36  2  38 81. the product of 5 and m, 1 plus n 14444244443 23 half 142of 43 1  5m n 2

1

3

odds of equal to or greater than 5 

1

80. 18 2 9  18 2  9

2

8

1 12 1 1  15t 1 5 2  25 1 5 2

79. 16 1 4  16 1  4

odds of divisible by 3  6

74.

 (15t  25)  (5)

 3t  (5)  3t  5 77. 17  9  17(10  1)  17(10)  17(1)  170  17  153 78. 13(101)  13(100  1)  13(100)  13(1)  1300  13  1313

The number of the number of models in 1990 1 plus equals models in 2000. 1442443 23 18 { 123 1442443



15t  25 5

 (15t  25) 5

The number of models in 1990 plus 18 equals the number of models in 2000. Variable: Let m  the number of models in 1990.

2 3

114 2 1 1  3a 1 4 2  16 1 4 2  4a  4

68. Words:

6

 (3a  16)  4  (3a  16)

1 2

1 2

6

3a  16 4

90

3 a  b2  Thus, the algebraic expression is 3a  b2. 5d  2d  (5  2)d 85. 11m  5m  (11  5)m  3d  6m 8t  6t  (8  6)t 87. 7g  15g  (7  15)g  14t  8g 9f  6f  (9  6)f  3f 3m  (7m)  [3  (7)]m  10m

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Solving Equations with the Variable on Each Side

3-5

Pages 151–152

3 8

2g 2



13 13

1

The solution is 7.



3a 3

13z 13

13 5

10.

1

3

8

1 1 2.

9

3



5d 5

1

c  1 8 c  1 8

c

4

2  814c 2

c  1  2c c  1  c  2c  c 1c Check:

 2t  4 1

 2t  4  2t 3

 4 3

3

4

2.6  d Check: 6  3  5(d  2) ? 6  3  5(2.6  2) ? 6  3  5(0.6) ? 633 66✓ The solution is 2.6.

60

3

3

8

11 12 2  34

a3 Check: 3(a  5)  6 ? 3(3  5)  6 ? 3(2)  6 6  6 ✓ The solution is 3. 8. 7  3r  r  4(2  r) 7  3r  r  8  4r 7  3r  8  3r 7  3r  3r  8  3r  3r 7  8 Since 7  8 is a false statement, this equation has no solution. 9. 6  3  5(d  2) 6  3  5d  10 6  5d  7 6  7  5d  7  7 13  5d

 15

1

1 12

3(a  5)  6 3a  15  6 3a  15  15  6  15 3a  9

12 2

c4 Check: 20c  5  5c  65 ? 20(4)  5  5(4)  65 ? 80  5  20  65 85  85 ✓ The solution is 4.

1

 4 12

3

00✓

1z The solution is 1. 2. If both sides of the equation are always equal, the equation is an identity. 3. Sample answer: 2x  5  2x  5 4a. Subtract 6n from each side. 4b. Simplify. 4c. Add 13 to each side. 4d. Simplify. 4e. Divide each side by 2. 4f. Simplify. 5. 20c  5  5c  65 20c  5  5c  5c  65  5c 15c  5  65 15c  5  5  65  5 15c  60

3 1  4t 8 3 1 1  4t  2t 8 3 3  4t 8 3 3 3  4t  8 8 3 4t 4 3 3 4 t

1

? 1 2 ? 3 4

 4t  2t  4

3 8

g6 The solution is 6. 1b. correct 1c. Incorrect; to eliminate 6z on the left side of the equals sign, 6z must be added to each side of the equation. 6z  13  7z 6z  13  6z  7z  6z 13  13z

6.

1

Check for Understanding

1a. Incorrect; the 2 must be distributed over both g and 5. 2(g  5)  22 2g  10  22 2g  10  10  22  10 2g  12

15c 15

3 8

Check:

c  1 c 4 8 1  1 ? 1 4 8 2 ? 1 4 8 1 1 4 4



The solution is 1. 11. 5h  7  5(h  2)  3 5h  7  5h  10  3 5h  7  5h  7 Since the expressions on each side of the equation are the same, this equation is an identity. The statement 5h  7  5(h  2)  3 is true for all values of h.

3

 4  8 9

 8

2  43 198 2 3

t2

1

t  12

91

Chapter 3

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12.

5.4w  8.2  9.8w  2.8 5.4w  8.2  9.8w  9.8w  2.8  9.8w 8.2  4.4w  2.8 8.2  4.4w  8.2  2.8  8.2 4.4w  11 4.4w 4.4

Exercises 16–39 For checks, see students’ work. 16. 3  4q  10q  10 3  4q  10q  10q  10  10q 3  14q  10 3  14q  3  10  3 14q  7

11

 4.4

14q 14

w  2.5 5.4w  8.2  9.8w  2.8 Check: ? 5.4(2.5)  8.2  9.8(2.5)  2.8 ? 13.5  8.2  24.5  2.8 21.7  21.7 ✓ The solution is 2.5. 13. D; Solve by substitution. A 75  9t  5(4  2t) ? 75  9(5)  5[4  2(5)] ? 75  (45)  5[4  (10)] ? 75  45  5[14] 120  70 B 75  9t  5(4  2t) ? 75  9(4)  5[4  2(4)] ? 75  (36)  5[4  (8)] ? 75  36  5[12] 111  60 C 75  9t  5(4  2t) ? 75  9(4)  5[4  2(4)] ? 75  36  5[4  8] ? 39  5[4] 39  20 D 75  9t  5(4  2t) ? 75  9(5)  5[4  2(5)] ? 75  45  5[4  10] ? 30  5[6] 30  30 ✓

7

 14 1

q  2 or 0.5 17.

3k  5  7k  21 3k  5  7k  7k  21  7k 4k  5  21 4k  5  5  21  5 4k  16 4k 4

18.



16 4

k4 5t  9  3t  7 5t  9  3t  3t  7  3t 8t  9  7 8t  9  9  7  9 8t  16 8t 8



16 8

t2 19. 8s  9  7s  6 8s  9  7s  7s  6  7s s96 s9969 s  3 3 n 4

20. 3 n 4

1

1

1

 16  8n  2  8n  8n 7 n 8

7 n 8

1

 16  2  8n  16  2

 16  16  2  16 7 n 8 8 7 n 7 8

 14

1 2  87 1142

The answer is D.

n  16

Pages 152–154

Practice and Apply

21.

14a. Multiply each side by 10. 14b. Simplify. 14c. Distributive Property 14d. Add 4 to each side. 14e. Simplify. 14f. Divide each side by 6. 14g. Simplify. 15a. Subtract v from each side. 15b. Simplify. 15c. Subtract 9 from each side. 15d. Simplify. 15e. Divide each side by 6. 15f. Simplify.

1 4

2  3y 1 2 1  3y  3y 4 1 1  3y 4 1 1 1  3y  4 4 1 3y 1 3 3y

1

1

3

1

1

 4  3y  3y 3

4 3

1

44 1

2

2  3112 2 3

y  2

1

y  1 2 22.

8  4(3c  5) 8  12c  20 8  20  12c  20  20 12  12c 12 12



12c 12

1  c

Chapter 3

3

 4  3y

92

23.

7(m  3)  7 7m  21  7 7m  21  21  7  21 7m  28 7m 7



30.

1 y 2

28 7



1

 y  4  2y



1 y 2 1 y 2

2

4 b 5 5 4 b 4 5

2

8

1 2  54(8) b  10

1

32.

8848 2x  4

2  2(4)

x8 26. 4(2a  1)  10(a  5) 8a  4  10a  50 8a  4  10a  10a  50  10a 18a  4  50 18a  4  4  50  4 18a  54

1 (7  3g) 4 7 3  4g 4 7 3 g  4g  8 4 7 7  8g 4 7 7 7  8g  4 4 7 g 8 8 7 g 7 8

g

 8 g

 8 g

g

 8  8 0 7

04 7

 4

1 2  87 174 2 g  2

33.

54

 18

a3 27. 4(f  2)  4f 4f  8  4f 4f  8  4f  4f  4f 8  0 Since 8  0 is a false statement, this equation has no solution. 28. 3(1  d)  5  3d  2 3  3d  5  3d  2 3d  2  3d  2 Since the expressions on each side of the equation are the same, this equation is an identity. The statement 3(1  d)  5  3d  2 is true for all values of d. 29. 2(w  3)  5  3(w  1) 2w  6  5  3w  3 2w  1  3w  3 2w  1  3w  3w  3  3w w  1  3 w  1  1  3  1 w  2 w 1

2

4

2x  8  4

18a 18

2

3  5b  3  11  3

1

1

2

4

5  2x  3  4

2

2

3  5b  11

18 6

r  3

1 2x

1

3  5b  5b  11  5b  5b

1

1

1

 4  2y  2y

3  5b  11  5b

31.

25. 5  2 (x  6)  4

1 2x

1

 4  2y

04 Since 0  4 is a false statement, this equation has no solution.

m4 24. 6(r  2)  4  10 6r  12  4  10 6r  8  10 6r  8  8  10  8 6r  18 6r 6

3 y 2

1 (a  4) 6 1 2 a3 6 1 2 2 a  3  3a 6 1 2 2a  3 1 2 2 2a  3  3 1 2a 1 2 2a

1

34.

1

 3(2a  4) 2

4

2

4

 3a  3 4

3 4

2

33 2

2  2(2)

a  4 28  2.2x  11.6x  262.6 28  2.2x  11.6x  11.6x  262.6  11.6x 28  13.8x  262.6 28  13.8x  28  262.6  28 13.8x  234.6 13.8x 13.8

35.

2

 1

234.6

 13.8

x  17 1.03p  4  2.15p  8.72 1.03p  4  2.15p  2.15p  8.72  2.15p 3.18p  4  8.72 3.18p  4  4  8.72  4 3.18p  12.72 3.18p 3.18

w2

2

 3a  3  3a



12.72 3.18

p4

93

Chapter 3

36.

18  3.8t  7.36  1.9t 18  3.8t  1.9t  7.36  1.9t  1.9t 18  1.9t  7.36 18  1.9t  18  7.36  18 1.9t  10.64 1.9t 1.9

2(n  2)  3n  13 2n  4  3n  13 2n  4  3n  3n  13  3n n  4  13 n  4  4  13  4 n  17

10.64 1.9



n 1

t  5.6 37. 13.7v  6.5  2.3v  8.3 13.7v  6.5  2.3v  2.3v  8.3  2.3v 16v  6.5  8.3 16v  6.5  6.5  8.3  6.5 16v  14.8 16v 16



14.8 16

Three greatest equals twice the least 1 plus 38. 123 times 123 the 14 4244 3 123 1442443 23 1 23 3  (n  4)  2 n  38

38. 2[s  3(s  1) ]  18 2[ s  3s  3]  18 2[ 4s  3]  18 8s  6  18 8s  6  6  18  6 8s  24 

3(n  4)  2n  38 3n  12  2n  38 3n  12  2n  2n  38  2n n  12  38 n  12  12  38  12 n  26 n  2  26  2 or 28 n  4  26  4 or 30 The consecutive even integers are 26, 28, and 30. 44. 0.8(220  a)  152 176  0.8a  152 176  0.8a  176  152  176 0.8a  24

24 8

s3 39. 3(2n  5)  0.5(12n  30) 6n  15  6n  15 Since the expressions on each side of the equation are the same, this equation is an identity. The statement 3(2n  5)  0.5(12n  30) is true for all values of n. 40. One half of increased two thirds of a number by 14243 14243 1 2

n

 1 n 2

1 n 2

16 123

{

16



is

the number minus 144244 3 123 2 3

n



0.8a 0.8

{

1

1

1

1

22  1  4(x  2)

45.

4

2 2

2

1

1

1

1

1

1

22  2  4x

1

1

6n  16  4 1

1

2  4x

1

4(2)  4 4x

6n  20

2

n  120 41. The sum of one half of a44244 number and 6 14 443 1 n 2 1 n 2

equals 123 

n6 6 1

 6  2n  6 6(6) 

1 n 3 1 1 n  2n 3 1 6n 1 6 6n

1

one third of the number. 14 424 43 1 3

n

0.425x 0.425

2.4

 0.425

x  5.647 It will take about 5.6 years.

2

36  n 42. Let n  the lesser odd integer. Then n  2  the greater odd integer. Twice the greater three times the odd integer is lesser number minus 13. 144424443 123 123 123 123 2(n  2)  3 n  13

Chapter 3

11 2

8x The name is 8-penny. 46. 4.9  0.275x  2.5  0.7x 4.9  0.275x  0.7x  2.5  0.7x  0.7x 4.9  0.425x  2.5 4.9  0.425x  4.9  2.5  4.9 0.425x  2.4

6 6n  6(20)

1 2

1

22  2  2  4x  2

6n  16  16  4  16 1

1

22  1  4x  2

 16  3n  3n  4  3n

1

24

 0.8

a  30 The age is 30 years.

4.

 16  3n  4 2

17 1

n  17 n  2  17  2 or 19 The consecutive odd integers are 17 and 19. 43. Let n  the least even integer. Then n  2  the next greater even integer, and n  4  the greatest of the three even integers.

v  0.925

8s 8



94

47. 2(3x  1)  2x  4(3x) 6x  2  2x  12x 2  8x  12x 2  8x  8x  12x  8x 2  4x 2 4 1 2



54. 7

4x 4

0.5  x 3x  1  3(0.5)  1 or 2.5 3x  3(0.5) or 1.5 The dimensions of the rectangle are 2.5 by 0.5, and the dimensions of the square are 1.5 by 1.5. 48. 0.60(10)(1000)  50(8)(tf  50) 6000  400(tf  50) 6000  400tf  20,000 6000  20,000  400tf  20,000  20,000 26,000  400tf 26,000 400



400t

The number of the number of the number Calories burned is Calories per minute times of minutes. 144 4244 43 123 144424443 123 14 424 43

f

Equation: C  4.5 The equation is C  4.5m. 57. 150  4.5m

400

150 4.5 100 3 1 333

m m

      

     10

12

14

16

18

20

59. The lowest value is 19, and the highest value is 28, so use a scale that includes those values. Place an  above each value for each occurrence.            

8

 2

18

20

22

24

26

28

60. 10  (17)  ( 010 0  017 0 )  (10  17)  27 61. 12  (8)  12  (8)  12  8  ( 012 0  08 0 )  (12  8)  4 62. 6  14  6  (14)  ( 014 0  06 0 )  (14  6)  8 63. Sample answer: 1 and 3 1  3  4 is even, but 1 and 3 are odd. 64. Sample answer: You could bake sugar cookies, which do not require chocolate chips.

Maintain Your Skills  6  14

 6  6  14  6 2 v 9 9 2 v 2 9

4.5m 4.5

58. The lowest value is 11, and the highest value is 17, so use a scale that includes those values. Place an  above each value for each occurrence.

Exercises 53–55 For checks, see students’ work.

2 v 9

m

1

x  4 5n  4  7(n  1)  2n 5n  4  7n  7  2n 5n  4  5n  7 5n  4  5n  5n  7  5n 47 Since 4  7 is a false statement, this equation has no solution.

2 v 9





It will take 333 minutes.

52. C;

53.

18

 9

w  2 56. Words: The number of Calories burned is the number of Calories per minute times the number of minutes. Variables: Let C  the number of Calories burned, and m  the number of minutes.

The final temperature is 65F. 49. Sample answer: 3(x  1)  x  1 50. Set two expressions equal to each other and solve the equation. Answers should include the following. • The steps used to solve the equation are (1) subtract 7.6x from each side, (2) subtract 6 from each side, and (3) divide each side by 0.4. • The number of male and female Internet users will be the same in 2010. • If two expressions that represent the growth in use of two items are set equal to each other, the solution to the equation can predict when the number of items in use will be equal. 51. D; 8x  3  5(2x  1) 8x  3  10x  5 8x  3  10x  10x  5  10x 2x  3  5 2x  3  3  5  3 2x  8

Page 154

2  7(2)

9w 9

65  tf

2x 2

 2

x  3  14 x  3  3  14  3 x  11 5  9w  23 5  9w  5  23  5 9w  18

55.

x

1

x  3 7 x  3 7

 20

1 2  92(20) v  90

95

Chapter 3

65.

3a2 b  c

3152 2

7.

87 75

87

3x 3

75 15

70.

72.

12  3  15  3 4 5 36  12 36  60  12 60 3 5 108 108  9  99 9 12  1 12 15

69.

71.

73.

9.

16 40

16  8

 40  8

75.

5.082 1.1

2

5 120

1750 120 175 12 7 1412

19  19

11.

 57  19

12.

6.

13.

g

 350



120g 120

g g

3 ? 21  14 2 ?

8 ? 12  18 9 ?

14.

16 17



15.

8 . 9

2.3 3.4

3.0

 3.6.

4.2 ? 1.68  2.24 5.6 ?

21.1 ? 1.1  1.2 14.4 ?

21.1(1.2)  14.4(1.1) 25.32  15.84 No, the cross products are not equal, so The ratios do not form a proportion. 16.



12

 18.

4.2(2.24)  5.6(1.68) 9.408  9.408 1.68 4.2 Yes, the cross products are equal, so 5.6  2.24. Since the ratios are equal, they form a proportion.

2.1 ? 0.5  0.7 3.5 ?

2.1 3.5

8 9

2.3 ? 3.0  3.6 3.4 ?

2.3(3.6)  3.4(3.0) 8.28  10.2 No, the cross products are not equal, so The ratios do not form a proportion.

16 ? 8 9 17 ?

Chapter 3

1.1n 1.1

No, the cross products are not equal, so The ratios do not form a proportion.

4 ? 12  33 11 ?

2.1(0.7)  3.5(0.5) 1.47  1.75 No, the cross products are not equal, so The ratios do not form a proportion.



8(18)  9(12) 144  108

Check for Understanding

16(9)  17(8) 144  136 No, the cross products are not equal, so The ratios do not form a proportion.



3(14)  2(21) 42  42 3 21 Yes, the cross products are equal, so 2  14. Since the ratios are equal, they form a proportion.

3

4(33)  11(12) 132  132 12 4 Yes, the cross products are equal, so 11  33. Since the ratios are equal, they form a proportion. 5.

a  15 n 8.47

Pages 158–159 Practice and Apply

1. See students’ work. 2. A ratio is a comparison of two numbers, and a proportion is an equation of two equal ratios. 3. Find the cross products and divide by the value with the variable. 4.

225 15

The Lehmans need about 14.6 gallons.

Ratios and Proportions

Page 158



5(350)  120(g) 1750  120g

28  7  49  7 4 7 8 8  8  120  8 120 1  15 28 28  14  42  14 42 2 3

1

5

3-6

15a 15

4.62  n 10. Let g represent the amount of gasoline needed for a 350-mile trip.

28 49

19 57

0.6 1.1

5

 15

a(15)  45(5) 15a  225

24 3



a 45

0.6(8.47)  1.1(n) 5.082  1.1n

 12 74.

8.

x8

5 66. x(a  2b)  y  2(5  2  8)  1  2(5  16)  1  2(21)  1  42  1  41 67. 5(x  2y)  4a  5(2  2  1)  4(5)  5(2  2)  4(5)  5(4)  4(5)  20  20 0 68.

6 x



3(x)  4(6) 3x  24

3(25)

87 

3 4

0.5 . 0.7

5 ? 4  1.6 2 ?

5(1.6)  2(4) 88

96

21.1 14.4

1.1

 1.2.

5

4

Yes, the cross products are equal, so 2  1.6. Since the ratios are equal, they form a proportion. 17. For each row in the table write the ratio of the number in the first column to the number in the 871 498 fourth column. USA: 2116; USSR/Russia: 1278;

5 3

30.

5x 5

374 180 ; Great Britain: 638; 1182 136 179 188 ; Italy: 479; Sweden: 469 598

19.

2 10



20.

4(10)  x(2) 40  2x 40 2

8

15 3

20  x 6 5

21.

x

 15

22.

23.





a 25.

1 0.19



56 6 28 3

152d 152

3y 3 n

 21



24.

16 7

248 4

28n 28

248h 248

9

b

or

1 93



b

63 16 63 16

2 0.21

2.5 10

30 10

1 1

64

28.



h

3.67s 3.67



3 20

x



x

n

 122

3(122)  20(n) 366  20n 366 20

7  3

116 6 58 or 3



20n 20

18.3  n You would expect about 18 animals.

6(x  3)  14(7) 6x  18  98 6x  18  18  98  18 6x  116 6x 6

1

or 46

35. Let n represent the number of pets from a breeder.

1.066z 1.066 6 14

25 6

1

63.37  z 29.

h

The car is 46 feet high.

z 9.65

7(9.65)  1.066(z) 67.55  1.066z 67.55 1.066

2 3

1(h)  64

1.23  s 7 1.066



1 1 2123 2 25 2 h  4 13 2

s 1.88

2.405(1.88)  3.67(s) 4.5214  3.67s 

10d 10



3d The wall in the blueprint is 3 in. long. 34. Let h represent the actual height.

n  0.84

4.5214 3.67

d

 12

2.5(12)  10(d) 30  10d

1.68 2



1

33. Let d represent the scale length. 15

or 316

8

27.

372

 248 1

12 n

2.405 3.67

93 h

It will take 12 hours.

n





3

2(n)  0.21(8) 2n  1.68 2n 2

2128 152

h  2 or 12

1(n)  0.19(12) n  2.28 26.



248(h)  4(93) 248h  372

16(b)  7(9) 16b  63 16b 16

532 d

d  14 It will take 14 days. 32. Let h represent the number of hours needed to drive 93 miles.

15  n

7 a

6(a)  8(7) 6a  56 6a 6

20 28

420 28

18  x 6 8





152(d)  4(532) 152d  2128

20(21)  28(n) 420  28n

5x 5



152 4

3 15

5y

6(15)  5(x) 90  5x 90 5



3

31. Let d represent the number of days needed to earn $532.

1(15)  y(3) 15  3y

2x 2



1 y

8

5

x  5 or 15

18. No; if two of these ratios formed a proportion, the two countries would have the same part of their medals as gold medals. 4 x

6  2

5(x  2)  3(6) 5x  10  18 5x  10  10  18  10 5x  8

Germany: France:

x

1

193

97

Chapter 3

36. Evaluate 2a  3b 4b  3c

The

2a  3b with 4b  3c 2(3)  3(1)

 4(1)

43.

a  3, b  1, and c  5.

 3(5)

5  9w  23 5  9w  5  23  5 9w  18 9w 9

6  3  4  15 9  19 9 value is 19.

w  2 m 5

44. m 5

37. Sample answer: Ratios are used to determine how much of each ingredient to use for a given number of servings. Answers should include the following. • To determine how much honey is needed if you use 3 eggs, write and solve the proportion 2 3  , where h is the amount of honey. 3 h

45. 5

2

2

48.

3

15  y 5

15 z

5

3

3

15(5)  z(3) 75  3z 75 3



3z 3

25  z

Page 159

2  5(3)

Maintain Your Skills

Temperature

Exercises 40–45 For checks, see students’ work. 40. 8y  10  3y  2 8y  10  3y  3y  2  3y 11y  10  2 11y  10  10  2  10 11y  12 11y 11

12

 11 12

1 21 2  499 3 3 7 7

49. (0.075)(5.5)  0.4125 50. 33 is thirty-three units from zero in the negative direction. 033 0  33 51. 77 is seventy-seven units from zero in the positive direction. 077 0  77 52. 2.5 is two and five tenths units from zero in the positive direction. 02.5 0  2.5 53. –0.85 is eighty-five hundredths unit from zero in the negative direction. |0.85|  0.85 54. The temperature is high when you enter the house, but decreases to a lower constant temperature due to the air conditioner.

2y 2

y z

1

y  11 or 111

Time

17  2n  21  2n 17  2n  2n  21  2n  2n 17  21 Since 17  21 is a false statement, this equation has no solution. 42. 7(d  3)  4 7d  21  4 7d  21  21  4  21 7d  25 41.

7d 7



d Chapter 3

 3

 1

3



1

z  7 5 z  7 5

47. 9

10(3)  y(2) 30  2y 30 2

1 2  5(25)

1 8 2198 2  7272

9(27)  12(18) 243  216 10 y

 25

z  7  15 z  7  7  15  7 z  8 46. (7)(6)  42

9 ? 18  27 12 ?

x y

m 5 m 5

m  125

4

39. C;

 6  31

 6  6  31  6 5

• To alter the recipe to get 5 servings, multiply 1 each amount by 14. 38. D;

18

 9

25 7 25 or 7

55. The base is 60, and the part is 18. Let p represent the percent. a b 18 60

p

 100 p

 100

18(100)  60(p) 1800  60p 1800 60



60p 60

30  p Eighteen is 30% of 60.

4

37

98

Find the percent using the original number, 72, as the base.

56. The base is 14, and the part is 4.34. Let p represent the percent. a b 4.34 14

p

36 72

 100 p

36(100)  72(r) 3600  72r

 100

4.34(100)  14(p) 434  14p 434 14



3600 72

14p 14

p

 100 p

 100

5 45

6(100)  15(p) 600  15p 600 15



500 45

p

 100 p

 100



2p 2

2 14

400  p Eight is 400% of 2.

Page 162



45r 45

r

 100

2(100)  14(r) 200  14r 200 14

3-7

r

 100

11  r The percent of increase is about 11%. 6. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 16  14  2 Find the percent using the original number, 14, as the base.

8(100)  2(p) 800  2p 800 2

72r 72

5(100)  45(r) 500  45r

15p 15

40  p Six is 40% of 15. 58. The base is 2, and the part is 8. Let p represent the percent. a b 8 2



50  r The percent of decrease is 50%. 5. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 50  45  5 Find the percent using the original number, 45, as the base.

31  p Four and thirty-four hundredths is 31% of 14. 57. The base is 15, and the part is 6. Let p represent the percent. a b 6 15

r

 100



14r 14

14  r The percent of increase is about 14%. 7. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 150  120  30 Find the percent using the original number, 150, as the base.

Percent of Change Check for Understanding

1. Percent of increase and percent of decrease are both percents of change. If the new number is greater than the original number, the percent of change is a percent of increase. If the new number is less than the original number, the percent of change is a percent of decrease. 2. Sample answer: If the original number is 10 and the new number is 30, the percent proportion is 30  10 r  100 and the percent of change is 200%, 10 which is greater than 100%. 3. Laura; Cory used the new number as the base instead of the original number. 4. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 72  36  36

30 150

r

 100

30(100)  150(r) 3000  150r 3000 150



150r 150

20  r The percent of decrease is 20%. 8. The tax is 6.5% of the price of the software. 6.5% of $39.50  0.065  39.50  2.5675 Round $2.5675 to $2.57 since tax is always rounded up to the nearest cent. Add this amount to the original price. $39.50  $2.57  $42.07 The total price of the software is $42.07.

99

Chapter 3

Find the percent using the original number, 25, as the base.

9. The tax is 5.75% of the price of the compact disc. 5.75% of $15.99  0.0575  15.99  0.919425 Round $0.919425 to $0.92 since tax is always rounded up to the nearest cent. Add this amount to the original price. $15.99  $0.92  $16.91 The total price of the compact disc is $16.91. 10. The discount is 25% of the original price. 25% of $45  0.25  45  11.25 Subtract $11.25 from the original price. $45.00  $11.25  $33.75 The discounted price of the jeans is $33.75. 11. The discount is 33% of the original price. 33% of $19.95  0.33  19.95  6.5835 Subtract $6.58 from the original price. $19.95  $6.58  $13.37 The discounted price of the book is $13.37. 12. Find the amount of change. 40,478  22,895  17,583 Write the equation using the original number, 22,895, as the base. 17,583 22,895

7 25

7(100)  25(r) 700  25r 700 25

36 66

17,583 22,895

3600 66

r

 100 .

r

 100



94 58

9400 58

77  r The percent of increase is about 77%.

Practice and Apply

r

2650 13.7

58r 58

r

 100



13.7r 13.7

193  r The percent of increase is about 193%. 19. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 15.6  11.4  4.2

50r 50

40  r The percent of increase is 40%. 15. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 25  18  7

Chapter 3



26.5(100)  13.7(r) 2650  13.7r

 100



r

 100

26.5 13.7

20(100)  50(r) 2000  50r 2000 50

66r 66

162  r The percent of increase is about 162%. 18. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 40.2  13.7  26.5 Find the percent using the original number, 13.7, as the base.

14. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 70  50  20 Find the percent using the original number, 50, as the base. 20 50



94(100)  58(r) 9400  58r

22,895r 22,895

Pages 162–164

r

 100

55  r The percent of decrease is about 55%. 17. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 152  58  94 Find the percent using the original number, 58, as the base.

17,583(100)  22,895(r) 1,758,300  22,895r 1,758,300 22,895

25r 25

36(100)  66(r) 3600  66r

r

17,583 22,895



28  r The percent of decrease is 28%. 16. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 66  30  36 Find the percent using the original number, 66, as the base.

 100

13. Find the percent by solving

r

 100

100

Find the percent using the original number, 9.8, as the base.

Find the percent using the original number, 15.6, as the base. 4.2 15.6

2.3 9.8

r

 100

2.3(100)  9.8(r) 230  9.8r

4.2(100)  15.6(r) 420  15.6r 420 15.6

230 9.8

15.6r 15.6



7.5 40

r

 100

750 40

132r 132



3.5 25

r

 100



350 25

85r 85



25r 25

r

 100

64,000,000(100)  253,000,000(r) 6,400,000,000  253,000,000r

r

 100



r

 100

64,000,000 253,000,000

2.5(100)  32.5(r) 250  32.5r 250 32.5

40r 40

14  r The percent of decrease is 14%. 26. Find the amount of change. 317,000,000  253,000,000  64,000,000 Find the percent using the original number, 253,000,000, as the base.

6r The percent of increase is about 6%. 22. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 32.5  30  2.5 Find the percent using the original number, 32.5, as the base. 2.5 32.5



3.5(100)  25(r) 350  25r

5(100)  85(r) 500  85r 500 85

r

 100

19  r The percent of decrease is about 19%. 25. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 25  21.5  3.5 Find the percent using the original number, 25, as the base.

14  r The percent of increase is about 14%. 21. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 90  85  5 Find the percent using the original number, 85, as the base. 5 85

9.8r 9.8

7.5(100)  40(r) 750  40r

18(100)  132(r) 1800  132r 1800 132



23  r The percent of increase is about 23%. 24. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 40  32.5  7.5 Find the percent using the original number, 40, as the base.

27  r The percent of decrease is about 27%. 20. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 150  132  18 Find the percent using the original number, 132, as the base. 18 132

r

 100

6,400,000,000 253,000,000



253,000,000r 253,000,000

25  r The percent of increase is about 25%. 27. Find the change. 2,000,000  1,400,000  600,000 Find the percent using the original number, 2,000,000, as the base.

32.5r 32.5

8r The percent of decrease is about 8%. 23. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 12.1  9.8  2.3

600,000 2,000,000

r

 100

600,000(100)  2,000,000(r) 60,000,000  2,000,000r 60,000,000 2,000,000



2,000,000r 2,000,000

30  r The percent of decrease is 30%.

101

Chapter 3

28. Let n  the original number. Since 16% is a percent of increase, the new number is more than the original number. Therefore, 522  n represents the amount of change. 522  n n

16

 100 34.

(522  n)(100)  n(16) 52,200  100n  16n 52,200  100n  100n  16n  100n 52,200  116n 52,200 116



116n 116

450  n The original number is 450. 29. Let f  the amount of fat is one ounce of regular chips. Since 25% is a percent of decrease, the amount of fat in regular chips is more than the amount of fat in reduced fat chips. Therefore, f  6 represents the amount of change. f  6 f

35.

25

 100

(f  6)(100)  f(25) 100f  600  25f 100f  600  100f  25f  100f 600  75f 600 75



36.

75f 75

8f The least amount of fat in one ounce of regular chips is 8 grams. 30. Let n  the number of internet hosts in 1996. Since 1054% is a percent of increase, the number of internet hosts in 1996 is less than the number of internet hosts in 2001. Therefore, 109,600,000  n represents the amount of change. 109,600,000  n n



37.

1054 100

(109,600,000  n)(100)  n(1054) 10,960,000,000  100n  1054n 10,960,000,000  100n  100n  1054n  100n 10,960,000,000  1154n 10,960,000,000 1154



38.

1154n 1154

9,497,400  n The number of internet hosts in 1996 was about 9.5 million. 31. The tax is 5.5% of the price of the umbrella. 5.5% of $14  0.055  14  0.77 Add this amount to the original price. $14.00  $0.77  $14.77 The total price of the umbrella is $14.77. 32. The tax is 7% of the price of the backpack. 7% of $35  0.07  35  2.45 Add this amount to the original price. $35.00  $2.45  $37.45 The total price of the backpack is $37.45. 33. The tax is 5.75% of the price of the candle. 5.75% of $7.50  0.0575  7.50  0.43125

Chapter 3

39.

40.

41.

102

Round $0.43125 to $0.43 since tax is always rounded to the nearest cent. Add this amount to the original price. $7.50  $0.43  $7.93 The total price of the candle is $7.93. The tax is 6.25% of the price of the hat. 6.25% of $18.50  0.0625  18.50  1.15625 Round $1.15625 to $1.16 since tax is always rounded to the nearest cent. Add this amount to the original price. $18.50  $1.16  $19.66 The total price of the hat is $19.66. The tax is 6.75% of the price of the clock radio. 6.75% of $39.99  0.0675  39.99  2.699325 Round $2.699325 to $2.70 since tax is always rounded to the nearest cent. Add this amount to the original price. $39.99  $2.70  $42.69 The total price of the clock radio is $42.69. The tax is 5.75% of the price of the sandals. 5.75% of $29.99  0.0575  29.99  1.724425 Round $1.724425 to $1.72 since tax is always rounded to the nearest cent. Add this amount to the original price. $29.99  $1.72  $31.71 The total price of the sandals is $31.71. The discount is 40% of the original price. 40% of $45  0.40  45  18 Subtract $18 from the original price. $45.00  $18.00  $27.00 The discounted price of the shirt is $27.00. The discount is 20% of the original price. 20% of $6  0.20  6  1.20 Subtract $1.20 from the original price. $6.00  $1.20  $4.80 The discounted price of the socks is $4.80. The discount is 35% of the original price. 35% of $37.55  0.35  37.55  13.1425 Subtract $13.14 from the original price. $37.55  $13.14  $24.41 The discounted price of the watch is $24.41. The discount is 33% of the original price. 33% of $24.25  0.33  24.25  8.0025 Subtract $8.00 from the original price. $24.25  $8.00  $16.25 The discounted price of the gloves is $16.25. The discount is 45% of the original price. 45% of $175.95  0.45  175.95  79.1775

42.

43.

44.

45.

46.

x  1.24 1.24

Subtract $79.18 from the original price. $175.95  $79.18  $96.77 The discounted price of the suit is $96.77. The discount is 30% of the original price. 30% of $79.99  0.30  79.99  23.997 Subtract $24.00 from the original price. $79.99  $24.00  $55.99 The discounted price of the coat is $55.99. The discount is 20% of the original price. 20% of $120  0.20  120  24 Subtract $24 from the original price. $120  $24  $96 The discounted price of the lamp is $96. The tax is 6% of the discounted price of the lamp. 6% of $96  0.06  96  5.76 Add this amount to the discounted price. $96.00  $5.76  $101.76 The total price of the lamp is $101.76. The discount is 30% of the original price. 30% of $70  0.30  70  21 Subtract $21 from the original price. $70  $21  $49 The discounted price of the dress is $49. The tax is 7% of the discounted price of the dress. 7% of $49  0.07  49  3.43 Add this amount to the discounted price. $49.00  $3.43  $52.43 The total price of the dress is $52.43. The discount is 25% of the original price. 25% of $58  0.25  58  14.50 Subtract $14.50 from the original price. $58.00  $14.50  $43.50 The discounted price of the camera is $43.50. The tax is 6.5% of the discounted price of the camera. 6.5% of $43.50  0.065  43.50  2.8275 Round $2.8275 to $2.83 since tax is always rounded up to the nearest cent. Add this amount to the discounted price. $43.50  $2.83  $46.33 The total price of the camera is $46.33. Let x  the population in China in 2050. Since 22.6% is a percent of increase, the population in 2050 will be greater than the population in 1997. Therefore, x  1.24 represents the amount of change.



22.6 100

(x  1.24)(100)  1.24(22.6) 100x  124  28.024 100x  124  124  28.024  124 100x  152.024 100x 100



152.024 100

x  1.52024 The projected population of China is about 1.52 billion people in 2050. Let x  the population in India in 2050. Since 57.8% is a percent of increase, the population in 2050 will be greater than the population in 1997. Therefore, x  0.97 represents the amount of change. x  0.97 0.97



57.8 100

(x  0.97)(100)  0.97(57.8) 100x  97  56.066 100x  97  97  56.066  97 100x  153.066 100x 100



153.066 100

x  1.53066 The projected population of India is about 1.53 billion people in 2050. Let x  the population in the U.S. in 2050. Since 44.4% is a percent of increase, the population in 2050 will be greater than the population in 1997. Therefore, x  0.27 represents the amount of change. x  0.27 0.27



44.4 100

(x  0.27)(100)  0.27(44.4) 100x  27  11.988 100x  27  27  11.988  27 100x  38.988 100x 100



38.988 100

x  0.38988 The projected population of the U.S. is about 0.39 billion people in 2050. 47. Since the projected population for China is 1.52 billion people, for India is 1.53 billion people, and for the U.S. is 0.39 billion people, India will be the most populous in 2050. 48. See students’ work. x

49. Always; x% of y 1 100  y

y% of x 1 100 

103

P y P x

xy

or P  100; xy

or P  100

Chapter 3

50. Find the amount of change and express this change as a percent of the original number. Answers should include the following. • To find the percent of increase, first find the amount of increase. Then find what percent the amount of increase is of the original number. • The percent of increase from 1996 to 1999 is about 67%. • An increase of 100 is a very large increase if the original number is 50, but a very small increase if the original number is 100,000. The percent of change will indicate whether the change is large or small relative to the original. 51. B; Find the amount of change. 910  840  70 Write the proportion using the original number, 840, as the base. 70 840

1

52. C; The discount is 30% of the original price. 30% of $249.00  0.30  249.00  74.70 Subtract $74.70 from the original price. $249.00  $74.70  $174.30

3t 3

67. 53.

Maintain Your Skills 

3 15

a(15)  45(3) 15a  135 15a 15



2 3

8

d

55.

2(d)  3(8) 2d  24 2d 2



5.22 13.92

250.56 13.92

t

 48



13.92t 13.92



6 6

5c 5

n  1 58. 18  4a  2 18  2  4a  2  2 20  4a 20 4





69.

30 5

c  6

2 5

70.

4a 4

2

1

 4  54

2

4

3

60. 5  3  5  2  10

1

 5 or 1 5

 10

12 6



104



2a 2

9.5  a

1



28 7

22 2

p  11 5  4  2(a  5) 5  4  2a  10 5  14  2a 5  14  14  2a  14 19  2a 19 2

2

 20

Chapter 3

4

12 4

d4 6(p  3)  4(p  1) 6p  18  4p  4 6p  18  4p  4p  4  4p 2p  18  4 2p  18  18  4  18 2p  22 2p 2

5a 59.



7d 7

d  12 18  t Exercises 56–58 For checks, see students’ work. 56. 57. 6n  3  3 7  5c  23 6n  3  3  3  3 7  5c  7  23  7 6n  6 5c  30 6n 6

45 3

y  3 68. 7(d  3)  2  5 7d  21  2  5 7d  23  5 7d  23  23  5  23 7d  28

5.22(48)  13.92(t) 250.56  13.92t

24 2



t  15 7y  7  3y  5 7y  7  3y  3y  5  3y 4y  7  5 4y  7  7  5  7 4y  12 4y 4

135 15

a9 54.

4

 27

64. False; 8y  y2  y  10 ? 8  4  42  4  10 ? 8  4  16  14 ? 32  16  14 16  14 65. True; 16p  p  15p ? 16(2.5)  2.5  15(2.5) ? 40  2.5  37.5 37.5  37.5 ✓ Exercises 66–71 For checks, see students’ work. 66. 43  3t  2  6t 43  3t  6t  2  6t  6t 43  3t  2 43  3t  43  2  43 3t  45

r

a 45

1 42

1

62. True; a2  5  17  a 63. False; 2v2  v  65 ? ? 32  5  17  3 2(5)2  5  65 ? ? 9  5  14 2(25)  5  65 ? 50  5  65 14  14 ✓ 55  65

 100

Page 164

1 32

61. 9  4  9  3

71.

8x  4  10x  50 8x  4  10x  10x  50  10x 18x  4  50 18x  4  4  50  4 18x  54 18x 18

5. 7  2(w  1)  2w  9 7  2w  2  2w  9 2w  9  2w  9 Since the expressions on each side of the equation are the same, this equation is an identity. The statement 7  2(w  1)  2w  9 is true for all values of w. 6. 8(4  9r)  7(2  11r) 32  72r  14  77r 32  72r  77r  14  77r  77r 5r  32  14 5r  32  32  14  32 5r  18

54

 18

x3

Page 164 1.

Practice Quiz 2

3x  7  18 3x  7  7  18  7 3x  25 3x 3

5r 5

25

 3 25

r

1

x   3 or 8 3

1

Check:

2

? ?

25  7  18 18  18 ✓ 1

The solution is 8 3. 5 4(5) 

m  5 4 m  5 4 4

1

7.

2

2a 2

5

m  5 4 ? 25  5  4 ? 20  4



4y 4

3

h  2 or 1.5

24 10

4h  5  11 ? 4(1.5)  5  11 ? 6  5  11 11  11 ✓ The solution is 1.5. 4. 5d  6  3d  9 5d  6  3d  3d  9  3d 2d  6  9 2d  6  6  9  6 2d  15 Check:

d

15 2 15 2



24 x

3(x)  5(24) 3x  120 3x 3



120 3

x  40 y  5 8



20 4



y5 10. Find the amount of change. 34  10  24 Find the percent using the original number, 10, as the base.

6



3 5

y(8)  4(y  5) 8y  4y  20 8y  4y  4y  20  4y 4y  20

4

2d 2

8.

10 2 y 4

9.

55✓ The solution is 25. 3. 4h  5  11 4h  5  5  11  5 4h  6 4h 4

1

a

a5

5 5

2 10

2(a)  10(1) 2a  10

20  m  5 20  5  m  5  5 25  m Check:

or 3.6

8(4  9r)  7(2  11r) ? 8[4  9(3.6)]  7[2  11(3.6)] ? 8[4  32.4]  7[2  39.6] ? 8[36.4]  7[41.6] 291.2  291.2 ✓ The solution is 3.6.

3 8 3  7  18

2.

18 5 18 5

Check:

3x  7  18 1



r

 100

24(100)  10(r) 2400  10r 2400 10



10r 10

240  r The percent of increase is 240%.

Page 165

Reading Mathematics

1. original number: 166 lb amount of change: 166  158 or 8 lb 166  158 166 8 166

or 7.5

5d  6  3d  9 ? 5(7.5)  6  3(7.5)  9 ? 37.5  6  22.5  9 31.5  31.5 ✓ The solution is 7.5. Check:

r

 100 r

 100

8(100)  166(r) 800  166r 800 166



166r 166

5r There was about a 5% decrease in Monsa’s weight.

105

Chapter 3

2. original number: 75 points amount of change: 94  75 or 19 points 94  75 75 19 75

6.

r

 100 r

 100

4z  b  2z  c 4z  b  2z  2z  c  2z 2z  b  c 2z  b  b  c  b 2z  c  b 2z 2

19(100)  75(r) 1900  75r 1900 75



z

75r 75

The value of z

25  r There was about a 25% increase in her score. 3. original number: 12 orders amount of change: 18  12 or 6 orders 18  12 12 6 12

7. 3

r

 100



p b  c p b  c

12r 12

50  r There was a 50% increase in production

Check for Understanding

1. (1) Subtract az from each side. (2) Add y to each side. (3) Use the Distributive Property to write ax  az as a(x  z). (4) Divide each side by x  z. 2. t can be any number except 2.

1

10. A  2bh 1

A  63 The area is 63 ft2. 12. h 

bh h

h

b

The value of b is

b 9 b 9



13.

x

1

2 1

 b 2bh h

2

5a  y  54 5a  y  y  54  y 5a  54  y

Practice and Apply

14. 5g  h  g 8t  r  12t 5g  h  5g  g  5g 8t  r  8t  12t  8t h  4g r  4t h 4 h 4

b

Chapter 3

2 (A) b 2A b

2A b 2(28) 8

Pages 168–170

9x 9

The value of x is 9.

The value of

1

A  2bh

h7 The height is 7 ft.

2A . h

3x  b  6x 3x  b  3x  6x  3x b  9x

a

11.

A  2(18) (7)

11 2

5a 5

5  t

m2 5  t

2A  bh

5.

mw  t  2w  5 mw  t  2w  2w  5  2w mw  t  2w  5 mw  t  2w  t  5  t mw  2w  5  t w(m  2)  5  t

The value of w is m  2. Since division by 0 is undefined, m  2  0 or m  2.

2A  2 2bh

4.

a

5  t

A  2bh



a(b  c) b  c

wm2

1

2A h 2A h



w(m  2) m  2

3. Sample answer: For a triangle, area is 2 times the product of the length of the base times the height 1 or A  2bh. Solve for b. 1

2  3(c)

p

Solving Equations and Formulas

Page 168

c

The value of a is b  c. Since division by 0 is undefined, b  c  0 or b  c. 9.

3-8

1

y  a 3 y  a 3

c  b 2 c  b 2 c  b is 2 .

y  a  3c y  a  a  3c  a y  3c  a The value of y is 3c  a. 8. p  a(b  c)

r

 100

6(100)  12(r) 600  12r 600 12





4g 4

g

The value of g is

54  y 5 54  y 54  y  5 or 5 54  y a is 5 .



106

h 4 .

r 4 r 4



4t 4

t r

The value of t is 4.

15.

y  mx  b y  b  mx  b  b y  b  mx y  b x y  b x



5

m y  b . x



3ax 3a

x

22.

a

a



y The value of y

19.

23.

am  z 7 am  z 7 am  z is 7 .

3

4b 4



b The value for

1

by  2 3 by  2 3

5x  y . 2

c

2  3(c)

3c  2 b 3c  2 y b 3c  2 value of y is b . by b



The undefined, b  0. 24. 7

1

6c  t 7 6c  t 7

Since division by 0 is

b

2  7(b)

6c  t  7b 6c  t  t  7b  t 6c  7b  t

km  5x  6y km  5x  5x  6y  5x km  6y  5x

6c 6



The value undefined, k  0. 20. 4b  5  t 4b  5  5  t  5 4b  t  5

a

by  2  3c by  2  2  3c  2 by  3c  2



6y  5x k 6y  5x m k 6y  5x of m is . k

2a 2



The value of a is

c  2b 5 c  2b 2b  c a  5 or 5 2b  c of a is 5 .

km k

2

2  a(2)

5x  y 2 5x  y 2

9a  2b  c  4a 9a  2b  4a  c  4a  4a 5a  2b  c 5a  2b  2b  c  2b 5a  c  2b 5a 5

1

5x  y a 5x  y a

by 0 is

5x  y  2a

v  r

7y 7

20  n 3a 20  n n  20  3a or 3a n  20 x is 3a . Since division



The value of undefined, a  0.

at t

The value

 4

2  5(4)

Since division by 0 is

The value of a is t . Since division by 0 is undefined, t  0. 17. 3y  z  am  4y 3y  z  4y  am  4y  4y 7y  z  am 7y  z  z  am  z 7y  am  z

18.

1

3ax  n 5 3ax  n 5

3ax  n  20 3ax  n  n  20  n 3ax  20  n

mx x

The value of m is undefined, x  0. 16. v  r  at v  r  r  at  r v  r  at v  r t v  r t

21.



c The value of Since division by 0 is

7b  t 6 7b  t 6 7b  t c is 6 .

3

c  4y  b

25.

3

c  b  4y  b  b 3

c  b  4y 4 (c 3 4 (c 3

t  5 4 t  5 5  t or 4 4 5  t b is 4 .

1 2

4 3

 b)  3 4y  b)  y

4

The value of y is 3(c  b). 3 m 5

26. 3 m 5

ab

aaba 3 m 5 5 3 m 3 5

ba

1 2  53(b  a) 5

m  3(b  a) 5

The value of m is 3(b  a).

107

Chapter 3

n

S  2 (A  t)

27.

2 (S) n 2S n 2S  t n 2S t n 2S  nt n

3

2 n

 n 2 (A  t) At

33. Five-eighths of

4

a number x 14 42443 5 8

Att A A

The value of A is undefined, n  0. 28. p(t  1)  2 p(t  1) p

2S  nt . n

2 Since division by 0 is

c  b t  r c  b c  b a  t  r or r  t c  b a is r  t . Since division

35.

1

3

2 2 1 (A)  a  b 2h(a a  b 2A h a  b 2A hab 2(60) h  8  12

S

by 0 is 37. e e

 b)

4

w  10e m w  10e m m

1

2

ms  w 10e  10 10 ms  w e 10 w  ms e 10 w  ms  10 410  5(76)  10

e3 Miguel made 3 errors. 38.



P H2 (P) 1.2

39.

1.2W H2

1

H 2 1.2W H2

 1.2

H 2P W 1.2 H2P W  1.2

W

2

32  30 1.2

W  225 The person weighs 225 lb. 40. Solve the formula for F. R (S  P)R 

six times twice a another minus p 1424 equals3 number plus one. 1 424 3 number 14243 1442443q 1 23 1 23 2p 6q    1

S  F  P S  P S  F  P (S  P) S  P

1

(S  P )R  S  F  P (S  P)R  S  S  F  P  S (S  P)R  S  F  P (S  P)R  S  P  F  P  P (S  P)R  S  P  F F  (S  P)R  S  P F  (900  900)6  900  900 F  9000 9000 grams of fuel should be loaded.

5  2p  6q  1 5  2p  5  6q  1  5 2p  6q  4 6q  4 2

p  3q  2 or 2  3q The value of p is 2  3q. Chapter 3

33

ms  w  10e ms  w  w  10e  w ms  w  10e

r6 r65  r  11 of t is r  11.



3

y

m(s) 

A number t 123 minus 1 five equals another number plus six. 14 4244 3 23 123 1444244 43r 1 23 1 23 t  5  r  6

2p 2



2 1 2

36.

The value of g division by 0 is undefined, 2  h  0 or h  2.

Five 1 23 5



y

h6 The height is 6 meters.

5  m 2  h 5  m g 2h 5  m is 2  h . Since

32.



1 y 2 1 y 2 1 y 2 1 2 2y

A  2h(a  b)



The value of undefined, r  t  0 or r  t. 30. 2g  m  5  gh 2g  m  gh  5  gh  gh 2g  m  gh  5 2g  m  gh  m  5  m 2g  gh  5  m g(2  h)  5  m

t5 t55 t The value



5 x3 8 5 x3 8 5 x3 8 5 x6 4

34.

Since division by 0 is undefined, p  0. 29. at  b  ar  c at  b  ar  ar  c  ar at  b  ar  c at  b  ar  b  c  b at  ar  c  b a(t  r)  c  b

31.



5 x 8

plus 1 23

5

2 p 2 t1 p 2 t11 p 1 2 t p 1 2  p t p 2  p The value of t is p .

g(2  h) 2  h

x

1 2

The value of y is 4x  6.



a(t  r) t  r

1

one-half of another number 14 44244 43y

is {

108

2

three. 1 23 3

Subtract $5.25 from the original price. $15.00  $5.25  $9.75 The discounted price of the scarf is $9.75. 48. The discount is 15% of the original price. 15% of $299  0.15  299  44.85 Subtract $44.85 from the original price. $299.00  $44.85  $254.15 The discounted price of the television is $254.15.

41. Solve for h. V  r2h V r2h  r2 r2 V h r2 V h  r2

h

1202 22

5453 

h  17.4 The height of the container should be about 17.4 cm. 42. The area of the arrow is the sum of the area of the triangle with base  3s and height  s and the square with side length  s.

2 9

49.

5

a

2(a)  9(5) 2a  45 2a 2



45 2 45 2

1

a

3

a  222 or 22.5

A  2(3s)(s)  s2

1

A  2s2  s2



x 4x  4

120 32 15 4 3 34

15 2 y 2

32t 32

t  t or 3.75  t 3

4

(x  1)(4)  8(3) 4x  4  24 4x  4  4  24  4 4x  20 4x 4



20 4

x5 52. Write each number as a decimal.

5  y . 2

1 4

 0.25

1 4

 0.5

3

1

0.5  0.555555 p or about 0.56 0.2  0.2 0.2 6 0.25 6 0.5 6 0.56 The numbers arranged in order from least to

1

greatest are 0.2, 4, 3 4, 0.5.

 2(5  y)  10  2y 45. C; A  2bh

1

A  2(16)(7)

1

53. Write each number as a decimal. 25  2.23606797 p or about 2.2. 3  3.0

A  56 The area of the triangle is 56 m2.

2 3

Page 170



x  1 8

51.

5  y 2 5  y 2

Now, replace x in 4x with

t

8

15(8)  32(t) 120  32t

43. Equations from physics can be used to determine the height needed to produce the desired results. Answers should include the following. • Use the following steps to solve for h. (1) Use the Distributive Property to write the 1 equation in the form 195g  hg  2mv2. (2) Subtract 195 from each side. (3) divide each side by g. • The second hill should be 157 ft. 44. B; First, solve 2x  y  5 for x. 2x  y  5 2x  y  y  5  y 2x  5  y 2x 2

15 32

50.

5

A  2 s2

 0.666666 p or about 0.7.

1.1  1.1 0.7 6 1.1 6 2.2 6 3.0 The numbers arranged in order from least to 2 greatest are 3, 1.1, 25, 3.

Maintain Your Skills

46. The discount is 20% of the original price. 20% of $85  0.20  85  17 Subtract $17 from the original price. $85  $17  $68 The discounted price of the camera is $68.00. 47. The discount is 35% of the original price. 35% of $15  0.35  15  5.25

54. 2.18  (5.62)  ( 05.62 0  02.18 0 )  (5.62  2.18)  3.44

109

Chapter 3

1 32

1

2

5.

3

55. 2  4  4  4  ° `  1

4 2

2

10

3 2 ` ` ` ¢ 4 4

10

The equation is 0.10(6  p)  1.00p  0.40(6). 6. 0.10(6  p)  1.00p  0.40(6) 0.6  0.1p  p  2.4 0.6  0.9p  2.4 0.6  0.9p  0.6  2.4  0.6 0.9p  1.8

134  24 2 6

56. 3  5  15  15

1

6

 15  15

2

0.9p 0.9

11015  156 2

16

1

 15 or 115 57. 58. 59. 60. 61.

Multiplicative Identity Property Symmetric Property of Equality Reflexive Property of Equality Substitution Property of Equality 6(2  t)  6(2)  6(t)  12  6t 62. (5  2m)3  (5)3  (2m)3  15  6m 63. 7(3a  b)  7(3a)  (7)(b)  21a  (7b)  21a  7b 2

2

Units(lb) Price per Unit (lb) Total Price

2

 4h  6 65.

3

Walnuts

10

$4.00

Cashews

c

$7.00

1 32

 5t)  5(15)  5 (5t)

 9  (3t)  9  3t 66. 0.25(6p  12)  0.25(6p)  0.25(12)  1.5p  3

7.00 c

Mixture

10  c

$5.50

5.50(10  c)

1.5c 1.5

15

 1.5

c  10 10 pounds of cashews should be mixed with 10 pounds of walnuts. 4(1)  3(1)  4(1)  3(1)  4

3-9

9.

Weighted Averages

Page 174

1

11112

1. Sample answer: grade point average 2. The formula d  rt is used to solve uniform motion problems. In the formula, d represents distance, r represents rate, and t represents time. 3. Let d  the number of dimes.

Quarters



Number of Coins

Value of Each Coin

d

$0.10

0.10d

d8

$0.25

0.25(d  8)

Total Value

First Cyclist Second Cyclist Distance traveled by first cyclist 1444244 43 20t

Quarts

Amount of Orange Juice

10% Juice

6p

0.10(6  p)

100% Juice

p

1.00p

40% Juice

6

0.40(6)

43432 1

42

16 412

 3.56 Her GPA was about 3.56. 10. Let t  the number of hours until the two cyclists are 15 miles apart.

4.

Chapter 3

112 2



Check for Understanding

Dimes

4.00(10)

Price of price of price of walnuts plus cashews equals mixture. 1 424 3 123 1 424 3 123 1 424 3 4.00(10)  7.00c  5.50(10  c) 4.00(10)  7.00c  5.50(10  c) 40  7c  55  5.5c 40  7c  5.5c  55  5.5c  5.5c 40  1.5c  55 40  1.5c  40  55  40 1.5c  15

64. 3(6h  9)  3(6h)  3(9) 3 5(15

1.8

 0.9

p2 2 quarts of pure orange juice 7. Replace p in 6  p with 2. 6p62 4 4 quarts of 10% juice 8. Let C  the number of pounds cashews in the mixture.

10 6  ° ` `  ` ` ¢ 15 15 

Amount of orange amount of orange amount of orange juice in 10% juice plus juice in 100% juice equals juice in 40% juice. 144424443 123 144424443 123 144424443 0.10(6  p)  1.00p  0.40(6)

minus 123 

r 20 14

t t t

distance traveled by second cyclist 1 444244 43 14t

d  rt 20t 14t equals 123 

15. 1 23 15

20t  14t  15 6t  15 6t 6



15 6

t  2.5 The cyclists will be 15 miles apart in 2.5 hours.

110

Pages 174–177

21. 40t  30t  245 70t  245

Practice and Apply

11. Number of Price per Dozens Dozen Peanut Butter Cookies Chocolate Chip Cookies 12.

$6.50

6.50p

p  85

$9.00

9.00(p  85)

1444442444443 123

t



sales of the chocolate chip cookies

equals

9.00(p  85)



1444442444443 1 424 3

15.5p 15.5

22. Let s  the number of rolls of solid gift wrap sold. Rolls Price per Roll Total Price Solid Wrap s $4.00 4.00s Print Wrap 480  s $6.00 6.00(480  s)

total sales. 123 4055.50

Price of solid wrap 1 plus of print wrap 123 equals $2340. 144424443 23 price 144424443 123 4.00s  6.00(480  s)  2340



4.00s  6.00(480  s)  2340 4s  2880  6s  2340 2880  2s  2340 2880  2s  2880  2340  2880 2s  540

4820.5 15.5

2s 2

p  311 311 doz peanut butter cookies were sold. 14. Replace p in p  85 with 311. p  85  311  85  226 226 doz chocolate chip cookies were sold. 15.

16.

Value of 43 gold 14424 270g

Number of Ounces g 15  g 15 plus 1 23 

Price per Ounce $270 $5 $164

value of silver 1442443 5(15  g)

equals 123 

265g 265



Value 270g 5(15  g) 164(15)

Units (lb)

Price per Unit (lb)

Total Price

$6.40 coffee

9

$6.40

6.40(9)

$7.28 coffee

p

$7.28

7.28p

$6.95 coffee

9p

$6.95

6.95(9  p)

0.33p 0.33

g9 9 oz of gold was used. 18. Replace g in 15  g with 9. 15  g  15  9 6 6 oz of silver was used. 19. r 40 30

540 2

Price of price of price of $6.40 coffee plus $7.28 coffee equals $6.95 coffee. 1442443 123 1442443 123 14243 6.40(9)  7.28p  6.95(9  p) 6.40(9)  7.28p  6.95(9  p) 57.6  7.28p  62.55  6.95p 57.6  7.28p  6.95p  62.55  6.95p  6.95p 57.6  0.33p  62.55 57.6  0.33p  57.6  62.55  57.6 0.33p  4.95

value of alloy. 14424 43 164(15)

2385 265

Eastbound Train Westbound Train



s  270 480  s  480  270 or 210 270 rolls of solid wrap and 210 rolls of print wrap were sold. 23. Let p  the number of pounds of the $7.28 coffee in the mixture.

The equation is 270g  5(15  g)  164(15). 17. 270g  5(15  g)  164(15) 270g  75  5g  2460 265g  75  2460 265g  75  75  2460  75 265g  2385

20.

245 70 1 32

The trains will be 245 miles apart in 32 hours.

The equation is 6.50p  9.00( p  85)  4055.50. 13. 6.50p  9.00( p  85)  4055.50 6.5p  9p  765  4055.5 15.5p  765  4055.5 15.5p  765  765  4055.5  765 15.5p  4820.5

Gold Silver Alloy



1

p

Sales of the peanut butter cookies plus 6.50p

70t 70

Total Price

4.95

 0.33

p  15 15 pounds of the $7.28 coffee should be mixed with 9 pounds of the $6.40 coffee.

t t t

d  rt 40t 30t

Distance traveled distance traveled by eastbound train 123 plus by westbound train equals 144424443 144424443 123 40t  30t 

245 1 23. 245

The equation is 40t  30t  245.

111

Chapter 3

27. Let x  the amount of 25% solution to be added.

24. Let w  the number of gallons of whipping cream.

Whipping Cream 2% Milk 4% Milk

Units (gal)

Percent Butterfat

Total Butterfat

w 35w 35

9% 2% 4%

0.09w 0.02(35  w) 0.04(35)

25% Solution 60% Solution 30% Solution



0.25x  0.60(140  x)  0.30(140) 0.25x  84  0.60x  42 84  0.35x  42 84  0.35x  84  42  84 0.35x  42 0.35x 0.35

w  10 35  w  35  10 or 25 10 gallons of whipping cream should be mixed with 25 gallons of 2% milk. 25. Let x  the amount of 25% copper to be added.

40% Glycol 60% Glycol 48% Glycol

Amount of Alloy Amount of Copper x

0.25x

50% Copper

1000  x

0.50(1000  x)

45% Copper

1000

0.45(1000)

Amount of copper in 25% copper

plus 14243 1 23 0.25x



amount of copper in 50% copper

amount of copper in equals 45% copper. 

0.45(1000)

0.20x 0.20

r 

 500

29.

85(3)  92(3)  82(1)  75(1)  95(1) 3  3  1  1  1 255  276  82  75  95  9 783  9

 87 The average grade is 87. 30. Let t  the number of hours until the helicopter reaches the trawler.

d r t 1000  3 1  3333 1 or 3333 miles

or 500 miles per hour per hour You must find the weighted average. 1 500(2)  3333(3) M 23 2000  5

Trawler Helicopter

r 30 300

t t t

d  rt 30t 300t

Distance traveled distance traveled by trawler plus by helicopter equals 660 144424443 123 144424443 123 1 42km. 43 30t  300t  660

 400 The airplane’s average speed was 400 mph. Chapter 3

12

 0.20

x  60 100  x  100  60 or 40 60 gallons of 40% antifreeze should be mixed with 40 gallons of 60% antifreeze.

50

 0.25

x  200 1000  x  1000  200 or 800 200 g of 25% copper alloy should be mixed with 800 g of 50% copper alloy. 26. To find the average speed for each leg of the trip, d rewrite d  rt as r  t . East South d t 1000 2

Amount of Glycol 0.40x 0.60(100  x) 0.48(100)

0.40x  0.60(100  x)  0.48(100) 0.40x  60  0.60x  48 60  0.20x  48 60  0.20x  60  48  60 0.20x  12

0.25x  0.50(1000  x)  0.45(1000) 0.25x  500  0.50x  450 500  0.25x  450 500  0.25x  500  450  500 0.25x  50 0.25x 0.25

Amount of Solution x 100  x 100

Amount of amount of amount of glycol in glycol in glycol in 40% solution 1 plus 60% solution 123 equals 1442443 48% solution. 1442443 23 1442443 0.40x  0.60(100  x)  0.48(100)

14243 123 14243

0.50(1000  x)

42

 0.35

x  120 140  x  140  120 or 20 120 mL of 25% solution should be mixed with 20 mL of 60% solution. 28. Let x  the amount of 40% glycol to be added.

0.7 0.07

25% Copper

Amount of Copper Sulfate 0.25x 0.60(140  x) 0.30(140)

Amount of amount of amount of copper sulfate copper sulfate copper sulfate in 25% solution 123 plus in 60% solution 123 equals in 30% solution. 1442443 1442443 1442443 0.25x  0.60(140  x)  0.30(140)

Amount of butterfat in amount of amount of whipping butterfat butterfat cream plus in 2% milk equals in 4% milk. 14243 123 144244 3 123 1 44 244 3 0.09w  0.02(35  w)  0.04(35) 0.09w  0.02(35  w)  0.04(35) 0.09w  0.7  0.02w  1.4 0.07w  0.7  1.4 0.07w  0.7  0.7  1.4  0.7 0.07w  0.7 0.07w 0.07

Amount of Solution x 140  x 140

112

30t  300t  660 330t  660 330t 330

0.25(16  x)  1.00x  0.40(16) 4  0.25x  x  6.4 4  0.75x  6.4 4  0.75x  4  6.4  4 0.75x  24

660

 330

t2 The helicopter will reach the trawler in 2 hours. 31. Let t  the number of seconds until the cheetah catches its prey. r 90 70

Cheetah Prey equals

distance traveled by prey

90t



70t

1442443

123

1444442444443

plus

300 ft.



300

14243

Express Train Local Train

123

90t  70t  300 90t  70t  70t  300  70t 20t  300 20t 20



t  15 The cheetah will catch the prey in 15 seconds. 32. Let t  the number of seconds until the sprinter catches his opponent. r 8.2 8

Distance traveled by sprinter equals 1442443 1 4243

8.2t  8.2t  8(t  1) 8.2t  8t  8 8.2t  8t  8t  8  8t 0.2t  8 0.2t 0.2

t t t1

32t 32

8(t  1)

8

 0.2

t  40 Distance traveled by sprinter is 8.2t  8.2(40) or 328 meters. Therefore, the sprinter would not catch his opponent in a 200-meter race. 33. Let x  the amount of 100% antifreeze to be added.

25% of Antifreeze 100% Antifreeze 40% Antifreeze

t d  rt t 80t t  2 48(t  2)

96

 32

t3 The distance from Ironton to Wildwood is 80t  80(3) or 240 km. 35. R  [50  2000(C  A)  8000(T  A) 10,000(I  A)  100(Y  A)]  24  [50  2000(297  474)  8000(33  474) 10,000(16  474)  100(3937  474)]  24  98.0 Daunte Culpepper’s rating was about 98.0. 36. Sample answer: How many grams of salt must be added to 40 grams of a 28% salt solution to obtain a 40% salt solution? 37. A weighted average is used to determine a skater’s average. Answers should include the following. • The score of the short program is added to twice the score of the long program. The sum is divided by 3.

d  rt 8.2t 8(t  1)

distance traveled by opponent. 1444244 43

Amount of Solution 16  x x 16

r 80 48

Distance traveled distance traveled by express train equals by local train. 144424443 123 144 424 443 80t  48(t  2) 80t  48(t  2) 80t  48t  96 80t  48t  48t  96  48t 32t  96

300 20

Sprinter Opponent

2.4

 0.75

x  3.2 3.2 quarts of pure antifreeze must replace 25% antifreeze solution. 34. Let t  the time the express train travels.

d  rt 90t 70t

t t t

Distance traveled by cheetah

0.75x 0.75

4.9(1)  5.2(2) 1  2



Amount of Antifreeze 0.25(16  x) 1.00x 0.40(16)

 5.1

38. D; Amount Amount at 4.5% d Amount at 6% 6000  d

Interest Amount of Rate Interest 4.5% 0.045d 6% 0.06(6000  d)

39. C;

Amount of amount of amount of antifreeze antifreeze antifreeze in 25% in 100% in 40% solution plus solution equals solution. 14243 123 14243 123 14243 0.25(16  x)  1.00x  0.40(16)

r

d t 616

 16  2 

616 14

 44 mph

113

Chapter 3

Page 177 40.

46. (2b)(3a)  2(3)ab  6ab 47. 3x(3y)  (6x)(2y)  3(3)xy  (6)(2)xy  9xy  12xy  (9  12)xy  3xy 48. 5s(6t)  2s(8t)  5(6)st  2(8)st  30st  (16)st  30st  16st  (30  16)st  46st 49. The bold arrow on the left means that the graph continues indefinitely in that direction. The coordinates are {. . ., 2, 1, 0, 1, 2, 3}. 50. The dots indicate each point on the graph. The coordinates are {0, 2, 5, 6, 8}.

Maintain Your Skills

3t  4  6t  s 3t  4  6t  6t  s  6t 3t  4  s 3t  4  4  s  4 3t  s  4 3t 3



t a6

41.

4(a  6) 

s  4 3 s  4 s  4 or 3 3 b  1 4 b  1 4 4

1

2

4a  24  b  1 4a  24  1  b  1  1 4a  25  b 42. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 25  14  11 Find the percent using the original number, 25, as the base. 11 25

Page 178 1.

r

 100

16.95(56)  12.59(97)  10.75(124)  10.15(71)  11.25(69)  9.95(45) 56  97  124  71  69  45 949.20  1221.23  1333  720.65  776.25  447.75  56  97  124  71  69  45 5448.08  462

11(100)  25(r) 1100  25r 1100 25



25r 25

 11.7923810 The average price was about $11.79.

44  r The percent of decrease is 44%. 43. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 42  35  7 Find the percent using the original number, 35, as the base. 7 35

2.



r

18.65(56)  13.85(97)  11.83(124)  11.17(71)  12.38(69)  10.95(45) 56  97  124  71  69  45

35r 35

20  r The percent of increase is 20%. 44. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 300  244  56 Find the percent using the original number, 244, as the base. 56 244



5994.81 462

4. 16.95(50)  12.59(50)  10.75(50)  10.15(50)  11.25(50)  9.95(50) 50  50  50  50  50  50 847.50  629.50  537.50  507.50  562.50  497.50  50  50  50  50  50  50 3582  300

r

 11.94

The weighted average is $11.94.

244r 244

16.95  12.59  10.75  10.15  11.25  9.95 6

23  r The percent of increase is about 23%.



71.64 6

 11.94 The average of the per-pound coffee prices is $11.94. The average of the prices per pound is the same as the weighted average if the same number of pounds of each type are sold. This is because each price is multiplied by the same weight, and then that weight is divided out.

2

45. The probability that the event will occur is 3, so 1 the probability that it will not occur is 3. 2 1 odds of event occurring  3:3 or 2:1 The odds that the event will occur are 2:1.

Chapter 3

1044.40  1343.45  1466.92  793.07  854.22  492.75 56  97  124  71  69  45

 12.9757792

 100





The weighted average increased by 10% to about $12.98.

56(100)  244(r) 5600  244r 5600 244

1005.20  1318.23  1457  791.65  845.25  492.75 56  97  124  71  69  45 5910.08 462

 12.7923809 The weighted average increases by $1.00 to about $12.79. 3. Multiply each price by 1.1.

 100



17.95(56)  13.59(97)  11.75(124)  11.15(71)  12.25(69)  10.95(45) 56  97  124  71  69  45



7(100)  35(r) 700  35r 700 35

Spreadsheet Investigation (Follow-Up of Lesson 3–9)

114

Chapter 3 Study Guide and Review Page 179 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

23.

Addition ratio different explore identity 3x  2 increase Dimensional analysis weighted average 8 and 9

25.

27.

29.

4   z 3z The equation is 4  3z  z  2. 13. The sum of the square of a and the cube of b is 16. 1444442444443 { { a2  b3  16



7

12r 2

12 4

4 4



4v 4

1v c 4

30. c 4

 8  8  42  8

 51

1 2  3(51)

 8  42

4

c 4 c 4

 34

1 2  4(34) c  136

7

2  7(7)

4d  5  49 4d  5  5  49  5 4d  44

2

32. 

{

8

r

{



44 4

1

7n  (1) 8 7n  (1) 8

8

2  8(8)

7n  (1)  64 7n  (1)  1  64  1 7n  65

Sixteen minus the product of 9 and a number r is equal to r.

7n 7



n 33.

65 7 65 7

2

or 97

n  2  4  2n n  2  2n  4  2n  2n 3n  2  4 3n  2  2  4  2 3n  6 3n 3

6

3

n2 34. 3t  2(t  3)  t 3t  2t  6  t t6t t6ttt 6  0 Since 6  0 is a false statement, this equation has no solution.

3

r  2  4 3

1

1

or 833

10  r 6  4v  2 6  2  4v  2 4  4v

y  153

31.

250 3

d  11

{

1



 6  6  45  6

4d  5 7 4d  5 7

5

y r

28.

 6  45 y 3 y 3

2

3

2(5)  2

a  10 4p  7  5 4p  7  7  5  7 4p  12

y 3

1

52

26.

1 2  25(25)

3

Exercises 15–38 For checks, see students’ work. 15. r  21  37 r  21  21  37  21 r  16 16. 14  c  5 14  c  14  5  14 c  19 17. 27  6  p 27  6  6  p  6 21  p 18. b  (14)  6 b  (14)  14  6  14 b  20 19. d  (1.2)  7.3 d  1.2  7.3 d  1.2  1.2  7.3  1.2 d  8.5

1 12

 25

4d 4

The equation is a2  b3  16. 14. 16  9r

1 12

5

3

5y  50

3 5y  3(50)

p3

11. Three times a number n decreased by 21 is 57. 1442443 1442443 { { {  21  57 3n The equation is 3n  21  57. 12. three is equal Four minus times to z decreased by 2. 123 14243 14243z 14243 { 144424443 {

20.

5 a 2 2 5 a 5 2

Lesson-by-Lesson Review

{

24.

1 2  43(30)

4p 4

y 3

{

 30

n  40

Vocabulary and Concept Check

Pages 179–184

3 n 4 4 3 n 3 4

1

r  2  2  4  2 21. 6x  42 6x 6



42 6

x  7

1

r  4 22. 7w  49 7w 7



49 7

w7

115

Chapter 3

5

1

1

1

3  6y  2  6y

35.

5

3y2 3y323 y  1

6s 6

1

 1

6

1

x  2 6 x  2 6

x

2

2  612x 2

x  2  3x x  2  x  3x  x 2  2x 2 2



8 40

2x 2

38.

800 40

3

b  3 8.3h  2.2  6.1h  8.8 8.3h  2.2  6.1h  6.1h  8.8  6.1h 2.2h  2.2  8.8 2.2h  2.2  2.2  8.8  2.2 2.2h  6.6 

6 15

n

 45

6(45)  15(n) 270  15n 270 15



3800 50

55x 55

18  n 41.

12 d



42.

20d 20

9d 2 3

43.





14 20

385 55



21 m

14(m)  20(21) 14m  420 14m 14



2.1 35



m  30

210 35

b  5 9



50r 50

r

 100



35r 35

6r The percent of increase is 6%. 48. The tax is 6.25% of the price of the book. 6.25% of $14.95  0.0625  14.95  0.934375 Round $0.934375 to $0.93 since tax is always rounded to the nearest cent. Add this amount to the original price. $14.95 + $0.93 = $15.88 The total price of the book is $15.88.

3b 3

1b

Chapter 3

r

 100

2.1(100)  35(r) 210  35r

420 14

2(9)  3(b  5) 18  3b  15 18  15  3b  15  15 3  3b 3 3

40r 40

76  r The percent of increase is 76%. 47. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 37.1  35  2.1 Find the percent using the original number, 35, as the base.

35

 55

x7

20

 15

12(15)  d(20) 180  20d 180 20

x 11

x (55)  11(35) 55x  385

15n 15



38(100)  50(r) 3800  50r

6.6 2.2

40.

r

 100

38 50

h  3 39.

96 6

20  r The percent of decrease is 20%. 46. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 88  50  38 Find the percent using the original number, 50, as the base.

 1

2.2h 2.2



8(100)  40(r) 800  40r

1  x 37. 2(b  3)  3(b  1) 2b  6  3b  3 2b  6  3b  3b  3  3b b  6  3 b  6  6  3  6 b  3 b 1

9  4

s  16 45. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 40  32  8 Find the percent using the original number, 40, as the base.

y1 36.

s

6(s  4)  8(9) 6s  24  72 6s  24  24  72  24 6s  96

1

3  6y  6y  2  6y  6y

y 1

6 8

44.

116

55. Let r  the speed of the slower airplane.

49. The discount is 20% of the original price. 20% of $12.99  0.20  12.99  2.598 Subtract $2.60 from the original price. $12.99  $2.60  $10.39 The discounted price of the T-shirt is $10.39. 50. 5x  y 5x 5

Slower Airplane Faster Airplane

r132  1r  8023  2940 3r  3r  240  2940 6r  240  2940 6r  240  240  2940  240 6r  2700

y

5 y

x5

6r 6

y

The value of x is 5. 51. ay  b  c ay  b  b  c  b ay  c  b 



1. 2. 3. 4.

number every greater than The sum of twice x and three times y 14243 is equal to 1 thirteen. 144424443 4243 2x  3y  13 The equation is 2x  3y  13. 5. Two thirds of a number is negative eight-fifths. 14243 { 14243 { 144 4424 4443 2 8  n   3 5

(c  y)x c  y

x x a

2y  a 3



a  3b 4

2

The value

15  k  8 15  k  15  8  15 k  23 15  k  8 Check: ? 15  23  8 88✓ The solution is 23. 7. 1.2x  7.2 6.

7a  9b 8 7a  9b y 8 7a  9b of y is . 8



54. Let x  the number of pounds of the $7.28 coffee to be added. Units (lb) $8.40 Coffee 9 $7.28 Coffee x $7.95 Coffee 9  x

Price per Unit (lb) $8.40 $7.28 $7.95

1.2x 1.2

7.2

 1.2

x  6 1.2x  7.2 Check: ? 1.2(6)  7.2 7.2  7.2 ✓ The solution is 6. 8. k  16  21 k  16  16  21  16 k  5 k  16  21 Check: ? 5  16  21 21  21 ✓ The solution is 5.

Total Price 8.40(9) 7.28x 7.95(9  x)

8.40(9)  7.28x  7.95(9  x) 75.60  7.28x  71.55  7.95x 75.60  7.28x  7.95x  71.55  7.95x  7.95x 75.60  0.67x  71.55 75.60  0.67x  75.60  71.55  75.60 0.67x  4.05 0.67x 0.67

8

The equation is 3 n  5.

4(2y  a)  3(a  3b) 8y  4a  3a  9b 8y  4a  4a  3a  9b  4a 8y  7a  9b 8y 8

2700 6

Page 185

The value of x is y  c. Since division by 0 is undefined, y  c  0 or y  c. 53.

d  rt r(3) 1r  8023

Chapter 3 Practice Test

The Since division by 0 is undefined, a  0. 52. yx  a  cx yx  a  yx  cx  yx a  cx  yx a  (c  y)x a c  y a c  y a y  c



t 3 3

r  450 r  80  450  80 or 530 The speed of the slower plane was 450 mph, and the speed of the faster plane was 530 mph.

c  b a c  b y a c  b value of y is a . ay a

r r r  80

4.05

 0.67

x  6.04477612 About 6 lb of $7.28 coffee should be mixed with 9 lb of $8.40 coffee.

117

Chapter 3

9. 4

1

t  7 4 t  7 4

14. 5a  125

 11

2  4(11)

5a 5

t  7  4 51  7 ?  4 44 ?  4

5a  125 ? 5(25)  125 125  125 ✓ The solution is 25.

11 11 11

r 5

2r 5 2r 5

3

3

2r 5



 16  16 

r

r

5  3  3  16  3

y  36

5 5  5(19)

1 2  43(27)

r

5  19

3 y 4

1 r2

r  95

 27

? 3 (36)  4

12  7  12  7  7  19 

3(19)  3 57  y Check:

y 3 y 3

0.1r 0.1

1 2 ?

0.1r  19 ?

0.1(190)  19

y 3 57 3

19  19 ✓ The solution is 190.

?

12  7  19

17.

12  12 ✓ The solution is 57. 12. t  (3.4)  5.3 t  (3.4)  (3.4)  5.3  (3.4) t  8.7 t  (3.4)  5.3 Check: ? 8.7  (3.4)  5.3 5.3  5.3 ✓ The solution is 8.7. 13. 3(x  5)  8x  18 3x  15  8x  18 3x  15  8x  8x  18  8x 11x  15  18 11x  15  15  18  15 11x  33 

19

 0.1

r  190 Check:

12  7 

3

3

1

2

4

3z  9 2

2

3

1 42

2 3z  2 9 2

z3

2

The solution

12

2 2 ?  3 4 9  2 is 3.

3

18.

4

3z  9

Check:

4

9 4

9 ✓

w  11  4.6 w  11  11  4.6  11 w  6.4 w 1



6.4 1

w  6.4 w  11  4.6 ? 6.4  11  4.6 4.6  4.6 ✓ The solution is 6.4. 19. 2p  1  5p  11 2p  1  5p  5p  11  5p 3p  1  11 3p  1  1  11  1 3p  12 Check:

33 11

x  3 3(x  5)  8x  18 Check: ? 3(3  5)  8(3)  18 ? 3(2)  24  18 6  6 ✓ The solution is 3.

3p 3



12 3

p4 Chapter 3

16

19  3  38  16 22  22 ✓ The solution is 95. 16. 0.1r  19

7

12  7 

2r  16 5 ? 2(95)  5 

3 ?

y 3

11x 11

r 5 95 5

Check:

27

27  27 ✓ The solution is 36.

y 3

2r 5

5  3  16

 27

3 y 4 4 3 y 3 4

Check:

11.

r 5

15.

11  11 ✓ The solution is 51. 10.

125 5

a  25 Check:

t  7  44 t  7  7  44  7 t  51 Check:



118

2p  1  5p  11 ? 2(4)  1  5(4)  11 ? 8  1  20  11 99✓ The solution is 4. 20. 25  7w  46 25  7w  25  46  25 7w  21 Check:

7w 7

26.

h  0.25vt2 t h  0.25vt2 t

21

 7

9

 11

22.

36(11)  t(9) 396  9t 396 9



9t 9

52n 52

1

x

2.00x  2.50(30  x)  178.50 2x  75  2.5x  178.50 4.5x  75  178.50 4.5x  75  75  178.50  75 4.5x  103.50

10  1

5(x  1)  12(10) 5x  5  120 5x  5  5  120  5 5x  125 5x 5



4.5x 4.5

125 5



r 100

36(100)  45(r) 3600  45r 3600 45



t t

The River Rover

10

t2

1

2

1

1

1

10 t  2

2

8t  10t  5 8t  10t  10t  5  10t 2t  5 2t 2

5

 2 5

t  2 or 2.5 The River Rover overtakes The Yankee Clipper 2.5 hours after 9:00 A.M. or at 11:30 A.M. 30. B;

r

 100



d  rt 8t

r 8

1

8(100)  12(r) 800  12r 800 12

103.50 4.5

The Yankee Clipper

8t  10 t  2

45r 45

80  r The percent of decrease is 80%. 25. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 20  12  8 Find the percent using the original number, 12, as the base. 8 12



x  23 30  x  30  23 or 53 23 cups of espresso were sold, and 53 cups of cappuccino were sold. 29. Let t  the number of hours the Yankee Clipper traveled.

x  25 24. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 45  9  36 Find the percent using the original number, 45, as the base. 36 45



Cups Price per Cup Total Price Espresso x $2.00 2.00x Cappuccino 30  x $2.50 2.50(30  x)

13

 52

n  4 or 0.25

5 12

Since division by 0 is

The Since division by 0 is undefined, a  0. 28. Let x  the number of cups of espresso sold.

3.25 52

n(52)  4(3.25) 52n  13

44  t 23.



a

b  a a b  a y a b  a value of y is a . ay a

n 4

at t

h  0.25vt2 . t

w  3 25  7w  46 ? 25  7(3)  46 ? 25  (21)  46 46  46 ✓ The solution is 3. 36 t



The value of a is undefined, t  0. 27. a(y  1)  b ay  a  b ay  a  a  b  a ay  b  a

Check:

21.

h  at  0.25vt2 h  0.25vt2  at  0.25vt2  0.25vt2 h  0.25vt2  at

4 5

20

3

2

x

4  54 12 20 12 20

2x

 20

1 2  201202x 2 12  2x

12r 12

12 2

67  r The percent of increase is about 67%.



2x 2

6x

119

Chapter 3

9

Chapter 3 Standardized Test Practice

10. F  5C  32 9

 5(5)  32  9  32  23 The temperature is 23F. 11. There are five pairs of black socks, and there are 3 pairs of socks that are not black. 5 odds of black  3

Pages 186–187 1. C; P  2/  2w  2(15)  2(6)  30  12  42 2. C; 65 percent is less than 100 percent, and 100 percent equals 1. Therefore, 65 percent of 20 is less than 1 times 20 or 20. 3. B; From the bar graph we see that the plant is about 4 cm tall at the end of the fifth week and growing about 1 cm a week. Therefore, at the end of the sixth week the plant’s height should be about 4  1 or 5 cm. 4. D; Write each probability as a decimal. 25%  0.25 1 in 4  0.25 1 5

The odds of choosing a black pair are 5:3. 12. Let n  age of the youngest sister. Then n  1  age of the next oldest sister, and n  2  age of the oldest sister. The sum of the ages of the three sisters { is 39. 1444444442444444443 { n  (n  1)  (n  2) n  (n  1)  (n  2)  39 3n  3  39 3n  3  3  39  3 3n  36

 0.20

3n 3

0.3  0.30 0.20 6 0.25 6 0.30 Therefore, 0.3 is the greatest probability. 5. A; Profit  Revenue minus costs. p  24.95n  (0.8n  575) p  24.95n  0.8n  575 p  (24.95  0.8)n  575 6. D; 8(x  2)  12 8x  16  12 1 (8x 4

Then Tyson drove

1

1

3 x 2



x 2

36  x 1

8. D; The sum of x and y is 144424443 { 1 xy  x x

1 y



{

0

0 0

1 y

1

x(xy  5) 4

    

Chapter 3

 500  435

 65 

290 2

 65

 145  65  210 Tyson drove 210 miles. 14. 7(x  2)  4(2x  3)  47 7x  14  8x  12  47 15x  2  47 15x  2  2  47  2 15x  45

0.

x  y 9.

 500

x  290

6x 6

1 y 1 y

 65 miles.

1 2  23(435)

8(27)  6(x) 216  6x 216 6

36 3

 65  65  500  65 3 x 2 2 3 x 3 2

x 27

x 2

2  500

x x  2  65 x x  2  65 3 x  65 2

 16)  4(12)



39

n  12 n  1  12  1 or 13 n  2  12  2  14 The middle sister is 13 years old. 13. Let x  the number of miles Pete drove.

2x  4  3 7. C; Let x  the amount needed for 27 servings. 8 6





15x 15

2[2(3)  5] 4 2[6  5] 4 2[1] 4 2 4 1 2 or 0.5

45

 15

x3 The solution is 3. 15. The discount is 45% of the original price. 45% of $22.95  0.45  22.95  10.3275 $22.95  $10.33  $12.62 The discounted price of the book is $12.62. 16. C; a and a are the same number of units from zero, but in opposite directions. Therefore, 0a 0  0a 0 .

120

17. B; 3x  7  10 3x  7  7  10  7 3x  3 3x 3

Kirby’s pickup travels at 36 mph. 15 mi 25 min

15(60)  25(x) 900  25x

3

3

x1 4y  2  6 4y  2  2  6  2 4y  8 4y 4

900 25

8

4

r

 100



75r 75

0.20(200  x)

r



0.80x



0.50(200)

The equation is 0.20(200  x)  0.80x  0.50(200). 20c. 0.20(200  x)  0.80x  0.50(200) 40  0.20x  0.80x  100 40  0.60x  100 40  0.60x  40  100  40 0.60x  60

r

 100

0.60x 0.60

50(100)  150(r) 5000  150r 

Liters of Acid 0.20(200  x) 0.80x 0.50(200)

20b.

1

5000 150 1 33 3

Liters of Solution 200  x x 200

The amount of acid the amount of acid the amount of acid in the 20% solution 1 plus in the 80% solution 123 equals 144424443 in the 50% solution 144424443 23 144424443

The rate of increase from 75 to 100 is 33 3 %. Find the amount of change. 200  150  50 Find the percent using the original number, 150, as the base. 50 150

25x 25

20% Solution 80% Solution 50% Solution

25(100)  75(r) 2500  75r 2500 75 1 33 3



36  x Nola’s SUV travels at 36 mph. Both are traveling at 36 mph, and therefore, neither is exceeding the speed limit. 20a.

y2 18. C; Find the amount of change. 100  75  25 Find the percent using the original number, 75, as the base. 25 75

x mi

 60 min

60

 0.60

x  100 200  x  200  100 or 100 100 L of the 20% solution and 100 L of the 80% solution are needed.

150r 150

r 1

The rate of increase from 150 to 200 is 33 3 %. 19. Calculate the miles per hour rate for each vehicle. 6 mi 10 min

x mi

 60 min

6(60)  10(x) 360  10x 360 10



10x 10

36  x

121

Chapter 3

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Chapter 4

Graphing Relations and Functions

Page 191

16. 2c  b  2(3)  (4)  6  4  10 17. c  3a  (3)  3(1)  3  (3)  3  3 0 18. 3a  6b  2c  3(1)  6(4)  2(3)  3  24  (6)  27  6  21

Getting Started

1. 1 0

1

2

3

4

5

6

7

8

9

5 4 3 2 1

0

1

2

3

4

5

8 7 6 5 4 3 2 1

0

1

2

2. 3. 4. 2 1 2 11 1  1 0 2

2

1 2

2

1 11 2 21 2

2

5. 3(7  t)  3  7  3  t  21  3t 6. 4(w  2)  (4)(w)  14)(2)  4w  (8)  4w  8 7. 5(3b  2)  (5)(3b)  (5)(2)  15b  (10)  15b  10 8.

1 (2z 2

1 2

1

1

19. 8a  2b  3c  8(1)  2 (4)  3(3)  8  2  (9)  6  9 3 2

2

20. 6a  8b  3c  6(1)  8(4)  3 (3)  6  32  (2)  26  2  24

1 2

 4)   2z   4

z2 2x  y  1 2x  y  2x  1  2x y  1  2x 10. x8y xy8yy xy8 xyx8x y8x 6x  3y  12 11. 6x  3y  6x  12  6x 3y  12  6x 9.

Page 194

Check for Understanding

1. Draw two perpendicular number lines. Label the horizontal line x since this is the x-axis, and label the vertical line y since this is the y-axis. Label the point where the two axes meet 0 since this is the origin. The axes divide the coordinate plane into four regions: the upper right region is quadrant I, the upper left region is quadrant II, the lower left region is quadrant III, and the lower right region is quadrant IV.

13 (3y)  13 (12  6x) y  4  2x or 2x  4 2x  3y  9 12. 2x  3y  2x  9  2x 3y  9  2x 1 (3y) 3

The Coordinate Plane

4-1

y

II

y  3  23x or 23x  3 9

13. 9

1 y 2

1

1 y 2

3

1

 4x

III

IV

 9  4x  9 1

2 y  4x  9 1

2

2. To graph (1, 4), move 1 unit left from the origin and 4 units up. This point is in quadrant II. To graph (4, 1), move 4 units right from the origin and 1 unit down. This point is in quadrant IV. 3. Sample answer: (3, 3) is in quadrant I since both coordinates are positive; (3, 3) is in quadrant II since the x-coordinate is negative and the y-coordinate is positive; (3, 3) is in quadrant III since both coordinates are negative; (3, 3) is in quadrant IV since the x-coordinate is positive and the y-coordinate is negative.

y  18  8x

y  5 3 y  5 3

x2

2  3(x  2)

y  5  3x  6 y  5  5  3x  6  5 y  3x  1 15. a  b  c  (1)  (4)  (3)  1  4  3 33 6

Chapter 4

x

O

2 2 y  2(4x  9) 14.

I

 13 (9  2x)

122

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11. M(2, 2) • Start at the origin. • Move left 2 units since the x-coordinate is 2. • Move down 2 units since the y-coordinate is 2. • Draw a dot and label it M. 8–11. y J K

4. • Follow along a vertical line through point E to find the x-coordinate on the x-axis. The x-coordinate is 2. • Follow along a horizontal line through point E to find the y-coordinate on the y-axis. The y-coordinate is 5. • So, the ordered pair for point E is (2, 5). • Since both coordinates are negative, point E is located in quadrant III. 5. • Follow along a vertical line through point F to find the x-coordinate on the x-axis. The x-coordinate is 1. • Follow along a horizontal line through point F to find the y-coordinate on the y-axis. The y-coordinate is 1. • So, the ordered pair for point F is (1, 1). • Since the x-coordinate is negative and the y-coordinate is positive, point F is located in quadrant II. 6. • Follow along a vertical line through point G to find the x-coordinate on the x-axis. The x-coordinate is 4. • Follow along a horizontal line through point G to find the y-coordinate on the y-axis. The y-coordinate is 4. • So, the ordered pair for point G is (4, 4). • Since both coordinates are positive, point G is located in quadrant I. 7. • Follow along a vertical line through point H to find the x-coordinate on the x-axis. The x-coordinate is 4. • Follow along a horizontal line through point H to find the y-coordinate on the y-axis. The y-coordinate is 2. • So, the ordered pair for point H is (4, 2). • Since both coordinates are negative, point H is located in quadrant III. 8. J(2, 5) • Start at the origin. • Move right 2 units since the x-coordinate is 2. • Move up 5 units since the y-coordinate is 5. • Draw a dot and label it J. (See coordinate plane after Exercise 11.) 9. K(1, 4) • Start at the origin. • Move left 1 unit since the x-coordinate is 1. • Move up 4 units since the y-coordinate is 4. • Draw a dot and label it K. (See coordinate plane after Exercise 11.) 10. L(0, 3) • Start at the origin. • Since the x-coordinate is 0, the point will be located on the y-axis. • Move down 3 units since the y-coordinate is 3. • Draw a dot and label it L. (See coordinate plane after Exercise 11.)

x

O

M L

12. Sketch the given figure on the coordinate plane by plotting point A at (40, 10). From point A, move 40 units right and label point B. From point B, move up 10 units and label point C. From point C, move left 20 units and label point D. From point A, move up 30 units and label point E. y

E

C D A

B

40 30 20 10

–20 –15–10–10 O10 20 30 40 x –10 –20 –30 –40

Point B has coordinates (0, 10), C(0, 20), D(20, 20), and E(40, 40).

Pages 195–196

Practice and Apply

xCoordiPoint nate 13. N 4 14. P 5 15. Q 1 16. R 5 17. S 3 18. T 2 19. U 2 20. V 4 21. W 0 22. Z 3

yCoordinate 5 3 3 2 3 0 1 2 4 3

Ordered Pair (4, 5) (5, 3) (1, 3) (5, 2) (3, 3) (2, 0) (2, 1) (4, 2) (0, 4) (3, 3)

Quadrant II IV III I II none IV III none I

23. A point 12 units down from the origin has y-coordinate 12. A point 7 units to the right of the origin has x-coordinate 7. So, the ordered pair is (7, 12).

123

Chapter 4

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24. A point 9 units to the left of the origin has x-coordinate 9. A point that lies on the x-axis has y-coordinate 0. So, the ordered pair is (9, 0). 25. A(3, 5) • Start at the origin. • Move right 3 units and up 5 units. • Draw a dot and label it A. (See coordinate plane after Exercise 36.) 26. B(2, 2) • Start at the origin. • Move left 2 units and up 2 units. • Draw a dot and label it B. (See coordinate plane after Exercise 36.) 27. C(4, 2) • Start at the origin. • Move right 4 units and down 2 units. • Draw a dot and label it C. (See coordinate plane after Exercise 36.) 28. D(0, 1) • Start at the origin. • Since the x-coordinate is 0, the point is on the y-axis. • Move down 1 unit. • Draw a dot and label it D. (See coordinate plane after Exercise 36.) 29. E(2, 5) • Start at the origin. • Move left 2 units and up 5 units. • Draw a dot and label it E. (See coordinate plane after Exercise 36.) 30. F(3, 4) • Start at the origin. • Move left 3 units and down 4 units. • Draw a dot and label it F. (See coordinate plane after Exercise 36.) 31. G(4, 4) • Start at the origin. • Move right 4 units and up 4 units. • Draw a dot and label it G. (See coordinate plane after Exercise 36.) 32. H(4, 4) • Start at the origin. • Move left 4 units and up 4 units. • Draw a dot and label it H. (See coordinate plane after Exercise 36.) 33. I(3, 1) • Start at the origin. • Move right 3 units and up 1 unit. • Draw a dot and label it I. (See coordinate plane after Exercise 36.) 34. J(1, 3) • Start at the origin. • Move left 1 unit and down 3 units. • Draw a dot and label it J. (See coordinate plane after Exercise 36.) Chapter 4

35. K(4, 0) • Start at the origin. • Move left 4 units. • Since the y-coordinate is 0, the point is on the x-axis. • Draw a dot and label it K. (See coordinate plane after Exercise 36.) 36. L(2, 4) • Start at the origin. • Move right 2 units and down 4 units. • Draw a dot and label it L. 25–36. y E A H G

B I K x

O

D C J F

L

37. Latitude lines run east and west. Sample answer: Louisville and Richmond. 38. Longitude lines run north and south. Sample answer: Austin and Oklahoma City. 39. xyOrdered Artifact Coordinate Coordinate Coins 3 5 Plate 7 2 Goblet 8 4 Vase 5 9

Pair (3, 5) (7, 2) (8, 4) (5, 9)

40. The Shapiro Undergraduate Library is in column C and in row 5, so is located in sector C5. 41. The Natural Science, Chemistry, and Natural Resources and Environment Buildings are in column C and in row 4, so are located in sector C4. 42. E. Huron St. runs horizontally from sector A2 to sector D2. 43. The four sectors that have bus stops are B5, C2, D4, and E1. 44a. xy 7 0 indicates that the product of x and y is positive. This is true if both x and y are positive, placing (x, y) in quadrant I; it is also true if both x and y are negative, placing (x, y) in quadrant III. 44b. xy 7 0 indicates that the product of x and y is negative. This is true if x is positive and y is negative, placing (x, y) in quadrant IV; it is also true if x is negative and y is positive, placing (x, y) in quadrant II. 44c. xy 7 0 indicates that the product of x and y is 0. This is true if both x and y are 0, placing (x, y) at the origin; it is also true if x is 0, placing (x, y) on the y-axis; it is also true if y is 0, placing (x, y) on the x-axis.

124

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The distance traveled by each plane is the same. Write an equation. 7r  2240

45. Archaeologists use coordinate systems as mapping guides and as systems to record locations of artifacts. Answers should include the following. • The grid gives archaeologists points of reference so they can identify and explain to others the location of artifacts in a site they are excavating. You can divide the space so more people can work at the same time in different areas. • Knowing the exact location of artifacts helps archaeologists reconstruct historical events. 46. C Vertex B is located 3 units right and 2 units up from the origin. Since the rectangle is centered at the origin, vertex A must be located 3 units left and 2 units up from the origin at (3, 2).

7r 7

47. B Vertex A is 2 units above the x-axis and vertex D is 2 units below the x-axis. So, the length of AD is 2  2  4 units. 48. Let (a, b)  (7, 1) and (c, d )  (3, 1).

1a 2 c, b 2 d 2  1 2  1 42, 22 2

7  132 1  1 , 2

4c 4

2

1

21  1 14 , 10 2 2 2

5  9 2  (8) , 2 2

3h 3

2

55. 4

Maintain Your Skills

t

58. You can use a calculator to find an approximation for 1180. 1180  13.41640786 p Rounded to the nearest hundredth, 1180 is about 13.42. 59. 1256 represents the negative square root of 256.

8t So, the first airplane arrived in Baltimore after 8 h. Since the second airplane left Tucson 45 min after the first airplane, and is scheduled to land in Baltimore 15 min before the first airplane, the second airplane travels 60 min, or 1 hr, less than the first airplane. So, the second airplane will travel for 7 h. Make a table of the information. t 8 7

11t 11

163  7.937253933 p Rounded to the nearest hundredth, 163 is about 7.94.

280t 280

r 280 r



56. 181 represents the negative square root of 81. 81  92 S 181  9 57. You can use a calculator to find an approximation for 163.

51. Since the first airplane traveled 2240 mi at 280 mph, substitute d  2240 and r  280 into the formula d  rt. d  rt 2240  280t

First Airplane Second Airplane

 2t

4  4(2t)

3a 11 3a 11

 (0, 0)



3

3(a  t) 4 3(a  t) 4

b  6w 3 6w  b 3

3(a  t)  8t 3a  3t  8t 3a  3t  3t  8t  3t 3a  11t

1a 2 c, b 2 d 2  14 2 4, 4 2(4) 2 0 0  1 2, 2 2

2240 280



h

 (7, 5) 50. Let (a, b)  (4, 4) and (c, d)  (4, 4).

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4d

 4 cd 54. 6w  3h  b 6w  3h  6w  b  6w 3h  b  6w

 (2, 1) 49. Let (a, b)  (5, 2) and (c, d )  (9, 8). a  c b  d , 2 2

2240

 7 r  320 The second plane must travel at 320 mph to arrive on schedule. 52. 3x  b  2x  5 3x  b  b  2x  5  b 3x  2x  5  b 3x  2x  2x  5  b  2x x5b 53. 10c  2(2d  3c) 10c  2  2d  2  3c 10c  4d  6c 10c  6c  4d  6c  6c 4c  4d

256  162 S 1256  16 60. 52  018  7 0 61. 081  47 0  17  52  011 0  034 0  17  52  11  34  17  63  51 62. 42  060  74 0 63. 36  015  21 0  42  014 0  36  06 0  42  14  36  6  28  30

d  rt 2240 7r

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64. 010  16  27 0 65. 038  65  21 0  06  27 0  027  21 0  021 0  048 0  21  48 66. 4(x  y)  4x  4y 67. 1(x  3)  (1)(x)  (1)(3)  x  3 68. 3(1  6y)  3  1  3  6y  3  18y 69. 3(2x  5)  (3)(2x)  (3)(5)  6x  (15)  6x  15 1

1

5. To reflect the triangle over the x-axis, multiply the y-coordinate of each vertex by 1. (x, y) S (x, y) P(1, 2) S P¿(1, 2) Q(4, 4) S Q¿(4, 4) R(2, 3) S R¿(2, 3) y

Q R’ P

P’

2

 3x  2y 1

x

O

1

70. 3(2x  6y)  3(2x)  3(6y) 1

R 1

Q’

71. 4(5x  2y)  4(5x)  4(2y) 

4-2

5 x 4



1 y 2

6. To translate the quadrilateral 3 units up, add 3 to the y-coordinate of each vertex. (x, y) S (x, y  3) A(4, 2) S A¿(4, 2  3) S A¿(4, 5) B(4, 2) S B¿(4, 2  3) S B¿(4, 1) C(1, 3) S C¿(1, 3  3) S C¿(1, 0) D(3, 2) S D¿(3, 2  3) S D¿(3, 5)

Transformations on the Coordinate Plane

Pages 200–201

Check for Understanding

1. Transformation Reflection Rotation Translation Dilation

Size same same same changes

Shape same same same same

Orientation changes changes same same

y

D’

A’ A

D

B’ x

C’ O 2. Sample answer: The preimage is a square centered at the origin with a side 4 units long. The image after a dilation that is an enlargement 3 having a scale factor of 2 is the square centered at the origin with a side 6 units long. The image after a dilation that is a reduction having a scale 1 factor of 2 is the square centered at the origin with a side 2 units long.

1

2

2

B C

1

7. To dilate the parallelogram, multiply the coordinates of each vertex by 2. (x, y) S (2x, 2y) E(1, 4) S E¿(2  (1), 2  4) S E¿(2, 8) F(5, 1) S F¿(2  5, 2  (1)) S F¿(10, 2) G(2, 4) S G¿(2  2, 2  (4)) S G¿(4, 8) H(4, 1) S H¿(2  (4), 2  1) S H¿(8, 2)

y

y x

E’ E H’ H

2 2

x

F’

3. The figure has been shifted horizontally to the right. This is a translation. 4. The figure has been turned around a point. This is a rotation.

Chapter 4

F

G G’

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17b.

8. To rotate the triangle 90 counterclockwise about the origin, switch the coordinates of each vertex and then multiply the new first coordinate by 1. (x, y) S (y, x) J(0, 0) S J¿(0, 0) K(2, 5) S K¿(5, 2) L(4, 5) S L¿(5, 4)

y

T’

T R’

x

R

O

y

L

S’

S

J’ J

18a. To reflect the trapezoid over the x-axis, multiply the y-coordinate of each vertex by 1. (x, y) S (x, y) A(2, 3) S A¿(2, 3) B(5, 3) S B¿(5, 3) C(6, 1) S C¿(6, 1) D(2, 1) S D¿(2, 1) 18b. y

x

K’ L’ K 9. Draw a dot and label it A. Move 10 units left (10 mi west) and 7 units down (7 mi south). Draw a dot and label it B.

A

B

D

10 mi

x

O

A

C’

D’ B’

A’

7 mi

19a. To translate the quadrilateral 8 units right, add 8 to the x-coordinate of each vertex. (x, y) S (x  8, y) R(6, 3) S R¿(6  8, 3) S R¿(2, 3) S(4, 2) S S¿(4  8, 2) S S¿(4, 2) T(1, 5) S T¿(1  8, 5) S T¿(7, 5) U(3, 7) S U¿(3  8, 7) S U¿(5, 7) 19b. U U’

B 10. To translate the point A(x, y) 10 units left, add 10 to the x-coordinate of A. To translate A(x, y) 7 units down, add 7 to the y-coordinate of A. (x, y) S (x  10, y  7) The ship’s current location is represented by (x  10, y  7).

y

T’

T

R

Pages 201–203

C

Practice and Apply

R’ S

11. The figure has been shifted horizontally to the right. This is a translation. 12. The figure has been turned around a point. This is a rotation. 13. The figure has been flipped over a line. This is a reflection. 14. The figure has been increased in size. This is a dilation. 15. The figure has been flipped over a line. This is a reflection. 16. The figure has been shifted vertically down. This is a translation. 17a. To reflect the triangle over the y-axis, multiply the x-coordinate of each vertex by 1. (x, y) S (x, y) R(2, 0) S R¿(2, 0) S(2, 3) S S¿(2, 3) T(2, 3) S T¿(2, 3)

S’ O

x

20a. To translate the parallelogram 3 units right, add 3 to the x-coordinate of each vertex. To translate the parallelogram 2 units down, add 2 to the y-coordinate of each vertex. (x, y) S (x  3, y  2) M(6, 0) S M¿(6  3, 0  2) S M¿(3, 2) N(4, 3) S N¿(4  3, 3  2) S N¿(1, 1) O(1, 3) S O¿(1  3, 3  2) S O¿(2, 1) P(3, 0) S P¿(3  3, 0  2) S P¿(0, 2)

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20b.

23b.

y

y

G

O

N

N’

O’

H F

M

P

O

M’

P’

x x

O

F’ H’ G’

21a. To dilate the trapezoid by a scale factor 1 1 of 2, multiply the coordinates of each vertex by 2.

11

1

(x, y) S 2x, 2y

24a. To rotate the quadrilateral 90 counterclockwise about the origin, switch the coordinates of each vertex and then multiply the new first coordinate by 1. (x, y) S (y, x) T(4, 2) S T¿(2, 4) U(2, 4) S U¿(4, 2) V(0, 2) S V¿(2, 0) W(2, 4) S W¿(4, 2) 24b. y U

2

1 1 2 1 1 K(2, 4) S K¿ 1 2  (2), 2  4 2 S K¿(1, 2) 1 1 L(4, 4) S L¿ 1 2  4, 2  4 2 S L¿(2, 2) 1 1 M(4, 4) S M¿ 1 2  (4), 2  (4) 2 S M¿(2, 2) 1 2

J(4, 2) S J¿  (4), 2  2 S J¿(2, 1)

21b.

y

K J

L

V

T

K’

V’

J’

O

L’

x

U’

x

O

W’

M’

W T’

M 25a. To reflect the parallelogram over the y-axis, multiply the x-coordinate of each point by 1. To then rotate the result 180 about the origin, multiply both coordinates of the reflected point by 1. (x, y) S (x, y) S (x, y) (x, y) S (x, y) W(1, 2) S W¿(1, 2) X(3, 2) S X¿(3, 2) Y(0, 4) S Y¿(0, 4) Z(4, 4) S Z¿(4, 4) 25b. y

22a. To dilate the square by a scale factor of 3, multiply the coordinates of each vertex by 3. (x, y) S (3x, 3y) A(2, 1) S A¿(3  (2), 3  1) S A¿(6, 3) B(2, 2) S B¿(3  2, 3  2) S B¿(6, 6) C(3, 2) S C¿(3  3, 3  (2)) S C¿(9, 6) D(1, 3) S D¿(3  (1), 3  (3)) S D¿(3, 9) 22b. 8

A’

y

B’

4

B

A 8

4

O

D

4

D’

8

4

C

8x

W

C’

X x

O

X’

W’

23a. To rotate the triangle 180 about the origin, multiply both coordinates of each vertex by 1. (x, y) S (x, y) F(3, 2) S F¿(3, 2) G(2, 5) S G¿(2, 5) H(6, 3) S H¿(6, 3)

Chapter 4

Y’

Z’

Z

128

Y

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31. The x-coordinate of each vertex of the dilated 1 triangle is 2 of the x-coordinate of the corresponding preimage vertex. The y-coordinate 1 of each vertex of the dilated triangle is 2 of the y-coordinate of the corresponding preimage vertex. So, triangle QRS was dilated by a scale 1 factor of 2. 32. The x-coordinate of each vertex of the image is 1 times the x-coordinate of the corresponding preimage vertex. The y-coordinate of each vertex of the image is equal to the y-coordinate of the corresponding preimage vertex. So, parallelogram WXYZ was reflected over the y-axis. 33. To x-coordinate of each vertex of the image is 1 times the y-coordinate of the corresponding preimage vertex. The y-coordinate of each vertex of the image is equal to the x-coordinate of the corresponding preimage vertex. That is, for each ordered pair, the coordinates were switched and the new first coordinate was multiplied by 1. So, triangle XYZ was rotated 90 counterclockwise about the origin.

26a. To reflect the pentagon over the x-axis, multiply the y-coordinate of each vertex by 1. To then translated the result as described, add 2 to the x-coordinate and 1 to the y-coordinate of each reflected vertex. (x, y) S (x, y) S (x  2, y  1) (x, y) S (x  2, y  1) P(0, 5) S P¿(0  2, 5  1) S P¿(2, 4) Q(3, 4) S Q¿(3  2, 4  1) S Q¿(1, 3) R(2, 1) S R¿(2  2, 1  1) S R¿(0, 0) S(2, 1) S S¿(2  2, 1  1) S S¿(4, 0) T(3, 4) S T¿(3  2, 4  1) S T¿(5, 3) 26b. y P T Q

S

R

O

S’

x

R’

T’

Q’ P’

1

34. Multiply each dimension by 22. 1

1800  22  4500

27. A(5, 1), B(3, 3), C(5, 5), D(5, 4), E(8, 4), F(8, 2), and G(5, 2) are the vertices of the arrow. 28. To translate the arrow 2 units right, add 2 to the x-coordinate of each vertex. To reflect the translated arrow across the x-axis, multiply the y-coordinate of each vertex by 1. (x, y) S (x  2, y) A(5, 1) S A¿(3, 1) B(3, 3) S B¿(1, 3) C(5, 5) S C¿(3, 5) D(5, 4) S D¿(3, 4) E(8, 4) S E¿(6, 4) F(8, 2) S F¿(6, 2) G(5, 2) S G¿(3, 2) 29. y C’

E’

1

1600  22  4000 The new digital photograph will be 4500 pixels wide by 4000 pixels high. 35–36. 4800 (0, 4000)

3200 2400 (0, 1600)

800 (0, 0) 0

B’ A’

F

O

x

G B

E

(1800, 0)

(4500, 0)

800 1600 2400 3200 4000 4800

35. The other three vertices of the 1800  1600 digital photograph are (0, 1600), (1800, 1600), and (1800, 0). 36. The vertices of the enlarged 4500  4000 photograph are (0, 0), (0, 4000), (4500, 4000), and (4500, 0). 37.

G’ A

(1800, 1600)

1600

D’

F’

(4500, 4000)

4000

D C

30. The x-coordinate of each vertex of the translated trapezoid is 3 more than the x-coordinate of the corresponding preimage vertex. The y-coordinate of each vertex of the translated trapezoid is 2 less than the y-coordinate of the corresponding preimage vertex. So, trapezoid JKLM was translated 3 units right and 2 units down. Sample answer: The pattern resembles a snowflake.

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45.

38. Yes, the same pattern could be drawn using translations since all octagons in the figure have the same size, shape, and orientation. 39. To rotate a point 90 clockwise about the origin, we switch the coordinates of the point and multiply the new second coordinate by 1. (x, y) S ( y, x) 40. To reflect the point A(x, y) over the x-axis, we multiply the y-coordinate by 1. Then, to reflect this image over the y-axis, we multiply the x-coordinate by 1. So, the final image is A¿(x, y). To rotate a point 180 about the origin, we multiply both coordinates by 1. So, A(x, y) S A¿(x, y). Since the final image in each case is the same, the statement is always true. 41. Artists use computer graphics to simulate movement, change the size of objects, and create designs. • Objects can appear to move by using a series of translations. Moving forward can be simulated by enlarging objects using dilations so they appear to be getting closer. • Computer graphics are used in special effects in movies, animated cartoons, and web design. 42. C; For the translations described, add 1 to each x-coordinate and add 3 to each y-coordinate. S(4, 2) S S¿(4  1, 2  3) S S¿(5, 1) 43. C 2

y3

46.

R x yx

The coordinates of the vertices of the image are R¿(3, 3), S¿(0, 4), and T¿(4, 1).

Page 203

84  z y

A’ B’ D’ x C’

x0

The coordinates of the vertices of the image are A¿(3, 4), B¿(2, 2), C¿(3, 2), and D¿(4, 0).

Chapter 4

Maintain Your Skills

47. A(2, 1) • Start at the origin. • Move right 2 units and down 1 unit. • Draw a dot and label it A. (See coordinate plane after Exercise 52.) 48. B(4, 0) • Start at the origin. • Move left 4 units. • Since the y-coordinate is 0, the point is on the x-axis. • Draw a dot and label it B. (See coordinate plane after Exercise 52.) 49. C(1, 5) • Start at the origin. • Move right 1 unit and up 5 units. • Draw a dot and label it C. (See coordinate plane after Exercise 52.) 50. D(1, 1) • Start at the origin. • Move left 1 unit and down 1 unit. • Draw a dot and label it D. (See coordinate plane after Exercise 52.) 51. E(2, 3) • Start at the origin. • Move left 2 units and up 3 units. • Draw a dot and label it E. (See coordinate plane after Exercise 52.)

1

C

S

O

T

4  21  4  4z

O

R’

T’

1

D

y

S’

If y  4z and y  21, then 21  4z.

B

L’

The coordinates of the vertices of the image are J¿(3, 5), K¿(2, 8), L¿(1, 8), and M¿(3, 5).

21  y

A

M’

K’

2

44.

M

J’

 14  2  3 y 1

x

O

2

3

L

J

If x  3y and x  14, then 14  3y. 3 2

y

K

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52. F(4, 3) • Start at the origin. • Move right 4 units and down 3 units. • Draw a dot and label it F. 47–52. y C

Page 204

Graphing Calculator Investigation (Preview of Lesson 4-3)

1. {(10, 10), (0, 6), (4, 7), (5, 2)} Step 1 Enter the data. • Enter the x-coordinates in L1 and the y-coordinates in L2.

STAT ENTER 10 ENTER 0 ENTER 4 ENTER 5 ENTER 10 ENTER 6 ENTER 7 ENTER 2 ENTER

E

KEYSTROKES:

O

x

B D

Step 2 Format the graph. • Turn on the statistical plot.

A F

KEYSTROKES:

30% Solution 15% Solution 25% Solution

Amount of Nitric Acid 0.30(20) 0.15 x 0.25(20  x)

ENTER



ENTER ENTER

2nd

L2

ENTER

Step 3 Choose the viewing window. • Be sure you can see all of the points. [10, 15] scl : 1 by [10, 15] scl: 1 WINDOW 10 ENTER 15 ENTER 1 ENTER 10 ENTER 15 ENTER 1

KEYSTROKES:

0.30(20)  0.15x  0.25(20  x) 6  0.15x  5  0.25x 6  0.15x  0.15x  5  0.25x  0.15x 6  5  0.10x 6  5  5  0.10x  5 1  0.10x 1 0.10

STAT PLOT

• Select the scatter plot, L1 as the Xlist and L2 as the Ylist. ENTER 2nd L1 KEYSTROKES:

53. Let x  the amount of 15% solution to be added. Amount of Solution (mL) 20 x 20  x

2nd

Step 4 Graph the relation. • Display the graph. KEYSTROKES:

GRAPH

0.10x 0.10

10  x Jamaal should add 10 mL of the 15% solution to the 20 mL of the 30% solution. 54. There are 26 favorable outcomes of the 36 total possible outcomes. 26

[10, 15] scl: 1 by [10, 15] scl: 1

13

P (sum is less than 9)  36  18 or about 72%

2. {(4, 1), (3, 5), (4, 5), (5, 1)} Step 1 Enter the data. • Enter the x-coordinates in L1 and the y-coordinates in L2.

55. There are three favorable outcomes of the 36 total possible outcomes. 3

1

P (sum is greater than 10)  36  12 or about 8%

STAT ENTER 4 ENTER ENTER ENTER 3 4 5 ENTER 1 ENTER 5 ENTER 5 ENTER 1 ENTER

KEYSTROKES:

56. There are 15 favorable outcomes of the 36 total possible outcomes. 15

15

P (sum is less than 7)  36  12 or about 42%

Step 2 (See Step 2 in Exercise 1.) Step 3 Choose the viewing window. • Be sure you can see all of the points. [10, 10] scl: 1 by [10, 10] scl: 1

57. There are 30 favorable outcomes of the 36 total possible outcomes. 30

5

P (sum is greater than 4)  36  6 or about 83% 58. The number of toppings is the independent variable, and the cost is the dependent variable. So the data in the table can be represented as the set of ordered pairs {(1, 9.95), (2, 11.45), (3, 12.95), (4, 14.45), (5, 15.95), (6, 17.45)}. 59. Time is the independent variable, and the temperature is the dependent variable. So the data in the table can be represented as the set of ordered pairs {(0, 100), (5, 90), (10, 81), (15, 73), (20, 66), (25, 60), (30, 55)}.

ENTER 10 ENTER 1 ENTER 10 ENTER 10 ENTER 1

KEYSTROKES:

Step 4

WINDOW 10

(See Step 4 in Exercise 1.)

[10, 10] scl: 1 by [10, 10] scl: 1

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3. {(12, 15), (10, 16), (11, 7), (14, 19)} Step 1 Enter the data. • Enter the x-coordinates in L1 and the y-coordinates in L2.

Page 207

Algebra Activity y

ENTER 12 ENTER 10 ENTER 11 ENTER 14 ENTER 15 ENTER 16 ENTER 7 ENTER 19 ENTER

KEYSTROKES:

STAT

O

x

Step 2 (See Step 2 in Exercise 1.) Step 3 Choose the viewing window. • Be sure you can see all of the points. [15, 15] scl: 2 by [20, 20] scl: 2

FOLD LINE

ENTER 15 ENTER 2 ENTER 20 ENTER 20 ENTER 2

KEYSTROKES: WINDOW 15

Step 4

1. The inverse of each point matches the point. 2. The inverse of each point is a reflection of the point across the fold. 3. Sample ordered pairs: (1, 1), (0, 0), (2, 2) For each ordered pair (x, y), x  y. 4. Reflect the points across the line in which the x-coordinate equals the y-coordinate.

(See Step 4 in Exercise 1.)

4-3

[15, 15] scl: 2 by [20, 20] scl: 2

4. {(45, 10), (23, 18), (22, 26), (35, 26)} Step 1 Enter the data. • Enter the x-coordinates in L1 and the y-coordinates in L2.

Page 208

Check for Understanding

1. A relation can be represented as a set of ordered pairs, a table, a graph, or a mapping. 2. Sample answer: {(1, 2), (3, 4), (5, 6), (7, 8), (9, 8)} has five elements in its domain {1, 3, 5, 7, 9} and four elements in its range {2, 4, 6, 8}. 3. The domain of a relation is the range of the inverse, and the range of a relation is the domain of the inverse. 4. Table Graph List the set of Graph each x-coordinates in the ordered pair on first column and the a coordinate corresponding y-coordinates plane. in the second column.

STAT ENTER 45 ENTER 23 ENTER 22 ENTER 35 ENTER 10 ENTER 18 ENTER 26 ENTER 26 ENTER

KEYSTROKES:

Step 2 (See Step 2 in Exercise 1.) Step 3 Choose the viewing window. • Be sure you can see all of the points. [2, 50] scl: 2 by [2, 30] scl: 2 WINDOW 2 ENTER 50 ENTER 2 ENTER 2 ENTER 30 ENTER 2

KEYSTROKES:

Step 4

Relations

(See Step 4 in Exercise 1.)

x 5 8 7

y 2 3 1

4 2

O 8642

[2, 50] scl: 2 by [2, 30] scl: 2

2

5. The scale of the x-axis should include the least and greatest values in the domain, and the scale of the y-axis should include the least and greatest values in the range.

Chapter 4

4

132

y

2 4 6 8x

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Mapping List the x values in set X and the y values in set Y. Draw an arrow from each x value in X to the corresponding y value in Y.

Mapping List the x values in set X and the y values in set Y. Draw an arrow from each x value in X to the corresponding y value in Y.

X

Y

X

Y

5 8 7

2 3 1

7 3 2

1 0 5

The domain for this relation is {7, 5, 8}. The range is {2, 1, 3}. 5. Table Graph List the set of Graph each x-coordinates in the ordered pair first column and the on a coordinate corresponding y-coordinates plane. in the second column. x 6 3 1 5

y 4 3 9 3

10 8 6 4 2 2

2 4

The domain for this relation is {2, 3, 7}. The range is {0, 1, 5}. 7. Table Graph List the set of Graph each x-coordinates in the ordered pair first column and the on a coordinate corresponding y-coordinates plane. in the second column. x 4 1 4 6

y

O

2

4

Y

6 3 1 5

4 3 9

y 1 0 5

8. 9. 10.

y

11. 12. 13.

O

4

2

2 4 6 8

y

x

O 2

4

6

Mapping List the x values in set X and the y values in set Y. Draw an arrow from each x value in X to the corresponding y value in Y.

The domain for this relation is {1, 3, 5, 6}. The range is {3, 4, 9}. 6. Table Graph List the set of Graph each x-coordinates in the ordered pair on first column and the a coordinate corresponding y-coordinates plane. in the second column. x 7 3 2

8 6 4 2

6x

Mapping List the x values in set X and the y values in set Y. Draw an arrow from each x value in X to the corresponding y value in Y.

X

y 8 9 7 9

x

14. 15.

133

X

Y

4 1 6

8 9 7

The domain for this relation is {4, 1, 6}. The range is {7, 8, 9}. relation: {(3, 2), (6, 7), (4, 3), (6, 5)} inverse: {(2, 3), (7, 6), (3, 4), (5, 6)} relation: {(4, 9), (2, 5), (2, 2), (11, 12)} inverse: {(9, 4), (5, 2), (2, 2), (12, 11)} relation: {(3, 0), (5, 2), (7, 4)} inverse: {(0, 3), (2, 5), (4, 7)} relation: {(2, 8), (3, 7), (4, 6), (5, 7)} inverse: {(8, 2), (7, 3), (6, 4), (7, 5)} relation: {(1, 2), (2, 4), (3, 3), (4, 1)} inverse: {(2, 1), (4, 2), (3, 3), (1, 4)} relation: {(4, 4), (3, 0), (0, 3), (2, 1), (2, 1)} inverse: {(4, 4), (0, 3), (3, 0), (1, 2), (1, 2)} Sample answer: (1989, 25), (1991, 20), (1996, 10) The domain of the relation is {1988, 1989, 1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999}.

Chapter 4

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16. The least value in the range is about 5.7 students, and the greatest value in the range is 25 students. 17. There are fewer students per computer in more recent years. So the number of computers in schools has increased.

Pages 209–210 18.

19.

x 4 1 1 2

8 6 4 2 4

X

Y

4 1 2

3 7 9

2

O 2

2 4 6 8

4x

22.

The domain of this relation is {1, 2, 4}. The range is {7, 3, 9}.

y

y 2 0 4 7

X

Y

5 5 6 2

2 0 4 7

The domain of this relation is {5, 2, 5, 6}. The range is {0, 2, 4, 7}.

4321 2 4 6 8

O

Y

0 6 5 4

0 1 6 2

Y

3 2 1

8 7 9

x 4 3 1 6

O 1 2 3 4x

The domain of this relation is {1, 2, 3}. The range is {9, 7, 8}.

y

y 2 4 2 4

x

O

X

Y

4 3 1 6

2

The domain of this relation is {1, 3, 4, 6}. The range is {2, 4}.

4

x 0 5 0 1

y 2 1 6 9

12 10 8 6 4 2

y

O 1 2x

4

y

X

X

y

6543 21 2

y 0 1 6 2

Chapter 4

8 6 4 2

x

23.

x 0 6 5 4

y 8 7 9 9

y

O

20.

x 3 3 2 1

Practice and Apply

y 3 7 3 9

x 5 5 6 2

21.

x

The domain of this relation is {0, 4, 5, 6}. The range is {1, 0, 2, 6}.

134

X

Y

0 5 1

2 1 6 9

The domain of this relation is {5, 1, 0}. The range is {1, 2, 6, 9}.

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25.

x 3 4 2 5 4

O

X

Y

3 4 2 5 4

4 3 2 4 5

x 7 3 4 2 3

35. relation: {(2, 0), (2, 4), (3, 7), (5, 0), (5, 8), (7, 7)} inverse: {(0, 2), (4, 2), (7, 3), (0, 5), (8, 5), (7, 7)} 36. relation: {(3, 3), (2, 2), (1, 1), (1, 1), (2, 2), (3, 3)} inverse: {(3, 3), (2, 2), (1, 1), (1, 1), (2, 2), (3, 3)} 37. relation: {(3, 1), (3, 3), (3, 5), (0, 3), (2, 3), (4, 3)} inverse: {(1, 3), (3, 3), (5, 3), (3, 0), (3, 2), (3, 4)} 38. y

y

y 4 3 2 4 5

x

The domain of this relation is {4, 2, 3, 4, 5}. The range is {4, 2, 3, 4, 5}.

222 218 Boiling Points (°F)

24.

y

y 6 4 5 6 2

214 210 206 202 198 194 190 186 0

x 1000 3000 5000 7000 9000 2000 4000 6000 8000 10,000 Altitude (ft)

O

X

Y

7 3 4 2 3

6 4 5 2

x

39. The inverse of the relation, as a set of ordered pairs, is {(212.0, 0), (210.2, 1000), (208.4, 2000), (206.5, 3000), (201.9, 5000), (193.7, 10,000)}. 40. Use the inverse relation to find the corresponding altitude for a given boiling point. 41. The domain of the relation is {1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000}. The range of the relation is approximately {6.3, 7.5, 9.2, 9.5, 9.8, 10, 10.4}. 42. The lowest production, about 6.3 billion bushels, occurred in 1993. The highest production, about 10.4 billion bushels, occurred in 2000. 43. Production seems to alternately increase and decrease each year. However, since 1995, the trend has shown an overall production increase. 44. Body Weight (lb) Muscle Weight (lb)

The domain of this relation is {3, 2, 3, 4, 7}. The range is {2, 4, 5, 6}.

26. relation: {(1, 2), (3, 4), (5, 6), (7, 8)} inverse: {(2, 1), (4, 3), (6, 5), (8, 7)} 27. relation: {(0, 3), (5, 2), (4, 7), (3, 2)} inverse: {(3, 0), (2, 5), (7, 4), (2, 3)} 28. relation: {(6, 2), (4, 5), (3, 3), (1, 7)} inverse: {(2, 6), (5, 4), (3, 3), (7, 1)} 29. relation: {(8, 4), (1, 1), (0, 6), (5, 4)} inverse: {(4, 8), (1, 1), (6, 0), (4, 5)} 30. relation: {(4, 2), (2, 1), (2, 4), (2, 3)} inverse: {(2, 4), (1, 2), (4, 2), (3, 2)} 31. relation: {(3, 3), (1, 3), (4, 2), (1, 5)} inverse: {(3, 3), (3, 1), (2, 4), (5, 1)} 32. relation: {(0, 0), (4, 7), (8, 10.5), (12, 13), (16, 14.5)} inverse: {(0, 0), (7, 4), (10.5, 8), (13, 12), (14.5, 16)} 33. relation: {(1, 16.50), (1.75, 28.30), (2.5, 49.10), (3.25, 87.60), (4, 103.40)} inverse: {(16.50, 1), (28.30, 1.75), (49.10, 2.5), (87.60, 3.25), (103.40, 4)} 34. relation: {(3, 2), (3, 8), (6, 5), (7, 4), (11, 4)} inverse: {(2, 3), (8, 3), (5, 6), (4, 7), (4, 11)}

100 105 110 115 120 125 130

40 42 44 46 48 50 52

45. The domain of the relation is {100, 105, 110, 115, 120, 125, 130}. The range is {40, 42, 44, 46, 48, 50, 52}.

135

Chapter 4

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46.

52. The inverse of the relation is {(4, 4), (2, 2), (0, 0), (2, 2), (4, 4)}.

y 52

y

Muscle Weight (lb)

50 48 O

46

x

44 42

53a. Sample answer:

40 100 105 110 115 120 125 130 Body Weight (lb)

x

47. The domain of the inverse of the relation is {40, 42, 44, 46, 48, 50, 52}. The range of the inverse is {100, 105, 110, 115, 120, 125, 130}. 48. 8

A’

y

4 4

O

4

C

D

4

D’

53b. Sample answer: [10, 10] scl: 1 by [10, 12] scl: 1 53c. The inverse of the relation is {(10, 0), (8, 2), (6, 6), (4, 9)}. Sample answer:

B

A 8

[10, 10] scl: 1 by [10, 12] scl: 1

B’

8x

C’

8

49. Sample answer: F  {(1, 1), (2, 2), (3, 3)} G  {(1, 2), (2, 3), (3, 1)} The elements in the domain and range of F should be paired differently in G. 50. Expressing real-world data as relations shows how the members of a domain relate to the members of the range. For example, a table helps to organize the data or a graph may show a pattern in the data. Answers should include the following. • Graph

Strikeouts

130

[10, 12] scl: 1 by [10, 10] scl: 1

53d.

y

Relation Point Quadrant (0, 10) none (2, 8) IV (6, 6) I (9, 4) IV

Inverse Point Quadrant (10, 0) none (8, 2) II (6, 6) I (4, 9) II

54a. Sample answer:

110 90 70 50 0

20

40 60 80 Home Runs

[10, 10] scl: 1 by [6, 40] scl: 2

x

54b. Sample answer: [10, 10] scl: 1 by [6, 40] scl: 2

• There seems to be a positive relationship between the number of home runs and strikeouts. In years when Griffey hit more home runs, he also struck out more. 51. B; The set of ordered pairs graphed is {(4, 4), (2, 2), (0, 0), (2, 2), (4, 4)}. The domain of this relation is {4, 2, 0, 2, 4}. The range is {0, 2, 4}.

Chapter 4

136

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54c. The inverse of the relation is {(18, 1), (23, 2), (28, 3), (33, 4)}. Sample answer:

56c. The inverse of the relation is {(77, 92), (200, 93), (50, 19)}. Sample answer:

[6, 40] scl: 2 by [10, 10] scl: 1

54d.

Relation Point Quadrant (1, 18) II (2, 23) II (3, 28) II (4, 33) II

[100, 250] scl: 10 by [100, 50] scl: 10

56d.

Inverse Point Quadrant (18, 1) IV (23, 2) IV (28, 3) IV (33, 4) IV

Relation QuadPoint rant (92, 77) III (93, 200) II (19, 50) IV

Inverse QuadPoint rant (77, 92) III (200, 93) IV (50, 19) II

55a. Sample answer:

Page 211

[10, 80] scl: 5 by [10, 60] scl: 5

55b. Sample answer: [10, 80] scl: 5 by [10, 60] scl: 5 55c. The inverse of the relation is {(12, 35), (25, 48), (52, 60)}. Sample answer:

xyCoordi- Coordi- Ordered Point nate nate Pair Quadrant 60. A 4 4 (4, 4) IV 61. K 3 2 (3, 2) I 62. L 1 3 (1, 3) III 63. W 1 1 (1, 1) IV 64. B 4 0 (4, 0) none 65. P 4 2 (4, 2) III 66. R 5 3 (5, 3) II 67. C 2 5 (2, 5) II

[10, 60] scl: 5 by [10, 80] scl: 5

55d.

Relation Point Quadrant (35, 12) I (48, 25) I (60, 52) I

Maintain Your Skills

57. The figure has been turned around a point. This is a rotation. 58. The figure has been flipped over a line. This is a reflection. 59. The figure has been shifted vertically down. This is a translation.

68. Find the amount of change. 10.15  9.75  0.40 Find the percent using the original number, 9.75, as the base.

Inverse Point Quadrant (12, 35) I (25, 48) I (52, 60) I

0.40 9.75

r

 100

0.40(100)  9.75 r 40  9.75 r

56a. Sample answer:

40 9.75



9.75 r 9.75

4.1  r The percent of increase in Dominique’s salary is about 4.1%. 69. 72  9  8 70. 105  15  7 3

1

9

71. 3  3  3  1  1  9

[100, 50] scl: 10 by [100, 250] scl: 10

1

4

72. 16  4  16  1 

56b. Sample answer: [100, 50] scl: 10 by [100, 250) scl: 10

137

64 1

 64

Chapter 4

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73.

54n  78 6

  98x  35y 7

12 12

1 (54n  78) 6 1 1 54n 6  78 6

12

 9n  13 74.

79.

 (54n  78)  6

 (98x  35y)  7

12 1 1  98x 1 7 2  35y 1 7 2 1 7

 (98x  35y)

75. a

a  15  20 ?

3  15  20 S 18 20

3

?

4

4  15  20 S 19 20

5

5  15  20 S 20  20

? ?

6  15  20 S 21 20

6

?

7  15  20 S 22 20

7

?

8  15  20 S 23 20

8

True or False? false

r

r62

3

3  6  2 S 3 2

? ?

4  6  2 S 2 2

4

?

5  6  2 S 1 2

5

80.

true ✓

3

9  5(6)  6 S 9 24

6

?

9  5(7)  6 S 9 29

7

?

9  5(8)  6 S 9 34

8

3 4 5 6 7 8

3  8w  35 ?

3  8(3)  35 S 27 35 ?

3  8(4)  35 S 35  35 ?

3  8(5)  35 S 43 35 ?

3  8(6)  35 S 51 35 ?

3  8(7)  35 S 59 35 ?

3  8(8)  35 S 67 35

2

false

8

8 3

 15  17 S 173 17

m 5

m 3 5 4 5

3

52

3 ?

false

2

false

3

false

4

false



3 ?  5

2 S 15 2

3 ?  5

2 S 15 2

 5  2 S 25 2

false

3 ? 3 ? 3 ?

True or False?

1

 5  2 S 15 2

8 5

1

true ✓ false

Since m  7 makes the equation true, the solution set is {7}.

Page 211

Practice Quiz 1

1. Q(2, 3) • Start at the origin. • Move right 2 units and up 3 units. • Draw a dot and label it Q. (See coordinate plane after Exercise 4.) 2. R(4, 4) • Start at the origin. • Move left 4 units and down 4 units. • Draw a dot and label it R. (See coordinate plane after Exercise 4.) 3. S(5, 1) • Start at the origin. • Move right 5 units and down 1 unit. • Draw a dot and label it S. (See coordinate plane after Exercise 4.)

false false false false false

True or False? false true ✓ false false false false

Since w  4 makes the equation true, the solution set is {4}. Chapter 4

?

 15  17 S 173 17

8

Since n  3 makes the equation true, the solution set is {3}. 78. w

false

7 3

false

true ✓

?

1

7

5  2S 22

9  5(3)  6 S 9  9

true ✓

?

7 5

True or False?

9  5(5)  6 S 9 19

5

 15  17 S 17  17

7

true ✓

?

false

?

false

22

9  5(4)  6 S 9 14

4

2

 5  2 S 15 2

false

?

6 3

?

6 5

1 2

?

 15  17 S 163 17

false

6

True or False?

9  5n  6

n

5 3

1



Since r  8 makes the equation true, the solution set is {8}. 77.

5

false

?

5 5

?

86 2S

8

 15  17 S 163 17

5

?

76 2S

4 3

4

false

false

7

4

3

false

0 2

66 2S

 15  17 S 16 17

false

?

6

?

3 3

True or False?

Since g  6 makes the equation true, the solution set is {6}.

false

Since a  5 makes the equation true, the solution set is {5}. 76.

 15  17

3

6

 14x  5y

g 3

g

138

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7. The domain of the relation is {1, 2, 4}. The range is {3, 5, 6}. The inverse of the relation is {(3, 1), (6, 4), (3, 2), (5, 1)}. 8. The domain of the relation is {2, 0, 4, 8}. The range is {5, 2, 3, 6}. The inverse of the relation is {(6, 2), (3, 0), (2, 4), (5, 8)}. 9. The domain of the relation is {8, 11, 15}. The range is {3, 5, 22, 31}. The inverse of the relation is {(5, 11), (3, 15), (22, 8), (31, 11)}. 10. The domain of the relation is {5, 1, 2, 6}. The range is {0, 3, 4, 7, 8}. The inverse of the relation is {(8, 5), (0, 1), (4, 1), (7, 2), (3, 6)}.

4. T(1, 3) • Start at the origin. • Move left 1 unit and up 3 units. • Draw a dot and label it T. y

Q

T

x

O

S R

Equations as Relations

4-4 5. To reflect the triangle over the x-axis, multiply the y-coordinate of each vertex by 1. (x, y) S (x, y) A(4, 8) S A¿(4, 8) B(7, 5) S B¿(7, 5) C(2, 1) S C¿(2, 1) The coordinates of the vertices of the image are A (4, 8), B (7, 5), and C (2, 1). 8

y

Pages 214–215

A B

4

C’ O

1

4

7x

5

3

C

B’

1

8

1

1

2

10

3

12

7

1

A’

6. To translate the quadrilateral 5 units to the left, add 5 to the x-coordinate of each vertex. To translate the quadrilateral 4 units down, add 4 to the y-coordinate of each vertex. (x, y) → (x  5, y  4) W(1, 0) → W (1  5, 0  4) → W (4, 4) X(2, 3) → X (2  5, 3  4) → X (3, 1) Y(4, 1) → Y (4  5, 1  4) → Y (1, 3) Z(3, 3) → Z (3  5, 3  4) → Z (2, 7) The coordinates of the vertices of the image are W (4, 4), X (3, 1), Y (1, 3), and Z (2, 7). y

X

x

6.

Y’ Z

W’

true ✓ true ✓ false false

x 7

y 3

7

3

2

1

2

1

2x  5y  1 2(7)  5(3)  1 11 2(7)  5(3)  1 1  1 2(2)  5(1)  1 1  1 2(2)  5(1)  1 11

True or False? true ✓ false false true ✓

The solution set is {(7, 3), (2, 1)}.

W O

1  3(1)  4 11 10  3(2)  4 10  10 12  3(3)  4 12  13 1  3(7)  4 1  25

The solution set is {(1, 1), (2, 10)}. 5.

Y X’

Check for Understanding

1. To find the domain of an equation if you are given the range, substitute the values for y and solve for x. 2. Sample answer: y  x  5 is an equation in two variables. Two solutions for this equation are (9, 4) and (10, 5). 3. Bryan is correct. x represents the domain and y represents the range. So, replace x with 5 and y with 1. 4. x y y  3x  4 True or False?

Z’

x 3 1 0 2

2x  1 2(3)  1 2(1)  1 2(0)  1 2(2)  1

y 7 3 1 3

(x, y) (3, 7) (1, 3) (0, 1) (2, 3)

The solution set is {(3, 7), (1, 3), (0, 1), (2, 3)}.

139

Chapter 4

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7.

4x 4  (3) 4  (1) 40 42

x 3 1 0 2

y 7 5 4 2

11. First solve the equation for y in terms of x. 2y  x  2

(x, y) (3, 7) (1, 5) (0, 4) (2, 2)

2y 2

y

The solution set is {(3, 7), (1, 5), (0, 4), (2, 2)}. 8. First solve the equation for y in terms of x. 2y  2x  12 2y  2x  2x  12  2x 2y  12  2x 2y 2



x2 2

y

(x, y)

4

4  2 2

1

(4, 1)

2

2  2 2

0

(2, 0)

0

0  2 2

1

(0, 1)

2

2  2 2

2

(2, 2)

4

4  2 2

3

(4, 3)

12  2x 2

6x 6  (3) 6  (1) 60 62

y 9 7 6 4

(x, y) (3, 9) (1, 7) (0, 6) (2, 4)

x  2 2 x  2 2

x

y6x x 3 1 0 2



Graph the solution set {(4, 1), (2, 0), (0, 1), (2, 2), (4, 3)}. y

The solution set is {(3, 9), (1, 7), (0, 6), (2, 4)}. 9. First solve the equation for y in terms of x. 3x  2y  13 3x  2y  3x  13  3x 2y  13  3x 

y

12. y

(x, y)

3

13  3x 2 13  3(3) 2

11

(3, 11)

1

13  3(1) 2

8

(1, 8)

0

13  3(0) 2

6.5

(0, 6.5)

2

13  3(2) 2

3.5

(2, 3.5)

x

10.

13  3x 2 13  3x 2

3x 3(3) 3(2) 3(1) 3(0) 3(1) 3(2) 3(3)

x 3 2 1 0 1 2 3

y 9 6 3 0 3 6 9

(x, y) (3, 9) (2, 6) (1, 3) (0, 0) (1, 3) (2, 6) (3, 9)

4

Chapter 4

2

2 4 6 8

2

g

10

25(10) 6

413

14

25(14) 6

583

18

25(18) 6

75

24

25(24) 6

100

2 1

y

80 70 60 50 40 0

y

O

25k 6

90

Graph the solution set {(3, 9), (2, 6), (1, 3), (0, 0), (1, 3), (2, 6), (3, 9)}. 8 6 4 2

k

100

Gold %

2y 2

x

O

4x

140

x 10 12 14 16 18 20 22 24 26 Number of Karats

(k, g)

110, 4123 2 114, 5813 2 (18, 75)

(24, 100)

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25k

13. Solve the equation g  6 for k since the value of k will depend on the given value of g. g 6g 

18. x 0

25k 6 25k 6 6

1 2

6g  25k 6g 25 6g 25



y 0.5

4

1

2

0.75

2

4

25k 25

k

For a ring that is 50% gold, g  50. So 6(50) k  25  12. A 12-karat ring is 50% gold.

14. x 2

y 1

1

5

9

2

0

1

0

True or False?

3

x 4

y 4

8

0

2

2

3

3

y  8  3x 4  8  3(4) 4  4 0  8  3(8) 0  16 2  8  3(2) 22 3  8  3(3) 3  1

true ✓

x 1

y 2

2

1

2

4

2

3

x  3y  7 1  3(2)  7 7  7 2  3(2)  7 4  7 2  3(4)  7 10  7 2  3(3)  7 7  7

x 3

y 0

2

1

2

1

4

1

2x  2y  6 2(3)  2(0)  6 66 2(2)  2(1)  6 66 2(2)  2(1)  6 6  6 2(4)  2(1)  6 66

false false false

3.5

1

2

2y  4x  8 2(2)  4(0)  8 48 2(0.5)  4(3)  8 11  8 2(3.5)  4(0.25)  8 88 2(2)  4(1)  8 88

True or False? false false true ✓ true ✓

The solution set is {(0.25, 3.5), (1, 2)}. true ✓

20.

True or False? true ✓ false

x 2 1 1 3 4

4  5x 4  5(2) 4  5(1) 4  5(1) 4  5(3) 4  5(4)

y 14 9 1 11 16

(x, y) (2, 14) (1, 9) (1, 1) (3, 11) (4, 16)

The solution set is {(2, 14), (1, 9), (1, 1), (3, 11), (4, 16)}.

true ✓

21.

false

True or False? true ✓

x 2 1 1 3 4

2x  3 2(2)  3 2(1)  3 2(1)  3 2(3)  3 2(4)  3

y 1 1 5 9 11

(x, y) (2, 1) (1, 1) (1, 5) (3, 9) (4, 11)

The solution set is {(2, 1), (1, 1), (1, 5), (3, 9), (4, 11)}. 22. First solve the equation for y in terms of x. xy4 x4y44 x4y

false false true ✓

x 2 1 1 3 4

The solution set is {(1, 2), (2, 3)}. 17.

0.5

0.25

false

The solution set is {(4, 4), (2, 2)}. 16.

y 2

false

The solution set is {(1, 5), (0, 1)}. 15.

true ✓

The solution set is {(0, 0.5)}.

Practice and Apply y  4x  1 1  4(2)  1 1  9 5  4(1)  1 55 2  4(9)  1 2  37 1  4(0)  1 11

True or False?

19. x

Pages 215–217

3x  8y  4 3(0)  8(0.5)  4 4  4 3(4)  8(1)  4 4  4 3(2)  8(0.75)  4 0  4 3(2)  8(4)  4 26  4

True or False? true ✓ true ✓

x4 2  4 1  4 14 34 44

y 6 5 3 1 0

(x, y) (2, 6) (1, 5) (1, 3) (3, 1) (4, 0)

The solution set is {(2, 6), (1, 5), (1, 3), (3, 1), (4, 0)}.

false true ✓

The solution set is {(3, 0), (2, 1), (4, 1)}.

141

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26. First solve the equation for y in terms of x. 8x  4y  12 8x  4y  8x  12  8x 4y  12  8x

23. First solve the equation for y in terms of x. x7y xy7yy xy7 xyx7x y7x 7x 7  (2) 7  (1) 71 73 74

x 2 1 1 3 4

y 9 8 6 4 3

 4 y  3  2x

(x, y) (2, 9) (1, 8) (1, 6) (3, 4) (4, 3)



3  2x 3  2(2) 3  2(1) 3  2(1) 3  2(3) 3  2(4)

x 2 1 1 3 4

The solution set is {(2, 9), (1, 8), (1, 6), (3, 4), (4, 3)}. 24. First solve the equation for y in terms of x. 6x  3y  18 6x  3y  6x  18  6x 3y  18  6x 3y 3

18  6x 3

6  2x 6  2(2) 6  2(1) 6  2(1) 6  2(3) 6  2(4)

2x 2

y 10 8 4 0 2

3  6x 3  6(2) 3  6(1) 3  6(1) 3  6(3) 3  6(4)

y 9 3 9 21 27



2y 2

x

y

(x, y)

2 1 1 3 4

2 1 1 3 4

(2, 2) (1, 1) (1, 1) (3, 3) (4, 4)

The solution set is {(2, 2), (1, 1), (1, 1), (3, 3), (4, 4)}. 28. First solve the equation for y in terms of x. 5x  10y  20 5x  10y  5x  20  5x 10y  20  5x 10y 10

(x, y) (2, 9) (1, 3) (1, 9) (3, 21) (4, 27)



20  5x 10

y  2  0.5x x 2 1 1 3 4

The solution set is {(2, 9), (1, 3), (1, 9), (3, 21), (4, 27)}.

Chapter 4

(x, y) (2, 7) (1, 5) (1, 1) (3, 3) (4, 5)

xy

(x, y) (2, 10) (1, 8) (1, 4) (3, 0) (4, 2)

The solution set is {(2, 10), (1, 8), (1, 4), (3, 0), (4, 2)}. 25. First solve the equation for y in terms of x. 6x  y  3 6x  y  6x  3  6x y  3  6x 1(y)  1(3  6x) y  3  6x x 2 1 1 3 4

y 7 5 1 3 5

The solution set is {(2, 7), (1, 5), (1, 1), (3, 3), (4, 5)}. 27. First solve the equation for y in terms of x. 2x  2y  0 2x  2y  2y  0  2y 2x  2y

y  6  2x

x 2 1 1 3 4

12  8x

4y 4

2  0.5x 2  0.5(2) 2  0.5(1) 2  0.5(1) 2  0.5(3) 2  0.5(4)

y 3 2.5 1.5 0.5 0

(x, y) (2, 3) (1, 2.5) (1, 1.5) (3, 0.5) (4, 0)

The solution set is {(2, 3), (1, 2.5), (1, 1.5), (3, 0.5), (4, 0)}.

142

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29. First solve the equation for y in terms of x. 3x  2y  14 3x  2y  3x  14  3x 2y  14  3x 2y 2



y

14  3x 2 14  3x 2

14  3x 2

y

(x, y)

2

14  3(2) 2

10

(2, 10)

1

14  3(1) 2

8.5

(1, 8.5)

1

14  3(1) 2

5.5

(1, 5.5)

3

14  3(3) 2

2.5

(3, 2.5)

4

14  3(4) 2

1

x

32.

2x  3 2(3)  3 2(2)  3 2(1)  3 2(1)  3 2(2)  3 2(3)  3

x 3 2 1 1 2 3

y 3 1 1 5 7 9

(x, y) (3, 3) (2, 1) (1, 1) (1, 5) (2, 7) (3, 9)

Graph the solution set {(3, 3), (2, 1), (1, 1), (1, 5), (2, 7), (3, 9)}. y

(4, 1)

The solution set is {(2, 10), (1, 8.5), (1, 5.5), (3, 2.5), (4, 1)}. 30. First solve the equation for y in terms of x.

x

O

1

x2y8 1

x  2y  x  8  x

2

1 y 2 1 y 2

33.

8x

1 2  2(8  x) y  16  2x

x

16  2x

y

(x, y)

2 1 1 3 4

16  2(2) 16  2(1) 16  2(1) 16  2(3) 16  2(4)

20 18 14 10 8

(2, 20) (1, 18) (1, 14) (3, 10) (4, 8)

2x  2x 

1 y 3

1

8 6 4 2 4

4

2

(3) 3 y  (3)(4  2x) x 2 1 1 3 4

y  12  6x 12  6x 12  6(2) 12  6(1) 12  6(1) 12  6(3) 12  6(4)

2

O

2

4

x

4 6 8 10 12 14 16

3 y  4  2x 1

(x, y) (5, 16) (2, 7) (1, 2) (3, 8) (4, 11)

y

 2x  4  2x 1

y 16 7 2 8 11

Graph the solution set {(5, 16), (2, 7), (1, 2), (3, 8), (4, 11)}.

The solution set is {(2, 20), (1, 18), (1, 14), (3, 10), (4, 8)}. 31. First solve the equation for y in terms of x. 1 y 3

3x  1 3(5)  1 3(2)  1 3(1)  1 3(3)  1 3(4)  1

x 5 2 1 3 4

y 24 18 6 6 12

(x, y) (2, 24) (1, 18) (1, 6) (3, 6) (4, 12)

The solution set is {(2, 24), (1, 18), (1, 6), (3, 6), (4, 12)}.

143

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34. First solve the equation for y in terms of x. 3x  2y  5 3x  2y  3x  5  3x 2y  5  3x  2y 2



y

Graph the solution set {(4, 7), (1, 3.25), (0, 2), (2, 0.5), (4, 3), (6, 5.5)}. y

5  3x 2 3x  5 2

3x  5 2

y

(x, y)

3

3(3)  5 2

7

(3, 7)

1

3(1)  5 2

4

(1, 4)

x

2

3(2)  5 2

0.5

(2, 0.5)

4

3(4)  5 2

3.5

(4, 3.5)

5

3(5)  5 2

5

x

O

36. First solve the equation for y in terms of x.

(5, 5)

1 x 2

Graph the solution set {(3, 7), (1, 4), (2, 0.5), (4, 3.5), (5, 5)}.

1 x 2

y2 1

1

 y  2x  2  2x 1

y  2  2x

y

x

x

O

35. First solve the equation for y in terms of x. 5x  4y  8 5x  4y  5x  8  5x 4y  8  5x 4y 4



y

(x, y)

1

4

(4, 4)

1

4

2  2(4)

1

2  2(1)

2.5

(1, 2.5)

1

2  2 112 1

1.5

(1, 1.5)

4

2  2(4)

1

0

7

2

1 (7) 2

1.5

8

2  2(8)

1

2

(4, 0) (7, 1.5) (8, 2)

y

8  5x 4 8  5x 4

y

(x, y)

4

8  5(4) 4

7

(4, 7)

1

8  5(1) 4

3.25

0

8  5(0) 4

2

2

8  5(2) 4

0.5

4

8  5(4) 4

3

6

8  5(6) 4

5.5

Chapter 4

y

Graph the solution set {(4, 4), (1, 2.5), (1, 1.5), (4, 0), (7, 1.5), (8, 2)}.

8  5x 4

x

1

2  2x

O

(1, 3.25) (0, 2) (2, 0.5) (4, 3) (6, 5.5)

144

x

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37.

x

1 x 4

3

4

1 (4) 4

3

4

2

1 (2) 4

3

3.5

0

1 (0) 4

3

3

2

1 (2) 4

4

1 (4) 4

3

2

6

1 (6) 4

3

1.5

3

y

2.5

41.

(x, y) (4, 4) (2, 3.5) (0, 3) (2, 2.5) (4, 2) (6, 1.5)

42.

x

8  3x 8  3(1) 8  3(2) 8  3(5) 8  3(8)

2y  6 2(4)  6 2(3)  6 2(1)  6 2(6)  6 2(7)  6

34

34  32 1.8

1.1

Chicago

23

23  32 1.8

San Francisco

55

55  32 1.8

12.7

72

72  32 1.8

22.2

40

40  32 1.8

4.4

5

P  2/  2w 24  2/  2w 24  2w  2/  2w  2w 24  2w  2/ 

2/ 2

12  w 12  1 12  2 12  3 12  4 12  5

w 1 2 3 4 5

 11 10 9 8 7

(w, ) (1, 11) (2, 10) (3, 9) (4, 8) (5, 7)

For the values of w chosen, the solution set is {(1, 11), (2, 10), (3, 9), (4, 8), (5, 7)}.

y 11 2 7 16

45.

x 14 12 4 6 8

Length of Tibia (cm) 30.5 34.8 36.3 37.9

Male Height (cm) 81.7  2.4(30.5) 81.7  2.4(34.8) 81.7  2.4(36.3) 81.7  2.4(37.9)

(T, H ) (30.5, 154.9) (34.8, 165.2) (36.3, 168.8) (37.9, 172.7)

Length of Tibia (cm) 30.5 34.8 36.3 37.9

Female Height (cm) 72.6  2.5(30.5) 72.6  2.5(34.8) 72.6  2.5(36.3) 72.6  2.5(37.9)

(T, H ) (30.5, 148.9) (34.8, 159.6) (36.3, 163.4) (37.9, 167.4)

176 172 168 164 160 156 152 148 144

The domain is {14, 12, 4, 6, 8}. 40. F  1.8C  32 F  32  1.8C  32  32 F  32  1.8C 

New York

12  w  / 43. Since the equation is solved for , the dependent variable is  and the independent variable is w. 44. Sample answer:

The range is {16, 7, 2, 11}. 39. Since the range is given, first solve the equation for x in terms of y. 2y  x  6 2y  x  2y  6  2y x  6  2y 1121x2  11216  2y2 x  2y  6

F  32 1.8 F  32 1.8

C

24  2w 2

38. First solve the equation for y in terms of x. 3x  y  8 3x  y  3x  8  3x y  8  3x

y 4 3 1 6 7

F  32 1.8

Washington, D.C.

y

x 1 2 5 8

F

Miami

Graph the solution set {(4, 4), (2, 3.5), (0, 3), (2, 2.5), (4, 2), (6, 1.5)}.

O

City

1.8C 1.8

C

0

H

Male

Female

T 30 31 32 33 34 35 36 37 38

46. See students’ work.

145

Chapter 4

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47c. Substitute each given range value for y and solve for x. y  04x  16 0 0  04x  16 0 The only number whose absolute value gives 0 is 0, so the expression 4x 16 must represent 0. 4x  16  0 4x  16  16  0  16 4x  16

47a. Substitute each given range value for y and solve for x. y  x2 0  x2 The only number that gives 0 when squared is 0, so x  0 when y  0. 16  x2 Both 42 and (4)2 give 16, so x  4 or x  4 when y  16.

4x 4

36  x2 Both 62 and (6)2 give 36, so x  6 or x  6 when y  36. Thus, if the range of y  x2 is {0, 16, 36}, then the domain is {6, 4, 0, 4, 6}. 47b. Substitute each given range value for y and solve for x. y  04x 0  16 0  04x 0  16 0  16  04x 0  16  16 16  04x 0 Since both 016 0 and 016 0 give 16, the expression 4x may represent either 16 or 16. 4x  16 4x 4



16 4



4x  16  16 4x  16  16  16  16 4x  32 4x 4



4x 4



16 4

4x 4

4x 4



0 4

x0



52 4

4x 4



20 4

x  13 x  5 So, x  13 or x  5 when y  36. Thus, if the range of y  04x  16 0 is {0, 16, 36}, then the domain is {5, 0, 4, 8, 13}. 48. Sample answer: x 1 2 3 4 5

x4 14 24 34 44 54

y 5 6 7 8 9

(x, y) (1, 5) (2, 6) (3, 7) (4, 8) (5, 9)

For the values of x chosen, the relation is {(1, 5), (2, 6), (3, 7), (4, 8), (5, 9)} and the inverse relation is {(5, 1), (6, 2), (7, 3), (8, 4), (9, 5)}. Since each y-coordinate is 4 less than each x-coordinate in each ordered pair of the inverse relation, the equation of the inverse relation is y  x  4.

52 4

x  13 x  13 So, x  13 or x  13 when y  36. Thus, if the range of y  04x 0  16 is {0, 16, 36}, then the domain is {13, 8, 4, 4, 8, 13}.

Chapter 4

32 4

x8

x8 x  8 So, x  8 or x  8 when y  16. 36  04x 0  16 36  16  04x 0  16  16 52  04x 0 Since both 052 0 and 052 0 give 52, the expression 4x may represent either 52 or 52. 4x  52 4x  52 52 4



4x  16  16 4x  16  16  16  16 4x  0

So, x  8 or x  0 when y  16. 36  04x  16 0 Since both 036 0 and 036 0 give 36, the expression 4x  16 may represent either 36 or 36. 4x  16  36 4x  16  36 4x  16  16  36  16 4x  16  16  36  16 4x  52 4x  20

x4 x  4 So, x  4 or x  4 when y  0. 16  04x 0  16 16  16  04x 0  16  16 32  04x 0 Since both 032 0 and 032 0 give 32, the expression 4x may represent either 32 or 32. 4x  32 4x  32 4x 32 4x 32   4 4 4 4

4x 4

16 4

x4 So, x  4 when y  0. 16  04x  16 0 Since both 016 0 and 016 0 give 16, the expression 4x 16 may represent either 16 or 16.

4x  16 4x 4



146

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49. When traveling to other countries, currency and measurement systems are often different. You need to convert these systems to the system with which you are familiar. Answers should include the following. • At the current exchange rate, 15 pounds is roughly 10 dollars and 10 pounds is roughly 7 dollars. Keeping track of every 15 pounds you spend would be relatively easy. • If the exchange rate is 0.90 compared to the dollar, then items will cost less in dollars. For example, an item that is 10 in local currency is equivalent to $9.00. If the exchange rate is 1.04, then items will cost more in dollars. For example, an item that costs 10 in local currency is equivalent to $10.40. 50. D; 3x  y  18 3x  3  18 3x  3  3  18  3 3x  21 3x 3

54.



X .4 .6 1.8 2.2 3.1

55.

TABLE ENTER 2nd TABLE 2.5 ENTER 1.75 ENTER 0 ENTER 1.25 ENTER 3.33 ENTER X 2.5 1.75 0 1.25 3.33

21

CLEAR 3 X,T,␪,n ENTER 2nd

X 11 15 23 44

Page 217

4

KEYSTROKES:

Y1 37 41 65 128

y

CLEAR 6.5 X,T,␪,n ENTER 2nd

X 8 5 0 3 7 12

X’

X

TblSet TABLE 8 ENTER 5 ENTER 0 ENTER 3 ENTER 7 ENTER 12 ENTER

42 2nd

Maintain Your Skills

56. relation: {(4, 9), (3, 2), (1, 5), (4, 2)} inverse: {(9, 4), (2, 3), (5, 1), (2, 4)} 57. relation: {(2, 7), (6, 4), (6, 1), (11, 8)} inverse: {(7, 2), (4, 6), (1, 6), (8, 11)} 58. relation: {(3, 2), (2, 3), (3, 3), (4, 2)} inverse: {(2, 3), (3, 2), (3, 3), (2, 4)} 59. To reflect the triangle over the y-axis, multiply the x-coordinate of each vertex by 1. (x, y) S (x, y) X(6, 4) S X¿(6, 4) Y(5, 0) S Y¿(5, 0) Z(3, 3) S Z¿(3, 3)

The solution set is {(11, 37), (15, 41) , (23, 65), (44, 128)}. 53.

Y1 4.26 3.21 .76 .99 3.902

The solution set is {(2.5, 4.26), (1.75, 3.21), (0, 0.76), (1.25, 0.99), (3.33, 3.90)}.

TblSet TABLE 11 ENTER 15 ENTER 23 ENTER 44 ENTER 2nd

CLEAR 1.4 X,T,␪,n

KEYSTROKES:

0.76 2nd

2/  2w 2

KEYSTROKES:

Y1 13.2 13.8 17.4 18.6 21.3

The solution set is {(0.4, 13.2), (0.6, 13.8), (1.8, 17.4), (2.2, 18.6), (3.1, 21.3)}.

7/w So, the sum of the length and width is 7, and the product of the length and width is 12. Of the dimensions given, this is true only for a 3  4 rectangle. 52.

12

TblSet TABLE 0.4 ENTER 0.6 ENTER 1.8 ENTER 2.2 ENTER 3.1 ENTER 2nd

 3 x7 51. C; P  2/  2w A  /w 14  2/  2w 12  /w 14 2

CLEAR 3 X,T,␪,n ENTER 2nd

KEYSTROKES:

Z’ Y

Y1 94 74.5 42 22.5 3.5 36

Z O

Y’

x

The solution set is {(8, 94), (5, 74.5), (0, 42), (3, 22.5), (7, 3.5), (12, 36)}.

147

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67. Hypothesis: it is hot Conclusion: we will go swimming 68. Hypothesis: you do your chores Conclusion: you get an allowance 69. Hypothesis: 3n  7  17 Conclusion: n  8 70. Hypothesis: a b and b c Conclusion: a c 71. 72. a  15  20 r  9  12 a  15  15  20  15 r  9  9  12  9 a5 r  21 73. 74. 4  5n  6 3  8w  35 4  6  5n  6  6 3  8w  3  35  3 10  5n 8w  32

60. To rotate the quadrilateral 90 counterclockwise about the origin, switch the coordinates of each vertex and then multiply the new first coordinate by 1. (x, y) S (y, x) Q(2, 2) S Q¿(2, 2) R(3, 3) S R¿(3, 3) S(1, 4) S S¿(4, 1) T(4, 3) S T¿(3, 4) y

R’ Q

Q’

S’ x

O

S

T

61.

10 5

R T’

g 4

75. g 4

6 ? 18  45 15 ?

4

11 ? 33  34 12 ?

11(34)  12(33) 374 396 The cross products are not equal, so ratios do not form a proportion.

11 12



33 . 34

The

8(55)  22(20) 440  440 20

The cross products are equal, so 22  55. Since the ratios are equal, they form a proportion. 64.

6 ? 3 4 8 ?

6(4)  8(3) 24  24 6

3

The cross products are equal, so 8  4. Since the ratios are equal, they form a proportion. 65.

3 ? 9  25 5 ?

3(25)  5(9) 75 45 The cross products are not equal, so ratios do not form a proportion. 66.

3 5

9

25. The

26 ? 12  15 35 ?

26(15)  35(12) 390 420 The cross products are not equal, so ratios do not form a proportion.

Chapter 4

26 35

76.

3

1 2  4(3)

5

1

3 m 5 5 m  3 5 m  3 5

2 2

2  5(2)

m  3  10 m  3  3  10  3 m7

4-5

Graphing Linear Equations

Page 221

Check for Understanding

1. The former will be a graph of four points, and the latter will be a graph of a line. 2a. Sample answer: y  8 is a linear equation in the form Ax  By  C, where A  0, B  1, and C  8. 2b. Sample answer: x  5 is a linear equation in the form Ax  By  C, where A  1, B  0, and C  5. 2c. Sample answer: x  y  0 is a linear equation in the form Ax  By  C, where A  1, B  1, and C  0. 3. Determine the point at which the graph intersects the x-axis by letting y  0 and solving for x. Likewise, determine the point at which the graph intersects the y-axis by letting x  0 and solving for y. Draw a line through the two points. 4. Since the term y2 has an exponent of 2, the equation cannot be written in the form Ax  By  C. Therefore, this is not a linear equation. 5. First rewrite the equation so that the variables are on one side of the equation and a constant is on the other side. 3y  2  0 3y  2  2  0  2 3y  2 The equation can be written as 0x  3y  2. Therefore, it is a linear equation in standard form where A  0, B  3, and C  2.

8 ? 20  55 22 ?

8

g 4 g 4

g  12

18

6

The cross products are equal, so 15  45. Since the ratios are equal, they form a proportion.

63.

25

32

 8 w  4

2252

6(45)  15(18) 270  270

62.

8w 8

5n

 5 2  n

12

15. The

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6. To write the equation with integer coefficients, multiply each term by 5. 3 x 5

5

10. Select five values for the domain and make a table.

135 2x  5125 2y  5152

3x  2y  25 The equation is now in standard form where A  3, B  2, and C  25. This is a linear equation. 1

y 1 0 1 2 4

y  2x  8

(x, y) (3, 1) (3, 0) (3, 1) (3, 2) (3, 4)

11. To find the x-intercept, let y  0. y  3  x 0  3  x 0  3  3  x  3 3  x 1(3)  1(x) 3  x The graph intersects the x-axis at (3, 0). To find the y-intercept, let x  0. y  3  x y  3  (0) y  3 The graph intersects the y-axis at (0, 3). Plot these points and draw the line that connects them.

x3

x

9. Solve the equation for y. xy0 xyx0x y  x 11y2  11x2 yx Select five values for the domain and make a table. y 2 1 0 2 4

x

O

y

x 2 1 0 2 4

(x, y) (4, 0) (3, 2) (1, 6) (0, 8) (1, 10)

y

Graph the ordered pairs and draw a line through the points.

O

y 0 2 6 8 10

Graph the ordered pairs and draw a line through the points.

7. Since the term y has a variable in the denominator, the equation cannot be written in the form Ax  By  C. Therefore, this is not a linear equation. 8. The only value in the domain is 3. Since there is no y in the equation, the value of y does not depend on the value on x. Therefore, y can be any real number. Select five values for the range and make a table. x 3 3 3 3 3

2x  8 2(4)  8 2(3)  8 2(1)  8 2(0)  8 2(1)  8

x 4 3 1 0 1

2

 5y  5

y

y  3  x

(x, y) (2, 2) (1, 1) (0, 0) (2, 2) (4, 4)

O

x

Graph the ordered pairs and draw a line through the points. y xy0

O

x

149

Chapter 4

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12. To find the x-intercept, let y  0. x  4y  10 x  4102  10 x  10 The graph intersects the x-axis at (10, 0). To find the y-intercept, let x  0. x  4y  10 0  4y  10 4y  10 4y 4



14. Select five values for the domain and make a table. 0.75m  2.25 0.75(0)  2.25 0.75(4)  2.25 0.75(8)  2.25 0.75(12)  2.25 0.75(16)  2.25

m 0 4 8 12 16

18 16 14 12 Cost ($) 10 8 6 4 2

y

x  4y  10

c

c  0.75m  2.25

x 0

O

12 4

x3 The graph intersects the x-axis at (3, 0). To find the y-intercept, let x  0. 4x  3y  12 4(0)  3y  12 3y  12 3y 3



Pages 221–223

12 3

y 4x  3y  12

Chapter 4

Practice and Apply

16. First rewrite the equation so that the variables are on the same side of the equation. 3x  5y 3x  5y  5y  5y 3x  5y  0 The equation is now in standard form where A  3, B  5, and C  0. This is a linear equation. 17. First rewrite the equation so that the variables are on the same side of the equation. 6  y  2x 6  y  y  2x  y 6  2x  y 2x  y  6 The equation is now in standard form where A  2, B  1, and C  6. This is a linear equation. 18. Since the term 6xy has two variables, the equation cannot be written in the form Ax  By  C. Therefore, this is not a linear equation.

y4 The graph intersects the y-axis at (0, 4). Plot these points and draw the line that connects them.

O

m

2 4 6 8 10 12 14 16 Miles

15. Use the graph to estimate the y-coordinate in the ordered pair that contains 18 as the x-coordinate. The ordered pair (18, 16) appears to be on the line, so an 18-mile taxi ride will cost about $16. To check this solution algebraically, substitute m  18 into the original equation: c  0.75m  2.25 c  0.751182  2.25 c  15.75 An 18-mile taxi ride will cost $15.75.

13. To find the x-intercept, let y  0. 4x  3y  12 4x  3(0)  12 4x  12 

(m, c) (0, 2.25) (4, 5.25) (8, 8.25) (12, 11.25) (16, 14.25)

Graph the ordered pairs and draw a line through the points.

10 4

y  2.5 The graph intersects the y-axis at (0, 2.5). Plot these points and draw the line that connects them.

4x 4

c 2.25 5.25 8.25 11.25 14.25

x

150

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19. First rewrite the equation so that the variables are on one side of the equation and a constant is on the other side. y50 y5505 y  5 The equation can be written as 0x  y  5. Therefore, it is a linear equation in standard form where A  0, B  1, and C  5. 20. First simplify. Then rewrite the equation so that the variables are on the same side of the equation. 7y  2x  5x 7y  7x 7y  7y  7x  7y 0  7x  7y 7x  7y  0 Since the GCF of 7, 7, and 0 is not 1, divide each side by 7. 7(x  y)  0 7(x  y2 7

25. First rewrite the equation so that the variables are on one side of the equation and a constant is on the other side. 3a  b  2  b 3a  b  2  b  b  b 3a  2  0 3a  2  2  0  2 3a  2 The equation can be written as 3a  0b  2. Therefore, it is a linear equation in standard form where A  3, B  0, and C  2. 26. The only value in the range is 1. Since there is no x in the equation, the value of x does not depend on y. Therefore, x can be any real number. Select five values for the domain and make a table. x 3 1 0 2 4

0

7

xy0 The equation is now in standard form where A  1, B  1, and C  0. This is a linear equation. 21. Since the term 4x2 has an exponent of 2, the equation cannot be written in the form Ax  By  C. Therefore, this is not a linear equation. 3 4 22. Since each of the terms x and y has a variable in the denominator, the equation cannot be written in the form Ax  By  C. Therefore, this is not a linear equation. 23. To write the equation with integer coefficients, multiply each term by 6.

6

x 2 x 2

 10 

(x, y) (3, 1) (1, 1) (0, 1) (2, 1) (4, 1)

y 1 1 1 1 1

Graph the ordered pairs and draw a line through the points. y

x

O

y  1

2y 3

1 2  6(10)  61 3 2 2y

27. Select five values for the domain and make a table.

3x  60  4y Then rewrite the equation so that the variables are on the same side of the equation. 3x  4y  60  4y  4y 3x  4y  60 The equation is now in standard form the A  3, B  4, and C  60. This is a linear equation. 24. First rewrite the equation so that the variables are on the same side of the equation. 7n  8m  4  2m 7n  8m  2m  4  2m  2m 7n  6m  4 6m  7n  4 1(6m  7n)  1(4) 6m  7n  4 The equation is now in standard form where A  6, B  7, and C  4. This is a linear equation.

2x 2(2) 2(1) 2(0) 2(1) 2(2)

x 2 1 0 1 2

y 4 2 0 2 4

(x, y) (2, 4) (1, 2) (0, 0) (1, 2) (2, 4)

Graph the ordered pairs and draw a line through the points. y y  2x O

151

x

Chapter 4

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Graph the ordered pairs and draw a line through the points.

28. Select five values for the domain and make a table. 5x 5  (1) 50 51 53 55

x 1 0 1 3 5

y 6 5 4 2 0

y

(x, y) (1, 6) (0, 5) (1, 4) (3, 2) (5, 0)

y  4  3x x

O

Graph the ordered pairs and draw a line through the points. y

31. Select five values for the domain and make a table.

x

O

2x  8 2(1)  8 2(0)  8 2(2)  8 2(4)  8 2(5)  8

y 10 8 4 0 2

x

O

yx6

32. Solve the equation for y. x  3y

y

1 1 (x)  3 (3) y 3 1 xy 3

x

y  2x  8

Select five values for the domain and make a table. 1 x 3

x

30. Select five values for the domain and make a table. x 1 0 1 2 3

4  3x 4  3(1) 4  3(0) 4  3(1) 4  3(2) 4  3(3)

y 7 4 1 2 5

(x, y) (1, 7) (0, 4) (1, 1) (2, 2) (3, 5)

y

(x, y) (3, 1)

3

1 (3) 3

1

1

1 (1) 3

1 3

0

1 (0) 3

0

1

1 (1) 3

1 3

3

1 (3) 3

1

11, 13 2

(0, 0)

11, 13 2

(3, 1)

Graph the ordered pairs and draw a line through the points. y

x  3y O

Chapter 4

(x, y) (1, 7) (0, 6) (2, 4) (4, 2) (6, 0)

y

(x, y) (1, 10) (0, 8) (2, 4) (4, 0) (5, 2)

Graph the ordered pairs and draw a line through the points.

O

y 7 6 4 2 0

Graph the ordered pairs and draw a line through the points.

29. Select five values for the domain and make a table. x 1 0 2 4 5

x6 1  6 06 26 46 66

x 1 0 2 4 6

y5x

152

x

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35. Solve the equation for y. x  3y  9 x  3y  x  9  x 3y  9  x

33. Solve the equation for y. x  4y  6 x  6  4y  6  6 x  6  4y x  6 4 x  6 4



3y 3

4y 4

y



y

Select five values for the domain and make a table. x  6 4

x

y

6

6  6 4

4

4  6 4

2

2  6 4

1

0

0  6 4

1 12

1

1  6 4

14

Select five values for the domain and make a table.

(x, y)

1 2

14, 2 1 2

(2, 1)

3

9  x 3

y

(x, y)

3

9  (3) 3

4

(3, 4)

2

9  (2) 3

2 33

0

9  0 3

3

1

9  1 3

23

3

9  3 3

2

x

(6, 0)

0

9  x 3 9  x 3

10, 2 11, 134 2 1 12

Graph the ordered pairs and draw a line through the points.

12, 323 2

(0, 3)

11, 223 2

2

(3, 2)

Graph the ordered pairs and draw a line through the points.

y

y

x  4y  6 x  3y  9 O

x

34. Solve the equation for y. x  y  3 x  y  x  3  x y  3  x 1(y)  1(3  x) y3x Select five values for the domain and make a table. 3x 3  (3) 3  (1) 30 31 32

x 3 1 0 1 2

x

O

y 0 2 3 4 5

36. To find the x-intercept, let y  0. 4x  6y  8 4x  6(0)  8 4x  8 4x 4

8

4

x2 The graph intersects the x-axis at (2, 0). To find the y-intercept, let x  0. 4x  6y  8 4(0)  6y  8 6y  8

(x, y) (3, 0) (1, 2) (0, 3) (1, 4) (2, 5)

6y 6

8

6 4

1

y  3 or 13

Graph the ordered pairs and draw a line through the points.

1

1

2

The graph intersects the y-axis at 0, 13 . Plot these points and draw the line that connects them.

y

y

x  y  3

4x  6y  8 O

x O

153

x

Chapter 4

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37. To find the x-intercept, let y  0. 3x  2y  15 3x  2(0)  15 3x  15 3x 3



39. Solve the equation for y. 2.5x  5y  75 2.5x  5y  2.5x  75  2.5x 5y  75  2.5x 5y 5

15 3

x5 The graph intersects the x-axis at (5, 0). To find the y-intercept, let x  0. 3x  2y  15 3(0)  2y  15 2y  15 2y 2

15

3x  2y  15

40. To find the x-intercept, let y  0.

38. To find the x-intercept, let y  0. 1.5x  y  4 1.5x  (0)  4 1.5x  4

1 2

1 2

xy4

x  102  4

4

 1.5

1

2

2

2

1 2 1 2

x4

1 2 x  2(4)

x8 The graph intersects the x-axis at (8, 0). To find the y-intercept, let x  0.

The graph intersects the x-axis at 23, 0 . To find the y-intercept, let x  0. 1.5x  y  4 1.5(0)  y  4 y4 The graph intersects the y-axis at (0, 4). Plot these points and draw the line that connects them.

1 2

xy4

1 (0) 2

y4

y4 The graph intersects the y-axis at (0, 4). Plot these points and draw the line that connects them.

y

y 1.5x  y  4 O

Chapter 4

(x, y) (2, 16) (0, 15) (2, 14) (6, 12) (10, 10)

y 16 14 12 10 8 2.5x  5y  75 6 4 2 O 42 2 4 6 8 10 12x

x

2

y 16 15 14 12 10

Graph the ordered pairs and draw a line through the points.

y

x  23

15  0.5x 15  0.5(2) 15  0.5(0) 15  0.5(2) 15  0.5(6) 15  0.5(10)

x 2 0 2 6 10

y  7.5 The graph intersects the y-axis at (0, 7.5). Plot these points and draw the line that connects them.

1.5x 1.5

75  2.5x 5

y  15  0.5x Select five values for the domain and make a table.

 2

O



x

O

154

1 xy4 2

x

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41. To find the x-intercept, let y  0.

Plot these points and draw the line that connects them.

2 3

x y1 2

x  3 (0)  1

2

x1 The graph intersects the x-axis at (1, 0). To find the y-intercept, let x  0.

y

4x 3y 41 3

1

2

2

x3y1

O

2x

1

1

2

(0)  3 y  1 2

3 y  1

3

1 22

3

43. Solve the equation for y.

2 3 y  2 (1)

1

1 2x 

x

3y 1 4 3(0) 1 4

10 3

(x, y)

y



10 3

7 312

1 (0) 4



10 3

33

2

1 (2) 4



10 3

26

4

1 (4) 4



10 3

23

6

1 (6) 4



10 3

16

1 0

1 5 1 5

11, 3127 2 10, 313 2 12, 256 2 14, 213 2 16, 156 2

y

3 (1) 4

 

0 01 1  4 (1) 3 4 3 1 13



Graph the ordered pairs and draw a line through the points.

x  0.75 The graph intersects the x-axis at (0.75, 0). To find the y-intercept, let x  0. 4x 3 4(0) 3

10

1 (1) 4

1

3 4 4 3

1

1 x 4

x

42. To find the x-intercept, let y  0. 

1 1

2



1

y  4x  3 Select five values for the domain and make a table.

x  3y  1

4x 3 4x 3 4x 3

1

y  3  3  4x  3  3

y

O

1

y  3  4x  3

y  1.5 The graph intersects the y-axis at (0, 1.5). Plot these points and draw the line that connects them.



3y 1 4 3y 1 4 3y 1 4 3y 1 4 3y 4 4 3 y 3 4

1

1

y  3  4x  3

44. To find the x-intercept, let y  0. 4x  7y  14 4x  7(0)  14 4x  14

1

12

4x 4



x

y y

x

O

1

1

14 4 7 2

The x-intercept is

2

7 2

1

or 32.

To find the y-intercept, let x  0. 4x  7y  14 4(0)  7y  14 7y  14

The graph intersects the y-axis at 0,  13 .

7y 7

14

 7

y  2. The y-intercept is 2.

155

Chapter 4

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45. Since the x-intercept is 3, the graph intersects the x-axis at (3, 0). Since the y-intercept is 5, the graph intersects the y-axis at (0, 5). Both points, (3, 0) and (0, 5), lie on the line, so must make Ax  By  C true. Substituting (3, 0) for (x, y), we have A(3)  B(0)  C, or 3A  C. Substituting (0, 5) for (x, y) we have A(0)  B(5)  C, or 5B  C. Since 3A  C and 5B  C, we can conclude that 3A  5B. Since A and B must be integers with a GCF of 1, A  5 and B  3. This means C  15. Therefore, the equation of the line with an x-intercept of 3 and a y-intercept of 5 in standard form is 5x  3y  15. 46. 2/  2w  P 2(2x)  2( y)  30 4x  2y  30 Since the GCF of 4, 2, and 30 is not 1, the equation is not written in standard form. Divide each side by the GCF. 2(2x  y)  30 2(2x  y) 2



49.



Distance (miles)

30 2

3 0.21

15 2

d  0.21t t 2 4 6 8 10 12 14 Time (seconds)



0.21t 0.21

14.29  t So, it will take about 14 s to hear the thunder from a storm 3 mi away. 52. Select five values for w and make a table. 0.07w 0.07(100) 0.07(140) 0.07(220) 0.07(260) 0.07(300)

w 100 140 220 260 300

y 7 9.8 15.4 18.2 21

(w, y) (100, 7) (140, 9.8) (220, 15.4) (260, 18.2) (300, 21)

Graph the ordered pairs in the table and draw a line through the points. 24

y

Pints of Blood

20

x

16 12 8 0

Chapter 4

d

51. Use the graph to estimate the value of the x-coordinate in the ordered pair that contains 3 as the y-coordinate. The ordered pair (14, 3) appears to be on the line. To check this solution algebraically, substitute d  3 into the original equation: d  0.21t 3  0.21t

y

O

5 4 3 2 1 0

x  7.5 The x-intercept of the graph is 7.5. To find the y-intercept, let x  0. 2x  y  15 2(0)  y  15 y  15 The y-intercept of the graph is 15. 48. Since the x-intercept is 7.5, the graph crosses the x-axis at (7.5, 0). Since the y-intercept is 15, the graph crosses the y-axis at (0, 15). Plot these points and draw the line that connects them. 18 16 14 12 10 8 6 4 2

(t, d) (0, 0) (2, 0.42) (4, 0.84) (6, 1.26) (8, 1.68) (10, 2.1) (12, 2.52) (14, 2.94) (16, 3.36)

d 0 0.42 0.84 1.26 1.68 2.1 2.52 2.94 3.36

50. Graph the ordered pairs in the table and draw a line through the points.

2x  y  15 47. To find the x-intercept, let y  0. 2x  y  15 2x  (0)  15 2x  15 2x 2

0.21t 0.21(0) 0.21(2) 0.21(4) 0.21(6) 0.21(8) 0.21(10) 0.21(12) 0.21(14) 0.21(16)

t 0 2 4 6 8 10 12 14 16

156

w 100 140 180 220 260 300 340 Weight (lb)

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58. You can graph an equation that represents how many Calories and nutrients your diet should contain. Since your diet is different every day, it is easier to use the graph to determine your goal instead of making calculations every day. Answers should include the following. • Nutrition information labels provide facts about how many grams of fat are in each serving and/or how many Calories are from fat. • The number of grams of protein would equal 10% of the total number of Calories divided by 4 Calories per gram or p  0.025C. 59. A; x 3x  5 y (x, y)

53. Use the graph to estimate the value of the x-coordinate in the ordered pair that contains 12 as the y-coordinate. The ordered pair (170, 12) appears to be on the line. To check this solution algebraically, substitute y  12 into the original equation: y  0.07w 12  0.07w 12 0.07



0.07w 0.07

171.43  w So, a person weighing about 171 lb will have 12 pt of blood. 54. Select five values for d and make a table. 0.43d  14.7 0.43(0)  14.7 0.43(20)  14.7 0.43(60)  14.7 0.43(80)  14.7 0.43(100)  14.7

d 0 20 60 80 100

p 14.7 23.3 40.5 49.1 57.7

(d, p) (0, 14.7) (20, 23.3) (60, 40.5) (80, 49.1) (100, 57.7)

1

2

(1, 2)

60. B; of the given points, only (2, 2) appears to lie on the line.

Page 223 61.

Graph the ordered pairs in the table and draw a line through the points. p 100 80

x 3 1 2 5 8

Maintain Your Skills x5 3  5 1  5 25 55 85

y 8 6 3 0 3

(x, y) (3, 8) (1, 6) (2, 3) (5, 0) (8, 3)

The solution set is {(3, 8), (1, 6), (2, 3), (5, 0), (8, 3)}.

60 40

62.

20 O

3(1)  5

20

40

60

80

100

d

55. Substitute d  400 into the equation: p  0.43d  14.7 p  0.43(4002  14.7 p  186.7 At a depth of 400 ft, the pressure is 186.7 psi. 56. The pressure at sea level is 14.7 psi. The pressure 186.7 at 400 ft, 186.7 psi, is 14.7  12.7 times as great. 57. Solve the equation for y in terms of x. 2x  y  8 2x  y  2x  8  2x y  8  2x 1(y)  1(8  2x) y  2x  8 Substitute the value for x into the expression 2x  8. If the value of y is less than the value of 2x  8, then the point lies below the line. If the value of y is greater than the value of 2x  8, then the point lies above the line. If the value of y is equal to the value of x, then the point lies on the line. Sample answers: (1, 5) Since 5 7 2(1)  8, the point (1, 5) lies above the line. (5, 1) Since 1 6 2(5)  8, the point (5, 1) lies below the line. (6, 4) Since 4  2(6)  8, the point (6, 4) lies on the line.

x 3 1 2 5 8

2x  1 2(3)  1 2(1)  1 2(2)  1 2(5)  1 2(8)  1

y 5 1 5 11 17

(x, y) (3, 5) (1, 1) (2, 5) (5, 11) (8, 17)

The solution set is {(3, 5), (1, 1), (2, 5), (5, 11), (8, 17)}. 63. First solve the equation for y in terms of x. 3x  y  12 3x  y  3x  12  3x y  12  3x x 3 1 2 5 8

12  3x 12  3(3) 12  3(1) 12  3(2) 12  3(5) 12  3(8)

y 21 15 6 3 12

(x, y) (3, 21) (1, 15) (2, 6) (5, 3) (8, 12)

The solution set is {(3, 21), (1, 15), (2, 6), (5, 3), (8, 12)}.

157

Chapter 4

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64. First solve the equation for y in terms of x. 2 x  y  3 2 x  y  2x  3  2x y  3  2x 1(y2  113  2x2 y  3  2x 3  2x

x 3 1 2 5 8

3 3 3 3 3

    

2(3) 2(1) 2(2) 2(5) 2(8)

y 3 1 7 13 19

67.

3x  3x 

1 y 2

Y

(3, 3) (1, 1) (2, 7) (5, 13) (8, 19)

3 4 3

5 1 2 1

The domain of this relation is {4, 3, 3}. The range is {1, 1, 2, 5}. 68.

6

 3x  6  3x 1

1 2y  2(6  3x)

x 4 2 1 4

1 2

6x  12 6(3)  12 6(1)  12 6(2)  12 6(5)  12 6(8)  12

y 30 18 0 18 36

(x, y) (3, 30) (1, 18) (2, 0) (5, 18) (8, 36) 69.

1

2x  3 y  4 1

2x  3 y  2x  4  2x

3

x 3 1 2 5 8

12

 4  2x  3(4  2x)

y  12  6x

12  6x 12  6(3) 12  6(1) 12  6(2) 12  6(5) 12  6(8)

y 6 6 24 42 60

(x, y) (3, 6) (1, 6) (2, 24) (5, 42) (8, 60)

Y

4 2 1

0 3 4

x 1 3 1 3

y

y 4 0 1 5

X

Y

1 3 1

4 0 1 5

O

The domain of the relation is {1, 1, 3}. The range is {1, 0, 4, 5}.

The solution set is {(3, 6), (1, 6), (2, 24), (5, 42), (8, 60)}.

Chapter 4

x

O

The domain of the relation is {1, 2, 4}. The range is {3, 0, 4}.

The solution set is {(3, 30), (1, 18), (2, 0), (5, 18), (8, 36)}. 66. First solve the equation for y in terms of x.

1 y 3 1 y 3

y

y 0 3 3 4

X

y  6x  12

x 3 1 2 5 8

x

O

X

2 y  6  3x

2

y

y 5 1 2 1

(x, y)

The solution set is {(3, 3), (1, 1), (2, 7), (5, 13), (8, 19)}. 65. First solve the equation for y in terms of x. 1 y 2

x 3 4 3 3

158

x

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70.

x 4 2 4 3

74.

y

y 5 5 1 2 O

X

Y

4 2 3

5 1 2

x

6(x  32  3x 6x  18  3x 6x  18  3x  3x  3x 3x  18  0 3x  18  18  0  18 3x  18 3x 3

?

6[ (6)  3]  3(6) ?

6(3)  18 18  18 ✓ 75. The lowest value is 1, and the highest value is 40, so use a scale that includes these values. Place an X above each value for each occurrence.                   

The domain of the relation is {2, 3, 4}. The range is {1, 2, 5}. 2(x  22  3x  (4x  52 2x  4  3x  4x  5 2x  4  x  5 2x  4  x  x  5  x 3x  4  5 3x  4  4  5  4 3x  9 3x 3

Check:

0

9

x3 2(x  2)  3x  (4x  5) ?

2(3  2)  3(3)  [4(3)  5] ?

2(1)  9  (12  5) ?

2  9  (7) 22✓ 72. 3a  8  2a  4 3a  8  2a  2a  4  2a a  8  4 a  8  8  4  8 a  12 Check: 3a  8  2a  4 ? 3(12)  8  2(12)  4 ?

36  8  24  4 28  28 ✓ 73. 3n  12  5n  20 3n  12  3n  5n  20  3n 12  2n  20 12  20  2n  20  20 8  2n

Check:



2

4

6

8 10 12 14 16 18 20 22 24

40

76. Looking at the line plot, we can see that there are 12 Xs above the numbers between 7 and 16, inclusive. So, 12 animals have average life spans between 7 and 16 years, inclusive. 77. Since there are more Xs above 15 than above any other number, we can easily see that an average life span of 15 years occurs most frequently. 78. Since there are 4 Xs above numbers 20 and greater, there are 4 animals with average life spans of at least 20 years. 79. 19  5  4  19  20  39 80. (25  4)  (22  13)  (25  4)  (4  1)  21  3 7 81. 12  4  15  3  3  45  48 82. 12(19  15)  3  8  12(4)  3  8  48  24  24 83. 6(43  22)  6(64  4)  6(68)  408 84. 7[43  2(4  3)]  7  2  7[43  2(4  3 )]  7  2

3

8 2

18 3

x  6 6(x  3)  3x

Check:

71.



     

2n 2

4n 3n  12  5n  20 ? 3(4)  12  5(4)  20 ?

12  12  20  20 00✓

159

7[64 2(7)]  7  2 7(64  14)  7  2 7(50)  7  2 350  7  2 50  2 52

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5. First solve the equation for y. x  y  2 x  y  x  2  x y  2  x

Pages 224–225 Graphing Calculator Investigation (Follow-Up of Lesson 4-5) 1.

KEYSTROKES:

CLEAR

X,T,␪,n

2

ZOOM 6

[⫺10, 10] scl: 1 by [⫺10, 10] scl: 1 [⫺10, 10] scl: 1 by [⫺10, 10] scl: 1

2.

KEYSTROKES:

CLEAR 2

KEYSTROKES:

CLEAR 4 X,T,␪,n

5

X,T,␪,n

ZOOM 6

ZOOM 6

6. First solve the equation for y. x  4y  8 x  4y  x  8  x 4y  8  x 4y 4



y

8  x 4 x  8 4

[⫺10, 10] scl: 1 by [⫺10, 10] scl: 1

3.

KEYSTROKES:

CLEAR 6

5 X,T,␪,n

ZOOM 6 [⫺10, 10] scl: 1 by [⫺10, 10] scl: 1 KEYSTROKES:

8 7.

[⫺10, 10] scl: 1 by [⫺10, 10] scl: 1

)

CLEAR

(

X,T,␪,n

⫼ 4 ZOOM 6

KEYSTROKES:

CLEAR 5 X,T,␪,n

9

ZOOM 6

4. First solve the equation for y. 2x  y  6 2x  y  2x  6  2x y  6  2x

Since the origin and the x- and y-intercepts of the graph are displayed, the graph is complete.

[⫺10, 10] scl: 1 by [⫺10, 10] scl: 1 [⫺10, 10] scl: 1 by [⫺10, 10] scl: 1 KEYSTROKES:

CLEAR 6

8.

KEYSTROKES:

CLEAR 10 X,T,␪,n

2 X,T,␪,n

Since the origin and the x- and y-intercepts of the graph are displayed, the graph is complete.

ZOOM 6

[⫺10, 10] scl: 1 by [⫺10, 10] scl: 1

Chapter 4

6

ZOOM 6

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9.

KEYSTROKES:

CLEAR 3 X,T,␪,n

11. First solve the equation for y. 4x  2y  21 4x  2y  4x  21  4x 2y  21  4x

18

ZOOM 6 Since the y-intercept is outside of the standard viewing window, the graph is not complete. Find the y-intercept. y  3x  18 y  3(0)  18 y  18

2y 2

CLEAR 10.5

KEYSTROKES:

2 X,T,␪,n

ZOOM 6

[10, 10] scl: 1 by [10, 10] scl: 1

One possible viewing window that will show a complete graph is [2, 10] scl: 1 by [20, 6] scl: 2.

Since the y-intercept is outside of the standard viewing window, the graph is not complete. Find the y-intercept. 4x  2y  21 4102  2y  21 2y  21 2y 2



21 2

y  10.5 One possible viewing window that will show a complete graph is [5, 10] scl: 1 by [5, 15] scl: 1.

[2, 10] scl: 1 by [20, 6] scl: 2

10. First solve the equation for y. 3x  y  12 3x  y  3x  12  3x y  12  3x 11y2  112  3x2 y  3x  12

CLEAR 3 X,T,␪,n

21  4x 2

y  10.5  2x

[10, 10] scl: 1 by [10, 10] scl: 1

KEYSTROKES:



12

ZOOM 6

[5, 10] scl: 1 by [5, 15] scl: 1

[10, 10] scl: 1 by [10, 10] scl: 1

Since the y-intercept is outside of the standard viewing window, the graph is not complete. Find the y-intercept. 3x  y  12 3102  y  12 y  12 11y2  11122 y  12 One possible viewing window that will show a complete graph is [2, 8] scl: 1 by [15, 5] scl: 1.

[2, 8] scl: 1 by [15, 5] scl: 1

161

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4. The mapping represents a function since, for each element of the domain, there is only one corresponding element in the range. It does not matter if two elements of the domain, 2 and 4, are paired with the same element in the range. 5. The table represents a relation that is not a function. The element 2 in the domain is paired with both 1 and 4 in the range. If you are given that x is 2, you cannot determine the value of y. 6. Since an element of the domain is paired with more than one element in the range, the relation is not a function. If you are given that x is 24, you cannot determine the value of y since both 1 and 5 in the range are paired with x  24. 7. Graph the equation using the x- and y-intercepts. For each value of x, a vertical line passes through no more than one point on the graph. Thus, the line represents a function.

12. First solve the equation for y. 3x  5y  45 3x  5y  3x  45  3x 5y  45  3x 

45  3x 5

y

45  3x 5

5y 5

CLEAR

KEYSTROKES:

)

X,T,␪,n

(

45

3

⫼ 5 ZOOM 6

[10, 10] scl: 1 by [10, 10] scl: 1

Since the x-intercept is outside of the standard viewing window, the graph is not complete. Find the x-intercept. 3x  5y  45 3x  5102  45 3x  45 3x 3

y

x  15 One possible viewing window that will show a complete graph is [20, 4] scl: 2 by [10, 5] scl: 1.

8. The vertical line x  1 intersects the graph at two points, (1, 2) and (1, 1). Thus, the relation graphed does not represent a function. 9. For each value of x, a vertical line passes through no more than one point on the graph. Thus, the relation graphed represents a function. 10. f(x)  4x  5 11. g(x)  x 2  1 f(2)  4(2)  5 g(1)  (1) 2  1 85 11 3 2

[20, 4] scl: 2 by [10, 5] scl: 1

13. See students’ work. 14. The complete graph is in the standard viewing window if 10 b 10. 15. b is the y-intercept of the graph.

4-6

x

O

 45

12. f(x)  4x  5 f(c)  4(c)  5  4c  5

g(x)  x 2  1

g(t)  4  [ (t) 2  1]  4  t2  1  4  t2  3 14. f(x)  4x  5 15. f(x)  4x  5 f(3a2 )  4(3a2 )  5 f(x  5)  4(x  5)  5  12a2  5  4x  20  5  4x  15 16. D;

Functions

Pages 228–229

13.

Check for Understanding

1. y is not a function of x since 3 in the domain is paired with 2 and 3 in the range. x is not a function of y since 3 in the domain of the inverse is paired with 4 and 3 in the range. 2. Sample answer: # x #  x  1 3. x  c, where c is any constant is a linear equation that is not a function, since the value of c in the domain is paired with every real number.

x**  2x  1 5**  2**  [2(5)  1]  [ 2(2)  1]  (10  1)  (4  1) 93 6

Pages 229–231

Practice and Apply

17. The mapping does not represent a function since the element 3 in the domain is paired with two elements in the range.

Chapter 4

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27. Graph the equation. For each value of x, a vertical line passes through no more than one point on the graph. Thus, the equation represents a function.

18. The mapping represents a function since, for each element in the domain, there is only one corresponding element in the range. 19. The table represents a relation that is a function since each element in the domain is paired with exactly one element in the range. 20. The table represents a relation that is not a function. The element 3 in the domain is paired with both 6 and 2 in the range. 21. For each value of x, a vertical line passes through no more than one point on the graph. Thus, the relation graphed represents a function. 22. For each value of x, a vertical line passes through no more than one point on the graph. Thus, the relation graphed represents a function. 23. Since each element of the domain is paired with exactly one element of the range, the relation is a function. 24. Since an element of the domain is paired with more than one element in the range, the relation is not a function. Both 5 and 7 in the range are paired with x  4. 25. Graph the equation. For each value of x, a vertical line passes through no more than one point on the graph. Thus, the equation represents a function. 16 12 8 4

y

28. Graph the equation. For each value of x, a vertical line passes through no more than one point on the graph. Thus, the equation represents a function. y

O

x

29. The line x  1 intersects the graph at (1, 1), (1, 2), (1, 1.5), and many other points. Thus, the relation graphed does not represent a function. 30. The vertical line x  2 intersects the graph at two points, (2, 2) and (2, 2). Thus, the relation graphed does not represent a function. 31. For each value of x, a vertical line passes through no more than one point on the graph. Thus, the relation graphed represents a function. 32. f(x)  3x  7 33. f(x)  3x  7 f(3)  3(3)  7 f(2)  3(2)  7 97  6  7  16 1

y

–16 –12 –8 –4 O 4 8 12 16 x –4 –8 –12 –16

y  8 26. Graph the equation. The vertical line x  15 intersects the graph at more than one point. Thus, the equation does not represent a function. 20 15 10 5

x

O

34. g(x)  x2  2x g(5)  (5) 2  2(5)  25  10  15

y

35. g(x)  x2  2x g(0)  (0) 2  2(0) 00 0

g(x)  x2  2x g(3)  1  [ (3) 2  2(3) ]  1  [ 9  (6) ]  1  15  1  16 37. f(x)  3x  7 f(8)  5  [ 3(8)  7]  5  (24  7)  5  31  5  26 36.

–20 –15–10 –5 O 5 10 15 20 x –5 –10 –15 –20

x  15

38. g(x)  x2  2x g(2c)  (2c) 2  2(2c)  4c2  4c 40. f(x)  3x  7 f(K  2)  3(K  2)  7  3K  6  7  3K  13

163

39. f(x)  3x  7 f(a2 )  3(a2 )  7  3a2  7

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f(x)  3x  7 f(2m  5)  3(2m  5)  7  6m  15  7  6m  8

42.

g(x)  x2  2x 3[g(x)  4]  3[ (x2  2x)  4]  3(x2  2x  4)  3x2  6x  12 f(x)  3x  7 2[f(x2 )  5]  2[ (3x2  7)  5]  2(3x2  2)  6x2  4 a; The cost is the same for 0–1 hour, excluding 0. Then it jumps after 1 hour and remains the same up to 2 hours. So the line is constant for x values between 0 and 1, then it jumps at 1. This trend continues at each hourly interval until 5 hours; then the cost is constant. This trend is best represented by graph a. Since, for each value of x, a vertical line passes through no more than one point on the graph, the relation represents a function. Since the equation t  77  0.005h is solved for t, h is the independent variable and t is the dependent variable. This means h represents the elements of the domain. Therefore, in function notation, the corresponding elements in the range are represented by f(h). Thus, the original equation can be written in function notation as f(h)  77  0.005h. f(100)  77  0.005(100)  77  0.5  76.5 f(200)  77  0.005(200)  77  1  76 f(1000)  77  0.005(1000)  77  5  72 Using the function values determined in Exercise 46, make a table.

43.

44.

45.

46.

47.

h 100 200 1000 2000 5000

f(h) 76.5 76 72 67 52

48. It appears that the ordered pair (4000, 55) represents a point that lies on the line. To check this solution algebraically, substitute h  4000 into the function: f(4000)  77  0.005(4000)  77  20  57 The temperature at 4000 ft is 57F. 49. Select five values for s and make a table.

Temperature (°F)

O

Chapter 4

f(s) 72 152 232 312 392

(s, f(s)) (0, 72) (100, 152) (200, 232) (300, 312) (400, 392)

Graph the ordered pairs and draw the line that connects them. 450 400 350 300 250 200 150 100 50

f (s )

O

f (s)  0.8s  72

100

200 300 400 s

Science Scores

50. It appears that (290, 308) is a point that lies on the line. To check this solution algebraically, substitute f(s)  308 into the function: f(s)  0.8s  72 308  0.8s  72 308  72  0.8s  72  72 236  0.8s 236 0.8



0.8s 0.8

295  s A science score of 295 corresponds to a math score of 308. 51. Krista’s math score is above the average because the point at (260, 320) lies above the graph of the line for f(s). 52. The set of ordered pairs {(1, 2), (3, 4), (5, 6)} represents a function since each element of the domain is paired with exactly one element of the range. The inverse of this function, {(2, 1), (4, 3), (6, 5)}, is also a function. However, {(1, 2), (3, 2), (5, 6)} is a function, while its inverse {(2, 1), (2, 3), (6, 5)} is not a function since the element 2 in the domain is paired with both 1 and 3 in the range. It is sometimes true that the inverse of a function is also a function.

(h, f(h)) (100, 76.5) (200, 76) (1000, 72) (2000, 67) (5000, 52)

Graph the ordered pairs and draw the line that connects them. 90 80 70 60 50 40 30 20 10

0.8s  72 0.8(0)  72 0.8(100)  72 0.8(200)  72 0.8(300)  72 0.8(400)  72

s 0 100 200 300 400

Math Scores

41.

t t  770.005h

1000 2000 3000 4000 h Height (ft)

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Graph the ordered pairs and draw a line through the points.

53. Functions can be used in meteorology to determine if there is a relationship between certain weather conditions. This can help to predict future weather patterns. Answers should include the following. • As barometric pressure decreases, temperature increases. As barometric pressure increases, temperature decreases. • The relation is not a function since there is more than one temperature for a given barometric pressure. However, there is still a pattern in the data and the two variables are related. 54. A; f(x)  20  2x f(7)  20  2(7)  20  14 6 55. A; I. If f(x)  2x, then f(3x)  2(3x)  6x and 3[f(x) ]  3(2x)  6x. So it is true that f(3x)  3[f(x) ]. II. If f(x)  2x, then f(x  3)  2(x  3)  2x  6 and f(x)  3  2x  3. So it is not true that f(x  3)  f(x)  3.

y

x

O

y  2x  4

58. To find the x-intercept, let y  0. 2x  5y  10 2x  5(0)  10 2x  10 2x 2



10 2

x5 The graph intersects the x-axis at (5, 0). To find the y-intercept, let x  0. 2x  5y  10 2(0)  5y  10 5y  10 5y 5

III. If f(x)  2x, then f(x2 )  2(x2 )  2x2 and [ f(x) ] 2  (2x) 2  4x2. So it is not true that f(x2 )  [f(x) ] 2. Thus, only statement I is true.



10 5

y2 The graph intersects the y-axis at (0, 2). Plot these points and draw the line that connects them. y

Page 231

Maintain Your Skills

56. Select five values for the domain and make a table. x3 3  3 1  3 03 13 23

x 3 1 0 1 2

y 0 2 3 4 5

2x  5y  10

(x, y) (3, 0) (1, 2) (0, 3) (1, 4) (2, 5)

O

59.

Graph the ordered pairs and draw a line through the points. y yx3

x

O

x 3

y 12

1

2

2

7

1

8

x

y  5x  3 12  5(3)  3 12  12 2  5(1)  3 2  2 7  5(2)  3 7  13 8  5(1)  3 8  8

True or False? true ✓ false false true ✓

The solution set is {(3, 12), (1, 8)}.

57. Select five values for the domain and make a table. x 0 1 2 3 4

2x  4 2(0)  4 2(1)  4 2(2)  4 2(3)  4 2(4)  4

y 4 2 0 2 4

(x, y) (0, 4) (1, 2) (2, 0) (3, 2) (4, 4)

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60.

x 3

y 0

1

4

6

0

5

1

y  2x  6 0  2(3)  6 0  12 4  2(1)  6 44 0  2(6)  6 0  18 1  2(5)  6 1  16

Page 231

True or False?

1. false true ✓ false false



2.

26 t

6.2(t)  0.75(26) 6.2t  19.5 6.2t 6.2



3 . 5

65. 12  16  12  (16)  ( 016 0  012 0 )  (16  12)  4 66. 5  (8)  5  [(8) ]  5  8  ( 08 0  05 0 )  (8  5) 3 67. 16  (4)  16  4  20 68. 9  6  9  (6)  ( 09 0  06 0 )  (9  6)  15 69.

3 4

1

6

1

888 5

8

1

y 5 1 4 10 16

(x, y) (3, 5) (1, 1) (0, 4) (2, 10) (4, 16)

8  x 2

y

(x, y)

3

8  (3)

2

5.5

(3, 5.5)

1

8  (1) 2

4.5

(1, 4.5)

0

8  0 2

4

(0, 4)

2

8  2 2

3

(2, 3)

4

8  4 2

2

(4, 2)

The solution set is {(3, 5.5), (1, 4.5), (0, 4), (2, 3), (4, 2)}. 4. Select five values for the domain and make a table. x2 2  2 02 12 22 42

x 2 0 1 2 4 2

1

2

3

4

 36  16 

y 4 2 1 0 2

(x, y) (2, 4) (0, 2) (1, 1) (2, 0) (4, 2)

Graph the ordered pairs and draw a line through the points.

70. 32  (13 )  32  13 

y

7 46 1 56

x O

yx2

Chapter 4

(x, y) (3, 2) (1, 4) (0, 5) (2, 7) (4, 9)

8  x 2 8  x 2

x

Multiplicative Identity Property 



y

3

n  1, since

3x  4 3(3)  4 3(1)  4 3(0)  4 3(2)  4 3(4)  4

x 3 1 0 2 4

2y 2

64. 5n  5 3 (1) 5

y 2 4 5 7 9

The solution set is {(3, 5), (1, 1), (0, 4), (2, 10), (4, 16)}. 3. First solve the equation for y in terms of x. x  2y  8 x  2y  x  8  x 2y  8  x

19.5 6.2

t  3.15 It will take Adam about 3.15 h, or approximately 3 h 9 min, to finish the marathon. 62. 16  n  16 Additive Identity Property n  0, since 16  0  16. 63. 3.5  6  n  6 Reflexive Property of Equality n  3.5, since 3.5  6  3.5  6. 3

x5 3  5 1  5 05 25 45

x 3 1 0 2 4

The solution set is {(3, 2), (1, 4), (0, 5), (2, 7), (4, 9)}.

The solution set is {(1, 4)}. 61. Let t represent the number of hours it will take Adam to finish the 26-mile marathon. (45 min  0.75 h) 6.2 0.75

Practice Quiz 2

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5. To find the x-intercept, let y  0. 3x  2y  6 3x  2(0)  6 3x  6 3x 3

3. a2  a1  4 is a formula for a2 in terms of a1. A formula for a3 in terms of a1 is a  a  4  3 1 4 or a3  a1  2(4). A formula for a4 in terms of a1 is a4  a1  4  4  4 or a4  a1  3(4). 4. Each term can be found by adding the first term a1 to the product of 4 and 1 less than the position n of the term in the sequence. an  a1  (n  1)4

6

3

x2 The graph intersects the x-axis at (2, 0). To find the y-intercept, let x  0. 3x  2y  6 3(0)  2y  6 2y  6 2y 2

5. The 21st term in the sequence is a21. Substitute n  21 and a1  7 into the equation from Exercise 4. a  a  (n  1)4 n 1

6

2

a 21 a21 a21 a21

y3 The graph intersects the y-axis at (0, 3). Plot these points and draw the line that connects them.

 7  (21  1)4  7  (20)4  7  80  87

The 21st term in the sequence is 87.

y 3x  2y  6

4-7 O

x

Page 236

Check for Understanding

1. Sample answer: If the first term is 2 and the common difference is 10, add 10 to each term of the sequence to get the next term of the sequence. 2 8 18 28

6. Since each element of the domain is paired with exactly one element of the range, this relation is a function. 7. Since an element of the domain is paired with more than one element in the range, the relation is not a function. Both 5 and 1 in the range are paired with x  2. 8. f(x)  3x  5 9. f(x)  3x  5 f(4)  3(4)  5 f(2a)  3(2a)  5  12  5  6a  5  7 10. f(x)  3x  5 f(x  2)  3(x  2)  5  3x  6  5  3x  11

Page 232

Arithmetic Sequences

  10 10

 10

Thus, one possible arithmetic sequence with a common difference of 10 is 2, 8, 10, 28 … . 2. If an  5n  2, then: a1  5(1)  2 a2  5(2)  2 52  10  2 7  12 a3  5(3)  2 a4  5(4)  2  15  2  20  2  17  22 The first term in the sequence, a1, is 7. 7 12 17 22

   5 5 5

The common difference, d, is 5. 3. Marisela is correct. To find the common difference, subtract the first term from the second term. 4. 24 16 8 0

Spreadsheet Investigation (Preview of Lesson 4-7)

1. Step 1: Enter the initial value 7 in cell A1. Step 2: Highlight the cells in column A. Under the Edit menu, choose the Fill option and then Series. Step 3: Enter 4 as the Step value and 63 as the Stop value, and click OK. The spreadsheet column A is filled with the numbers in the sequence: 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63. 2. The last number in the sequence is in cell A15, so there are 15 numbers in the sequence.

   8 8 8

This is an arithmetic sequence because the difference between terms is constant. The common difference is 8. 5. 3 6 12 24

   3 6 12

This is not an arithmetic sequence because the difference between terms is not constant.

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6. 7

14

21

28

?

?

?

12. 6

    7 7 7 7 7 7

12

The common difference is 7. Add 7 to the last term of the sequence and continue adding 7 until the next three terms are found. 28 35 42 49

24

The first term is 6. The common difference is 6. Use the formula for the nth term to write an equation with a1  6 and d  6. an  a1  (n  1)d an  6  (n  1)6 an  6  6n  6 an  6n

   7 7 7

The next three terms are 35, 42, 49. 7. 34 29 24 19 ? ? ?

     5 5 5 5 5 5

The common difference is 5. Add 5 to the last term of the sequence and continue adding 5 until the next three terms are found. 19 14 9 4

   5 5 5

The next three terms are 14, 9, 4. 8. Use the formula for the nth term of an arithmetic sequence with a1  3, d  4, and n  8. an  a1  (n  1)d a8  3  (8  1)4 a8  3  28 a  31 8

n

6n

an

(n, an)

1 2 3 4 5

6(1) 6(2) 6(3) 6(4) 6(5)

6 12 18 24 30

(1, 6) (2, 12) (3, 18) (4, 24) (5, 30)

32 28 24 20 16 12 8 4

The 8th term of the arithmetic sequence is 31. 9. Use the formula for the nth term of an arithmetic sequence with a1  10, d  5, and n  21. a  a  (n  1)d n 1 a21  10  (21  1) (5) a21  10  (100) a21  90

an

O

21

13. 12

1 2 3 4 5 6n

17

22

27

   5 5 5

The first term is 12. The common difference is 5. Use the formula for the nth term to write an equation with a1  12 and d  5. an  a1  (n  1)d a  12  (n  1)5 n an  12  5n  5 an  5n  7

The 21st term of the arithmetic sequence is 90. 10. 23 25 27 29

   2 2 2

The first term is 23. The common difference is 2. Use the formula for the nth term of an arithmetic sequence with a1  23, d  2, and n  12. an  a1  (n  1)d a12  23  (12  1)2 a  23  22 12 a12  45

5n  7 5(1)  7 5(2)  7 5(3)  7 5(4)  7 5(5)  7

n 1 2 3 4 5

The 12th term of the arithmetic sequence is 45. 11. 27 19 11 3

   8 8 8

32 28 24 20 16 12 8 4

The first term is 27. The common difference is 8. Use the formula for the nth term of an arithmetic sequence with a1  27, d  8, and n  17. an  a1  (n  1)d a17  27  (17  1)8 a17  27  128 a17  101

21

The 17th term of the arithmetic sequence is 101.

Chapter 4

18

   6 6 6

168

an

O 1 2 3 4 5 6n

an 12 17 22 27 32

(n, (1, (2, (3, (4, (5,

an) 12) 17) 22) 27) 32)

23. 66

14. For the first week, Latisha walks 20 min/day, so a1  20. She must increase each week by 7 min/day, so the common difference, d, is 7. We are to find the first value of n for which an is more than 1 h, or 60 min. 20 27 34 41 48 55 62

Practice and Apply 4

6

5

78

?

?

?

   4 4 4

a1 a2 a3 a4 a5 a6 a7 Since a7  62 is the first time an is more than 60, Latisha will begin walking over an hour a day during the seventh week of her exercise program.

15. 7

74

The common difference is 4. Add 4 to the last term of the sequence and continue adding 4 until the next three terms are found. 78 82 86 90

      7 7 7 7 7 7

Pages 236–238

70

     4 4 4 4 4 4

The next three terms are 82, 86, 90. 24. 31 22 13 4 ? ? ?

     9 9 9 9 9 9

The common difference is 9. Add 9 to the last term of the sequence and continue adding 9 until the next three terms are found. 4 5 14 23

 9 9 9

 1 1 1

The next three terms are 5, 14, 23.

This is an arithmetic sequence because the difference between terms is constant. The common difference is 1. 16. 10 12 15 18

1

2

25. 23

23

1

33

3

?

?

?

 1 1 1 1 1 1 3

   2 3 3

3

3

3

3

3

1 . 3

1

2

1

The common difference is Add 3 to the last term 1 of the sequence and continue adding 3 until the next three terms are found.

This is not an arithmetic sequence because the difference between terms is not constant. 17. 9 5 1 5

1

2

33

  4 6 4

33

1

43

4

 1 1 1 3

This is not an arithmetic sequence because the difference between terms is not constant. 18. 15 11 7 3

3

3

The next three terms are 33, 4, 43. 26.

  4 4 4

7 12

1

5

212

26

?

?

?

 3 3 3 3 3 3 4

This is an arithmetic sequence because the difference between terms is constant. The common difference is 4. 19. 0.3 0.2 0.7 1.2

1

13

4

4

4

4

4

3

3

The common difference is 4. Add 4 to the last term 3 of the sequence and continue adding 4 until the next three terms are found.

  0.5 0.5 0.5

5

26

7

1

312

43

1

512

   3 3 3

This is an arithmetic sequence because the difference between terms is constant. The common difference is 0.5. 20. 2.1 4.2 8.4 17.6

4

4

4

7

1

1

The next three terms are 312, 43, 512.

 2.1 4.2 9.2

27. Use the formula for the nth term of an arithmetic sequence with a1  5, d  5, and n  25. an  a1  (n  1)d a25  5  (25  1)5 a25  5  120 a25  125

This is not an arithmetic sequence because the difference between terms is not constant. 21. 4 7 10 13 ? ? ?

  3 3 3 3 3 3

The common difference is 3. Add 3 to the last term of the sequence and continue adding 3 until the next three terms are found. 13 16 19 22

The 25th term of the arithmetic sequence is 125. 28. Use the formula for the nth term of an arithmetic sequence with a1  8, d  3, and n  16. an  a1  (n  1)d a16  8  (16  1)3 a16  8  45 a16  53 The 16th term of the arithmetic sequence is 53.

  3 3 3

The next three terms are 16, 19, 22. 22. 18 24 30 36 ? ? ?

    6 6 6 6 6 6

The common difference is 6. Add 6 to the last term of the sequence and continue adding 6 until the next three terms are found. 36 42 48 54

  6 6 6

The next three terms are 42, 48, 54.

169

Chapter 4

35. 0.5

29. Use the formula for the nth term of an arithmetic sequence with a1  52, d  12, and n  102. an  a1  (n  1)d a102  52  (102  1)12 a102  52  1212 a102  1264 The 102nd term of the arithmetic sequence is 1264. 30. Use the formula for the nth term of an arithmetic sequence with a1  34, d  15, and n  200. an  a1  (n  1)d a200  34  (200  1)15 a200  34  2985 a200  3019 The 200th term of the arithmetic sequence is 3019. 31. Use the formula for the nth term of an arithmetic 5 1 sequence with a  8, d  8, and n 22.

The first term is 5.3. The common difference is 0.6. Use the formula for the nth term of an arithmetic sequence with a1  5.3, d  0.6, and n  12. an  a1  (n  1)d a12  5.3  (12  1)(0.6) a12  5.3  6.6 a12  11.9 The 12th term of the arithmetic sequence is 11.9. 37. 24 35 46 57

a22  a22 

12

21 8

  11 11 11

13 4 1 34

The first term is 24. The common difference is 11. Use the formula for the nth term of an arithmetic sequence with a1  24 and d  11 to find n when an  200. an  a1  (n  1)d 200  24  (n  1)11 200  24  11n  11 200  11n  13 200  13  11n  13  13 187  11n

1

The 22nd term of the arithmetic sequence is 34. 32. Use the formula for the nth term of an arithmetic 1 1 sequence with a  12, d  24, and n  39. 1

an  a1  (n  1)d 1 1 a39  12  (39  1) 24 a

39



1 12



1 852

1 2

a39  87

187 11

The 39th term of the arithmetic sequence is 87. 33. 9 7 5 3



11n 11

17  n 200 is the 17th term of the arithmetic sequence. 38. 30 22 14 6

  2 2 2

The first term is 9. The common difference is 2. Use the formula for the nth term of an arithmetic sequence with a1  9, d  2, and n  18. an  a1  (n  1)d a18  9  (18  1)2 a18  9  34 a18  25 The 18th term of the arithmetic sequence is 25. 34. 7 3 1 5

   8 8 8

The first term is 30. The common difference is 8. Use the formula for the nth term of an arithmetic sequence with a1  30 and d  8 to find n when an  34. an  a1  (n  1)d 34  30  (n  1)(8) 34  30  8n  8 34  38  8n 34  38  38  8n  38 72  8n

 4 4 4

The first term is 7. The common difference is 4. Use the formula for the nth term of an arithmetic sequence with a1  7, d  4, and n  35. an  a1  (n  1)d a35  7  (35  1)4 a35  7  136 a35  129 The 35th term of the arithmetic sequence is 129.

Chapter 4

2

   0.6 0.6 0.6

1

5

1.5

The first term is 0.5. The common difference is 0.5. Use the formula for the nth term of an arithmetic sequence with a1  0.5, d  0.5, and n  50. an  a1  (n  1)d a50  0.5  (50  1)(0.5) a50  0.5  24.5 a50  25 The 50th term of the arithmetic sequence is 25. 36. 5.3 5.9 6.5 7.1

an  a1  (n  1)d 5 1 a22  8  (22  1) 8 a22  8 

1

  0.5 0.5 0.5

72 8



8n 8

9n 34 is the 9th term of the arithmetic sequence.

170

39. 3 6 9 12

41. 2

The first term is 3. The common difference is 3. Use the formula for the nth term to write an equation with a1  3 and d  3. an  a1  (n  1)d an  3  (n  1)(3) an  3  3n  3 an  3n n

3n

an

1 2 3 4 5

3(1) 3(2) 3(3) 3(4) 3(5)

3 6 9 12 15

2 2

2 4 6 8 10 12 14

40. 8

9

8

10

2

4

20

The first term is 2. The common difference is 6. Use the formula for the nth term to write an equation with a1  2 and d  6. an  a1  (n  1)d an  2  (n  1)6 an  2  6n  6 an  6n  4

(n, an) (1, 3) (2, 6) (3, 9) (4, 12) (5, 15)

n

6n  4

an

(n, an)

1 2 3 4 5

6(1)  4 6(2)  4 6(3)  4 6(4)  4 6(5)  4

2 8 14 20 26

(1, 2) (2, 8) (3, 14) (4, 20) (5, 26)

an O

14

 6 6 6

  3 3 3

26 24 22 20 18 16 14 12 10 8 6 4 2

6n

11

 1 1 1

The first term is 8. The common difference is 1. Use the formula for the nth term to write an equation with a1  8 and d  1.

2

n7

an

(n, an)

1 2 3 4 5

17 27 37 47 57

8 9 10 11 12

(1, 8) (2, 9) (3, 10) (4, 11) (5, 12)

2

O

42. 18

16

6n

4

14

12

   2 2 2

an  a1  (n  1)d an  8  (n  1)1 an  8  n  1 an  n  7 n

an

The first term is 18. The common difference is 2. Use the formula for the nth term to write an equation with a1  18 and d  2. an  a1  (n  1)d an  18  (n  1)2 an  18  2n  2 an  2n  20

an

n

2n  20

an

1 2 3 4 5

2(1)  20 2(2)  20 2(3)  20 2(4)  20 2(5)  20

18 16 14 12 10

(n, an) (1, (2, (3, (4, (5,

18) 16) 14) 12) 10)

an 2

O

O

2

4

6n

4 6 8 10 12 14 16 18

n

171

Chapter 4

43. y  4, 6, y, . . . is an arithmetic sequence only if the difference between the second and first terms is the same as the difference between the third and second terms. a2  a1  a3  a2 6  ( y  4)  y  6 6y4y6 2yy6 2yyy6y 2  2y  6 2  6  2y  6  6 8  2y 8 2



47. The number of seats in each row form an arithmetic sequence. The last three terms in the sequence are 60, 68, 76. There are seven rows, so there are seven terms in the sequence. This means a5  60, a6  68, and a7  76. 60 68 76

  8 8

The common difference is 8. Use the formula for the nth term to write an equation with d  8, n  7, and an  76 to find a1. an  a1  (n  1)d 76  a1  (7  1)8 76  a1  48 76  48  a1  48  48 28  a1 Use the formula for the nth term to write an equation with a1  28 and d  8. an  a1  (n  1)d an  28  (n  1)8 an  28  8n  8 an  8n  20 The formula to find the number of seats in any given row is an  8n  20. 48. Use the formula from Exercise 47 with n  1. an  8n  20 a1  8(1)  20 a1  8  20 a1  28 There are 28 seats in the first row. 49. n 8n  20 a

2y 2

4y The sequence is arithmetic if y  4. 44. y  8, 4y  6, 3y, . . . is an arithmetic sequence only if the difference between the second and first terms is the same as the difference between the third and second terms. a2  a1  a3  a2 (4y  6)  ( y  8)  3y  (4y  6) 4y  6  y  8  3y  4y  6 3y  2  y  6 3y  2  y  y  6  y 4y  2  6 4y  2  2  6  2 4y  4 4y 4



4 4

y  1 The sequence is arithmetic if y  1. 45. 5 8 11 14

n

1 2 3 4 5 6 7

 3 3 3

The first term is 5. The common difference is 3. Use the formula for the nth term to write an equation with a1  5 and d  3. Pn  a1  (n  1)d Pn  5  (n  1)3 Pn  5  3n  3 Pn  3n  2 The formula for the perimeter of a pattern containing n trapezoids is Pn  3n  2. 46. Use the formula from Exercise 45 with n  12. an  3n  2 a12  3(12)  2 a12  36  2 a12  38 The perimeter of the pattern containing 12 trapezoids is 38 units.

Chapter 4

8(1)  20 8(2)  20 8(3)  20 8(4)  20 8(5)  20 8(6)  20 8(7)  20

28 36 44 52 60 68 76

The total number of seats in the section is: a1  a2  a3  a4  a5  a6  a7  28  36  44  52  60  68  76  364 If 368 tickets were sold for the orchestra section having 364 seats, the section was oversold by 4 seats. 50. 9 13 17 21 25 29

     4 4 4 4 4

The distances form an arithmetic sequence because the difference between terms is constant. The common difference is 4. 51. The first term is 9. The common difference is 4. Use the formula for the nth term of an arithmetic sequence with a1  9 and d  4. an  a1  (n  1)d an  9  (n  1)4 an  9  4n  4 an  4n  5

172

52. Use the formula from Exercise 51 with n  35. an  4n  5 a35  4(35)  5 a35  140  5 a35  145 The ball will travel 145 cm during the 35th s. 53. an

57. Between 29 and 344, the least multiple of 7 is 35 and the greatest multiple of 7 is 343. The common difference between multiples of 7 is 7. Use the formula for the nth term of an arithmetic sequence with a1  35, an  343, and d  7 to find the value of n. an  a1  (n  1)d 343  35  (n  1)7 343  35  7n  7 343  7n  28 343  28  7n  28  28 315  7n

Distance Traveled (cm)

26 24 22 20 18 16 14 12 10 8 6 4 2 O

315 7

2 4 Time (s)

6

n

n

1500n  1000

1 2 3 4 5 6 7 8 9 10

1500(1)  1000 1500(2)  1000 1500(3)  1000 1500(4)  1000 1500(5)  1000 1500(6)  1000 1500(7)  1000 1500(8)  1000 1500(9)  1000 1500(10)  1000

an 2500 4000 5500 7000 8500 10,000 11,500 13,000 14,500 16,000

9 3

 2x



3d 3

3d Use the formula for the nth term with a1  2, d  3, n  20. an  a1  (n  1)d a20  2  (20  1)3 a20  2  (19)3 a20  2  57 a20  59

The total won for ten correct answers would be: a1  a2  a3  a4  a5  a6  a7  a8  a9  a10  2500  4000  5500  7000  8500  10,000  11,500  13,000  14,500  16,000  92,500. The contestant would win $92,500 for ten correct answers. 56. 2x  5 4x  5 6x  5 8x  5

 2x

7n 7

45  n There are 45 multiples of 7 between 29 and 344. 58. By finding a pattern in a sequence of numbers, scientists can predict results of large numbers that they are not able to observe. Answers should include the following. • The formula at  8.2t  1.9 represents the altitude at of the probe after t seconds. • Replace t with 15 in the equation for at to find that the altitude of the probe after 15 seconds is 121.1 feet. 59. C; Total amount saved  amount in savings now  (weekly deposit)(number of weeks)  350  (25)(12)  350  300  650 60. B; Use the formula for the nth term of an arithmetic sequence with a1  2, a4  11, and n  4 to find d. an  a1  (n  1)d 11  2  (4  1)d 11  2  3d 11  2  2  3d  2 9  3d

54. Use the formula for the nth term to write an equation with a1  2500 and d  1500. Then find a10. an  a1  (n  1)d an  2500  (n  1)1500 an  2500  1500n  1500 an  1500n  1000 a10  1500(10)  1000 a10  15,000  1000 a10  16,000 The value of the 10th question is $16,000. 55.



Page 238

Maintain Your Skills

61. f(x)  3x  2 f(4)  3(4)  2  12  2  10

 2x

63.

The expressions form an arithmetic sequence because the difference between terms is constant. The common difference is 2x.

173

62. g(x)  x2  5  (3)2  5 95 4

f(x)  3x  2 2[f(6)]  2[3(6)  2]  2(18  2)  2(16)  32 Chapter 4

3. Inductive reasoning; you are observing specific pairs of terms and discovering a common difference, and you conclude that the common difference applies to the sequence in general. 4. Deductive reasoning; you are using the general formula for the nth term and applying it to a particular term of a particular series. 5a. 37 38 39 31 32 33 34 35 36 3 9 27 81 243 729 2187 6561 19,683

64. Since the term x2 has an exponent of 2, the equation cannot be written in the form Ax  By  C. Therefore, this is not a linear equation. 65. First rewrite the equation so that the variables are on one side of the equation and a constant is on the other side. y  8  10  x y  8  x  10  x  x x  y  8  10 x  y  8  8  10  8 x  y  18 The equation is now in standard form where A  1, B  1, and C  18. This is a linear equation. 66. First rewrite the equation so that the variables are on the same side of the equation. 2y  y  2x  3 2y  y  y  2x  3  y y  2x  3 y  2x  2x  3  2x 2x  y  3 1( 2x  y)  1(3) 2x  y  3 The equation is now in standard form where A  2, B  1, and C  3. This is a linear equation.

5b. The ones digits in the numbers in the second row of the table are: 3, 9, 7, 1, 3, 9, 7, 1, 3. 5c. The pattern repeats every fourth digit. Therefore, digits 4, 8, 12, 16, and so on, will all be the same. According to the pattern, all powers in the first row of the table with exponents divisible by 4 have 1 in the ones place. Since the exponent of 3100 is 100, which is divisible by 4, the value of 3100 will have a 1 in the ones place. Since the answer was obtained by observing a pattern in order to make a conjecture, the answer was reached by inductive reasoning. 6. Since the conclusion was based on a given rule, it was reached by deductive reasoning.

hundred times x is equal to nine. 67. Two 144244 3 minus 123 three 14424 43 14243 123 

200



3x

9

The equation is 200  3x  9. 68. Rewrite the sentence so it is easier to translate.

4-8

Twice plus times s is identical to thirteen. 123r 1 23 three 14424 43 14 42443 14243 2r  3s  13

Page 241

73.

75. 76. 77. 78. 79. 80.

1 21 2  4 7

Point H J K L M N

Page 239

20 56

or

5 14

12

1

7

74. 5  32  5  2 

x-Coordinate 2 3 4 3 3 5

y-Coordinate 2 0 2 4 5 1

35 2

1

or 172

Ordered Pair (2, 2) (3, 0) (4, 2) (3, 4) (3, 5) (5, 1)

Page 243

Check for Understanding

1. Once you recognize a pattern, you can find a general rule that can be written as an algebraic expression. 2. Sample answer: 4, 8, 16, 32, 64, . . . ; each successive term doubles. 3. Test the values of the domain in the equation. If the resulting values match the range, the equation is correct.

Reading Mathematics

1. Sample answer: Inductive reasoning uses examples or past experience to make conclusions; deductive reasoning uses rules to make conclusions. Looking at a pattern of numbers to decided the next number is an example of inductive reasoning. Using the formula and the length and width of a rectangle to find the area of a rectangle is an example of deductive reasoning. 2. Deductive reasoning; he is applying a general rule about men’s heights to a specific case.

Chapter 4

Algebra Activity

• When the string makes 1 loop around the scissors, you end up with 3 pieces as a result of the cut. • When the string makes 2 loops around the scissors, you end up with 4 pieces as a result of the cut. 1. The number of pieces is 2 more than the number of loops; 3, 4, 5, 6, 7,… . 2. n  2 3. n  2  20  2  22

The equation is 2r  3s  13. 69. 7(3)  21 70. 11  15  165 2 12 71. 8(1.5)  12 72. 6 3  3 or 4 5 8

Writing Equations from Patterns

174

4. The pattern consists of squares with one corner of each shaded. The corner that is shaded is rotated in a clockwise direction. The next two figures in the pattern are shown.

If x  3, then y  2(3) or 6. But the y value for x  3 is 5. This is a difference of 1. Try other values in the domain to see if the same difference occurs.

Check:

;

2 4 3

 1 2 3 4

The difference between each pair of terms increases by 1 for each successive pair. Continue increasing each successive difference by 1. Add 5, 6, and 7. 1 2 4 7 11 16 22 29

9.

y

Temperature (ßC)

The next three terms are 16, 22, and 29. 6. 5 9 6 10 7 11

  4 3 4 3 4

200 150 100 50

The difference between terms alternates between 4 and 3. Continue the alternating pattern. Add 3, 4, and 3. 5 9 6 10 7 11 8 12 9

0

2

 4 3 4 3 4 3 4 3

x y

1

1

1

1

x 1 35x 35 y 55

2 70 90

3 4 5 6 105 140 175 210 125 160 195 230

y is always 20 more than 35x.



For each pair, the difference in the y values is the same as the difference in the x values. This suggests that y  x. Check: If x  4, then y  4. If x  1, then y  1. For all values in the domain, y is always equal to x. Thus, y  x or f(x)  x describes this relation. 8. Make a table of ordered pairs for several points on the graph.

x y

1

The difference of the values for x is 1, and the difference of the values for y is 35. This suggests that y  35x. Check: If x  1, then y  35(1) or 35. But the y value for x  1 is 55. This is a difference of 20. Try other values in the domain to see if the same difference occurs.

    1 4 2 1

2

x

   35 35 35 35 35

1

    4 3 1 3 4 4 3 1 3 4

1

4 6 Depth (km)

    10. Depth (km) 1 2 3 4 5 6 Temperature (C) 55 90 125 160 195 230

The next three terms are 8, 12, and 9. 7. Make a table of ordered pairs for several points on the graph. 2

y is always 1 more than 2x.

250

   1 2 3 4 5 6 7

4

1 2 3

Check y  2x  1. If x  2, then y  2(2)  1 or 3. If x  0, then y  2(0)  1 or 1. If x  1, then y  2(1)  1 or 3. Thus, y  2x  1 or f(x)  2x  1 describes this relation.

The pattern repeats every fourth design. Therefore, designs 4, 8, 12, 16, and so on, will all be the same. So the 16th figure in the pattern will be the same as the fourth square. 5. 1 2 4 7 11

1

0 0 1



3 6 5

x 2x y

Check y  35x  20. If x  2, then y  35(2)  20 or 90. If x  5, then y  35(5)  20 or 195. Thus, y  35x  20 or f(x)  35x  20 describes this relation. 11. Use the function from Exercise 10 with x  10. f(x)  35x  20 f(10)  35(10)  20 f(10)  350  20 f(10)  370 The temperature of a rock that is 10 km below the surface is 370C.

1

   3 2 0 1 5 3 1 3

   2 4 2

When the difference of the x values is 1, the difference of the y values is 2. When the difference of the x values is 2, the difference of the y values is 4. The difference in y values is twice the difference of x values. This suggests that y  2x.

175

Chapter 4

Pages 244–245

16. 1

Practice and Apply

4

9

16

  3 5 7

12. The pattern consists of triangles that are alternately inverted (or rotated 180) with every other pair shaded. The next two figures in the pattern are shown.

The difference between each pair of terms increases by 2 for each successive pair. Continue increasing each successive difference by 2. Add 9, 11, and 13. 1 4 9 16 25 36 49

     3 5 7 9 11 13

The next three terms are 25, 36, and 49. 17. 0 2 5 9 14 20

    2 3 4 5 6

The pattern repeats every fourth design. Therefore, designs 4, 8, 12, 16, and so on, will all be the same. Since the 20th figure will be the same as the fourth triangle, the 21st figure will be the same as the first triangle.

The difference between each pair of terms increases by 1 for each successive pair. Continue increasing each successive difference by 1. Add 7, 8, and 9. 0 2 5 9 14 20 27 35 44

     2 3 4 5 6 7 8 9

The next three terms are 27, 35, and 44. 18. a  1 a2 a3

 1

13. The pattern consists of circles with one-eighth shaded. The section that is shaded is the third section in a clockwise direction from the previously-shaded section. The next two figures in the pattern are shown.

 1

The difference between each pair of terms is always 1. The sequence is arithmetic with a common difference of 1. Each term is 1 more than the term before it. Add 1, 1, and 1. a1 a2 a3 a4 a5 a6

 1

 1

 1

 1

 1

The next three terms are a  4, a  5, and a  6. 19. x  1 2x  1 3x  1

 x

The pattern repeats every eighth design. Therefore, designs 8, 16, 24, and so on, will all be the same. Since 16 is the greatest number less than 21 that is a multiple of 8, the 17th circle in the pattern will be the same as the first circle.

The difference between each pair of terms is always x. The sequence is arithmetic with a common difference of x. Each term is x more than the term before it. Add x, x, and x. x  1 2x  1 3x  1 4x  1 5x  1 6x  1

 x

17

14. 0

18

2

6

19

12

20

 x

 x

 x

 x

The next three terms are 4x  1, 5x  1, and 6x  1. 20. Make a table of ordered pairs for several points on the graph.

21

20

   2 4 6 8

The difference between each pair of terms increases by 2 for each successive pair. Continue increasing each successive difference by 2. Add 10, 12, and 14. 0 2 6 12 20 30 42 56

x y

       2 4 6 8 10 12 14

1

1

1

1

2

2

2

2

    2 1 0 1 2 4 2 0 2 4    

The difference of the x values is 1, and the difference of the y values is 2. The difference in y values is twice the difference of the opposite of the x values. This suggests y  2x. Check: If x  2, then y  2(2) or 4. If x  1, then y  2(1) or 2. Thus, y  2x or f(x)  2x describes the relation.

The next three terms are 30, 42, and 56. 15. 9 7 10 8 11 9 12

      2 3 2 3 2 3

The difference between terms alternates between 2 and 3. Continue alternating the difference. Add 2, 3, and 2. 9 7 10 8 11 9 12 10 13 11

  2 3 2 3 2 3 2 3 2

The next three terms are 10, 13, and 11.

Chapter 4

 x

176

Check y  x  6. If x  2, then y  (2)  6 or 4. If x  6, then y  (6)  6 or 0. Thus, y  x  6 or f(x)  6  x describes this relation. 24. Make a table of ordered pairs for several points on the graph.

21. Make a table of ordered pairs for several points on the graph. 2

1

2

1

When the difference of x values is 2, the difference of the y values is 1. When the difference of x values is 4, the difference of the y values is 2. The difference in y values is one-half 1 the difference of x values. This suggests y  2 x.

x y

1

If x  4, then y  2(4)  2.

Check:

1

1

Thus, y  2x or f(x)  2x describes this relation. 22. Make a table of ordered pairs for several points on the graph. x y

2

3

1

2

3

1

   4 2 1 2 2 0 3 4   

2 0

1 3

2 4

x y

2

2

2

2

2

4 4 2

6 6 0

2

4

0

3

6

y

6

3

0

y is always 6 more 3

than 2x.

3

If x  4, then y  2(4)  6 or 0. 3

3

Thus, y  2x  6 or f(x)  6  2x describes this relation. 25. Make a table of ordered pairs for several points on the graph.

y is always 2 more than x.

x y

2

2

6

6

 0 2 4 12 6 0 

For each pair, the difference in the y values is three times the opposite of the difference of the x values. This suggests y  3x. Check: If x  2, then y  3(2) or 6. But the y value for x  2 is 6. This is a difference of 12. Try other values in the domain to see if the same difference occurs. x 3x y

0 0 12

2 6 6

4 12 0

y is always 12 more than 3x.

Check y  3x  12. If x  2, then y  3(2)  12 or 6. If x  4, then y  3(4)  12 or 0. Thus, y  3x  12 or f(x)  12  3x describes this relation.

y is always 6 more than x.



2 2 4

0

3 2x

3

   0 2 4 6 6 4 2 0   

0 0 6

x

If x  2, then y  2(2)  6 or 3.

For each pair, the difference in the y values is the opposite of the difference in x values. This suggests y  x. Check: If x  2, then y  (2) or 2. But the y value for x  2 is 4. This is a difference of 6. Try other values in the domain to see if the same difference occurs. x x y

3

If x  2, then y  2(2) or 3. But the y value for x  2 is 3. This is a difference of 6. Try other values in the domain to see if the same difference occurs.

3

Check y  x  2. If x  4, then y  (4)  2 or 2. If x  1, then y  (1)  2 or 3. Thus, y  x  2 or f(x)  x  2 describes this relation. 23. Make a table of ordered pairs for several points on the graph. 2

3

Check y  2x  6.



4 2

3

Check:

For each pair, the difference in y values is the same as the difference in the x values. This suggests y  x. Check: If x  4, then y  4. But the y value for x  4 is 2. This is a difference of 2. Try other values in the domain to see if the same difference occurs. x y

2

The difference of the x values is 2, and the difference of the y values is 3. The difference in 3 the y values is 2 times the opposite of the 3 difference in the x values. This suggests y  2x.

If x  2, then y  2(2)  1. 1

2

  0 2 4 6 3 0  



4



x y

2

   4 2 2 4 2 1 1 2   

177

Chapter 4

1

26. The pattern (3 red, 3 blue, 3 green) repeats every ninth chain. Therefore, chains 9, 18, 27, and so on, will all be the same. Since 45 is the greatest number less than 50 that is a multiple of 9, the 46th chain will be the same as the first chain. red red red blue blue 46 47 48 49 50 The 50th person will receive a blue flower chain. 27. 1 1 211 321 532 853 13  8  5 21  13  8 34  21  13 55  34  21 89  55  34 144  89  55 28. 3, 21, 144; every fourth term is divisible by 3. 5, 55; every fifth term is divisible by 5. 10

29.

a p





10



10



10

10

9

9

9

9

9

32. n f(n)

3

50 45 148

60 54 139

70 63 130

1 3 5

2 6 8

3 9 11

4 12 14

f(n) is always 2 more than 3n.

Check f(n)  3n  2. If n  1, then f(1)  3(1)  2 or 5. If n  4, then f(4)  3(4)  2 or 14. Thus, f(n)  3n  2 represents this function. 33. Use the function from Exercise 32 with n  24. f(n)  3n  2. f(24)  3(24)  2 f(24)  72  2 f(24)  74 The perimeter of the arrangement if 24 pentagons are used is 74 cm. 34. In scientific experiments you try to find a relationship or develop a formula from observing the results of your experiment. Answers should include the following. • For every 11 cubic feet the volume of water increases, the volume of ice increases 12 cubic feet. • The container should have a volume of at least 108 cubic feet. 35. B; 3 4 6 9

 1 2 3

p is always 193 more than 0.9a. Check p  0.9a  193. If a  30, then p  0.9 (30)  193 or 166. If a  50, then p  0.9(50)  193 or 148. Thus, p  0.9a  193 or f(a)  0.9a  193 describes this relation. 30. Use the function from Exercise 29 with a  10, then with a  80. f(a)  0.9a  193 f(a)  0.9a  193 f(10)  0.9(10)  193 f(80)  0.9(80)  193 f(10)  9  193 f(80)  72  193 f(10)  184 f(80)  121 A 10-year old should maintain a maximum heart rate of 184 beats/min and an 80-year old should maintain a maximum heart rate of 121 beats/min in aerobic training. 31. Number of Pentagons 1 2 3 4

The difference between each pair of terms increases by 1 for each successive pair. Continue increasing each successive difference by 1. Add 4 and 5. 3 4 6 9 13 18

  1 2 3 4 5

The next two terms are 13 and 18. 36. D; Let n represent the number of pieces given to each child. Then 5n represents the total number of pieces given to 5 children. Since there were 4 pieces remaining, there were originally 5n  4 pieces. Thus, P  5n  4. If you then had P  4 pieces to distribute and gave n pieces to each of the 5 children, you would have 4 more pieces left than you had previously, or 8 pieces remaining. However, that would mean you would have enough pieces to give each child 1 additional piece, leaving 3 pieces of candy remaining.

Perimeter of Arrangement (cm) 5 8 11 14

Chapter 4

3



n 3n f(n)



40 36 157

3

For each pair, the difference in f(n) values is three times the difference in the n values. This suggests f(n)  3n. Check: If n  2, then f(n)  3(2) or 6. But the f(n) value for n  2 is 8. This is a difference of 2. Try other values in the domain to see if the same difference occurs.

The difference of the a values is 10, and the difference of the p values is 9. The difference in 9 the p values is 10 of the opposite of the x values. This suggests p  0.9a. Check: If a  20, then p  0.9(20) or 18. But the y value for a  20 is 175. This is a difference of 193. Try other values in the domain to see if the same difference occurs. 30 27 166

1

  

 20 30 40 50 60 70 175 166 157 148 139 130     

a 20 0.9a 18 p 175

1

   1 2 3 4 5 8 11 14

178

Page 245 37. 1

7. 8. 9. 10.

Maintain Your Skills 4

7

10

?

?

?

    3 3 3 3 3 3

The common difference is 3. Add 3 to the last term of the sequence and continue adding 3 until the next three terms are found. 10 13 16 19

Pages 246–250

   3 3 3

Lesson-by-Lesson Review

11. A(4, 2) • Start at the origin. • Move right 4 units and up 2 units. • Draw a dot and label it A. (See coordinate plane after Exercise 16.) 12. B(1, 3) • Start at the origin. • Move left 1 unit and up 3 units. • Draw a dot and label it B. (See coordinate plane after Exercise 16.) 13. C(0, 5) • Start at the origin. • Since the x-coordinate is 0, the point is on the y-axis. • Move down 5 units. • Draw a dot and label it C. (See coordinate plane after Exercise 16.) 14. D(3, 2) • Start at the origin. • Move left 3 units and down 2 units. • Draw a dot and label it D. (See coordinate plane after Exercise 16.) 15. E(4, 0) • Start at the origin. • Move left 4 units. • Since the y-coordinate is 0, the point is on the x-axis. • Draw a dot and label it E. (See coordinate plane after Exercise 16.) 16. F(2, 1) • Start at the origin. • Move right 2 units and down 1 unit. • Draw a dot and label it F. 11–16. y

The next three terms are 13, 16, 19. 38. 9 5 1 3 ? ? ?

    4 4 4 4 4 4

The common difference is 4. Add 4 to the last term of the sequence and continue adding 4 until the next three terms are found. 3 7 11 15

   4 4 4

The next three terms are 7, 11, 15. 39. 25 19 13 7 ? ? ?

    6 6 6 6 6 6

The common difference is 6. Add 6 to the last term of the sequence and continue adding 6 until the next three terms are found. 7 1 5 11

   6 6 6

The next three terms are 1, 5, 11. 40. 22 34 46 58 ? ? ?

    12 12 12 12 12 12

The common difference is 12. Add 12 to the last term of the sequence and continue adding 12 until the next three terms are found. 58 70 82 94

   12 12 12

The next three terms are 70, 82, 94. 41. The vertical line x  2 intersects the graph at two points, (2, 2) and (2, 3). Thus, the relation graphed does not represent a function. 42. Let T  the height of Tulega Falls. The height of the height of Angel Falls is 102 ft higher than Tulega Falls. 14424 43 { 123 1 4424 43 14 424 43 3212  102  3212  102  T 3212  102  102  T  102 3110  T Tulega Falls is 3110 ft high.

c; linear function a; domain b; dilation i; y-axis

T

B (1, 3)

A (4, 2)

E (4, 0) x

O

Chapter 4 Study Guide and Review

F ( 2,1) D (3, 2)

Page 246 1. 2. 3. 4. 5. 6.

C (0, 5)

Vocabulary and Concept Check

e; origin g; relation d; reflection h; x-axis k; y-coordinate f; quadrants

179

Chapter 4

17. To reflect the triangle over the x-axis, multiply the y-coordinate of each vertex by 1. (x, y) S (x, y) A(3, 3) S A¿(3, 3) B(5, 4) S B¿(5, 4) C(4, 3) S C¿(4, 3) y

C’

20. To rotate the trapezoid 90 counterclockwise about the origin, switch the coordinates of each vertex and then multiply the new first coordinate by 1. (x, y) S (y, x) M(2, 0) S M¿(0, 2) N(4, 3) S N¿(3, 4) O(6, 3) S O¿(3, 6) P(8, 0) S P¿(0, 8)

B

A

y

P’ O’

x

O

A’ C

N

N’ B’

M’

18. To translate the quadrilateral 3 units down, add 3 to the y-coordinate of each vertex. (x, y) S (x, y  3) P(2, 4) S P¿(2, 4  3) S P¿(2, 1) Q(0, 6) S Q¿(0, 6  3) S Q¿(0, 3) R(3, 3) S R¿(3, 3  3) S R¿(3, 0) S(1, 4) S S¿(1, 4  3) S S¿(1, 7)

P

M

x

O

21.

Q y

x 2 3 3 4

O

y

y 6 2 0 6

X

Y

2 3 4

2 0 6

O

x

P R

Q’ P’

R’ x

O

The domain of this relation is {2, 3, 4}. The range is {2, 0, 6}.

S 22.

S’ 1

19. To dilate the parallelogram by a scale factor of 2, 1 multiply the coordinates of each vertex by 2.

112 x, 12 y2 1 1 G(2, 2) S G¿ 1 2  2, 2  2 2 S G¿(1, 1) 1 1 H(6, 0) S H¿ 1 2  6, 2  0 2 S H¿(3, 0) 1 1 I(6, 2) S I¿ 1 2  6, 2  2 2 S I¿(3, 1) 1 1 J(2, 4) S J¿ 1 2  2, 2  4 2 S J¿(1, 2)

x 1 3 6

X

y

y 0 0 2

Y

(x, y) S

O 1 3 6

J

I’

G’ O

Chapter 4

I

G H’

2

The domain of this relation is {1, 3, 6}. The range is {0, 2}.

y

J’

0

H x

180

x

23.

x 3 9 3 5

y

y 8 3 8 3

26.

X

Y

12

3 3 5 9

3

8

X

Y

2 3 4

5 1 2 3

x9 4  9 2  9 09 29 49

x 4 2 0 2 4

4

2

x

y 13 11 9 7 5

2 4 6 8 10 12 14

2

2

4x

4x  5 4(4)  5 4(2)  5 4(0)  5 4(2)  5 4(4)  5

x 4 2 0 2 4

y 11 3 5 13 21

(x, y) (4, 11) (2, 3) (0, 5) (2, 13) (4, 21)

Graph the solution set {(4, 11), (2, 3), (0, 5), (2, 13), (4, 21)}.

(x, y) (4, 13) (2, 11) (0, 9) (2, 7) (4, 5)

21 18 15 12 9 6 3 4321

y O

O

27. First solve the equation for y in terms of x. 4x  y  5 4x  y 4x  5  4x y  5  4x 1(y)  1(5  4x) y  4x  5

Graph the solution set {(4, 13), (2, 11), (0, 9), (2, 7), (4, 5)}. 2

2

4

y

y 5 1 2 3 O

25.

y

4 4

x 2 3 4 2

(x, y) (4, 12) (2, 8) (0, 4) (2, 0) (4, 4)

8

The domain of this relation is {3, 3, 5, 9}. The range is {3, 8}. 24.

y 12 8 4 0 4

Graph the solution set {(4, 12), (2, 8), (0, 4), (2, 0), (4, 4)}.

x

O

4  2x 4  2(4) 4  2(2) 4  2(0) 4  2(2) 4  2(4)

x 4 2 0 2 4

y

O 1 2 3 4x

6 9 12

4x

181

Chapter 4

28. First solve the equation for y in terms of x. 2x  y  8 2x  y  2x  8  2x y  8  2x x 4 2 0 2 4

8  2x

y

(x, y)

8  2(4) 8  2(2) 8  2(0) 8  2(2) 8  2(4)

16 12 8 4 0

(4, 16) (2, 12) (0, 8) (2, 4) (4, 0)

30. First solve the equation for y in terms of x. 4x  3y  0 4x  3y  4x  0  4x 3y  4x 3y 3

y

4

1 53

2

4(2) 3

 23

0

4(0) 3

0

2

4(2) 3

2 23

4

4(4) 3

53

12

51

2

O

2

8 6 4 2

4x

29. First solve the equation for y in terms of x. 3x  2y  9 3x  2y  3x  9  3x 2y  9  3x 2y 2



y

4

y

4

9  3(4) 2

1 102

2

9  3(2) 2

72

0

9  3(0) 2

42

2

9  3(2) 2

12

4

9  3(4) 2

12

(x, y)

1 1 1 1

14, 1012 2 12, 712 2 10, 412 2 12, 112 2 14, 112 2

514, 1012 2, 12, 712 2, 10, 412 2, 12, 112 214, 112 26.

2

1

21

12, 223 2 14, 513 2 26

4x

y  x  2

O

2

2

y

12 10 8 6 4 2

Chapter 4

1

2

y

O

y

O

(0, 0)

31. To find the x-intercept, let y  0. y  x  2 0  x  2 0  x  x  2  x x2 The graph intersects the x-axis at (2, 0). To find the y-intercept, let x  0. y  x  2 y  (0)  2 y2 The graph intersects the y-axis at (0, 2). Plot these points and draw the line that connects them.

Graph the solution set

2

2

4 6 8

9  3x 2 9  3x 2 9  3x 2

x

4

14, 513 2 12,223 2

Graph the solution set 1 2 2 1 4, 53 , 2, 23 , (0, 0), 2, 23 , 4, 53 .

4 4

21

(x, y)

y

4(4) 3

y

8

4x 3 4x 3

4x 3

x

Graph the solution set {(4, 16), (2, 12), (0, 8), (2, 4), (4, 0)}. 16



x

182

x

34. To find the x-intercept, let y  0. 5x  2y  10 5x  2(0)  10 5x  10

32. To find the x-intercept, let y  0. x  5y  4 x  5(0)  4 x4 The graph intersects the x-axis at (4, 0). To find the y-intercept, let x  0. x  5y  4 0  5y  4 5y  4 5y 5

5x 5

4 4

1 42

2y 2

The graph intersects the y-axis at 0, 5 . Plot these points and draw the line that connects them.



10 2

y5 The graph intersects the y-axis at (0, 5). Plot these points and draw the line that connects them.

y

y

x  5y  4

5x  2y  10

x

O

10 5

x2 The graph intersects the x-axis at (2, 0). To find the y-intercept, let x  0 5x  2y  10 5(0)  2y  10 2y  10

5

y5



x

O

33. To find the x-intercept, let y  0. 2x  3y  6 2x  3(0)  6 2x  6 2x 2



35. To find the x-intercept, let y  0. 1 x 2

6 2

1 x 2

x3 The graph intersects the x-axis at (3, 0). To find the y-intercept, let x  0. 2x  3y  6 2(0)  3y  6 3y  6 3y 3

1

 3(0)  3

2

1 x 2 1 x 2

3

1 2  2(3)

x6 The graph intersects the x-axis at (6, 0). To find the y-intercept, let x  0.

6

 3

y  2 The graph intersects the y-axis at (2, 0). Plot these points and draw the line that connects them.

1 x 2

 3y  3

1

1 (0) 2

 3y  3

1

3

y

O

1

 3y  3

1 y 3 1 y 3

3

1 2  3(3)

y9 The graph intersects the y-axis at (0, 9). Plot these points and draw the line that connects them. x

y 14 12 10 8 6 4 2

2x  3y  6

2 2

183

1 x  1y  3 2 3

O 2 4 6 8 10 12 14x

Chapter 4

36. To find the x-intercept, let y  0. 1 y3 1 03 1 3 1 2 3  3

   

1  3(1) 

1 x 3 1 x 3 1 x 3 1 x 3 1 x 3 1 3 3x

2 3 2 3 2 3 2 3



1

1

1

1

1 3 1 3

2 3 2 3

g(x)  x2  x  1 g(2a)  (2a) 2  (2a)  1  4a2  2a  1 46. 9 18 27 36 ?

2 2

 

   9 9 9

The next three terms are 45, 54, 63. 47. 6 11 16 21 ? ?

y

The common difference is 5. Add 5 to the last term of the sequence and continue adding 5 until the next three terms are found. 21 26 31 36

y  3  3x  3

1 3

2

1

O

   5 5 5

1x

1

The next three terms are 26, 31, 36. 48. 10 21 32 43 ? ?

The common difference is 11. Add 11 to the last term of the sequence and continue adding 11 until the next three terms are found. 43 54 65 76

   11 11 11

The next three terms are 54, 65, 76. 49. 14 12 10 8 ? ?

      2 2 2 2 2 2

2

40. g(x)  x  x  1 g(2)  22  2  1 421 21 3

   2 2 2

The next three terms are 6, 4, 2. 50. 3 11 19 27 ?

 8

112 2  112 22  112 2  1

 8

1

421

 

Chapter 4

1 4 3 4

 8

?

    8 8 8 8

The common difference is 8. Add 8 to the last term of the sequence and continue adding 8 until the next three terms are found. 27 35 43 51

42. g(x)  x2  x  1 1

?

The common difference is 2. Add 2 to the last term of the sequence and continue adding 2 until the next three terms are found. 8 6 4 2

g(x)  x2  x  1 g(1)  (1) 2  (1)  1 111 21 3 g

?

      11 11 11 11 11 11

37. The mapping represents a function since, for each element of the domain, there is only one corresponding element in the range. 38. The table represents a relation that is not a function. The element 1 in the domain is paired with both 4 and 6 in the range. 39. Since each element of the domain is paired with exactly one element of the range, the relation is a function.

41.

?

      5 5 5 5 5 5

2 2

1

?

The common difference is 9. Add 9 to the last term of the sequence and continue adding 9 until the next three terms are found. 36 45 54 63

1

3

3

?

      9 9 9 9 9 9

y1 The graph intersects the y-axis at (0, 1). Plot these points and draw the line that connects them.

1

g(a)  x2  x  1 g(a  1)  (a  1) 2  (a  1)  1  (a2  2a  1)  (a  1)  1  a2  2a  1  a  1  1  a2  a  1

45.

y  3  3(0)  3 y

44.

1 2

y  3  3x  3

1 3

g(x)  x2  x  1 g(5)  3  (52  5  1)  3  (25  5  1)  3  (20  1)  3  21  3  18

2

3

3  x The graph intersects the x-axis at (3, 0). To find the y-intercept, let x  0.

y

43.

1

 8

 8

The next three terms are 35, 43, 51.

184

?

51. 35

29

23

17

?

?

      6 6 6 6 6 6

• Move down 5 units. • Draw a dot and label it K. M(3, 5) • Start at the origin. • Move right 3 units and down 5 units. • Draw a dot and label it M. N(2, 3) • Start at the origin. • Move left 2 units and down 3 units. • Draw a dot and label it N.

?

The common difference is 6. Add 6 to the last term of the sequence and continue adding 6 until the next three terms are found. 17 11 5 1

   6 6 6

The next three terms are 11, 5, 1. 52. Make a table of ordered pairs for several points on the graph. 1



x y

2

1

1

  2 1 1 2 3 6 3 3 6 9     3



6

3

y

3

N

The difference in y values is three times the difference of x values. This suggests that y  3x. Check: If x  2, then y  3(2) or 6. If x  3, then y  3(3) or 9. Thus, y  3x or f(x)  3x describes the relation. 53. Make a table of ordered pairs for several points on the graph. 1

1

 x y

2 1

1

 1 0

 1



1

 0 1

1



1 2

1

 1

5. Since both coordinates are positive, P(25, 1) lies in quadrant I. 6. To reflect the parallelogram over the y-axis, multiply the x-coordinate of each vertex by 1. (x, y) S (x, y) H(2, 2) S H¿(2, 2) I(4, 6) S I¿(4, 6) J(5, 5) S J¿(5, 5) K(3, 1) S K¿(3, 1)

3 4  1

The difference in y values is the opposite of the difference of x values. This suggests that y  x. Check: If x  2, then y  (2)  2. But the y value for x  2 is 1. This is a difference of 1. Try other values in the domain to see if the same difference occurs. 2 2 1

1 1 0

0 0 1

1 1 2

2 2 3

3 3 4

y

K

K’ x

O

H

H’ J’

J 

x x y

M

K

 2 3

(3, 5)

(0, 5)

1



x

O (2, 3)

I

y is always 1 less than x.

I’

7. To translate the parallelogram up 2 units, add 2 to the y-coordinate of each vertex. (x, y) S (x, y  2) H(2, 2) S H¿(2, 2  2) S H¿(2, 0) I(4, 6) S I¿(4, 6  2) S I¿(4, 4) J(5, 5) S J¿(5, 5  2) S J¿(5, 3) K(3, 1) S K¿(3, 1  2) S K¿(3, 1)

Check y  x  1. If x  1, then y  (1)  1 or 0. If x  2, then y  (2)  1 or 3. Thus, y  x  1 or f(x)  x  1 describes this relation.

y

Chapter 4 Practice Test

K’ H’ K

Page 251 1. 2. 3. 4.

J

x

H

J’

b; rotation c; translation a; reflection K(0, 5) • Start at the origin. • Since the x-coordinate is 0, the point is on the y-axis.

O

I’ I

8. relation: {(0, 1), (2, 4), (4, 5), (6, 10)} inverse: {(1, 0), (4, 2), (5, 4), (10, 6)} 9. relation: {(1, 2), (2, 2), (3, 2)} inverse: {(2, 1), (2, 2), (2, 3)}

185

Chapter 4

10. relation: {(1, 1), (0, 3), (1, 0), (4, 2)} inverse: {(1, 1), (3, 0), (0, 1), (2, 4)} 11. x 4x  10 y (x, y) 2 4(2)  10 18 (2, 18) 1 4(1)  10 14 (1, 14) 0 4(0)  10 10 (0, 10) 2 4(2)  10 2 (2, 2) 4 4(4)  10 6 (4,6)

13. First solve the equation for y in terms of x. 1 x 2 1 x 2

2

y

2

2

Chapter 4

2 4 6 8 8 10 12 14 16

2

5

y

(x, y) (2, 6)

6

1

1 (1) 2

5

52

0

1 (0) 2

5

5

2

1 (2) 2

5

4

(2, 4)

4

1 (4) 2

5

3

(4, 3)

y 16 13 10 4 2

y

1

11, 512 2

(0, 5)

2

x

O

14. To find the x-intercept, let y  0. yx2 0x2 02x22 2  x The graph intersects the x-axis at (2, 0). To find the y-intercept, let x 0. yx2 y02 y2 The graph intersects the y-axis at (0, 2). Plot these points and draw the line that connects them.

(x, y) (2, 16) (1, 13) (0, 10) (2, 4) (4, 2)

y O

5

2

5

1

Graph the solution set {(2, 16), (1, 13), (0, 10), (2, 4), (4, 2)}.

4

1

1 (2) 2

12. First solve the equation for y in terms of x. 3x  y  10 3x  y  3x  10  3x y  10  3x 1(y)  1(10  3x) y  3x  10

2

1

Graph the solution set 1 {(2, 6), 1, 52 , (0, 5), (2, 4), (4, 3)}.

4x

3x  10 3(2)  10 3(1)  10 3(0)  10 3(2)  10 3(4)  10

1

2

4

x 2 1 0 2 4

1 x 2

1 x 2

x

y

O

1

1(y)  1 5  2x

12 10 8 6 4 2 4

1

 y  2x  5  2x y  5  2x

Graph the solution set {(2, 18), (1, 14), (0, 10), (2, 2), (4, 6)}. 18 16

y5

4x

y yx2

O

186

x

15. To find the x-intercept, let y  0. x  2y  1 x  2(0)  1 x  1 The graph intersects the x-axis at (1, 0). To find the y-intercept, let x  0. x  2y  1 0  2y  1 2y  1 2y 2



y

1 2 1 2

1

1

19. Graph the equation. For each value of x, a vertical line passes through no more than one point on the graph. Thus, the equation represents a function. y

O

2

The graph intersects the y-axis at 0,  2 . Plot these points and draw the line that connects them.

x

g(x)  x2  4x  1

20.

g(2)  (2) 2  4(2)  1  4  4(2)  1  4  (8)  1 481  12  1  13 21. f(x)  2x  5

y

O

x x  2y  1

f

112 2  2112 2  5  1  5 4

g(x)  x2  4x  1 g(3a)  1  [ (3a) 2  4(3a)  1]  1  [9a2  4(3a)  1]  1  (9a2  12a  1)  1  9a2  12a  2 23. f(x)  2x  5 f(x  2)  2(x  2)  5  2x  4  5  2x  1 24. 16 24 32 40 22.

16. To find the x-intercept, let y  0. 3x  5  y 3x  5  0 3x  5 3x 3

5

 3 5

x  3

1

5

2

The graph intersects the x-axis at 3, 0 . To find the y-intercept, let x  0. 3x  5  y 3(0)  5  y 05y 0y 5yy y5 The graph intersects the y-axis at (0, 5). Plot these points and draw the line that connects them.

   8 8 8

This is an arithmetic sequence because the difference between terms is constant. The common difference is 8. 25. 99 87 76 65

   12 11 11

This is not an arithmetic sequence because the difference between terms is not constant. 26. 5 17 29 41

y

   12 12 12

This is an arithmetic sequence because the difference between terms is constant. The common difference is 12. 27. 5 10 15 20 25

3x  5  y

O

x

    15 25 35 45

The difference between each pair of terms alternates in sign and the absolute value of each difference is increased by 10 for each successive pair. Continue the pattern. Add 55, 65, and 75. 5 10 15 20 25 30 35 40

17. Since each element of the domain is paired with exactly one element of the range, the relation is a function. 18. The relation is not a function. The element 3 in the domain is paired with both 1 and 2 in the range.

       15 25 35 45 55 65 75

The next three terms are 30, 35, and 40.

187

Chapter 4

5

6

8

11

3. B; Let x represent the number of students who eat five servings.

15

     0 1 2 3 4

The difference between each pair of terms increases by 1 for each successive pair. Continue increasing each successive difference by 1. Add 5, 6, and 7. 5 5 6 8 11 15 20 26 33

2 5

5(x)  2(470) 5x  940

        0 1 2 3 4 5 6 7

5x 5

1a 2 c, b 2 d 2  11  2(5) , 4 2 4 2 4 8  1 2 , 22

 (2, 4) The center of the circle is at (2, 4). 6. B; The relation would not be a function if an element in the domain were paired with more than one element in the range. This would occur for x  2 or x  7. 7. D;

1. B; Use the percent proportion. You know the percent, 2, and the base, 315. Let a represent the part. p

 100 2

 100

x 4

100(a)  315(2) 100a  630 630 100

1 1

p p

 100

50,800 2183

p  23.3 About 23% of the total distance remained.

Chapter 4

True or False? true ✓

y 3 1 ? 3

2

For this relation to be linear, the differences of the y values must be the same when the differences of the x values are the same. Therefore, the difference between 1 and ? must be 2, and the difference between ? and 3 must also be 2. Thus, the missing value must be 1.

 100



x 1 2 3 4



1

2183( p)  508(100) 2183p  50,800 2183p 2183

3x  4y  12 3(4)  4(0)  12 12  12

y 0

8. C;

a  6.3 There will be about 6 new students, so a total of 315  6, or 321, students will attend next year. 2. C; The total distance to be traveled is 1675  508, or 2183, mi. Use the percent proportion. You know the part, 508, and the base, 2183. Let p represent the percent. a b 508 2183

6

3

x2 5. B; The center is the midpoint of (1, 4) and (5, 4). Let (a, b)  (1, 4) and (c, d)  (5, 4). The midpoint is

Pages 252–253



940 5

3x 3

Chapter 4 Standardized Test Practice

100a 100



x  188 You can expect 188 students to eat five servings of fruits and vegetables daily. 4. B; 13x  2(5x  3) 13x  10x  6 13x  10x  10x  6  10x 3x  6

The next three terms are 20, 26, and 33. 29. K  C  273 K  273  C  273  273 K  273  C Since this equation is solved for C in terms of K, K is the independent variable and C is the dependent variable. For five values of K and corresponding values of C, see students’ work. 30. D; f(x)  3x  2 f(8)  f(5)  [3(8)  2 ]  [3(5)  2]  (24  2)  (15  2)  22  (17)  22  17  39

a b a 315

x

 470

  

28. 5

188

14. original: $160 new: $120 160  120  40

3 3 2

x 2 1 4 6

y 5 2 1 3

  

  

9. C; 3

40 160

3

160(r)  40(100) 160r  4000

2

160r 160

The difference in y values is the opposite of the difference in x values. This suggests y  x. Check: If x  2, then y  (2)  2. But the y value for x  2 is 5. This is a difference of 3. Try other values in the domain to see if the same difference occurs. 2 2 5

1 1 2

4 4 1

6 6 3



x x y

15 4

V /h V /h

y is always 3 more than x.



/wh /h

w

1800 (20) (6) 1800 120

w w

15  w The pool is 15 ft wide. 16. 1 triangle 2 triangles 3 toothpicks 5 toothpicks

6

w



4000 160

Substitute V  1800, h  6, and /  20.

4(6)  15(w) 24  15w 24 15 8 5



r  25 There was a 25% decrease in price. 15. Since the unknown quantity is width, solve the formula for w. V  /wh

Check y  x  3. If x  1, then y  (1)  3  2. If x  6, then y  (6)  3  3. Thus, y  x  3 describes this relation. 10.

r

 100

 2

15w 15

The number of toothpicks form an arithmetic sequence since the number added each time is constant. The first term is 3. The common difference is 2. Use the formula for the nth term to write an equation with a1  3 and d  2. an  a1  (n  1)d an  3  (n  1)2 an  3  2n  2 an  2n  1

w 8

So, the width is 5, or 1.6, cm. 25

11. P(winning a television)  2000  0.0125 The probability of winning is 0.0125, or 1.25%. 12. Let n  the first integer. Then n  3  the second integer, and 2n  1  the third integer. n  (n  3)  (2n  1)  52 4n  4  52 4n  4  4  52  4 4n  48 4n 4



To find the number of toothpicks used to make 7 triangles, find a7. a7  2(7)  1 a7  14  1 a7  15 So, 15 toothpicks will be used to make 7 triangles. 17. C;

48 4

n  12 n  3  12  3 or 15 2n  1  2(12)  1 or 25 The integers are 12, 15, and 25. 13. 5(x  2)  3(x  4)  10 5x  10  3x  12  10 2x  22  10 2x  22  22  10  22 2x  32 2x 2



 2

3 triangles 7 toothpicks

42  16(2  5)  3  42  16(7)  3  16  16(7)  3  1(7)  3  73  21

32 2

60  23  3  6 3

4  62 60  8  3  6  64  62 60  24  6  2 36  6  2 42  2

 21

x  16

The two quantities are equal.

189

Chapter 4

18. B;

123 21158 2119 2 30 1  1 24 21 9 2 5 1  1 4 21 9 2

7

If the width must be a whole number, the domain of w is {1, 2, 3, . . . . , 24}. For each of these widths, determine the area of the garden.

21

w (ft)

24  w

 (ft)

3

1

24  1

23

23

2

24  2

22

44

3

24  3

21

63

4

24  4

20

80

5

24  5

19

95

6

24  6

18

108

7

24  7

17

119

8

24  8

16

128 135

 56

5

 36 5

134 21141 2 21 1  1 4 21 14 2 8

10

3

27

Since 36  72 and 8  72, the quantity in Column B is greater. 19. B; 6x  15  3x  75 6x  15  3x  3x  75  3x 9x  15  75 9x  15  15  75  15 9x  90 9x 9



90 9

x  10 3y  32  7y  74 3y  32  3y  7y  74  3y 32  4y  74 32  74  4y  74  74 42  4y 42 4



4y 4

10.5  y The quantity in Column B is greater. 20. A; f(x)  37  10x f(10)  37  10(10)  37  100  63 g(x)  9x  7 g(15)  9(15)  7  135  7  142 The quantity in Column A is greater. 21a. P  2/  2w Since the perimeter is 48 ft, 2/  2w  48. 21b. A  /w Use the formula from part a to solve for one variable in terms of the other. 2/  2w  48 2(/  w)  48 2(/  w) 2



9

24  9

15

10

24  10

14

140

11

24  11

13

143

12

24  12

12

144

13

24  13

11

143

14

24  14

10

140

15

24  15

9

135

16

24  16

8

128

17

24  17

7

119

18

24  18

6

108

19

24  19

5

95

20

24  20

4

80

21

24  21

3

63

22

24  22

2

44

23

24  23

1

23

24

24  24

0

0

The greatest possible area is 144 ft2.

48 2

/  w  24 /  w  w  24  w /  24  w

Chapter 4

A  w (ft2)

190

— GREATEST ←

Chapter 5 Page 255 1.

3.

5.

7.

Analyzing Linear Equations 2.

10.

a  b c  d

a  b c  d

y2  y1

8  4

8 12

 12  4

mx

2

2

3 4.

6.

8.

3 9.

6. Let (0, 0)  (x1, y1) and (5, 4)  (x2, y2).

Getting Started

2  2  10  2 1 5 2 2  2  8  2 8 1  4 5 5  5  15  5 15 1 3 9 9  3 33 3 3 1 2 10



4  4  84 1  2 7  7 7  28  7 28 1 4 18  6 18  12  6 12 3 2 1  12 4 8



7. Let (2, 2)  (x1, y1) and (1, 2)  (x2, y2). y2  y1

mx

2

 

y2  y1

mx

6  5

 

5 2 5 2 6 3

 

2

    

(12 (12 1 1

11.

a  b c  d

  



2  1 4  0 3 4 3 4

a  b c  d

  

8  (22 1  1 8  2 2 10 2

13.

a  b c  d

 

 5 14.

a  b c  d



1 2





y2  y1

mx

2



3  (32 4  7 3  3 3 0 3



 x1

5  5 2  3 0 5

0 10. Let (1, 3)  (x1, y1) and (1, 0)  (x2, y2). y2  y1

mx

0

2

 x1

0  3  (12 3 0

 1

3 2



79 1  2

Since division by zero is undefined, the slope is undefined. 11. Let (6, 2)  (x1, y1) and (r, 6)  (x2, y2).

1

 2 1

2

y2  y1

16. (3, 2) 19. (2, 2)

15. (1, 2) 18. (0, 3)

 x1

1  (42 9  7 3 2

9. Let (3, 5)  (x1, y1) and (2, 5)  (x2, y2).

2 12.

 x1

2  2 1  (22 4 1

 4 8. Let (7,4)  (x1, y1) and (9, 1)  (x2, y2).

84 1 4

 x1

4  0 5  0 4 5

mx

17. (2, 3) 20. (3, 0)

2

4 4

4(r  62  4 4r  24  4 4r  24  24  4  24 4r  20

Slope

5-1

Pages 259–260

Check for Understanding

4r 4

1. Sample answer: Use (1,3) as (x1, y1) and (3,5) as (x2, y2) in the slope formula. 2. See students’ work. 3. The difference in the x values is always 0, and division by 0 is undefined. 4. Carlos; Allison switched the order of the x-coordinates, resulting in an incorrect sign. 5. Let (1, 1)  (x1, y1) and (3, 4)  (x2, y2). 2

 



20 4

r5 12. Let (9, r)  (x1, y1) and (6, 3)  (x2, y2). y2  y1

mx 1 3 1 3

 

2

 x1

3  r 6  9 3  r 3

1(32  3(3  r2 3  9  3r 3  9  9  3r  9 6  3r

y2  y1

mx

 x1

6  (22 r  6 4 r  6

 x1

4  1 3  1 3 2

6 3



3r 3

2r

191

Chapter 5

22. Let (3, 6)  (x1, y1) and (2, 4)  (x2, y2).

13. Use the formula for slope. rise run

y2  y1

change in quantity change in time 55  52 1992  1990 3 or 1.5 2

  

mx

2

 

Over this 2-year period, the number of subscribers increased by 3 million, for a rate of change of 1.5 million per year. 14. Sample answer: ’92–’94; steeper segment means greater rate of change.

Pages 260–262



23. Let (3, 4)  (x1, y1) and (5, 1)  (x2, y2). y2  y1

mx

2



Practice and Apply



15. Let (2, 4)  (x1, y1) and (2, 1)  (x2, y2). 2

 

y2  y1

 x1

mx

1  (4) 2  (2) 3 4

2

 

16. Let (0, 3)  (x1, y1) and (3, 2)  (x2, y2). y2  y1

mx

2



 x1



y2  y1

mx

1 3 1 3

2



Since division by zero is undefined, the slope is undefined. 26. Let (2, 6)  (x1, y1) and (1, 3)  (x2, y2).

y2  y1 2

 

 x1

3  (12 3  (42 2 1

y2  y1

mx

2



 2 18. Let (3, 3)  (x1, y1) and (1, 3)  (x2, y2).



y2  y1

mx

2

 

y2  y1

mx

2



0 19. Let (2, 1)  (x1, y1) and (2, 3)  (x2, y2).



y2  y1 2

 

y2  y1

mx

2

 

y2  y1  x1

29. Let (8, 3)  (x1, y1) and (6, 2)  (x2, y2).

7  3

92 

y2  y1

mx

4 7

2



21. Let (5, 7)  (x1, y1) and (2, 3)  (x2, y2).



y2  y1

mx

2

 x1



3  7

 2  5  

10 7 10 7

Chapter 5

 x1

6  9  (32 3 4 3 4

 7

Since division by zero is undefined, the slope is undefined. 20. Let (2, 3)  (x1, y1) and (9, 7)  (x2, y2). 2

 x1

3  3 8  (22 0 10

0 28. Let (3, 9)  (x1, y1) and (7, 6)  (x2, y2).

 x1

3  1 2  (22 2 0

mx

 x1

3  6 1  2 3 3

1 27. Let (2, 3)  (x1, y1) and (8, 3)  (x2, y2).

 x1

3  3 1  (32 0 4

mx

 x1

1  4  (52 5 0

 5

17. Let (4, 1)  (x1, y1) and (3, 3)  (x2, y2). mx

 x1

3  (1) 5  2 2 3 2 3

25. Let (5, 4)  (x1, y1) and (5, 1)  (x2, y2).

2  3

30 

 x1

1  (4) 5  (3) 3 8

24. Let (2, 1)  (x1, y1) and (5, 3)  (x2, y2).

y2  y1

mx

 x1

4  6 2  (32 2 5 2 5

192

 x1

2  3 6  (82 1 2 1 2

30. Let (2, 0)  (x1, y1) and (1, 1)  (x2, y2).

37. Let (0, 0)  (x1, y1) and (r, s)  (x2, y2).

y2  y1

mx   

y2  y1

mx

 x1

2

 

2

y2  y1



 x1

2

2  (12

 5.3  4.5  



3 0.8 15 4

39.

rise run

40.

rise run

 

 x1

1  1 0.75  0.75 2 0

1

1

2

1

1

33. Let 22,12  (x , y ) and 2, 1 1 y2  y1

mx

2



1 2

 x1

1

1

 12 1

1

2  22

1 2

2  (x2,

1

y2 ).

2

y2  y 1

mx 1 

2

(x1, y1 ) and

1

1 2,

2

1  (x2, y2 ).

 x1

y2  y 1

1  14 1

mx

3

2  4

8

9



4

8

5

4

35. Sample answer: The rise in the picture is about 16 mm, and the run is about 22 mm.  

rise run 16 22 8 11

y2  y 1

mx 4 3 4 3

36. Sample answer: The rise in the picture is about 10 mm, and the run is about 30 mm. slope   

 x1

2

r  (52 3  4 r  5 1

8(12  r  5 8  r  5 8  5  r  5  5. 13  r 43. Let (5, r)  (x1, y1) and (2, 3)  (x2, y2).

9

5

slope 

 x1

2

r  2 9  6 r  2 3

1(32  r  2 3  r  2 3  2  r  2  2 1  r 42. Let (4, 5)  (x1, y1) and (3, r)  (x2, y2).

1



change in quantity change in time 5.15  4.25 1997  1991 0.9 6

1 

2  3 3 1 Let 4, 14  y2  y1 2

16 4

 0.15 Over this 6-year period, the minimum wage increased by $0.90, for a rate of change of $0.15 per year. 41. Let (6, 2)  (x1, y1) and (9, r)  (x2, y2).

2

mx





 3

34.





Since division by zero is undefined, the slope is undefined.

1

1

b  b a  a 2b 0

4

y2  y1 2

0

Since division by zero is undefined, the slope is undefined.

32. Let (0.75, 1)  (x1, y1) and (0.75, 1)  (x2, y2). mx

s  0 r  0 s if r r

38. Let (a, b)  (x1, y1) and (a,  b)  (x2, y2). y2  y1 mx x

31. Let (4.5, 1)  (x1, y1) and (5.3, 2)  (x2, y2). mx

 x1

2

1  0 1  (22 1 3 1 3

 

2

 x1

3  r 2  5 3  r 3

4(32  3(3  r2 12  9  3r 12  9  9  3r  9 3  3r

rise run 10 30 1 3

3 3



3r 3

1r

193

Chapter 5

48. Let (r, 5)  (x1, y1) and (2, r)  (x2, y2).

44. Let (2, 7)  (x1, y1) and (r, 3)  (x2, y2). y2  y1

mx 4 3 4 3

 

y2  y1

mx

 x1

2

2

3  7 r  (22 4 r  2

2 9

212  r2  91r  52 4  2r  9r  45 4  2r  9r  9r  45  9r 4  7r  45 4  7r  4  45  4 7r  49

4(r  22  3(42 4r  8  12 4r  8  8  12  8 4r  20 4r 4

45. Let

1



20 4

7r 7

r  5

1 , 2

1 4

2  (x1, y1) and 1r, 54 2  (x2, y2). y2  y1

 x1 5 1 4  (4 2 1 r2 2

r

4(r  2 2  1

1 2

1

4r  2  1 4r  2  2  1  2 4r  1 4r 4

1

1 r2  (x1, y1) and 11, 2  (x2, y2). y2  y1  x1

1 2

r

2

1

1 2



1 2



r

113 2  21 1 3

1 3

2

13 1 2

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r

 1  2r

2

54.

 1  1  2r  1 2 3 2 3

2 1 3

r

47. Let (4, r)  (x1, y1) and (r, 2)  (x2, y2). y2  y1 2



5 3



 x1

2  r r  4

5(r  4)  3(2  r) 5r  20  6  3r 5r  20  3r  6  3r  3r 2r  20  6 2r  20  20  6  20 2r  14 2r 2



14 2

r7 Chapter 5

change in enrollment change in time

11.3  12.4

 1990  1985 1.1 5

 0.22 Between 1985 and 1990, the rate of change was 0.22 million students per year. 55. The negative slope represents a decline in enrollment. 56. See students’ work.

2r 2

mx

12 14 16 18 20 Age (years)



 2r 

60

51. Karen grew the fastest in the two-year period from age 12 to age 14. The line segment representing this two-year period is the steepest part of the graph. 52. There was no change in height in that two-year period. 53. The steepest segment is between ’90 and ’95. Thus, the rate of change was the greatest during this five-year period. The least steep segment is between ’80 and ’85. Thus, the rate of change was the least during this five-year period.

1 2

mx

62

0

1

r4

46. Let

66 64

58

4

2 , 3

49 7

68

1 Height (in.)

4



r7 49. (4, 5) is in Quadrant III and (4, 5) is in Quadrant I. The segment connecting them goes from lower left to upper right, which is a positive slope. 50. Karen’s Height

mx 4



 x1

r  5 2  r

194

Let (2, 2)  (x1, y1) and (1, 1)  (x2, y2).

57. In the picture, the stairs rise 7 in. for each 11 in. of run. slope  

y2  y1

mx

rise run 7 11

2

 

7

Use the formula for slope with slope  11, rise  8 feet 9 inches or 105 in. and run  r.

Let (4, 0)  (x1, y1) and (2, 2)  (x2, y2). y2  y1

mx

rise run 105 r

slope  7 11



2



7(r)  11(105) 7r  1155 7r 7



 x1

1  (2) 1  (2) 1 3

 

1155 7

 x1

2  0 2  4 2 6 1 3

The slope is the same regardless of points chosen. 62. Let (2, 3)  (x1, y1) and (1, 1)  (x2, y2).

r  165 The total run would be 165 in. or 13 feet 9 inches. 58. Sample answer: Analysis of the slope of a roof might help to determine the materials of which it should be made and its functionality. Answers should include the following. • To find the slope of the roof, find a vertical line that passes through the peak of the roof and a horizontal line that passes through the eave. Find the distances from the intersection of those two lines to the peak and to the eave. Use those measures as the rise and run to calculate the slope. • A roof that is steeper than one with a rise of 6 and a run of 12 would be one with a rise greater than 6 and the same run. A roof with a steeper slope appears taller than one with a less steep slope. 59. D; Let (5,4)  (x1, y1) and (5, 10)  (x2, y2).

y2  y1

mx

2

 x1

1  3  2 4 3 4 3

 1  

Let (1, 1)  (x1, y1) and (4, 2)  (x2, y2). y2  y1

mx

2

 x1

2  (1)  (1) 1 3 1 3

 4  

4

No, they do not. Slope of QR is 3 and slope of RS 1 is 3. If they lie on the same line, the slopes should be the same.

y2  y1

mx

2

 

 x1

Page 262

10  (4) 5  5 6 0

1  63.

Since division by zero is undefined, the slope is undefined. 60. B; Let (c, d)  (x1, y1) and (a, b)  (x2, y2). 2



 x1

b  d a  c

61. Let (5, 3)  (x1, y1) and (2, 2)  (x2, y2). y2  y1

mx

2

 x1

2  (3)  (5) 1 3

 2 

x f(x)

1 5

1  2 10

1  3 15

1  4 20

5 25

    5 5 5 5 The difference of the values for x is 1, and the difference of the values for f(x) is 5. This suggests that f(x)  5x. Check this equation. Check: If x  1, then f(x)  5(1) or 5. If x  2, then f(x)  5(2) or 10. If x  3, then f(x)  5(3) or 15. The equation checks. We can write the equation as f(x)  5x.

y2  y1

mx

Maintain Your Skills

Let (5, 3)  (x1, y1) and (4, 0)  (x2, y2). y2  y1

mx

2

  

 x1

0  (3) 4  (5) 3 9 1 3

195

Chapter 5

1 2 1 2     64. x 2 1 1 2 4 f(x) 13 12 10 9 7     1 2 1 2 The difference in f(x) values is 1 times the difference in x values. This suggests that f(x)  x. Check this equation. Check: If x  2, then f(x)  (2) or 2. But the f(x) value for x  2 is 13. This is a difference of 11. Try some other value in the domain to see if the same difference occurs. 2 2 13

x x f(x)

1 1 12

1 1 10

2 2 9

69. Solve the equation for y. xy0 xyx0x y  x (1)(y)  (1)(x) yx Select five values for the domain and make a table. Then graph the ordered pairs and draw a line through the points. x 3 1 0 2 4

4 4 7

y

f(x) is always 11 more than x. This pattern suggests that 11 should be added to one side of the equation in order to correctly describe the relation. Check f(x)  x  11. Check: If x  2, then f(x)  (2)  11 or 13. If x  1, then f(x)  (1)  11 or 12. The equation checks. We can write the equation as f(x)  11  x. 65. Graph the equation. No vertical line passes through more than one point on the graph. Thus, the line represents a function. 6 3

x y  0

O

a b

 100

a 37.5

 100

y

p

40

100a  37.5(40) 100a  1500 100a 100



1500 100

a  15 Forty percent of 37.5 is 15. 71. 7(3)  21 72. (4)(2)  8 73. (9)(4)  36 74. (8)(3.7)  29.6

–18

66. Graph the equation. The vertical line x  5 passes through more than one point on the graph. Thus, the line does not represent a function.

1 7 2113 2  247

75. 8

y

2

6

3

77. 6  3  1  2

x5



76.

114 2112 2 (14)  118 21142 7

81. 67. This set of ordered pairs represents a relation that is not a function. The element 1 in the domain is paired with both 0 and 4 in the range. 68. This set of ordered pairs represents a function since, for each element of the domain, there is only one corresponding element in the range.

83.

85.

196

3 8

1

78. 12  4  12  4  48

18 2

10 8  1 3 80 2  3 or 26 3 3 1 3 6  6  41 4 18  4 9 1  2 or 42 7 18 8 18  8  1  7 144 4  7 or 20 7 2 1 8 4 23  4  3  1 32 2  3 or 103



3

 4 or 14

9

x

79. 10 

Chapter 5

x

70. The percent is 40, and the base is 37.5. Let a represent the part.

–12 –9 –6 –3 O 3 6 9 12 x –3 –6 –9 y  15 –12

O

(x, y) (3, 3) (1, 1) (0, 0) (2, 2) (4, 4)

y 3 1 0 2 4

80.

1 2

1

1

3

 3  21 3

1

 2 or 1 2 82.

3 4

3

1

 6  46 3

 24 1

8 84.

3 8

2

3

5

 5  82 15

 16

Page 263

6. Step 1:

Reading Mathematics

Term 2a. slope

2b. intercept

2c. parallel

Everyday Meaning 1. to diverge from the vertical or horizontal; incline 2. to move on a slant; ascend or descend to stop, deflect, or interrupt the progress or intended course of of, relating to, or carrying out the simultaneous performance of separate tasks

Write the slope as a ratio. 2

21 Step 2: Graph (0, 0). Step 3: From the point (0, 0), move up 2 units and right 1 unit. Draw a dot. Step 4: Draw a line containing the points.

1. Sample answer: The mathematical meaning of function is most closely related to the third definition in the everyday meanings. Mathematical Meaning the ratio of the rise to the run

y

x

the coordinate at which a graph intersects an axis 7. Step 1: lines that never intersect; nonvertical lines that have the same slope

Write the slope as a ratio. 3 

3 1

Step 2: Graph (0, 0). Step 3: From the point (0, 0), move down 3 units and right 1 unit. Draw a dot. Step 4: Draw a line containing the points. y

Slope and Direct Variation

5-2

y  2x

O

y  3 x

Page 265

Graphing Calculator Investigation

1. All the graphs pass through the origin. 2. None of the graphs have the same slope. 3. Sample answer: y  5x; See students’ graphs.

x

O

3

4. Sample answer: y  2x; See students’ graphs. 5. This family of graphs has a y-intercept of 0. Their slopes are all different. 6. As|m|increases the graph becomes more steep.

Page 267

8. Step 1:

1 2

Step 2: Graph (0, 0). Step 3: From the point (0, 0), move up 1 unit and right 2 units. Draw a dot. Step 4: Draw a line containing the points.

Check for Understanding

1. y  kx, where k is a constant of variation. 2. b; 4a  b means b varies directly as a, and 4 is the constant of variation. 1

c; z  3 x means z varies directly as x, and constant of variation. 3. They are equal.

1 3

y

is the

y  12 x

1

4. The constant of variation is 3.

O

x

y2  y1

mx

2

Write the slope as a ratio.

 x1

0  1 (3) 1 3

m0 m

1

The slope is 3. 5. The constant of variation is 1. y2  y1

mx

2

 x1

2  0

m20 m1 The slope is 1.

197

Chapter 5

13. The graph of y  6x passes through the origin with slope 6.

9. Find the value of k. y  kx 27  k(6) 27 6 9 2



6

m1

k(6) 6

y

k 9

Therefore, y  2 x.

y  6x

Now find x when y  45. 9

y  2x 9

O

45  2 x 2 (45) 9



1 2

x

2 9 x 9 2

10  x 10. Find the value of k. y  kx 10  k(9) 10 9 10 9



14. y  6x y  6(30) y  180 You will earn $180 if you work 30 hours.

k(9) 9

k

Therefore, y 

Pages 268–270

10 x. 9

y2  y1

Now find x when y  9. y 9 9 (9) 10 81 10



mx

2

10 x 9 10 x 9 9 10 x 10 9

x

m



1 2

y2  y1

mx

2

m

 x1

0  (4) 0  (1)

m4 The slope is 4.

k(14) 14

1

17. The constant of variation is 2.

k

y2  y1

mx

1

2

Therefore, y  2 x.

m

Now find y when x  20.

 x1

1  0 2  0 1 2

1

m

1

The slope is 2.

y  2x

1

y  2 (20) y  10 12. Words: Variables:

18. The constant of variation is 1. y2  y1

mx

Your pay for 7.5 hours is $45. Let k  pay rate.

2

m

Amount of Pay equals pay rate times time.

$45 Equation: Solve for the rate. 45  k(7.5)





k



 x1

2  0 2  0

m  1 The slope is 1.

424 3 144424443 14243 14243 1 424 3 1

45 7.5

 x1

4  0 2  0

m2 The slope is 2. 16. The constant of variation is 4.

8.1  x 11. Find the value of k. y  kx 7  k(14) 7 14 1 2

Practice and Apply

15. The constant of variation is 2.

7.5h

3

19. The constant of variation is 2. y2  y1

mx

k(7.5) 7.5

2

m

6k Therefore, the direct variation equation is y  6x.

m

 x1

3  0 2  0 3 2 3

The slope is 2.

Chapter 5

198

1

24. Write the slope as a ratio.

20. The constant of variation is 4. y2  y1

mx

2

m m

4 

 x1

4 1

Graph (0, 0). From the point (0, 0), move down 4 units and right 1 unit. Draw a dot. Draw a line containing the points.

1  0 4  0 1 4

y

1

The slope is 4. 21. Write the slope as a ratio.

y  4 x

1

11 Graph (0, 0). From the point (0, 0), move up 1 unit and right 1 unit. Draw a dot. Draw a line containing the points.

x

O

y

25. Write the slope as a ratio. yx

O

1 4

x

Graph (0, 0). From the point (0, 0), move up 1 unit and right 4 units. Draw a dot. Draw a line containing the points. y y  14 x

22. Write the slope as a ratio. 3

31

O

x

Graph (0, 0). From the point (0, 0), move up 3 units and right 1 unit. Draw a dot. Draw a line containing the points. y

26. Write the slope as a ratio.

y  3x

3 5

O

Graph (0, 0). From the point (0, 0), move up 3 units and right 5 units. Draw a dot. Draw a line containing the points.

x

y

y  35 x

23. Write the slope as a ratio. 1 

1 1

O

x

Graph (0, 0). From the point (0, 0), move down 1 unit and right 1 unit. Draw a dot. Draw a line containing the points. y y  x

27. Write the slope as a ratio. 5 2

O

Graph (0, 0). From the point (0, 0), move up 5 units and right 2 units. Draw a dot. Draw a line containing the points.

x

y

O

y  52 x x

199

Chapter 5

32. Write the slope as a ratio.

28. Write the slope as a ratio.

9

9 2

7 5

2 

Graph (0, 0). From the point (0, 0), move up 7 units and right 5 units. Draw a dot. Draw a line containing the points.

Graph (0, 0). From the point (0, 0), move down 9 units and right 2 units. Draw a dot. Draw a line containing the points.

y

y x

O

y   92 x y  75 x O

x

29. Write the slope as a ratio.

33. Find the value of k. y  kx 8  k(4)

1 5

Graph (0, 0). From the point (0, 0), move up 1 unit and right 5 units. Draw a dot. Draw a line containing the points.

8 4

y  15 x O

x

36 6

30. Write the slope as a ratio. 2



Graph (0, 0). From the point (0, 0), move down 2 units and right 3 units. Draw a dot. Draw a line containing the points. y

42 6



x

16 4

4 3

Graph (0, 0). From the point (0, 0), move down 4 units and right 3 units. Draw a dot. Draw a line containing the points.

20 4

y   43 x

Chapter 5



k(4) 4

4x 4

5  x

y

O



4  k Therefore, y  4x. Now find x when y  20. y  4x 20  4x

31. Write the slope as a ratio. 4

6x 6

7x 35. Find the value of k. y  kx 16  k(4)

y   23 x

3 

k(6) 6

6k Therefore, y  6x. Now find x when y  42. y  6x 42  6x

2 3

O

k(4) 4

2k Therefore, y  2x. Now find y when x  5. y  2x y  2(5) y  10 34. Find the value of k. y  kx 36  k(6)

y

3 



x

200

36. Find the value of k. y  kx 18  k(6) 18 6



40. Find the value of k. y  kx 6.6  k(9.9) 6.6 9.9 2 3

k(6) 6

3  k Therefore, y  3x. Now find x when y  6. y  3x 6  3x 6 3



Now find y when x  6.6. 2

y  3 x

3x 3

2

y  3 (6.6) y  4.4 41. Find the value of k. y  kx

114 2 4 8 4 1  1 1k  4 2 1 13 2 2

23  k

k

32 3

1

Therefore, y  3 x.

32 x. 3 1

1

Now find y when x  18.

1

y

y  3x y  3 (24) y  8 38. Find the value of k. y  kx 12  k(15) 

9k Therefore, y  9x. Now find x when y  12. y  9x 12  9x

4

y  5x 21  5x 4

105 4

x

1 2

5 4 x 5

12 9 4 3

26.25  x 39. Find the value of k. y  kx 2.5  k(0.5) 

1 2 12

123 2 3 3 2 (6)  2 1 k  3 2 2

Now find x when y  21.

5 (21) 4

y

32 9 3 8

6k

4

Therefore, y  5x.

4

y

32 x 3 32 1 18 3

y  12 42. Find the value of k. y  kx

k(15) 15

k

2.5 0.5

k

Therefore, y 

Now find y when x  24.

12 15 4 5

k 2

k(12) 12



k(9.9) 9.9

Therefore, y  3 x.

2  x 37. Find the value of k. y  kx 4  k(12) 4 12 1 3





9x 9

x

43. The the diameter d. circumference C is 3.14 times 1444 442444 443 { 123 1 424 3 1444 424444 3 C

k(0.5) 0.5

 3.14



d

The direct variation equation is C  3.14d. The graph of C  3.14d passes through the origin with slope 3.14.

5k Therefore, y  5x. Now find y when x  20. y  5x y  5(20) y  100

m

3.14 1

C

C  3.14 d

0

201

d

Chapter 5

48. Line 4 passes through the points (0, 0) and (1, 25).

The perimeter P is 4 times the length of a side s. 3 144444424444443 44. 14444244443 { { 1424 P 4  s The direct variation equation is P  4s. The graph of P  4s passes through the origin with slope 4.

y2  y1

mx

2

m

 x1

25  0 1  0

m  25 The slope of line 4 is 25. Therefore, line 4 represents the sprinting speed of the elephant. 49. Line 2 passes through the points (0, 0) and (1, 32).

4

m1 P

y2  y1

mx

P  4s

2

m

m  32 The slope of line 2 is 32. Therefore, line 2 represents the sprinting speed of the reindeer. 50. Line 1 passes through the points (0, 0) and (1, 50).

s

0

is 0.99 times the number of yards n. The total cost C 444424444 3 { 123 1 424 3 1444442444443 45. 1  0.99  C n

y2  y1

mx

The direct variation equation is C  0.99n. The graph of C  0.99n passes through the origin with slope 0.99. m

2

m

y2  y1

mx

C  0.99 n

2

m

The total cost C is 14.49 times the number of pounds p. { 123 1 424 3 1 4444442444444 3 C  14.49  p

1 444424444 3

The direct variation equation is C  14.49p. The graph of C  14.49p passes through the origin with slope 14.49.

1442443 1 424 3 1442443 1 424 3 1442443

60  k  Solve for the constant of variation. 60  k(360)

14.49 1

C

60 360 1 6

40 30

k(360) 360

k

1

53. m  6 e

10

1

p 2

4

6

m  6 (138)

8

m  23 If you weigh 138 pounds on Earth, you would weigh 23 pounds on the moon.

y

47. It also doubles. If x  k, and x is multiplied by 2, y must also be multiplied by 2 to maintain the value of k.

Chapter 5



360

Therefore, the direct variation equation is 1 m  6 e.

C  14.49 p

20

 x1

30  0 1  0

m  30 The slope of line 3 is 30. Therefore, line 3 represents the sprinting speed of the grizzly bear. 52. The weight on the moon is 60 pounds, and the weight on Earth is 360 pounds. Let k  constant of variation. the weight The weight the constant equals times on Earth. on the moon of variation

n

0

m

 x1

50  0 1  0

m  50 The slope of line 1 is 50. Therefore, line 1 represents the sprinting speed of the lion. 51. Line 3 passes through the points (0, 0) and (1, 30).

0.99 1

C

46.

 x1

32  0 1  0

202

54. The age of a human is 3 years old, and the age of a horse is 1 year old. Let k  constant of variation. The age of the constant the age of equals times a human of variation a horse.

Page 270 63. m  x

2

m

y2  y1

3  k  1 Solve for the constant of variation. 3  k(1) 3k Therefore, the direct variation equation is y  3x. 55. y  3x 16  3x 

 x1

0  3 2  1

m  3

1442443 1 424 3 1442443 1 424 3 1 44244 3

16 3 1 53

Maintain Your Skills

y2  y1

64. m  x

2

m m

 x1

2  (2) 2  2 4 0

Since division by zero is undefined, the slope is undefined. y2  y1

65. m  x

3x 3

2

m

x

2

3  1 2  (3)

2

m2

The equivalent horse age for a human who is 1 16 years old is 53 years old or 5 years 4 months.

2

2r 2

m m



2 2

r  1 1

67.

1

 x y

0 1

1

 1

1

 2 9

y2  y1 2

 x1

3  7 r  1 4 r  1

21r  12  4 2r  2  4 2r  2  2  4  2 2r  2

56. The slope of the equation that relates time and water use is the number of gallons used per minute in the shower. Answers should include the following. • y  2.5x • Less steep; the slope is less than the slope of the graph on page 268. 57. D; The graph passes through the points (0, 0) and (2, 1). mx

y2  y1

mx

66.

 x1

1



3 13 

 4

5 21

4

 x1

1  0 2  0 1 2

The difference in the x values is 1, and the difference in the y values is 4. Since 1  4  5, and 13  4  17, the numbers 5 and 17 are inserted. 1

Therefore, the equation is y  2 x.

x y

58. C; The ordered pair (0, 0) is a solution of a direct variation equation. Check (0, 0) in the equation y  3x  1. Check: y  3x  1 ? 0  3(0)  1 01 59. The calculator screen shows the graphs of y  1x, y  2x and y  4x. [10, 10] scl: 1 by [10, 10] scl: 1

0 1

1 5 2

68.

2

 x y

2

2 9 2

 4

3 13 

6 4

4 17 2

5 21 2

  8 10 12 2 2

 2

The difference in the x values is 2, and the difference in the y values is 2. Since 4  2  6, 6  2  8, and 2  2  0, the numbers 8, 6, and 0 are inserted. x y

2 8

4 6

6 4

8 2

10 0

12 2

69. 15  1122  1|15|  |12|2  115  122 3 70. 8  152  8  5  13 71. 9  6  9  162  (|9|  |6|)  19  62  15 72. 18  12  18  1122  1|18|  |12|2  118  122  30

60. They all pass through (0, 0), but these have negative slopes. 61. Sample answer: y  5x. 62. Sample answer: Find the absolute value of k in each equation. The one with the greatest value of |k| has the steeper slope.

203

Chapter 5

3x  y  8 3x  y  3x  8  3x y  3x  8 74. 2x  y  7 2x  y  2x  7  2x y  2x  7 75. 76. 4x  y  3 2y  4x  10 1 1 4x  3  y  3  3 (2y)  2 (4x  10) 2 4x  3  y y  2x  5 77. 9x  3y  12 9x  3y  9x  12  9x 3y  9x  12 73.

1 (3y) 3

78.

7. Write the slope as a ratio. 7 

Graph (0, 0). From the point (0, 0), move down 7 units and right 1 unit. Draw a dot. Draw a line containing the points. y y  7 x

1

 3 (9x  12) 8. Write the slope as a ratio. 3 4

Graph (0, 0). From the point (0, 0), move up 3 units and right 4 units. Draw a dot. Draw a line containing the points.

5  x 2 x  5 2



y

Page 270

y y  34 x

Practice Quiz 1

y2  y1

1. m  x

2

y2  y1

2. m  x

 x1

2

8  (6) (4)

m  2

m  19

0

m0

y2  y1

m m

x

3  3

m  11  8

2

O

 x1

m  3 

3. m  x

y2  y1

4. m  x

 x1

2

9  8 5  (4) 1 9

m m y2  y1

mx

5.

2

2 2

9. Find the value of k. y  kx 24  k182

 x1

11  1 7  0 10 7

24 8

2r 2

8

2

10 15 2 3

r4

y2  y1

mx

2

3 2 3 2

 

 x1

9  r 4  6 9  r 10





k(15) 15

k 2

2

y  3x 2

6  3x 3

3

1

2

2 (6)  2 3 x 9x

2r 2

24  r

Chapter 5

k182 8

Therefore, y  3x. Now find x when y  6.

3(10)  2(9  r) 30  18  2r 30  18  18  2r  18 48  2r 48 2



3k Therefore, y  3x. Now find y when x  3. y  3x y  3(3) y  9 10. Find the value of k. y  kx 10  k(15)

 x1

5  (3) r  5 2 r  5

2(r  5)  2 2r  10  2 2r  10  10  2  10 2r  8

6.

x

O

y  3x  4 x  2y  5 x  2y  x  5  x 2y  5  x 2y 2

7 1

204

2

Page 271

5-3

See students’ work. Sample answer: It is a linear pattern. It is the y-intercept. See students’ work. Sample answer: 1.5 The slope represents the rate of change. Add 2.5 units to the y coordinate of each data point. Plot points at (0, 10.5), (1, 11.75), (2, 13), (3, 14.5), and (4, 15.5). Length (cm)

1. 2. 3. 4. 5. 6.

Algebra Activity (Preview of Lesson 5-3)

16 14

Page 275

12 10 8

x

0

1 2 3 4 5 Number of Washers

y2  y1

mx

Length (cm)

7. The sample data has a distance of 8 cm for 0 washers and an approximate change of 1.25 cm for each washer added. Therefore, the new graph should have a point at (0, 8), then increase the distance more than 1.25 cm for each washer added. Sample answer: plot points at (0, 8), (1, 10), (2, 12), (3, 14), and (4, 16). 16 14

2

m m

12 10 8

x

0

1 2 3 4 5 Number of Washers

8. The sample data has a distance of 8 cm for 0 washers and an approximate change of 1.25 cm for each washer added. Therefore, the new graph should have a point at (0, 8), then increase the distance less than 1.25 cm for each washer added. Sample answer: plot points at (0, 8), (1, 9), (2, 10), (3, 11), and (4, 12).

Length (cm)

12

 x1

3  (1) 2  0 4 2

m2 The slope is 2. Step 2: The line crosses the y-axis at (0, 1). So, the y-intercept is 1. Step 3: Finally, write the equation. y  mx  b y  2x  112 y  2x  1 The equation of the line is y  2x 1. 7. Step 1: You know the coordinates of two points on the line. Find the slope. Let (x1, y1)  (0, 2) and (x2, y2)  (2, 1).

y

6

Check for Understanding

1. Sample answer: y  7x  2 2. Vertical lines have undefined slope. Horizontal lines have a slope of 0. 3. The rate of change is described by the slope. 4. Replace m with 3 and b with 1. y  mx  b y  3x  1 5. Replace m with 4 and b with 2. y  mx  b y  4x  122 y  4x  2 6. Step 1: You know the coordinates of two points on the line. Find the slope. Let (x1, y1)  (0, 1) and (x2, y2)  (2, 3).

y

6

Slope-Intercept Form

y2  y1

mx

2

m m m

y

 x1

1  2 2  0 3 2 3 2 3

The slope is 2.

11 10 9

Step 2: The line crosses the y-axis at (0, 2). So, the y-intercept is 2. Step 3: Finally, write the equation. y  mx  b

8 7 6

3

y  2x  2

5

x

3

The equation of the line is y  2x  2.

0

1 2 3 4 5 Number of Washers

205

Chapter 5

8. Step 1: The y-intercept is 3. So, graph (0, 3). 2 Step 2: The slope is 1. From (0, 3), move up 2 units and right 1 unit. Draw a dot. Step 3: Draw a line containing the points.

12. The graph passes through (0, 50) with slope 5. T 80 60

T  50  5w

y 40

y  2x  3

20 0

x

O

2

Pages 275–277

y

y  3x  1

16. Replace m with y  mx  b

10. Step 1: Solve for y to find the slope-intercept form. 2x  y  5 2x  y  2x  5  2x y  2x  5 Step 2: The y-intercept is 5. So, graph (0, 5). 2 Step 3: The slope is 1 . From (0, 5), move down 2 units and right 1 unit. Draw a dot. Step 4: Draw a line containing the points.

3

y  5 x  0 3

y  5 x 18. Replace m with 1 and b with 10. y  mx  b y  1x  10 y  x  10 19. Replace m with 0.5 and b with 7.5. y  mx  b y  0.5x  7.5 20. Find the slope. y2  y1

mx m

11. Words:

The amount you save increases $5 per week, so the rate of change is $5 per week. You start with $50. Variables: T is the total amount. W is the number of weeks. Equation: Total rate of number of weeks equals times amount change from now

plus

w



 x1

4  1 1  0

m3 The line crosses the y-axis at (0, 1). So, the y-intercept is 1. Finally, write the equation. y  mx  b y  3x  1

amount at start.

14243 1 424 3 14243 1 424 3 14444244443 1 424 3 14243

50

The equation is T  5w  50.

Chapter 5

and b with 3.

3

2



1 2

17. Replace m with 5 and b with 0. y  mx  b

x

5

Practice and Apply

1

2x  y  5



w

y  2x  3

y

T

8

14. Replace m with 2 and b with 6. y  mx  b y  2x  162 y  2x  6 15. Replace m with 3 and b with 5. y  mx  b y  3x  152 y  3x  5

x

O

6

13. After 7 weeks, w  7. T  5w  50 T  5172  50 T  85 So, the total amount saved after 7 weeks is $85.

9. Step 1: The y-intercept is 1. So, graph (0, 1). 3 Step 2: The slope is 1 . From (0, 1), move down 3 units and right 1 unit. Draw a dot. Step 3: Draw a line containing the points.

O

4

206

So, the y-intercept is 2. Finally, write the equation. y  mx  b y  0x  2 y2 26. The slope of a horizontal line is 0. Since the line crosses the y-axis at (0, 5), the y-intercept is 5. Replace m with 0 and b with 5. y  mx  b y  0x  (5) y  5 27. Since the line passes through the origin, the y-intercept is 0. Replace m with 3 and b with 0. y  mx  b y  3x  0 y  3x 28. The y-intercept is 1. So, graph (0, 1). The slope is 3 . From (0, 1), move up 3 units and right 1 unit. 1 Draw a dot. Draw a line containing the points.

21. Find the slope. y2  y1

mx

2

m m

 x1

1  (4) 2  0 3 2

The line crosses the y-axis at (0, 4). So, the y-intercept is 4. Finally, write the equation. y  mx  b 3

y  2 x  (4) 3

y  2x  4 22. Find the slope. y2  y1

mx

2

m m

 x1

2  2 1  0 4 1

m  4 The line crosses the y-axis at (0, 2). So, the y-intercept is 2. Finally, write the equation. y  mx  b y  4x  2 23. Find the slope.

y y  3x  1

O

x

y2  y1

mx

2

m m m

 x1

1  1 3  0 2 3 2 3

29. The y-intercept is 2. So, graph (0, 2). The slope 1 is 1. From (0, 2), move up 1 unit and right 1 unit. Draw a dot. Draw a line containing the points.

The line crosses the y-axis at (0, 1). So, the y-intercept is 1. Finally, write the equation. y  mx  b

y

yx2

2

y  3 x  1

O

24. Find the slope.

x

y2  y1

mx

2

m m

 x1

3  0 2  0 3 2

30. The y-intercept is 1. So, graph (0, 1). The slope 4 is 1 . From (0, 1) move down 4 units and right 1 unit. Draw a dot. Draw a line containing the points.

The line crosses the y-axis at (0, 0). So, the y-intercept is 0. Finally, write the equation. y  mx  b

y

3

y  2x  0 3

y  2x 25. Find the slope.

O

y2  y1

mx

2

m m

x

y  4x  1

 x1

2  2 2  0 0 2

m0 The line crosses the y-axis at (0, 2).

207

Chapter 5

35. Solve for y to find the slope-intercept form. 2x  y  3 2x  y  2x  3  2x y  2x  3 1(y)  1(2x  3) y  2x  3 The y-intercept is 3. So, graph (0, 3). The slope is 2 . From (0, 3) move up 2 units and right 1 unit. 1 Draw a dot. Draw a line containing the points.

31. The y-intercept is 2. So, graph (0, 2). The slope is 1 . From (0, 2) move down 1 unit and right 1 unit. 1 Draw a dot. Draw a line containing the points. y

x

O

y  x  2

y

32. The y-intercept is 4. So, graph (0, 4). The slope is 1 . From (0, 4) move up 1 unit and right 2 units. 2 Draw a dot. Draw a line containing the points.

2x  y  3

y

36. Solve for y to find the slope-intercept form. 3y  2x  3

1

y  2x  4

3y 3 3y 3

x

O

x

O

 

y 33. The y-intercept is 3. So, graph (0,3). The slope 1 is 3 . From (0, 3) move down 1 unit and right 3 units. Draw a dot. Draw a line containing the points.

2x  3 3 2x 3 3 3 2 x1 3

The y-intercept is 1. So, graph (0, 1). The slope is 2 . From (0, 1) move up 2 units and right 3 units. 3 Draw a dot. Draw a line containing the points. y

y

x

O

x

O 3y  2x  3 1

y 3x  3

37. Solve for y to find the slope-intercept form. 2y  6x  4

34. Solve for y to find the slope-intercept form. 3x  y  2 3x  y  3x  2  3x y  3x  2 The y-intercept is 2. So, graph (0, 2). The slope 3 is 1 . From (0, 2) move down 3 units and right 1 unit. Draw a dot. Draw a line containing the points.

2y 2 2y 2

 

6x  4 2 6x 4  2 2

y  3x  2 The y-intercept is 2. So, graph (0, 2). The slope 3 is 1 . From (0, 2) move down 3 units and right 1 unit. Draw a dot. Draw a line containing the points.

y O

y x

O 3x  y  2 2y  6x  4

Chapter 5

208

x

41. The cost increases $25 per hour, so the rate of change is $25 per hour. The initial cost is $50. Let C  total cost. Let h  number of hours used for repair. number of hours Total rate of used for initial cost equals change times repair plus cost. 123 123 1 424 3 123 1 424 3 123 123

38. Solve for y to find the slope-intercept form. 2x  3y  6 2x  3y  2x  6  2x 3y  2x  6 3y 3 3y 3

 

y

2x  6 3 2x 6 3 3 2 3x  2

C  25  The equation is C  25h  50.

The y-intercept is 2. So, graph (0, 2). The slope is 2 . From (0, 2) move down 2 units and right 3 3 units. Draw a dot. Draw a line containing the points.

x

C



y

The y-intercept is 1. So, graph (0, 1). The slope 4 is 3. From (0, 1) move up 4 units and right 3 units. Draw a dot. Draw a line containing the points.

x

4x  3y  3



t



6

 6.

S  1  t  16 The equation is S  t  16. 46. The year 2005 is 14 years after 1991. So, t  14. S  t  16 S  14  16 S  30 So, the total sales will be $30 billion in 2005.

40. The cost increases $2 per hour, so the rate of change is $2 per hour. The initial cost is $20. Let C  total cost. Let t  number of hours bicycle is rented. number Total rate of of hours initial cost change times rental plus cost. 1 23 equals 123 1 424 3 12 3 of 1 424 31 23 123 t

 1 2 t

 T  2 h  15 The equation is T  2h  15. 44. They all have a y-intercept of 3. 45. The amount of sales increased $1 billion per year, so the rate of change is $1 billion per year. In 1991 there were $16 billion in book sales. S is the total sales. t is the number of years after 1991. number rate of years amount Total of after at sales equals change times 1991 plus start. 123 123 1 424 3 123 1 424 3 123 1 424 3

y

C  2  The equation is C  2t  20.

1

2

43. The temperature will fall 2 each hour, so the rate of change is 2 each hour. The temperature is currently 15. Let T  temperature. Let h  number of hours passed during the night. rate number temperTemperof of hours ature ature equals change times passed plus at start. 1 424 3 123 1 424 3 123 1 424 3 123 1 424 3

4x  3 3 4x 3  3 3 4 x1 3

O



The equation is H 

39. Solve for y to find the slope-intercept form. 4x  3y  3 4x  3y  4x  3  4x 3y  4x  3 

50

1

2x  3y  6

3y 3 3y 3



42. The candle decreases in height 2 in. per hour, so 1 the rate of change is 2 in. per hour. The candle is 6 in. tall when lit. Let H  height of the candle. Let t  number of hours the candle burns. number Height rate of hours height of of candle at candle equals change times burns plus start. 123 123 1 424 3 123 1 424 3 123 123

y

O

h

20

209

Chapter 5

55a. Replace A with 2, B with 1, and C with 4.

47. The fatality rate decreased 0.12 each year, so the rate of change is 0.12 each year. In 1966 there were 5.5 fatalities per 100 million vehicle miles. R is the fatality rate. t is the number of years after 1966. number rate of years amount Fatality of after at rate equals change times 1966 plus start. 14243 123 1 424 3 123 1 424 3 123 1 424 3

A

2

B  1  2 The slope is 2. C B



4 1

 4 The y-intercept is 4. 55b. Replace A with 3, B with 4, and C with 12.

R  0.12  t  5.5 The equation is R  0.12t  5.5. 48. The graph passes through (0, 5.5) with slope 0.12.

A

3

B  4 3

The slope is 4. C B

R

12 4



3 The y-intercept is 3. 55c. Replace A with 2, B with 3, and C with 9. A

2

B  3

R  5.5  0.12t

2

3 2

0

The slope is 3.

t

C B

49. The year 1999 is 33 years after 1966. So, t  33. R  0.12t  5.5 R  0.12(33)  5.5 R  1.54 So, the fatality rate was 1.54 fatalities per 100 million vehicle miles in 1999. 50. The y-intercept is the flat fee in an equation that represents a price. Answers should include the following. • The graph crosses the y-axis at 5.99. • Sample answer: A mechanic charges $25 plus $40 per hour to work on your car. 51. D; Solve for y to find the slope-intercept form. 2x  y  5 2x  y  2x  5  2x y  2x  5 (1)(y)  (1)(2x  5) y  2x  5 The y-intercept is 5. 52. B; The y-intercept is 100, which indicates that you already have $100. The slope is 5 which indicates that you are adding or saving $5 for each time frame. 53. Ax  By  C Ax  By  Ax  C  Ax By  Ax  C Ax  C B Ax C  B B A C y  B x  B, where B  0. A C A slope of y  B x  B is B, where B C y-intercept is B. By B By B

54. The The

Chapter 5

9

 3

 3 The y-intercept is 3.

Page 277

Maintain Your Skills

56. Find the value of k. y  kx 45  k(60) 45 60 3 4



k(60) 60

k 3

Therefore, y  4x. Now find x when y  8. 3

y  4x 3

8  4x 4 (8) 3 32 3 2 103

1 2

4 3 x 4

3

x x

57. Find the value of k. y  kx 15  k(4) 15 4 15 4





k(4) 4

k

Therefore, y 

15 x. 4

Now find y when x  10. y  0.

y y

210

15 x 4 15 (10) 4 75 1 or 372 2

y2  y1

58. m  x

2

 x1 6  0  (3)

m  4 6

y2  y1

59. m  x

2

m

Notice that the graph of y  4 is the same as the graph of y  0, moved 4 units down. Also, the graph of y  7 is the same as the graph of y  0, moved 7 units up. All graphs have the same slope and different intercepts. Because they all have the same slope, this family of graphs can be described as linear graphs with a slope of 0. 2. Enter the equations in the Y  list and graph.

 x1

4  (1) 3  3 3 0

m  1

m

m  6

Since division by zero is undefined, the slope is undefined.

y2  y1

60. m  x

2

m m

 x1

2  (5) 9  5 7 4

61. Write each number as a decimal. 2.5 3 4

 0.75

[10, 10] scl: 1 by [10, 10] scl: 1

0.5 7 8

• The graph of y  x  1 has a slope of 1 and a y-intercept of 1. • The graph of y  2x  1 has a slope of 2 and a y-intercept of 1. 1 1 • The graph of y  4 x  1 has a slope of 4 and a y-intercept of 1. These graphs have the same intercept and different slopes. This family of graphs can be described as linear graphs with a y-intercept of 1. 3. Enter the equations in the Y  list and graph.

 0.875

0.5 0.75 0.875 2.5 The numbers arranged in order from least to 3 7 greatest are 0.5, 4, 8, 2.5. 62. x  x

15  9 2 6 2

63. 3(7)  2  b 21  2  b

x3 64. q  62  22 q  36  4 q  32

y2  y1

65. m  x

2

m m

y2  y1

66. m  x

2

 x1

8  8  5 0 7

y2  y1 2

m

m

m

Page 279

y  x 4

m  2 67. m  x

m  2 m0

 x1

2  2 1  (1) 4 2

m

y  2x  4

 x1

13  (1) 10  1 12 9 4 3

y  2x  4

[10, 10] scl: 1 by [10, 10] scl: 1

• The graph of y  x  4 has a slope of 1 and a y-intercept of 4. • The graph of y  2x  4 has a slope of 2 and a y-intercept of 4. • The graph of y  2x  4 has a slope of 2 and a y-intercept of 4. These equations are similar in that they all have positive slope. However, since the slopes are different and the y-intercepts are different, these graphs are not all in the same family.

Graphing Calculator Investigation (Follow-Up of Lesson 5-3)

1. Enter the equations in the Y  list and graph.

[10, 10] scl: 1 by [10, 10] scl: 1

• The graph of y  4 has a slope of 0 and a y-intercept of 4. • The graph of y  0 has a slope of 0 and a y-intercept of 0. • The graph of y  7 has a slope of 0 and a y-intercept of 7.

211

Chapter 5

4. Enter the equations in the Y  list and graph.

Because they all have the same slope, this family of graphs can be described as linear graphs with a slope of 3. 7. Sample answer: The value of m determines the steepness and direction of the graph. If the graph has a positive slope, it slants upward from left to right. A graph with a negative slope slants downward from left to right. The greater the absolute value of the slope, the steeper the line. Lines with the same slope are parallel. The value of b determines where the line crosses the y-axis. Lines with the same value of b form a family of lines that intersect at that intercept. The values of b in a family of parallel lines determine how far the lines are apart on the y-axis. 8. This class of functions has graphs that are lines with slope 1. Their y-intercepts are all different. 9. See students’ graphs. The graph of y  0 x 0  c is the same as the graph of y  0 x 0 translated vertically c units. If c is positive, the translation is up; if c is negative, the translation is down. The graph of y  0 x  c 0 is the same as the graph of y  0 x 0 , translated horizontally c units. If c is positive, the translation is to the left; if c is negative, the translation is to the right.

1

y 4x4

1

y 3x3

1

y 2x2

[10, 10] scl: 1 by [10, 10] scl: 1 1

1

• The graph of y  2x  2 has a slope of 2 and a y-intercept of 2. 1 1 • The graph of y  3x  3 has a slope of 3 and a y-intercept of 3. 1 1 • The graph of y  4x  4 has a slope of 4 and a y-intercept of 4. These equations are similar in that they all have positive slope. However, since the slopes are different and the y-intercepts are different, these graphs are not all in the same family. 5. Enter the equations in the Y  list and graph. y  2x  2

y  2x  2

1

y 2x2 [10, 10] scl: 1 by [10, 10] scl: 1

• The graph of y  2x  2 has a slope of 2 and a y-intercept of 2. • The graph of y  2x  2 has a slope of 2 and a y-intercept of 2. 1

Writing Equations in SlopeIntercept Form

Page 283

Check for Understanding

1. When you have the slope and one point, you can substitute these values in for x, y, and m to find b. When you are given two points, you must first find the slope and then use the first procedure. 2. Sample answer: y  2x  3 3. Sometimes; if the x- and y-intercepts are both zero, you cannot write the equation of the graph. 4. Step 1: The line has slope 2. To find the y-intercept, replace m with 2 and (x, y) with (4, 2) in the slope-intercept form. Then solve for b. y  mx  b 2  2(4)  b 2  8  b 2  8  8  b  8 10  b Step 2: Write the slope-intercept form using m  2 and b  10. y  mx  b y  2x  (10) y  2x  10 Therefore, the equation is y  2x  10.

1

• The graph of y  2x  2 has a slope of 2 and a y-intercept of 2. These graphs have the same intercept and different slopes. This family of graphs can be described as linear graphs with a y-intercept of 2. 6. Enter the equations in the Y  list and graph. y  3x  6 y  3x

y  3x  7 [10, 10] scl: 1 by [10, 10] scl: 1

• The graph of y  3x has a slope of 3 and a y-intercept of 0. • The graph of y  3x  6 has a slope of 3 and a y-intercept of 6. • The graph of y  3x  7 has a slope of 3 and a y-intercept of 7. Notice that the graph of y  3x  6 is the same as the graph of y  3x, moved 6 units up. Also, the graph of y  3x  7 is the same as the graph of y  3x, moved 7 units down. All graphs have the same slope and different intercepts.

Chapter 5

5-4

212

5. Step 1: The line has slope 3. To find the y-intercept, replace m with 3 and (x, y) with (3, 7) in the slope-intercept form. Then solve for b. y  mx  b 7  3(3)  b 7  9  b 7  9  9  b  9 16  b Step 2: Write the slope-intercept form using m  3 and b  16. y  mx  b y  3x  16 Therefore, the equation is y  3x  16. 6. Step 1: The line has slope 1. To find the y-intercept, replace m with 1 and (x, y) with (3, 5) in the slope-intercept form. Then solve for b. y  mx  b 5  1(3)  b 53b 533b3 2b Step 2: Write the slope-intercept form using m  1 and b  2. y  mx  b y  1x  2 y  x  2 Therefore, the equation is y  x  2. 7. Step 1: Find the slope of the line containing the points. Let (x1, y1)  (5, 1) and (x2, y2)  (8, 2).

8. Step 1: Find the slope of the line containing the points. Let (x1, y1)  (6, 0) and (x2, y2)  (0, 4). y2  y1

mx

2

m m m

Step 2: You know the slope and two points. The point (0, 4) lies on the y-axis. Thus, the y-intercept is 4. Step 3: Write the slope-intercept form using 2 m  3 and b  4. y  mx  b 2

y  3x  4 2

Therefore, the equation is y  3x  4. 9. Step 1: Find the slope of the line containing the points. Let (x1, y1)  (5, 2) and (x2, y2)  (7, 4). y2  y1

mx

2

m m m

2

m m

 x1

4  2 7  5 6 12 1 2

Step 2: You know the slope and two points. Choose one point and find the y-intercept. In this case, we choose (5, 2). y  mx  b 1

2  2 (5)  b

y2  y1

mx

 x1

4  0 0  6 4 6 2 3

 x1

5

22b

2  1 8  5 3 3

5

5

5

222b2 1

2  b

m  1 Step 2: You know the slope and two points. Choose one point and find the y-intercept. In this case, we choose (5, 1). y  mx  b 1  1(5)  b 1  5  b 1  5  5  b  5 6b Step 3: Write the slope-intercept form using m  1 and b  6. y  mx  b y  1x  6 y  x  6 Therefore, the equation is y  x  6.

Step 3:

Write the slope-intercept form using 1

1

m  2 and b  2. y  mx  b 1

1 12

y  2x  2 1

1

y  2x  2

1

1

Therefore, the equation is y  2x  2.

213

Chapter 5

14. Find the y-intercept. y  mx  b 4  5(5)  b 4  25  b 4  25  25  b  25 29  b Write the slope-intercept form. y  mx  b y  5x  29 15. Find the y-intercept. y  mx  b 0  2(3)  b 0  6  b 0  6  6  b  6 6b Write the slope-intercept form. y  mx  b y  2x  6 16. Find the y-intercept. y  mx  b

10. A; Read the Test Item The table represents the ordered pairs (5, 2) and (0, 7). Solve the Test Item Step 1: Find the slope of the line containing the points. Let (x1, y1)  (5, 2) and (x2, y2)  (0, 7). y2  y1

mx

2

m

 x1

7  2 0  (5) 5

m5 m1 Step 2: You know the slope and two points. The point (0, 7) lies on the y-axis. Thus, the y-intercept is 7. Step 3: Write the slope-intercept form using m  1 and b  7. y  mx  b y  1x  7 yx7 Therefore, the equation is y  x  7.

Pages 284–285

1

3  2 (5)  b 5

32b

Practice and Apply

11. Find the y-intercept. y  mx  b 2  3(1)  b 23b 233b3 1  b Write the slope-intercept form. y  mx  b y  3x  (1) y  3x  1 12. Find the y-intercept. y  mx  b 1  1(4)  b 1  4  b 1  4  4  b  4 3b Write the slope-intercept form. y  mx  b y  1x  3 y  x  3 13. Find the y-intercept. y  mx  b 2  3(5)  b 2  15  b 2  15  15  b  15 17  b Write the slope-intercept form. y  mx  b y  3x  (17) y  3x  17

Chapter 5

5

5

5

1 2

b

322b2 Write the slope-intercept form. y  mx  b 1

1

y  2x  2 17. Find the y-intercept. y  mx  b 2

1  3 (3)  b 1  2  b 1  2  2  b  2 3  b Write the slope-intercept form. y  mx  b 2

y  3x  (3) 2

y  3x  3 18. Find the y-intercept. y  mx  b 5

5  3 (3)  b 5  5  b 5  5  5  b  5 10  b Write the slope-intercept form. y  mx  b 5

y  3x  (10) 5

y  3x  10

214

Write the slope-intercept form. y  mx  b y  2x  (8) y  2x  8 23. Find the slope.

19. Find the slope. y2  y1

mx

2

m m

 x1

2  1 5  4 1 1

y2  y1

mx

m1 Find the y-intercept. y  mx  b 1  1(4)  b 14b 144b4 3  b Write the slope-intercept form. y  mx  b y  1x  (3) yx3 20. Find the slope.

2

m m

m  2 Find the y-intercept. y  mx  b 3  2(1)  b 32b 322b2 1b Write the slope-intercept form. y  mx  b y  2x  1 24. Find the slope.

y2  y1

mx

2

 x1

0  2

m20 m

y2  y1

2 2

mx

2

m  1 From the graph, we see that the y-intercept is 2. Write the slope-intercept form. y  mx  b y  1x  2 y  x  2 21. Find the slope.

m m

2

m m

 x1

4  2 2  4 6 6

m1 Find the y-intercept. y  mx  b 2  1(4)  b 24b 244b4 2  b Write the slope-intercept form. y  mx  b y  1x  (2) yx2 22. Find the slope.

y2  y1

mx

2

m m

2

m m

 x1

2  (2) 4  7 0 11

m0 Find the y-intercept. y  mx  b 2  0(7)  b 2  0  b 2  b Write the slope-intercept form. y  mx  b y  0x  (2) y  2

y2  y1

mx

 x1

2  (2) 3  2 4 1

m4 Find the y-intercept. y  mx  b 2  4(3)  b 2  12  b 2  12  12  b  12 10  b Write the slope-intercept form. y  mx  b y  4x  (10) y  4x  10 25. Find the slope.

y2  y1

mx

 x1

3  3 2  (1) 6 3

 x1

4  (2) 6  3 6 3

m2 Find the y-intercept. y  mx  b 4  2(6)  b 4  12  b 4  12  12  b  12 8  b

215

Chapter 5

Find the y-intercept. y  mx  b

26. Find the slope. y2  y1

mx

2

m m

 x1

1

5

1  16  b

m m m

1

30. Find the slope of the line containing the points (3, 0) and (0, 5). y2  y1

 x1

mx

2

m m



5

y  3x  5

b b

31. Find the slope of the line containing the points (3, 0) and (0, 4).

1 2

y2  y1

mx

b

2

m

Write the slope-intercept form. y  mx  b 1

m

1

y  2x  2

m

28. Find the slope. 2

m m m

 x1

6  7 0  5 1 5 1 5

4

y  3x  4 32. Find the slope of the line containing the points (6, 0) and (0, 3). y2  y1

The point (0, 6) lies on the y-axis. Thus, the y-intercept is 6. Write the slope-intercept form. y  mx  b

mx

2

m m

1

y  5x  6

m

29. Find the slope. 2

m

 x1

3  4 1 4  1

m

4

1

154 2

1

y  2x  3

1 1 m  4

Chapter 5

 x1

3  0 0  6 3 6 1 2

The y-intercept is 3. Write the slope-intercept form. y  mx  b

y2  y1

mx

 x1

4  0 0  3 4 3 4 3

The y-intercept is 4. Write the slope-intercept form. y  mx  b

y2  y1

mx

 x1

5  0 0  (3) 5 3

The y-intercept is 5. Write the slope-intercept form. y  mx  b

1 1 2 1 2

11

y  4x  16

1  2 (1)  b 1

b

y  mx  b

Find the y-intercept. y  mx  b

1 2 1 2

11 16

5

1

4  1 7  1 3 6 1 2

1

5

Write the slope-intercept form using m  4 and 11 b  16.

y2  y1 2

5

1  16  16  b  16

m0 The point (0, 5) lies on the y-axis. Thus, the y-intercept is 5. Write the slope-intercept form. y  mx  b y  0x  5 y5 27. Find the slope. mx

1 52

1  4 4  b

5  5 3  0 0 3

216

PQ249-6481F-05[217-243] 7/31/02 10:04 PM Page 217

Write the slope-intercept form. W  mt  b

33. Find the slope of the line containing the points (2, 0) and (0, 2). y2  y1

mx

2

m

3

W  20t  (274.7)

 x1

3

W  20t  274.7

2  0 0  2

3

37. W  20t  274.7

2

m  2

3

W  20 (2005)  274.7

m1 The y-intercept is 2. Write the slope-intercept form. y  mx  b y  1x  (2) yx2 34. Write an equation of the line that passes through (1970, 23.2) and (1998, 26.7). Find the slope.

38.

y2  y1

mx

2

m m m

 x1

26.7  23.2 1998  1970 3.5 28 1 8

Find the intercept. M  mt  b

39.

1

23.2  8 (1970)  b 23.2  246.25  b 23.2  246.25  246.25  b  246.25 223.05  b Write the slope-intercept form. M  mt  b

40.

1

M  8t  (223.05) 1

M  8t  223.05 1

35. M  8t  223.05 1

M  8 (2005)  223.05 M  27.575 The median age of men who marry for the first time in 2005 should be about 27.6 years old. 36. Write an equation of the line that passes through (1970, 20.8) and (1998, 25). Find the slope.

41.

y2  y1

mx

2

y2  y1

mx

2

m

 x1

m

25  20.8

m  1998  1970 m m

W  26.05 The median age of women who marry for the first time in 2005 should be about 26.05 years old. Write an equation of the line that passes through (1995, 175,000) with slope 2000. Find the y-intercept. y  mx  b 175,000  2000(1995)  b 175,000  3,990,000  b 175,000  3,990,000  3,990,000  b  3,990,000 3,815,000  b Write the slope-intercept form. y  mx  b y  2000x  (3,815,000) y  2000x  3,815,000 y  2000x  3,815,000 y  2000(2010)  3,815,000 y  205,000 The population of Orlando, Florida, in 2010 should be 205,000. Write an equation of the line that passes through (3, 45) with slope 10. Find the intercept. C  mh  b 45  10(3)  b 45  30  b 45  30  30  b  30 15  b Write the slope-intercept form. C  mh  b C  10h  15 Find the slope.

m

4.2 28 3 20

 x1

4  2 7  14 6 21 2 7

Find the y-intercept. y  mx  b

Find the intercept. W  mt  b

2

2  7 (14)  b 24b 244b4 2  b

3

20.8  20 (1970)  b 20.8  295.5  b 20.8  295.5  295.5  b  295.5 274.7  b

Write the slope intercept form. y  mx  b 2

y  7 x  (2) 2

y  7x  2

217

Chapter 5

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42. The slope-intercept form of the line is y

2 x 7

Page 285

 2. 2

The slope is 7. 43. To find the x-intercept, let y  0. 22

y  13x  22

0  13x  22

22

0  13x  13x  22 13x 13 22 22 13x

1



2

x

282 13 282 13 282 13

282  13 13 282 22  13 141 11

1

y

22

y  3x  2

 13x

2

282

The y-intercept is  13 .

1

2.

Therefore, line  intersects the x-axis at 282  13

49. Solve for y. xy6 xyx6x y  x  6 The y-intercept is 6. So, graph (0, 6). The slope is 1 . From (0, 6), move down 1 unit and right 1 1 unit. Draw a dot. Draw a line connecting the points.

, 02 1141 11

44. Find the slope of the line containing the points ( p, 0) and (0, q). y2  y1

mx

 x1

2

x

O

The slope-intercept form of the line is 22 282 y  13 x  13 .

and the y-axis at 0,

Maintain Your Skills

48. The y-intercept is 2. So, graph (0, 2). The slope 3 is 1. From (0, 2), move up 3 units and right 1 unit. Draw a dot. Draw a line connecting the points.

y

q  0

m0p q

xy6

m  p q

m  p The y-intercept is q. Write the slope-intercept form. y  mx  b

x

O

q

y  p x  q, where p  0.

50. Solve for y. x  2y  8 x  2y  x  8  x 2y  x  8

45. Answers should include the following. • Linear extrapolation is when you use a linear equation to predict values that are outside of the given points on the graph. • You can use the slope-intercept form of the equation to find the y value for any requested x value. 46. B; Find the y-intercept. y  mx  b

2y 2 2y 2

1

2

1  3  b 2

y

2

1  3  3  b  3 5 3



x  8 2 x 8 2 2 1 2x 

y 4 The y-intercept is 4. So, graph (0, 4). The slope is 1 . From (0, 4), move down 1 unit and right 2 2 units. Draw a dot. Draw a line connecting the points.

1  3 (2)  b 2



b

x  2y  8

Write the slope-intercept form. y  mx  b 1

x

5

y  3x  3

O

47. B; Write an equation of the line that passes through (0, 560) with slope 20. y  mx  b y  20x  560

Chapter 5

equals rate times number of beats. Volume 51. 1 4243 123 123 123 144424443 2.5  r  b The equation is V  2.5b.

218

PQ249-6481F-05[217-243] 7/31/02 10:04 PM Page 219

52. 53. 54. 55.

The domain for this relation is {0, 4, 9}. The domain for this relation is {2, 0, 5}. 3  5. You can use a calculator to find an approximation 16 for 3 .

7.

4  4.0 16 3

 5.33333333...

Therefore, 4 

16 . 3

4y  12  3(x  1) 4y  12  3x  3 4y  12  12  3x  3  12 4y  3x  9 4y  3x  3x  9  3x 3x  4y  9 9. y  3  2.5(x  1) 2(y  3)  2(2.5)(x  1) 2y  6  5(x  1) 2y  6  5x  5 2y  6  6  5x  5  6 2y  5x  11 2y  5x  5x  11 5x 5x  2y  11 (1)(5x  2y)  (1)11 5x  2y  11 10. y  6  2(x  2) y  6  2x  4 y  6  6  2x  4  6 y  2x  10

 0.66666666... 3 4

1 32

4(y  3)  4 4 (x  1)

 0.75

Therefore,

3

y  3  4 (x  1)

8.

56. You can use a calculator to find an approximation 2 for 3. 3 4 2 3

y  5  4(x  2) y  5  4x  8 y  5  5  4x  8  5 y  4x  13 y  4x  4x  13  4x 4x  y  13 (1)(4x  y)  (1)13 4x  y  13

2

7 3.

57. 4  7  4  (7)  (|7||4|)  (7  4)  3 58. 5  12  5  (12)  (|12||5|)  (12  5)  7 59. 2  (3)  2  3 5 60. 1  4  1  (4)  (|1|  |4|)  (1  4)  5 61. 7  8  7  (8)  (|7|  |8|)  (7  8)  15 62. 5  (2)  5  2  (|5||2|)  (5  2)  3

11.

2

y  3  3 (x  6) 2

y  3  3 x  4 2

y  3  3  3 x  4  3 2

y  3 x  1 12.

1

7

1

7

1

y  2  2x  2 7

5-5

7

y  2  2 (x  4) 7

y  2  2  2x  2  2

Writing Equations in Point-Slope Form

1

3

y  2x  2 13. Step 1:

First find the slope of AD. y2  y1

Page 289

mx

Check for Understanding

2

1. They are the coordinates of a given point on the graph of the equation. 2. Akira; (2, 6) and (1, 6) are both on the line, so either could be substituted into point-slope form to find a correct equation. 3. Sample answer: y  2  4(x  1); y  4x  6 4. y  y1  m(x  x1) y  3  2(x  1) 5. y  y1  m(x  x1) y  (2)  3[x  (1)] y  2  3(x  1) 6. y  y1  m(x  x1) y  (2)  0(x  2) y20

m m

 x1

1  3 3  112 4 2

m2 Step 2: You can use either point for (x1, y1) in the point-slope form. We chose (1, 3). y  y1  m(x  x1) y  3  2[x  (1)] y  3  2(x  1) or y  1  2(x  3)

219

Chapter 5

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14.

y  5  2(x  6) y  5  2x  12 y  5  5  2x  12  5 y  2x  7 y  2x  2x  7  2x 2x  y  7 32. y  3  5(x  1) y  3  5x  5 y  3  3  5x  5  3 y  5x  8 y  5x  5x  8  5x 5x  y  8

y  3  2(x  1) y  3  2x  2 y  3  3  2x  2  3 y  2x  5 y  2x  2x  5  2x 2x  y  5 1(2x  y)  1(5) 2x  y  5

Pages 289–291

31.

Practice and Apply

15. y  y1  m(x  x1) y  8  2(x  3) 16. y  y1  m(x  x1) 17. y  y1  m(x  x1) y  (3)  1[x  (4)] y  4  3[x  (2)] y3x4 y  4  3(x  2) 18. y  y1  m(x  x1) 19. y  y1  m(x  x1) y  1  4[x  (6)] y  6  0[x  (3)] y  1  4(x  6) y60 20. y  y  m(x  x ) 21. y  y  m(x  x ) 1 1 1 1 2

33.

2(y  7)  2

3

3

y  3  4 (x  8)

34.

22. y  y  m(x  x ) 23. y  y1  m(x  x1 ) 1 1 2 5 y  3  3 [x  (6) ] y  (3)  8 (x  1) y3

2 3 (x

y3

 6)

y  y1  m(x  x1) y  (5)  0(x  9) y50 25. y  y  m(x  x ) 1 1

5 8 (x

y8 y8

26.

156 2 (x  4)

6y  6  5(x  4) 6y  6  5x  20 6y  6  6  5x  20  6 6y  5x  14 6y  5x  5x  14  5x 5x  6y  14 (1)(5x  6y)  (1) (14) 5x  6y  14

 1)

y  y1  m(x  x1 ) 8 y  (4)  3 (x  1) 35.

8

y  4  3 (x  1)

2

y  2  5 (x  8)

1 22

5(y  2)  5 5 (x  8)

27. The slope of a horizontal line is zero. y  y1  m(x  x1) y  (9)  0(x  5) y90 28. The slope of a horizontal line is zero. y  y1  m(x  x1) y  7  0(x  0) y70 29. y  13  4(x  2) y  13  4x  8 y  13  13  4x  8  13 y  4x  5 y  4x  4x  5  4x 4x  y  5 (1)(4x  y)  (1)5 4x  y  5 30. y  3  3(x  5) y  3  3x  15 y  3  3  3x  15  3 y  3x  12 y  3x  3x  12  3x 3x  y  12 (1)(3x  y)  (1)12 3x  y  12

Chapter 5

5

y  1  6 (x  4) 6(y  1)  6

24.

7 [x  (4) ] 2 7 (x  4) 2

112 2 (x  2)

2y  14  x  2 2y  14  14  x  2  14 2y  x  12 2y  x  x  12  x x  2y  12 (1)(x  2y)  (1) (12) x  2y  12

y  (3)  4 (x  8)

y  1  3 (x  9)

1

y  7  2 (x  2)

5y  10  2(x  8) 5y  10  2x  16 5y  10  10  2x  16  10 5y  2x  26 5y  2x  2x  26  2x 2x  5y  26 36.

1

y  4  3 (x  12)

1 12

3(y  4)  3 3 (x  12) 3y  12  1(x  12) 3y  12  x  12 3y  12  12  x  12  12 3y  x 3y  x  x  x x  3y  0 37.

5

y  2  3 (x  6) 3(y  2)  3

153 2 (x  6)

3y  6  5(x  6) 3y  6  5x  30 3y  6  6  5x  30  6 3y  5x  24 3y  5x  5x  24  5x 5x  3y  24 (1)(5x  3y)  (1)24 5x  3y  24

220

PQ249-6481F-05[217-243] 7/31/02 10:04 PM Page 221

38.

3

y  6  2 (x  4) 2(y  6)  2

39.

40.

41.

43.

44.

45.

132 2 (x  4)

2

2y  12  3(x  4) 2y  12  3x  12 2y  12  12  3x  12  12 2y  3x  24 2y  3x  3x  24  3x 3x  2y  24 (1)(3x  2y)  (1)(24) 3x  2y  24 y  6  1.3(x  7) 10(y  6)  10(1.3)(x  7) 10y  60  13(x  7) 10y  60  13x  91 10y  60  60  13x  91  60 10y  13x  151 10y  13x  13x  151  13x 13x  10y  151 (1)(13x  10y)  (1)151 13x  10y  151 y  2  2.5(x  1) 2(y  2)  2(2.5)(x  1) 2y  4  5(x  1) 2y  4  5x  5 2y  4  4  5x  5  4 2y  5x  9 2y  5x  5x  9  5x 5x  2y  9 y  2  3(x  1) 42. y  5  6(x  1) y  2  3x  3 y  5  6x  6 y  2  2  3x  3  2 y  5  5  6x  6  5 y  3x  1 y  6x  11 y  2  2(x  5) y  2  2x  10 y  2  2  2x  10  2 y  2x  8 y  1  7(x  3) y  1  7x  21 y  1  1  7x  21  1 y  7x  22

2

y  5  5  5 x  6  5 2

y  5 x  1 1

y



1

y y

1 4



1 4 1 4

 

y 3

y5

52.

y y

1

3 5



3 5 3 5

2

2 2x  3 2 2x  3  1 2x  3 1 3 x  2 3 3x  2 3 3x  2  7 3x  4 1 4 x  2

1

1 3

2

1

 4x  2

1 4

2

3

 4x  2  5 7

y  4x  5 53. Point-slope form: y  y1  m(x  x1) y  (3)  10(x  5) y  3  10(x  5) Slope-intercept form: y  3  10(x  5) y  3  10x  50 y  3  3  10x  50  3 y  10x  53 Standard form: y  10x  53 y  10x  10x  53  10x 10x  y  53 (1)(10x  y)  (1)(53) 10x  y  53 Slope-intercept form: 54. Point-slope form: 3

y  y1  m(x  x1 ) 3 y  (6)  2 (x  1)

2

y  1  3 (x  9)

y





y4

51.

1

y33

1 3

1 3 1 3

y

y  2x  1

y3

1

1

1

y

1

y3

1

y  3  2 x  3

50.

y  3  3  2x  2  3

47.

1

yx1

1

y

1

y22x22

1

y11

1

y2x2

49.

y  3  2 (x  4)

y1

2

y  5  5 x  6

y  3  2x  2

46.

y  5  5 1x  152

48.

y  6  2 (x  1)

3

y  6  2 (x  1)

2 x6 3 2 x61 3 2 x7 3 1 4 1x  22 1 1 4 x  2 1 1 4 x  2  3 1 7 4 x  2

3

3

3

3

3

15 2

y  6  2x  2 y  6  6  2x  2  6 y  2x 

Standard form: 3

y  2x 

13

15 2

2(y)  2 2 x 

15 2

2

2y  3x  15 2y  3x  3x  15  3x 3x  2y  15 (1)(3x  2y)  (1) (15) 3x  2y  15

221

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55. The rate of change is $5 per week, so m  5. Write the point-slope form using m  5 and (x1, y1)  (12, 210). y  y1  m(x  x1) y  210  5(x  12) 56. y  210  5(x  12) y  210  5x  60 y  210  210  5x  60  210 y  5x  150 57. The flat fee is the y-intercept of the equation. The y-intercept of y  5x  150 is 150. Therefore, the flat fee for installation is $150. 58. The rate of change is 1500 each year, so m  1500. Write the point-slope form using m  1500 and (x1, y1)  (1996, 29,690). y  y1  m(x  x1) y  29,690  1500(x  1996) 59. y  29,690  1500(x  1996) y  29,690  1500x  2,994,000 y  29,690  29,690  1500x  2,994,000  29,690 y  1500x  2,964,310 60. y  1500x  2,964,310 y  1500(2005)  2,964,310 y  43,190 The number of movie screens in the United States should be 43,190 in 2005. 61. RQ: First, find the slope.

Method 1

2

1

62. RQ: y  3  2 (x  1)

m

1

2

m

2

or

63. RQ:

Method 2 y  y1  m(x  x1 ) 1 y  (1)  2 (x  3)

PS:

2 Method 2 y  y1  m(x  x1 ) y  3  2(x  1)

1 2

Method 2 y  y1  m(x  x1 ) 1 y  1  2 [x  (3) ]

 x1

Chapter 5

or

5

11

5

2

1

5

y  2x  2

11

5

2

2y  x  5 2y  x  x  5  x x  2y  5 x  2y  5 RS: y  2x  5 y  2x  2x  5  2x 2x  y  5 64. Sample answer: The point-slope form of the equation is y  1  (x  9). Let x  10 and y  0. The equation becomes 0  1  (10  9) or 1  1. Since the equation holds true, (10, 0) is a point on the line passing through (9, 1) and (5, 5). 65. Answers should include the following. • Write the definition of the slope using (x, y) as one point and (x1, y1) as the other. Then solve the equation so that the ys are on one side and the slope and xs are on the other.

y2  y1

m

1

y  2x  2

2(y)  2 2 x  2

RS: First, find the slope. 2

5

2y  x  5 2y  x  x  5  x x  2y  5 x  2y  5 QP: y  2x  5 y  2x  2x  5  2x 2x  y  5

1

1  (3) 3  (1)

1

2(y)  2 2 x  2

y  1  2 (x  3) mx

1

RS: y  3  2(x  1) y  3  2x  2 y  3  3  2x  2  3 y  2x  5

1

Method 1 or y  y1  m(x  x1 ) 1 y  3  2 (x  1)

1

y  2x  2

 x1

1  3  1

m  3

1

y  3  3  2x  2  3

y2  y1

mx

1

y  3  2x  2

y  1  2 (x  3)

Method 1 or y  y1  m(x  x1 ) y  (1)  2(x  3) y  1  2(x  3) PS: First, find the slope.

5

1

 x1

or

1

PS: y  3  2 (x  1)

y2  y1 3  (1) 1  3

1

QP: y  1  2(x  3) y  1  2x  6 y  1  1  2x  6  1 y  2x  5

QP: First, find the slope. mx

1

y  2x  2

1

y  3  2 (x  1)

1

y  3  3  2x  2  3

or 2

Method 1 or y  y1  m(x  x1 ) 1 y  (3)  2 [x  (1) ]

1

y  3  2x  2

 x1

1  (3) 3  (1)

Method 2

y  y1  m(x  x1 ) y  y1  m(x  x1 ) y  (3)  2[x  (1) ] y  1  2[x  (3) ] y  3  2(x  1) y  1  2(x  3)

y2  y1

mx

or

2

222

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66. D; x  3y  15 x  3y  x  15  x 3y  x  15 3y 3

x  15 3 1 y  3x  5 1 Therefore, m  3 and

Page 291



the line passes through (0, 5).

67. Write the point-slope form. Then solve for y. y  y1  m(x  x1) y  (5)  m(x  2) y  5  m(x  2) y  5  mx  2m y  5  5  mx  2m  5 y  mx  2m  5 68. (1, 3) and (0, 1): First, find the slope.

y2  y1

mx

2

m

y2  y1

mx

2

 x1

1  3 (1)

m0

m

or 2

y2  y1 2

m

 x1

1  1 1  0

or 2

Write the point-slope form. Then solve for y. y  y1  m(x  x1) y  1  2(x  0) y  1  2x y  1  1  2x  1 y  2x  1 (1, 1) and (2, 3): First, find the slope.

a5

y2  y1

mx

2

m

2 v 9

78.

 x1

3  (1) 2  1

 x1

6  (4) 0  2 10 2

m  5 The point (0, 6) lies on the y-axis. The y-intercept is 6. Write the slope-intercept form. y  mx  b y  5x  6 75. The line has slope 0. Find the y-intercept. y  mx  b 1  0(1)  b 1  0  b 1  b Write the slope-intercept form. y  mx  b y  0x  (1) y  1 76. 77. 4a  5  15 7  3c  11 4a  5  5  15  5 7  3c  7  11  7 4a  20 3c  18 4a 20 3c 18   3 4 4 3

Write the point-slope form. Then solve for y. y  y1  m(x  x1) y  1  2(x  0) y  1  2x y  1  1  2x  1 y  2x  1 (0, 1) and (1, 1): First, find the slope. mx

Maintain Your Skills

72. y  mx  b y  2x  (5) y  2x  5 73. Find the y-intercept. y  mx  b 4  3(2)  b 4  6  b 4  6  6  b  6 10  b Write the slope-intercept form. y  mx  b y  3x  10 74. Find the slope.

2 v 9

or 2

 6  6  14  6 2 v 9 9 2 v 2 9

Write the point-slope form. Then solve for y. y  y1  m(x  x1) y  (1)  2(x  1) y  1  2(x  1) y  1  2x  2 y  1  1  2x  2  1 y  2x  1 69. All of the equations are the same. 70. The equation will be y  2x  1; see students’ work. 71. Regardless of which two points on a line you select, the slope-intercept form of the equation will always be the same.

c  6

 6  14  20

1 2  92 (20)

v  90 79. (25  4) (22  13)  21 (4  1)  21 3 7 2 # 1 1 80. 1

2

1

The reciprocal of 2 is 2. 81.

10 1

1

 10  1

The reciprocal of 10 is

1 . 10

82. 1  1  1 The reciprocal of 1 is 1.

223

Chapter 5

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83.

1 1

The reciprocal of 1 is 84.

2 3



5. The line parallel to y  x  5 has the same slope, 1. Replace m with 1, and (x1, y1) with (2, 3) in the point-slope form. y  y1  m(x  x1 ) y  3  1(x  2) y3x2 y33x23 yx1 6. The line parallel to y  2x  1 has the same slope, 2. Replace m with 2, and (x1, y1) with (1, 3) in the point-slope form. y  y1  m(x  x1 ) y  (3)  2(x  1) y  3  2x  2 y  3  3  2x  2  3 y  2x  5 7. Write the given equation in slope-intercept form. 3x  y  4 3x  y  3x  4  3x y  3x  4 The line parallel to y  3x  4 has the same slope, 3. Replace m with 3, and (x1, y1) with (2, 2) in the point-slope form. y  y1  m(x  x1 ) y  2  3[x  (2) ] y  2  3(x  2) y  2  3x  6 y  2  2  3x  6  2 y  3x  8 8. Find the slope of each segment.

1

 1  1 3 2

or 1.

1

The reciprocal of

1 92

1

1 1

2 3

3

is 2.

85. 9  1  1 1

The reciprocal of 9 is 9. 86.

5 2

2

51

The reciprocal of 2

1 32

5 2

2

is 5.

87. 3  2  1 2

3

The reciprocal of 3 is 2.

Geometry: Parallel and Perpendicular Lines

5-6 Page 293

Algebra Activity

1. Sample answer: B is 3 units right of the origin and 6 units up. Therefore, B  (3, 6). 2. Sample answer: Find the slope of the line containing (0, 0) and (3, 6). y2  y1

mx

2

 x1

6  0

m  3  0 or 2 3. Sample answer: B is 6 units left of the origin and 3 units up. Therefore, B  (6, 3). 4. Sample answer: Find the slope of the line containing (0, 0) and (6, 3). y2  y1

mx

2

Slope of BD: m 

 x1

3  0  0

m  6

6 7

5. See students’ work. 6. They are perpendicular. 7. The x- and y-coordinates are reversed, and the x-coordinate is multiplied by 1. 8. They are opposite reciprocals. 9. Their product is 1.

1. The slope is 2 which is 3.

1

1

Step 2: The slope of the given line is 3. So, the slope of the line perpendicular to this 1 line is the opposite reciprocal to 3 or 3. Step 3: Use the point-slope form to find the equation. y  y  m(x  x ) 1 1 y  1  3[ x  (3) ] y  1  3(x  3) y  1  3x  9 y  1  1  3x  9  1 y  3x  8

3

so find the opposite reciprocal of 2, 1

2. Sample answer: 2, 2 3. Parallel lines lie in the same plane and never intersect. Perpendicular lines intersect at right angles. 4. The line parallel to y  2x  4 has the same slope, 2. The point (0, 1) lies on the y-axis, so the y-intercept is 1. Replace m with 2, and b with 1 in the slope-intercept form. y  mx  b y  2x  (1) y  2x  1

Chapter 5

1

9. Step 1: The slope of y  3x  2 is 3.

Check for Understanding 3 , 2

123 2  1.

The line segments are not perpendicular because

1

or 2

Pages 295–296

7  1 6 or 7 (2) 5  3 2 or 3 0  3

Slope of AC: m  5 

224

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10. Step 1:

3

15. The line parallel to y  x  6 has the same slope, 1. y  y1  m(x  x1 ) y  2  1[x  (3) ] y2x3 y22x32 yx5 16. The line parallel to y  2x  1 has the same slope, 2. y  y1  m(x  x1 ) y  (1)  2(x  4) y  1  2x  8 y  1  1  2x  8  1 y  2x  9

3

The slope of y  5x  4 is 5. 3 . 5

Step 2: The slope of the given line is So, the slope of the line perpendicular to this 3 5 line is the opposite reciprocal of 5 or 3. Step 3: Use the point-slope form to find the equation. y  y  m(x  x ) 1 1 5

y  (2)  3 (x  6) 5

y  2  3 (x  6) 5

y  2  3x  10 5

y  2  2  3x  10  2 5

y  3x  8

1

17. The line parallel to y  2x  1 has the same 1 slope, 2.

11. Step 1:

Find the slope of the given line. 2x  y  5 2x  y  2x  5  2x y  2x  5 Step 2: The slope of the given line is 2. So, the slope of the line perpendicular to this 1 line is the opposite reciprocal of 2 or 2.

y  y1  m(x  x1 ) 1 y  (4)  2 [ x  (5) ] 1

y  4  2 (x  5) 5

1

3 2

18. The line parallel to y  3x  1 has the same 2 slope, 3.

1 1

y  2  2x  1

y  y1  m(x  x1 ) 2 y  3  3 (x  3)

1

y  2  2  2x  1  2 1

y  2x  3

2

y  3  3x  2

12. The slope of the given line is 3. So, the slope of the line perpendicular to this line is the opposite 1 reciprocal of 3 or 3. Use the point-slope form to find the equation. y  y1  m(x  x1 )

2

y  3  3  3x  2  3 2

y  3x  1 1

19. The line parallel to y  3x  3 has the same 1 slope, 3.

1

y  5  3 (x  3)

y  y1  m(x  x1 ) 1 y  (3)  3 [ x  (4) ]

1

y  5  3x  1 1

y  5  5  3x  1  5

Pages 296–297

1

y  2x  2

y  (2)  2 (x  2)

y

5

y  4  4  2x  2  4

Step 3: Use the point-slope form to find the equation. y  y1  m(x  x1 )

1 3x

1

y  4  2x  2

1

y  3  3 (x  4)

6

1

4

1

4

1

13 3

y  3  3x  3 y  3  3  3x  3  3

Practice and Apply

y  3x 

13. The line parallel to y  x  2 has the same slope, 1. y  y1  m(x  x1 ) y  (7)  1(x  2) y7x2 y77x27 yx9 14. The line parallel to y  2x  2 has the same slope, 2. y  y1  m(x  x1 ) y  (1)  2(x  2) y  1  2x  4 y  1  1  2x  4  1 y  2x  5

1

20. The line parallel to y  2x  4 has the same 1 slope,  2 . y  y1  m(x  x1 ) 1 y  2  2 [ x  (1) ] 1

y  2  2 (x  1) 1

1

1

1

1

3

y  2  2x  2 y  2  2  2x  2  2 y  2 x  2

225

Chapter 5

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.25. The lines for x  3 and x  1 are parallel because all vertical lines are parallel. The lines 2 2 for y  3x  2 and y  3x  3 are parallel because they have the same slope. Thus, both pairs of opposite sides are parallel, and the figure is a parallelogram. 26. The line parallel to y  5x  3 has the same slope, 5. Since the line passes through the origin, the y-intercept is 0. Use the slope-intercept form. y  mx  b y  5x  0 y  5x 27. Write the given equation in slope-intercept form. x  3y  8 x  3y  x  8  x 3y  x  8

21. Write the given equation in slope-intercept form. 2y  x  1 2y 2



y

x  1 2 1 1 x 2 2 1

1

The line parallel to y  2x  2 has the same 1 slope, 2. y  y1  m(x  x1 ) 1 y  0  2 [x  (3) ] 1

y  2 (x  3) 1

3

y  2x  2 22. Write the given equation in slope-intercept form. 3y  2x  6 3y 3



y

2x  6 3 2 3x  2

3y 3

y

2

The line parallel to y  3x  2 has the same 2 slope, 3.

1

2

4

y  3x  (6)

2

4

y  3x  6

2 3x

10 3

y  2  3x  3 

1

28. The slope of the given line is 1. So, the slope of the line perpendicular to this line is the opposite reciprocal of 1, or 1. Use the point-slope form. y  y  m(x  x ) 1 1 y  0  1[x  (2) ] y  (x  2) y  x  2 29. The slope of the given line is 4. So, the slope of the line perpendicular to this line is the opposite 1 reciprocal of 4, or 4. Use the point-slope form.

23. Write the given equation in slope-intercept form. 6x  y  4 6x  y  6x  4  6x y  6x  4 The line parallel to y  6x  4 has the same slope, 6. y  y1  m(x  x1 ) y  3  6[ x  (2) ] y  3  6(x  2) y  3  6x  12 y  3  3  6x  12  3 y  6x  9 24. Write the given equation in slope-intercept form. 3x  4y  4 3x  4y  3x  4  3x  4y  3x  4 4y 4



y

y  y1  m(x  x1 ) 1 y  1  4 (x  1)

y

Chapter 5



1

5

y  y1  m(x  x1 ) 1 y  1  3 [x  (3) ] 1

y  1  3 (x  3) 1

y  1  3x  1

y  2  4x  2 

1

30. The slope of the given line is 3. So, the slope of the line perpendicular to this line is the opposite 1 reciprocal of 3, or 3. Use the point-slope form.

y  y1  m(x  x1 ) 3 y  2  4 (x  2) 3 2 1 2

1

y  4x  4

3

3 x 4 3 x 4

1

y  1  1  4x  4  1

The line parallel to y  4x  1 has the same 3 slope, 4.

3

1

y  1  4x  4

3x  4 4 3 x1 4

3

8

1

y  2  2  3x  3  2

y22

x  8 3 1 8 x3 3

The line parallel to y  3x  3 has the same 1 slope, 3. The y-intercept is 6. Use the slopeintercept form. y  mx  b

y  y  m(x  x ) 1 1 2 y  2  3 (x  2)

y



1

y  1  1  3x  1  1

2

1

y  3x  2

226

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31. The slope of the given line is 8. So, the slope of the line perpendicular to this line is the opposite 1 reciprocal of 8, or 8. Use the point-slope form. y  y1  m(x  x1 )

36. Find the slope of the given line. 3y  x  3 3y  x  x  3  x 3y  x  3 3y 3

1

y  5  8 (x  0)

y

1

y  5  8x 1

y  8x  5 1

32. The slope of the given line is 2. So, the slope of the line perpendicular to this line is the opposite 1 reciprocal of 2, or 2. Use the point-slope form. y  y1  m(x  x1 ) y  (3)  2(x  1) y  3  2x  2 y  3  3  2x  2  3 y  2x  1 2 33. The slope of the given line is 3. So, the slope of the line perpendicular to this line is the opposite 2 3 reciprocal of 3, or 2. Use the point-slope form. y  y1  m(x  x1 ) 3

y  7  2 (x  4) 3

y  7  2x  6 3

y  7  7  2x  6  7

1

y  5x  (1)

3

y  2x  13

1

y  5x  1

34. Find the slope of the given line. 3x  8y  4 3x  8y  3x  4  3x 8y  3x  4 

y

38. Find the slope of the given line. 5x  7  3y 3y  5x  7 3y 3

 3x  4 8 3 1 8x  2

y



y

5x  7 3 5 7 x 3 3

The slope of the line perpendicular to this line is 5 3 the opposite reciprocal of 3, or 5. Use the pointslope form. y  y  m(x  x ) 1 1

The slope of the line perpendicular to this line is 3 8 the opposite reciprocal of 8, or 3. The point (0, 4) lies on the y-axis. Thus, the y-intercept is 4. Use the slope-intercept form. y  mx  b 8 x 3

 x  3 3 1 3x  1

The slope of the line perpendicular to this line is 1 the opposite reciprocal of 3, or 3. Use the pointslope form. y  y1  m(x  x1 ) y  (1)  3(x  6) y  1  3x  18 y  1  1  3x  18  1 y  3x  19 37. Find the slope of the given line. 5x  y  3 5x  y  5x  3  5x y  5x  3 (1)(y)  (1) (5x  3) y  5x  3 The slope of the line perpendicular to this line is 1 the opposite reciprocal of 5, or 5. The point (0, 1) lies on the y-axis. Thus, the y-intercept is 1. Use the slope-intercept form. y  mx  b

1

y  5  5  8x  5

8y 8



3

y  (2)  5 (x  8)

4

3

y  2  5x 

35. Find the slope of the given line. 2x  5y  3 2x  5y  2x  3  2x 5y  2x  3 5y 5



y

3

y  2  2  5x  y

3 5x



24 5 24 5 14 5

2

39. Find the slope of the given line. 3x  7  2x 3x  7  2x  2x  2x x70 x7707 x  7 This line is vertical. A line perpendicular to a vertical line is horizontal and has slope 0. Use the point-slope form. y  y  m(x  x ) 1 1 y  (3)  0(x  3) y30 y3303 y  3

2x  3 5 2 3 x 5 5

The slope of the line perpendicular to this line is 2 5 the opposite reciprocal of 5, or 2. Use the pointslope form. y  y  m(x  x ) 1 1 5

y  7  2 [x  (2) ] 5

y  7  2 (x  2) 5

y  7  2x  5 5

y  7  7  2x  5  7 5

y  2x  2

227

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46. Write each equation in slope-intercept form. y  ax  5 2y  (a  4)x  1

40. Find the slope of the given line. 3x  6y  2 3x  6y  3x  2  3x 6y  3x  2 6y 6



y

2y 2

y

3x  2 6 1 1 2x  3

2  10 3  9

a (2)a 



or 2

2x 2

4x The line perpendicular to y  2x  8 and passing 1 through its x-intercept has slope 2 and passes through (4, 0). Use the point-slope form. 1

2x  14 3 2

y  3x 

Page 297

y

y  7  5[x  (4) ] y  7  5(x  4) y  y1  m(x  x1 ) 52.

2x  2 3 2 2 x 3 3

1

y  (3)  2 [x  (1) ]

2 , 3

The slopes are both so the two graphs are parallel. 44. The slope of y  5x is 5. The slope of y  5x  18 is 5. The slopes are not the same, and 5(5)  1. Therefore, the two graphs are neither parallel nor perpendicular. 45. Find the slope of each segment. 3  (3) (1) 1  1 3  (3)

Slope of AC : m  1 

or 3

Slope of BD : m 

or 3

1

y  3  2 (x  1)

1

The diagonals AC and BD are perpendicular 1 because 3 (3)  1.

Chapter 5

Maintain Your Skills

50. y  y1  m(x  x1 ) y  5  2(x  3) 51. y  y1  m(x  x1 )

3y  2x  2 

3x  24 4 3 4x  6

49. C; The y-intercept of y  3x  4 is 4. The y-intercept of y  3x  2 is 2. The slope of both lines is 3. Therefore, to graph y  3x  2 change the y-intercept from 4 to 2.

2x  3y  2x  2  2x 3y 3



y

1

14 3

2

The slope of the line perpendicular to this line is 3 4 the opposite reciprocal of 4 or 3.

42. Write each equation in slope-intercept form. y  2x  11 y  2x  23 y  2x  2x  23  2x y  2x  23 The slopes are both 2, so the two graphs are parallel. 43. Write each equation in slope-intercept form. 3y  2x  14 2x  3y  2 

1

4y 4

y  0  2 (x  4) or y  2x  2

3y 3

a  4 2 a  4 (2) 2

2a  a  4 2a  a  a  4  a a4 47. If two equations have the same slope, then the lines are parallel. Answers should include the following. • Sample answer: y  5x  1; The graphs have the same slope. 1 • Sample answer: y  5x; The slopes are negative reciprocals of each other. 48. D; Write the given equation in slope-intercept form. 3x  4y  24 3x  4y  3x  24  3x 4y  3x  24

Use the point-slope form to write its equation. y  10  2(x  9) or y  2x  8 To find the x-intercept, let y  0. 0  2x  8 0  8  2x  8  8 8  2x 8 2

(a  4)x  1 2 a  4 1 x 2 2

The lines are parallel if the slopes are equal.

The slope of the line perpendicular to this line is 1 the opposite reciprocal of 2, or 2. The y-intercept is 2. Use the slope-intercept form. y  mx  b y  2x  (2) y  2x  2 41. Find the slope of the line through (9, 10) and (3, 2). m



228

PQ249-6481F-05[217-243] 7/31/02 10:05 PM Page 229

58. Find the slope of the line.

53. Write an equation of the line that passes through (10, 3.19) and (15, 4.29). Find the slope. m m

1  5

m  8  5 or 2 Find the y-intercept. y  mx  b 5  2(5)  b 5  10  b 5  10  10  b  10 15  b Write the slope-intercept form. y  mx  b y  2x  15 59. Find the slope of the line.

4.29  3.19 15  10 1.1 5

m  0.22 Find the y-intercept. y  mx  b 3.19  0.22(10)  b 3.19  2.2  b 3.19  2.2  2.2  b  2.2 0.99  b Write the slope-intercept form. y  mx  b y  0.22x  0.99 Therefore, using m for the number of minutes and C for the total cost, the equation is C  0.22m  0.99. 54. y  0.22x  0.99 y  0.22(12)  0.99 y  3.63 The cost of a 12-minute call is $3.63. 55. Find the slope of the line. m

3  (1) 3  5

9  9

m  4  6 or 0 Find the y-intercept. y  mx  b 9  0(6)  b 9b Write the slope-intercept form. y  mx  b y  0x  9 y9 60. Find the slope of the line. 2  4 (6)

m2

1

or 2

or

3 4

Find the y-intercept. y  mx  b

Find the y-intercept. y  mx  b

3

4  4 (6)  b

1  2 (5)  b

1

42b

5

422b2

9

1  2  b 5

5

1  2  2  b  3 2

9

5 2

9

1

2  b

b

Write the slope-intercept form. y  mx  b

Write the slope-intercept form. y  mx  b 1

9

3

1 12

y  4x  2

3

y  2x  2

3

1

y  4x  2

56. Find the slope of the line. 0  2

1

m  8  0 or 4

Page 297

The point (0, 2) lies on the y-axis. The y-intercept is 2. Write the slope-intercept form. y  mx  b 1

y  4x  2 57. Find the slope of the line. m

4  1 3  2

Practice Quiz 2

1. Use the slope-intercept form. y  mx  b y  4x  (3) y  4x  3 2. Find the y-intercept. y  mx  b 3  2(1)  b 3  2  b 3  2  2  b  2 5  b Write the slope-intercept form. y  mx  b y  2x  (5) y  2x  5

or 5

Find the y-intercept. y  mx  b 1  5(2)  b 1  10  b 1  10  10  b  10 11  b Write the slope-intercept form. y  mx  b y  5x  11

229

Chapter 5

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2a. Sample answer:

3. Find the slope. m

3  (2) 1  (1)

y

5 2

or

Find the y-intercept. y  mx  b 5

3  2 (1)  b 5

32b 5

5

1 2

b

2b. Sample answer: y

Write the slope-intercept form. y  mx  b 5

x

O

5

322b2

1

y  2x  2 4. The line parallel to y  2x  2 has the same slope, 2. Use the point-slope form. y  y1  m(x  x1 ) y  3  2[x  (2) ] y  3  2(x  2) y  3  2x  4 y  3  3  2x  4  3 y  2x  7 5. Standard form:

x

O

2c. Sample answer: y

1

y  4  2 (x  3) 2(y  4)  2

112 2 (x  3)

2y  8  1(x  3) 2y  8  x  3 2y  8  8  x  3  8 2y  x  11 2y  x  x  11  x x  2y  11 (1)(x  2y)  (1)11 x  2y  11 Slope-intercept form:

3. Linear extrapolation predicts values outside the range of the data. Linear interpolation predicts values inside the range of the data. 4. The graph shows a positive correlation. As the number of hours of study increases, the test score increases. 5. The graph shows a negative correlation. As the number of hours of TV increases, the number of hours of exercise decreases. 6. Let the independent variable x be the air temperature, and let the dependent variable y be the body temperature. The scatter plot seems to indicate that as the air temperature increases, the body temperature increases. There is a positive correlation between the two variables.

1

y  4  2 (x  3) 1

3

1

3

1

11 2

y  4  2x  2 y  4  4  2x  2  4 y  2x 

Statistics: Scatter Plots and Lines of Fit

Page 299

Body Temperature (°C)

5-7

Algebra Activity

1. See students’ work. 3. See students’ work.

Pages 301–302

2. See students’ work.

Check for Understanding

40 30 20 10 0

1. If the data points form a linear pattern such that y increases as x increases, there is a positive correlation. If the linear pattern shows that y decreases as x increases, there is a negative correlation.

Chapter 5

x

O

230

10 20 30 40 Air Temperature (°C)

15. Use the equation y  2.15x  4285.45. y  2.15x  4285.45 y  2.15(2002)  4285.45 y  18.85 The number of bushels of apples in storage in 2002 will be about 18.85 million. 16. Find the slope.

Body Temperature (˚C)

7. No one line will pass through all of the data points. Draw a line that passes close to the points. 35 30 25 20 15

y2  y1

mx

10 0

m

15 20 25 30 35 Air Temperature (˚C)

m

Use the point-slope form. y  y1  m(x  x1 )

8. The line of fit passes through the data points (26.2, 25.6) and (31.2, 31.0). Step 1: Find the slope.

y  6000  1200(x  5) y  6000  1200x  6000 y  6000  6000  1200x  6000  6000 y  1200x  12,000 17. Use the equation y  1200x  12,000. y  1200x  12,000 y  1200(7)  12,000 y  3600 The price of a 7-year-old car should be about $3600. 18.

y2  y1

mx

2

m m

 x1

31.0  25.6 31.2  26.2 5.4 or 1.08 5

1 2 3 4 5 6 7 8 Number of Carbon Atoms

19. Draw a line that passes close to the points. 120 90 60 30 0 ⫺30 ⫺60 ⫺90 ⫺120

Practice and Apply

10. The graph shows a negative correlation. As the year increases, the percentage of forms returned decreases. 11. The graph shows no correlation. 12. The graph shows a positive correlation. As the year increases, the number of electronic tax returns increases. 13. The graph shows a positive correlation. As the amount of sugar increases, the number of calories increases. 14. Find the slope.

1 2 3 4 5 6 7 8 Number of Carbon Atoms

20. We use the data points (3, 42) and (6, 69). Find the slope. y2  y1

y2  y1

mx

2

mx

 x1

2

m

12.4  8.1

m  1999  1997 4.3 2

120 90 60 30 0 –30 –60 –90 –120

Boiling Point (˚C)

Pages 302–305

Boiling Point (°C)

Step 2: Use m  1.08 and the point-slope form to write the equation. You can use either data point. We chose (26.2, 25.6). y  y1  m(x  x1 ) y  25.6  1.08(x  26.2) y  25.6  1.08x  28.296 y  25.6  25.6  1.08x  28.296  25.6 y  1.08x  2.696 9. Use the equation y  1.08x  2.696, where x is the air temperature and y is the body temperature. y  1.08x  2.696 y  1.08(40.2)  2.696 y  40.72 The body temperature would be about 40.1F.

m

 x1

2

6000  9600 5  2 3600 or 1200 3

m

or 2.15

 x1

69  (42) 6  3 111 or 37 3

Use the point-slope form. y  y  m(x  x ) 1 1 y  69  37(x  6) y  69  37x  222 y  69  69  37x  222  69 y  37x  153

Use the point-slope form. y  y1  m(x  x1 ) y  8.1  2.15(x  1997) y  8.1  2.15x  4293.55 y  8.1  8.1  2.15x  4293.55  8.1 y  2.15x  4285.45

231

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27. Use the equation y  0.5125x  4.5. The year 2005 is 25 years after 1980. y  0.5125x  4.5 y  0.5125(25)  4.5 y  17.3125 The federal spending on space and other technologies in 2005 will be about $17.3 billion. 28. Sample answer: The government’s prediction is $17.3  $14.3 or $3 billion less. 29. 800

251.4 37



Acres Burned (thousands)

21. Use the equation y  37x  153. y  37x  153 y  37(1)  153 y  116 The boiling point for methane should be about 116 C. 22. Use the equation y  37x  153. y  37x  153 y  37(5)  153 y  32 The boiling point for pentane should be about 32 C. 23. Use the equation y  37x  153. y  37x  153 98.4  37x  153 98.4  153  37x  153  153 251.4  37x

Acres Burned (thousands)

Spending (billions of dollars) Spending (billions of dollars)

0

10 20 30 Rainfall (in.)

800 600 400 200 0

10 20 30 Rainfall (in.)

31. We use the points (12.7, 340) and (17.5, 194). Find the slope. y2  y1

mx

’79 ’80’85’90’95’00 Year

2

m m

12 10 8 6 4 0

‘80 ‘85 ‘90 ‘95 ‘00 Year

26. We use the points (0, 4.5) and (16, 12.7). Find the slope. y2  y1

mx

2

 x1

12.7  4.5 16  0 8.2 or 0.5125 16

The y-intercept is 4.5. Use the slope-intercept form. y  mx  b y  0.5125x  4.5

Chapter 5

 x1

194  340 17.5  12.7 146 4.8

m  30.4 Use the point-slope form. y  y  m(x  x ) 1 1 y  194  30.4(x  17.5) y  194  30.4x  532 y  194  194  30.4x  532  194 y  30.4x  726 32. Use the equation y  30.4x  726. y  30.4x  726 y  30.4(8.25)  726 y  475.2 The number of acres burned in 2000 should be about 475 thousand acres. 33. The data point lies beyond the main grouping of data points. It can be ignored as an extreme value.

14

m

200

30. Draw a line that passes close to the points.

25. Draw a line that passes close to the points.

m

400

37x 37

7x The number of carbon atoms in heptane should be about 7. 24. The scatter plot seems to indicate that as the year increases, federal spending increases. There is a positive correlation between the two variables. 18 16 14 12 10 8 6 4 2 0

600

232

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34.

48.

Incorrect Answers

20 16

49.

12 8

50.

4 0

4 8 12 16 Correct Answers

20

35. See students’ work. 36. See students’ work. 37. You can visualize a line to determine whether the data has a positive or negative correlation. Answers should include the following. y Height

51.

x

O Age

38. 39.

40. 41. 43.

• Write a linear equation for the line of fit. Then substitute the person’s height and solve for the corresponding age. D; Graph D shows a negative correlation. As the value of x increases, the value of y decreases. B; The equation y  2x  3 represents a line that passes through the points (1, 5), (2, 7), and (4, 11), and lies close to the point (3, 7). Each of the other three equations represents a line that does not pass as close to the four points. Sample answer: Cities with greater north latitudes have lower January temperatures. See students’ work. 42. See students’ work. See students’ work. 44. See students’ work.

Page 305

52.

y  y1  m(x  x1 ) y  (2)  3(x  1) y  2  3(x  1) y  y1  m(x  x1 ) y  (3)  1[ x  (3) ] y3x3 To find the x-intercept, let 3x  4y  12 3x  4(0)  12 3x  12 x4 To find the y-intercept, let 3x  4y  12 3(0)  4y  12 4y  12 y3 To find the x-intercept, let 2x  5y  8 2x  5(0)  8 2x  8 x4 To find the y-intercept, let 2x  5y  8 2(0)  5y  8 5y  8 y  1.6 To find the x-intercept, let y  3x  6 0  3x  6 3x  6 x  2 To find the y-intercept, let y  3x  6 y  3(0)  6 y6

53. 12

1

r  7 4 r  7 4



y  0.

x  0.

y  0.

x  0.

y  0.

x  0.

r  2 6

2  121r 6 2 2

3(r  7)  2(r  2) 3r  21  2r  4 3r  21  2r  2r  4  2r 5r  21  4 5r  21  21  4  21 5r  25

Maintain Your Skills

45. The line parallel to y  4x  5 has the same slope, 4. Use the point-slope form. y  y1  m(x  x1 ) y  5  4[x  (2) ] y  5  4(x  2) y  5  4x  8 y  5  5  4x  8  5 y  4x  3 46. The slope of the given line is 2. So, the slope of the line perpendicular to this line is the opposite 1 reciprocal of 2, or 2. The y-intercept is 0. Use the slope-intercept form. y  mx  b

5r 5



25 5

r  5 54. 3

1

n  (4) 3 n  (4) 3

7

2  3 (7)

n  (4)  21 n  (4)  (4)  21  (4) n  25

1

y  2x  0 1

y  2x 47. y  y  m(x  x ) 1 1 y  3  2[x  (2) ] y  3  2(x  2)

233

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55. 35

1

2x  1 5 2x  1 5



4x  5 7 4x  5 35 7

2 1

4. • Enter the number of votes cast in 1996 in L1 and the number of votes cast in 2000 in L2.

2

STAT ENTER 14,447 ENTER ENTER 19,458 28,674 ENTER 31,658 ENTER 32,739 ENTER 46,543 ENTER 49,186 ENTER

KEYSTROKES:

7(2x  1)  5(4x  5) 14x  7  20x  25 14x  7  20x  20x  25  20x 6x  7  25 6x  7  7  25  7  6x  18 6x 6



69,208 ENTER 103,429 ENTER 103,574

ENTER

18 6

30,921 ENTER 38,545 ENTER 38,626 ENTER

x3

Page 307

52,457 ENTER 53,907 ENTER 80,787 ENTER 126,911 ENTER 123,466 • Find the regression equation by selecting LinReg (ax  b) on the STAT CALC menu.

Graphing Calculator Investigation (Follow-Up of Lesson 5-7)

4 ENTER The regression equation is about y  1.23x  3414.80. • The data are already in Lists 1 and 2. Find the median-fit equation by using Med-Med on the STAT CALC menu. KEYSTROKES:

1. • Enter the number of years since 1985 in L1 and the number of bald eagle pairs in L2.

STAT ENTER 3 ENTER 5 ENTER 7 ENTER 9 ENTER 11 ENTER 14 ENTER 2500 ENTER 3000 ENTER 3700

KEYSTROKES:

• Find the regression equation by selecting LinReg(ax  b) on the STAT CALC menu. 4 ENTER The regression equation is about y  309.48x  1555.88. • The data are already in Lists 1 and 2. Find the median-fit equation by using Med-Med on the STAT CALC menu.

STAT

STAT

According to the median-fit equation, the number of votes in that county in 2000 was about 23,882. The estimates both show the number of votes increasing from 1996 to 2000. But both are far from the actual number.

Chapter 5 Study Guide and Review Page 308

Vocabulary and Concept Check

1. direct variation 3. parallel 5. slope-intercept

Pages 308–312

2. rise; run 4. point-slope 6. y-intercept

Lesson-by-Lesson Review

7. Let (1, 3)  (x1, y1) and (2, 6)  (x2, y2). y2  y1 mx x 2

m

234

1

6  3  1 9 or 3 3

m  2

Chapter 5

STAT

According to the regression equation, the number of votes in that county in 2000 was about 24,753. Graph the median-fit equation. • Find y when x  22,839 using value on the CALC menu. CALC 1 22,839 ENTER KEYSTROKES: 2nd

3 ENTER The median-fit equation is about y  311.76x  1537.25. 2. The correlation coefficient of y  309.48x  1555.88 is 0.9970385087. The data are very nearly linear. 3. Graph the regression equation. • Find y when x  13 using value on the CALC menu. CALC 1 13 ENTER KEYSTROKES: 2nd According to the regression equation, the number of bald eagle pairs was about 5579. Graph the median-fit equation. • Find y when x  13 using value on the CALC menu. CALC 1 13 ENTER KEYSTROKES: 2nd According to the median-fit equation, the number of bald eagle pairs was about 5590. Both predictions are close to the prediction in Example 4. But these predictions show fewer pairs of eagles in 1998. KEYSTROKES:

STAT

3 ENTER The median-fit equation is about y  1.24x  4454.74. 5. Graph the regression equation. • Find y when x  22,839 using value on the CALC menu. CALC 1 22,839 ENTER KEYSTROKES: 2nd KEYSTROKES:

ENTER 4500 ENTER 5000 ENTER 5800 ENTER

KEYSTROKES:

16,284 ENTER 19,281 ENTER

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8. Let (0, 5)  (x1, y1) and (6, 2)  (x2, y2). y2  y1 mx x 2

14. Write the slope as a ratio. 4 

1

Graph (0, 0). From the point (0, 0), move down 4 units and right 1 unit. Draw a dot. Draw a line containing the points.

2  5

m60 m

3 6

4 1

1

or 2

9. Let (6, 4)  (x1, y1) and (6, 2)  (x2, y2).

y

y2  y1

mx

2

m m

 x1

2  4 6  (6) 6 0

O

y  4x

x

Since division by zero is undefined, the slope is undefined. 10. Let (8, 3)  (x1, y1) and (2, 3)  (x2, y2). y2  y1

mx

2

m m

15. Write the slope as a ratio.

 x1

1 3

3  (3) 2  8 0 or 0 10

Graph (0, 0). From the point (0, 0), move up 1 unit and right 3 units. Draw a dot. Draw a line containing the points.

11. Let (2.9, 4.7)  (x1, y1) and (0.5, 1.1)  (x2, y2). y2  y1 mx x 2

y

1

y  13 x

1.1  4.7

m  0.5  2.9 3.6

m  2.4 or 1.5 12. Let

112, 12  (x1, y1) and 11, 23 2  (x2, y2).

x

O

y2  y1

mx

2

m m

2 3

 x1

1 1

1  2 1 3 3 2

16. Write the slope as a ratio. 1

4 

2

or 9

1 4

Graph (0, 0). From the point (0, 0), move down 1 unit and right 4 units. Draw a dot. Draw a line containing the points.

13. Write the slope as a ratio. 2

21

y

Graph (0, 0). From the point (0, 0), move up 2 units and right 1 unit. Draw a dot. Draw a line containing the points. y

y   14 x

y  2x O

x

O

x

17. Write the slope as a ratio. 3 2

Graph (0, 0). From the point (0, 0), move up 3 units and right 2 units. Draw a dot. Draw a line containing the points. y y  32 x O

235

x

Chapter 5

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31. The y-intercept is 1. So, graph (0, 1). The slope is 2 2 or 1. From (0, 1) move up 2 units and right 1 unit. Draw a dot. Draw a line containing the points.

18. Write the slope as a ratio. 4 3



4 3

Graph (0, 0). From the point (0, 0), move down 4 units and right 3 units. Draw a dot. Draw a line containing the points.

y

y y  2x  1 x

O

x

O

y   43 x

19. Find the value of k. y  kx 6  k(9) 6 9 2 3



k(9) 9

15 2 15 2

k

Therefore, y 

2 3x.

21. Find the value of k. y  kx 4  k(4)



k(2) 2

y

k

Therefore, y 

22. Find the value of k. y  kx 6  k(18)

1  k Therefore, y  x.

Therefore, y  3x.



k(4) 4

23. Find the value of k. y  kx 10  k(5) 10 5



k(5) 5

2  k

y  x  5

15 x. 2

6 18 1 3

4 4

32. The y-intercept is 5. So, graph (0, 5). The slope is 1 1 or 1 . From (0, 5) move down 1 unit and right 1 unit. Draw a dot. Draw a line containing the points.

20. Find the value of k. y  kx 15  k(2)



k

33. The y-intercept is 3. So, graph (0, 3). The slope is 1 . From (0, 3) move up 1 unit and right 2 units. 2 Draw a dot. Draw a line containing the points.

1

24. Find the value of k. y  kx 7  k(14) 7 14 1 2



x

O

k(18) 18

y

k(14) 14

y  1 x 3 2

k 1

25.

26.

27.

28.

Therefore, y  2x. Therefore, y  2x. Replace m with 3 and b with 2. y  mx  b y  3x  2 Replace m with 1 and b with 3. y  mx  b y  1x  (3) yx3 Replace m with 0 and b with 4. y  mx  b y  0x  4 y4 1 Replace m with 3 and b with 2. y  mx  b

34. The y-intercept is 1. So, graph (0, 1). The slope 4 4 is 3 or 3. From (0, 1) move up 4 units and left 3 units. Draw a dot. Draw a line containing the points. y

O

1

y   4x 1

y3x2

3

29. Replace m with 0.5 and b with 0.3. y  mx  b y  0.5x  (0.3) y  0.5x  0.3 30. Replace m with 1.3 and b with 0.4. y  mx  b y  1.3x  0.4 Chapter 5

x

O

236

x

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39. Find the y-intercept. y  mx  b

35. Solve for y. 5x  3y  3 5x  3y  5x  3  5x 3y  5x  3 3y 3



y

1

6  2(1)  b 1

62b

5x  3 3 5 x 1 3

1

1

1

11 2

b

622b2

The y-intercept is 1. So, graph (0, 1). The slope is 5 . From (0, 1) move up 5 units and right 3 units. 3 Draw a dot. Draw a line containing the points.

Write the slope-intercept form. y  mx  b 1

y  2x 

y

11 2

40. Find the y-intercept. y  mx  b 3

3  5(4)  b 5x  3y  3

12

3   5  b

x

O

12 5 3 5

3 



The y-intercept 3

1 2

3

3

y  5x  5

41. Find the slope. y2  y1

1 2. The slope is

So, graph 0,

mx

9 2

2

m

3 or 1 . From move down 3 units and right 1 unit. Draw a dot. Draw a line containing the points.

m

 x1

12  2 1  (4) 10 or 2 5

Find the y-intercept. y  mx  b 12  2(1)  b 12  2  b 12  2  2  b  2 10  b Write the slope-intercept form. y  mx  b y  2x  10 42. Find the slope.

y 6x  2 y  9

O

1 32

3

9

9 is 2. 9 0, 2

b

y  5x  5

6x  9 2

y  3x  2

12 5

Write the slope-intercept form. y  mx  b

36. Solve for y. 6x  2y  9 6x  2y  6x  9  6x 2y  6x  9 2y 2

12

 5  b 

x

y2  y1

37. Find the y-intercept. y  mx  b 3  1(3)  b 3  3  b 3  3  3  b  3 6b Write the slope-intercept form. y  mx  b y  1x  6 yx6 38. The point (0, 6) lies on the y-axis. The y-intercept is 6. y  mx  b y  2x  6

mx

2

 x1

5  0

m45 5

m  1 or 5 Find the y-intercept. y  mx  b 0  5(5)  b 0  25  b 0  25  25  b  25 25  b Write the slope-intercept form. y  mx  b y  5x  25

237

Chapter 5

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54. The line parallel to y  3x  2 has the same slope, 3. y  y1  m(x  x1 ) y  6  3(x  4) y  6  3x  12 y  6  6  3x  12  6 y  3x  6 55. The line parallel to y  2x  4 has the same slope, 2. y  y1  m(x  x1 2 y  (62  2(x  62 y  6  2x  12 y  6  6  2x  12  6 y  2x  6 56. The line parallel to y  6x  1 has the same slope, 6. y  y  m(x  x ) 1 1 y  2  6(x  1) y  2  6x  6 y  2  2  6x  6  2 y  6x  8

43. Find the y-intercept. y  mx  b 1  0(8)  b 1  b Write the slope-intercept form. y  mx  b y  0x  (1) y  1 44. Find the y intercept. y  mx  b 6  0(4)  b 6b Write the slope-intercept form. y  mx  b y  0x  6 y6 Therefore, the equation is y  6. 45. y  y  m(x  x ) 46. y  y  m(x  x ) 1 1 1 1 y  6  5(x  4) y  4  2[ x  (1) ] y  4  2(x  1) 47. y  y1  m(x  x1 ) 48. y  y1  m(x  x1 ) 1

y  (4)  2 (x  1)

1

y  4  2 (x  1)

y  (3)  2 (x  5) y  3  2 (x  5) 49.

5

57. The line parallel to y  12 x  2 has the same 5 slope, 12. The point (0, 4) lies on the y-axis. The y-intercept is 4. y  mx  b

5 5

y  y1  m(x  x1 ) 1 y  (2)  3 x  4

1 2 1 y  2  31x  4 2

5

y  12 x  4 58. Solve the equation for y. 4x  y  7 4x  y  4x  7  4x y  4x  7 (1)(y)  (1)(4x  7) y  4x  7 The line parallel to y  4x  7 has the same slope, 4. y  y1  m(x  x1 ) y  (1)  4(x  2) y  1  4x  8 y  1  1  4x  8  1 y  4x  9 59. Solve the equation for y. 3x  9y  1 3x  9y  3x  1  3x 9y  3x  1

y  y1  m(x  x1 ) y  (2)  0(x  4) y20 51. y  1  2(x  1) y  1  2x  2 y  1  1  2x  2  1 y  2x  3 y  2x  2x  3  2x 2x  y  3 (1)(2x  y)  (1)3 2x  y  3 50.

52.

1

y  6  3(x  9) 3( y  6)  3

113 2(x  9)

3y  18  x  9 3y  18  18  x  9  18 3y  x  27 3y  x  x  27  x x  3y  27 (1)(x  3y)  (1)(27) x  3y  27 53. y  4  1.5(x  4) 2( y  4)  2(1.5)(x  4) 2y  8  3(x  4) 2y  8  3x  12 2y  8  8  3x  12  8 2y  3x  20 2y  3x  3x  20  3x 3x  2y  20 (1)(3x  2y)  (1)(20) 3x  2y  20 Chapter 5

9y 9



y

3x  1 9 1 1 3x  9 1

1

The line parallel to y  3x  9 has the same 1 slope, 3. y  y1  m(x  x1 ) 1 y  0  3(x  3) 1

y  3x  1

238

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65. Find the slope of the given line. 5y  x  1

60. The slope of the given line is 4. So, the slope of the line perpendicular to this line is the opposite 1 reciprocal of 4 or 4. y  y1  m(x  x1 )

5y 5

y

1

y  3  4(x  1) 1

1

1

1

1

13 4

The slope of the line perpendicular to this line is 1 the opposite reciprocal of 5 or 5. y  y1  m(x  x1 ) y  (5)  5(x  2) y  5  5x  10 y  5  5  5x  10  5 y  5x  15

y  3  4x  4 y  3  3  4x  4  3 y  4x 

61. The slope of the given line is 2. So, the slope of the line perpendicular to this line is the opposite 1 reciprocal of 2 or 2. The point (0,3) lies on the y-axis. The y-intercept is 3. y  mx  b

Weight (long tons)

66.

1

y  2x  (3) 1

y  2x  3 2

62. The slope of the given line is 5. So, the slope of the line perpendicular to this line is the opposite 2 5 reciprocal of 5 or 2. y  y  m(x  x ) 1 1

Weight (long tons)

5

y  5  2x  5 5

y  5  5  2x  5  5 5

y  2x y  2.5x 63. Find the slope of the given line. 2x  7y  1 2x  7y  2x  1  2x 7y  2x  1 

y

60 50 40 30 20 10 10 0 0 30 35 40 45 50 55 60 Length (ft)

67. Draw a line that passes close to the points.

5

y  (5)  2(x  2)

7y 7

x  1 5 1 1 5x  5



60 50 40 30 20 10

68. We use points (40, 25) and (52, 45). Find the slope.

2x  1 7 2 1 x7 7

y2  y1

mx

m

Use the point-slope form. y  y  m(x  x )

7

y  0  2 [x  (4) ]

1

7

y  2(x  4) 7

5

y  25  3x 

64. Find the slope of the given line. 8x  3y  7 8x  3y  8x  7  8x 3y  8x  7 y

5

y  25  25  3x  y

5

5

y  3x  5

3

3

13 2

125 3

1

y  383

3

3

125 . 3

125 3

y  3(482 

y  5  8(x  4) 3



 25

69. Use the equation y  3x 

The slope of the line perpendicular to this line is 8 3 the opposite reciprocal of 3 or 8. y  y  m(x  x ) 1 1 3

5 x 3

200 3 200 3 125 3

Using W for weight and / for length, the equation is 5 125 W  3x  3 .

8x  7 3 8 7 x3 3

The weight of a 48-foot humpback whale is about 1 383 long tons.

y  5  8x  2 y  5  5  8x  2  5 y  8x 

1

5

y  25  3(x  40)

y  2x  14



 x1

2

45  25  40 20 5 or 3 12

m  52

The slope of the line perpendicular to this line is 2 7 the opposite reciprocal of 7 or 2. y  y  m(x  x ) 1 1

3y 3

35 40 45 50 55 60 Length (ft)

0

239

Chapter 5

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5

70. Use the equation y  3x  5

y  3x 

125 . 3

9. The y-intercept is 3. So, graph (0, 3). The slope is 2 2 or 1. From (0, 3) move up 2 units and right 1 unit. Draw a dot. Draw a line containing the points.

125 3

5

y  3(12) 

125 3

2

y  213 The weight of a 12-foot humpback whale is about 2 213 long tons. The prediction is not accurate. A negative weight is not reasonable.

y y  2x  3

x

O

Chapter 5 Practice Test Page 313 1. Vertical lines have no slope. 2. Sample answer:

10. Solve for y. 2x  3y  9 2x  3y  2x  9  2x 3y  2x  9

y

3y 3



y

2x  9 3 2 3x  3 2

The y-intercept of y  3x  3 is 3. So, graph 2 2 (0, 3). The slope is 3 or 3 . From (0, 3) move down 2 units and right 3 units. Draw a dot. Draw a line containing the points.

x

O

3. The slope represents the rate of change. 4. Let (5, 8)  (x1, y1) and (3, 7)  (x2, y2).

y

y2  y1

mx

2

m m

 x1

2x  3y  9

78 3  5 1 1 or 8 8

O

x

5. Let (5, 2)  (x1, y1) and (3, 2)  (x2, y2). y2  y1

mx

2

m m

 x1

2  (2) 35 0 or 0 2

11. The temperature is expected to fall 2 each hour, so the rate of change is 2 each hour. At midnight the temperature is 16 F.

6. Let (6, 3)  (x1, y1) and (6, 4)  (x2, y2). y2  y1 mx x 2

m m

number of hours temperature rate of Temperature equals change times after42444 midnight plus at44244 midnight. 14 44244 43 1 424 3 14243 123 1444 43 123 14 43

1

6 9 2 3

8. The y-intercept is 1. So, graph (0, 1). The slope 3 is 3 or 1. From (0, 1) move up 3 units and right 1 unit. Draw a dot. Draw a line containing the points.



k(92





h

12 4

9

16

Therefore, y 

8 8

y  3x  1



2 x. 3

k(8) 8

1  k Therefore, y  x.

x

240



k(4) 4

3  k

k

14. Find the value of k. y  kx 8  k(8)

y

Chapter 5

2

Therefore, the equation is T  2h + 16. 12. Find the value of k. 13. Find the value of k. y  kx y  kx 6  k(9) 12  k(4)

Since division by zero is undefined, the slope is undefined. 7. The rate is $9.95 per month. Therefore, the direct variation equation is C  0.95m.

O



T

4  (3) 66 7 0

Therefore, y  3x. 15. y  mx  b y  4x  3

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16. Find the slope. m

3  (5) 8  (2)

or

21. The scatter plot seems to indicate that as the number of dog years increases, the number of human years increases. There is a positive correlation between the two variables.

1 5

Find the y-intercept. y  mx  b

y

1

3  5(8)  b 8

3  5  b 8

Human Years

3  5  b 23

5  b Write the slope-intercept form. y  mx  b 1

1

23

y  5x   5 1 x 5

23 5

2

 y 17. Solve the given equation for y. 3x  7y  4 3x  7y  3x  4  3x 7y  3x  4 

y

3x  4 7 3 4 7x  7 3

4

The line parallel to y  7x  7 has the same 3 slope, 7. y  y1  m(x  x1 ) 3 y  (2)  7(x  5) 3

y  2  7x  3

y  2  2  7x  3

y  7x 

15 7 15 7 1 7

2



y

4 6 Dog Years

x

8

y 50 45 40 35 30 25 20 15 10 5 0

2

x

4 6 Dog Years

8

23. We use the points (1, 15) and (7, 47). Find the slope.

18. The line has slope 0. y  y1  m(x  x1 ) y  (8)  0(x  5) y80 y8808 y  8 19. Find the slope of the given line. 5x  3y  9 5x  3y  5x  9  5x  3y  5x  9 3y 3

2

22. Draw a line that passes close to the points.

Human Years

7y 7

50 45 40 35 30 25 20 15 10 5 0 0

m

47  15 7  1

or

16 3

Use the point-slope form to write the equation. y  y  m(x  x ) 1

1

16 (x  1) 3 16 16 y  15  3 x  3 16 16 y  15  15  3 x  3  15 16 29 y 3x 3 16 29 Use the equation y  3 x  3 . 16 29 y 3x 3 16 29 y  3 (13)  3

y  15 

24.

5x  9 3 5 x 3 3

The slope of the line perpendicular to this line is 5 3 the opposite reciprocal of 3 or 5. The y-intercept is 0. y  mx  b

y  79 The number of human years comparable to 13 dog years is 79. 4

25. B; The y-intercept of y  3x  3 is 3. Therefore, the line represented by this equation passes through (0, 3) not (0, 4).

3

y  5x  0 3

y  5x 20. y  y1  m(x  x1 ) y  3  2[ x  (4) ] y  3  2(x  4)

241

Chapter 5

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Find the y-intercept. y  mx  b 7  2(3)  b 76b 766b6 1b Write the slope-intercept form. y  mx  b y  2x  1 9. D; The line parallel to y  3x  4 has the same slope, 3. The slope of y  3x  5 is 3. 10. Let x  the lowest score. Twice the lowest score minus 12 is 98. 14444244443 123 { { 123 2x  12  98

Chapter 5 Standardized Test Practice Pages 314–315 1. B; The amount of the person’s salary that is spent is $(x  y). The fraction of the salary that is spent is the ratio of the amount spent to the total salary x  y or x . 2. A; 2x  7y  2(5)  7(4)  10  7(4)  10  28  38 3. D; 5x  6  10 5x  6  6  10  6 5x  4 5x 5

4

5

2x  12  98 2x  12  12  98  12 2x  110

4

x5 4. C; Mean age   

8(12  10(32  14(22  16(12  17(22 1  3  2  1  2

2x 2

116 9 8 129 8

To find the median age, order the numbers from least to greatest. 8, 10, 10, 10, 14, 14, 16, 17, 17 The median age is 14. The mode is 10. Therefore, mean age  median age. 5. D; When the difference in the x values is 1, the difference in the y values is 1. When the difference in the x values is 2, the difference in the y values is 2. This suggests that y  x. Check this equation. If x  3, then y  (3) or 3. But the y value for x  3 is 4. This is a difference of 1. Try some other values in the domain to see if the same difference occurs. 2 2 3

0 0 1

2y 2

The y-intercept

4  6

m  0  (12 m

1 3 5 1 3 5 0 2 4

y y

21 y

2  1

1 2

13. 14.

y

2 2  2(1) y  2 7  4

7. A; m  4  2

15. 1

m  6 or 2 8. C; Find the slope. 21  7

16.

14 7

17.

m  10  3 m

Chapter 5

3x  3 2 3 3 2x  2 3 is 2 or 1.5.

12. Use the points (1, 6) and (0, 4) to find the equation of the line. Find the slope.

21

3



y

y is always 1 more than the opposite of x. 6. B; To find the y-intercept, let x  0. x 3 0 3

110 2

x  55 The lowest score was 55. 11. Solve for y. 3x  2y  3 3x  2y  3x  3  3x 2y  3x  3

The mean age is 129.

x x y



or 2

242

2 1

or 2

The point (0, 4) lies on the y-axis. The y-intercept is 4. Write the slope-intercept form. y  mx  b y  2x  4 Use the equation y  2x + 4 with x  5. y  2x  4 y  2(5)  4 y  6 The y-coordinate of the point (5, y) is 6. The slope is 2. A; 2(x  6)  2x  12  2x  6  6 Therefore, 2(x  6) is 6 more than 2x  6. D; Let x  3. The |x| 3 and |x  1|  4. Let x  3. Then |x|  3 and |x  1| 2. Therefore, the relationship cannot be determined. C; Two nonvertical lines are parallel if they have the some slope. B; The slope of y  2x is 2. The slope of the line perpendicular to y  2x is the opposite 1 reciprocal of 2 or 2.

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18c. Evaluate each equation when m  100. For Plan 1: C  0.59m C  0.59 (1002 C  59 For Plan 2: C  0.39m  10 C  0.39 (1002  10 C  49 For Plan 3: C  59.95 Your friend should enroll in Plan 2. When m  100, the cost is least for Plan 2.

18a. For Plan 1: Monthly rate of number of monthly cost plus fee. equals change times minutes 14243 123 14243 123 14243 123 14243 .  C 0.59 m  0

The equation for Plan 1 is C  0.59m. For Plan 2: monthly Monthly rate of number of fee. cost plus equals change times minutes 123 14243 123 14243 123 14243 .  C 0.39 m  10

14243

The equation for Plan 2 is C  0.39m  10. For plan 3: Monthly rate of number of monthly cost plus fee. equals change times minutes 123 14243 123 14243 123 14243 .  C 0 m  59.95

14243

The equation for Plan 3 is C  59.95. 18b. The graph of C  0.59m passes through (0, 0) with slope 0.59. The graph of C  0.39m  10 passes through (0, 10) with slope 0.39. The graph of C  59.95 passes through (0, 59.95) with slope 0. C 60 50

C  59.95

40

C  0.59m

30

C  0.39m  10 20 10

0

20

40

60

80

100

m

243

Chapter 5

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Chapter 6 Page 317

Solving Linear Inequalities Getting Started

t  31  84 t  31  31  84  31 t  53 b  17  23 2. b  17  17  23  17 b  40 3. 18  27  f 18  27  27  f  27 9  f 1.

2

1

2

1

2(n  3) 

13.

d32

4.

2

14.

2

d3323 7

1

d  6 or 1 6

15.

3r  45  4r 3r  45  3r  4r  3r 45  r 6. 5m  7  4m  12 5m  7  4m  4m  12  4m m  7  12 m  7  7  12  7 m  19 7. 3y  4  16 3y  4  4  16  4 3y  12 5.

3y 3



16.

17.

18.

12 3

y4 8. 2a  5  3a  4 5a4 5a545 a  1 (1)(a)  (1)(1) a1 1 k 2

9. 1 k 2

 11

1 2  2(11)

21.

k  22 4.3b  1.8  8.25 4.3b  1.8  1.8  8.25  1.8 4.3b  6.45 4.3b 4.3

11.

20.

4474

2 10.

19.

47 1 k 2 1 k 2





1

2



6  2x 2

y3x Select five values for the domain and make a table.

6.45 4.3

x 2 0 1 3 5

16 4

s4

Chapter 6

n  1 2 n  1 2 2

2n  6  n  1 2n  6  n  n  1  n n61 n6616 n7 8 is eight units from zero in the negative direction. |8|  8 20 is twenty units from zero in the positive direction. |20|  20 30 is thirty units from zero in the negative direction. |30|  30 1.5 is one and five-tenths units from zero in the negative direction. |1.5|  1.5 |14  7|  |7| 7 is seven units from zero in the positive direction. |14  7|  |7|  7 |1  16|  |15| 15 is fifteen units from zero in the negative direction. |1  16|  |15|  15 |2  3|  |1| 1 is one unit from zero in the negative direction. |2  3|  |1|  1 |7  10|  |3| 3 is three units from zero in the negative direction. |7  10|  |3|  3 Solve the equation for y. 2x  2y  6 2x  2y  2x  6  2x 2y  6  2x 2y 2

b  1.5 6s  12  2(s  2) 6s  12  2s  4 6s  12  2s  2s  4  2s 4s  12  4 4s  12  12  4  12 4s  16 4s 4

n3

12.

244

3 3 3 3 3

3x  (2) 0 1 3 5

y 5 3 2 0 2

(x, y) (2, 5) (0, 3) (1, 2) (3, 0) (5, 2)

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Graph the ordered pairs and draw a line through the points.

Graph the ordered pairs and draw a line through the points. y

y

2x  2y  6

x

O

y  2x  3

x

O

24. The only value in the range is 4. Since there is no x in the equation, the value of x does not depend on y. Therefore, x can be any real number. Select five values for the domain and make a table.

22. Solve the equation for y. x  3y  3 x  3y  x  3  x 3y  3  x 3y 3

3  x 3 1 1  3x



y

x 3 1 0 2 4

Select five values for the domain and make a table. 1

x

1  3x

y

(x, y)

6

1  3 (6)

1

1

(6, 1)

4

1  3 (4)

1

3

3

1  3 (3)

1

0

1

1

1

13

0

1  3 (0)

1

1  3 (1)

1

1

14, 13 2

(x, y) (3, 4) (1, 4) (0, 4) (2, 4) (4, 4)

y 4 4 4 4 4

Graph the ordered pairs and draw a line through the points. y

(3, 0)

x

O

(0, 1)

13, 1 13 2

y  4

Graph the ordered pairs and draw a line through the points. y

25. Solve the equation for y.

x  3y  3

1

x  2 y O

x

1

1

(2)(x)  (2) 2 y

2

2x  y Select five values for the domain and make a table. x 2 1 0 1 2

23. Select five values for the domain and make a table. x 1 0 1 2 3

2x  3 2(1)  3 2(0)  3 2(1)  3 2(2)  3 2(3)  3

y 5 3 1 1 3

(x, y) (1, 5) (0, 3) (1, 1) (2, 1) (3, 3)

245

2x 2(2) 2(1) 2(0) 2(1) 2(2)

y 4 2 0 2 4

(x, y) (2, 4) (1, 2) (0, 0) (1, 2) (2, 4)

Chapter 6

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The graph intersects the y-axis at (0, 5). Plot these points and draw the line that connects them.

Graph the ordered pairs and draw a line through the points. y

y x   12 y

15  3(x  y )

x

O

x

O

26. To find the x-intercept, let y  0. 3x  6  2y 3x  6  2(0) 3x  6  0 3x  6  6  0  6 3x  6 3x 3

28. To find the x-intercept, let y  0. 2  x  2y 2  x  2 (0) 2x0 2xx0x 2x The graph intersects the x-axis at (2, 0). To find the y-intercept, let x  0. 2  x  2y 2  0  2y 2  2y

6

3

x2 The graph intersects the x-axis at (2, 0). To find the y-intercept, let x  0. 3x  6  2y 3(0)  6  2y 6  2y 6 2



2 2



2y 2

1y The graph intersects the y-axis at (0, 1). Plot these points and draw the line that connects them.

2y 2

3  y The graph intersects the y-axis at (0, 3). Plot these points and draw the line that connects them.

y

y 2  x  2y O

x

x

O

3x  6  2y

27. To find the x-intercept, let y  0. 15  3 (x  y) 15  3 (x  0) 15  3x 15 3



3x 3

Page 321

5x The graph intersects the x-axis at (5, 0). To find the y-intercept, let x  0. 15  3 (x  y) 15  3 (0  y) 15  3y 15 3



Check for Understanding

1. Sample answers: y  1  2, y  1  4, y30 2.

3y 3

5y

Chapter 6

Solving Inequalities by Addition and Subtraction

6-1

–2

–1

0

1

2

3 4 a4

5

6

7

8

–2

–1

0

1

2

3 4 a4

5

6

7

8

In both graphs, the line is darkened to the left. In the graph of a  4, there is a circle at 4 to indicate that 4 is not included in the graph. In the graph of a  4, there is a dot at 4 to indicate that 4 is included in the graph.

246

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3. {b|b  5} is the set of all numbers b such that b is greater than or equal to 5. 4. m3 7 7 m33 7 73 m 7 4 The solution set is {m|m  4}. The graph must have a heavy arrow pointing to the right to show that the inequality includes all numbers greater than 4. The graph must have a circle at 4 to show that 4 is not included in the inequality. This is represented by graph a. 5. a4 6 2 a44 6 24 a 6 2 Check: Substitute 2, a number less than 2, and a number greater than 2. Let a  2. Let a  5. Let a  6. ? ? ? 2  4 6 2 5  4 6 2 64 6 2 2 2 1 6 2 ✓ 10 2 The solution set is {a | a  2}.

9.

5.2r  6.7  6.2r 5.2r  6.7  5.2r  6.2r  5.2r 6.7  r 6.7  r is the same as r  6.7. Check: Substitute 6.7, a number less than 6.7, and a number greater than 6.7. Let r  6.7. ? 5.2 (6.7)  6.7  6.2 (6.7) ?

34.84  6.7  41.54 41.54  41.54 ✓ Let r  1. ? 5.2 (1)  6.7  6.2 (1) 11.9  6.2 ✓ Let r  10. ? 5.2 (10)  6.7  6.2 (10) ?

52  6.7  62 58.7 62 The solution set is {r | r  6.7}. 0 1 2 3 4 5 6 7 8

8 765 4 321 0

6.

10.

9b4 94b44 5b 5  b is the same as b  5. Check: Substitute 5, a number less than 5, and a number greater than 5. Let b  5. Let b  0. Let b  10. ? ? ? 954 904 9  10  4 99✓ 9 4 9  14 ✓ The solution set is {b | b  5}.

7p  6p  2 7p  6p  6p  2  6p p  2 Check: Substitute 2, a number less than 2, and a number greater than 2. Let p  2. ? 7 (2)  6(2)  2 ?

14  12  2 14  14 ✓ Let p  5. ? 7 (5)  6 (5)  2 ?

35  30  2 35  32 ✓ Let p  1. ? 7(1)  6(1)  2 7 4 The solution set is { p|p  2}.

0 1 2 3 4 5 6 7 8

7.

t75 t7757 t  12 Check: Substitute 12, a number less than 12, and a number greater than 12. Let t  12. Let t  2. Let t  15. ? ? ? 12  7  5 275 15  7  5 55✓ 5 5 85✓ The solution set is {t | t  12}.

432 1 0 1 2 3 4

11. Let n  the number. decreased A number by 8 14 424 43

y  2.5 7 3.1 y  2.5  2.5 7 3.1  2.5 y 7 5.6 Check: Substitute 5.6, a number less than 5.6, and a number greater than 5.6. Let y  5.6. Let y  1. Let y  8. ?

?

1 424 3

14.

123

n 8  14 n  8  14 n  8  8  14  8 n  22 Check: Substitute 22, a number less than 22, and a number greater than 22. Let n  22. Let n  10. Let n  30. ? ? ? 22  8  14 10  8  14 30  8  14 22 14 14  14 ✓ 2  14 ✓ The solution set is {n|n  22}.

0 2 4 6 8 10 12 14 16

8.

14 424 43

is at most

?

5.6  2.5 7 3.1 1  2.5 7 3.1 8  2.5 7 3.1 3.1 3.1 1.5 3.1 5.5 7 3.1 ✓ The solution set is { y | y  5.6}. 0 1 2 3 4 5 6 7 8

247

Chapter 6

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12. Let n  the number. A number

plus

is greater 7 than

14 424 43

123

{

14 424 43

2. {

n



7

7

2

Exercises 20–37 For checks, see students’ work. 20. t  14  18 t  14  14  18  14 t4 The solution set is {t|t  4}.

n7 n77 n Check:

7 2 7 27 7 5 Substitute 5, a number less than 5, and a number greater than 5. Let n  5. Let n  10. Let n  3. ? ? ? 5  7 7 2 10  7 7 2 37 7 2 2 2 3 2 10 7 2 ✓ The solution set is {n|n  5}. 13. Let g  the number of grams of fat Chapa can have during the rest of the day. 2132  21  g  60 6  21  g  60 27  g  60 27  g  27  60  27 g  33 Chapa can have no more than 33 g of fat during the rest of the day.

Pages 321–323

0 1 2 3 4 5 6 7 8

21.

0 1 2 3 4 5 6 7 8

22.

16.

17.

18.

19.

Practice and Apply

s  5  1 s  5  5  1  5 s4 The solution set is {s|s  4}. 0 1 2 3 4 5 6 7 8

x  3  2 x  3  3  2  3 x1 The solution set is {x|x  1}. The corresponding graph is d. f; x76 x7767 x  1 The solution set is {x|x  1}. The corresponding graph is f. a; 4x 7 3x  1 4x  3x 7 3x  1  3x x 7 1 The solution set is {x|x  1}. The corresponding graph is a. c; 8x 6 9 8x8 6 98 x 6 1 The solution set is {x|x  1}. The corresponding graph is c. e; 5x6 56x66 1  x 1  x is the same as x  1. The solution set is {x|x  1}. The corresponding graph is e. b; x1 7 0 x11 7 01 x 7 1 The solution set is {x|x  1}. The corresponding graph is b.

Chapter 6

n  7  3 n  7  7  3  7 n4 The solution set is {n|n  4}. 0 1 2 3 4 5 6 7 8

23.

14. d;

15.

d57 d5575 d2 The solution set is {d|d  2}.

24.

53g 533g3 2g 2  g is the same as g  2. The solution set is { g|g  2}. 0 1 2 3 4 5 6 7 8

25.

48r 488r8 4  r 4  r is the same as r  4. The solution set is {r|r  4}. 8 765 4 321 0

26.

3  q  7 3  7  q  7  7 4q 4  q is the same as q  4. The solution set is {q|q  4}. 0 1 2 3 4 5 6 7 8

27.

2m1 21m11 3m 3  m is the same as m  3. The solution set is {m|m  3}. 0 1 2 3 4 5 6 7 8

248

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28.

2y  8  y 2y  y  8  y  y y  8 The solution set is { y|y  8}.

1

3f  3  2f 3f  2f  3  2f  2f f  3 The solution set is { f|f  3}.

1

6

432 1 0 1 2 3 4 2

4

2

4

p39

37.

2

2

p3393 1

p  19

3b  2b  5 3b  2b  2b  5  2b b  5 The solution set is {b|b  5}.

The solution set is

5 p|p  1 19 6.

432 1 0 1 2 3 4

d  5  17 d  5  5  17  5 d  12 38b. d  5  17 d  5  3  17  3 d  8  20 38c. d  5  17 d  5  10  17  10 d57 39a. z  2  10 z  2  2  10  2 z  12 39b. z  2  10 z  2  5  10  5 z75 39c. z  2  10 z  2  6  10  6 z  4  16 Exercises 40–45 For checks, see students’ work. 40. Let n  the number. n  13  27 n  13  13  27  13 n  14 The solution set is {n|n  14}. 41. Let n  the number. n  5  33 n  5  5  33  5 n  38 The solution set is {n|n  38}. 42. Let n  the number. 30  n  (8) 30  n  8 30  8  n  8  8 38  n 38  n is the same as n  38. The solution set is {n|n  38}. 43. Let n  the number. 2n  n  14 2n  n  n  14  n n  14 The solution set is {n|n  14}. 38a.

4w  3w  1 4w  3w  3w  1  3w w1 The solution set is {w|w  1}. 432 1 0 1 2 3 4

432 1 0 1 2 3 4

33. a  (2)  3 a  2  3 a  2  2  3  2 a  5 The solution set is {a|a  5}. 8 765 4 321 0

0.23  h  (0.13) 0.23  h  0.13 0.23  0.13  h  0.13  0.13 0.36  h 0.36  h is the same as h  0.36. The solution set is {h|h  0.36}. 432 1 0 1 2 3 4

35.

5

The solution set is a|a 7 8 .

32. v  (4)  3 v43 v4434 v  1 The solution set is {v|v  1}.

34.

1

4 1

8 765 4 321 0

31.

1 8 1 8

a 7 8

8 7654 32 1 0

30.

1

a44 7

8 765 4 321 0

29.

1

a4 7

36.

x  1.7  2.3 x  1.7  1.7  2.3  1.7 x  0.6 The solution set is {x|x  0.6}. 432 1 0 1 2 3 4

249

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53a. If a  b, then a  c  b  c by the Addition Property of Inequalities. If c  d, then b  c  b  d by the Addition Property of Inequalities. Since a  c  b  c and b  c  b  d, it follows that a  c  b  d by the Transitive Property of Inequalities (If x  y and y  z, then x  z.). Therefore, the given statement is always true. 53b. If x  y is always true, then x  y is never true. In Exercise 53, it was shown that, if a  b and c  d, the statement a  c  b  d is always true. Therefore, the statement a  c  b  d is never true. 53c. If a  1, b  2, c  3, and d  5, then a  c  1  3 or 2, but b  d  2  5 or 3. In this case, a  c  b  d. However, if a  1, b  2, c  3, and d  4, then a  c  1  3 or 2, and b  d  2  4 or 2. In this case, a  c  b  d. Therefore, the given statement is sometimes true. 54. 225  p  200 55. 225  p  200 225  p  p  200  p 225  200  p 225  200  200  p  200 25  p 25  p is the same as p  25. The solution set is {p|p  25}. 56. Inequalities can be used to compare the number of schools participating in certain sports, to compare the number of participating schools if sports are added or discontinued in a certain number of schools, and to determine how many schools need to add a certain sport to surpass the number participating in another sport. Answers should include the following. • To find how many schools must add girls’ track and field to surpass the current number of schools participating in girls’ basketball, solve 16,526  14,587  x. More than 1939 schools must add girls’ track and field. 57. C; x  1  13 x  1  1  13  1 x  14 is at The sum of a 58. A; least five. number and 6 144424443 123 123  n6 5

44. Let n  the number. n  (7)  18 n  7  18 n  7  7  18  7 n  25 The solution set is {n|n  25}. 45. Let n  the number. 4n  3n  (2) 4n  3n  3n  (2)  3n n  2 The solution set is {n|n  2}. 46. Let n  the number of pounds a Nile crocodile might be expected to gain. n  157  2200 n  157  157  2200  157 n  2043 The crocodile would be expected to gain no more than 2043 lb. 47. Let s  the number of stars that cannot be seen without a telescope. s  1100  200,000,000,000 s  1100  1100  200,000,000,000  1100 s  199,999,998,900 There are at least 199,999,998,900 stars that cannot be seen without a telescope. 48. Let n  the number of species of insects that are not bees. n  3500  600,000 n  3500  3500  600,000  3500 n  596,500 There are more than 596,500 species of insects that are not bees. 49. Let m  the amount of money in the account. m  1300  947  1500 m  2247  1500 m  2247  2247  1500  2247 m  3747 Mr. Hayashi should have at least $3747 in his account before writing the checks. 50. 12  4  x 12  4  4  x  4 8 x 8  x is the same as x  8. The value of x must be more than 8 in. 51. Let j  the cost of a pair of jeans. 18  14  j  65 32  j  65 32  j  32  65  32 j  33 The jeans must cost no more than $33. 52. Let w  the number of wins to meet the goal. w  4  0.60(18) w  4  10.8 w  4  4  10.8  4 w  6.8 The team must win at least 7 more games to meet their goal.

Chapter 6

Page 323

Maintain Your Skills

59. Since a person’s height is not related to his or her grade on a math test, a scatter plot for the relationship would show no correlation. 60. The line parallel to y  3x  2 has the same slope, 3. Replace m with 3, and (x1, y1) with (1, 3) in the point-slope form. y  y1  m(x  x1) y  (3)  3(x  1) y  3  3x  3 y  3  3  3x  3  3 y  3x  6

250

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66.

61. Find the slope of the given line. x  y  3 x  y  x  3  x y  x  3 The slope of the given line is 1, so the slope of the line parallel to y  x  3 has the same slope, 1. Replace m with 1 and (x1, y1) with (0, 4) in the point-slope form. y  y1  m(x  x1) y  4  1(x  0) y  4  x y  4  4  x  4 y  x  4 62. Find the slope of the given line. 2x  y  1 2x  y  2x  1  2x y  2x  1 1(y)  1(2x  1) y  2x  1 The slope of the given line is 2, so the slope of the line parallel to y  2x  1 has the same slope, 2. Replace m with 2 and (x1, y1) with (1, 2) in the point-slope form. y  y1  m(x  x1) y  2  2[x  (1)] y  2  2(x  1) y  2  2x  2 y  2  2  2x  2  2 y  2x  4 63. 7 13 19 25

y 8 4 2

(x, y) (1, 8) (3, 4) (5, 2)

2x  6 2(1)  6 2(3)  6 2(5)  6

y 8 0 4

(x, y) (1, 8) (3, 0) (5, 4)

The solution set is {(1, 8), (3, 0), (5, 4)}. 69. 6g  42 6g 6



42 6

g7 t 9 t 9

70. 192 71.

 14

 19214

t  126

2 y 3 3 2 y 2 3

 14

1 21 2  132 214

y  21 72. 3m  435

The next two terms are 31 and 37. 64. 243 81 27 9

3m 3

   162 54 18

435 3



m  145 1

The difference between each pair of terms is 3 the 1 difference of the previous pair. Continue taking 3 of each successive difference. Add 6, then add 2. 9

7x 7  (1) 73 75

x 1 3 5

x 1 3 5

     6 6 6 6 6

27

(x, y) (1, 2) (3, 6) (5, 10)

The solution set is {(1, 8), (3, 4), (5, 2)}. 68. First solve the equation for y in terms of x. 2x  y  6 2x  y  2x  6  2x y  6  2x 1(y)  1(6  2x) y  2x  6

The difference between each pair of terms is always 6. The sequence is arithmetic with a common difference of 6. Add 6, then add 6. 7 13 19 25 31 37

81

y 2 6 10

The solution set is {(1, 2), (3, 6), (5, 10)}. 67.

   6 6 6

243

2x 2(1) 2(3) 2(5)

x 1 3 5

3

73.

4 x 7 7 4 x 4 7

 28

1 21 2  174 228

x  49 74. 5.3g  11.13

1

    162 54 18 6 2

5.3g 5.3

The next two terms are 3 and 1. 65. 3 6 12 24



11.13 5.3

g  2.1

   3 6 12

75. 13.52

The difference between each pair of terms doubles for each successive pair. Continue doubling each successive difference. Add 24, then add 48. 3 6 12 24 48 96

76.

a 3.5 a 3.5

 13.52 7

a  24.5 8p  35 8p 8

     3 6 12 24 48

7



p

The next two terms are 48 and 96.

251

35 8 35 8

or 4.375

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Page 324

Step 4 Separate the tiles into 4 groups.

Algebra Activity (Preview of Lesson 6-2)

1. 4x  12 Step 1 Write a  symbol and model the inequality.

x

1

1

1

x

1

1

1

1

1

1

1

1

1



x x

1

x

1

1

1

x

1

1

1

x

1

1

1

x

3  x or x  3

The solution set is {x|x  3}. 2. 2x  8 Step 1 Write a  symbol and model the inequality.

Step 2 Since you do not want to solve for a negative x-tile, eliminate the negative by adding 4 positive x-tiles to each side. Remove the zero pairs. x

1

1

1

x

1

1

1

x

1

1

1

x

1

1

1



1



4x  12

x

1

x

 x

x

x

x

x

x

x

x

1

1

1

1

1

1

1

1

2x  8

Step 2 Since you do not want to solve for a negative x-tile, eliminate the negative x-tiles by adding 2 positive x-tiles to each side. Remove the zero pairs.

4x  4x  12  4x

Step 3 Add 12 negative 1-tiles to each side to isolate the x-tiles. Remove the zero pairs.

x

x 1 1

1 1 1



1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1



1

1

1

1

1

1

x

x

x 2x  2x  8  2x Step 3 Add 8 negative 1-tiles to each side to isolate the x-tiles. Remove the zero pairs.

x

1 1 1

x

1 1 1

1

x

x

1 1 1

x 12  4x

252

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1

1

1

1

1

1

x

1

1

x



8  2x

Chapter 6

1

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Step 4 Separate the tiles into 3 groups.

Step 4 Separate the tiles into 2 groups. 1

1

1

1

1

x



 1

1

1

1

x

1

x

4  x or x  4 The solution set is {x|x  4}. 3. 3x  6 Step 1 Write a  symbol and model the inequality. Use a self-adhesive note to cover the equals sign on the equation mat. Then write a  symbol on the note. Model the inequality.

1

1

x

1

1

x

2  x or x  2

The solution set is {x|x  2}. 4. 5x  5 Step 1 Write a  symbol and model the inequality. x

1

x

1



x x



x x

1

1

1

1

1

1



x

x

1

x

1

x

1

1

x

1

1

1

x

1

1

1

x

1

x

1

Step 2 Since you do not want to solve for a negative x-tile, eliminate the negative x-tiles by adding 5 positive x-tiles to each side. Remove the zero pairs.

Step 2 Since you do not want to solve for a negative x-tile, eliminate the negative x-tiles by adding 3 positive x-tiles to each side. Remove the zero pairs.

x

x

5x  5

3x  6

x

1

x

x



1

x

x

x x

x

x

x

x

x

x

x

x

3x  3x  6  3x

Step 3 Add 6 positive 1-tiles to each side to isolate the x-tiles. Remove the zero pairs.

x

5x  5x  5  5x



1 1

1 1

1 1

1 1

1 1

1 1

1

1

1

1

x

1

1

x

x

6  3x

253

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There are no negative x-tiles, so the variable remains on the left and the symbol remains .

Step 3 Add 5 positive 1-tiles to each side to isolate the x-tiles. Remove the zero pairs.



1

1

1

1

1

1

1

1

1

1

Pages 328–329

1 1 1

1

2. Sample answer: Three fourths of a number is greater than 9. 3. Ilonia is correct since when you divide each side of an inequality by a negative number, you must reverse the direction of the inequality symbol. 4. a; since the phrase is at least is translated as , statement a represents 7n  14. 5. c; since the phrase is less than is translated as , inequality c represents the statement. 6. 15g 7 75

5  5x

Step 4 Separate the tiles into 5 groups. 1

x

1

x

15g 15

 1

x

1

x

1

x

1

?

7. 192

108 9

1

1

1

180

?

18 9

?

6 12 2 12

2

3b  9

132 2 123b2  132 2 (9)

1

1

1

1

1

1

3 113.52  9 2

?

3 1122  9 3 1182  9 2

?

9  9 ✓ 8 9 The solution set is {b|b  13.5}.

 

x3

Chapter 6

?

Check:

Step 2 Since there is no need to eliminate x-tiles or isolate x-tiles, the only step remaining is to group the tiles.

x

6 1921122

b  13.5 Substitute 13.5, a number less than 13.5, and a number greater than 13.5. Let b  13.5. Let b  12. Let b  18.

2x  6

x

?

6 12

6 12 6 12 9 12 12 20 12 ✓ The solution set is {t|t  108}. 8.

1

t 9 t 9

?

t 6 108 Check: Substitute 108, a number less than 108, and a number greater than 108. Let t  108. Let t  180. Let t  18.

 

x

75 15

15 152 7 75 15 1102 7 75 15 152 7 75 75 75 150 7 75 ✓ 75 75 The solution set is { g|g  5}.

The solution set is {x|x  1}. 5. In Exercises 1–4, the coefficients of x are 4, 2, 3, and 5, respectively. Thus, all coefficients are negative. 6. The symbols in the solutions point in the opposite direction with relationship to the variable than the symbols in the original problem. 7. 2x  6 Step 1 Model the inequality. 1

6

g 6 5 Check: Substitute 5, a number less than 5, and a number greater than 5. Let g  5. Let g  10. Let g  5.

1  x or x  1

x

Check for Understanding

1. You could solve the inequality by multiplying each 1 side by 7 or by dividing each side by 7. In either case, you must reverse the direction of the inequality symbol.

x x x x x

1

Solving Inequalities by Multiplication and Division

6-2

254

2

?

12  9 ✓

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9. 25f  9 25f 25



Examples 19–34 For checks, see students’ work. 19. 6g  144

9 25

6g 6

f  0.36 Check: Substitute 0.36, a number less than 0.36, and a number greater than 0.36. Let f  0.36. Let f  0. Let f  1. ? ? ? 25 10.362  9 25 102  9 25 112  9 99✓ 0 9 25  9 ✓ The solution set is {f|f  0.36}. 10. Let n  the number.

g  24 The solution set is {g|g  24}. 20. 7t 7 84 7t 7

The opposite of 4 times a number is more than 12.  4n 7 12

14d 14

4n 7 12 6

16z 16

23.

24.

 (2) 26

?

6x 6

1

?

7  (5)7

b 10 b (10) 10

5  (10)5

b  50 The solution set is {b|b  50}. r

7 6 7

25.

1 r2

(7) 7 7 (7) (7)

? 1 1602  2

r 7 49 The solution set is {r|r  49}.

26 30  26 ✓ 26.

1

a

11 7 9 a

2

(11) 11 6 (11)9

24 6

a 6 99 The solution set is {a|a  99}.

6 x 6 4

27.

Pages 329–331

m 5 m (5) 5

m  35 The solution set is {m|m  35}.

 26

1522  26 1402  26 2 26  26 ✓ 20 26 The solution set is {n|n  52}. 12. B; 6x 6 24

64

 16

z4 The solution set is {z|z  4}.

n  52 Check: Substitute 52, a number less than 52, and a number greater than 52. Let n  52. Let n  40. Let n  60. 1 2

84

 14

d  6 The solution set is {d|d  6}. 22. 16z  64

12 4

n 6 3 Check: Substitute 3, a number less than 3, and a number greater than 3. Let n  3. Let n  5. Let n  1. ? ? ? 4 132 7 12 4 152 7 12 4 112 7 12 12 12 20 7 12 ✓ 4 12 The solution set is {n|n  3}. 11. Let n  the number. Half of a number is at least 26. 123 123 14424 43 1442443 123 1 n

 26 2 1 n 2 1 (2) 2 n

84 7

7

t 7 12 The solution set is {t|t  12}. 21. 14d  84

144424443 14444244443 1442443 123

4n 4

144 6



Practice and Apply

1

5 y 8 8 5 y 5 8

12

 15 

185 2 (15)

y  24 The solution set is {y ƒ y  24} .

13. d; 5 n 7 10 can be translated One fifth of a number is greater than ten. 14. a; 5n  10 can be translated Five times a number is less than or equal to ten. 15. e; 5n  10 can be translated Five times a number is greater than ten. 16. f; 5n  10 can be translated Negative five times a number is less than ten.

28.

2 v 3 3 2 v 2 3

12

6 6

132 26

6

v 6 9 The solution set is {v ƒ v  9}. 29.

1

17. b; 5 n  10 can be translated One fifth of a number is no less than ten. 18. c; 5n  10 can be translated Five times a number is less than ten.

3

4q  33

143 2134q2  143 2 (33)

q  44 The solution set is {q ƒ q  44}.

255

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30.

2

38c.

5 p 7 10

152 2125 p2 6 152 210

p 6 25 The solution set is {p ƒ p  25}. 31. 2.5w 6 6.8 2.5w 2.5

Exercises 39–44 For checks, see students’ work. 39. Let n  the number. 7n 7 28

6.8 2.5

7

7n 7

w 7 2.72 The solution set is {w ƒ w 7 2.72}. 32. 0.8s 7 6.4 0.8s 0.8

7n 7

s 6 8 33.

3 14 7 3 15 14

1 21 2 6 1 21 2 7

c 6

1 10

1

24  3n

5

The solution set is c ƒ c 6 34.

4m 5 5 4m 4 5

3 15 5 3 4 15 1 4

1 21 2 6 1 21 2 6

m 6

5

1 10

The solution set is m ƒ m 6 35. (8)

y 8 y 8

1 26 7

1 2

1

6.

1 4

(3)24  (3) 3n 72  n 72  n is the same as n  72. The solution set is {n ƒ n  72}. 42. Let n  the number.

6

2 n 3 3 2 n 2 3

36. (9)



0

1

0.25n 0.25

m3 The solution set is {m ƒ m  3}. 1

2

3

4

5

6

7

0.40n 0.40

8

7

2

a  3.5 37b.

37c.

2a  7 (2)2a  (2)7 4a  14

20w 20

1

46. Let b  the number of bags of mulch to be sold. 2.50b  2000 2.50b 2.50

2 4

4t 6 2 (2)4t 7 (2)(2) 8t 7 4

Chapter 6



2000 2.50

b  800 The band should sell at least 800 bags of mulch.

t 6 0.5 38b.

85 20 1 44

The width is less than 44 ft.

38a. 4t 6 2 6

6

w 6

2a  7 (3)2a  (3)7 6a  21 4t 4

45

 0.40

n  112.5 The solution set is {n ƒ n  112.5}. 45. Let w  the width of the rectangle. A 6 85 /w 6 85 20w 6 85

37a. 2a  7 2a 2

90

 0.25

n  360 The solution set is {n ƒ n  360}. 44. Let n  the number. 0.40n  45

2

1 3

1 2  (9) 113 2

0

132 2 (15)

6

n 6 22.5 The solution set is {n ƒ n  22.5}. 43. Let n  the number. 0.25n  90

1 (8) 2

6 5 4 3 2 1

6 15

12

.

y 6 4 The solution set is {y ƒ y  4}.

m 9 m 9

14

 7

n  2 The solution set is {n ƒ n  2}. 41. Let n  the number.

The solution set is {s ƒ s  8}. 15c 7 7 15c 15 7

28 7

7

n 7 4 The solution set is {n ƒ n 7 4}. 40. Let n  the number. 7n  14

6.4 0.8

6

4t 6 2 (7)4t 7 (7) (2) 28t 7 14

256

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54. Let s  the number of signatures to be sought. 0.85s  6000

47. Let m  the number of minutes Juan can talk. 0.09m  2.50 0.09m 0.09

2.50

0.85s 0.85

 0.09

m  27. 7 Juan can talk to his friend for no more than 27 min. 48. Let p  the number of people attending the reunion. 28.95p  4000 28.95p 28.95

6

4000

 28.95

x

56. B; 5  1 x

(5) 5  (5) (1)

38 2

x  5

r 6 6.0 (to the nearest tenth) The garden can have a radius of up to about 6 ft. 50. Let d  the distance that can be traveled legally. d 6 65 1 12

11 12 2 1d1

d 6

16

t 7 15

The solution set is

132 265

Page 331

5t 0 t 7 1615 6 .

Maintain Your Skills

Exercises 58–60 For checks, see students’ work. 58. s  7 6 12 s  7  7 6 12  7 s 6 19 The solution set is {s ƒ s  19}.

1 97 2

1

A person can legally travel no more than 972 mi. 51. Let n  the number of visits to the zoo in one year (assuming two adults and two children go to the zoo each visit). 144 6 [ 2(18)  2(8) ] n 144 6 52n

15 16 17 18 19 20 21 22 23

59.

52n 52

2.8 6 n (to the nearest tenth) The yearly membership will be less expensive than regular admission if the family (of two adults and two children) visits the zoo at least 3 times during the year. 52a. Sample answer: Let a  2 and b  3. It is true that a  b since 2  3. However, it is not true that a2  b2 since 4 9. 52b. Sample answer: Let a  1, b  2, c  3, and d  2. It is true that a  b since 1  2. It is also true that c  d since 3  2. However, it is not true that ac  bd since ac  (1)(3) or 3, bd  (2)(2) or 4, and 3 4. 53. Let s  the number of parking spaces in the lot. 0.20s  35 0.20s 0.20

14 15

187 2178t2 7 187 211415 2

1 12

2

6

7

8 t 6

57. C;

6 1 2 65

d 6

144 52

6000 0.85

s  7058.8 (to the nearest tenth) The candidate should seek at least 7059 signatures on the petition. 55. Inequalities can be used to compare the heights of walls. Answers should include the following. • If x represents the number of bricks and the wall must be no higher than 4 ft or 48 in., then 3x  48. • To solve this inequality, divide each side by 3 and do not change the direction of the inequality symbol. The wall must be no more than 16 bricks high.

p  138.2 (to the nearest tenth) At least 139 people must attend the reunion to avoid a rental fee. 49. Let r  the radius of the garden. C 6 38 2r 6 38 2r 2



g  3  4 g  3  3  4  3 g  7 The solution set is {g ƒ g  7}. 8 7 6 5 4 3 2 1

60.

0

7 7 n2 72 7 n22 5 7 n 5  n is the same as n  5. The solution set is {n ƒ n  5}. 0

1

2

3

4

5

6

7

8

35

 0.20

s  175 The parking lot must have at least 175 spaces.

257

Chapter 6

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Choose (3, 3) and find the y-intercept of the line. y  mx  b

61. Sample answer: Since y increases as x increases, there is a positive correlation between x and y in the graph shown.

1

3  4 (3)  b 3

34b

y

3

3

3

9 4

b

344b4 1

Write the slope-intercept form using m  4 and 9 b  4.

x

O

y  mx  b

62. Find the slope. Let (x1, y1)  (1, 3) and (x2, y2)  (2, 4).

1

h(x)  3x  2 h(4)  3(4)  2  12  2  10 67. h(x)  3x  2 h(w)  3(w)  2  3w  2 65.

y2  y1

mx

2

9

y  4x  4

 x1

4  3 (1)

m2 1

m3 Choose (2, 4) and find the y-intercept of the line. y  mx  b 1

4  3 (2)  b

69.

2

43b 2

2

10 3

b

24 4



71.

y  mx  b

m m

1.6t 1.6

 x1

72.

m m

x 3

12x 12

25 5



x  5 15

15

 12 5

1

x  4 or 1 4 73.

5x  3  32 5x  3  3  32  3 5x  35 5x 5



74.

4t  9  14 4t  9  9  14  9 4t  5

35 5

x7 75.

 x1

2  3 1  3 1 4 1 4

Chapter 6



15(x)  3 (x  5) 15x  3x  15 15x  3x  3x  15  3x 12x  15

y2  y1

m



7 5

w5

m0 Choose (5, 2) and find the y-intercept of the line. y  mx  b 2  0(5)  b 2  b Write the slope-intercept form using m  0 and b  2. y  mx  b y  (0)x  (2) y  2 The standard form of the equation is y  2. 64. Find the slope. Let (x1, y1)  (3, 3) and (x2, y2)  (1, 2). 2

3.6

 1.6

t  2.25 w  2 5

5w 5

2  (2) 1  5 0 6

mx

t(1.6)  1.5(2.4) 1.6t  3.6

4x 4

y2  y1 2

2.4

 1.6

5(w  2)  5(7) 5w  10  35 5w  10  10  35  10 5w  25

10 3

63. Find the slope. Let (x1, y1)  (5, 2) and (x2, y2)  (1, 2). mx

t 1.5

70.

6x

1

Write the slope-intercept form using m  3 and 10 b 3. 1

x

8

3(8)  4(x) 24  4x

2

433b3

y  3x 

3 4

66. h(x)  3x  2 h(2)  3(2)  2 62 8 68. h(x)  3x  2 h(r  6)  3(r  6)  2  3r 18  2  3r  16

6y  1  4y  23 6y  1  4y  4y  23  4y 2y  1  23 2y  1  1  23  1 2y  24 2y 2



24 2

y  12

258

4t 4

5

4 5

t  4 or 1.25

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14g  5 6

76. 6

Substitute 5, a number less than 5, and a number greater than 5. Let p  5. Let p  8. Let p  2. ? ? ? 4  5  9 4  8  9 4  2  9 44✓ 41✓ 4 7 The solution set is { p|p  5}. Check:

9

114g6 5 2  6 (9)

14g  5  54 14g  5  5  54  5 14g  49 14g 14

49

 14

8 7 6 5 4 3 2 1

7

g  2 or 3.5 77.

5a  6  9a  (7a  18) 5a  6  9a  7a  18 5a  6  2a  18 5a  6  2a  2a  18  2a 3a  6  18 3a  6  6  18  6 3a  24 3a 3

78.

4.



24 3

a  8 2( p  4)  7 ( p  3) 2p  8  7p  21 2p  8  2p  7p  21  2p 8  5p  21 8  21  5p  21  21 29  5p 29 5 29 5



0

5.

5p 5

p

1.

1

2

3

4

5

7

6

8

?

?

Practice Quiz 1

7(2)  6(2)  1 14 11 The solution set is {g|g  1}.

3

4

5

6

7(5)  6(5)  1 35  31 ✓

?

7

4 3 2 1

0

1

2

3

4

6. 15z  105 15z 15



105 15

z7 Check:

Substitute 7, a number less than 7, and a number greater than 7. Let z  7. Let z  0. Let z  10. ? ? ? 15(7)  105 15(0)  105 15(10)  105 105  105 ✓ 0 105 150  105 ✓ The solution set is {z|z  7}.

8

r  3  1 r  3  3  1  3 r  4 Check: Substitute 4, a number less then 4, and a number greater than 4. Let r  4. Let r  10. Let r  1. ? ? ? 4  3  1 10  3  1 1  3  1 1  1 ✓ 7  1 ✓ 4 1 The solution set is {r|r  4}. 8 7 6 5 4 3 2 1

3.

2

7(1)  6(1)  1 7  7 ✓ Let g  2.

h  16 7 13 h  16  16 7 13  16 h 7 3 Check: Substitute 3, a number less than 3, and a number greater than 3. Let h  3. Let h  1. Let h  7. ? ? ? 3  16 7 13 1  16 7 13 7  16 7 13 13 13 15 13 9 7 13 ✓ The solution set is {h|h  3}. 0

2.

1

7g  6g  1 7g  6g  6g  1  6g g  1 Check: Substitute 1, a number less than 1, and a number greater than 1. Let g  1. Let g  5.

p   5 or 5.8

29

Page 331

0

3  a  5 3  5  a  5  5 2a 2  a is the same as a  2. Check: Substitute 2, a number less than 2, and a number greater than 2. Let a  2. Let a  0. Let a  10. ? ? ? 3 6 2  5 3 6 0  5 3 6 10  5 3 3 3 5 3 6 5 ✓ The solution set is {a|a  2}.

7.

v 5 v (5) 5

6 7 6 (5)7

v 6 35 Check: Substitute 35, a number less than 35, and a number greater than 35. Let v  35. Let v  30. Let v  70. 35 5

0

4p9 4  9  p  9 9 5  p 5  p is the same as p  5.

?

6 7

30 5

?

6 7

7 7 6 6 7✓ The solution set is {v|v  35}.

259

70 5

?

6 7

14 7

Chapter 6

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8.

3

7 q 7 15

173 2137q2 6 173 215

q 6 35 Substitute 35, a number less than 35, and a number greater than 35. Let q  35. Let q  70. Let q  0. ? ? ? 3 3 3 7 (0) 7 15 7 (35) 7 15 7 (70) 7 15

Pages 334–335

Check:

15 15 30 7 15 ✓ The solution set is {q|q  35}. 9. 156 6 12r 156 12

6

0 15

13 6 r 13  r is the same as r  13. Check: Substitute 13, a number less than 13, and a number greater than 13. Let r  20.

?

Let r  10.

?

156 6 12(13) 156 156

4y 4

?

156 6 12(20) 156 6 12(10) 156 240 156 6 120 ✓

2

1

5w  2

152 2125w2  152 2112 2 5

Let w  0.

1 2  12

2 5 4 1 2

5

?

2

?

Let w  5. 1

5 (0)  2

1

1

 2 ✓

5

0 2 5

6

?

2

1

80  23 6 19 57 19

?

4  23 6 19 27 6 19 ✓ The solution set is {y|y  10.5}.

1

2  2 ✓

2 r 3

5. 2 r 3

 9  3

 9  9  3  9 2 r 3 3 2 r 2 3

12

29 GRAPH

Check:

A portion of a horizontal line is shown. It is part of the graph of y  1. 2. For all values of x less than 2, the y value is 1. For all values of x greater than or equal to 2, the y value is 0. That is, y  1 if x  2, and otherwise y  0. 3. 6x  9 6 4x  29 6x  9  4x 6 4x  29  4x 10x  9 6 29 10x  9  9 6 29  9 10x 6 20

 12 

132 2 (12)

r  18 Substitute 18, a number less than 18, and a number greater than 18.

Let r  18.

Let r  24.

2 (18) 3

2 (24) 3

?

 9  3 ?

12  9  3 3  3 ✓

?

 9  3 ?

16  9  3 7 3

The solution set is {r|r  18}.

20 10

x 6 2 The solution set is {x|x  2}. y  1 for those values of x for which the inequality is true; y  0 for those values of x for which the inequality is not true. Chapter 6

?

?

5 (5)  2

1. KEYSTROKES: Y CLEAR 6 X,T,␪,n TEST 5 4 X,T,␪,n 9 2nd

6

?

?

Graphing Calculator Investigation

10x 10

4(20)  23 6 19

4(1)23 6 19

The solution set is w ƒ w  4 .

Page 333

?

42  23 6 19 19 19 Let y  1.

5

5

Substitute 4, a number less than 4, and 5 a number greater than 4.

Let w  4.

42 4

4(10.5)  23 6 19

5

w4

Check:

7

y 7 10.5 Check: Substitute 10.5, a number less than 10.5, and a number greater than 10.5. Let y  10.5. Let y  20.

The solution set is {r|r  13}. 10.

Check for Understanding

1. To solve both the equation and the inequality, you first subtract 6 from each side and then divide each side by 5. In the equation, the equal sign does not change. In the inequality, the inequality symbol is reversed because you divided by a negative number. 2. Sample answer: 2x  4  2 3a. Distributive Property 3b. Add 12 to each side. 3c. Divide each side by 3. 4. 4y  23 6 19 4y  23  23 6 19  23 4y 6 42

12r 12

Let r  13.

Solving Multi-Step Inequalities

6-3

260

Let r  3. 2 (3) 3

?

 9  3 ?

2  9  3 7  3 ✓

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6.

7b  11 7b  11  7b 11 11  13 24

7 7 7 7 7

9b  13 9b  13  7b 2b  13 2b  13  13 2b

24 2

7

2b 2

8.

12 7 b 12  b is the same as b  12.

8 2

Substitute 12, a number less than 12, and a number greater than 12. Let b  12. ?

7(12)  11 7 9(12)  13 ?

3  5 (4)  3(4  1)  4 (2  4) ?

3  20  3 (5)  4 (2) ?

23  15  8 23  23 ✓ Let t  1.

?

7(10)  11 7 9(10)  13 ?

70  11 7 90  13 81 7 77 ✓ Let b  20.

?

3  5 (1)  3(1  1)  4 (2  1) ?

3  5  3 (2)  4 (1) ? 864 8 2 Let t  5.

?

7(20)  11 7 9(20)  13 ?

140  11 7 180  13 151 167 The solution set is {b|b  12}.

8 8

?

3  5 (5)  3 (5  1)  4 (2  5)

7 3(g  4) 7 3g  12 7 3g  12  5g 7 8g  12 7 8g  12  12 7 8g 7

2t 2

?

84  11 7 108  13 95 95 Let b  10.

5(g  4) 5g  20 5g  20  5g 20 20  12 8



4t 4  t is the same as t  4. Check: Substitute 4, a number less than 4, and a number greater than 4. Let t  4.

Check:

7.

3  5t  3 (t  1)  4 (2  t) 3  5t  3t  3  8  4t 3  5t  7t  5 3  5t  5t  7t  5  5t 3  2t  5 3  5  2t  5  5 8  2t

?

3  25  3 (6)  4 (3) ?

28  18  12 28  30 ✓ The solution set is {t|t  4}. 9. Let n  the number.

8g 8

two times is less three times Seven minus a number than the number plus thirty-two.

1 424 3 1 424 3 1 4424 43 1 424 3

1 7 g 1  g is the same as g  1. Check: Substitute 1, a number less than 1, and a number greater than 1. Let g  1.

7

?

5(1  4) 7 3(1  4) ?

5(3) 7 3(5) 15 15 Let g  4. ? 5(4  4) 7 3(4  4)



2n

1 44424 43 123

6

7  2n 7  2n  2n 7 7  32 25

6 6 6 6 6

3n  32 3n  32  2n 5n  32 5n  32  32 5n

25 5

6

5n 5

3n



14 424 43

32

5 6 n 5  n is the same as n  5. Check: Substitute 5, a number less than 5, and a number greater than 5. Let n  5. ? 7  2(5) 6 3(5)  32

?

5(0) 7 3(8) 0 7 24 ✓ Let g  4. ? 5(4  4) 7 3(4  4)

?

7  10 6 15  32 17 17 Let n  10.

?

5(8) 7 3(0) 40 0

?

7  2(10) 6 3(10)  32

The solution set is {g|g  1}.

?

7  20 6 30  32 27 2 Let n  2. ? 7  2(2) 6 3(2)  32 ?

7  4 6 6  32 3 6 38 ✓ The solution set is {n|n  5}.

261

Chapter 6

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10. Let s  the amount of sales. 22,000  0.05s 7 35,000 22,000  0.05s  22,000 7 35,000  22,000 0.05s 7 13,000 0.05s 0.05

7

2 

17.

2 

1 d2

(5) 5 7 (5)25 d 7 125 The solution set is {d|d  125}. w 8

18.

11a. 11b. 12a. 12b. 12c. 13.

Subtract 7 from each side. Multiply each side by 52. Multiply each side by 3 and change  to . Add 2m to each side. Multiply each side by 1 and change  to . Original inequality 4 (t  7)  2 (t  9) Distributive 4t  28  2t  18 Property 4t  28  2t  2t  18  2t Subtract 2t from each side. Simplify. 2t  28  18 Add 28 to each side. 2t  28  28  18  28 Simplify. 2t  46 

t  23 The solution set is {t|t  23}. 14.

5k  20 7 3k  12 5k  20  5k 7 3k  12  5k 20 7 8k  12 20  12 7 8k  12  12 8 7 8k 8 8

9q 9

3t 3

2a 2

Original inequality Distributive Property Add 5k to each side. Simplify. Add 12 to each side. Simplify.

30 9 10 3

1

or 33

5

1

6

18

 2

a  9 The solution set is {a|a  9}. 21. 9r  15  24  10r 9r  15  9r  24  10r  9r 15  24  r 15  24  24  r  24 9  r 9  r is the same as r  9. The solution set is {r|r  9}. 22. 13k  11 7 7k  37 13k  11  7k 7 7k  37  7k 6k  11 7 37 6k  11  11 7 37  11 6k 7 48

8k 8

6k 6

7

48 6

k 7 8 The solution set is {k|k  8}. 23.

9

2v  3 5 2v  3 (5) 5

7  (5)7

2v  3  35 2v  3  3  35  3 2v  38 2v 2

64 8



38 2

v  19 The solution set is {v|v  19}.

f 6 8 The solution set is {f|f  8}.

Chapter 6



20. 8a  2  10a  20 2a  2  20 2a  2  2  20  2 2a  18

 3

6

7 (8)7

The solution set is q|q  33 .

t3 The solution set is {t|t  3}. 16. 5  8f 7 59 5  8f  5 7 59  5 8f 7 64 8f 8

7 7

q

Divide each side by 8. Simplify. 1 7 k 1  k is the same as k  1. The solution set is {k|k  1}. Exercises 15–32 For checks, see students’ work. 15. 3t  6  3 3t  6  6  3  6 3t  9 7

 13  13 7 6  13

w 7 56 The solution set is {w|w  56}. 19. 7q  1  2q  29 9q  1  29 9q  1  1  29  1 9q  30

Divide each side by 2. Simplify.

5(k  4) 7 3 (k  4)

 13 7 6

w 8 w (8) 8

Practice and Apply

46 2

 2 6 23  2 d

w 8

2t 2

6 23

5 6 25

13,000 0.05

s 7 260,000 The salesperson would need more than $260,000 in sales to have an annual income greater than $35,000.

Pages 335–337

d 5

d 5

262

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24.

3a  8 2 3a  8 (2) 2

Since the inequality results in a statement that is always true, the solution set is {b|b is a real number.}. 31. 3.1v  1.4  1.3v  6.7 3.1v  1.4  1.3v  1.3v  6.7  1.3v 1.8v  1.4  6.7 1.8v  1.4  1.4  6.7  1.4 1.8v  8.1

6 10 6 (2)10

3a  8 6 20 3a  8  8 6 20  8 3a 6 12 3a 3

6

12 3

a 6 4 The solution set is {a|a  4}. 25.

3w  5 4 3w  5 (4) 4

1.8v 1.8

v  4.5 The solution set is {v|v  4.5}. 32. 0.3(d  2)  0.8d 7 4.4 0.3d  0.6  0.8d 7 4.4 0.5d  0.6 7 4.4 0.5d  0.6  0.6 7 4.4  0.6 0.5d 7 5

 2w  (4)2w

3w  5  8w 3w  5  3w  8w  3w 5  5w 5 5



5w 5

0.5d 0.5

1w 1  w is the same as w  1. The solution set is {w|w  1}. 26.

5b  8 3 5b  8 (3) 3

6 3b 6 (3)3b

6

4b 4

22 2

2 6 b

27.

28.

29.

30.

6

5 0.5

d 6 10 The solution set is {d|d  10}. 33. 4(y  1)  3(y  5)  3(y  1) 4y  4  3y  15  3y  3 y  19  3y  3 y  19  y  3y  3  y 19  2y  3 19  3  2y  3  3 22  2y

5b  8 6 9b 5b  8  5b 6 9b  5b 8 6 4b 8 4

8.1

 1.8



2y 2

11  y 11  y is the same as y  11. The solution set is { y|y  11}.

2  b is the same as b  2. The solution set is {b|b  2}. 7  3t  2(t  3)  2(1  t) 7  3t  2t  6  2  2t 7  3t  4t  8 7  3t  3t  4t  8  3t 7t8 78t88 1  t 1  t is the same as t  1. The solution set is {t|t  1}. 5(2h  6)  7(h  7)  4h 10h  30  7h  49  4h 3h  79  4h 3h  79  3h  4h  3h 79  h 79  h is the same as h  79. The solution set is {h|h  79}. 3y  4  2(y  3)  y 3y  4  2y  6  y 3y  4  3y  6 3y  4  3y  3y  6  3y 46 Since the inequality results in a false statement, the solution set is the empty set . 3  3(b  2) 6 13  3(b  6) 3  3b  6 6 13  3b  18 3b  9 6 3b  31 3b  9  3b 6 3b  31  3b 9 6 31

5

6

7

8

9

10 11 12 13

34. 5(x  4)  2(x  6)  5(x  1)  1 5x  20  2x  12  5x  5  1 3x  8  5x  4 3x  8  3x  5x  4  3x 8  2x  4 8  4  2x  4  4 4  2x 4 2



2x 2

2x 2  x is the same as x  2. The solution set is {x|x  2}. 4 3 2 1

0

1

2

3

4

Exercises 35–38 For checks, see students’ work. 35. Let n  the number. 1 n 8 1 n 8

 5  30

 5  5  30  5 1 n 8 1 (8) 8n

 35  (8)35

n  280 The solution set is {n|n  280}.

263

Chapter 6

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36. Let n  the number. 2 n 3 2 n 3

44.

 8 7 12

 8  8 7 12  8 2 n 3 3 2 n 2 3

12

132 24

n 7 6 The solution set is {n ƒ n 7 6}. 37. Let n  the number. 4n  9  n  21 4n  9  n  n  21  n 5n  9  21 5n  9  9  21  9 5n  30 5n 5



7

6

42.

91  95  88  s 4 274  s 4 274  s (4) 4

 92

520w 520

Chapter 6

4.375

 1.4375



1040 520

w2 The union worker can strike for no more than 2 weeks. 49. Let w  the number of weeks the worker could strike. [600  0.04(600) ] (52  w)  (600)52 (600  24)(52  w)  31,200 624(52  w)  31,200 32,448  624w  31,200 32,448  624w  32,448  31,200  32,448 624w  1248

 92  (4)92

274  s  368 274  s  274  368  274 s  94 To earn an A in math, Carmen must score at least 94 on the test. 43. Mercury is a solid until it reaches its melting point at 38°C. 5(F  32) 9

1

122

t  3.0 (to the nearest tenth) Nicholas can order 3 or fewer toppings. 48. Let w  the number of weeks the worker could strike. [500  0.04(500) ] (52  w)  (500)52 (500  20)(52  w)  26,000 520(52  w)  26,000 27,040  520w  26,000 27,040  520w  27,040  26,000  27,040 520w  1040

105 3

 92

25 2 25 or 2

1.4375t 1.4375

a 6 35 The solution set is {a ƒ a 6 35}. 41.

7

Keith should stay on the diet for more than 1 122 weeks. 46. Sample answers: 2x  5  2x  3; 2x  5  2x  3 47. Let t  the number of toppings Nicholas can order. 7.50  1.25t  0.15(7.50  1.25t)  13.00 7.50  1.25t  1.125  0.1875t  13.00 1.4375t  8.625  13.00 1.4375t  8.625  8.625  13.00  8.625 1.4375t  4.375

2n 2

91  95  88  s 4

195 2 (38)

w 7

30 5

17 7 n 17  n is the same as n  17. The solution set is {n ƒ n 6 17}. 39. 3a  15 6 90 40. 3a  15 6 90 3a  15  15 6 90  15 3a 6 105 3a 3

6

2w 2

n6 The solution is {n ƒ n  6}. 38. Let n  the number. 3(n  7) 7 5n  13 3n  21 7 5n  13 3n  21  3n 7 5n  13  3n 21 7 2n  13 21  13 7 2n  13  13 34 7 2n 34 2

12

6 38

F  32 6 68.4 F  32  32 6 68.4  32 F 6 36.4 Mercury is a solid for temperatures less than 36.4°F. 45. Let w  the number of weeks Keith should stay on the diet. 200  2w 6 175 200  2w  200 6 175  200 2w 6 25

7 4 7

5(F  32) 9 9 5(F  32) 5 9

624w 624



1248 624

w2 The number of weeks the worker can strike does not change if the worker makes $600 per week.

6 38

264

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50. Let w  the number of weeks the worker could strike. 150w  [500  0.04(500) ] (52  w)  (500)52 150w  (500  20)(52  w)  26,000 150w  520(52  w)  26,000 150w  27,040  520w  26,000 370w  27,040  26,000 370w  27,040  27,040  26,000  27,040 370w  1040  370w 370



55. C;



 1040 370

57.

6

8 2

CLEAR 13 X,T,␪,n TEST 6 7 X,T,␪,n

KEYSTROKES:

37 GRAPH y  1 for those values of x for which the inequality is true. Since y  1 for values of x less than or equal to 8, the solution set is {x|x  8}. 58.

CLEAR 2 ( 5 3 ( TEST

KEYSTROKES:

)

3

2nd

X,T,␪,n 2

X,T,␪,n

GRAPH 2 ) y  1 for those values of x for which the inequality is true. Since y  1 for values of x greater than 3, the solution set is {x|x  3}.

Page 337

Maintain Your Skills

59. Let m  the number of miles Mrs. Ludlow can drive. 0.12m  50

1

or 113 1

0.12m 0.12

The positive even integers less than 113 are 2, 4, 6, 8, and 10. So the sets of three consecutive positive even integers whose sum is less than 40 are 2, 4, and 6; 4, 6, and 8; 6, 8, and 10; 8, 10, and 12; 10, 12, and 14. 53. Inequalities can be used to describe the temperatures for which an element is a gas or a solid. Answers should include the following. • The inequality for temperatures in degrees Celsius for which bromine is a gas is 9 C 5

2t 2

11 2nd

n8 The positive odd integers less than or equal to 8 are 1, 3, 5, and 7. So the pairs of consecutive positive odd integers whose sum is no greater than 18 are 1 and 3, 3 and 5, 5 and 7, 7 and 9. 52. Let n  the first positive even integer. Then n  2 and n  4 represent the next two consecutive positive even integers, respectively. n  (n  2)  (n  4) 6 40 3n  6 6 40 3n  6  6 6 40  6 3n 6 34 3n 34 6 3 3 34 3

8t  (6t  10) 8t  6t  10 2t  10 2t  10  2t 10 10  2 8

9 GRAPH y  1 for those values of x for which the inequality is true. Since y  1 for values of x less than 2, the solution set is {x|x  2}.

16 2

n 6

6 6 6 6 6 6 6

t 6 4 The solution set is {t|t  4}. CLEAR 3 X,T,␪,n 56. KEYSTROKES: TEST 3 4 X,T,␪,n 7 2nd

w  2.8 (to the nearest tenth) The union worker can strike for up to 2.8 weeks. 51. Let n  the first positive odd integer. Then n  2 represents the next consecutive positive odd integer. n  1n  22  18 2n  2  18 2n  2  2  18  2 2n  16 2n 2

4t  2 4t  2 4t  2 4t  2  2t 2t  2 2t  2  2 2t

50

 0.12 2

m  4163 If Mrs. Ludlow is charged $0.12 per mile or any part of one mile, she can travel up to 416 mi without going over her budget. Exercises 60–62 For checks, see students’ work. 60. d  13  22 d  13  13  22  13 d9 The solution set is {d|d  9}.

 32 7 138.

• Sample answer: Scientists may use inequalities to describe the temperatures for which an element is a solid.

5

61.

y  5

54. D; To solve 9  13, first eliminate fractions by multiplying both sides of the inequality by 9.

6

8

t5 6 t55 6 t 6 The solution 5

265

7

6

7

9

10 11 12 13

3 35 8 set is {t|t < 8}. 8

9

10 11 12 13

Chapter 6

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62.

4 7 y7 47 7 y77 3 7 y 3  y is the same as y  3. The solution set is {y|y  3}. 8 7 6 5 4 3 2 1

69. First rewrite the equation so that the variables are on one side of the equation. 4x  7  2y 4x  2y  7  2y  2y 4x  2y  7 The equation is now in standard form where A  4, B  2, and C  7. This is a linear equation. 70. Since the term 2x2 has an exponent of 2, the equation cannot be written in the form Ax  By  C. Therefore, this is not a linear equation. 71. The equation can be written as x  0y  12. Therefore, it is a linear equation in standard form where A  1, B  0, and C  12. 72. 21x  22  3x  14x  52 2x  4  3x  4x  5 2x  4  x  5 2x  4  x  x  5  x 3x  4  5 3x  4  4  5  4 3x  9

0

y  y1  m(x  x1) y  (3)  2(x  1) y  3  2x  2 y  3  3  2x  2  3 y  2x  5 y  2x  2x  5  2x 2x  y  5 (1)(2x  y)  (1)(5) 2x  y  5 The standard form of the equation is 2x  y  5. The point-slope form of the equation is y  3  2(x  1). 64. y  y  m(x  x ) 1 1 63.

3x 3

2

y  (1)  3 x  (2)

x3 Check: 2(x  2)  3x  (4x  5) ? 2(3  2)  3(3)  [ 4(3)  5] ? 2(1)  9  (12  5) ? 297 22✓ The solution is 3. 73. 5t  7  t  3 5t  7  t  t  3  t 4t  7  3 4t  7  7  3  7 4t  10

2

y  1  3 (x  2)

1 22

3(y  1)  3 3 (x  2) 3y  3  2(x  2) 3y  3  2x  4 3y  3  3  2x  4  3 3y  2x  7 3y  2x  2x  7  2x 2x  3y  7 The standard form of the equation is 2x  3y  7. The point-slope form of the equation is 2

y  1  3 (x  2).

4t 4

y  y1  m(x  x1) y  6  0(x  3) y60 y6606 y6 The standard form of the equation is y  6. The point-slope form of the equation is y  6  0. 66. Let (3, 1)  (x1, y1) and (4, 6)  (x2, y2). 65.

Check:

2

 

321 0 1 2 3 4 5

 x1

75. 321 0 1 2 3 4 5

76. 654321 0 1 2

77.

y2  y1

 

321 0 1 2 3 4 5

 x1

3  (4) 1  (2) 7 3

78. 4321 0 1 2 3 4

79.

68. Let (0, 3)  (x1, y1) and ( 2, 5)  (x2, y2).

0 1 2 3 4 5 6 7 8

y2  y1

mx

2

80.

 x1

54321 0 1 2 3

5  3  0 8 or 4 2

 2 

Chapter 6

t  2.5 5t  7  t  3 ? 5(2.5)  7  2.5  3

74.

67. Let (2, 4)  (x1, y1) and (1, 3)  (x2, y2). 2

10 4

?

6  (1) 4  3 5 or 5 1

mx



12.5  7  5.5 5.5  5.5 ✓ The solution is 2.5.

y2  y1

mx

9

3

81. 54321 0 1 2 3

266

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12. For a compound statement connected by the word and to be true, both simple statements must be true. Since it is not true that 0  3, the compound statement is false.

82. 21 0 1 2 3 4 5 6

Page 338

Reading Mathematics

1. For a compound statement connected by the word or to be true, at least one of the simple statements must be true. Since it is true that a hexagon has six sides, the compound statement is true. 2. For a compound statement connected by the word and to be true, both simple statements must be true. It is false that a pentagon has six sides; it has five. Thus, the compound statement is false. 3. For a compound statement connected by the word and to be true, both simple statements must be true. It is true that a pentagon has five sides. It is also true that a hexagon has six sides. Thus, the compound statement is true. 4. For a compound statement connected by the word or to be true, at least one of the simple statements must be true. It is true that an octagon does not have seven sides; it has eight. Thus, the compound statement is true. 5. For a compound statement connected by the word or to be true, at least one of the simple statements must be true. In this case, it is not true that a pentagon has three sides; it has five. It is also not true that an octagon has ten sides; it has eight. Since neither simple statement is true, the compound statement is false. 6. For a compound statement connected by the word or to be true, at least one of the simple statements must be true. Since it is true that a square has four sides, the compound statement is true. 7. For a compound statement connected by the word or to be true, at least one of the simple statements must be true. In this case, it is not true that 5  4. It is also not true that 8  6. Since neither simple statement is true, the compound statement is false. 8. For a compound statement connected by the word and to be true, both simple statements must be true. Since it is not true that 1  0, the compound statement is false. 9. For a compound statement connected by the word and to be true, both simple statements must be true. In this case, it is true that 4  0. It is also true that 4  0. Thus, the compound statement is true. 10. For a compound statement connected by the word or to be true, at least one of the simple statements must be true. Since it is true that 0  0, the compound statement is true. 11. For a compound statement connected by the word or to be true, at least one of the simple statements must be true. It is true that 1  4. Thus, the compound statement is true.

Solving Compound Inequalities

6-4 Page 341

Check for Understanding

1. A compound inequality containing and is true if and only if both inequalities are true. A compound inequality containing or is true if and only if one of the inequalities is true. 2. 7  t  12 3. Sample answer: x  2 and x  3 4. Graph a  6. Graph a  2. Find the intersection. –3

–2

–1

0

1

2

3

4

5

6

7

–3

–2

–1

0

1

2

3

4

5

6

7

–3

–2

–1

0

1

2

3

4

5

6

7

5. Graph y  12. Graph y  9. Find the union. 5

6

7

8

9

10

11

12

13

14

15

5

6

7

8

9

10

11

12

13

14

15

5

6

7

8

9

10

11

12

13

14

15

6. 3 6 x  1 7. x  1 or x  5 8. and w  3 6 11 6 6 w3 w  3  3 6 11  3 63 6 w33 3 6 w w 6 8 The solution set is the intersection of the two graphs. Graph 3  w or w  3. Graph w  8. Find the intersection. 0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

The solution set is 5w|3 6 w 6 86.

267

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9.

n  7  5 n  7  7  5  7 n2

or

12. Let n  the number. 5 6 3n  7 6 17 First express 5  3n  7  17 using and. Then solve each inequality. and 3n  7 6 17 5 6 3n  7 3n  7  7 6 17  7 5  7 6 3n  7  7 12 6 3n 3n 6 24

n71 n7717 n8

The solution set is the union of the two graphs. Graph n  2. Graph n  8. Find the union.

10.

0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

12 3

6

12 3

20 4.5

z 6 4 The solution set is the union of the two graphs. Graph z  4. Graph z  1. Find the union. –2

–1

0

1

2

3

4

5

6

7

–1

0

1

2

3

4

5

6

7

8

–2

–1

0

1

2

3

4

5

6

7

8

–6

–5

–4

–3

–2

–1

0

1

2

3

–7

–6

–5

–4

–3

–2

–1

0

1

2

3

–7

–6

–5

–4

–3

–2

–1

0

1

2

3

6

24 3

4.5x 4.5

4.5x 4.5

30

 4.5

Practice and Apply

14. Graph x  5. Graph x  9. Find the intersection. 0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

15. Graph s  7. Graph s  0. Find the intersection.

The solution set is {x|4 6 x  1}.

Chapter 6



Pages 342–343

Notice that the graph of z  4 contains every point in the graph of z  1. So, the union is the graph of z  4. The solution set is 5z|z 6 46. 11. 8 6 x  4  3 First express 8  x  4  3 using and. Then solve each inequality. and x  4  3 8 6 x  4 x  4  4  3  4 8  4 6 x  4  4 4 6 x x1 The solution set is the intersection of the two graphs. Graph 4  x or x  4. Graph x  1. Find the intersection. –7

3n 3

4.4  x x  6.6 The lengths of the stretched spring will be between about 4.44 and 6.67 in., inclusive, (to the nearest hundredth).

8

–2

3n 3

4 6 n n 6 8 The solution set is the intersection of 4  n and n  8. The solution set is 5n|4 6 n 6 86. 13. If forces are between 20 and 30 lb, inclusive, then 20  F  30. Since F  4.5x, 20  4.5x  30. Express 20  4.5x  30 using and. Then solve each inequality. 20  4.5x and 4.5x  30

The solution set is {n|n  2 or n  8}. or z1 3z  1 6 13 3z  1  1 6 13  1 3z 6 12 3z 3

6

268

–10 –9

–8

–7

–6

–5

–4

–3

–2

–1

0

–10 –9

–8

–7

–6

–5

–4

–3

–2

–1

0

–10 –9

–8

–7

–6

–5

–4

–3

–2

–1

0

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16. Graph r  6. Graph r  6. Find the union.

28.

0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

17. Graph m  4. Graph m  6. Find the union. –10

–8

–6

–4

–2

0

2

4

6

8

10

–10

–8

–6

–4

–2

0

2

4

6

8

10

–10

–8

–6

–4

–2

0

2

4

6

8

10

29.

18. First express 7  d  11 using and. 7 6 d and d 6 11 Graph 7 6 d or d 7 7. Graph d  11. Find the intersection. 2

3

4

5

6

7

8

9

10

11

12

2

3

4

5

6

7

8

9

10

11

12

2

3

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5

6

7

8

9

10

11

12

30.

19. First express 1  g  3 using and. 1  g and g 6 3 Graph 1  g or g  1. Graph g  3. Find the intersection. –5

–5

–5

20. 21. 22. 23. 24. 25. 26. 27.

–4

–4

–4

–3

–3

–3

–2

–2

–2

–1

–1

–1

0

0

0

1

1

1

2

2

2

3

3

3

4

4

4

and k  2  18 k  2 7 12 k  2  2  18  2 k  2  2 7 12  2 k 7 10 k  16 The solution set is the intersection of the two graphs. Graph k  10. Graph k  16. Find the intersection. 8

9

10

11

12

13

14

15

16

17

18

8

9

10

11

12

13

14

15

16

17

18

8

9

10

11

12

13

14

15

16

17

18

The solution set is 5k|10 6 k  166. and f  9  4 f83 f  9  9  4  9 f8838 f  5 f  13 The solution set is the intersection of the two graphs. Graph f  5. Graph f  13. Find the intersection. –14 –13 –12 –11 –10

–9

–8

–7

–6

–5

–4

–14 –13 –12 –11 –10

–9

–8

–7

–6

–5

–4

–14 –13 –12 –11 –10

–9

–8

–7

–6

–5

–4

The solution set is 5f|13  f  56. or d41 d4 7 3 d4414 d44 7 34 d 7 7 d5 The solution set is the union of the two graphs. Graph d  7. Graph d  5. Find the union.

5 0

1

2

3

4

5

6

7

8

9

10

0

1

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8

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10

5

5

2  x  2 7 6 x 6 3 x  12 or x 7 15 x  7 or x  6 x  0 or x  4 x  2 or x 7 5 158  w  206 t  18 or t  22

The solution set is {d|d  5 or d 7 7}.

269

Chapter 6

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31.

or h3 6 2 h  10 6 21 h33 6 23 h  10  10 6 21  10 h 6 11 h 6 1 The solution set is the union of the two graphs. Graph h  11. Graph h  1. Find the union. –16 –14 –12 –10

–8

–6

–4

–2

0

2

4

–16 –14 –12 –10

–8

–6

–4

–2

0

2

4

–16 –14 –12 –10

–8

–6

–4

–2

0

2

34.

3t 3

6

2x 2

2x 2

6

1

2

3

4

5

6

7

8

9

3

4

5

6

7

8

9

10

0

1

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9

10

The solution set is 5x|3 6 x 6 96. 33. First express 4  2y  2  10 using and. and 2y  2 6 10 4 6 2y  2 2y  2  2 6 10  2 4  2 6 2y  2  2 6 6 2y 2y 6 12 2y 2

6

36.

12 2

3 6 y y 6 6 The solution set is the intersection of the two graphs. Graph 3  y or y  3. Graph y  6. Find the intersection. 0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

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7

8

9

10

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

6

3q 3

3q 3

6

18 3

–3

–2

–1

0

1

2

3

4

5

6

7

–3

–2

–1

0

1

2

3

4

5

6

7

–3

–2

–1

0

1

2

3

4

5

6

7

The solution set is 5q|1 6 q 6 66. or x  4 1  x  3 (1) (x)  (1) (4) 1  x  1  3  1 x4 x4 The solution set is the union of the two graphs. Graph x  4. Graph x  4. Find the union. –5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

The solution set is 5x|x is a real number.6.

The solution set is 5y|3 6 y 6 66. Chapter 6

6

2

1 6 q q 6 6 The solution set is the intersection of the two graphs. Graph 1  q or q  1. Graph q  6. Find the intersection.

10

2

2y 2

2t 2

–5

3 3

1

6

12 3

Since the graphs do not intersect, the solution set is the empty set . 35. and 5  3q 7 13 8 7 5  3q 5  3q  5 7 13  5 8  5 7 5  3q  5 3 7 3q 3q 7 18

18 2

0

6 2

2t  6  12 2t  6  6  12  6 2t  6

4

3 6 x x 6 9 The solution set is the intersection of the two graphs. Graph 3  x or x  3. Graph x  9. Find the intersection. 0



and

t4 t3 The solution set is the intersection of the two graphs. Graph t  4. Graph t  3. Find the intersection.

Notice that the graph of h  1 contains every point in the graph of h  11. So, the union is the graph of h  1. The solution set is 5h|h 6 16. 32. First express 3  2x  3  15 using and. and 2x  3 6 15 3 6 2x  3 2x  3  3 6 15  3 3  3 6 2x  3  3 6 6 2x 2x 6 18 6 2

3t  7  5 3t  7  7  5  7 3t  12

270

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3n  11  13

37.

3n  12

or

3n 3

3n  11  11  13  11 3n  2 3n 3



Graph g  3. Graph 12  g or g  12. Find the intersection.

12 3

n4

2

3

n

–6

–4

–2

0

2

4

6

8

10

12

14

–6

–4

–2

0

2

4

6

8

10

12

14

–6

–4

–2

0

2

4

6

8

10

12

14

2 3

The solution set is the union of the two graphs. 2

Graph n  3. Graph n  4. Find the union. 0

1

2

3

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5

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7

8

9

10

0

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7

8

9

Since the graphs do not intersect, the solution set is the empty set . 40. 4c 6 2c  10 or 3c 6 12 2c 6 10 2c 2

10

2p  2  4p  8

4p  8  3p  3

and

4p  8  3p  3p  3  3p

2  2p  8

p  8  3

2  8  2p  8  8

p  8  8  3  8

6  2p 6 2



3p

2

3

4

5

6

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8

9

10

0

1

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10

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

3g  12  6  g 3g  12  g  6  g  g

and

6  g  3g  18

2g  12  6

6  2g  18

2g  12  12  6  12

6  18  2g  18  18 24  2g 

6 0.5

3b  16  b 6 8  b  b 2b  16 6 8 2b  16  16 6 8  16 2b 6 24 6

24 2

b 6 12 The solution set is the union of the two graphs. Graph b  12. Graph b  12. Find the union.

6  g  g  3g  18  g

24 2

7

2b 2

The solution set is 5p|3  p  56. 39. First express 3g  12  6  g  3g  18 using and.

g  3

–4

b 7 12

1

6 2

–5

0.5b 0.5

0



c 7 4

10 2

The solution set is {c|c 6 5 or c 7 4}. 41. 0.5b 7 6 or 3b  16 6 8  b

The solution set is the intersection of the two graphs. Graph 3  p or p  3. Graph p  5. Find the intersection.

2g 2

12 3

p5

2p 2

2g  6

6

7

c 6 5 The solution set is the union of the two graphs. Graph c  5. Graph c  4. Find the union.

Notice that the graph of n  4 contains every 2 point in the graph of n  3. So, the union is the graph of n  4. The solution set is 5n|n  46. 38. First express 2p  2  4p  8  3p  3 using and. 2p  2  2p  4p  8  2p

3c 3

4c  2c 6 2c  10  2c

–18 –16 –14 –12 –10

–8

–6

–4

–2

0

2

–18 –16 –14 –12 –10

–8

–6

–4

–2

0

2

–18 –16 –14 –12 –10

–8

–6

–4

–2

0

2

The solution set is {b|b 6 12 or b 7 12}.

2g 2

12  g

The solution set is the intersection of the two graphs.

271

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47. Let p  the price of a color printer. Then p  30 represents the price of a printer after a $30 rebate. If 175  p  260, then 175  p and p  260. and p  260 175  p 175  30  p  30 p  30  260  30 145  p  30 p  30  230 So, 145  p  30  230. After the rebate, Luisana can expect to spend between $145 and $230, inclusive. 48. Let s  the total amount of sales needed to earn a prize in category D. Then s  70 represents the amount of sales Rashid still needs after selling $70 worth of chocolates. If 121  s  180, then 121  s and s  180. 121  s and s  180 s  70  180  70 121  70  s  70 51  s  70 s  70  110 So, 51  s  70  110. Rashid must have additional sales between $51 and $110, inclusive, to earn a category D prize. 49a. Graph x  5 or x  8.

42. Let n  the number. 5  n  8  14 First express 5  n  8  14 using and. and n  8  14 5n8 n  8  8  14  8 58n 13  n n  22 The solution set is the intersection of 13  n and n  22. The solution set is 5n|13  n  226. 43. Let n  the number. 8 6 3n  4 6 10 First express 8  3n  4  10 using and. and 3n  4 6 10 8 6 3n  4 8  4 6 3n  4  4 3n  4  4 6 10  4 12 6 3n 3n 6 6 12 3

6

3n 3

3n 3

6

6 3

4 6 n n 6 2 The solution set is the intersection of 4  n and n  2. The solution set is 5n|4 6 n 6 26. 44. Let n  the number. 5n 7 35 or 5n 6 10 5n 5

6

35 5

5n 5

7

10 5

n 6 7 n 7 2 The solution is the union of n  7 and n  2. The solution set is 5n|n 6 7 or n 7 26. 45. Let n  the number. 1

1

First express 0 6 2n  1 using and. 1

(2)0 6

1 (2) 2n

and

1 n 2 1 (2) 2n

3

4

5

6

7

8

9

10

11

12

2

3

4

5

6

7

8

9

10

11

12

Find the points not graphed. These values make the given compound inequality false. So, x  5 and x  8. This compound inequality may also be expressed as 5  x  8. 49b. Graph x  6 and x  1.

0 6 2n  1 0 6 2n

2

1  (2)1

0 6 n n2 The solution set is the intersection of 0  n and n  2. The solution set is 5n|0 6 n  26. 46. Let h  the number of hours an adult sleeps. Then 0.20 h represents the number of hours an adult spends in REM sleep. If 7  h  8 then 7  h and h  8. and h8 7h (0.20)7  (0.20)h (0.20)h  (0.20)8 1.4  0.20h 0.20h  1.6 So, 1.4  0.20h  1.6. An adult spends between 1.4 and 1.6 h, inclusive, in REM sleep.

–1

0

1

2

3

4

5

6

7

8

9

–1

0

1

2

3

4

5

6

7

8

9

Find the points not graphed. These values make the given compound inequality false. So, x  1 or x  6. 50. Let h  the number of hertz heard by humans. 20  h  20,000 Let d  the number of hertz heard by dogs. 15  d  50,000 51. Graph 20  h  20,000 Graph 15  h  50,000.

20

15

20,000

50,000

Find the union. Notice that the graph of 15  h  50,000 contains every point in the graph of 20  h  20,000. So, the union is the graph of 15  h  50,000. Find the intersection. The two graphs intersect for values of x between 20 and 20,000, inclusive. So, the intersection is the graph of 20  h  20,000. Chapter 6

272

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52. Of the number of hertz a dog can hear, humans cannot hear sounds between 15 and 20 hertz, including 15 hertz, or sounds between 20,000 and 50,000 hertz, including 50,000 hertz. So, 15  h  20 or 20,000  h  50,000. 53. Let a  the altitude in miles. Sample answers: troposphere: a  10 stratosphere: 10  a  30 mesosphere: 30  a  50 thermosphere: 50  a  400 exosphere: a  400 54. The tax table gives intervals of income and how much a taxpayer with taxable income in each interval must pay in taxes. Each interval can be expressed as compound inequalities. Answers should include the following. • The incomes are in $50 intervals. • 41,100  x  41,150 represents the possible incomes of a head of a household paying $7024 in taxes. 55. A; Let c  the number of cups of cooked tomatoes c made from ten lb of fresh tomatoes. Then 10 represents the number of cups of cooked tomatoes made from one lb of fresh tomatoes. If 10  c  15, then 10  c and c  15. 10 6 c and c 6 15 10 10

6

1 6

c 10 c 10

So, 1 6

c 10 c 10 c 10

6 6

15 10 3 2

Page 344

1.40r 1.40

18d 18

7v 7

61.

1

or 12

6 (13)13

3

183 2138b2 6 183 29



k(3) 3

k 8

Therefore, y  3x. Use the direct variation equation to find x when y  6.

2

8

y  3x 8

6  3x

138 2 6  138 2 83 x

y  1 for those values of x for which the compound inequality is true. Since y  1 for values of x that are either less than 6 or greater than 1, the solution set is {x|x  6 or x  1}. CLEAR X,T,␪,n 57b. KEYSTROKES: 3

X,T,␪,n

6 13

8b 7 9

8 3 8 3

GRAPH

TEST

t 13 t (13) 13

b 6 24 The solution set is {b|b  24}. 63. Write a direct variation equation that relates x and y. Find the value of k. y  kx 8  k(3)

1

56. B; First express 7  x  2  4 using and. 7 6 x  2 and x2 6 4 7  2 6 x  2  2 x22 6 42 9 6 x x 6 2 The solution set is 9  x and x  2, which can be expressed as 9  x  2. CLEAR X,T,␪,n 57a. KEYSTROKES: 4

2nd

91 7

7

t 6 169 The solution set is {t|t  169}.

Between 1 and cups of cooked tomatoes are made from one lb of fresh tomatoes.

X,T,␪,n

90

 18

v 7 13 The solution set is {v|v  13}.

62.

5 2 2nd TEST 3 3 TEST 4 2nd

800,000 1.40

d5 The solution set is {d|d  5}. 60. 7v 6 91

6 12.

TEST



r  571,428.57 (to the nearest hundredth) The school must raise at least $571,428.57 from other sources. Examples 59–62 For checks, see students’ work. 59. 18d  90

1 12

2nd

Maintain Your Skills

58. Let r  the amount the school must raise from other sources. The 0.40 r represents the amount promised by the corporation. r  0.40r  800,000 1.40r  800,000

9 4

x

9

Therefore, x  4 or 2.25 when y  6.

6 5 2nd 1 TEST TEST 4 4 GRAPH 6 2nd

y  1 for those values of x for which the compound inequality is true. Since y  1 for values of x between 2 and 8, inclusive, the solution set is {x|2  x  8}.

273

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76. 3.5 is three and one-half units from zero in the negative direction. |3.5|  3.5 77. |12  6|  |6| 6 is six units from zero in the positive direction. |12  6|  |6|  6 78. |5  9|  |4| 4 is four units from zero in the negative direction. |5  9|  |4|  4 79. |20  21|  |1| 1 is one unit from zero in the negative direction. |20  21|  |1|  1 80. |3  18|  |15| 15 is fifteen units from zero in the negative direction. |3  18|  |15|  15

64. Write a direct variation equation that relates x and y. Find the value of k. y  kx 2.5  k(0.5) 2.5 0.5

65.

66.

67.

68.



k(0.5) 0.5

5k Therefore, y  5x. Use the direct variation equation to find y when x  20. y  5x y  5(20) y  100 Therefore, y  100 when x  20. The relation, as a set of ordered pairs, is {(6, 0), (3, 5), (2, 2), (3, 3)}. The domain of this relation is {3, 2, 6}. The range is {2, 0, 3, 5}. Exchange x and y in each ordered pair to write the inverse relation. The inverse of this relation is {(0, 6), (5, 3), (2, 2), (3,3)}. The relation, as a set of ordered pairs, is {(5, 2), (3, 1), (2, 2), (1, 7)}. The domain of this relation is {3, 1, 2, 5}. The range is {1, 2, 7}. Exchange x and y in each ordered pair to write the inverse relation. The inverse of this relation is {(2, 5), (1, 3), (2, 2), (7, 1)}. The relation, as a set of ordered pairs, is {(3, 4), (3, 2), (2, 9), (5, 4), (5, 8), (7, 2)}. The domain of this relation is {7, 2, 3, 5}. The range is {2, 4, 8, 9}. Exchange x and y in each ordered pair to write the inverse relation. The inverse of this relation is {(4, 3), (2, 3), (9, 2), (4, 5), (8, 5), (2, 7)}. There are 6 possible outcomes, 4 are successes and 2 are failures. So, the odds of rolling a number greater than 2 4 2 are 2 or 1 or 2:1.

Page 344 1.

4b 4

1 22

10

Let b  10. ? 5  4(10) 7 23 ? 5  40 7 23 35 23 The solution set is {b|b  7}. 1 n 2

 3  5

 3  3  5  3 1 n 2 1 (2) 2n

 8  (2) (8)

n  16 Check: Substitute 16, a number less than 16, and a number greater than 16. Let n  16. Let n  20.

1

1 2134 2  12 134 2

7 6 7

1 (16) 2

3 8

?

 3  5 ?

1 (20) 2

8  3  5 5  5 ✓ Let n  0.

73. 7 is seven units from zero in the negative direction. |7|  7 74. 10 is ten units from zero in the positive direction. |10|  10 75. 1 is one unit from zero in the negative direction. |1|  1

Chapter 6

1 n 2

2.

71. 100(4.7)  470



28 4

b 6 7 Substitute 7, a number less than 7, and a number greater than 7. Let b  7. Let b  1. ? ? 5  4(7) 7 23 5  4(1) 7 23 ? ? 5  28 7 23 5  4 7 23 23 23 1 7 23 ✓

70. 6 5  30 or 3 72. 12

6

Check:

69. There are 6 possible outcomes, 5 are successes and 1 is a failure. So, the odds of rolling a number that is not a 3 5 are 1 or 5:1. 5

Practice Quiz 2

5  4b 7 23 5  4b  5 7 23  5 4b 7 28

1 (0) 2

?

 3  5 ?

0  3  5 3  5 ✓ The solution set is {n|n  16}.

274

?

 3  5 ?

10  3  5 7 5

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3.

3(t  6) 3t  18 3t  18  18 3t

6 6 6 6

9 9 9  18 9

3t 3

6

9 3

t 6 3 Substitute 3, a number less than and a number greater than 3. Let t  3. Let t  10. ? ? 3(10  6) 6 3(3  6) 6 9 ? ? 3(3) 6 9 3(4) 6 99 12 6 Let t  4. ? 3(4  6) 6 9 ? 3(10) 6 9 30  9 The solution set is {t|t  3}. 4. 9x  2 7 20 9x  2  2 7 20  2 9x 7 18 Check:

9x 9

7

6.

(3)a 6

3,

9 9 9✓

? 2(15)  15 3 ? 30  15 6 3 ? 45 6 3

? 2(18)  15 3 ? 36  15 6 3 ? 51 6 3

15 6

18 6

15

18

15

18

15  15

18 6 17 ✓

Let a  12. ? 2(12)  15 3 ? 24  15 6 3 ? 39 6 3

12 6 12 12

12  13 The solution set is {a|a  15}. 7. x2 6 7 and x2 7 5 x22 6 72 x22 7 52 x 6 9 x 7 3 The solution set is the intersection of the two graphs. Graph x  9. Graph x  3. Find the intersection.

18 9



2a  15 3 2a  15 (3) 3

3a 6 2a  15 3a  2a 6 2a  15  2a a 6 15 Check: Substitute 15, a number less than 15, and a number greater than 15. Let a  15. Let a  18.

x 7 2 Check: Substitute 2, a number less than 2, and a number greater than 2. Let x  2. Let x  3. ? ? 9(2)  2 7 20 9(3)  2 7 20 ? ? 18  2 7 20 27  2 7 20 20 20 25 20 Let x  5. ? 9(5)  2 7 20 ? 45  2 7 20 47 7 20 ✓ The solution set is {x|x  2}. 5. 2m  5  4m  1 2m  5  2m  4m  1  2m 5  2m  1 5  1  2m  1  1 6  2m 6 2

a 6

8.

2m 2

3m 3  m is the same as m  3. Check: Substitute 3, a number less than 3, and a number greater than 3. Let m  3. Let m  1. ? ? 2(3)  5  4(3)  1 2(1)  5  4(1)  1 ? ? 6  5  12  1 2541 11  11 ✓ 7 3 Let m  6. ? 2(6)  5  4(6)  1 ? 12  5  24  1 17  23 ✓ The solution set is {m|m  3}.

0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

The solution set is {x|3  x  9}. 2b  5  1 or b  4  4 2b  5  5  1  5 b  4  4  4  4 2b  6 b0 2b 2



6 2

b  3 The solution set is the union of the two graphs. Graph b  3. Graph b  0. Find the union. –7

–6

–5

–4

–3

–2

–1

0

1

2

3

–7

–6

–5

–4

–3

–2

–1

0

1

2

3

–7

–6

–5

–4

–3

–2

–1

0

1

2

3

The solution set is {b|b  3 or b  0}.

275

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9.

4m  5 7 7 4m  5  5 7 7  5 4m 7 12 4m 4

7

or

4m  5 6 9 4m  5  5 6 9  5 4m 6 4

12 4

4m 4

6

So, the absolute error is less than 6 sec when the student’s estimated time is greater than 54 s and less than 66 s. 5. See students’ work.

4 4

m 6 1

m 7 3

The solution set is the union of the two graphs. Graph m  3. Graph m  1. Find the union.

6-5

Solving Open Sentences Involving Absolute Value

Pages 348–349

10.

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

The solution set is {m|m  1 or m  3}. a4 6 1 or a2 7 1 a44 6 14 a22 7 12 a 6 5 a 7 1 The solution set is the intersection of the two graphs. Graph a  5. Graph a  1. Find the intersection. –3

–2

–1

0

1

2

3

4

5

6

7

–3

–2

–1

0

1

2

3

4

5

6

7

–3

–2

–1

0

1

2

3

4

5

6

7

5 432 1 0 1 2 3 4 5

3. Leslie is correct. You need to consider the case when the value inside the absolute value symbols is positive and the case when the value inside the absolute value symbols is negative. So x  3  2 or x  3  2. 4. a; Write |k|  3 as k  3 and k  3. The solution set is {k| 3  k  3}. The graph of this solution set is graph a. 5. c; Write |x  4|  2 as x  4  2 or x  4  2. Case 1: Case 2: x4 7 2 x  4 6 2 x44 7 24 x  4  4 6 2  4 x 7 6 x 6 2 The solution set is {x|x  2 or x  6}. The graph of this solution set is graph c. 6. The difference between the guess and the actual number, 832, is within 46. This statement can be expressed as |g  832|  46, where g represents Amanda’s guess. 7. Write |r  3|  10 as r  3  10 or r  3  10. Case 1: Case 2: r  3  10 r  3  10 r  3  3  10  3 r  3  3  10  3 r7 r  13 The solution set is {13, 7}.

The solution set is {a|1  a  5}.

Page 347

Algebra Activity

1. See students’ work. 2. A negative error indicates that the time guessed was less than 1 min. A positive error indicates that the time guessed was more than 1 min. 3. If the absolute error is 6, then the error is either 6 or 6. Let t  the student’s time. Then t  60 represents the error. So, either t  60  6 or t  60  6. t  60  6 t  60  60  6  60 t  66

or

t  60  6 t  60  60  6  60 t  54

1412108642 0 2 4 6 8

If the absolute error is 6, the student’s estimated time is 54 s or 66 s. 4. Let t  the student’s estimated time. If the error is represented by t  60, then the absolute error is represented by |t  60|. The absolute error is less than 6 sec when |t  60|  6. Write |t  60|  6 as t  60  6 and t  60  6. Case 1: Case 2: t  60 6 6 t  60 7 6 t  60  60 6 6  60 t  60  60 7 6  60 t 6 66 t 7 54 Chapter 6

Check for Understanding

1. |x  2|  6 means that the distance between x and 2 is greater than 6 units. |x  2|  6 means that the distance between x and 2 is less than 6 units. The solution of |x  2|  6 includes all values that are less than 4 or greater than 8. The solution of |x  2|  6 includes all values that are greater than 4 and less than 8. 2. Sample answer: |x|  2

8. Write |c  2|  6 as c  2  6 and c  2  6. Case 1: Case 2: c2 6 6 c  2 7 6 c22 6 62 c  2  2 7 6  2 c 6 8 c 7 4 The solution set is {c|4  c  8}. 10864 2 0 2 4 6 8 10

276

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9. Write |10  w|  15 as 10  w  15 or 10  w  15. Case 1: Case 2:

Pages 349–351

10  w 7 15

10  w 6 15

10  w  10 7 15  10

10  w  10 6 15  10

w 7 5

w 6 25

(1) (w) 6 (1)5

(1)(w) 6 (1)(25)

w 6 5

w 7 25

The solution set is {w|w  5 or w  25}. 15105 0 5 10 15 20 25 30 35

10. Write |2g  5|  7 as 2g  5  7 or 2g  5  7. Case 1: Case 2: 2g  5  7 2g  5  7 2g  5  5  7  5 2g  5  5  7  5 2g  2 2g  12 2g 2

2

2g 2

2



12 2

g1 g  6 The solution set is {g|g  6 or g  1}. 8 765 4 321 0 1 2

2x 2

11. Find the point that is the same distance from 2 as the distance from 4. The midpoint between 2 and 4 is 1. 3 units

3 units

17.

–3

–2

–1

0

1

2

3

4

5

6

7

So, an open sentence is |x  1|  3. 12. Find the point that is the same distance from 4 as the distance from 12. The midpoint between 4 and 12 is 8. 4 units

18.

4 units

19.

3

4

5

6

7

8

9

10

11

12

13

The distance from 8 to any point on the graph is greater than 4 units. So, an open sentence is |x  8|  4. 13. The difference between the actual diameter and 1.5 cm is within 0.001 cm. This statement can be expressed as |d  1.5|  0.001, where d represents the actual diameter. Write |d  1.5|  0.001 as d  1.5  0.001 and d  1.5  0.001. Case 1: Case 2: d  1.5  0.001

d  1.5  0.001

d  1.5  1.5  0.001  1.5

d  1.5  1.5  0.001  1.5

d  1.501

Practice and Apply

14. c; Write |x  5|  3 as x  5  3 and x  5  3. Case 1: Case 2: x53 x  5  3 x5535 x  5  5  3  5 x  2 x  8 The solution set is {x|8  x  2}. The graph of this solution set is graph c. 15. f; Write |x  4|  4 as x  4  4 or x  4  4. Case 1: Case 2: x4 7 4 x  4 6 4 x44 7 44 x  4  4 6 4  4 x 7 8 x 6 0 The solution set is {x|x  0 or x  8}. The graph of this solution set is graph f. 16. a; write |2x  8|  6 as 2x  8  6 or 2x  8  6. Case 1: Case 2: 2x  8  6 2x  8  6 2x  8  8  6  8 2x  8  8  6  8 2x  14 2x  2

20.

21.

d  1.499

22.

The solution set is {d|1.499  d  1.501}. The diameter of the bolts must be between 1.499 and 1.501 cm, inclusively.

277



14 2

2x 2

2

2

x7 x1 The solution set is {1, 7}. The graph of this solution set is graph a. b; Write |x  3|  1 as x  3  1 or x  3  1. Case 1: Case 2: x  3  1 x31 x  3  3  1  3 x3313 x  4 x  2 The solution set is {x|x is a real number.}. The graph of this solution set is graph b. e; Write |x|  2 as x  2 and x  2. The solution set is {x| 2  x  2}. The graph of this solution set is graph e. d; Write |8  x|  2 as 8  x  2 or 8  x  2. Case 1: Case 2: 8x2 8  x  2 8x828 8  x  8  2  8 x  6 x  10 (1)(x)  (1) (6) (1) (x)  (1) (10) x6 x  10 The solution set is {6, 10}. The graph of this solution set is graph d. The difference between the pH and 7.3 must be within 0.002. This statement can be expressed as |p  7.3|  0.002, where p represents the pH of the eye solution. The difference between the temperature and 38 should be within 1.5. This statement can be expressed as |t  38|  1.5, where t represents the temperature inside the refrigerator. The difference between the score and 98 was within 6. This statement can be expressed as |s  98|  6, where s represents Ramona’s bowling score.

Chapter 6

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The solution set is {t| 10 6 t 6 6}.

23. The difference between the speed and 55 should be within 3. This statement can be expressed as |s  55|  3, where s represents the speed of the car. 24. Write |x  5|  8 as x  5  8 or x  5  8. Case 1: Case 2: x58 x  5  8 x5585 x  5  5  8  5 x  13 x  3 The solution set is {3, 13}.

10 9876 5 4 32 1 0

30. Write |v  3| 7 1 as v  3 7 1 or v  3 6 1. Case 1: Case 2: v3 7 1 v  3 6 1 v33 7 13 v  3  3 6 1  3 v 7 2 v 6 4 The solution set is {v 0v 6 4 or v 7 2}. 8 7654 321 0 1 2

31. Write 0w  6 0  3 as w  6  3 or w  6  3. Case 1: Case 2: w63 w  6  3 w6636 w  6  6  3  6 w9 w3 The solution set is {w 0w  3 or w  9}.

642 0 2 4 6 8 10 12 14

25. Write |b  9|  2 as b  9  2 or b  9  2. Case 1: Case 2: b92 b  9  2 b9929 b  9  9  2  9 b  7 b  11 The solution set is {11, 7}.

0 1 2 3 4 5 6 7 8 9 10

1312111098 765 4 3

32. Write |3s  2| 7 7 as 3s  2 7 7 or 3s  2 6 7. Case 1: Case 2: 3s  2 7 7 3s  2 6 7 3s  2  2 7 7  2 3s  2  2 6 7  2 3s 7 9 3s 6 5

26. Write 02p  3 0  17 as 2p  3  17 or 2p  3  17. Case 1: Case 2: 2p  3  17 2p  3  17 2p  3  3  17  3 2p  3  3  17  3 2p  20 2p  14 2p 2



20 2

2p 2

p  10 The solution set is {7, 10}.



3s 3

14 2

20 5

c4 The solution set is {0.8, 4}.

3s 3

5 3

6

2

5 432 1 0 1 2 3 4 5

33. Write |3k  4|  8 as 3k  4  8 or 3k  4  8. Case 1: Case 2: 3k  4  8 3k  4  8 3k  4  4  8  4 3k  4  4  8  4 3k  4 3k  12

27. Write 05c  8 0  12 as 5c  8  12 or 5c  8  12. Case 1: Case 2: 5c  8  12 5c  8  12 5c  8  8  12  8 5c  8  8  12  8 5c  20 5c  4 

9 3

s 7 3 s 6 13 The solution set is {s|s is a real number.}.

p  7

8 642 0 2 4 6 8 10 12

5c 5

7

5c 5



c

4 5 4 5

3k 3

4

3k 3

3 1

k  13

5

The solution set is k|k  4 or k 

or 0.8



12 3

k  4 1 13

6.

5 432 1 0 1 2 3 4 5 5 4321 0 1 2 3 4 5

34. Write |2n  1| 6 9 as 2n  1 6 9 and 2n  1 7 9. Case 1: Case 2: 2n  1 6 9 2n  1 7 9 2n  1  1 6 9  1 2n  1  1 7 9  1 2n 6 8 2n 7 10

28. Write 0z  2 0  5 as z  2  5 and z  2  5. Case 1: Case 2: z25 z  2  5 z2252 z  2  2  5  2 z7 z  3 The solution set is {z|3  z  7}.

2n 2

Chapter 6

8 2

2n 2

7

10 2

n 6 4 n 7 5 The solution set is {n|5 6 n 6 4} .

32 1 0 1 2 3 4 5 6 7

29. Write |t  8| 6 2 as t  8 6 2 and t  8 Case 1: Case 2: t8 6 2 t8 7 t88 6 28 t88 7 t 6 6 t 7

6

7 2.

5 432 1 0 1 2 3 4 5

2 2  8 10

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35. Write |6r  8| 6 4 as 6r  8 6 4 and 6r  8 7 4. Case 1: Case 2: 6r  8 6 4 6r  8 7 4 6r  8  8 6 4  8 6r  8  8 7 4  8 6r 6 12 6r 7 4 6r 6

6

12 6

6r 6

r 6 2

39. Write

Case 1:

r 7

Case 2:

2  3x 5 2  3x (5) 5

3x 3

The solution set is the empty set .

 (5)2

3x 3

8

x  23

36. Write |6  (3d  5)|  14 as 6  (3d  5)  14 and 6  (3d  5)  14. Case 1: Case 2: 6  (3d  5)  14 6  (3d  5)  14 6  3d  5  14 6  3d  5  14 11  3d  14 11  3d  14 11  3d  11  14  11 11  3d  11  14  11 3d  3 3d  25 3d 3

3

 3

d  1

5

The solution set is d|1  d 

1 83

6.



d

5h  2 6

Case 1:

6

or x  4 .

5 units

25 3 1 83

–5

–4

–3

–2

–1

0

5h  2 6

1

2

3

4

5

So, an equation is |x| 5. 41. Find the point that is the same distance from 2 as the distance from 8. The midpoint between 2 and 8 is 3. 5 units

–2

–1

0

5 units

1

2

3

4

5

6

7

8

So, an equation is |x  3| 5. 42. Find the point that is the same distance from 3 as the distance from 3. The midpoint between 3 and 3 is 0. 3 units

 7 or

12 3

x4 2 23

5 units

0 2 4 6 8 10 12 14 16 18 20

`  7 as



40. Find the point that is the same distance from 5 as the distance from 5. The midpoint between 5 and 5 is 0.

1 0 1 2 3 4 5 6 7 8 9

5h  2 6

 (5) (2)

5 432 1 0 1 2 3 4 5

37. Write 08  (w  1) 0  9 as 8  (w  1)  9 and 8  (w  1)   9. Case 1: Case 2: 8  (w  1)  9 8  (w  1)  9 8w19 8  w  1  9 9w9 9  w  9 9w999 9  w  9  9  9 w  0 w  18 (1)(w)  (1)0 (1)(w)  (1)(18) w0 w  18 The solution set is {w|0  w  18} .

38. Write `

5

The solution set is x 0x 

5 432 1 0 1 2 3 4 5

 2

2  3x  10 2  3x  2  10  2 3x  12

 3 2

3d 3

2  3x 5 2  3x (5) 5

2

2  3x  10 2  3x  2  10  2 3x  8

4 6 2 3

7

0 2 5 3x 0  2 as 2 5 3x  2 or 2 5 3x  2.

3 units

 7.

Case 2:

5h  2 6 5h  2 (6) 6

5h  2 6 5h  2 (6) 6

7  (6)7

5h  2  42 5h  2  2  42  2 5h  40 5h 5



The solution set is

5h 5

5

4 85,

–5

6



h

–4

–3

–2

–1

0

1

2

3

4

5

The distance from 0 to any point on the graph is less than or equal to 3 units. So, an inequality is |x|  3. 43. Find the point that is the same distance from 7 as the distance from 1. The midpoint between 7 and 1 is 3.

 (6)(7)

5h  2  42 5h  2  2  42  2 5h  44

40 5

h8

 7

44 5 4 85

4 units

4 units

8 .

10864 2 0 2 4 6 8 10 –8

–7

–6

–5

–4

–3

–2

–1

0

1

2

The distance from 3 to any point on the graph is less than 4 units. So, an inequality is |x  (3)|  4, or |x  3|  4.

279

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44. Find the point that is the same distance from 1 as the distance from 3. The midpoint between 1 and 3 is 1. 2 units

–5

–4

–3

–2

–1

0

49. The difference between the temperature and 68 should be within 3. This statement can be expressed as |t  68|  3, where t represents the temperature in the house. Write |t  68|  3 as t  68  3 and t  68  3. Case 1: Case 2: t  68  3 t  68  3 t  68  68  3  68 t  68  68  3  68 t  71 t  65 The solution set is {t|65  t  71}. The temperature in the house should be between 65 F and 71 F, inclusive. 50. The difference between the percent of people who say protection of the environment should have priority and 52 is within 3. This statement can be expressed as |p  52|  3, where p represents the percent of people described. Write |p  52|  3 as p  52  3 and p  52  3. Case 1: Case 2: p  52  3 p  52  3 p  52  52  3  52 p  52  52  3  52 p  55 p  49 The solution set is { p|49  p  55}. The percent of people who say protection of the environment should have priority is between 49% and 55%, inclusive. 51. The difference between the pressure and 30 should be within 2. This statement can be expressed as |p  30|  2, where p represents the tire pressure in psi. Write |p  30|  2 as p  30  2 and p  30  2. Case 1: Case 2: p  30  2 p  30  2 p  30  30  2  30 p  30  30  2  30 p  32 p  28 The solution set is { p|28  p  32}. The tire pressure should be between 28 and 32 psi, inclusive. 52a. Since absolute value represents a distance, it cannot be negative. So, there is no value of x that will make |x  3| less than 5. Therefore, |x  3| 5 is never true. 52b. Since absolute value represents a distance, it cannot be negative. So, any value of x will make |x  6| greater than 1. Therefore, |x  6|  1 is always true. 52c. |x  2|  0 is true only if x  2. Therefore, |x  2|  0 is sometimes true.

2 units

1

2

3

4

5

The distance from 1 to any point on the graph is greater than 2 units. So, an inequality is |x  1|  2. 45. Find the point that is the same distance from 12 as the distance from 8. The midpoint between 12 and 8 is 10. 2 units

–15 –14 –13 –12 –11 –10

2 units

–9

–8

–7

–6

–5

The distance from 10 to any point on the graph is greater than or equal to 2 units. So, an inequality is |x  (10)|  2, or |x  10|  2. 46. The difference between the number of days and 280 should be within 14. This statement can be expressed as |d  280|  14, where d represents the number of days of the pregnancy. 47. Write |d  280|  14 as d  280  14 and d  280  14. Case 1: d  280  14 d  280  280  14  280 d  294 Case 2: d  280  14 d  280  280  14  280 d  266 The solution set is {d|266  d  294}. The length of a full-term pregnancy should be between 266 and 294 days, inclusive. 48. The difference between the pressure and 195 should be within 25. This statement can be expressed as |p  195|  25, where p represents the pressure in pounds per square inch. Write |p  195|  25 as p  195  25 and p  195  25. Case 1: p  195 6 25 p  195  195 6 25  195 p 6 220 Case 2: p  195 7 25 p  195  195 7 25  195 p 7 170 The solution set is { p|170  p  220}. The pressure of a fire extinguisher should be between 170 and 220 psi.

Chapter 6

280

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56. Inequalities involving absolute value are used to represent margin of error. Answers should include the following. • The inequality representing the people who are against the tax levy is |x  45|  3. To solve this inequality, find the intersection of x  45  3 and x  45  3. To solve these inequalities, add 45 to each side of each inequality. The solution set is {x|42  x  48}. • The votes for the tax levy can be between 44% and 50%. The votes against the tax levy can be between 42% and 48%. Depending on where the actual votes are in each range, it could either pass or fail. 57. B; Write |x  5| 2 as x  5  2 or x  5  2. Case 1: Case 2: x52 x  5  2 x5525 x  5  5  2  5 x  3 x  7 The solution set is {7, 3}. 58. C; Write 6  |x|  6 as 6  |x| and |x|  6. Case 1: 6  |x| is the same as |x|  6. Write |x|  6 as x  6 or x  6. The solution set is {x|x is a real number.}. Case 2: Write |x|  6 as x  6 and x  6. The solution set is {x| 6  x  6}. The final solution is the intersection of {x|x is a real number.} and {x| 6  x  6}. Therefore, the solution set is {x| 6  x  6}. If 6  x  6, then 6  x. 6 6 x (1)(6) 7 (1) (x) 6 7 x 6  x is the same as x  6.

53. The difference between the amount of sodium chloride and 3.0 must be within 0.5. This statement can be represented by |a  3.0|  0.5, where a represents the amount of sodium chloride to be added. Write |a  3.0|  0.5 as a  3.0  0.5 and a  3.0  0.5. Case 1: a  3.0  0.5 a  3.0  3.0  0.5  3.0 a  3.5 Case 2: a  3.0  0.5 a  3.0  3.0  0.5  3.0 a  2.5 The solution set is {a|2.5  a  3.5}. The amount of sodium chloride added must be between 2.5 and 3.5 mm, inclusive. 54. The difference between the guess and 18,000 must be within 1500 without going over 18,000. This statement can be expressed as the compound inequality |p  18,000|  1500 and p  18,000, where p represents the price guessed by Luis. Write |p  18,000|  1500 as p  18,000  1500 and p  18,000  1500. Case 1: p  18,000  1500 p  18,000  18,000  1500  18,000 p  19,500 Case 2: p  18,000  1500 p  18,000  18,000  1500  18,000 p  16,500 The solution set is the intersection of 16,500  p  19,500 and p  18,000. This set is { p|16,500  p  18,000}. Luis will win the vehicle if his guess is between $16,500 and $18,000, inclusive. 55a. Write x  3  1.2 as x  3  1.2 or x  3  1.2. Case 1: Case 2: x  3  1.2 x  3  1.2 x  4.2 x  1.8 55b. Find the point that is the same distance from 1.8 as the distance from 4.2. The midpoint between 1.8 and 4.2 is 3. 1.2 units

1

1.4

1.8 2.2

2.6

Page 351

1.2 units

3

3.4 3.8 4.2 4.6

Maintain Your Skills

59. Let t  number of beats per minute in Rafael’s target zone. 0.60(190) 6 t 6 0.80(190) 114 6 t 6 152 Rafael’s target zone is between 114 and 152 beats per min. Exercises 60–62 For checks, see students’ work. 60. 2m  7 7 17 2m  7  7 7 17  7 2m 7 10 2m 2

5

So, an equation is |x  3|  1.2.

7

10 2

m 7 5 The solution set is {m|m  5}.

281

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61.

2  3x  2 2  3x  2  2  2 3x  4 3x 3

70. 13.2  6.1  13.2  (6.1)  (|13.2|  |6.1|)  (13.2  6.1)  19.3 71. 4.7  (8.9)  4.7  8.9  (|8.9|  |4.7|)  (8.9  4.7)  4.2 72. Distributive Property 73. Substitution Property 74. To find the x-intercept, let y  0. y  3x  4 0  3x  4 0  4  3x  4  4 4  3x

4

 3 1

x  13

5

1

6

The solution set is x|x  13 . 2 w 3

62. 2 w 3

37

3373 2 w 3 3 2 w 2 3

12

 10 

132 210

w  15 The solution set is {w|w  15}. 63. To express the equation in slope-intercept form, solve the equation for y in terms of x. 2x  y  4 2x  y  2x  4  2x y  2x  4 The equation is now of the form y  mx  b, so the slope m is 2 and the y-intercept b is 4. 64. To express the equation in slope-intercept form, solve the equation for y in terms of x. 2y  3x  4 2y  3x  3x  4  3x 2y  3x  4 2y 2



y

4 3 4 3



3x 3

x

1

4

2

The graph intersects the x-axis at 3, 0 . To find the y-intercept, let x  0. y  3x  4 y  3(0)  4 y4 The graph intersects the y-axis at (0, 4). Plot these points and draw the line that connects them. y

3x  4 2 3x 4 2 2 3 x2 2

y  3x  4

y The equation is now of the form y  mx  b, so 3 the slope m is 2 and the y-intercept b is 2.

x

O

65. To express the equation in slope-intercept form, solve the equation for y in terms of x. 1 x 2



1 x 2 3 y 4

3

 4y  0 1

75. Select five values for the domain and make a table. The only value in the range is 2. Since there is no x in the equation, the value of x does not depend on y. Therefore, x can be any real number.

1

 2x  0  2x 3 y 4 4 3 y 3 4

12

1

 2x 

143 2112x2

x 3 1 0 2 4

2

y  3x

2

This equation can be written as y  3x  0. The equation is now of the form y  mx  b, so the 2 slope m is 3 and the y-intercept is 0. 66. I  prt I pt I pt



67.

prt pt

r

ex  2y  3z ex  2y  2y  3z  2y ex  3z  2y ex e



x 68.

a  5 3 1 a  5 7 3 a  5 21

12

(x, y) (3, 2) (1, 2) (0, 2) (2, 2) (4, 2)

Graph the ordered pairs and draw a line through the points.

3z  2y e 3z  2y e

y

 7x 

117 27x

x

O

x

69. 13  8  (|13|  |8|)  (13  8)  5 Chapter 6

y 2 2 2 2 2

y  2

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76. In order to find values for y more easily, solve the equation for y. xy3 xyx3x y3x Select five values for the domain and make a table. 3x 3  (1) 30 31 32 33

x 1 0 1 2 3

The graph intersects the x-axis at (6, 0). To find the y-intercept, let x  0. 2y  x  6 2y  (0)  6 2y  6 2y 2

6 2

y  3 The graph intersects the y-axis at (0, 3). Plot these points and draw the line that connects them.

(x, y) (1, 4) (0, 3) (1, 2) (2, 1) (3, 0)

y 4 3 2 1 0



y

x

O

Graph the ordered pairs and draw a line through the points.

2y  x  6

y xy3

79. To find the x-intercept, let y  0. 2(x  y)  10 2(x  0)  10 2x  10

x

O

2x 2

77. To find the x-intercept, let y  0. y  2x  1 (0)  2x  1 2x  1 2x 2

1 1

The graph intersects the x-axis at

1 2 1 , 2

10 2

x 5 The graph intersects the x-axis at (5, 0). To find the y-intercept, let x  0. 2(x  y)  10 2(0  y)  10 2y  10

 2

x2



2y 2

0.



10 2

y 5 The graph intersects the y-axis at (0, 5). Plot these points and draw the line that connects them.

To find the y-intercept, let x = 0 y  2x  1 y  2(0)  1 y  1 The graph intersects the y-axis at (0, 1). Plot these points and draw the line that connects them.

y 2(x  y )  10

y

O O

x

x y  2x  1

6-6 78. To find the x-intercept, let y  0. 2y  x  6 2(0)  x  6 x  6 (1)(x)  (1)(6) x6

Page 355

Graphing Inequalities in Two Variables Check for Understanding

1. The graph of y  x  2 is a line. The graph of y  x  2 does not include the boundary y  x  2, and it includes all ordered pairs in the half-plane that contains the origin.

283

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2. Sample answer: x  y Step 1 Solve for y in terms of x. xy yx Step 2 Graph y  x. Since the inequality includes y values less than x, but not equal to x, the boundary is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (1, 3). yx 3  1 false Since the statement is false, the half-plane containing (1, 3) is not part of the solution. Shade the other half-plane.

6. b; Since y  2x  2 means y  2x  2 or y 2x  2, the boundary line is included in the solution set. So, the boundary should be drawn as a solid line. When (0, 0) is used as a test point in the original inequality, the resulting statement is 0  2. Since this statement is false, the half-plane containing the origin is not part of the solution. So, the other half-plane should be shaded. This describes graph b. 7. Step 1 The inequality y  4 is already solved for y in terms of x. Step 2 Graph y  4. Since y  4 means y  4 or y  4, the boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y4 0  4 false Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane.

y xy

O

x

y

y4

3. If the test point results in a true statement, shade the half-plane that contains the point. If the test point results in a false statement, shade the other half-plane. 4. Use a table to substitute the x and y values of each ordered pair into the inequality. x

y

1

0

3

2

2

5

2

1

yx1 0  1  1 00 231 24 521 53 1  2  1 1  1

8. Step 1 The inequality y  2x  3 is already solved for y in terms of x. Step 2 Graph y  2x  3. Since y  2x  3 means y 6 2x  3 or y  2x  3, the boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y  2x  3 0  2(0)  3 0  3 false Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane.

True or False true true false false

The ordered pairs {(1, 0), (3, 2)} are part of the solution set. 5. Use a table to substitute the x and y values of each ordered pair into the inequality. x

y

2

6

0

1

3

5

1

2

y  2x 6  2(2) 64 1  2(0) 1  0 5  2(3) 56 2  2(1) 2  2

True or False

y y  2x  3

true false

O

false false

The ordered pair {(2, 6)} is part of the solution set.

Chapter 6

x

O

284

x

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11. Step 1 Let x  the number of pizzas ordered. Let y  the number of pitchers of soft drink ordered.

9. Step 1 Since there is no y in the inequality, solve the inequality for x. 4  2x 6 2 4  2x  4 6 2  4 2x 6 6 2x 2

the cost the number the of one of pitchers The cost number pitcher of soft of one of pizzas of soft drink pizza times ordered plus drink times ordered 14243 123 14243 123 14243 123 1442443

6 2

7 x 7 3 Step 2 Graph x  3. Since the inequality includes x values greater than 3, but not equal to 3, the boundary is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). x 7 3 0 7 3 false Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane.

Step 2



3



123 123



y

60

Solve for y in terms of x. 12x  3y  60 12x  3y  12x  60  12x 3y  60  12x 

y

60  12x 3 60 12x  3 3

y  20  4x Step 3 Since the open sentence includes the equation, graph y  20  4x as a solid line. Test a point in one of the halfplanes, for example (0, 0). Shade the half-plane containing (0, 0) since 0  20  4(02 is true. Step 4 Examine the solution. • Coach Riley cannot order a negative number of pizzas or a negative number of pitchers of soft drinks. Therefore, the domain and range contain only nonnegative numbers. • Coach Riley cannot order a fraction of a pizza or a portion of a pitcher of soft drink. Thus, only points in the shaded half-plane whose x-and y-coordinates are whole numbers are possible solutions.

4  2x  2

x

10. Step 1

Solve for y in terms of x. 1y 7 x 1y1 7 x1 y 7 x  1 (1)(y) 6 (1)(x  1) y 6 1x Step 2 Graph y  1  x. Since the inequality includes y values less than 1  x, but not equal to 1  x, the boundary is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y 6 1x 0 6 10 0 6 1 true Since the statement is true, the half-plane containing the origin is part of the solution. Shade the half-plane containing (0, 0).

y 24 20 16 12 8 4 21 O 4

1 2 3 4 5 6x

Pages 356–357

Practice and Apply

12. Use a table to substitute the x and y values of each ordered pair into the inequality.

y

1 y  x O

x

3y 3

y

O



12

is not more than $60 .

x

x 0

y 4

1

3

6

8

4

5

y  3  2x 4  3  2(02 43 3  3  2(12 35 8  3  2(62 8  9 5  3  2(42 5  11

True or False false true false true

The ordered pairs {(1, 3), (4, 5)} are part of the solution set.

285

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17. Use a table to substitute the x and y values of each ordered pair into the inequality.

13. Use a table to substitute the x and y values of each ordered pair into the inequality. x 3

y 1

3

2

1

1

1

2

1 1 2 2 1 1 2 2

y  3x 6 3132 6 9 6 3132 6 9 6 3112 6 3 6 3112 6 3

True or False

x 1

y 1

false

2

1

true

1

1

true

2

4

false

y 7

13

10

4

4

6

2

x  y  11 5  7 6 11 12 6 11 13  10 6 11 3 6 11 4  4 6 11 8 6 11 6  122 6 11 8 6 11

True or False false

x 6

y 4

true

1

8

true

3

2

true

5

7

The ordered pairs {( 13, 10), (4, 4), (6, 2)} are part of the solution set. 15. Use a table to substitute the x and y values of each ordered pair into the inequality. x 3

y 2

2

4

6

2

5

1

2x  3y  6 2132  3122 7 0 7 2122  3142 7 8 7 2162  3122 7 6 7 2152  3112 7 7 7

false

x 2

y 4

true

1

5

false

6

7

true

0

0

The ordered pairs {( 2, 4), (5, 1)} are part of the solution set. 16. Use a table to substitute the x and y values of each ordered pair into the inequality. x 5

y 1

0

2

2

5

2

0

4y  8  0 4112  8  0 12  0 4122  8  0 00 4152  8  0 12  0 4102  8  0 8  0

true true false

|x  3| y 06  3 0  4 34 01  3 0  8 48 03  3 0  2 62 05  3 0  7 27

True or False false false true false

|y  2|  x 04  2 0 6 2 2 6 2 05  2 0 6 1 3 6 1 07  2 0 6 6 5 6 6 00  2 0 6 0 2 6 0

True or False false false true false

The ordered pair {(6, 7)} is part of the solution set.

True or False false true true false

The ordered pairs {(0, 2), (2, 5)} are part of the solution set.

Chapter 6

false

The ordered pair {(3, 2)} is part of the solution set. 19. Use a table to substitute the x and y values of each ordered pair into the inequality.

True or False 6 6 6 6 6 6 6 6

True or False 7 7 7 7 7 7 7 7

The ordered pairs {(2, 1), (1, 1)} are part of the solution set. 18. Use a table to substitute the x and y values of each ordered pair into the inequality.

The ordered pairs {(1, 1), (1, 2)} are part of the solution set. 14. Use a table to substitute the x and y values of each ordered pair into the inequality. x 5

3x  4y  7 3112  4112 6 7 6 3122  4112 6 2 6 3112  4112 6 1 6 3122  4142 6 10 6

286

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20. c; Step 1

1

Step 2 Graph y  3  2x. Since the inequality 1 includes values of y greater than 3  2x, 1 but not equal to 3  2x, the boundary is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0).

6  x 2 6  x y 2 6  x 6  x Graph y  2 . Since y  2 means 6  x 6  x y 6 2 or y  2 , the boundary is 2y 2

Step 2

22. d; 1 Step 1 The inequality y 7 3  2x is already solved for y in terms of x.

Solve for y in terms of x. 2y  x  6 2y  x  x  6  x 2y  6  x 

included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y 0

1

y 7 3  2x 1

0 7 3  2 (0) false 0 7 3 Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane. The graph described is graph d. 23. b; Step 1 Solve for y in terms of x. 4y  2x  16 4y  2x  2x  16  2x 4y  16  2x

6  x 2 6  0 2

true 03 Since the statement is true, the half-plane containing the origin is part of the solution. Shade the half-plane containing (0, 0). The graph described is graph c. 21. a; Step 1 Solve for y in terms of x. 1 x 2 1 x 2

4y 4

y 7 4 1

y 1

 y  2x 7 4  2x y 7 4 



16  2x 4 1 4  2x

1

1

Step 2 Graph y  4  2x. Since y  4  2x

1 x 2

1

1

means y 7 4  2x or y  4  2x, the boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0).

1

(1)(y) 6 (1)(4  2x) 1

y 6 2x  4 1

Step 2 Graph y  2x  4. Since the inequality 1 includes values of y less than 2x  4, but 1 not equal to 2x  4, the boundary is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0).

1

y  4  2x 1

0  4  2 (0) 04 false Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane. The graph described is graph b. 24. Substitute x  2 into the equation 2x  3y  5 to find the corresponding value of y. 2x  3y  5 2122  3y  5 4  3y  5 4  3y  4  5  4 3y  9

1

y 6 2x  4

0 6 2 102  4 1

false 0 6 4 Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane. The graph described is graph a.

3y 3

9

3

y3 The point A (2, 3) is on the graph of 2x  3y  5.

287

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25. Substitute x  0 into the equation 4x  3y  4 to find the corresponding value of y. 4x  3y  4 4102  3y  4 3y  4 3y 3



10y 10

1 1 13

1

2

1

2

1

y 7 2x

1 7 2 112 1 1

1 7 2

true

Since the statement is true, the half-plane containing (1, 1) is part of the solution. Shade the half-plane containing (1, 1). y 5x  10y  0

x

O

x

29. Step 1 The inequality y 6 x is already solved for y in terms of x. Step 2 Graph y  x. Since the inequality includes y values less than x, but not equal to x, the boundary is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (2, 0). y 6 x 0 6 2 true Since the statement is true, the half-plane containing (2, 0) is part of the solution. Shade the half-plane containing (2, 0).

y  3

27. Step 1 There is no y in the inequality. The inequality x  2 is already solved for x. Step 2 Graph x  2. Since x  2 means x 7 2 or x  2, the boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). x2 0  2 false Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane.

y yx

y

O

5x 10 1 2x

Step 2 Graph y  2x. Since the inequality 1 includes y values greater than 2x, but 1 not equal to 2x, the boundary line is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (1, 1).

y

O

7

y 7

The point 0,  is on the graph of 2x  3y  5. Since the graph of (0, 1) is above 1 the graph of 0,  13 , the point B (0, 1) is above the graph of 4x  3y  4. 26. Step 1 The inequality y 6 3 is already solved for y in terms of x. Step 2 Graph y  3. Since the inequality includes y values less than 3, but not equal to 3, the boundary is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y 6 3 0 6 3 false Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane.

O

x

x2

Chapter 6

Solve for y in terms of x. 5x  10y 7 0 5x  10y  5x 7 0  5x 10y 7 5x

4 3

y  13

1

28. Step 1

288

x

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30. Step 1

6  x 2 6  x y 2 6  x 6  x Graph y  2 . Since y  2 means 6  x 6  x y 6 2 or y  2 , the boundary is 2y 2

Step 2

32. Step 1

Solve for y in terms of x. 2y  x  6 2y  x  x  6  x 2y  6  x

12  4x 3 12  4x y 3 12  4x 12  4x . Since y  3 3 12  4x 12  4x or y  , the 3 3

3y 3



0



Step 2 Graph y  means y 7 boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0).

included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y

Solve for y in terms of x. 3y  4x  12 3y  4x  4x  12  4x 3y  12  4x

6  x 2 6  0 2

y 0

true 03 Since the statement is true, the half-plane containing the origin is part of the solution. Shade the half-plane containing (0, 0).

12  4x 3 12  4102 3

false 04 Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane.

y 2y  x  6

y

x

O

x

O 3y  4x  12

31. Step 1

Solve for y in 6x  3y 6x  3y  6x 3y 3y 3

terms of x. 7 9 7 9  6x 7 9  6x 7

33. Step 1 The inequality y  2x  4 is already solved for y in terms of x. Step 2 Graph y  2x  4. Since y  2x  4 means y 6 2x  4 or y  2x  4, the boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y  2x  4 0  2(0)  4 false 0  4 Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane.

9  6x 3

y 7 3  2x Step 2 Graph y  3  2x. Since the inequality includes y values greater than 3  2x, but not equal to 3  2x, the boundary is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y 7 3  2x 0 7 3  2102 false 0 7 3 Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane.

y y  2x  4

O

y

O

x

x

6x  3y  9

289

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34. Step 1

Solve for y in 8x  6y 8x  6y  8x 6y 6y 6

36. Step 1

terms of x. 6 10 6 10  8x 6 10  8x 7

y 7 y 7 4

3(x  2y) 3

2y 2

y

0 7

5

0 7



18 3

6  x 2 6  x 7 2 6  x . Since 2

7

Step 2 Graph y  the inequality 6  x includes y values greater than 2 , but 6  x not equal to 2 , the boundary is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0).

y 7 3x  3 4 102 3 5 3

7

x  2y 7 6 x  2y  x 7 6  x 2y 7 6  x

10  8x 6 8x  10 6 5 4 x 3 3 5 . Since 3

Step 2 Graph y  3x  the inequality 5 4 includes y values greater than 3x  3, 5 4 but not equal to 3x  3, the boundary is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). 4

Solve for y in terms of x. 3(x  2y) 7 18

y 7 5 3

0 7

6  x 2 6  0 2

true 0 7 3 Since the statement is true, the half-plane containing the origin is part of the solution. Shade the half-plane containing (0, 0).

true Since the statement is true, the half-plane containing the origin is part of the solution. Shade the half-plane containing (0, 0).

y

y

8x  6y  10 O

x

O

x 3(x  2y )  18

35. Step 1

Solve for y in terms of x. 3x  1  y y  3x  1 Step 2 Graph y  3x  1. Since y  3x  1 means y 6 3x  1 or y  3x  1, the boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y  3x  1 0  3(0)  1 0  1 false Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane. y

O

x

3x  1  y

Chapter 6

290

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37. Step 1

42. Let s  the number of tickets sold to a single person. Let c  the number of tickets sold to couples.

Solve for y in terms of x. 1 (2x 2 1 (2) 2 (2x

 y) 6 2  y) 6 (2)2

The cost of one ticket sold to a single person

2x  y 6 4 2x  y  2x 6 4  2x y 6 4  2x Step 2 Graph y  4  2x. Since the inequality includes y values less than 4  2x, but not equal to 4  2x, the boundary is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y 6 4  2x 0 6 4  2(0) 0 6 4 true Since the statement is true, the half-plane containing the origin is part of the solution. Shade the half-plane containing (0, 0).

14444244443

5

the number of tickets sold to a single person

times 1 424 3 

14 444244 443

s

plus

1 424 3



the number of the cost of tickets sold to one ticket sold times couples to a couple is at least $1200 424 3 14 1444 4244443 1 444244 443 1 4 44244 4 3 1 424 3  c  1200 8

This situation is represented by the inequality 5s  8c  1200. 43. Step 1 Solve the inequality from Exercise 42 for c in terms of s. 5s  8c  1200 5s  8c  5s  1200  5s 8c  1200  5s 8c 1200  5s  8 8 c  150 

y

Step 2

5 s 8

5

5

Graph c  150  8s. Since c  150  8s 5

1( 2x  y )  2 2

x

O

5

means c 7 150  8s or c  150  8s, the boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). 5

c  150  8s 38. Let /  the length of the package. Let d  the distance around the thickest part of the package.

5

0  150  8 (0) false 0  150 Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane. Step 4 Examine the solution. • It is not possible to sell a negative number of tickets. Therefore, the domain and range contain only nonnegative numbers. • It is not possible to sell a portion of a ticket. Thus, only points in the shaded half-plane whose x- and y-coordinates are whole numbers are possible solutions.

the distance around is less The length the thickest part than or of the of the package equal to 108 inches. longest side plus

1442443 1 424 3 14 44442444 443 14243

/





d

1442443

108

This situation is represented by the inequality /  d  108. 39. The solution set is limited to pairs of positive numbers. 40. Let t  the number of televisions. Let m  the number of microwaves. The weight the number the weight the number of one of of one of television times televisions plus microwave times microwaves

1442443 1 424 3 1442443 1 424 3 1442443 1 424 3 1442443

77



is less than or equal to

4000 lb.



4000

t



55



m

160 140 120 100 80 60 40 20

14243 1442443

This situation is represented by the inequality 77t  55m  4000. 41. Substitute the t and m values of the ordered pair (35, 25) into the inequality from Exercise 40. 77t  55m  4000 77(35)  55(25)  4000 2695  1375  4000 4070  4000 false Since the ordered pair does not make the inequality true, the truck will not be able to deliver 35 televisions and 25 microwaves at once. The total weight would be greater than 4000 lb.

O

c

5s  8c  1200

20

60 40

291

80

100 140 180 220 s 120 160 200 240

Chapter 6

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46. The amount of money spent in each category must be less than or equal to the budgeted amount. How much you spend on individual items can vary. Answers should include the following. • The domain and range must be positive integers. • Sample answers: Hannah could buy 5 cafeteria lunches and 3 restaurant lunches, 2 cafeteria lunches and 5 restaurant lunches, or 8 cafeteria lunches and 1 restaurant lunch. 47. D; y  2x 6 5 6  2152 6 5 6  10 6 5 4 6 5 false 48. B; Since 2x  y 7 1 does not include 2x  y  1, the boundary should be drawn as a dashed line. Test (2, 2). 2x  y 7 1 2(2)  2 7 1 6 7 1 true The half-plane containing (2, 2) should be shaded.

44. Since the graph of the ordered pair (100, 125) is in the shaded half-plane of the graph from Exercise 43, the values s  100 and c  125 make the inequality true. So, the committee would cover its expenses if 100 single tickets and 125 couple tickets are sold. To check algebraically, substitute s  100 and c  125 into the inequality from Exercise 42. 5s  8c  1200 5(100)  8(125)  1200 500  1000  1200 1500  1200 true Since the ordered pair makes the inequality true, the conclusion reached using the graph is correct. 45. First graph y  x  1. Since the inequality includes y  x  1, the boundary should be drawn as a solid line. Test (0, 0). 001 false 0  1 Shade the half-plane that does not contain the origin. y y  x 1

Page 357 x

O

Now graph y  x. Since the inequality includes y  x, the boundary should be drawn as a solid line. Test (1, 1). true 1  1 Shade the half-plane that contains (1, 1).

2t 2

8

14 2

t  7

50. Write ƒ x  8 ƒ 6 6 as x  8 6 6 and x  8 Case 1: Case 2: x8 6 6 x8 7 x88 6 68 x88 7 x 6 2 x 7 The solution set is {x ƒ 14 6 x 6 2}.

y  x

18 1614 12 10 8 6 4 2

0

7 6. 6 6  8 14

2

51. Write ƒ 2y  5 ƒ  3 as 2y  5  3 or 2y  5  3. Case 1: Case 2: 2y  5  3 2y  5  3 2y  5  5  3  5 2y  5  5  3  5 2y  2 2y  8

To graph the intersection of the graphs of y  x  1 and y  x, graph both inequalities above on the the same coordinate plane. The region where the two shaded half-planes overlap represents the intersection of the graphs. Only points in this region will make both inequalities true.

2y 2

y



2 2

2y 2



8 2

y  1 y  4 The solution set is { y ƒ y  4 or y  1}.

yx1

8 7 6 5 4 3 2 1 0 1 2

x

y  x

Chapter 6



108 6 4 2 0 2 4 6 8 10

x

O

2t 2

2

t4 The solution set is {7, 4}.

y

O

Maintain Your Skills

49. Write ƒ 3  2t 0  11 as 3  2t  11 or 3  2t  11. Case 1: Case 2: 3  2t  11 3  2t  11 3  2t  3  11  3 3  2t  3  11  3 2t  8 2t  14

292

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52.

and y  6 7 1 y2 6 4 y  6  6 7 1  6 y22 6 42 y 7 7 y 6 6 The solution set is the intersection of the two graphs. Graph y 7 7. Graph y 6 6. Find the intersection. –10

–8

–6

–4

–2

0

2

4

6

8

56. Find the amount of change. Since the new amount is greater than the original amount, the percent of change is a percent of increase. 75  53  22 Find the percent using the original number, 53, as the base. 22 53

22(100)  53(r) 2200  53r

10

–10

–8

–6

–4

–2

0

2

4

6

8

10

–10

–8

–6

–4

–2

0

2

4

6

8

10

r

 100

2200 53



53r 53

41.51  r To the nearest whole percent, the percent of increase is 42%. 57.

The solution set is {y ƒ 7 6 y 6 6}. 53. or m4 6 2 m2 7 1 m44 6 24 m22 7 12 m 6 2 m 7 3 The solution set is the union of the two graphs. Graph m 6 2. Graph m 7 3. Find the union.

3

1

d  2 3 d  2 3

7

2  3(7)

d  2  21 d  2  2  21  2 d  23 58. 3n  6  15 3n  6  6  15  6 3n  21 3n 3

–5

–4

–3

–2

–1

0

1

2

3

4

5

59. –5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5



n  7 35  20h  100 35  20h  35  100  35 20h  65



60.

64 4

61.

27c 9

62.

12a  14b 2

 27c  192  3c  112a  14b2  122

1 12

 112a  14b2 2

1 12

 6a  7b

200r 200

63.

1 12

18y  9 3

 118y  92  3

113 2 1 1  18y 1 3 2  9 1 3 2 1 1  18y 1 3 2  9 1 3 2  118y  92

 6y  3

Page 358

r

 100



or 3.25

 12a 2  114b2 2

4211002  1001r2 4200  100r 4200 100

13 4

 64  4  16

r 100

14  r The percent of decrease is 14%. 55. Find the amount of change. Since the new amount is greater than the original amount, the percent of change is a percent of increase. 142  100  42 Find the percent using the original number, 100, as the base. 42 100

65

 20

h

2811002  2001r2 2800  200r 2800 200

21 3

20h 20

The solution set is {m ƒ m 6 2 or m 7 3}. 54. Find the amount of change. Since the new amount is less than the original amount, the percent of change is a percent of decrease. 200  172  28 Find the percent using the original number, 200, as the base. 28 200



Graphing Calculator Investigation (Follow-Up of Lesson 6-6)

1. y  3x  1 is shaded below the line. y  3x  1. y  3x  1 is shaded above the line. y  3x  1.

100r 100

42  r The percent of increase is 42%.

293

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2.

KEYSTROKES:

4 , 10

2nd

)

3. 4. 5. 6. 7. 8.

DRAW 7 ( ) 2 X,T,␪,n ENTRY

d; a; c; g; h; b;

intersection Addition Property of Inequalities half-plane Subtraction Property of Inequalities union Division Property of Inequalities

Pages 359–362

2a. The lower boundary is y  2x  4. The upper boundary is Ymax or 10. 2b. The coordinates of any point in the shaded region are solutions of the inequality. Sample answer: {(0, 4), (1, 7), (2, 6), (4.2, 1.5)} 3a. The number of the number of plus student tickets adult tickets 1442443 123 1442443 x  y is at least 10. 14243 123  10

25

10.

The cost the number the cost of one of of one student student adult ticket times tickets plus ticket 14243 123 1442443 123 14243 4  x  8 the number of times 14 adult tickets no more than 1 $80. 123 424 43 is 1442443 23

23

11.

19

17

15

13

11

7

9

5

w  14  23 w  14  14  23  14 w  37 The solution set is {w ƒ w  37}. 35 36 37 38 39 40 41 42 43 44 45

12.

a6 7 a66 7 a 7 The solution

10 10  6 4 set is {a ƒ a 7 4}.

7 6 5 4 3 2 1 0 1 2 3

3c. KEYSTROKES: 2nd DRAW 7 ( ) X,T,␪,n 10 , ( ) 0.5 X,T,␪,n 10 )

13.

ENTRY

0.11  n  10.042 0.11  n  0.04 0.11  0.04  n  0.04  0.04 0.15  n 0.15  n is the same as n  0.15. The solution set is {n ƒ n  0.15}. 2

14. 3d. The coordinates of any point in the shaded region are solutions of the inequality. Sample answer: {(8, 5), (10, 4), (14, 2), (20, 0)} where (8, 5) represents the purchase of 8 student tickets and 5 adult tickets, for example. 15.

Chapter 6 Study Guide and Review Vocabulary and Concept Check

1. f; set-builder notation 2. e; Multiplication Property of Inequalities

1

0

1

2

2.3 6 g  12.12 2.3 6 g  2.1 2.3  2.1 6 g  2.1  2.1 0.2 6 g 0.2 6 g is the same as g 7 0.2. The solution set is { g ƒ g 7 0.2}. 2

Chapter 6

21

r  7 7 5 r  7  7 7 5  7 r 7 12 The solution set is {r ƒ r 7 12}. 15

 y  80 So, x  y  10 represents the total number of tickets, and 4x  8y  80 represents the total cost of the tickets. 3b. Rewrite x  y  10 as y  x  10. Rewrite 4x  8y  80 as y  0.5x  10. The lower boundary is y  x  10. The upper boundary is y  0.5x  10.

Page 359

Lesson-by-Lesson Review

Exercises 9–17 For checks, see students’ work. 9. c  51 7 32 c  51  51 7 32  51 c 7 19 The solution set is {c ƒ c 7 19}.

1

0

1

2

7h  6h  1 7h  6h  6h  1  6h h  1 The solution set is {h ƒ h  1}. 5 4 3 2 1 0 1 2 3 4 5

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16.

5b 7 4b  5 5b  4b 7 4b  5  4b b 7 5 The solution set is {b ƒ b 7 5}.

Exercises 27–35 For checks, see students’ work. 4h  7 7 15 27. 4h  7  7 7 15  7 4h 7 8 4b 4

2 1 0 1 2 3 4 5 6 7 8

h 6 2

17. Let n  the number. 21  n  (2) 21  n  2 21  2  n  2  2 23  n 23  n is the same as n  23. The solution set is {n ƒ n  23}. Exercises 18–26 For checks, see students’ work. 18. 15v 7 60 15v 15

7

The solution set is {h ƒ h 6 2}. 28.

60 15

16 8

72

75

 15

z5 The solution set is {z ƒ z  5}. 21. 9m 6 99 7

99 9

b 12 b (12) 12

d 13 d (13) 13

3  (12)3

7 5

56 8

6 (13)(5)

2 w 3 3 2 w 2 3

12

7 22 7

132 2 (22)

w 7 33 The solution set is {w ƒ w 7 33}. 25.

3 p 5 5 3 p 3 5

12

 15 

153 2 (15)

33.

p  25 The solution set is {p ƒ p  25}. 26. Let n  the number. 0.80n  24 0.80n 0.80

7

72 8

6

8q 8

7 6 q 7 6 q is the same as q 7 7. The solution set is 5q ƒ q 7 76. 7(g  8) 6 3(g  2)  4g 32. 7g  56 6 3g  6  4g 7g  56 6 7g  6 7g  56  7g 6 7g  6  7g 56 6 6 Since the inequality results in a false statement, the solution set is the empty set .

d 6 65 The solution set is {d ƒ d 6 65}. 24.

8x 8

b 7 9 The solution set is {b ƒ b 7 9}. 5(q  122 6 3q  4 31. 5q  60 6 3q  4 5q  60  5q 6 3q  4  5q 60 6 8q  4 60  4 6 8q  4  4 56 6 8q

b  36 The solution set is {b ƒ b  36}. 23.

6

8b 8

m 7 11 The solution set is {m ƒ m 7 11}. 22.

 24 6

2 6 x 2 x is the same as x 2. The solution set is {x ƒ x 7 2}. 15b  12 7 7b  60 30. 15b  12  7b 7 7b  60  7b 8b  12 7 60 8b  12  12 7 60  12 8b 7 72

r6 The solution set is {r ƒ r  6}. 20. 15z  75

9m 9

6

n 6 4 The solution set is 5n ƒ n 6 46. 5x  3 6 3x  19 29. 5x  3  5x 6 3x  19  5x 3 6 8x  19 3  19 6 8x  19  19 16 6 8x

 12

15z 15

5  6n 7 19 5  6n  5 7 19  5 6n 7 24 6n 6

v 7 4 The solution set is 5v ƒ v 7 46 . 19. 12r  72 12r 12

8 4

6

12

2(x  22

3 3 2(x  2) 2 3

4 

132 24

x26 x2262 x4 The solution set is {x ƒ x  4}.

24

 0.80

n  30 The solution set is {n ƒ n  30}.

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1  7n 5 1  7n (52 5

34.

38.

7 10 7 5(102

or 3w  8 6 2 w  12 7 2  w 3w  8  8 6 2  8 w  12  w 7 2  w  w 3w 6 6 2w  12 7 2

1  7n 7 50 1  7n  1 7 50  1 7n 7 49 7n 7

6

3w 3

2 n 3



132 236

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

39.

The solution set is 5p|4 6 p 6 26. 37. First express 3 6 2k  1 6 5 using and. 3 6 2k  1 2k  1 6 5 and 3  1 6 2k  1  1 2k  1  1 6 5  1 2 6 2k 2k 6 6 6

7

10 2

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

 36

n  54 The solution set is 5n ƒ n  546. 36. First express 1 6 p  3 6 5 using and. 1 6 p  3 p3 6 5 and 1  3 6 p  3  3 p33 6 53 4 6 p p 6 2 The solution set is the intersection of the two graphs. Graph 4 6 p or p 7 4. Graph p 6 2. Find the intersection.

2 2

2w 7 10

w 7 5 The solution set is the union of the two graphs. Graph w 6 2. Graph w 7 5. Find the union.

 27  27  9  27

12

2w  12  12 7 2  12 2w 2

49 7

 27  9 2 n 3 3 2 n 2 3

6 3

w 6 2

n 6 7 The solution set is {n ƒ n 6 7}. 35. Let n  the number. 2 n 3

6

2k 2

2k 2

1 6 k

6

40.

6 2

k 6 3

The solution set is the intersection of the two graphs. Graph 1 6 k or k 7 1. Graph k 6 3. Find the intersection.

The solution set is {w|w is a real number.}. or a38 a  5  21 a3383 a  5  5  21  5 a  11 a  16 The solution set is the union of the two graphs. Graph a  11. Graph a  16. Find the union. 9

10

11

12

13

14

15

16

17

18

19

9

10

11

12

13

14

15

16

17

18

19

9

10

11

12

13

14

15

16

17

18

19

The solution set is {a ƒ a  11 or a  16}. and 3m 6 5 m8 6 4 3m3 6 53 m88 6 48 m 6 4 m 6 2 (12(m2 7 (122 m 7 2 The solution set is the intersection of the two graphs. Graph m 6 4 Graph m 7 2. Find the intersection. –5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

Since the graphs do not intersect, the solution set is the empty set .

The solution set is 5k|1 6 k 6 36.

Chapter 6

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41.

and 10  2y 7 12 7y 6 4y  9 10  2y  10 7 12  10 7y  4y 6 4y  9  4y 2y 7 2 3y 6 9 2y 2

6

3y 3

2 2

6

46. Write ƒ r  10 ƒ 6 3 as r  10 6 3 and r  10 7 3. Case 1: Case 2: r  10 6 3 r  10 7 3 r  10  10 6 3  10 r  10  10 7 3  10 r 6 7 r 7 13 The solution set is 5r ƒ 13 6 r 6 76.

9 3

y 6 1 y 6 3 The solution set is the intersection of the two graphs. Graph y 6 1. Graph y 6 3. Find the intersection.

1514131211109 8 7 6 5

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5

47. Write ƒ t  4 ƒ  3 as t  4  3 and t  4  3. Case 1: Case 2: t43 t  4  3 t434 t  4  4  3  4 t  1 t  7 The solution set is 5t ƒ 7  t  16. 9 8 7 6 5 4 3 2 1 0 1

48. Write 0 2x  5 0 6 4 as 2x  5 6 4 and 2x  5 7 4. Case 1: Case 2: 2x  5 6 4 2x  5 7 4 2x  5  5 6 4  5 2x  5  5 7 4  5 2x 6 1 2x 7 9

The solution set is 5y ƒ y 6 16. 42. Write ƒ w  8 ƒ  12 as w  8  12 or w  8  12. Case 1: Case 2: w  8  12 w  8  12 w  8  8  12  8 w  8  8  12  8 w  20 w  4 The solution set is 54, 206.

2x 2

1 2 1 2

6

x 6

2x 2

7

x 7

5

1

1

6

9 2 9 2

1

or 42

The solution set is x ƒ 42 6 x 6 2 .

128 4 0 4 8 12 16 20 24 28

43. Write ƒ q  5 ƒ  2 as q  5  2 or q  5  2. Case 1: Case 2: q52 q  5  2 q5525 q  5  5  2  5 q  3 q  7 The solution set is 57, 36.

8 7 6 5 4 3 2 1 0 1 2

49. Write 0 3d  4 0 6 8 as 3d  4 6 8 and 3d  4 7 8. Case 1: Case 2: 3d  4 6 8 3d  4 7 8 3d  4  4 6 8  4 3d  4  4 7 8  4 3d 6 4 3d 7 12

9 8 7 6 5 4 3 2 1 0 1

44. Write ƒ h  5 ƒ 7 7 as h  5 7 7 or h  5 6 7. Case 1: Case 2: h5 7 7 h  5 6 7 h55 7 75 h  5  5 6 7  5 h 7 2 h 6 12 The solution set is 5h ƒ h 6 12 or h 7 26.

3d 3

6

d 6

4 3 4 3

3d 3 1

12 3

d 7 4

or 13

5

7

The solution set is d ƒ 4 6 d 6

1 13

6.

6 5 4 3 2 1 0 1 2 3 4

50. Use a table to substitute the x and y values of each ordered pair into the inequality.

16141210 8 6 4 2 0 2 4

45. Write ƒ w  8 ƒ  1 as w  8  1 or w  8  1. Case 1: Case 2: w81 w  8  1 w8818 w  8  8  1  8 w  7 w  9 The solution set is {w ƒ w  9 or w  7}.

x

109 8 7 6 5 4 3 2 1 0

y

1

3

3

2

2

7

4

11

3x  2y < 9 3(12  2(32 9 3(32  2(22 13 3(22  2(72 8 3(42  2(112 10

6 6 6 6 6 6 6 6

True or False 9 9 9 9 9 9 9 9

false false true false

The ordered pair 5(2, 726 is part of the solution set.

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51. Use a table to substitute the x and y values of each ordered pair into the inequality. x

y

5  y  4x

True or False

2

5

5  (52  4(22 10  8

true

1 2

7

574 2  2

false

1

6

5  6  4(12 1  4

true

3

20

5  20  4(32 15  12

false

112 2

54. Since the boundary is not included in the solution, draw a dashed line. Test the point (0, 0). y  2x 6 3 0  2(0) 6 3 0 6 3 false The half plane that does not contain (0, 0) should be shaded. y y  2x  3

The ordered pairs {(2, 5), (1, 6)} are part of the solution set. 52. Use a table to substitute the x and y values of each ordered pair into the inequality. x

y

4

15

5

3

1

8

1 y 2 1 (152 2 1 72

6x  6  (42  10

1 (12 2 1 2

1

1 (82 2

63

65

43 2

25

1 (252 2 1 122

 6  (22 8

True or False

55. Since the boundary is included in the solution, draw a solid line. Test the point (0, 0). x  2y  4 0  2(0)  4 0  4 false The half plane that does not contain (0, 0) should be shaded.

true

true

false

y

false

x  2y  4 x

The ordered pairs {(4, 15), (5, 1)} are part of the solution set. 53. Use a table to substitute the x and y values of each ordered pair into the inequality. x

y

2x  8  y

true

6

2(3) 6 8  6 6 6 2

true

4

0

2(42 6 8  0 8 6 8

false

3

6

2(32 6 8  6 6 6 2

false

10

3

O

True or False

2(5) 6 8  10 10 6 2

5

56. Since the boundary is included in the solution, draw a solid line. Test the point (0, 0). y  5x  1 0  5(0)  1 01 true The half plane that contains (0, 0) should be shaded. y

The ordered pairs 5(5, 102, (3, 626 are part of the solution set.

y  5x  1

Chapter 6

x

O

298

O

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6.

57. Since the boundary is not included in the solution, draw a dashed line. Test the point (0, 0). 2x  3y 7 6 2(0)  3(0) 7 6 0 7 6 false The half plane that does not contain (0, 0) should be shaded.

9p 6 8p  18 9p  8p 6 8p  18  8p p 6 18 Check: Substitute 18, a number less than 18, and a number greater than 18. Let p  18 Let p  20 ?

9(20) 6 8(20)  18

?

180 6 160  18 180 6 178 ✓

9(18) 6 8(18)  18

? ?

162 6 144  18 162 162

y

Let p  10. ?

9(10) 6 8(10)  18 ?

90 6 80  18 90 98 The solution set is 5p ƒ p 6 186. 7. d  5 6 2d  14 d  5  d 6 2d  14  d 5 6 d  14 5  14 6 d  14  14 9 6 d 9 6 d is the same as d 7 9. Check: Substitute 9, a number less than 9, and a number greater than 9. Let d  9. Let d  0.

x

O 2x  3y  6

Chapter 6 Practice Test Page 363

1. 5t ƒ t  176 2. 6(a  52 6 2a  8 6a  30 6 2a  8 6a  30  2a 6 4a  30 6 4a  30  30 6 4a 6 4a 4

6

Original equation Distributive Property 2a  8  2a Subtract 2a from each side. Simplify. 8 Subtract 30 from 8  30 each side. Simplify. 22 22 Divide each side 4 by 4.

?

23  17  6 23  23 ✓ Let g  10.

0  5 6 2(02  14

?

4 6 18  14 4 4

5 14

Let d  12. ? 12  5 6 2(12)  14 ?

7 6 24  14 7 6 10 ✓ The solution set is 5d|d 7 96.

a 6 5.5 Simplify. The solution set is 5a ƒ a 6 5.56. 3. Sample answer: 2 6 x 6 8 is a compound inequality that is an intersection since it can be written using and as 2 6 x and x 6 8. Sample answer: x 6 2 or x 7 8 is a compound inequality that is a union since it contains the word or. 4. Both graphs have dots at 3 and 3. The graph of ƒ x ƒ  3 is darkened between the two dots. The graph of ƒ x ƒ  3 is darkened to the right of the dot at 3 and to the left of the dot at 3. 5. 23  g  6 23  6  g  6  6 17  g 17  g is the same as g  17. Check: Substitute 17, a number less than 17, and a number greater than 17. Let g  17. Let g  20. ?

?

9  5 6 2(9)  14

8.

7 w 8 8 7 w 7 8

12

 21 

187 2(21)

w  24 Check: Substitute 24, a number less than 24, and a number greater than 24. Let w  24. Let w  40. Let w  16. ? 7 (242  8

21

? 7 (402  8

21  21 ✓

21

35 21

The solution set is 5w ƒ w  246.

? 7 (162  8

21

14  21 ✓

9. 22b  99 22b 22

99

 22

b  4.5 Check: Substitute 4.5, a number less than 4.5, and a number greater than 4.5. Let b  4.5. Let b  6. ?

22(4.5)  99 99  99 ✓

?

23  20  6 23  26 ✓

?

22(6)  99 132 99

Let b  1. ?

22(1)  99 22  99 ✓ The solution set is 5b ƒ b  4.56.

?

23  10  6 23 4 The solution set is {g ƒ g  17}.

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10.

4m  11  8m  7 4m  11  4m  8m  7  4m 11  4m  7 11  7  4m  7  7 18  4m 18 4



13.

4m 4

0.14y 0.14

4.5  m 4.5  m is the same as m  4.5. Check: Substitute 4.5, a number less than 4.5, and a number greater than 4.5. Let m  4.5. ? 4(4.52  11  8(4.52  7

?

0.3(16)  0.8(4  2) ?

4.8  0.8(6) 4.8  4.8 ✓ Let y  10. ? 0.3(10  4)  0.8[0.2(10)  2]

Let m  0. ? 4(02  11  8(02  7

?

?

0.3(6)  0.8(2  2) ?

1.8  0.8(4) 1.8  3.2 ✓ Let y  30. ?

0.3(30  4)  0.8[0.2(30)  2] ?

0.3(26)  0.8(6  2)

6 3

?

7.8  0.8(8) 7.8  6.4 The solution set is { y|y  20}. 14. Let p  the selling price of the house.

k 6 2 Check: Substitute 2, a number less than 2, and a number greater than 2. Let k  2. Let k  5. ? ? 3(2  2) 7 12 3(5  2) 7 12 ?

The selling price minus 1 7% times the selling price 14243 is at least 14243 $110,000. 14243 123 23 123 1442443 p – 0.07  p  110,000

?

3(4) 7 12 12  12 Let k  3. ? 3(3  2) 7 12

3(7) 7 12 21 7 12 ✓

p  0.07p  110,000 0.93p  110,000 0.93p 0.93

3(1) 7 12 3  12 The solution set is 5k|k 6 26. f  5 3 f  5 (3) 3

6  ƒrƒ  3 6 ƒrƒ  6  3  6 ƒ r ƒ  3 ƒ r ƒ  3 means that the distance between r and 0 is 3 units. Since distance cannot be negative, there is no real number that makes this statement true. The solution set is the empty set . 16. Write |d| 2 as d 2 or d 2. The solution set is {d|d is a real number}. 15.

7 3 7 (3)(3)

7 9 7 9  5 7 4 Substitute 4, a number less than 4, and a number greater than 4. Let f  4. Let f  10. Let f  5. ?

7 3 ?

7 3

10  5 3 15 3

?

7 3 ?

7 3

3  3 5  3 The solution set is { f|f  4}.

Chapter 6

110,000 0.93

The selling price must be at least $118,280.

f5 f55 f Check:

4  5 3 9 3



p  118,280 1to the nearest dollar2

?

12.

Substitute 20, a number less than 20, and a number greater than 20.

Let y  20. ? 0.3(20  4)  0.8[0.2(20)  2]

24  11  48  7 11 7 35  41 ✓ The solution set is {m ƒ m  4.5}. 3(k  2) 7 12 11. 3k  6 7 12 3k  6  6 7 12  6 3k 7 6 6

2.8

 0.14

y  20 Check:

?

18  11  36  7 29  29 ✓ Let m  6. ? 4(62  11  8(62  7

3k 3

0.3( y  4)  0.8(0.2y  2) 0.3y  1.2  0.16y  1.6 0.3y  1.2  0.16y  0.16y  1.6  0.16y 0.14y  1.2  1.6 0.14y  1.2  1.2  1.6  1.2 0.14y  2.8

5  5 3 0 3

?

7 3 ?

7 3

0 7 3 ✓

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r3 7 2

17.

20. Write |2a  5| 6 7 as 2a  5 6 7 2a  5 7 7. Case 1: Case 2: 2a  5 6 7 2a  5 2a  5  5 6 7  5 2a  5  5 2a 6 12 2a

4r 6 12

and

4r 4

12 4

6 r33 7 23 r 6 3 r 7 1 The solution set is the intersection of the two graphs. Graph r 1. Graph r 3. Find the intersection. –5

–4

–3

–2

–1

0

1

2

3

4

2a 2

6

2a 2

12 2

–4

–3

–2

–1

0

1

2

3

4

5

–5

–4

–3

–2

–1

0

1

2

3

4

5



3s 3

3n  2  1 3n  2  2  1  2 3n  3 3n 3

15 3



5

–1

0

1

2

3

4

5

6

7

–3

–2

–1

0

1

2

3

4

5

6

7

–3

–2

–1

0

1

2

3

4

5

6

7

7

6 2

16 8

7

s3

5

3 21 0 1 2 3 4

6

2

5 6 7

22. Write |7  5z| 7 3 as 7  5z 7 3 or 7  5z 6 3. Case 1: Case 2: 7  5z 7 3 7  5z 6 3 7  5z  7 7 3  7 7  5z  7 6 3  7 5z 7 4 5z 6 10 5z 5

6

5z 5

4 5

23. Let n  the number. 1 n 4 1 (4) 4n

 3

 (4) (3) n  12 Check: Substitute 12, a number less than 12, and a number greater than 12. Let n  12. Let n  20. Let n  4.

p 7 3 2 7 p The solution set is the intersection of the two graphs. Graph p 7 3. Graph p 6 2. Find the intersection.

? 1 1122  4

3

? 1 1202  4

3

3  3 ✓ 5 3 The solution set is {n|n  12}.

–6

–5

–4

–3

–2

–1

0

1

2

3

–7

–6

–5

–4

–3

–2

–1

0

1

2

3

–7

–6

–5

–4

–3

–2

–1

0

1

2

3

10 5

543 2 1 0 1 2 3 4 5

8p 8

–7

7

z 6 0.8 z 7 2 The solution set is {z|z 6 0.8 or z 7 2}.

The solution set is {n|n  1 or n  5}. 19. and 9  2p 7 3 13 7 8p  3 9  2p  9 7 3  9 13  3 7 8p  3  3 2p 7 6 16 7 8p 2p 2

2

9

 3

The solution set is s|s  13 or s  3 .

n5 n  1 The solution set is the union of the two graphs. Graph n  5. Graph n  1. Find the union. –2

3s 3

5

 3

s  3 or 13

3 3

–3

5 6 7

21. Write |7  3s|  2 as 7  3s  2 or 7  3s  2. Case 1: Case 2: 7  3s  2 7  3s  2 7  3s  7  2  7 7  3s  7  2  7 3s  5 3s  9

The solution set is {r|  1 r 3}.

3n 3

2 2

7

5

–5

or 3n  2  17 3n  2  2  17  2 3n  15

7 7 7 7  5 7 2

a 6 6 a 7 1 The solution set is 5a|1 6 a 6 66. 3 21 0 1 2 3 4

18.

and

? 1 142  4

3

1  3 ✓

The solution set is { p|3 6 p 6 2}.

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24. Let n  the number. 14  3n 6 2 14  3n  14 6 2  14 3n 6 12 3n 3

The half plane that contains (0, 0) should be shaded. y

12 3

7

n 7 4 Substitute 4, a number less than 4, and a number greater than 4. Let n  4. Let n  2. Let n  6. Check:

?

?

14  3(4) 6 2

14  3(2) 6 2

?

y  3x  2

?

14  3(6) 6 2

?

?

14  12 6 2 14  6 6 2 14  18 6 2 2 2 8 2 4 6 2 ✓ The solution set is 5n|n 7 46. 25. Let n  the number. 13 6 2n  5 6 21 First express 13 6 2n  5 6 21 using and. and 13 6 2n  5 2n  5 6 21 13  5 6 2n  5  5 2n  5  5 6 21  5 18 6 2n 2n 6 26 18 2

6

2n 2

2n 2

6

28. Since the boundary is not included in the solution, draw a dashed line. Test the point (0, 0). 2x  3y 6 6 2(0)  3(0) 6 6 0 6 6 true The half plane that contains (0, 0) should be shaded. y

26 2

9 6 n n 6 13 The solution set is the intersection of 9 6 n and n 6 13. Check: Substitute a number less than 9, 9, a number between 9 and 13, 13, and a number greater than 13. Let n  2. Let n  9. ?

?

13 6 2(2)  5 6 21 ?

?

? ?

Let n  13.

?

?

?

?

13 6 2(13)  5 6 21 ?

13 6 22  5 6 21 13 6 17 6 21 ✓

29. Since the boundary is not included in the solution, draw a dashed line. Test the point (0, 0). x  2y 7 4 0  2102 7 4 0 7 4 false The half plane that does not contain (0, 0) should be shaded.

?

Let n  11.

13 6 2(11)  5 6 21

2x  3y  6

?

13 6 18  5 6 21 13 13 6 21

?

x O

13 6 2(9)  5 6 21

13 6 4  5 6 21 13 1 6 21 ?

?

13 6 26  5 6 21 13 6 21 21

Let n  15. ?

x

O

y

?

13 6 2(15)  5 6 21 ?

?

13 6 30  5 6 21 13 6 25 21 The solution set is 5n|9 6 n 6 136. 26. Let m  the number of miles Megan drives.

x

O

x  2y  4

m

Then 15 represents the number of miles per gallon Megan’s car gets. If 18 6 18

m 6 15 m 6 15

21, then 18 6 and

115218 6 1152 15 m

m 15

and

m 15 m 1152 15

m 15

6 21.

30. B; Find the point that is the same distance from 3 as the distance from 7. The midpoint of 3 and 7 is 2.

6 21 6 115221

5 units

270 6 m m 6 315 So, 270 6 m 6 315. Megan can drive between 270 and 315 miles. 27. Since the boundary is included in the solution, draw a solid line. Test the point (0, 0). y  3x  2 0  3(0)  2 true 0  2 Chapter 6

–3

–2

–1

0

5 units

1

2

3

4

5

6

7

All points on the graph are at least 5 units from 2. So, an inequality is |x  2|  5.

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9. C; The graph is the intersection of x  2 and x 6 3. The compound inequality x  2 and x 6 3 can be expressed as 2  x 6 3. 10. There are 6 possible outcomes; 4 are successes and 2 are failures. So, the odds of rolling a number less than five are 4 2 or 1 or 2:1. 2

Chapter 6 Standardized Test Practice Pages 364–365 3

?

9

1. B; 9 7 3 1

3 7 3 ✓ 2. D; (6) (7)  42 3. C; V  r2h 5625  r2 1252 5625 25

11. Since 1 h  60 min, 54 miles per hour is the same 54 mi as 54 miles per 60 minutes, or 60 min. Let m  the number of minutes it will take to travel 117 mi.

 r2

225  r2 15  r The radius is 15 cm. 4. B; Let h  the number of hours worked. 80  65h  177.50 80  65h  80  177.50  80 65h  97.50 65h 65



54 60

 

54m 54

4(220  15) 5 412052

12 48

5 820 5

3

4 16

3

1 4

 12

3

  12

2 8

1200 48

The sum of the grades divided by (78  82  75  x)

So, the inequality is



8 32

m

m  1 Step 2 You know the slope and two points. Choose one point and find the y-intercept. In this case, (2, 1) is used. y  mx  b 1  1(2)  b 1  2  b 1  2  2  b  2 3b Step 3 Write the slope-intercept form using m  1 and b  3. y  mx  b y  1x  3 Therefore, the equation is y  x  3.

the number must be of4244 grades at least 80. 14 3 14243 123 4



1  4 (1) 3 3

m2

  12 12

78  82  75  x 4

48r 48

y2  y1

The difference of x values is 3, and the difference of y values is 12. The difference of y values is four times the difference of x values. This suggests y  4x. Check: If x  1, then y  4(1) or 4. If x  8, then y  4(8) or 32. Thus, y  4x describes this set of data. 8. B; 1444442444443 14 4244 3



mx x 2 1

 5 20

r

 100

25  r The percent of decrease is 25%. 13. Select two points on the line, for example (1, 4) and (2, 1). Step 1 Find the slope of the line containing the points. Let (x , y )  (1, 4) and (x , y )  (2, 1). 1 1 2 2

3



7020 54

12(100)  48(r) 1200  48r

 164 Cameron’s recommended maximum pulse rate is 164. 6. A; For the function y  2x  1, every ordered pair that satisfies the equation can be written as (x, 2x  1) . To translate the graph of this function 3 units up, add 3 to the y-coordinate of each point on the graph. P(x, 2x  1) S P¿(x, (2x  1)  3) S P¿(x, 2x  2) So, the equation that represents the new line, or image, is y  2x  2. 7. D; x y



m  130 It will take 130 min to travel 117 mi. 12. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 48  36  12 Find the percent using the original number, 48, as the base.

97.50 65



117 m

54(m)  60(117) 54m  7020

h  1.5 The technician worked 1.5 h. 4(220  A) 5. B; P  5 P



80

 80.

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1

19. A; Since x 7 5 or x 6 7, the distance from 0 to x is greater than 5 units. Since 3 y 4, the distance from 0 to y is less than 4 units. A number |x| that is greater than 5 is greater than a number |y| that is less than 4. So, the quantity in Column A is greater. 20a. 10 ft

2

14. The line parallel to 3y  3x  1 has the same slope. Rewrite the equation of the line in slopeintercept form, y  mx  b. 1 y 3

2

 3x  1

1

12

(3) 3y  (3) 3x  1

2

y  2x  3 The slope of the line is 2. 1

15. 2 (10x  8)  3(x  1)  15 5x  4  3x  3  15 2x  1  15 2x  1  1  15  1 2x  16 2x 2



16 2

10 ft

x8 The solution set is 5x|x  86. 16. Write |x  3| 7 5 as x  3 7 5 and x  3 6 5. Case 1: Case 2: x3 7 5 x  3 6 5 x33 7 53 x  3  3 6 5  3 x 7 8 x 6 2 The solution set is {x|x 6 2 and x 7 8}. 17. Since the boundary is not included in the solution, draw a dashed line. Test the point (0, 0). y 6 2x  4 0 6 2102  4 true 0 6 4 The half plane that contains (0, 0) should be shaded.

91 ft

Let /  the length of the house. The length cannot be the length decreased the number of the house greater than of the lot by 10 ft times of edges. 1442443 1442443 1442443 1442443 123 123 1442443 

/

91



10



2

So, the inequality for the possible lengths of the house is /  91  20, or /  71. Therefore, the length of the house is at most 71 ft. 20b. If A  the area of the house, then 2800  A  3200. Since A  /w, then 2800  /w  3200. If the house has the maximum possible length, then /  71, so 2800  71w  3200. First express 2800  71w  3200 using and. Then solve each inequality. and 2800  71w 71w  3200

y  2x  4

x

2800 71

18. B; Use a calculator to find an approximation for 168. 168  8.246211251... Therefore, 168 6 9. So, the quantity in Column B is greater.

Chapter 6

158 ft

10 ft

y

O

10 ft



71w 71

39  w

(to the nearest ft)

71w 71



3200 71

w  45

The solution set is 5w|39  w  456. The width of the house can be between 39 and 45 ft, inclusive.

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Chapter 7 Page 367

Solving Systems of Linear Equations and Inequalities Graph the ordered pairs and draw a line through the points.

Getting Started

1. The only value in the range is 1. Select five values for the domain and make a table. x 3 1 0 2 4

y y4x

(x, y) (3, 1) (1, 1) (0, 1) (2, 1) (4, 1)

y 1 1 1 1 1

Graph the ordered pairs and draw a line through the points.

4. To find the x-intercept, let y  0. y  2x  3 0  2x  3 0  3  2x  3  3 3  2x

y y1

3 2

x

O

2x 2(2) 2(1) 2(0) 2(1) 2(2)

y 4 2 0 2 4



2x 2

1.5  x The graph intersects the x-axis at (1.5, 0). To find the y-intercept, let x  0. y  2x  3 y  2(0)  3 y3 The graph intersects the y-axis at (0, 3). Plot these points and draw the line that connects them.

2. Select five values for the domain and make a table. x 2 1 0 1 2

x

O

(x, y) (2, 4) (1, 2) (0, 0) (1, 2) (2, 4)

y

Graph the ordered pairs and draw a line through the points.

y  2x  3

y y  2x

O

5. To find the x-intercept, let y  0. y  5  2x 0  5  2x 0  2x  5  2x  2x 2x  5

x

2x 2

3. Select five values for the domain and make a table. x 1 0 2 4 5

x

O

4x 4  (1) 40 42 44 45

y 5 4 2 0 1

5

2

x  2.5 The graph intersects the x-axis at (2.5, 0). To find the y-intercept, let x  0. y  5  2x y  5  2(0) y5 The graph intersects the y-axis at (0, 5).

(x, y) (1, 5) (0, 4) (2, 2) (4, 0) (5, 1)

305

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Plot these points and draw the line that connects them.

9.

y

7bc  d 10 7bc  d (10) 10

 (10)12

7bc  d  120 7bc  d  d  120  d 7bc  120  d

y  5  2x

120  d 7c 120  d b  7c 120  d of b is 7c .

7bc 7c

x

O

 12



The value Since division by 0 is undefined, 7c  0 or c  0. 10.

6. To find the x-intercept, let y  0. 1

y  2x  2

7m  n 2m 7m  n 2m

1

0  2  2x  2  2 1

2  2x

1

y  2x  2 1

y  2 (0)  2 y2 The graph intersects the y-axis at (0, 2). Plot these points and draw the line that connects them. y 1

4x  a  6x 4x  a  4x  6x  4x a  2x 2x 2

x a

The value of x is 2 . 8a  y  16 8a  y  y  16  y 8a  16  y

The

16  y 8 16  y a 8 16  y value of a is 8 .

Chapter 7

q

14. (8x  4y)  (8x  5y)  8x  4y  8x  5y  8x  8x  4y  5y  (8  8)x  (4  5)y  0x  (1)y y 15. 4(2x  3y)  (8x  y)  4 # 2x  4 # 3y  8x  (y)  8x  12y  8x  y  8x  8x  12y  y  (8  8)x  (12  1)y  0x  13y  13y 16. 3(x  4y)  (x  12y)  3  x  3  4y  x  12y  3x  12y  x  12y  3x  x  12y  12y  (3  1)x  (12  12)y  4x  0y  4x 17. 2(x  2y)  (3x  4y)  2  x  2  2y  3x  4y  2x  4y  3x  4y  2x  3x  4y  4y  (2  3)x  (4  4)y  5x  0y  5x

x

O

8a 8

2qm 2m

 16x  11x  3y  3y  (16  11)x  (3  3)y  27x  0y  27x

y 2x2

8.



7m  n

4  x The graph intersects the x-axis at (4, 0). To find the y-intercept, let x  0.



 (q)2m

The value of q is 2m . Since division by 0 is undefined, 2m  0 or m  0. 11. (3x  y)  (2x  y)  3x  y  2x  y  3x  2x  y  y  (3  2)x  (1  1)y  1x  0y x 12. (7x  2y)  (7x  4y)  7x  2y  7x  4y  7x  7x  2y  4y  (7  7)x  (2  4)y  0x  (6)y  6y 13. (16x  3y)  (11x  3y)  16x  3y  11x  3y

1

(2)(2)  (2) 2x

a 2 a 2

 2m

7m  n  2qm

1

0  2x  2

7.

7m  n q 7m  n (q) q



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18. 5(2x  y)  2(5x  3y)  5  2x  5  y  2  5x  2  3y  10x  5y  10x  6y  10x  10x  5y  6y  (10  10)x  (5  6)y  0x  (11)y  11y 19. 3(x  4y)  2(2x  6y)  3  x  3  4y  2  2x  2  (6y)  3x  12y  4x  (12)y  3x  4x  12y  12y  (3  4)x  (12  12)y  7x  0y  7x

Page 368

7-1

Pages 371–372

y

y (2, 3)

True or False?

y  375  0.15x

The system of equations has one solution at (2, 3) since the graphs intersect at (2, 3). 2. Always; if a system of linear equations has 2 solutions, the graphs are the same line and there are infinitely many solutions. 3. Sample answer: The graphs of the equations x  y  3 and 2x  2y  6 have a slope of 1. Since the graphs of the equations coincide, there are infinitely many solutions.

True or False?

true ✓

410  375  0.15(100) 410  390

false true ✓

100

410 410  400  0.1(100) 410  410

300

420 420  400  0.1(300) 420  430

false

420  375  0.15(300) 420  420

500

450 450  400  0.1(500) 450  450

true ✓

450  375  0.15(500) 450  450

true ✓

900

510 900  400  0.1(900) 900  490

false

510  375  0.15(900) 510  510

true ✓

1

4. Since the graphs of y  x  4 and y  3 x  2 intersect, there is one solution. 1

Job Salaries

Series 1

450 400 350

y  2x

(0, 0) Series 2

O

0 20 0 40 0 60 0 80 0 10 00

Total Weekly Salary

1

7. Since the graphs of x  y  4 and y  3x  4 intersect, there is one solution. 8. y

550 500

1

5. Since the graphs of y  3 x  2 and y  3 x  2 are parallel, there are no solutions. 6. Since the graphs of x  y  4 and y  x  4 coincide, there are infinitely many solutions.

Since only the ordered pair (500, 450) makes both statements true, the correct choice is c. 4.

x

O

Spreadsheet Investigation (Preview of Lesson 7-1)

y  400  0.1x

Check for Understanding

1. Sample answer:

1. y  400  0.1x 2. y  375  0.15x 3. Make a table. Substitute each ordered pair into both equations. x

Graphing Systems of Equations

x

y  x

Weekly Sales

The two lines intersect at (500, 450). If Mr. Winter makes $500 in sales, he will make $450 for either job. 5. Sample answer: Write and graph two linear equations. Find the point where the graphs intersect.

The graphs appear to intersect at the point with coordinates (0, 0). Check this estimate by replacing x with 0 and y with 0 in each equation. Check: y  x y  2x ? ? 0  2(0) 0  0 00✓ 00✓ There is one solution. It is (0, 0).

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13.

9. y

y

xy8

xy2

xy2

(1, 3)

(5, 3)

y  4x  7 x

x

O

O

The graphs appear to intersect at the point with coordinates (5, 3). Check this estimate by replacing x with 5 and y with 3 in each equation. Check: xy8 xy2 ? ? 538 532 88✓ 22✓ There is one solution. It is (5, 3). 10.

The graphs appear to intersect at the point with coordinates (1, 3). Check this estimate by replacing x with 1 and y with 3 in each equation. Check: xy2 y  4x  7 ? ? 3  4(1)  7 1  3  2 ? 22✓ 3  4  7 33✓ There is one solution. It is (1, 3). 14. Let a  the price for an adult, and let c  the price for a child. Rodriguez family: 2a  3c  40.50 Wong family: 3a  c  38.00 Graph the equations 2a  3c  40.50 and 3a  c  38.

y 2x  4y  2

x

O 3x  6y  3

11.

Price of Child Ticket

The graphs of the equations coincide. Since every point is a point of intersection, there are infinitely many solutions to this system of equations. y xy4

y

3y  2x  9

(10.5, 6.5) 2a  3c  40.50

The graphs appear to intersect at the point with coordinates (10.5, 6.5). Check this estimate by replacing a with 10.5 and c with 6.5 in each equation. Check: 2a  3c  40.50 3a  c  38.00 ? ? 2(10.5)  3(6.5)  40.50 3(10.5)  6.5  38.00 ? ? 31.5  6.5  38.00 21  19.5  40.50 40.5  40.50 ✓ 38  38.00 ✓ The price for an adult was $10.50, and the price for a child was $6.50.

The graphs of the equations are parallel lines. Since they do not intersect, there are no solutions to this system of equations. 12.

3a  c  38

0 4 8 12 16 20 24 28 32 36 a Price of Adult Ticket

x

O xy1

c 36 32 28 24 20 16 12 8 4

(3, 1) x

O

xy2

Pages 372–374

The graphs appear to intersect at the point with coordinates (3, 1). Check this estimate by replacing x with 3 and y with 1 in each equation. Check: xy2 3y  2x  9 ? ? 312 3(1)  2(3)  9 ? 22✓ 369 99✓ There is one solution. It is (3, 1). Chapter 7

Practice and Apply

15. Since the graphs of x  3 and y  2x  1 intersect, there is one solution. 16. Since the graphs of y  x  2 and y  2x  4 intersect, there is one solution. 17. Since the graphs of y  x  2 and y  x  2 coincide, there are infinitely many solutions. 18. Since the graphs of y  2x  1 and y  2x  4 are parallel, there are no solutions.

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19. Since the graphs of y  3x  6 and y  2x  4 intersect, there is one solution. 20. Since the graphs of 2y  4x  2 and y  2x  4 are parallel, there are no solutions. 21. Since the graphs of 2y  4x  2 and y  3x  6 intersect, there is one solution. 22. Since the graphs of 2y  4x  2 and y  2x  1 coincide, there are infinitely many solutions. 23. y

26.

x (2, 2)

y  x

The graphs appear to intersect at (2, 2). Check in each equation. y  x y  2x  6 Check: ? ? 2  (2) 2  2(2)  6 ? 2  2 ✓ 2  4  6 2  2 ✓ There is one solution. It is (2, 2).

4x  y  2

y  6 (2, 6)

27.

The graphs appear to intersect at (2, 6). Check in each equation. y  6 4x  y  2 Check: ? 6  6 ✓ 4(2)  (6)  2 ? 8  (6)  2 22✓ There is one solution. It is (2, 6). 24.

y  2x  6

O

x

O

y

y

x

O

y  3x  4 y  3x  4 (0, 4)

y

The graphs appear to intersect at (0, 4). Check in each equation. y  3x  4 y  3x  4 Check: ? ? 4  3(0)  4 4  3(0)  4 4  4 ✓ 4  4 ✓ There is one solution. It is (0, 4).

x2 3x  y  8

x

O (2, 2)

28. The graphs appear to intersect at (2, 2). Check in each equation. x2 3x  y  8 Check: ? 22✓ 3(2)  (2)  8 ? 628 88✓ There is one solution. It is (2, 2). 25.

y  2x  6 (3, 0)

y (4, 2)

O

x

The graphs appear to intersect at (3, 0). Check in each equation. y  2x  6 y  x  3 Check: ? ? 0  (3)  3 0  2(3)  6 ? ? 0  6  6 033 00✓ 00✓ There is one solution. It is (3, 0).

y  1x 2

x

O

y y  x  3

2x  y  10

The graphs appear to intersect at (4, 2). Check in each equation. Check:

1

y  2x 2

? 1  2 (4)

22✓

2x  y  10 ?

2(4)  2  10 ?

8  2  10 10  10 ✓

There is one solution. It is (4, 2).

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29.

33.

y

16

3x  y  6

y

12

(2, 0 )

O

5x  3y  6 (6, 8)

x

4 2x  y  4

x  2y  2

16 12 8

The graphs appear to intersect at (2, 0). Check in each equation. Check: x  2y  2 3x  y  6 ? ? 2  2(0)  2 3(2)  0  6 22✓ 66✓ There is one solution. It is (2, 0). 30.

8

xy 2

O x

4

The graphs appear to intersect at (6, 8). Check in each equation. Check: 2x  y  4 5x  3y  6 ? ? 2(6)  8  4 5(6)  3(8)  6 ? ? 12  8  4 30  24  6 4  4 ✓ 6  6 ✓ There is one solution. It is (6, 8).

y

34.

y

5x  8y  17

(2, 4) (3, 4)

2y  x  10

x O O

The graphs appear to intersect at (2, 4). Check in each equation. Check: xy2 2y  x  10 ? ? 2  4  2 2(4)  (2)  10 ? 22✓ 8  2  10 10  10 ✓ There is one solution. It is (2, 4). 31.

35.

3x  2y  12

y 2y  6x  6

3x  2y  6

3x  y  3

x

O

O

The graphs of the equations are parallel lines. Since they do not intersect, there are no solutions to this system of equations. y

2x  3y  4 O

x

The graphs of the equations coincide. Since every point is a point of intersection, there are infinitely many solutions to this system of equations.

4x  6y  8

x

The graphs of the equations coincide. Since every point is a point of intersection, there are infinitely many solutions to this system of equations.

Chapter 7

x

The graphs appear to intersect at (3, 4). Check in each equation. Check: 4x  3y  24 5x  8y  17 ? ? 4(3)  3(4)  24 5(3)  8(4)  17 ? ? 12  12  24 15  32  17 24  24 ✓ 17  17 ✓ There is one solution. It is (3, 4).

y

32.

4x  3y  24

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36.

40.

y

12 10 8 6 4 2

3y  x  5

yx3

(1, 2)

x

O

2 2 4

The graphs appear to intersect at (1, 2). Check in each equation. Check: yx3 3y  x  5 ? ? 2  1  3 3(2)  (1)  5 ? 22✓ 615 55✓ There is one solution. It is (1, 2). 37.

x O 2 4 6 8 10 12 14 1 1 x  3y  6 2

1 x 2

1

 3y  6 ?

1

 3 (6)  6 ?

426 66✓ There is one solution. It is (8, 6).

1

y  2x  2 ? 1

6  2 (8)  2 ?

642 66✓

41. Let   the length of the rectangle, and let w  the width of the rectangle. /  2w  1 2/  2w  40 Graph the equations /  2w  1 and 2/  2w  40.

yx4

Length (m)

The graphs appear to intersect at (1, 5). Check in each equation. Check: 2x  3y  17 yx4 ? ? 2(1)  3(5)  17 5  1  4 ? 2  (15)  17 5  5 ✓ 17  17 ✓ There is one solution. It is (1, 5). y

l 18 16 14 12 10 8 6 4 2 0

(7, 13) 2l  2w  40

l  2w  1 0 2 4 6 8 10 12 14 16 18 w Width (m)

The graphs appear to intersect at (7, 13). Check in each equation. /  2w  1 2/  2w  40 Check: ? ? 2(13)  2(7)  40 13  2(7)  1 ? ? 13  14  1 26  14  40 13  13 ✓ 40  40 ✓ The length of the rectangle is 13 m, and the width of the rectangle is 7 m.

x

O

(8, 6)

1 (8) 2

x

38.

2

Check:

O

(1, 5)

y  1x  2

The graphs appear to intersect at (8, 6). Check in each equation.

y 2x  3y  17

y

3y  2x 2

y 3x5

The graphs of the equations are parallel lines. Since they do not intersect, there are no solutions to this system of equations. 39. 16 14 12 10 8 6 4 2 2

y 3

6  8y  x 2 1 x  4y  4 3

x O 2 4 6 8 10 12 14

The graphs of the equations coincide. Since every point is a point of intersection, there are infinitely many solutions to this system of equations.

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42. Graph the equations y  2x  6, 3x  2y  19, and y  2.

3x  2y  19

y2

y y  2x  6

10 8 6

Height (m)

14

Graph the equations h  10  15m and h  150  20m.

(1, 8)

(–2, 2)

(5, 2) O

–8 –6 –4 –2–2

2 4 6 8x

The graphs of y  2x  6 and 3x  2y  19 appear to intersect at (1, 8). Check in each equation. Check: y  2x  6 3x  2y  19 ? ? 8  2(1)  6 3(1)  2(8)  19 ? ? 826 3  16  19 88✓ 19  19 ✓ The graphs of y  2x  6 and y  2 appear to intersect at (2, 2). Check in each equation. Check: y2 y  2x  6 ? 22✓ 2  2(2)  6 ? 2  4  6 22✓ The graphs of 3x  2y  19 and y  2 appear to intersect at (5, 2). Check in each equation. y2 3x  2y  19 Check: ? 22✓ 3(5)  2(2)  19 ? 15  4  19 19  19 ✓ Therefore, the coordinates of the vertices of the triangle are (1, 8), (2, 2), and (5, 2). 43. The base of the triangle from Exercise 42 is the segment joining the points with coordinates (2, 2) and (5, 2). The length of this segment is 7 units, so b  7. The height of the triangle is 6 units, the length of the segment joining the points with coordinates (1, 2) and (1, 8). Thus, h  6. Substitute b  7 and h  6 in the formula for the

h  150  20m h  10  15m (4, 70)

0 1 2 3 4 5 6 7 8 9m Minutes

Total Amount Saved

The graphs appear to intersect at (4, 70). Check in each equation. Check: h  10  15m h  150  20m ? ? 70  10  15(4) 70  150  20(4) ? ? 70  150  80 70  10  60 70  70 ✓ 70  70 ✓ The balloons will be the same height in 4 min. 45. Substitute m  4 in each equation from Exercise 44. h  10  15m h  150  20m h  10  15(4) h  150  20(4) h  10  60 h  150  80 h  70 h  70 In 4 min, both balloons will be 70 m high. 46. Let a  the total amount saved, and let w  the number of weeks. Monica: a  25  5w Michael: a  16  8w Graph the equations a  25  5w and a  16  8w.

1

area of a triangle, A  2bh. 1

A  2bh

a 45 40 35 30 25 20 15 10 5 0

a  25  5w (3, 40)

a  16  8w

0

1

A  276 A  21 So, the area of the triangle is 21 units2. 44. Let h  the height of the balloon in meters, and let m  the number of minutes. Balloon 1: h  10  15m Balloon 2: h  150  20m

Chapter 7

h 180 160 140 120 100 80 60 40 20 0

1

2 3 4 Number of Weeks

w

The graphs appear to intersect at the point with coordinates (3, 40). Check this estimate by replacing w with 3 and a with 40 in each equation. Check: a  25  5w a  16  8w ? ? 40  25  5(3) 40  16  8(3) ? ? 40  16  24 40  25  15 40  40 ✓ 40  40 ✓ Monica and Michael will have saved the same amount in 3 weeks.

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Substitute A  4 into the equation 2A  3B  20, and solve for B. 2A  3B  20 2(4)  3B  20 8  3B  20 8  3B  8  20  8 3B  12

Population (millions of people)

47. Substitute w  3 in each equation from Exercise 46. a  25  5w a  16  8w a  25  5(3) a  16  8(3) a  25  15 a  16  24 a  40 a  40 Monica and Michael will each have saved $40. 48. For each of the years shown, the profit of the Widget Company is greater than the profit of the Gadget Company. For example, the profit of the Widget Company in year 1 appears to be $6,000,000. This is greater than the profit of the Gadget Company in year 1, which appears to be $3,000,000. 49. Since the slopes of the graphs for both companies appear to be the same, the rates of growth appear to be the same. Thus, neither company has a greater rate of growth. 50. The graphs are parallel, so the lines will never meet. Thus, there is no year when the profits of the two companies will be equal if the profit patterns continue. 51. Let t  the number of years since 1990, and let p  the population of the Midwest in millions of people. p  60  0.4t 52. Let t  the number of years since 1990, and let p  the population of the West in millions of people. p  53  1t or p  53  t 53. p 80

3B 3

B  4 So, A  4 and B  4. 56. Graphs can show when the sales of one item are greater than the sales of the other item and when the sales of the items are equal. Answers should include the following. • The sales of cassette singles equaled the sales of CD singles in about 5 years or by the end of 1996. • The graph of each equation contains all of the points whose coordinates satisfy the equation. If a point is contained in both lines, then its coordinates satisfy both equations. 1

57. B; Since the graphs of y  3 x  2 and 1 y  3 x  1 are parallel, the system of equations has no solution. 58. B; Since the graphs of 4x  y  7 and 3x  y  0 intersect, there is one solution. y

3x  y  0

4x  y  7

p  60  0.4t

60

12

 3

(1, 3)

(11.7, 64.7)

p  53  t

40

x

20

t O

2

4 6 8 10 Years Since 1990

Page 374

12

54. The graphs appear to intersect at (11.7, 64.7). Check in each equation. Check: p  60  0.4t p  53  t ? ? 64.7  60  0.4(11.7) 64.7  53  11.7 ? 64.7  60  4.68 64.7  64.7 ✓ ? 64.7  64.68 64.7  64.7 (to the nearest tenth) Since t  11.7, the populations will be the same about 11.7 yr after 1990, or sometime in 2001. 55. Since (2, 3) is the solution of the system of equations, replace x with 2 and y with 3 in each equation. Ax  y  5 Ax  By  20 A(2)  (3)  5 A(2)  B(3)  20 2A  3  5 2A  3B  20 2A  3  3  5  3 2A  8 2A 2

Maintain Your Skills

59. Use a table to substitute the x and y values of each ordered pair into the inequality. x 1

y 4

1

5

5

6

7

0

y  2x 4  2(1) 42 5  2(1) 5  2 6  2(5) 6  10 0  2(7) 0  14

True or False false false true ✓ false

The ordered pair {(5, 6)} is part of the solution set of y  2x.

8

2

A4

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60. Use a table to substitute the x and y values of each ordered pair into the inequality. x 4

y 2

3

0

1

4

1

8

y  8  3x 2  8  3(4) 2  20 0  8  3(3) 0  17 4  8  3(1) 45 8  8  3(1) 85

66.

True or False true ✓

4a 4

true ✓

1

b

true ✓

b

67.

false

7m  n q 7m  n (q) q

 10  (q)10

7m  n  10q 7m  n 10 7m  n 10



10q 10

q

The value of q is 68.

5tz  s 2 5tz  s (2) 2

7m  n . 10

6  (2)6

5tz  s  12 5tz  s  s  12  s 5tz  12  s 12  s 5t 12  s z  5t 12  s value of z is 5t . 5tz 5t



The Since division by 0 is undefined, 5t  0 or t  0.

Page 375

Graphing Calculator Investigation (Follow-Up of Lesson 7-1)

1. Step 1 Each of the equations is already solved for y. y  3x  4 y  0.5x  6 Step 2 Enter these equations in the Y  list and graph. Step 3 Use the CALC menu to find the point of intersection. KEYSTROKES: 2nd [CALC] 5 ENTER

113 2 (x  3)

3y  6  x  3 3y  6  3  x  3  3 3y  3  x 3y  3  3y  x  3y 3  x  3y The standard form of the equation is x  3y  3. 64. y  4  6 (x  2) y  4  6x  12 y  4  4  6x  12  4 y  6x  8 y  6x  6x  8  6x 6x  y  8 The standard form of the equation is 6x  y  8. 65. 12x  y  10x 12x  y  10x  10x  10x 2x  y  0 2x  y  y  0  y 2x  y The value of y is 2x. Chapter 7

1

The value of a is 4 or 4 b.

y  2  3 (x  3) 3( y  2)  3

b

 4

a  4

The ordered pairs {(4, 2), (3, 0), (1, 4)} are part of the solution set of y  8  3x. 61. Let n  the number of ounces of perfume in a bottle. The difference between the actual number and 2 is less than 0.05. |n  2| 6 0.05 Write |n  2| 6 0.05 as n  2 6 0.05 and n  2 7 0.05. Case 1: n  2 6 0.05 n  2  2 6 0.05  2 n 6 2.05 Case 2: n  2 7 0.05 n  2  2 7 0.05  2 n 7 1.95 The solution set is {n|1.95  n  2.05}. A bottle is accepted if it contains between 1.95 and 2.05 oz of perfume. 62. y  1  4(x  5) y  1  4x  20 y  1  1  4x  20  1 y  4x  19 y  4x  4x  19  4x 4x  y  19 (1)(4x  y)  (1) (19) 4x  y  19 The standard form of the equation is 4x  y  19. 63.

6a  b  2a 6a  b  2a  2a  2a 4a  b  0 4a  b  b  0  b 4a  b

ENTER ENTER The solution is approximately (2.86, 4.57). 2. Step 1 Each of the equations is already solved for y. y  2x  5 y  0.2x  4 See Steps 2 and 3 in Exercise 1. The solution is approximately (4.09, 3.18).

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8. Step 1 Solve each equation for y to enter them into the calculator. 2x  3y  11 4x  y  6 2x  3y  2x  11  2x 4x  y  4x  6  4x 3y  11  2x y  6  4x

3. Step 1 Solve each equation for y to enter them into the calculator. x  y  5.35 x  y  x  5.35  x y  5.35  x 3x  y  3.75 3x  y  3x  3.75  3x y  3.75  3x (1)(y)  (1)(3.75  3x) y  3.75  3x See Steps 2 and 3 in Exercise 1. The solution is approximately (2.28, 3.08). 4. Step 1 Solve each equation for y to enter them into the calculator. 0.35x  y  1.12 0.35x  y  0.35x  1.12  0.35x y  1.12  0.35x (1)(y)  (1)(1.12  0.35x) y  1.12  0.35x 2.25x  y  4.05 2.25x  y  2.25x  4.05  2.25x y  4.05  2.25x See Steps 2 and 3 in Exercise 1. The solution is approximately (1.13, 1.51). 5. Step 1 Solve each equation for y to enter them into the calculator. 1.5x  y  6.7 1.5x  y  1.5x  6.7  1.5x y  6.7  1.5x 5.2x  y  4.1 5.2x  y  5.2x  4.1  5.2x y  4.1  5.2x (1)(y)  (1)(4.1  5.2x) y  4.1  5.2x See Steps 2 and 3 in Exercise 1. The solution is approximately (1.61, 4.28). 6. Step 1 Solve each equation for y to enter them into the calculator. 5.4x  y  1.8 5.4x  y  5.4x  1.8  5.4x y  1.8  5.4x (1)(y)  (1)(1.8  5.4x) y  1.8  5.4x 6.2x  y  3.8 6.2x  y  6.2x  3.8  6.2x y  3.8  6.2x See Steps 2 and 3 in Exercise 1. The solution is approximately (0.17, 2.73). 7. Step 1 Solve each equation for y to enter them into the calculator. 5x  4y  26 4x  2y  53.3 5x  4y  5x  26  5x 4x  2y  4x  53.3  4x 4y  26  5x 2y  53.3  4x 4y 4



y

26  5x 4 26 5x 4  4

y  6.5  1.25x See Steps 2 and 3 in Exercise 1. The solution is (10.2, 6.25).

2y 2



y

3y 3



y

11  2x 3 11 2  3x 3

See Steps 2 and 3 in Exercise 1. The solution is (2.9, 5.6). 9. Step 1 Solve each equation for y to enter them into the calculator. 0.22x  0.15y  0.30 0.22x  0.15y  0.22x  0.30  0.22x 0.15y  0.30  0.22x 0.15y 0.15



y y

0.30  0.22x 0.15 0.30 0.22x  0.15 0.15 22 2  15x

0.33x  y  6.22 0.33x  y  0.33x  6.22  0.33x y  6.22  0.33x See Steps 2 and 3 in Exercise 1. The solution is approximately (2.35, 5.44). 10. Step 1 Solve each equation for y to enter them into the calculator. 125x  200y  800 125x  200y  125x  800  125x 200y  800  125x 200y 200



y

800  125x 200 800 125x 200  200

y  4  0.625x 65x  20y  140 65x  20y  65x  140  65x 20y  140  65x 20y 20



y

140  65x 20 140 65x  20  20

y  7  3.25x See Steps 2 and 3 in Exercise 1. The solution is approximately (1.14, 3.29).

53.3  4x 2 53.3 4x  2 2

y  26.65  2x

315

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Page 379

Substitution

7-2 Page 376

Algebra Activity

1. Step 1 Model the equation. Since y  x  4, use 1 positive x tile and 4 negative 1 tiles to represent y.

3x

x

y

x

x



1 1

x

1 1

1

1

1

1

1

1

1

1

10y 10

3x  y  8

x

x

x



1

1

1

1

1

1

1 1

1 1

1

1

1 1

1 1

1

1

1

3

1

Separate the tiles into 4 groups.

x

1

1

1

4x 4

x

1

1

1

x

1

1

1

x

1

1

1

x3

3. 4.

5.

So, the value of x is 3. Replace x with 3 in the equation and solve for y. yx4 y34 y  1 The value of y is 1. Since x  3 and y  1, the solution of the system of equations is the ordered pair (3, 1). On an equation mat, use algebra tiles to model 4x  3y  10 using 1 positive x tile and 1 positive 1 tile to represent each y. Use what you know about equation mats to solve for x. Use the value of x and y  x  1 to solve for y. The solution is (1, 2). This method is called substitution since you substitute a representation of y for y.

Chapter 7



12 4

x3 Substitute 3 for x in either equation to find the value of y. y4x y43 y1 The solution is (3, 1). 6. Since x  1  4y, substitute 1  4y for x in the first equation. 2x  7y  3 2(1  4y)  7y  3 2  8y  7y  3 2y3 2y232 y  1 (1) (y)  (1)1 y  1 Substitute 1 for y in either equation to find the value of x. x  1  4y x  1  4(1) x14 x5 The solution is (5, 1).



2.

or 1.5

Use x  2y to find the value of x. x  2y x  2(1.5) x3 The solution is (3, 1.5). 5. Since y  3x  8, substitute 3x  8 for y in the second equation. y4x 3x  8  4  x 3x  8  8  4  x  8 3x  12  x 3x  x  12  x  x 4x  12

4x  12

Step 3

15

 10

y2

Step 2 Add 4 positive 1 tiles to each side to isolate the x tiles. Remove zero pairs.

x

Check for Understanding

1. Substitution may result in a more accurate solution than graphing. 2. Since the statement 4  2 is false, there are no solutions of the system of equations. This means that the graphs do not intersect, so they are graphs of parallel lines. 3. Sample answer: y  x  3 and 2y  2x  6 4. Since x  2y, substitute 2y for x in the second equation. 4x  2y  15 4(2y)  2y  15 8y  2y  15 10y  15

316

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Since both vehicles travel the same distance, t  c. Use substitution to solve this system.

7. Since y  3x  2, substitute 3x  2 for y in the first equation. 6x  2y  4 6x  2(3x  2)  4 6x  6x  4  4 4  4 The statement 4  4 is true. This means that there are infinitely many solutions of the system of equations. 8. Solve the second equation for x since the coefficient of x is 1. xy8 xyy8y x8y Find the value of y by substituting 8  y for x in the first equation. x  3y  12 (8  y)  3y  12 8  4y  12 8  4y  8  12  8 4y  4 4y 4

c 200 c 200 c (152,600) 200

563c 563

17x 17

13 2

3x  5 5x  15 3x  3x  15 0  15 The statement 0  15 is false. This means that there is no solution of the system of equations. 10. Let t  the distance traveled by the Thrust SSC in miles, and let c  the distance traveled by the car in miles. Use a table to organize the information.

763

t 763

t

Car

200

c 200

c

11y 11

t

76,300 563

Practice and Apply

34

 17

44

 11

y4 Use x  4y to find the value of x. x  4y x  4(4) x  16 The solution is (16, 4). 13. Since x  4y  5, substitute 4y  5 for x in the second equation. x  3y  2 4y  5  3y  2 4y  5  5  3y  2  5 4y  3y  7 4y  3y  3y  7  3y y  7 Use x  4y  5 to find the value of x. x  4y  5 x  4(7)  5 x  28  5 x  23 The solution is (23, 7).

Since the car had a head start of one-half hour, c 200



x2 Use y  5x to find the value of y. y  5x y  5(2) y  10 The solution is (2, 10). 12. Since x  4y, substitute 4y for x in the second equation. 2x  3y  44 2(4y)  3y  44 8y  3y  44 11y  44

3

Thrust SSC

1 763c  12 2

11. Since y  5x, substitute 5x for y in the second equation. 2x  3y  34 2x  3(5x)  34 2x  15x  34 17x  34

9. Since y  5 x, substitute 5 x for y in the second equation. 3x  5y  15

Distance

1

 (152,600)

Pages 379–381

y1 Substitute 1 for y in either equation to find the value of x. xy8 x18 x1181 x9 The solution is (9, 1).

Time

c

 763  2

c  135.5 (to the nearest tenth) The Thrust SSC will pass the car at about 135.5 mi.

4

Rate

1

763c  200c  76,300 763c  200c  200c  76,300  200c 563c  76,300

4

3

t

 763  2

1

 763  2.

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14. Since y  2x  3, substitute 2x  3 for y in the second equation. y  4x  1 2x  3  4x  1 2x  3  3  4x  1  3 2x  4x  4 2x  4x  4x  4  4x 2x  4 2x 2

18. Solve the second equation for y since the coefficient of y is 1. 3x  y  9 3x  y  3x  9  3x y  9  3x Find the value of x by substituting 9  3x for y in the first equation. 2x  y  4 2x  (9  3x)  4 2x  9  3x  4 9  x  4 9  x  9  4  9 x  13 (1) (x)  (1) (13) x  13 Substitute 13 for x in either equation to find the value of y. 3x  y  9 3(13)  y  9 39  y  9 39  y  39  9  39 y  30 The solution is (13, 30). 19. Solve the second equation for x since the coefficient of x is 1. x  3y  1 x  3y  3y  1  3y x  1  3y Find the value of y by substituting 1  3y for x in the first equation. 3x  5y  11 3(1  3y)  5y  11 3  9y  5y  11 3  4y  11 3  4y  3  11  3 4y  8

4

 2

x2 Use y  2x  3 to find the value of y. y  2x  3 y  2(2)  3 y43 y7 The solution is (2, 7). 15. Since c  d  1, substitute d  1 for c in the first equation. 4c  3d  3 4(d  1)  3d  3 4d  4  3d  3 4d  4  4  3d  3  4 4d  3d  7 4d  3d  3d  7  3d d7 Use c  d  1 to find the value of c. cd1 c71 c6 The solution is (6, 7). 16. Since y  3x  13, substitute 3x  13 for y in the first equation. 4x  5y  11 4x  5(3x  13)  11 4x  15x  65  11 19x  65  11 19x  65  65  11  65 19x  76 19x 19

4y 4

y2 Substitute 2 for y in either equation to find the value of x. x  3y  1 x  3(2)  1 x61 x6616 x7 The solution is (7, 2).

76

 19

x4 Use y  3x  13 to find the value of y. y  3x  13 y  3(4)  13 y  12  13 y  1 The solution is (4, 1). 17. Solve the second equation for y since the coefficient of y is l. 4x  y  11 4x  y  4x  11  4x y  11  4x Find the value of x by substituting 11  4x for y in the first equation. 8x  2y  13 8x  2(11  4x)  13 8x  22  8x  13 22  13 The statement 22  13 is false. This means that there is no solution of this system of equations. Chapter 7

8

4

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Find the value of r by substituting 5  5r for s in the second equation. 4r  5s  17 4r  5(5  5r)  17 4r  25  25r  17 21r  25  17 21r  25  25  17  25 21r  42

20. Solve the second equation for y since the coefficient of y is 1. 3x  y  15 3x  y  3x  15  3x y  15  3x Find the value of x by substituting 15  3x for y in the first equation. 2x  3y  1 2x  3(15  3x)  1 2x  45  9x  1 11x  45  1 11x  45  45  1  45 11x  44 11x 11



21r 21

r2 Substitute 2 for r in either equation to find the value of s. 5r  s  5 5(2)  s  5 10  s  5 10  s  10  5  10 s  5 (1)(s)  (1) (5) s5 The solution is (2, 5). 23. Solve the second equation for x since the coefficient of x is 1. x  2y  6 x  2y  2y  6  2y x  6  2y Find the value of y by substituting 6  2y for x in the first equation. 3x  2y  12 3(6  2y)  2y  12 18  6y  2y  12 18  8y  12 18  8y  18  12  18 8y  6

44 11

x  4 Substitute 4 for x in either equation to find the value of y. 3x  y  15 3(4)  y  15 12  y  15 12  y  12  15  12 y3 The solution is (4, 3). 21. Solve the first equation for c since the coefficient of c is 1. c  5d  2 c  5d  5d  2  5d c  2  5d Find the value of d by substituting 2  5d for c in the second equation. 2c  d  4 2(2  5d )  d  4 4  10d  d  4 4  11d  4 4  11d  4  4  4 11d  0 11d 11

42

 21

8y 8

6

 8 3

y4 3

0

Substitute 4 for y in either equation to find the value of x. x  2y  6

 11

d0 Substitute 0 for d in either equation to find the value of c. c  5d  2 c  5(0)  2 c02 c2 The solution is (2, 0). 22. Solve the first equation for s since the coefficient of s is 1. 5r  s  5 5r  s  5r  5  5r s  5  5r (1)(s)  (1)(5  5r) s  5  5r

x2

134 2  6 3

x26 3

3

3

x2262 1

x  42

1

1 3

2

The solution is 42, 4 .

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Find the value of x by substituting 104  2x for y in the first equation. 0.5x  2y  17 0.5x  2(104  2x)  17 0.5x  208  4x  17 4.5x  208  17 4.5x  208  208  17  208 4.5x  225

24. Solve the first equation for x since the coefficient of x is 1. x  3y  0 x  3y  3y  0  3y x  3y Find the value of y by substituting 3y for x in the second equation. 3x  y  7 3(3y)  y  7 9y  y  7 10y  7 10y 10

4.5x 4.5

x  50 Substitute 50 for x in either equation to find the value of y. 2x  y  104 2(50)  y  104 100  y  104 100  y  100  104  100 y4 The solution is (50, 4).

7

 10 7

y  10 7

Substitute 10 for y in either equation to find the value of x. x  3y  0 x3

1107 2  0

1

21

21

x  10  10  0  10 21

1

x  10 or 210 The solution is

1

1 7 210, 10

1 x 2

2.

1 x 2

0.41x 0.41



 3  2x  1

 3  2x  2x  1  2x 3

25. Solve the first equation for y since the coefficient of y is 1. 0.3x  y  0.5 0.3x  y  0.3x  0.5  0.3x y  0.5  0.3x Find the value of x by substituting 0.5  0.3x for y in the second equation. 0.5x  0.3y  1.9 0.5x  0.3(0.5  0.3x)  1.9 0.5x  0.15  0.09x  1.9 0.41x  0.15  1.9 0.41x  0.15  0.15  1.9  0.15 0.41x  2.05

2x  3  1 3

2x  3  3  1  3 3

2x  4

123 2132x2  123 2142 8

2

x  3 or 23

8

Substitute 3 for x in either equation to find the value of y. y  2x  1 y2 y

2.05 0.41

y

x5 Substitute 5 for x in either equation to find the value of y. 0.3x  y  0.5 0.3(5)  y  0.5 1.5  y  0.5 1.5  y  1.5  0.5  1.5 y2 The solution is (5, 2). 26. Solve for y in the second equation since the coefficient of y is 1. 2x  y  104 2x  y  2x  104  2x y  104  2x

Chapter 7

1

27. Since y  2x  3, substitute 2x  3 for y in the second equation. y  2x  1

21

x  10  0 21

225 4.5



183 2  1

16 3 13 3

1

1

or 43

1

2

1

2

The solution is 23, 43 . 1 y 2

1

28. Since x   3, substitute 2 y  3 for x in the second equation. 2x  y  6

11

2

2 2y  3  y  6 y6y6 66 The statement 6  6 is true. This means that there are infinitely many solutions of the system of equations.

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The system of equations is a  b  500 and 0.25a  0.50b  0.34(500). Use substitution to solve this system. a  b  500 a  b  b  500  b a  500  b

29. Let y  the length of the base of the triangle, and let x  the length of each of the other sides. Since the base is 4 in. longer than the length of one of the other sides, y  x  4. Since the perimeter is 46 in., x  x  y  46, or 2x  y  46. Use substitution to solve this system. 2x  y  46 2x  (x  4)  46 3x  4  46 3x  4  4  46  4 3x  42 3x 3



0.25a  0.50b  0.34(500) 0.25(500  b)  0.50b  0.34(500) 125  0.25b  0.50b  170 125  0.25b  170 125  0.25b  125  170  125 0.25b  45

42 3

x  14 yx4 y  14  4 y  18 The length of the base is 18 in., and the length of each of the other sides is 14 in. 30. Let r  the number of lb of raisins, and let s  the number of lb of sunflower seeds. Use a table to organize the information. Number of Cost per Pounds Pound Raisins r 1.50 Sunflower Seeds s 4.00

0.25b 0.25

b  180 a  b  500 a  180  500 a  180  180  500  180 a  320 320 gal of 25% acid solution and 180 gal of 50% acid solution should be used. 32. Let x  the measure of angle X, and let y  the measure of angle Y. Since angles X and Y are supplementary, x  y  180. Since the measure of angle X is 24 degrees greater than the measure of angle Y, x  y  24. Use substitution to solve this system. x  y  180 ( y  24)  y  180 2y  24  180 2y  24  24  180  24 2y  156

Total Cost 1.50r 4.00s

Since the mix will have three times the number of lb of raisins as sunflower seeds, r  3s. Since the total cost is $34.00, 1.50r  4.00s  34.00. Use substitution to solve this system. 1.50r  4.00s  34.00 1.50(3s)  4.00s  34.00 4.50s  4.00s  34.00 8.50s  34.00 8.50s 8.50



45

 0.25

2y 2

34.00 8.50



156 2

y  78

s4

x  y  24 x  78  24 x  102 The measure of angle X is 102 degrees, or mX  102. The measure of angle Y is 78 degrees, or mY  78. 33. Let y  the number of World Series won by the Yankees, and let r  the number of World Series won by the Reds. Since the total number of games won by both teams is 31, y  r  31. Since the Yankees won 5.2 times as many World Series as the Reds, y  5.2r. Use substitution to solve this system. y  r  31 5.2r  r  31 6.2r  31

r  3s r  3(4) r  12 The club should purchase 4 lb of sunflower seeds and 12 lb of raisins. 31. Let a  the number of gallons of 25% acid solution, and let b  the number of gallons of 50% solution. Use a table to organize the information. 25% 50% 34% Solution Solution Solution Total Gallons a b 500 Gallons of Acid 0.25a 0.50b 0.34 (500)

6.2r 6.2

31

 6.2

r5 y  5.2r y  5.2(5) y  26 The Yankees won 26 World Series, and the Reds won 5 World Series.

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34. Let p  the total price of the automobiles Shantel sells, and let t  her total income. At the first dealership, t  600  0.02p. At the second dealership, t  1000  0.015p. Use substitution to solve this system. t  600  0.02p 1000  0.015p  600  0.02p 1000  0.015p  0.015p  600  0.02p  0.015p 1000  600  0.005p 1000  600  600  0.005p  600 400  0.005p 400 0.005



The number of tourists would be expected to be the same about 23.3 yr after 2000, or some time during the year 2023. 38. See students’ work. 39. Since y  3z  7, substitute 3z  7 for y in the first equation. 2x  3y  z  17 2x  3(3z  7)  z  17 2x  9z  21  z  17 2x  10z  21  17 2x  10z  21  21  17  21 2x  10z  38 Next, use this equation and the third equation in the given system to write a system of two equations containing only x and z. 2x  10z  38 2x  z  2 Solve the second equation for z since the coefficient of z is 1. 2x  z  2 2x  2  z  2  2 2x  2  z Since z  2x  2, substitute 2x  2 for z in the first equation. 2x  10z  38 2x  10(2x  2)  38 2x  20x  20  38 18x  20  38 18x  20  20  38  20 18x  18

0.005p 0.005

80,000  p Shantel will make the same income from either dealership if the total price of the automobiles she sells is $80,000. 35. The second offer is better if she sells less than $80,000. The first offer is better if she sells more than $80,000. For example, suppose the total price of the automobiles she sells is $50,000. At the first dealership, her income will be 600  0.02(50,000) or $1600; at the second dealership, her income will be 1000  0.015(50,000) or $1750. Thus, the second dealership’s offer is better. However, if the total price is $100,000, Shantel’s income at the first dealership will be 600  0.02(100,000) or $2600; her income at the second dealership will be 1000  0.015(100,000) or $2500. Thus, the first dealership’s offer is better. 36. Let y  the number of years, and let h  the height of the tree in inches. Since the blue spruce grows 6 in. per yr and is 4 ft, or 48 in., tall, h  6y  48. Since the hemlock grows 4 in. per yr and is 6 ft, or 72 in., tall, h  4y  72. Use substitution to solve this system. h  6y  48 4y  72  6y  48 4y  72  4y  6y  48  4y 72  2y  48 72  48  2y  48  48 24  2y 24 2



18x 18

x  1 Use 2x  z  2 to find the value of z. 2x  z  2 2(1)  z  2 2  z  2 2  2  z  2  2 4  z Use y  3z  7 to find the value of y. y  3z  7 y  3(4)  7 y  12  7 y5 Since x  1, y  5 and z  4, the solution is (1, 5, 4).

2y 2

12  y The trees will be the same height in 12 yr. 37. Let y  the number of years since 2000, and let t  the number of tourists in millions. Since the number of tourists visiting South America and the Caribbean was 40.3 million, and tourism is increasing at a rate of 0.8 million per year, t  40.3  0.8y. Since the number of tourists to the Middle East was 17.0 million, and tourism is increasing at a rate of 1.8 million tourists per year, t  17.0  1.8y. Use substitution to solve this system. t  40.3  0.8y 17.0  1.8y  40.3  0.8y 17.0  1.8y  0.8y  40.3  0.8y  0.8y 17.0  y  40.3 17.0  y  17.0  40.3  17.0 y  23.3

Chapter 7

18

 18

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40. When problems about technology involve a system of equations, the problem can be solved by substitution. Answers should include the following. • To solve a system of equations using substitution, solve one equation for one unknown. Substitute this value for the unknown in the other equation and solve the equation. Use this number to find the other unknown. Since y  2.8x  170, substitute 2.8x  170 for y in the second equation. y  14.4x  2 2.8x  170  14.4x  2 2.8x  170  2.8x  14.4x  2  2.8x 170  17.2x  2 170  2  17.2x  2  2 168  17.2x 9.8  x • The number of hours will be the same about 9.8 years after 1993. That represents the year 2002. 41. B; Solve the first equation for x since the coefficient of x is 1. x  4y  1 x  4y  4y  1  4y x  1  4y Since x  1  4y, the expression 1  4y could be substituted for x in the second equation. 42. C; Solve the first equation for x since the coefficient of x is 1. x  3y  9 x  3y  3y  9  3y x  9  3y Find the value of y by substituting 9  3y for x in the second equation. 5x  2y  7 5(9  3y)  2y  7 45  15y  2y  7 45  13y  7 45  13y  45  7  45 13y  52 13y 13

Page 381

Maintain Your Skills

43.

y xy4

xy3 x O

The graphs of the equations are parallel lines. Since they do not intersect, there are no solutions to this system of equations. 44.

y

x  2y  1

O

2x  y  5

(3, 1) x

The graphs appear to intersect at (3, 1). Check in each equation. x  2y  1 2x  y  5 Check: ? ? 3  2(1)  1 2(3)  (1)  5 ? ? 3  (2)  1 6  (1)  5 11✓ 55✓ There is one solution. It is (3, 1). 45.

y

2x  y  3 4x  2y  6 O

x

The graphs of the equations coincide. Since every point is a point of intersection, there are infinitely many solutions to this system of equations.

52

 13

y4 Substitute 4 for y in either equation to find the value of x. x  3y  9 x  3(4)  9 x  12  9 x  12  12  9  12 x3 So, the value of x is 3.

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Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y 7 6  2x 0 7 6  2(0) false 0 7 6 Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane.

46. Step 1 The inequality is already solved for y in terms of x. Step 2 Graph y  5. Since the inequality includes y values less than 5, but not equal to 5, the boundary is not included in the solution set. The boundary should be drawn as a dashed line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y 6 5 0 6 5 false Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane.

y

2x  y  6

y x

O

x

O

49. Let d represent the number of pounds of denim left. number of pounds left → 1 d ← number of pounds left  number of pairs of jeans → 5 250 ← number of pairs of jeans 1(250)  5(d) 250  5(d)

y  5

250 5

47. Step 1 There is no y in the inequality. The inequality is already solved for x. Step 2 Graph x  4. Since x 4 means x 4 or x  4, the boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). x 4 0 4 false Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane.

50. 51. 52. 53.

y



5d 5

50  d There would be 50 lb of denim left. 6a  9a  (6  9)a  3a 8t  4t  (8  4)t  12t 7g  8g  (7  8)g  15g 7d  (2d  b)  7d  2d  b  5d  b

Page 381 1.

Practice Quiz 1 y

x4 xy3 O

x xy1

Solve for y in terms of x. 2x  y 7 6 2x  y  2x 7 6  2x y 7 6  2x Step 2 Graph y  6  2x. Since the inequality includes values greater than 6  2x, but not equal to 6  2x, the boundary is not included in the solution set. The boundary should be drawn as a dashed line.

O

(2, 1)

x

48. Step 1

Chapter 7

The graphs appear to intersect at (2, 1). Check in each equation. Check: xy3 xy1 ? ? 211 213 33✓ 11✓ There is one solution. It is (2, 1).

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2.

The statement 2  2 is true. This means that there are infinitely many solutions of the system of equations.

y 3x  2y  6

x

O

7-3

Elimination Using Addition and Subtraction

3x  2y  6

Pages 384–385

The graphs of the equations are parallel lines. Since they do not intersect, there are no solutions to this system of equations. 3. Solve the first equation for y since the coefficient of y is 1 (in both equations). xy0 xyx0x y  x Find the value of x by substituting x for y in the second equation. 3x  y  8 3x  (x)  8 2x  8 2x 2



Check for Understanding

1. Sample answer: 2a  b  5, a  b  4 Since the coefficients of the b terms are additive inverses, this system can be solved by using addition to eliminate one variable. 2. Subtraction can be used to eliminate one variable for a system in which one variable has the same coefficient in both equations. 3. Michael’s method is correct since, in order to eliminate the s terms, you must add the two equations. 4. Since the coefficients of the y terms, 1 and 1, are additive inverses, eliminate the y terms by adding the equations. x  y  14 () x  y  20 2x  34

8 2

x  4 Substitute 4 for x in either equation to find the value of y. xy0 (4)  y  0 4  y  4  0  4 y4 The solution is (4, 4). 4. Solve the first equation for x since the coefficient of x is 1. x  2y  5 x  2y  2y  5  2y x  5  2y Find the value of y by substituting 5  2y for x in the second equation. 3x  5y  8 3(5  2y)  5y  8 15  6y  5y  8 15  y  8 15  y  15  8  15 y  7 Substitute 7 for y in either equation to find the value of x. x  2y  5 x  2(7)  5 x  (14)  5 x  14  5 x  14  14  5  14 x  9 The solution is (9, 7). 5. Since y  2  x, substitute 2  x for y in the first equation. xy2 x  (2  x)  2 22

2x 2



34 2

x  17 Now substitute 17 for x in either equation to find the value of y. x  y  20 17  y  20 17  y  17  20  17 y3 The solution is (17, 3). 5. Since the coefficients of the b terms, 3 and 3, are additive inverses, eliminate the b terms by adding the equations. 2a  3b  11 8 () a  3b  3a  3 3a 3



3 3

a  1 Now substitute 1 for a in either equation to find the value of y. a  3b  8 1  3b  8 1  3b  1  8  1 3b  9 3b 3

9

3

b3 The solution is (1, 3).

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9. Since the coefficients of the m terms, 4 and 4, are the same, eliminate the m terms by subtracting the equations. 4m  2n  6 () 4m  n  8 n  2 Now substitute 2 for n in either equation to find the value of m. 4m  n  8 4m  (2)  8 4m  2  8 4m  2  2  8  2 4m  10

6. Since the coefficients of the x terms, 4 and 4, are the same, eliminate the x terms by subtracting the equations. 4x  y  9 () 4x  2y  10 y  1 (1)(y)  (1)1 y  1 Now substitute 1 for y in either equation to find the value of x. 4x  y  9 4x  (1)  9 4x  1  9 4x  1  1  9  1 4x  8 4x 4



4m 4



1

0

6x 6

y  5 The solution is (0, 5). 8. Since the coefficients of the a terms, 2 and 2, are additive inverses, eliminate the a terms by adding the equations. 2a  4b  30 () 2a  2b  21.5 2b  8.5 8.5 2

b  4.25 Now substitute 4.25 for b in either equation to find the value of a. 2a  4b  30 2a  4(4.25)  30 2a  17  30 2a  17  17  30  17 2a  13 2a 2



a

13 2 13 2

or 6.5

The solution is (6.5, 4.25).

Chapter 7

2



36 6

x6 Now substitute 6 for x in either equation to find the value of y. x  y  24 6  y  24 6  y  6  24  6 y  18 The numbers are 6 and 18. 11. D; Since the coefficients of the x terms, 2 and 2, are the same, eliminate the x terms by subtracting the equations. 2x  7y  17 () 2x  5y  11 2y  6 The value of 2y is 6.

10 2



1

10. Let x represent the first number and let y represent the second number. x  y  24 5x  y  12 Use elimination to solve the system. x  y  24 ()5x  y  12 6x  36

4

2b 2

1

The solution is 22, 2 .

x0 Now substitute 0 for x in either equation to find the value of y. 6x  2y  10 6(0)  2y  10 2y  10 2y 2

5

m  2 or 22

8 4

x  2 The solution is (2, 1) . 7. Since the coefficients of the y terms, 2 and 2, are the same, eliminate the y terms by subtracting the equations. 6x  2y  10 () 2x  2y  10 4x 0 4x 4

10

 4

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Pages 385–386

15. Since the coefficients of the x terms, 4 and 4, are additive inverses, eliminate the x terms by adding the equations. 4x  2y  8 () 4x  3y  10 y  2 (1)(y)  (1)(2) y2 Now substitute 2 for y in either equation to find the value of x. 4x  3y  10 4x  3(2)  10 4x  6  10 4x  6  6  10  6 4x  4

Practice and Apply

12. Since the coefficients of the y terms, 1 and 1, are additive inverses, eliminate the y terms by adding the equations. x  y  3 () x  y  1 2x  2 2x 2



2 2

x  1 Now substitute 1 for x in either equation to find the value of y. x  y  3 (1)  y  3 1  y  1  3  1 y  2 The solution is (1, 2) . 13. Since the coefficients of the t terms, 1 and 1, are additive inverses, eliminate the t terms by adding the equations. st4 () s  t  2 2s 6 2s 2

4x 4

6

s3 Now substitute 3 for s in either equation to find the value of t. st4 (3)  t  4 3t343 t  1 1

 1

t  1 The solution is (3, 1) . 14. Since the coefficients of the n terms, 2 and 2, are additive inverses, eliminate the n terms by adding the equations. 3m  2n  13 () m  2n  7 4m  20 4m 4



2n 2

8

2

n4 Now substitute 4 for n in either equation to find the value of m. 2m  5n  6 2m  5(4)  6 2m  20  6 2m  20  20  6  20 2m  14

20 4

m5 Now substitute 5 for m in either equation to find the value of n. m  2n  7 5  2n  7 5  2n  5  7  5 2n  2 2n 2

4 4

x  1 The solution is (1, 2) . 16. Since the coefficients of the b terms, 1 and 1, are the same, eliminate the b terms by subtracting the equations. 3a  b  5 () 2a  b  10 a  5 Now substitute 5 for a in either equation to find the value of b. 2a  b  10 2(5)  b  10 10  b  10 10  b  10  10  10 b  20 The solution is (5, 20) . 17. Since the coefficients of the m terms, 2 and 2, are the same, eliminate the m terms by subtracting the equations. 2m  5n  6 () 2m  7n  14 2n  8

2

t 1



2m 2



14 2

m7 The solution is (7, 4).

2

2

n1 The solution is (5, 1).

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18. Since the coefficients of the s terms, 5 and 5, are the same, eliminate the s terms by subtracting the equations. 3r  5s  35 () 2r  5s  30 r  5 Now substitute 5 for r in either equation to find the value of s. 3r  5s  35 3(5)  5s  35 15  5s  35 15  5s  15  35  15 5s  20 5s 5



21. Since the coefficients of the s terms, 6 and 6, are the same, eliminate the s terms by subtracting the equations. 6s  5t  1 () 6s  5t  11 10t  10 10t 10



20 5

6s 6



18 6

6y 6

26 13

4x 4

The

12 6

18 4 9 1 x  2 or 42 1 solution is 42, 2



1

2.

23. Since the coefficients of the b terms, 2 and 2, are the same, eliminate the b terms by subtracting the equations. a  2b  5 () 3a  2b  9 2a  4

6

 3

y  2 Now substitute 2 for y in either equation to find the value of x. 3x  5y  16 3x  5(2)  16 3x  10  16 3x  10  10  16  10 3x  6 3x 3



y2 Now substitute 2 for y in either equation to find the value of x. 4x  3y  12 4x  3(2)  12 4x  6  12 4x  6  6  12  6 4x  18

a  2 The solution is (2, 3) . 20. Since the coefficients of the x terms, 3 and 3, are additive inverses, eliminate the x terms by adding the equations. 3x  5y  16 () 3x  2y  10 3y  6 3y 3

6

6

s1 The solution is (1, 1) . 22. Since the coefficients of the x terms, 4 and 4, are the same, eliminate the x terms by subtracting the equations. 4x  3y  1 () 4x  3y  24 6y  12

b3 Now substitute 3 for b in either equation to find the value of a. 13a  5b  11 13a  5(3)  11 13a  15  11 13a  15  15  11  15 13a  26 13a 13

10 10

t  1 Now substitute 1 for t in either equation to find the value of s. 6s  5t  1 6s  5(1)  1 6s  5  1 6s  5  5  1  5 6s  6

s4 The solution is (5, 4) . 19. Since the coefficients of the a terms, 13 and 13, are the same, eliminate the a terms by subtracting the equations. 13a  5b  11 () 13a  11b  7 6b  18 6b 6



2a 2

4

 2

a2 Now substitute 2 for a in either equation to find the value of b. a  2b  5 2  2b  5 2  2b  2  5  2 2b  3

6

3

x2 The solution is (2, 2) .

2b 2

3

 2 3

1

b  2 or 12

1

1

2

The solution is 2, 12 .

Chapter 7

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Now substitute 1.75 for x in either equation to find the value of y. 1.44x  3.24y  5.58 1.4411.752  3.24y  5.58 2.52  3.24y  5.58 2.52  3.24y  2.52  5.58  2.52 3.24y  8.1

24. Since the coefficients of the y terms, 5 and 5, are the same, eliminate the y terms by subtracting the equations. 4x  5y  7 () 8x  5y  9 4x  2 4x 4

2

 4

3.24y 3.24

1

x2 Now substitute the value of y. 4x  5y  7 4

1 2

y  2.5 The solution is (1.75, 2.5). 27. Since the coefficients of the m terms, 7.2 and 7.2, are the same, eliminate the m terms by subtracting the equations. 7.2m  4.5n  129.06 () 7.2m  6.7n  136.54 2.2n  7.48

for x in either equation to find

112 2  5y  7

2  5y  7 2  5y  2  7  2 5y  5 5y 5

2.2n 2.2

5

5

y1 The solution is

1 12.

2 4 1 b  2 1 substitute 2

for b in either equation to find

1 12

1

()

8a  2  1 1

1

8a  2  2  1  2

3 1 c  5d 5 7 1 c  5d 5

 9

2c

 20 2c 2

3

8a  2

118 28a  118 232 3

a  16

The solution is

113.76 7.2

 11 

20 2

c  10 Now substitute 10 for c in either equation to find the value of d.

1163 , 12 2.

26. Since the coefficients of the y terms, 3.24 and 3.24, are additive inverses, eliminate the y terms by adding the equations. 1.44x  3.24y  5.58 () 1.08x  3.24y  9.99 2.52x  4.41 2.52x 2.52



m  15.8 The solution is (15.8, 3.4). 1 1 28. Since the coefficients of the d terms, 5 and 5, are additive inverses, eliminate the d terms by adding the equations.

8a  2  1 1

7.48 2.2

7.2m 7.2



Now the value of a. 8a  b  1



n  3.4 Now substitute 3.4 for n in either equation to find the value of m. 7.2m  4.5n  129.06 7.2m  4.513.42  129.06 7.2m  15.3  129.06 7.2m  15.3  15.3  129.06  15.3 7.2m  113.76

1 , 2

25. Since the coefficients of the a terms, 8 and 8, are the same, eliminate the a terms by subtracting the equations. 8a  b  1 () 8a  3b  3 4b  2 4b 4

8.1

 3.24

3 c 5

 5d  9

1

3 (10) 5

 5d  9

1 1

6  5d  9 1

6  5d  6  9  6

1

1

5d  3 1

2

(5) 5d  (5)3

4.41

 2.52

d  15 The solution is (10, 15).

x  1.75

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1

1

32. Let x represent the first number and let y represent the second number. 3x  y  18 2x  y  12 Use elimination to solve the system. 3x  y  18 () 2x  y  12 5x  30

29. Since the coefficients of the y terms, 2 and 2, are the same, eliminate the y terms by subtracting the equations. 2 1 x  2y 3 5 1 () 6 x  2 y 1 6x 1 (6) 6x

1

 14  18  4

2  (6)(4)

5x 5

x  24 Now substitute 24 for x in either equation to find the value of y. 2 x 3

 2y  14

2 (24) 3

 2y  14

1 1

16  2y  14 1

16  2y  16  14  16

1

2y  2 1

2

(2) 2y  (2)(2) y4 The solution is (24, 4). 30. Let x represent the first number and y represent the second number. x  y  48 x  y  24 Use elimination to solve the system. x  y  48 () x  y  24 2x  72 2x 2



3x 3



72 2

2y 2

15 3



18 2

y9 The numbers are 5 and 9.

64 2

x  32 Now substitute 32 for x in either equation to find the value of y. x  y  51 32  y  51 32  y  32  51  32 y  19 The numbers are 32 and 19.

Chapter 7



x5 Now substitute 5 for x in either equation to find the value of y. x  2y  23 5  2y  23 5  2y  5  23  5 2y  18

x  36 Now substitute 36 for x in either equation to find the value of y. x  y  48 36  y  48 36  y  36  48  36 y  12 The numbers are 36 and 12. 31. Let x represent the first number and let y represent the second number. x  y  51 x  y  13 Use elimination to solve the system. x  y  51 () x  y  13 2x  64 2x 2

30 5

x6 Now substitute 6 for x in either equation to find the value of y. 3x  y  18 3(6)  y  18 18  y  18 18  y  18  18  18 y0 The numbers are 6 and 0. 33. Let x represent the first number and let y represent the second number. x  2y  23 4x  2y  38 Use elimination to solve the system. x  2y  23 () 4x  2y  38 3x  15

1

1



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Rewrite t  d  0.467 so that the system can be written in column form. t  d  0.467 t  d  d  0.467  d t  d  0.467 Use elimination to solve the system. t  d  0.467 () t  d  12.867 2t  13.334

34. Let u represent the number of motor vehicles produced in the United States in millions, and let j represent the number of motor vehicles produced in Japan in millions. uj2 u  j  22 Rewrite u  j  2 so that the system can be written in column form. uj2 ujj2j uj2 Use elimination to solve the system. uj 2 () u  j  22 2u  24 2u 2



2t 2

24 2



y2  y1

mx

2

m m

18 2



 x1

1.52  1.28 50  0 0.24 50

m  0.0048 The slope is 0.0048. Choose (0, 1.28) and find the y-intercept of the line. y  mx  b 1.28  0.0048(0)  b 1.28  b Write the slope-intercept form using m  0.0048 and b  1.28. y  mx  b y  0.0048x  1.28 Therefore, the equation is y  0.0048x  1.28.

s9 Now substitute 9 for s in either equation to find the value of a. 2a  5s  77 2a  5(9)  77 2a  45  77 2a  45  45  77  45 2a  32 2a 2

13.334 2

t  6.667 Now substitute 6.667 for t in either equation to find the value of d. t  d  0.467 6.667  d  0.467 6.667  0.467  d  0.467  0.467 6.200  d In 1999, Troy Aikman earned $6.667 million and Deion Sanders earned $6.200 million. 37. China’s population in 2000, when x  0, is 1.28 billion people. This means (0, 1.28) is in the solution set for the equation. China’s population in 2050, when x  50, is 1.52 billion people. This means (50, 1.52) is in the solution set for the equation. Let (x1, y1)  (0, 1.28) and (x2, y2)  (50, 1.52).

u  12 Now substitute 12 for u in either equation to find the value of j. uj2 12  j  2 12  2  j  2  2 10  j In 1999, the U.S. produced about 12 million motor vehicles and Japan produced about 10 million motor vehicles. 35. Let a represent the adult price and let s represent the student price. 2a  5s  77 2a  7s  95 Use elimination to solve the system. 2a  5s  77 () 2a  7s  95 2s  18 2s 2



32 2

a  16 The adult price of the tour is $16 and the student price of the tour is $9. 36. Let t represent the amount earned by Troy Aikman in millions and let d represent the amount earned by Deion Sanders in millions. t  d  0.467 t  d  12.867

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38. India’s population in 2000, when x  0, is 1.01 billion people. This means (0, 1.01) is in the solution set for the equation. India’s population in 2050, when x  50, is 1.53 billion people. This means (50, 1.53) is in the solution set for the equation. Let (x1, y1)  (0, 1.01) and (x2, y2)  (50, 1.53).

Use elimination to solve the system. 2A  B  15 () 2 A  B  9 4A  24 4A 4

2

m m

 x1

1.53  1.01 50  0 0.52 50

m  0.0104 The slope is 0.0104. Choose (0, 1.01) and find the y-intercept of the line. y  mx  b 1.01  0.0104(0)  b 1.01  b Write the slope-intercept form using m  0.0104 and b  1.01. y  mx  b y  0.0104x  1.01 Therefore, the equation is y  0.0104x  1.01. 39. From Exercise 37, China’s population is represented by y  0.0048x  1.28. From Exercise 38, India’s population is represented by y  0.0104x  1.01. Use elimination to solve the system. y  0.0048x  1.28 () y  0.0104x  1.01 0  0.0056x  0.27 0  0.0056x  0.0056x  0.27  0.0056x 0.0056x  0.27 0.0056x 0.0056

24 4

A6 Now substitute 6 for A in either equation to find the value of B. 2A  B  15 2(6)  B  15 12  B  15 12  B  12  15  12 B3 So, the value of A is 6 and the value of B is 3. 41. Elimination can be used to solve problems about meteorology if the coefficients of one variable are the same or are additive inverses. Answers should include the following. • The two equations in the system of equations are added or subtracted so that one of the variables is eliminated. You then solve for the remaining variable. This number is substituted into one of the original equations, and that equation is solved for the other variable. • Write the equations in n  d  24 column form and add. () n  d  12 Notice the d variable 2n  36 is eliminated.

y2  y1

mx



2n 2



36 2

n  18 n  d  24 18  d  24 18  d  18  24  18

Simplify. First equation n  18 Subtract 18 from each side. Simplify. d6 On the winter solstice, Seward, Alaska, has 18 hours of darkness and 6 hours of daylight. 42. B; Use elimination to solve the system. 2x  3y  9 () 3x  3y  12 x 3 (1)(x)  (1)3 x  3 Now substitute 3 for x in either equation to find the value of y. 2x  3y  9 2(3)  3y  9 6  3y  9 6  3y  6  9  6 3y  3

0.27

 0.0056

x  48 Now substitute 48 for x in either equation to find the value of y. y  0.0048x  1.28 y  0.0048(48)  1.28 y  0.2304  1.28 y  1.5104 The populations of China and India are predicted to be the same about 48 yr after 2000, or in 2048. At that time, the predicted population is about 1.51 billion people. 40. Since the graphs intersect at (2, 1), (x, y)  (2, 1) must make both equations true. Substitute 2 for x and 1 for y in both equations to find the values of A and B. Ax  By  15 Ax  By  9 A(2)  B(1)  15 A(2)  B(1)  9 2A  B  15 2A  B  9

3y 3

3

 3

y1 The value of y is 1.

Chapter 7

Divide each side by 2.

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Since x  2y  5, substitute 2y  5 for x in the second equation. 4y  3x  1 4y  3(2y  5)  1 4y  6y  15  1 2y  15  1 2y  15  15  1  15 2y  14

43. C; Use elimination to solve the system. 4x  2y  8 () 2x  2y  2 2x 6 2x 2

6

2

x3 Now substitute 3 for x in either equation to find the value of y. 4x  2y  8 4(3)  2y  8 12  2y  8 12  2y  12  8  12 2y  4 2y 2



2y 2

y  7 Substitute 7 for y in either equation to find the value of x. 2y  x  5 2(7)  x  5 14  x  5 14  x  14  5  14 x  9 (1)(x)  (1)(9) x  9 The solution is (9, 7).

4 2

y  2 The solution is (3, 2).

Page 386

Maintain Your Skills

44. Since y  5x, substitute 5x for y in the second equation. x  2y  22 x  2(5x)  22 x  10x  22 11x  22 11x 11

47.

y

3x  y  1

xy3

22

x2 Use y  5x to find the value of y. y  5x y  5(2) y  10 The solution is (2, 10). 45. Since x  2y  3, substitute 2y  3 for x in the second equation. 3x  4y  1 3(2y  3)  4y  1 6y  9  4y  1 10y  9  1 10y  9  9  1  9 10y  10 

x

O

 11

10y 10

14

 2

(1, 2)

The graphs appear to intersect at (1, 2). Check in each equation. Check: xy3 3x  y  1 ? ? 1  (2)  3 3(1)  (2)  1 ? ? 123 3  (2)  1 33✓ 11✓ There is one solution. It is (1, 2). 48.

y 3y  7  2x

10 10

O

y  1 Use x  2y  3 to find the value of x. x  2y  3 x  2(1)  3 x  2  3 x1 The solution is (1, 1). 46. Solve the first equation for x since the coefficient of x is 1. 2y  x  5 2y  x  x  5  x 2y  5  x 2y  5  5  x  5 2y  5  x

x 2x  3y  7

The graphs of the equations are parallel lines. Since they do not intersect, there are no solutions to this system of equations.

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49.

4. Multiply the first equation by 4 so the coefficients of the y terms are additive inverses. Then add the equations. Multiply by 4. 2x  y  6 8x  4y  24 3x  4y  2 () 3x  4y  2 11x  22

y 12

4x  y  12

8 4

x

O

x  3  1y 4

4

11x 11

8

x2 Now substitute 2 for x in either equation to find the value of y. 2x  y  6 2(2)  y  6 4y6 4y464 y  2 (1)(y)  (1)2 y  2 The solution is (2, 2). 5. Multiply the first equation by 3 so the coefficients of the x terms are additive inverses. Then add the equations. Multiply by 3. 3x  15y  12 x  5y  4 3x  7y  10 () 3x  7y  10 22y  22

The graphs of the equations coincide. Since every point is a point of intersection, there are infinitely many solutions to this system of equations. 5

5

50. The line parallel to y  4x  3 has slope 4. 5 Replace m with 4, and (x1, y1 ) with (0, 0) in the point-slope form. y  y1  m(x  x1 ) 5

y  0  4 (x  0) 5

y  4x 5

Therefore, the equation is y  4x. 51. 2(3x  4y)  2  3x  2  4y  6x  8y 52. 6(2a  5b)  6  2a  6  5b  12a  30b 53. 3(2m  3n)  (3)(2m)  (3)(3n)  6m  (9n)  6m  9n 54. 5(4t  2s)  (5)(4t)  (5)(2s)  20t  (10s)  20t  10s

7-4

22y 22

4x  7y  6 6x  5y  20

Check for Understanding

1. If one of the variables cannot be eliminated by adding or subtracting the equations, you must multiply one or both of the equations by numbers so that a variable will be eliminated when the equations are added or subtracted. 2. Sample answer: 3x  2y  5 and 4x  10y  6 If the first equation is multiplied by 5, the coefficients of the y terms will be 10 and 10, which are additive inverses. Thus, when the two equations are added together, the y variable will be eliminated. 3. Sample answer: (1) You could solve the first equation for a and substitute the resulting expression for a in the second equation. Then find the value of b. Use this value for b and one of the equations to find the value of a. (2) You could multiply the first equation by 3 and add this new equation to the second equation. This will eliminate the b term. Find the value of a. Use this value for a and one of the equations to find the value of b. See students’ work for their preferences and explanations.

Chapter 7

22

 22

y1 Now substitute 1 for y in either equation to find the value of x. x  5y  4 x  5(1)  4 x54 x5545 x  1 The solution is (1, 1). 6. Eliminate x.

Elimination Using Multiplication

Pages 390–391

22

 11

Multiply by 3. Multiply by 2.

12x  21y  18 () 12x  10y  40 11y  22 11y 11



22 11

y  2

Now substitute 2 for y in either equation to find the value of x. 4x  7y  6 4x  7(2)  6 4x  14  6 4x  14  14  6  14 4x  20 4x 4



20 4

x5 The solution is (5, 2).

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7. Multiply the second equation by 2 so the coefficients of the x terms are additive inverses. Then add the equations.

Now substitute 2 for x in either equation to find the value of y. 3x  7y  6 3(2)  7y  6 6  7y  6 6  7y  6  6  6 7y  0

4x  2y  10.5 4x  2y  10.5 2x  3y  10.75 Multiply by 2. () 4x  6y  21.5 4y  11 4y 4



11 4

7y 7

y  2.75

y0 The solution is (2, 0). 10. • For an exact solution, an algebraic method is best. • Since neither the coefficients of x nor the coefficients of y are the same or additive inverses, you cannot use elimination using addition or subtraction. • Since the coefficient of y in the first equation is 1, you can use the substitution method. Since y  4x  11, substitute 4x  11 for y in the second equation. 3x  2y  7 3x  2 (4x  11)  7 3x  8x  22  7 5x  22  7 5x  22  22  7  22 5x  15

Now substitute 2.75 for y in either equation to find the value of x. 4x  2y  10.5 4x  2(2.75)  10.5 4x  5.5  10.5 4x  5.5  5.5  10.5  5.5 4x  5 4x 4

5

4

x  1.25 The solution is (1.25, 2.75). 8. • For an exact solution, an algebraic method is best. • Since neither the coefficients of x nor the coefficients of y are the same or additive inverses, you cannot use elimination using addition or subtraction. • Since no coefficient of x or y is 1 or 1, substitution is not the best method. • Since it is necessary to multiply each equation by a different number, solve the system by elimination using multiplication. Eliminate x.

5x 5

y1

Now substitute 1 for y in either equation to find the value of x. 4x  3y  19 4x  3(1)  19 4x  3  19 4x  3  3  19  3 4x  16 

16 4

2x 2

x4 The solution is (4, 1). 9. • For an exact solution, an algebraic method is best. • Since the coefficients of the y terms, 7 and 7, are additive inverses, solve by elimination using addition. 3x  7y  6 () 2x  7y  4 5x  10 5x 5



15

 5

x  3 Use y  4x  11 to find the value of y. y  4x  11 y  4(3)  11 y  12  11 y  1 The solution is (3, 1). 11. • For an exact solution, an algebraic method is best. • Since the coefficients of the y terms, 2 and 2, are the same, you can eliminate the y terms by subtracting the equations. 5x  2y  12 () 3x  2y  2 2x  14

4x  3y  19 Multiply by 3. 12x  9y  57 3x  4y  8 Multiply by 4. () 12x  16y  32 25y  25 25y 25  25 25

4x 4

0

 7



14 2

x7 Now substitute 7 for x in either equation to solve for the value of y. 5x  2y  12 5 (7)  2y  12 35  2y  12 35  2y  35  12  35 2y  23 2y 2



23 2

y  11.5 The solution is (7, 11.5).

10 5

x2

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12. Let t  the number of two-seat tables and f  the number of four-seat tables. t  f  17 2t  4f  56 Multiply the first equation by 2 so the coefficients of the t terms are additive inverses. Then add the equations. t  f  17 Multiply by 2. 2t  (2f )  34 2t  4f  56 () 2t  4f  56 2f  22 2f 2



15. Multiply the first equation by 2 so the coefficients of the y terms are additive inverses. Then add the equations. 2x  y  5 Multiply by 2. 4x  2y  10 () 3x  2y  4 3x  2y  4 7x  14 7x 7

22 2

4x  3y  12 x  2y  14

y  13 Now substitute 13 for y in either equation to find the value of x. xy4 x  (13)  4 x  13  4 x  13  13  4  13 x  9 The solution is (9, 13). 14. Multiply the first equation by 3 so the coefficients of the y terms are additive inverses. Then add the equations. xy3 Multiply by 3. 3x  3y  9 () 2x  3y  16 2x  3y  16 5x  25

23x 23



23 23

x  1 Now substitute 1 for x in either equation to find the value of y. 5x  2y  15 5(1)  2y  15 5  2y  15 5  2y  5  15  5 2y  10

25 5

x5 Now substitute 5 for x in either equation to find the value of y. xy3 5y3 5y535 y  2 The solution is (5, 2).

Chapter 7

12

() 4x  8y  56 11y  44 11y 44  11 11

Now substitute 4 for y in either equation to find the value of x. x  2y  14 x  2(4)  14 x  8  14 x  8  8  14  8 x6 The solution is (6, 4). 17. Multiply the first equation by 4 so the coefficients of the y terms are additive inverses. Then add the equations. 5x  2y  15 Multiply by 4. 20x  8y  60 () 3x  8y  37 3x  8y  37 23x  23

26



4x  3y  Multiply by 4.

y4

 2

5x 5

14 7

x2 Now substitute 2 for x in either equation to find the value of y. 2x  y  5 2(2)  y  5 4y5 4y454 y1 The solution is (2, 1). 16. Multiply the second equation by 4 so the coefficients of the x terms are additive inverses. Then add the equations.

f  11 Now substitute 11 for f in either equation to find the value of t. t  f  17 t  11  17 t  11  11  17  11 t6 The owners should purchase 6 two-seat tables and 11 four-seat tables. 13. Multiply the second equation by 5 so the coefficients of the x terms are additive inverses. Then add the equations. 5x  3y  6 5x  3y  6 x  y  4 Multiply by 3. () 5x  5y  20 2y  26 2y 2



2y 2



10 2

y5 The solution is (1, 5).

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21. Multiply the second equation by 1.8 so the coefficients of the x terms are additive inverses. Then add the equations.

18. Multiply the second equation by 4 so the coefficients of the x terms are additive inverses. Then add the equations. 8x  3y  11 2x  5y  27 Multiply by 4.

8x  3y  11 () 8x  20y  108

1.8x  0.3y  14.4

1.8x  0.3y 

x  0.6y  2.8 Multiply by 1.8.

0.78y  9.36

17y  119 17y 17



119 17

0.78y 0.78

y  7

Now substitute 12 for y in either equation to find the value of x. x  0.6y  2.8 x  0.6(12)  2.8 x  7.2  2.8 x  7.2  7.2  2.8  7.2 x  10 The solution is (10, 12). 22. Multiply the first equation by 3 so the coefficients of the x terms are additive inverses. Then add the equations.

32 8



x  4 The solution is (4, 7). 19. Eliminate y. 4x  7y  10

Multiply by 2. Multiply by 7.

3x  2y  7

0.4x  0.5y  2.5

5y 5



0.4x 0.4

2

 0.4

x5 The solution is (5, 1). 23. Multiply the second equation by 2 so the coefficients of the y terms are additive inverses. Then add the equations.

4 2

1

1

3x  2 y  10

Multiply by 5. 10x  15y  10 Multiply by 2. () 10x  8y  56 23y  46 23y 46  23 23

1

5x  4 y  8

3x  2 y  10 Multiply by 2.

1

() 10x  2 y  16  26

13x 13x 13

y  2

26

 13

x2 Now substitute 2 for x in either equation to find the value of y.

Now substitute 2 for y in either equation to find the value of x. 5x  4y  28 5x  4(2)  28 5x  8  28 5x  8  8  28  8 5x  20 

5

 5

Now substitute 1 for y in either equation to find the value of x. 0.4x  0.5y  2.5 0.4x  0.5(1)  2.5 0.4x  0.5  2.5 0.4x  0.5  0.5  2.5  0.5 0.4x  2

29 29

y  2 The solution is (1, 2). 20. Eliminate x.

5x 5

2.5

y1

 29 x  1

2x  3y  2 5x  4y  28

1.2x  3.5y 

5y  5

Now substitute 1 for x in either equation to find the value of y. 3x  2y  7 3(1)  2y  7 3  2y  7 3  2y  3  7  3 2y  4 

1.2x  1.5y  7.5 ()

8x  14y  20 () 21x  14y  49 29x 29

2y 2

Multiply by 3.

1.2x  3.5y  2.5

29x

9.36

 0.78

y  12

Now substitute 7 for y in either equation to find the value of x. 8x  3y  11 8x  3(7)  11 8x  (21)  11 8x  21  11 8x  21  21  11  21 8x  32 8x 8

14.4

() 1.8x  1.08y  5.04

1

3x  2y  10 1

3(2)  2y  10 1

6  2y  10 1

20 5

6  2y  6  10  6

x4 The solution is (4, 2).

1

1

2y  4 1

2

(2) 2y  (2)4 y  8 The solution is (2, 8).

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24. Multiply the second equation by 2 so the coefficients of the x terms are additive inverses. Then add the equations. 2

2

2x  3 y  4 x

1 y 2

7

2x  3 y  Multiply by 2.

Now substitute 4 for x in either equation to find the value of y. xy3 4y3 4y434 y  1 The numbers are 4 and 1. 27. • For an exact solution, an algebraic method is best. • Since neither the coefficients of x nor the coefficients of y are the same or additive inverses, you cannot use elimination using addition or subtraction. • Since no coefficient of x or y is 1 or 1, substitution is not the best method. • Since it is necessary to multiply at least one equation by a number to eliminate a variable, solve the system by elimination using multiplication. Multiply the first equation by 2 so the coefficients of the y terms are additive inverses. Then add the equations. 3x  4y  10 Multiply by 2. 6x  8y  20 5x  8y  2 () 5x  8y  2 11x  22

4

() 2x  y  14

12

5 y 3

3 5 y 5 3

 10

135 21102



y  6

Now substitute 6 for y in either equation to find the value of x. 1

x  2y  7 1

x  2 (6)  7 x  (3)  7 x37 x3373 x4 The solution is (4, 6). 25. Let x represent the first number and let y represent the second number. 7x  3y  1 x  y  3 Multiply the second equation by 7 so the coefficients of the x terms are additive inverses. Then add the equations. 7x  3y  1 x  y  3

Multiply by 7.

11x 11

4y  20 20

 4

y  5

Now substitute 5 for y in either equation to find the value of x. x  y  3 x  (5)  3 x  5  3 x  5  5  3  5 x2 The numbers are 2 and 5. 26. Let x represent the first number and let y represent the second number. 5x  2y  22 xy3 Multiply the second equation by 2 so the coefficients of the y terms are additive inverses. Then add the equations. 5x  2y  22 5x  2y  22 () 2x  2y  6 x  y  3 Multiply by 2. 7x  28 7x 7



8y 8

8

8

y1 The solution is (2, 1). 28. • For an exact solution, an algebraic method is best. • Since the coefficients of the y terms, 8 and 8, are additive inverses, solve by elimination using addition. 9x  8y  42 () 4x  8y  16 13x  26 13x 13

26

 13

x2 Now substitute 2 for x in either equation to find the value of y. 9x  8y  42 9(2)  8y  42 18  8y  42 18  8y  18  42  18 8y  24

28 7

x4

8y 8

24

 8

y  3 The solution is (2, 3). Chapter 7

22 11

x  2 Now substitute 2 for x in either equation to find the value of y. 5x  8y  2 5(2)  8y  2 10  8y  2 10  8y  10  2  10 8y  8

7x  3y  1 () 7x  7y  21 4y 4



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32. • For an exact solution, an algebraic method is best. • Since the coefficient of y in the second equation is 1, you can use the substitution method. Since y  x, substitute x for y in the first equation. 4x  2y  14 4x  2(x)  14 4x  2x  14 2x  14

29. • For an exact solution, an algebraic method is best. • Since the coefficient of y in the first equation is 1, you can use the substitution method. Since y  3x, substitute 3x for y in the second equation. 3x  4y  30 3x  4(3x)  30 3x  12x  30 15x  30 15x 15

30

 15

2x 2

x2 Use y  3x to find the value of y. y  3x y  3(2) y6 The solution is (2, 6). 30. • For an exact solution, an algebraic method is best. • Since the coefficient of x in the first equation is 1, you can use the substitution method. Since x  4y  8, substitute 4y  8 for x in the second equation. 2x  8y  3 2(4y  8)  8y  3 8y  16  8y  3 16  3 The statement 16  3 is false. This means that there is no solution of the system of equations. 31. • For an exact solution, an algebraic method is best. • Since the coefficients of the y terms, 3 and 3, are additive inverses, you can eliminate the y terms by adding the equations. 2x  3y  12 () x  3y  12 3x  24 3x 24  3 3

8y 8

4 4

8

8

y1 Now substitute 1 for y in either equation to find the value of x. xy2 x12 x1121 x3 The solution is (3, 1). 34. • For an exact solution, an algebraic method is best. • Since the coefficient of y in the first equation is 1, you can use the substitution method. Since y  2x  9, substitute 2x  9 for y in the second equation. 2x  y  9 2x  (2x  9)  9 2x  2x  9  9 9  9 The statement 9  9 is true. This means that there are infinitely many solutions of the system of equations.

3

y3

14 2

x7 Use y  x to find the value of y. yx y7 The solution is (7, 7). 33. • For an exact solution, an algebraic method is best. • Since the coefficient of x in the first equation is 1, you can use the substitution method. Solve the first equation for x. xy2 xyy2y x2y Since x  2  y, substitute 2  y for x in the second equation. 5x  3y  18 5(2  y)  3y  18 10  5y  3y  18 10  8y  18 10  8y  10  18  10 8y  8

x8 Now substitute 8 for x in either equation to find the value of y. x  3y  12 8  3y  12 8  3y  8  12  8 3y  4 3y 3



1 42

The solution is 8, 3 .

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38. • For an exact solution, an algebraic method is best. • Since neither the coefficients of x nor the coefficients of y are the same or additive inverses, you cannot use elimination using addition or subtraction. • Since no coefficient x or y is 1 or 1, substitution is not the best method. • Since it is necessary to multiply each equation by a different number, solve the system by elimination using multiplication. Eliminate y.

35. • For an exact solution, an algebraic method is best. • Since the coefficients of the x terms, 6 and 6, are the same, solve by elimination using subtraction. 6x  y  9 () 6x  y  11 0  2 The statement 0  2 is false. This means that there is no solution of the system of equations. 36. • For an exact solution, an algebraic method is best. • Since the coefficient of x in the first equation is 1, you can use the substitution method. Since x  8y, substitute 8y for x in the second equation. 2x  3y  38 2(8y)  3y  38 16y  3y  38 19y  38 19y 19

1 x 2 3 x 2

1

1

 2 y  18  4

2  (6)(4)

x  24 Now substitute 24 for x in either equation to find the value of y.  2 y  14

2 (24) 3

 2 y  14 1

16  2 y  14 1

16  2 y  16  14  16 2 y  2 1

2

(2) 2 y  (2)(2) y4 The solution is (24, 4).

Chapter 7

 18 

18 3

 2y  25

3 (6) 2

 2y  25



16 2

y 61 Now substitute 61 for y in either equation to find x. x  y  701 x  61  701 x  61  61  701  61 x  640 So Bryant had 640 2-point field goals and 61 3-point field goals.

1

1

 2y  25

x  y  701 Multiply by 2. 2x  2y  1402 2x  3y  1463 () 2x  3y  1463

1

1

7

y  8 The solution is (6, 8). 39. Since Bryant made 475 free throws and each is 1 point, he scored 475 points in free throws. Therefore, Bryant scored 1938  475, or 1463, points in 2-point and 3-point field goals. Let x  the number of 2-point field goals, and let y  the number of 3-point field goals Bryant made. x  y  701 2x  3y  1463 Eliminate x.

1

2 x 3

3 x 2

2y 2

 2 y  14

1

()

 2y 

9  2y  25 9  2y  9  25  9 2y  16

• Since the coefficients of the y terms, 2 and 2 , are the same, solve by elimination using subtraction.

(6)

 2y  25

3 x 2 3 x 2

x  6 Now substitute 6 for x in either equation to find the value of y.

38

1 6x

Multiply by 3.

3x 3

y2 Use x  8y to find the value of x. x  8y x  8(2) x  16 The solution is (16, 2). 37. • For an exact solution, an algebraic method is best.

2 x 3 5 () 6 x 1 6x

7

3x

 19

1

2

 3y  3

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42. Let x  the tens digit of the original number, and let y  the ones digit of the original number. Then the original number is represented by 10x  y, and the number with the digits reversed is represented by 10y  x. Since the sum of the digits is 14, x  y  14. Since the number with the digits reversed is 18 less than the original number, 10y  x  (10x  y)  18. Rewrite 10y  x  (10x  y)  18 so the system can be written in column form. 10y  x  (10x  y)  18 10y  x  10x  10x  y  18  10x 10y  9x  y  18 10y  9x  y  y  18  y 9x  9y  18 Eliminate x. x  y  14 Multiply by 9. 9x  9y  126 9x  9y  18 () 9x  9y  18 18y  108

40. Since (3, a) is the solution of the given system, substituting 3 for x and a for y will make both equations in the system true. 4x  5y  2 6x  2y  b 4(3)  5(a)  2 6(3)  2(a)  b 12  5a  2 18  2a  b 12  5a  12  2  12 5a  10 5a 5



10 5

a  2 Since a  2, substitute 2 for a in the second equation to find b. 18  2a  b 18  2(2)  b 18  (4)  b 18  4  b 22  b The values of a is 2 and the value of b is 22. 41. Let x  the tens digit of the student’s actual score, and let y  the ones digit of the student’s actual score. Then the actual score is given by 10x  y, and the accidentally-reversed score is given by 10y  x. Since the actual score is 36 points greater than the accidentally-reversed score, 10x  y  (10y  x)  36. Since the sum of the digits of the score is 14, x  y  14. Rewrite 10x  y  (10y  x)  36 so the system can be written in column form. 10x  y  (10y  x)  36 10x  y  x  10y  x  36  x 9x  y  10y  36 9x  y  10y  10y  36  10y 9x  9y  36 Eliminate y. x  y  14 Multiply by 9. 9x  9y  126 9x  9y  36 (  ) 9x  9y  36 18x  162 18x 18



18y 18



108 18

y6 Now substitute 6 for y in either equation to find the value of x. x  y  14 x  6  14 x  6  6  14  6 x8 So, the tens digit of the original number is 8, and the ones digit of the original number is 6. This means the original number is 86. 43. Let r  the rate of the plane in still air, and let w  the rate of the wind. Use the formula rate  time  distance, or rt  d. 2

(4 h 40 min  43 h)

162 18

x9 Now substitute 9 for x in either equation to find the value of y. x  y  14 9  y  14 9  y  9  14  9 y5 So, the tens digit of the actual score was 9 and the ones digit of the actual score was 5. This means the student’s actual score was 95.

rt  d

r

t

Against the wind

rw

2 43

d 2100

With a wind that is twice as fast

r  2w

4

2100

1 2 (r  w)  2100 2 43

4(r  2w)  2100

1423 2 (r  w)  2100 or 143 (r  w)  2100

4(r  2w)  2100 Use the distributative property to rewrite each equation so the system is in column form. 14 r 3



14 w 3

 2100

4r  8w  2100 Eliminate w. 14 r 3



14 w 3

 2100

4r  8w  2100

Multiply by

12 . 7

8r  8w  3600

()4r  8w  2100  5700

12r 12r 12



5700 12

r  475

The rate of the plane in still air is 475 mph.

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Now substitute 6 for x in either equation to find the value of y. xy8 6y8 6y686 y2 The solution is (6, 2). 48. Since the coefficients of the s terms, 1 and 1, are additive inverses, you can eliminate the s terms by adding the equations. 2r  s  5 () r  s  1 3r 6

44. By having two equations that represent the time constraints, a manager can determine the best use of employee time. Answers should include the following. • 20c  10b  800 S 20c  10b  800 10c  30b  900 S 20c  60b  1800 50b  1000 50b 50



1000 50

b  20 20c  10b  800 20c  10(20)  800 20c  200  800 20c  200  200  800  200 20c  600 20c 20



3r 3

600 20

r2 Now substitute 2 for r in either equation to find the value of s. rs1 2s1 2s212 s  1 (1)(s)  (1)(1) s1 The solution is (2, 1). 49. Since the coefficients of the x terms, 1 and 1, are the same, you can eliminate the x terms by subtracting the equations. x  y  18 () x  2y  25 y  7 (1)(y)  (1)(7) y7 Now substitute 7 for y in either equation to find the value of x. x  y  18 x  7  18 x  7  7  18  7 x  11 The solution is (11, 7). 50. Since x  3y, substitute 3y for x in the first equation. 2x  3y  3 2(3y)  3y  3 6y  3y  3 3y  3

c  30 • In order to make the most of the employee and oven time, the manager should make assignments to bake 30 batches of cookies and 20 loaves of bread. 45. A; Eliminate y. 5x  3y  12 Multiply by 5. 25x  15y  60 4x  5y  17 Multiply by 3. ()12x  15y  51 37x  111 37x 37



111 37

x3 Now substitute 3 for x in either equation to find the value of y. 5x  3y  12 5(3)  3y  12 15  3y  12 15  3y  15  12  15 3y  3 3y 3



3 3

y  1 The value of y is 1. 46. D; Multiply the first equation by 2 so the coefficients of the x terms are additive inverses. Then add the equations. x  2y  1 Multiply by 2. 2x  4y  2 2x  4y  2 ()2x  4y  2 0 0 The statement 0  0 is true. This means that there are infinitely many solutions of the system of equations.

Page 392

3y 3

Maintain Your Skills



12 2

x6

Chapter 7

3

 3

y  1 Use x  3y to find the value of x. x  3y x  3(1) x3 The solution is (3, 1).

47. Since the coefficients of the y terms, 1 and 1, are additive inverses, you can eliminate the y terms by adding the equations. xy8 ()x  y  4 2x  12 2x 2

6

3

342

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Since the statement is true, the half-plane containing the origin is part of the solution. Shade the half-plane containing (0, 0).

51. Solve the first equation for x since the coefficient of x is 1. xy0 xyy0y x  y Since x  y, substitute y for x in the second equation. 3x  y  8 3(y)  y  8 3y  y  8 2y  8 2y 2

y x

O

yx7

8

 2

y4 Substitute 4 for y in either equation to find the value of x. xy0 x40 x4404 x  4 The solution is (4, 4). 52. Solve the first equation for x since the coefficient of x is 1. x  2y  7 x  2y  2y  7  2y x  7  2y Since x  7  2y, substitute 7  2y for x in the second equation. 3x  6y  21 3(7  2y)  6y  21 21  6y  6y  21 21  21 The statement 21  21 is true. This means that there are infinitely many solutions of the system of equations. 53. Let a  the amount of revenue above quota. 32,000  0.04a 7 45,000 32,000  0.04a  32,000 7 45,000  32,000 0.04a 7 13,000 0.04a 0.04

7

55. Step 1 Solve for y in terms of x. x  3y  9 x  3y  x  9  x 3y  9  x 9  x 3 9 x y33 1 y  3  3x 1 1  3  3 x. Since y  3  3 x 1 1 7 3  3 x or y  3  3 x, the

3y 3

Step 2 Graph y



means y boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). 1

y  3  3x 1

0  3  3 (0) false 03 Since the statement is false, the half-plane containing the origin is not part of the solution. Shade the other half-plane. 8 6 4 2

13,000 0.04

42 2 4 6 8

a 7 325,000 The store must make more than $325,000 above quota. 54. Step 1 The inequality is already solved for y in terms of x. Step 2 Graph y  x  7. Since y  x  7 means y  x  7 or y  x  7, the boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). yx7 007 true 0  7

343

y x  3y  9 O 2 4 6 8 10

x

Chapter 7

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56. Step 1 Solve for y in terms of x. y x (1) (y)  (1)x y  x Step 2 Graph y  x. Since y  x means y  x or y  x, the boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (2, 1). y  x 1  (2) true 1  2 Since the statement is true, the half-plane containing (2, 1) is part of the solution. Shade the half-plane containing (2, 1).

Page 392

Practice Quiz 2

1. Since the coefficients of the y terms, 4 and 4, are additive inverses, you can eliminate the y terms by adding the equations. 5x  4y  2 () 3x  4y  14 8x  16 8x 8

16 8

x2 Now substitute 2 for x in either equation to find the value of y. 5x  4y  2 5(2)  4y  2 10  4y  2 10  4y  10  2  10 4y  8

y

4y 4



8 4

y  2 The solution is (2, 2). 2. Since the coefficients of the x terms, 2 and 2, are the same, you can eliminate the x terms by subtracting the equations. 2x  3y  13 () 2x  2y  2 5y  15

y  x

x

O



57. Step 1 Solve for y in terms of x. 3x  y  1 3x  y  3x  1  3x y  3x  1 Step 2 Graph y  3x  1. Since y  3x  1 means y  3x  1 or y  3x  1, the boundary is included in the solution set. The boundary should be drawn as a solid line. Step 3 Select a point in one of the half-planes and test it. Try (0, 0). y  3x  1 0  3(0)  1 0  1 true Since the statement is true, the half-plane containing the origin is part of the solution. Shade the half-plane containing (0, 0).

5y 5

15

 5

y  3 Now substitute 3 for y in either equation to find the value of x. 2x  3y  13 2x  3(3)  13 2x  (9)  13 2x  9  13 2x  9  9  13  9 2x  4 2x 2

4

2

x2 The solution is (2, 3). 3. Eliminate y. 6x  2y  24 Multiply by 2. 12x  4y  48 3x  4y  27 () 3x  4y  27 15x  75

y 3x  y  1

15x 15

O

x5 Now substitute 5 for x in either equation to find the value of y. 6x  2y  24 6(5)  2y  24 30  2y  24 30  2y  30  24  30 2y  6

x

2y 2

6

 2

y3 The solution is (5, 3).

Chapter 7

75

 15

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5.

4. Eliminate x. 5x  2y  4 Multiply by 2. 10x  4y  8 10x  4y  9 () 10x  4y  9 0 1 The statement 0  1 is false. This means that there is no solution of the system of equations. 5. Let p  the rate per minute for peak time, and let n  the rate per minute for nonpeak time. Kelsey’s charges: 45p  50n  27.75 Mitch’s charges: 70p  30n  36 Eliminate n.

System of Equations one variable has a coefficient of 1 or 1 Substitution

opposite signs

 96.75 215 p 215



Elimination

Variables with Same Coefficients

45p  50n  27.75 Multiply by 3. 135p  150n  83.25 70p  30n  36 Multiply by 5. () 350p  150n  180 215p

variables have any coefficient

Variables with Different Coefficients

same signs

Add

Subtract

96.75  215

Multiply by a Factor

p  0.45

Now substitute 0.45 for p in either equation to find the value of n. 70p  30n  36 70(0.45)  30n  36 31.5  30n  36 31.5  30n  31.5  36  31.5 30n  4.5 30n 30



Graphing Systems of Inequalities

Page 395

Graphing Calculator Investigation

1. To use the SHADE feature in the DRAW menu, you must enter the lower boundary of the region to be shaded first. Since, when both inequalities are solved for y, the lower boundary is the inequality having  or , you must enter y  2x  5 first. 2. KEYSTROKES: 2nd DRAW 7 ( ) 2 X,T,␪,n

4.5 30

n  0.15 The rate for peak time is $0.45 per min and the rate for nonpeak time is $0.15 per min.

Page 393

7-5

,

5

Reading Mathematics

3 X,T,␪,n

)

1

ENTER

1. There are two types of systems of equations, consistent and inconsistent. Consistent systems have one or more solutions and inconsistent systems have no solutions. If consistent systems have one solution, they are called independent. If consistent systems have infinite solutions, they are called dependent. 2. Use substitution if an expression of one variable is given or if the coefficient of a variable is 1. Use elimination if both equations are written in standard form. Sample answers: system to solve using system to solve using substitution elimination y  3x  3 4x  3y  9 5x  2y  6 6x  y  10 3. Multiply by a factor if neither variable has the same or opposite coefficients in the two equations. 4. Add if one of the variables has opposite coefficients that are additive inverses in the two equations. Subtract if one of the variables has the same coefficient in the two equations.

[10, 10] scl: 1 by [10, 10] scl: 1

3. Solve each inequality for y. Then enter the function that is the lower boundary ( y  7  2x), a comma, and the function that is the upper boundary ( y 0.5x  2.5). 2 X,T,␪,n 4. KEYSTROKES: 2nd DRAW 7 7

, )

.

5 X,T,␪,n

2

.

5

ENTER

[10, 10] scl: 1 by [10, 10] scl: 1

345

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Pages 396–397

4. The solution includes the ordered pairs in the intersection of the graphs of x  5 and y  4. The region is shaded. The graphs of x  5 and y  4 are boundaries of this region. The graph of x  5 is dashed and is not included in the graph of x  5. The graph of y  4 is solid and is included in the graph of y  4.

Check for Understanding

1. Sample answer: y



y  2x O

x

y x5

y  2  x

y4

Because the regions have no points in common, the system of inequalities graphed above, y  2  x and y  2  x, has no solution. 2a. Since the point with coordinates (3, 1) lies in the shaded region that represents the intersection of the graphs of the inequalities, the ordered pair (3, 1) does represent a solution of the system. 2b. Since the point with coordinates (1, 3) does not lie in the shaded region that represents the intersection of the graphs of the inequalities, the ordered pair (1, 3) does not represent a solution of the system. 2c. The point with coordinates (2, 3) lies on the boundary of the region that represents the intersection of the graphs of the inequalities. Since the portion of the boundary on which (2, 3) lies is drawn as a solid line, this point is included in the solution set. So the ordered pair (2, 3) does represent a solution of the system. 2d. Since the point with coordinates (4, 2) lies in the shaded region that represents the intersection of the graphs of the inequalities, the ordered pair (4, 2) does represent a solution of the system. 2e. The point with coordinates (3, 2) lies on the boundary of the region that represents the intersection of the graphs of the inequalities. Since the portion of the boundary on which (3, 2) lies is drawn as a dashed line, this point is not included in the solution set. So the ordered pair (3, 2) does not represent a solution of the system. 2f. Since the point with coordinates (1, 4) does not lie in the shaded region that represents the intersection of the graphs of the inequalities, the ordered pair (1, 4) does not represent a solution of the system. 3. Kayla’s solution is correct. The graph of x  2y  2 is the line representing x  2y  2 and the region above it. Sonia’s solution indicates that she shaded the region below this boundary.

Chapter 7

x

O

5. The solution includes the ordered pairs in the intersection of the graphs of y  3 and y  x  4. The region is shaded. The graphs of y  3 and y  x  4 are boundaries of this region. The graphs of both boundaries are dashed and are not included in the solution. y

y 3 O

y  x  4

x

6. The solution includes the ordered pairs in the intersection of the graphs of y  x  3 and y  x  3. The region is shaded. The graphs of y  x  3 and y  x  3 are boundaries of this region. The graphs of both boundaries are solid and are included in the solution. y y  x  3 y  x 3 x O

346

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10. Let x  the number of minutes Natasha spends walking, and let y  the number of minutes she spends jogging. Since her total exercise time is at most 1 halfhour, or 30 min, x  y  30. Since her total 4 8 distance is at least 3 mi, 60 x  60 y  3. The solution includes the ordered pairs in the intersection of the graphs of x  y  30 and 4 8 x  60 y  3. The region is shaded. The graphs of 60 4 8 x  y  30 and 60 x  60 y  3 are boundaries of this region. The graphs of both boundaries are solid and are included in the solution.

7. The graphs of 2x  y  4 and y  2x  1 are parallel lines. Because the two regions have no points in common, the system of inequalities has no solution. y



2x  y  4

y  2x  1

x

O

Natasha’s Daily Exercise y Minutes Jogging

8. The solution includes the ordered pairs in the intersection of the graphs of 2y  x 6 6 and 3x  y 7 4. The region is shaded. The graphs of 2y  x  6 and 3x  y  4 are boundaries of this region. The graphs of both boundaries are dashed and are not included in the solution. y

40 30

x  y  30

20

4 x 8 y3 60 60

10

3x  y  4

O

2y  x  6

x

11. Sample answers: Natasha could walk 15 min and jog 15 min, or walk 10 min and jog 20 min, or she could walk 5 min and jog 25 min since the points with coordinates (15, 15), (10, 20), and (5, 25) lie in the region that represents the intersection of the graphs of the inequalities.

x O

9. The solution includes the ordered pairs in the intersection of the graphs of x  2y  2, 3x  4y  12, and x  0. The region is shaded. The graphs of x  2y  2, 3x  4y  12, and x  0 are boundaries of this region. The graphs of these boundaries are solid and are included in the solution. y

10 20 30 40 Minutes Walking

Pages 397–398 12.

Practice and Apply

y x

O

y0

x0 3x  4y  12

x0

x O

x  2y  2

13.

y x  4 O

x

y  1

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14.

20.

y

y

y  x 1 x

O

y x 3 y x 2

y  2

x

O

15.

21.

y

y

x 2 3x  y  6

x 2x  y  4

O

x

O

y x 5

16.

22.

y

y

x 3

x y 2

3x  4y  1

x  2y  7

x

O

x O

17.

23.

y

y xy 4

y  2x  1

x

y  x  1

O

x

O

2x  3y  12

18.

24.

y

y

y  x  3

O

2x  y  4

x

O

5x  2y  1

y  2x  1

19.

25.

y



y

O

y  x 1

x x 2x  7y  4

Chapter 7

3x  2y  6

y  x 3

y x 3

O

x

348

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26.

The solution is the set of all ordered pairs whose graphs are in the intersection of the graphs of these inequalities. This region is shaded. Only the portion of the region in the first quadrant is used since x  0 and y  0.

x  3y  9 y x 2

2x  y  9

Green Paint

x 4y  x

y Dark Green

O

27. The equation of the graph of the solid boundary is y  x. Since the shaded region that represents the intersection of the graphs of the two inequalities is below this boundary, this region is represented by y  x. The equation of the graph of the dashed boundary is y  x  3. Since the shaded region that represents the intersection of the graphs of the two inequalities is above this boundary, this region is represented by y 7 x  3. So, the system graphed is y  x . y 7 x3 28. The equation of the graph of the solid boundary is 2 y  3x  2. Since the shaded region that represents the intersection of the graphs of the two inequalities is below this boundary, this 2 region is represented by y  3x  2. The equation of the graph of the dashed boundary is y  x  3. Since the shaded region that represents the intersection of the graphs of the two inequalities is above this boundary, this region is represented by y 7 x  3. So, the 2 system graphed is y  3x  2 . y 7 x  3 29. Let x  the number of gal of light green dye to be made. Then, for the light green dye, 4x represents the number of units of yellow dye required and 1x represents the number of units of blue dye required. Let y  the number of gal of dark green dye to be made. Then, for the dark green dye, 1y represents the number of units of yellow dye required and 6y represents the number of units of blue dye required. Since the total number of units of yellow dye cannot be greater than 32, 4x  1y  32. Since the total number of units of blue dye cannot be greater than 54, 1x  6y  54. The following system of inequalities can be used to represent the conditions of the problem. x0 y0 4x  y  32 x  6y  54

30 4x  y  32

20

x  6y  54

10

x O

10

20 30 40 Light Green

50

30. Sample answers: The painter could make 2 gal of light green dye and 8 gal of dark green dye, or 6 gal of light green dye and 8 gal of dark green dye, or 7 gal of light green dye and 4 gal of dark green dye since the points with coordinates (2, 8), (6, 8), and (7, 4) lie in the region that represents the intersection of the graphs of the inequalities. 31. Let x  the level of LDL a teenager should have, and let y  the level of HDL a teenager should have. The following system of inequalities can be used to represent the conditions of the problem. 0  x  110 35  y  59 The solution is the set of all ordered pairs whose graphs are in the intersection of the graphs of these inequalities. This region is shaded. Only the portion of the region in the first quadrant is used since there can be only positive levels of LDL and HDL.

Appropriate Cholesterol Levels y

HDL

60 40 20 O

20

40

60 80 LDL

100 120 x

32. Sample answer: x  4 and x  4

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33. Let x  the number of desks made. Then, 2x represents the number of hours of sanding required and 1.5x represents the number of hours of varnishing required for the desks. Let y  the number of tables made. Then, 1.5y represents the number of hours of sanding required and 1y represents the number of hours of varnishing required for the tables. Since the total number of hours available each week for sanding is 31, 2x  1.5y  31 . Since the total number of hours available for varnishing each week is 22, 1.5x  1y  22. The following system of inequalities can be used to represent the conditions of the problem. x0 y0 2x  1.5y  31 1.5x  y  22 The solution is the set of all ordered pairs whose graphs are in the intersection of the graphs of these inequalities. This region is shaded. Only the portion of the region in the first quadrant is used since x  0 and y  0. However, since the number of items produced are whole numbers, only wholenumber solutions make sense for this problem.

36. Use the SHADE feature in the DRAW menu with y  x  4 as the lower boundary and y  x  9 as the upper boundary. DRAW 7 ( ) X,T,␪,n KEYSTROKES: 2nd

,

4

)

X,T,␪,n

9

ENTER

[10, 10] scl: 1 by [10, 10] scl: 1

37. Use the SHADE feature in the DRAW menu with y  7x  15 as the lower boundary and y  2x  10 as the upper boundary. KEYSTROKES:

2nd

DRAW 7 7 X,T,␪,n 15 , 2 X,T,␪,n 10

)

ENTER

Furniture Manufacturing

Tables

y

20

1.5x  y  22

[10, 10] scl: 1 by [10, 10] scl: 1

38. Solve each inequality for y. 10

O

3x  y  6 3x  y  3x  6  3x y  6  3x

2x  1.5y  31

10

20

(1)(y)  (1)(6  3x)

x

(1)(y)  (1)(1  x)

y  3x  6

Desks

y1x

Use the SHADE feature in the DRAW menu with y  3x  6 as the lower boundary and y  1  x as the upper boundary. DRAW 7 3 X,T,␪,n KEYSTROKES: 2nd

34. Sample answers: The company can make 8 desks and 10 tables, or 6 desks and 12 tables, or 4 desks and 14 tables since the points with coordinates (8, 10), (6, 12), and (4, 14) lie in the region that represents the intersection of the graphs of the inequalities. 35. By graphing a system of equations, you can see the appropriate range of Calories and fat intake. Answers should include the following. • Two sample appropriate Calorie and fat intakes are 2200 Calories and 60 g of fat and 2300 Calories and 65 g of fat since the points with coordinates (2200, 60) and (2300, 65) lie in the shaded region representing appropriate eating habits. • The graph represents 2000  c  2400 and 60  f  75, where c represents the number of Calories and f represents the number of grams of fat consumed per day.

Chapter 7

x  y  1 x  y  x  1  x y  1  x

6

,

1

X,T,␪,n

[10, 10] scl: 1 by [10, 10] scl: 1

350

)

ENTER

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42. Multiply the first equation by 3 so the coefficients of the y terms are additive inverses. Then add the equations. 5x  2y  3 Multiply by 3. 15x  6y  9 () 3x  6y  9 3x  6y  9 18x  18

39. D; Use a table to substitute the x and y values of each ordered pair into both inequalities. x 3 0 1 0

y

x  2y  5

3  2(7) 11 5 0  2(5) 10 4 1  2(4) 7 2.5 0  2(2.5) 5 7

5 5 5 5 5 5 5 5

True or False true true true false

3x  y  2 3(3)  7 16 3(0)  5 5 3(1)  4 7 3(0) 2.5 2.5

True or False

 2  2  2  2  2  2  2  2

true

18x 18

true

true

2y 2

Multiply by 2.

2x  3y  13

Multiply by 3.

6x  4y 

24

() 6x  9y  39 5y  15 

15 5

y3

Now substitute 3 for y in either equation to find the value of x. 3x  2y  12 3x  2(3)  12 3x  6  12 3x  6  6  12  6 3x  6 3x 3

6

 3

x  2 The solution is (2, 3). 44. Eliminate y.

11 11

y  1 Now substitute 1 for y in either equation to find the value of x. 2x  3y  1 2x  3(1)  1 2x  3  1 2x  3  3  1  3 2x  4 2x 2

3x  2y  12

5y 5

Maintain Your Skills



2

 2

y  1 The solution is (1, 1). 43. Eliminate x.

41. Multiply the first equation by 2 so the coefficients of the x terms are additive inverses. Then add the equations. 2x  3y  1 Multiply by 2. 4x  6y  2 () 4x  5y  13 4x  5y  13 11y  11 11y 11

18 18

x  1 Now substitute 1 for x in either equation to find the value of y. 5x  2y  3 5(1)  2y  3 5  2y  3 5  2y  5  3  5 2y  2

true

Since it does not make both inequalities true, the ordered pair (0, 2.5) is not part of the solution set of the system. 40. A; The equation of the graph of the solid boundary is y  2x  2. Since the shaded region representing the intersection of the graphs of the two inequalities is below this boundary, this region is represented by y  2x  2. The equation of the graph of the dashed boundary is y  x  1. Since the shaded region representing the intersection of the graphs of the two inequalities is above this boundary, this region is represented by y 7 x  1. So, the system graphed is y  2x  2 . y 7 x  1

Page 398



6x  2y  4 5x  3y  2

Multiply by 3. Multiply by 2.

18x  6y  12 () 10x  6y  4 8x  16 8x 8



16 8

x2

Now substitute 2 for x in either equation to find the value of y. 6x  2y  4 6(2)  2y  4 12  2y  4 12  2y  12  4  12 2y  8

4

2

x2 The solution is (2, 1).

2y 2

8

 2

y4 The solution is (2, 4).

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48. Step 1 The line has slope 6. To find the y-intercept, replace m with 6 and (x, y) with (1, 0) in the slope-intercept form. Then, solve for b. y  mx  b 0  6(1)  b 0  6  b 0  6  6  b  6 6b Step 2 Write the slope-intercept form using m  6 and b  6. Therefore, the equation is y  6x  6.

45. Since the coefficients of the y terms, 5 and 5, are additive inverses, you can eliminate the y terms by adding the equations. 2x  5y  13 () 3x  5y  18 5x  5 5x 5

5 5



x  1 Now substitute 1 for x in either equation to find the value of y. 2x  5y  13 2(1)  5y  13 2  5y  13 2  5y  2  13  2 5y  15 5y 5



1

49. Step 1 The line has slope 3. To find the 1 y-intercept, replace m with 3 and (x, y) with (5, 2) in the slope-intercept form. Then, solve for b. y  mx  b

15 5

y3 The solution is (1, 3). 46. Since the coefficients of the x terms, 3 and 3, are the same, you can eliminate the x terms by subtracting the equations. 3x  y  6 () 3x  2y  15 3y  9 3y 3

1

2  3 (5)  b 5

2  3  b 5

5

11

3  b Step 2 Write the slope-intercept form using 1 11 . m  3 and b   3

9

 3

1

Therefore, the equation is y  3x 

y3 Now substitute 3 for y in either equation to find the value of x. 3x  y  6 3x  3  6 3x  3  3  6  3 3x  9 3x 3

5

2  3  3  b  3

11 . 3

Chapter 7 Study Guide and Review Page 399 1. 2. 3. 4. 5. 6.

9

3

x3 The solution is (3, 3). 47. Step 1 The line has slope 2. To find the y- intercept, replace m with 2 and (x, y) with (4, 1) in the slope-intercept form. Then, solve for b. y  mx  b 1  2(4)  b 1  8  b 1  8  8  b  8 9  b Step 2 Write the slope-intercept form using m  2 and b  9. y  mx  b y  2x  (9) Therefore, the equation is y  2x  9.

Vocabulary and Concept Check

independent inconsistent dependent parallel lines infinitely many consistent

Pages 399–402 7.

y 8 6 4 2 2 2 4 6 8

Lesson-by-Lesson Review x  y  11

(10, 1) O 2 4 6 8 10 12 14 x

xy9

The graphs appear to intersect at (10, 1). Check in each equation. Check: xy9 x  y  11 ? ? 10  1  9 10  1  11 99✓ 11  11 ✓ There is one solution. It is (10, 1).

Chapter 7

352

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8. y  3x  8

Substitute 5 for n in either equation to find the value of m. mn8 m  (5)  8 m58 m5585 m3 The solution is (3, 5). 12. Since x  3  2y, substitute 3  2y for x in the second equation. 2x  4y  6 2(3  2y)  4y  6 6  4y  4y  6 66 The statement 6  6 is true. This means that there are infinitely many solutions of the system of equations. 13. Solve the first equation for y since the coefficient of y is 1. 3x  y  1 3x  y  y  1  y 3x  1  y 3x  1  1  y  1 3x  1  y Since y  3x  1, substitute 3x  1 for y in the second equation. 2x  4y  3 2x  4(3x  1)  3 2x  12x  4  3 14x  4  3 14x  4  4  3  4 14x  7

y

x

O

9x  2  3y

The graphs of the equations are parallel lines. Since they do not intersect, there are no solutions to this system of equations. 9.

y

6y  4x  8

x

O

2x  3y  4

The graphs of the equations coincide. Since every point is a point of intersection, there are infinitely many solutions to this system of equations. 10.

y 3x  y  8

x

O (2, 2)

14x 14

3x  4  y

7

 14 1

x2 1

The graphs appear to intersect at (2, 2). Check in each equation. Check: 3x  y  8 3x  4  y ? ? 3(2)  (2)  8 3(2)  4  (2) ? ? 628 642 88✓ 66✓ There is one solution. It is (2, 2). 11. Solve the second equation for m since the coefficient of m is 1. mn8 mnn8n m8n Since m  8  n, substitute 8  n for m in the first equation. 2m  n  1 2(8  n)  n  1 16  2n  n  1 3n  16  1 3n  16  16  1  16 3n  15 3n 3



Substitute 2 for x in either equation to find the value of y. y  3x  1 y3 3

112 2  1

y21 1

y2 The solution is

112, 12 2.

15 3

n  5

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14. Since n  4.5  3m, substitute 4.5  3m for n in the first equation. 0.6m  0.2n  0.9 0.6m  0.2(4.5  3m)  0.9 0.6m  0.9  0.6m  0.9 1.2m  0.9  0.9 1.2m  0.9  0.9  0.9  0.9 1.2m  1.8 1.2m 1.2

Now substitute 4 for x in either equation to find the value of y. xy5 4y5 4y454 y1 The solution is (4, 1). 18. Rewrite the first equation so the system is in column form. 3x  1  7y 3x  1  1  7y  1 3x  7y  1 3x  7y  7y  1  7y 3x  7y  1 Since the coefficients of the y terms, 7 and 7, are the same, you can eliminate the y terms by subtracting the equations. 3x  7y  1 () 6x  7y  0 3x  1

1.8

 1.2

m  1.5 Use n  4.5  3m to find the value of n. n  4.5  3m n  4.5  3(1.5) n  4.5  4.5 n0 The solution is (1.5, 0). 15. Since the coefficients of the x terms, 1 and 1, are the same, you can eliminate the x terms by subtracting the equations. x  2y  6 () x  3y  4 5y  10 5y 5



3x 3

1

x3

10 5

Now substitute the value of y. 6x  7y  0

y2 Now substitute 2 for y in either equation to find the value of x. x  2y  6 x  2(2)  6 x46 x4464 x2 The solution is (2, 2). 16. Since the coefficients of the n terms, 1 and 1, are additive inverses, you can eliminate the n terms by adding the equations. 2m  n  5 () 2m  n  3 4m 8 4m 4

6



for x in either equation to find

113 2  7y  0

2 7 2 y  7 1 solution is 3, 7y 7

The



1

2

2

7 .

19. Multiply the first equation by 2 so the coefficients of the x terms are additive inverses. Then add the equations. x  5y  0 Multiply by 2. 2x  10y  0 2x  3y  7 () 2x  3y  7 7y  7

8

4

7y 7

7

7

y1 Now substitute 1 for y in either equation to find the value of x. x  5y  0 x  5(1)  0 x50 x5505 x5 The solution is (5, 1).

16 4

x4 Chapter 7

1 3

2  7y  0 2  7y  2  0  2 7y  2

m2 Now substitute 2 for m in either equation to find the value of n. 2m  n  5 2(2)  n  5 4n5 4n454 n  1 (1)(n)  (1)1 n  1 The solution is (2, 1). 17. Since the coefficients of the y terms, 1 and 1, are additive inverses, you can eliminate the y terms by adding the equations. 3x  y  11 () x  y  5 4x  16 4x 4

1

 3

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20. Multiply the first equation by 3 so the coefficients of the x terms are additive inverses. Then add the equations. x  2y  5 Multiply by 3. 3x  6y  15 3x  5y  8 () 3x  5y  8 y  7 Now substitute 7 for y in either equation to find the value of x. x  2y  5 x  2(7)  5 x  (14)  5 x  14  5 x  14  14  5  14 x  9 The solution is (9, 7) . 21. Multiply the second equation by 3 so the coefficients of the y terms are additive inverses. Then add the equations. 2x  3y  8 2x  3y  8 x  y  2 Multiply by 3. () 3x  3y  6 5x  14 5x 5



x Now substitute the value of y. xy2 14 5 14 5

14 5

14 5 14 5

23. • For an exact solution, an algebraic method is best. • Since the coefficient of y in the first equation is 1, you can use the substitution method. Since y  2x, substitute 2x for y in the second equation. x  2y  8 x  2(2x)  8 x  4x  8 5x  8 5x 5

8

14 5

2 4

y  5

Use y  2x to find the value of y. y  2x y2 y

4

1

4

or 25

14 5

1 42 2

The solution is 25, 5 . 22. Multiply the first equation by 2 so the coefficients of the x terms are additive inverses. Then add the equations. 5x  8y  21 Multiply by 2. 10x  16y  42 10x  3y  15 () 10x  3y  15 19y  57 19y 19

1

or 35

1

3

1

2

31y 31

0

 31

y0 Now substitute 0 for y in either equation to find the value of x. 9x  8y  7 9x  8(0)  7 9x  7

57

 19

y3 Now substitute 3 for y in either equation to find the value of x. 5x  8y  21 5x  8(3)  21 5x  24  21 5x  24  24  21  24 5x  3 5x 5

16 5

24. • For an exact solution, an algebraic method is best. • Since neither the coefficients of x nor the coefficients of y are the same or additive inverses, you cannot use elimination using addition or subtraction. • Since no coefficient of x or y is 1 or 1, substitution is not the best method. • Since it is necessary to multiply at least one equation by a number to eliminate a variable, solve the system by elimination using multiplication. Multiply the first equation by 2 so the coefficients of the x terms are additive inverses. Then add the equations. Multiply by 2. 18x  16y  14 9x  8y  7 18x  15y  14 () 18x  15y  14 31y  0

for x in either equation to find

4 4

185 2

The solution is 15 , 35 .

(1)(y)  (1) 5 y5

3

x  5 or 15

y2

y

8

5

9x 9

7

9 7

x9 The solution is

179, 02.

3

 5 3

x5 The solution is

135, 32.

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27. The solution includes the ordered pairs in the intersection of the graphs of y  3x and x  2y  21. The region is shaded. The graphs of y  3x and x  2y  21 are boundaries of this region. The graph of y  3x is dashed and is not included in the graph of y  3x. The graph of x  2y  21 is solid and is included in the graph of x  2y  21.

25. • For an exact solution, an algebraic method is best. • Since the coefficient of x in the second equation is 1, you can use the substitution method. Solve the second equation for x. x  3y  y x  3y  3y  y  3y x  2y Since x  2y, substitute 2y for x in the first equation. 3x  5y  2x 3(2y)  5y  2(2y) 6y  5y  4y y  4y y  4y  4y  4y 3y  0 3y 3

y 2 12 8

2 4 6 8 x  2y  21 10 12 y  3x 14



4x

y y  2x  1 x

O

y  x  1

29. The solution includes the ordered pairs in the intersection of the graphs of 2x  y  9 and x  11y  6. The region is shaded. The graphs of 2x  y  9 and x  11y  6 are boundaries of this region. The graphs of both boundaries are dashed and are not included in the solution. y 2x  y  9

O

5x 5

x

13  x Substitute 13 for x in either equation to find the value of y. y  x  15 y  13  15 y  2 The solution is (13, 2).

Chapter 7

O

28. The solution includes the ordered pairs in the intersection of the graphs of y  x  1 and y  2x  1. The region is shaded. The graphs of y  x  1 and y  2x  1 are boundaries of this region. The graph of y  x  1 is dashed and is not included in the graph of y  x  1. The graph of y  2x  1 is solid and is included in the graph of y  2x  1.

0

3

y0 Substitute 0 for y in either equation to find the value of x. x  2y x  2(0) x0 The solution is (0, 0). 26. • For an exact solution, an algebraic method is best. • Since the coefficient of y in the first equation is 1, you can use the substitution method. Solve the first equation for y. 2x  y  3x  15 2x  y  2x  3x  15  2x y  x  15 Since y  x  15, substitute x  15 for y in the second equation. x  5  4y  2x x  5  4(x  15)  2x x  5  4x  60  2x x  5  6x  60 x  5  x  6x  60  x 5  5x  60 5  60  5x  60  60 65  5x 65 5

4

x  11y  6

356

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6.

30. The solution includes the ordered pairs in the intersection of the graphs of x  1 and y  x  3. The region is shaded. The graphs of x  1 and y  x  3 are boundaries of this region. The graphs of both boundaries are solid and are included in the solution.

y

2y  10  6x 3x  y  5

y

x

O

x1

The graphs of the equations coincide. Since every point is a point of intersection, there are infinitely many solutions to this system of equations. 7. Eliminate y. 2x  5y  16 Multiply by 2. 4x  10y  32 5x  2y  11 Multiply by 5. () 25x  10y  55 29x  87

x

O

yx3

29x 29

Chapter 7 Practice Test

x3 Now substitute 3 for x in either equation to find the value of y. 2x  5y  16 2(3)  5y  16 6  5y  16 6  5y  6  16  6 5y  10

Page 403 1. c; inconsistent 2. a; consistent 3. b; elimination 4.

y O

5y 5

x

y  2x  7

The graphs appear to intersect at (5, 3). Check in each equation. Check: yx2 y  2x  7 ? ? 3  5  2 3  2(5)  7 ? 3  3 ✓ 3  10  7 3  3 ✓ There is one solution. It is (5, 3). 7 6 5 4 3 2 1 2 1



10 5

y2 The solution is (3, 2). 8. Solve the second equation for y. y  4  2x y  4  4  2x  4 y  2x  4 Since y  2x  4, substitute 2x  4 for y in the first equation. y  2x  1 (2x  4)  2x  1 4  1 The statement 4  1 is false. This means there is no solution of the system of equation. 9. Multiply the first equation by 3 so the coefficients of the y terms are additive inverses. Then add the equations. 2x  y  4 Multiply by 3. 6x  3y  12 () 5x  3y  6 5x  3y  6 x  6 (1) (x)  (1)6 x  6 Now substitute 6 for x in either equation to find the value of y. 2x  y  4 2(6)  y  4 12  y  4 12  y  12  4  12 y8 The solution is (6, 8).

yx2 (5, 3)

5.

87

 29

y x  2y  11 x  14  2y

x O 2 4 6 8 10 12 14

The graphs of the equations are parallel lines. Since they do not intersect, there are no solutions to this system of equations.

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Now substitute 2 for x in either equation to find the value of y. 3x  y  11 3(2)  y  11 6  y  11 6  y  6  11  6 y  17 (1)(y)  (1)17 y  17 The solution is (2, 17). 14. Since the coefficients of the x terms, 3 and 3, are the same, you can eliminate the x terms by subtracting the equations. 3x  y  10 () 3x  2y  16 3y  6

10. Since y  7  x, substitute 7  x for y in the second equation. x  y  3 x  (7  x)  3 x  7  x  3 2x  7  3 2x  7  7  3  7 2x  4 2x 2

4

2

x2 Use y  7  x to find the value of y. y7x y72 y5 The solution is (2, 5). 11. Since x  2y  7, substitute 2y  7 for x in the second equation. y  3x  9 y  3(2y  7)  9 y  6y  21  9 5y  21  9 5y  21  21  9  21 5y  30 5y 5



3y 3



30 5

3x 3

3x 3

3y 3

9

3

3

 3

y1 The solution is (3, 1).

14 7

x  2

Chapter 7

12 3

x3 Now substitute 3 for x in either equation to find the value of y. 5x  3y  12 5(3)  3y  12 15  3y  12 15  3y  15  12  15 3y  3

x6 Now substitute 6 for x in either equation to find the value of y. x  y  10 6  y  10 6  y  6  10  6 y4 The solution is (6, 4). 13. Multiply the first equation by 2 so the coefficients of the y terms are additive inverses. Then add the equations. Multiply by 2. 3x  y  11 6x  2y  22 x  2y  36 () x  2y  36 7x  14 



x4 The solution is (4, 2). 15. Since the coefficients of the y terms, 3 and 3, are additive inverses, you can eliminate the y terms by adding the equations. 5x  3y  12 () 2x  3y  3 3x  9

12 2

7x 7

6 3

y  2 Now substitute 2 for y in either equation to find the value of x. 3x  y  10 3x  (2)  10 3x  2  10 3x  2  2  10  2 3x  12

y6 Use x  2y  7 to find the value of x. x  2y  7 x  2(6)  7 x  12  7 x5 The solution is (5, 6). 12. Since the coefficients of the y terms, 1 and 1, are additive inverses, you can eliminate the y terms by adding the equations. x  y  10 () x  y  2 2x  12 2x 2



358

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19. Let t  the tens digit, and let u  the units digit. u  2t  1 u  t  10 Since u  2t  1, substitute 2t  1 for u in the second equation. u  t  10 (2t  1)  t  10 3t  1  10 3t  1  1  10  1 3t  9

16. Multiply the second equation by 2 so the coefficients of the x terms are additive inverses. Then add the equations. 2x  5y  12 2x  5y  12 x  6y  11 Multiply by 2. () 2x  12y  22 17y  34 17y 17

34

 17

y2 Now substitute 2 for y in either equation to find the value of x. x  6y  11 x  6(2)  11 x  12  11 x  12  12  11  12 x1 The solution is (1, 2). 17. Multiply the first equation by 3 so the coefficients of the x terms are additive inverses. Then add the equations. x  y  6 Multiply by 3. 3x  3y  18 3x  3y  13 () 3x  3y  13 6y  5 6y 6

3t 3

t3 Use u  2t  1 to find the value of u. u  2t  1 u  2(3)  1 u61 u7 The tens digit is 3 and the units digit is 7. So, the number is 37. 20. Let /  the length of the rectangle and let w  the width of the rectangle. /w7 2/  2w  50 Solve the first equation for /. /w7 /ww7w /7w Since /  7  w, substitute 7  w for / in the second equation. 2/  2w  50 2(7  w)  2w  50 14  2w  2w  50 14  4w  50 14  4w  14  50  14 4w  36

5

 6 5

y6 Now substitute the value of x. xy6

5 6

for y in either equation to find

5

x66 5

5

5

x6666 1

x  56

1

1

The solution is 56 ,

5 6

2.

18. Multiply the first equation by 5 so the coefficients of the y terms are additive inverses. Then add the equations. 1

15x  3y  50

5

() 2x  3 y  35

3x  3 y  10 Multiply by 5.

4w 4

 85

17x 17x 17



36 4

w9 Substitute 9 for w in either equation to find the value of /. /7w /79 /  16 The length of the rectangle is 16 cm and the width of the rectangle is 9 cm.

5

5

2x  3 y  35

9

3

85

 17

x5 Now substitute 5 for x in either equation to find the value of y. 1

3x  3 y  10 1

3(5)  3 y  10 1

15  3 y  10 1

15  3 y  15  10  15 1 y 3 1 (3) 3 y

 5  (3)(5)

y  15 The solution is (5, 15).

359

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Since x  10,000  y, substitute 10,000  y for x in the second equation. 0.06x  0.08y  760 0.06(10,000  y)  0.08y  760 600  0.06y  0.08y  760 600  0.02y  760 600  0.02y  600  760  600 0.02y  160

21. The solution includes the ordered pairs in the intersection of the graphs of y 7 4 and y 6 1. The region is shaded. The graphs of y  4 and y  1 are boundaries of this region. The graphs of both boundaries are dashed and are not included in the solution. y

0.02y 0.02

y  8000 Substitute 8000 for y in either equation to find the value of x. x  10,000  y x  10,000  8000 x  2000 $2000 was invested at 6% and $8000 was invested at 8%. 25. D; For the system y 7 2x  1 , both boundary lines should be y 6 x  2 dashed. The region representing the intersection of the graphs of these two inequalities is above the graph of y  2x  1 and below the graph of y  x  2. To check, test an ordered pair in this region to verify that the coordinates satisfy both inequalities. For example, check (3, 2). Check: y 7 2x  1 y 6 x  2 ? ? 2 6 (3)  2 2 7 2(3)  1 ? ? 2 6 3  2 2 7 6  1 2 7 5 ✓ 2 6 1 ✓

x

O

y  4

y  1

22. The solution includes the ordered pairs in the intersection of the graphs of y  3 and y 7 x  2. The region is shaded. The graphs of y  3 and y  x  2 are boundaries of this region. The graph of y  3 is solid and is included in the graph of y  3. The graph of y  x  2 is dashed and is not included in the graph of y 7 x  2. y

y3

y  x  2 x

O

Chapter 7 Standardized Test Practice

23. The solution includes the ordered pairs in the intersection of the graphs of x  2y and 2x  3y  7. The region is shaded. The graphs of x  2y and 2x  3y  7 are boundaries of this region. The graphs of both boundaries are solid and are included in the solution.

Pages 404–405 1. B; 4x  2(x  2)  8  0 4x  2x  4  8  0 2x  4  0 2x  4  4  0  4 2x  4

y 2x  3y  7

2x 2

x O

4

2

x2 2. C; Let p  the price of the CD before tax. p  0.07p  17.11 1.07p  17.11

x  2y

24. Let x  the amount invested at 6% and let y  the amount invested at 8%. x  y  10,000 0.06x  0.08y  760 Solve the first equation for x. x  y  10,000 x  y  y  10,000  y x  10,000  y

1.07p 1.07



17.11 1.07

p  15.99 (to the nearest hundredth) The price of the CD before tax was $15.99. 3. B; f(x)  2x  3 f(3)  2(3)  3 63 3

f(x)  2x  3 f(4)  2(4)  3 83 5

The range is {3, 5, 7}.

Chapter 7

160

 0.02

360

f(x)  2x  3 f(5)  2(5)  3  10  3 7

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4. D;

2

1

7. D;

2



Number of hours, x Number of birds, y

1 6

3 14

4 18

Tamika’s her car current her payment balance plus deposit minus withdrawal is at least $200. 14243 123 14243 1 424 3 1442443 14 424 43 1 424 3

6 26

185

  8 4 8

The difference in y values is four times the difference of the x values. This suggests that y  4x. Check this equation. Check: If x  1, then y  4(1) or 4. But the y value for x  1 is 6. This is a difference of 2. Try some other values in the domain to see if the same difference occurs. 1 4 6

3 12 14

4 16 18

6 24 26

y is always 2 more than 4x.



x 4x y

This pattern suggests that 2 should be added to one side of the equation in order to correctly describe the relation. Check y  4x  2. If x  3, then y  4(3)  2 or 14. If x  6, then y  4(6)  2 or 26. Thus, y  4x  2 correctly describes the relation. 5. C; Step 1 The coordinates of two points on the line are (3, 0) and (0,4). Find the slope. Let (x1, y1)  (3, 0) and (x2, y2)  (0, 4). 1

4  0  (3) 4 3

m0

4

The slope is 3. Step 2 The line crosses the y-axis at (0, 4). So, the y-intercept is 4. Step 3 Write the equation. y  mx  b

1

4

(3)y  (3) 3x  4

200

58 2



10 2

y5 The value of y is 5. 11. Use the formula V  / w h.

4 4



2y 2

y  3x  (4) y  3x  4



230

s  29 Use j  s  6 to find the value of j. js6 j  29  6 j  35 The jeans cost $35. 10. C; Since the coefficients of the x terms, 3 and 3, are the same, you can eliminate x by subtracting the equations. 3x  4y  8 () 3x  2y  2 2y  10

y  y

m



d

2s 2

m  x2  x1 2



8. B; Since the perimeter is 68 ft, 2/  2w  68. Since the length is 4 more than twice the width, /  2w  4. 9. C; Let j  the cost of the jeans and let s  the cost of the shirt. Since Ernesto’s total cost is $64, j  s  64. Since the jeans cost $6 more than the shirt, j  s  6. Since j  s  6, substitute s  6 for j in the first equation. j  s  64 (s  6)  s  64 2s  6  64 2s  6  6  64  6 2s  58

1

2

1

2 1

1

Volume  242 8 9  22 8 6

2

 1764  120  1644 The volume is 1644 ft3. 12. (5x  6)  (5x  6)  (5x  6)  (5x  6)  204 20x  24  204 20x  24  24  204  24 20x  180

3y  4x  12 3y  4x  4x  12  4x 3y  4x  12 The equation is 3y  4x  12. 6. B; A line parallel to the graph of y  3x  6 has the same slope. y  3x  6 y  3x  3x  6  3x y  3x  6 The slope of the line is 3.

20x 20



180 20

x9 The value of x is 9.

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13. To find the x-intercept, let y  0. 4x  3y  12 4x  3(0)  12 4x  12 4x 4



Column B: Let (2, 1)  (x1, y1) and (5, 3)  (x2, y2). y  y

m  x2  x1 2

5

12 4

2

x3 The x-intercept is 3. 14. Write the equation in slope-intercept form y  mx  b. 4x  2y  5 4x  2y  4x  5  4x 2y  5  4x 2y 2



y y y

7 1

15. Multiply the first equation by 2 so the coefficients of the y terms are additive inverses. Then add the equations. 10x  2y  20 () 7x  2y  11 3x  9 3x 3

9

 3

x3 Now substitute 3 for x in either equation to find the value of y. 5x  y  10 5(3)  y  10 15  y  10 15  y  15  10  15 y  5 (1)(y)  (1) (5) y5 The solution is (3, 5). 16. C;

3x  2y  19 5x  4y  17

2

Multiply by 2. ()

50

 10

6x  4y  38 5x  4y  17 11x  55 11x 11

55

 11

x5 Now substitute 5 for x in either equation to find the value of y. 5x  4y  17 5(5)  4y  17 25  4y  17 25  4y  25  17  25 4y  8

92  9 9 34  3 3 3 3  81  81 Since 34  81 and 92  81, the two quantities are equal. 17. A; Column A: Let (2, 4)  (x1, y1) and (1, 3)  (x2, y2).

4y 4



8 4

y  2 Since the value of x, 5, is greater than the value of y, 2, the quantity in Column A is greater.

y  y

m  x2  x1 1

3  4

 1  2 1

 3 or 3

Chapter 7

1

x5 Now substitute 5 for x in either equation to find the value of y. 3x  y  13 3(5)  y  13 15  y  13 15  y  15  13  15 y  2 The value of y is 2. Since 0 is greater than 2, the quantity in Column B is greater. 19. A; Multiply the first equation by 2 so the coefficients of the y terms are additive inverses. Then add the equations.

5

1

6

10x 10

The slope m is 2 and the y-intercept b is 2.

2

2

7

Since 3  21 and 7  21, 3 is greater than 7. So, the quantity in Column A is greater. 18. B; Multiply the second equation by 3 so the coefficients of the y terms are additive inverses. Then add the equations. x  3y  11 x  3y  11 3x  y  13 Multiply by 3. () 9x  3y  39 10x  50

5  4x 2 5 4x  2 2 5 2  2x 5 2x  2

5x  y  10 Multiply by 2. 7x  2y  11

1

3  1  (2)

362

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20a. Let A  the number of adult tickets sold and let C  the number of children’s tickets sold.

20b.

The number of the number of 650. adult tickets plus children’s 4443 tickets is 144424443 123 1444424 { 123 

A

7.5A  4.5C  3675 7.5(650  C)  4.5C  3675 4875  7.5C  4.5C  3675 4875  3C  3675 4875  3C  4875  3675  4875 3C  1200

 650

C

The the the the cost number cost number of an of of a of adult adult children’s children’s ticket times tickets plus ticket times tickets is $3675. 1 424 3 123 14243 123 14 4244 3 123 14 4244 3 { 1 424 3 7.50



A



4.50



C



A  C  650 A  C  C  650  C A  650  C

3C 3

3675

The equations are A  C  650 and 7.5 A  4.5 C  3675.



1200 3

C  400 A  650  C A  650  400 A  250 250 adult tickets and 400 child tickets were sold.

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Chapter 8 Page 409

Polynomials 25

1. 2  2  2  2  2  3. 5  5  52 5. a  a  a  a  a  a  a6 7.

1 2

1

1

1

1

2222 

112 25

34

2. 3  3  3  3  4. x  x  x  x3 6. x  x  y  y  y  x2y3 8.

a b

a

c

c

c

bddd 

1ab 221dc 23

9. 32  3  3 9 11. 52  5  5  25 13. (6) 2  (6)(6)  36

10. 43  4  4  4  64 12. 104  10  10  10  10  10,000 14. (3) 3  (3)(3) (3)  27

15.

7 7 7 16. 8 2  8 8

17.

3. Poloma; when finding the product of powers with the same base, keep the same base and add the exponents. Do not multiply the bases. 4. No; 5  7d shows subtraction, as well as multiplication. 4a 5. No; 3b shows division, as well as multiplication. 6. Yes; a single variable is a monomial. 7. x (x4 )(x6 )  x146  x11 4 8. (4a b)(9a2b3 )  (4)(9) (a4  a2 ) (b  b3 )  36 (a42 ) (b13 )  36a6b4 3 2 3 32 9. [ (2 ) ]  [2 ] 3  [26 ] 3  263  218 or 262,144 10. (3y5z) 2  32  (y5 ) 2  z2  9  y52  z2  9y10z2

Getting Started

123 24  23  23  23  23

16  81 1 Area  2bh 1  2 14  9

1 2



1 21 2

49 64

 79  63 The area of the triangle is 63 yd2. 18. Area  r2  (62 )  36  113.04 The area of the circle is 36 m2 or about 113.04 m2. 19. Volume  /wh  734  21  4  84 The volume is 84 ft3. 20. Volume  /wh  555  25  5  125 The volume is 125 cm3.

11. (4mn2 )(12m2n)  (4  12)(m  m2 ) (n2  n)  48(m12 ) (n21 )  48m3n3 3 4 3 3 12. (2v w ) (3vw ) 2  [ (2) 3 (v3 ) 3 (w4 ) 3 ] [ (3) 2v2 (w3 ) 2 ]  [ 8  v33  w43 ] [ 9  v2  w32 ]  [8v9w12 ] [ 9v2w6 ]  (8  9)(v9  v2 ) (w12  w6 )  72(v92 ) (w126 )  72v11w18 1

13. Area  2bh 1

 2 (5n3 )(2n2 ) 1

 2 (5  2)(n3  n2 ) 1

 2 (10)(n32 )  5n5 1

14. Area  2bh

Page 413

Check for Understanding

1

 2 (3a4b)(4ab5 )

1a. Sample answer: n2(n5)  n2 + 5  n7 2 5 1b. Sample answer: (n )  n2  5  n10 1c. Sample answer: (nm2)5  n5(m2)5  n5  m2  5  n5m10 2 2 2 2a. No; (5m)  5  m  25m2 2b. Yes; the power of a product is the product of the powers. 2c. No; (3a)2  (3)2  a2  9a2 7 3 2d. No; 2(c )  2  c7  3  2c21

1

 2 (3  4)(a4  a) (b  b5 ) 1

 2 (12)(a41 )(b15 )  6a5b6

Pages 413–415

19. No;

x y2

shows division, not multiplication of

variables. Chapter 8

Practice and Apply

15. Yes; 12 is a real number and therefore a monomial. 16. Yes; 4x3 is the product of a number and three variables. 17. No; a  2b shows subtraction, not multiplication of variables. 18. No; 4n  5m shows addition, not multiplication of variables.

364

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1

1

37. (2ag2 ) 4 (3a2g3 ) 2  [ (24 ) (a4 ) (g2 ) 4 ][ (32 ) (a2 ) 2 (g3 ) 2 ]  [ 16a4g24 ] [ 9a22g32 ]  [ 16a4g8 ] [ 9a4g6 ]  (16) (9) (a4  a4 ) (g8  g6 )  144(a44 ) (g68 )  144a8g14 2 3 3 3 38. (2m n ) (3m n) 4  [ (23 )(m2 ) 3 (n3 ) 3 ] [ (34 ) (m3 ) 4 (n4 ) ]  [8m23n33 ] [ 81m34n4 ]  [8m6n9 ] [ 81m12n4 ]  (8)(81) (m6  m12 ) (n9  n4 )  648(m612 ) (n94 )  648m18n13

20. Yes; 5abc14 is the product of a number, 5, and several variables. 21. (ab4 )(ab2 )  (a  a)(b4  b2 )  (a11 )(b42 )  a2b6 5 4 2 22. ( p q )( p q)  ( p5  p2 )(q4  q)  ( p52 )(q41 )  p7q5 23. (7c3d4 ) (4cd3 )  (7  4)(c3  c) (d4  d3 )  (28) (c31 ) (d43 )  28c4d7 7k5 )(8jk8 )  (3)(8)( j7  j)(k5  k8 ) 24. (3j  (24)( j71 )(k58 )  24j8k13 2 3 4 3 4 2 25. (5a b c )(6a b c )  (5  6) (a2  a3 )(b3  b4 ) (c4  c2 )  (30)(a23 )(b34 )(c42 )  30a5b7c6 5 3 4 6 3 26. (10xy z )(3x y z )  (10  3)(x  x4)(y5  y6)(z3  z3)  (30)(x1  4)(y5  6)(z3  3)  30x5y11z6 7 2 2 2 27. (9pq )  9  p  (q7 ) 2  81  p2  q72  81p2q14 28. (7b3c6 ) 3  73  (b3 ) 3  (c6 ) 3  343  b33  c63  343b9c18 2 4 2 29. [ (3 ) ]  [324 ] 2  [38 ] 2  382  316 or 43,046,721 2 3 2 30. [ (4 ) ]  [423 ] 2  [46 ] 2  462  412 or 16,777,216 31. (0.5x3 ) 2  (0.5) 2  (x3 ) 2  0.25  x32  0.25x6 5 3 32. (0.4h )  (0.4) 3  (h5 ) 3  0.064  h53  0.064h15

1

2

1 2

3 3 33. 4c 3  4 3  c3 27

 64 c3

34.

13 2

39. (8y3 ) (3x2y2 ) 8 xy4  (8) (3)

40.

1 2

1

 24

1 38 2 (x2  x) (y3  y2  y4 )

138 2 (x21) (y324)

 9x3y9

2

4 1 m 2 (49m) (17p) 34p5 7 4 1  7 2 (m2 ) (49m) (17p) 34p5 16 1  49 (49) (17) 34 (m2  m) ( p  p5 ) 1  (16) 2 (m21 ) ( p15 )

31 2 1 2

12

4

1 2

1

2

3 6

 8m p

41. (2b3 ) 4  3(2b4 ) 3  (2) 4 (b3 ) 4  3(2) 3 (b4 ) 3  16b34  3(8)b43  16b12  24b12  (16  24)b12  40b12 3 2 3 3 42. 2(5y )  (3y )  2(5) 2 (y3 ) 2  (3) 3 (y3 ) 3  2(25) (y32 )  (27)(y33 )  50y6  27y9 43. Area  /w  (5f 4g3 ) (3fg2 )  (5) (3) ( f 4  f ) ( g3  g2 )  15( f 41 ) (g32 )  15f 5g5 44. Area  s2 45. Area  r2 2 2  (a b)  (7x4 ) 2 2 2 2  (a ) (b )  [ 72 (x4 ) 2 ]  22 2 a b  (49x4 2 ) 4 2 a b  (49x8 ) 3 46. Volume  s 47. Volume  /wh  14k3 2 3  (xy3 ) (y) (x2y) 3 3 3  4 1k 2  (x  x2 ) (y3  y  y) 33  64k  (x12 ) (y311 )  64k9  x3y5 48. Volume  r2h  (2n) 2 (4n3 )  (22  n2 ) (4n3 )  (4n2 ) (4n3 )  (4  4) (n2  n3 )  16n23  16 n5

145a222  145 22  (a2)2 16

 25  a22 16

 25 a4 35. (4cd) 2 (3d2 ) 3  [ 42  c2  d2 ] [ (3) 3  (d2 ) 3 ]  [ 16c2d2 ] [27  d23 ]  [ 16c2d2 ] [27d6 ]  (16)(27)(c2 )(d2  d6 )  432  c2  d26  432c2d8 36. (2x5 ) 3 (5xy6 ) 2  [ (2) 3 (x5 ) 3 ] [ (5) 2 (x2 )(y6 ) 2 ]  [ 8x53 ] [ 25x2y62 ]  [ 8x15 ] [ 25x2y12 ]  (8)(25)(x15  x2 )(y12 )  200 x152y12  200x17y12

365

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49. 106  106  1066  1012 or 1 trillion 6 4 50. (10 )  1064  1024 51. For 1 km/min: I  2s2  2(1) 2 or 2 For 2 km/min: I  2s2  2122 2 or 8 For 4 km/min: I  2s2  2(4) 2 or 32 52. The collision impact quadruples, since 2(2s)2 is 4(2s2). 53. 212  210  21210  222 or 4,194,304 ways 54. There is 1 way of answering all questions correctly out of a total of 4,194,304 ways to answer all questions.

62.

y y  2x  1 x

O

yx2

63.

y x  2

yx3 O

x

1

P(answers all correct)  4,194,304

64. 4x  5y  2 x  2y  6 Multiply the second equation by 4. Then add. 4x  5y  2 () 4x  8y  24 13y  26

55. False. If a  4, then (4)2  16 and 42  16. 56. True. (ambn)p  (am) p(bn) p Power of a Product  ampbnp Power of a Power 57. False. Let a  3, b  4, and n  2. Then (a  b)n  (3  4)2 or 49 and an  bn  32  42 or 25. 58. Answers should include the following.

13y 13

80 feet

• the ratio 320 feet’ which simplifies to a ratio of 1 to 4 • If s is replaced by 2s in the formula for the braking distance required for a car traveling 1 s miles per hour the result is 20 (2s) 2. Using the Power of a Product and Power of a Power 1 Properties, this simplifies to 4  20s2 . This

1

y2 Substitute 2 for y in the second equation. x  2y  6 x  2122  6 x46 x4464 x2 The solution is (2, 2). 65. 3x  4y  25 2x  3y  6 Multiply the first equation by 3 and the second equation by 4. Then add. 9x  12y  75 () 8x  12y  24 17x  51

2

means that doubling the speed of a car multiplies the braking distance by 4. 59. D; 42  45  425 60. D; Volume  s3 7 4  15x2 3  53x3  125x3

Page 415

Maintain Your Skills

61.

y

17x 17



51 17

x  3 Substitute 3 for x in the second equation. 2x  3y  6 2(3)  3y  6 6  3y  6 6  3y  6  6  6 3y  12

y  2x  2 O

y  x  1

26

 13

x

3y 3

12

 3

y  4 The solution is (3, 4).

Chapter 8

366

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66. x  y  20 0.4x  0.15y  4 Multiply the first 0.15x  0.15y () 0.4x  0.15y 0.25x 0.25x 0.25

74. Let x be the number of hours. Train Northbound Southbound

equation by 0.15. Then add.  3  4  1 

6

4

2

0

70x 70

6 6 6 6

4a  12 6 24 4a  12 4a  12 6 24 and 4a  12  12 4a  12  12 6 24  12 4a 4a 6 12

8 4

6

4a 4

4a 4

6

4

2

0

2

14 14 14  2 12

6 6 6 6

3h  2 6 2 3h  2 3h  2  2 3h

12 3

6

3h 3

4

75.

78.

81.

4 6 h The solution set is the empty set, . 4

70.

2

0

2

2m  3 7 7 2m  3  3 7 7  3 2m 7 10 2m 2

7

4

or

2

h 6 0

4

3 15

3 3

 15 3

7

77.

10 5

80.

9 48

1

5 79.

14 36

82.

45 18

   

 

14 2 36 2 7 18 45 9 18 9 5 1 or 22 2

 

10 5 5 5 2 or 2 1 9 3 48 3 3 16

Algebra Activity (Follow-Up of Lesson 8-1)

Surface

2m 2

2

0 3

2m  7 7 9 2m  7  7 7 9  7 2m 7 2

10 2

0

76.

Volume Ratio

Original

2 by 5 by 3

62

30

A

4 by 10 by 6

248

240

248 62

810

558 62

B

2 2

Surface

Dimensions

Area

Volume

Area Ratio

Prism

6

m 7 5 The solution set is {m|m 1}. 4

6

Collect the data. • Method 1 By counting, we find that there are 62 squares. Method 2 SA  2w/  2wh  2/h  2(3) (5)  2(3) (2)  2(5) (2)  30  12  20  62 The surface area is 62 cm2. • V  /wh  532  30 The volume is 30 cm3. 1.

12 4

6

2 2

2 1 3 27 27 9  9 9 9 3  1 or 3 44 44 4  32 4 32 11 3  8 or 18 2 6

Page 416

3h  2 6 2 3h  2  2 6 2  2 3h 6 0 3h 3

1

32 1

6

and

245 70 7 or 2

The trains will be 245 miles apart in 32 hours.

2 6 a a 6 3 The solution set is 5a|2 6 a 6 36 .

69.



x

2

4 4 4  12 8

68.

Distance (d = rt) 40x 30x

40x  30x  245 70x  245

1 0.25

x4 Substitute 4 for x in the first equation. x  y  20 4  y  20 4  y  4  20  4 y  16 The solution is (4, 16). 4  h  3 4h5 67. or 4  h  4  3  4 4h454 h  7 h54 h1 The solution set is {h|h  7 or h  1}. 8

Rate 40 30

6 by 15 by 9

558

New New 1SASAofofOriginal 2 1V VofofOriginal 2

4 9

240 30 810 30

8  27

2. Sample answer:

m 7 1

Surface

Volume Ratio

Original

4 by 6 by 9

228

216

A

8 by 12 by 18

912

1728

912 228

4

B

12 by 18 by 27

2052

5832

2052 228

9

6

71. The figure has been enlarged. This is a dilation. 72. The figure has been moved around a point. This is a rotation. 73. The figure has been flipped over a line. This is a reflection.

Surface

Dimensions

Area

Volume

Area Ratio

Prism

New New 1SASAofofOriginal 2 1V VofofOriginal 2

1728 216 5832 216

8  27

3. The ratio of the surface areas is 4, and the ratio of the volumes is 8. 4. The ratio of the surface areas is 9, and the ratio of the volumes is 27.

367

Chapter 8

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5. The ratio of the surface areas is a2, and the ratio of the volumes is a3. 6. Yes, the conjectures hold. If the ratio of the dimensions of two cylinders is a, then the ratio of the surface areas is a2 and the ratio of the volumes is a3.

8. 132  132 1



7

10.

5pq 10p6q3

Dividing Monomials

  

Page 418 1.

Graphing Calculator Investigation

KEYSTROKES:

4 ENTER ; 2

2

8

4

2

1

1 4

1 8

11.

KEYSTROKES:

5

p q 15 10 21 p 21 q 2 7

6

3

1 16 1p 21q73 2 2 1 5 1p 21q4 2 2 1 1 (q4 ) 2 p5

1 2

(cd2 ) 3 (c4d9 ) 2

3

2

c (d ) 3 (c4 ) 2 (d9 ) 2



3 6

1cc 21dd 2



3

6

8

18

 1c3182 21d61182 2  c11d12

1 16

(4m3n5 ) 0 mn

13. C;

is the

1

 mn

Volume of cylinder Volume of sphere

1 ENTER

5



 

The value of 51 is 0.2 or 5. The conjecture is correct.

r2h 4 r3 3

 (r) 2 (2r) 4 (r) 3 3

1

4.

KEYSTROKES:



0 ENTER

5

0 0 ENTER An error message appears.

KEYSTROKES:

 

Page 421

3 5

a b ab2



3

a 1



5

b 1

1

Pages 421–423 12

14.

8

7 72

5.

12c7z d 23  (2c(7z d)) 3

3

 

Chapter 8

3

2 3

2

3

3 3 3

2 (c ) d 73 (z2 ) 3 9 3

8c d 343z6

4 42

3 3

2

3 2 3  (1)r33 1 4 3 0 3 r or 2 2

x y x2y7



Practice and Apply 13

 4122

7 3

8 12

 782  76 or 117,649

6.

1

1 21rr 2

15.

3 37

 3137

 410 or 1,048,576

1

 a  b2 16.

 a3  b5  a1  b2  a3(1)  b5(2)  a2b3 3. Jamal; a factor is moved from the numerator of a fraction to the denominator or vice versa only if 1 the exponent of the factor is negative; 4 4. 4.

4 r3 3

Check for Understanding

1. Sample answer: 9xy and 6xy2 (9xy)(6xy2 )  (9)(6)(x  x)(y  y2 )  54x2y3 2.

2 r3

2  a4b

The value of 50 is 1. 5.

3

g8 c5d3

c d

2a–d. The numbers are reciprocals of one another. 1 5

8

 c8d18

12. 1 3. The fractional value of 51 is 5, since reciprocal of 5 or 51.

8

3

4

3

1 2

5

q

Power 24 23 22 21 20 21 22 23 24 16

1c1 21d1 21g1 2 1 1 g  1 c 21 d 21 1 2 

 2p5

ENTER ; 2 2 ENTER ; 2 1 ENTER ; 2 0 ENTER ; 2 1 ENTER ; 2 2 ENTER ; 2 3 ENTER ; 2 4 ENTER

Value

c5 d3g8

1 169



8-2

9.

12

2

7

18.



1 y4

15b2a n 2 4

2

6

  

20.

2a3 10a8



7

3

4

2

3 9

17.

368

15b4n2 2 12a6 2 2

y z yz2



1yy 21zz 2

8

2

 (y  y2z7

19.

14x3my 2 7

4

5 3

25b8n2 4a12 3



13m7 2 4 14x5y3 2 4

34 1m7 2 4

 44 1x5 2 4 1y3 2 4 81m28

 256x20y12 21.

15b 45b5



11545 21bb 2 5



 

1 1 5 a5

3



1 5a5

1 2

9

31 ) (z92 )

52 1b4 2 2n2 22 1a6 2 2

a 12 10 21 a 2

3

1 38 (a ) 5 1 5 a 5



7. y0 1y5 21y9 2  y05192 

1pp 21nn 2

  p3n

 (x82 ) (y127 )  x6y5

y4



(p74 )(n32 )

1xx 21yy 2 8

p n p4n2

 36 or 729

1 3 1 3

(b15 )

 b 4 1



1b1 2

1 3b4

4

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22. x3y0x7  x3(7)y0  x 4 (1)

33.

p  4q  3 (p5q2 )  1



1

 x4



23. n2 ( p 4 )(n 5 )  n2(5)p 4  n 3p 4  

4

1

2

27.

2

2

2 3

3

52



1 21 21 21 2



 30.

18x3y4z7 2x2yz



3

3

14

3

37.

161 21h1 21k1 2

12a3ab 2

19y0z4 3z16

2

17

32.

1 21 21 21 2

3

3

3

3

4

9

9

16

 38. / 

1 2

 

9 3

27a c 8b9

Area w 24x5y3 8x3y2 24 x5 y3 8 x3 y2

1 21 21 2

 3(x53 ) (y32 )  3x2y The length is 3x2y units. Area 39. h  1 b 2

5  2r4 22r6 5  2 1 r4 1 22 r6

 

2



23 (a2 ) 3 (b3 ) (c1 ) 3 33 (a3 ) (b2 ) 3 2  3a6b  3c3 3  3a  3b6 23 1 a6 b  3 c3 1 3  3 a  3 b6 1 3

1 21 21 2 1 1  1 25 21 4 2 (r46 ) 1  1 100 2 (r 2 ) 1 1  1 100 21 r 2





(2a2bc1 ) 3 (3ab2 ) 3

1 21 21 21 21 2 1 3 c  1 2 21 1 2 (a6(3) ) (b 36 ) 1 1 2 1 27 c  1 8 21 1 2 (a9 ) (b 9 ) 1 1 2 27c a 1  8 1 1 21 b 2

19 (z12 ) 3 19 1  3 z12 19  3z12 (5r  2 ) 2 5  2 (r  2 )  2  22 (r3 ) 2 (2r3 ) 2







z (y0 ) 1 z 2 119 3 2 19  1 3 2 (1)(z416 )



2

b2 5n2z3



3

6k17 h3 18 x3 y4 z7 2 x2 y z





bc1 3

2

 9(x32 )(y41 )(z71 )  9xy3z6

31.

51b2 n4 n2 z3 51 b2 n4 1 1 1 n2 z3

2

 6(h21 ) (k14(3) )  6h3k17 

51 (b2 ) 1 (n4 ) 1 (n2 ) 1 (z3 ) 1

3

1305 21hh 21kk 2 2

1

2

  4a4c4 30h  2k14 5hk  3

0

2

8

 27

4(a73 )(1)(c 4(8) )

29.

2

2 3 1

1 21 21 21 2 1 b 1  1 5 21 1 2 (n4(2) ) 1 z 2 b  1 5z 2 (n2 ) b 1  1 5z 21 n 2 

23

 33

28 a7 1 c  4 7 a3 b0 c  8



1b 4cc dd 2

2 3 1



3

 42

28.

3

35.

2 4 1

1

3

2

25

0

1

3

2

 16

132 23  32 3 1  1 1 21 2 2 1 2  1 3 21 1 2

2 5

2 4

3

2

28a7c  4 7a3b0c  8

34.

1

 125

145 2 2  45 4 1  1 1 21 5 2 1 5  1 4 21 1 2

p q

11q 2

1r t t 2  1 5b n (5b n ) 36. 1 n z 2  (n z )

1

 36

26.



25. 5 3  53

1

1 21 2

p

1 n3p4

24. 6 2  62

p  4q  3 (p5 )  1 (q2 ) 1 p  4q  3 p  5q  2 p4 q3 p5 q2

 ( p4(5) ) (q3(2) )  p(q1 )

1n1 21p1 2 3



1 100r2



100 a3b 1 (20a2 ) 2 100a3b

1 21 21 2 10a2

100 a3 b 10 a2 1

 10 (a32 ) (b)  10ab The height is 10ab units.

369

Chapter 8

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40.

heavy traffic normal conversation

10  3

51. You can compare pH levels by finding the ratio of one pH level to another written in terms of the 1 concentration c of hydrogen ions, c  10 pH. Answers should include the following. • Sample answer: To compare a pH of 8 with a pH of 9 requires simplifying the quotient of powers,

 10  6

1 2

 103(6)  103 or 1000 The sound from heavy traffic is 103 or 1000 times more intense than normal conversation. 41. 10,000 noisy kitchen  104  10 2  104(2)  102 The sound of a jet plane (30 m away) is 10,000 times more intense than a noisy kitchen. 42.

whisper normal conversation

1101 2 1101 2

 109(6)  10 3



The intensity of a whisper is that of normal conversation. 43. If you toss a coin n times, the probability of 1 getting n heads is 2 n.

12 2

1



10 4  104 or 1

1 10,000

1 105

to

1 104

cm or

Page 423

1 100,000

1

55. to

1 109

cm or

1 10,000,000

to 56.

an3

47.  48. (54x3 )(52x1 )  5(4x3)(2x1)  56x2 c x7 c x4

50.

58. 59.

 c(x7)(x4)  c x7x4  c11

3b2n9 b3(n3)



131 21bb

2n9

3(n3)

2

60.

(2n9)3(n3)

)  3(b  3(b2n93n9 )  3bn  

Maintain Your Skills

 (m3  m) (n  n2 )  (m31) (n12 )  m4n3 4y3 )(4x4y)  (3  4) (x4  x4 ) (y3  y) (3x  12(x44) (y31 )  12x8y4 3 2 4 3 4 2 4 (a x )  (a ) (x ) 57. 13cd5 2 2  32 1c2 21d5 2 2 12 8 a x  9c2d10 3 123 2 2 4 2  3 26 4 2  212 or 4096 3 (2b3 ) 2  (3) 3 (a3 ) (b3 ) (22 ) (b3 ) 2 (3ab)  (27) (a3 ) (b3 ) (4) (b6 )  108a3 (b36 )  108a3b9 The following system of inequalities can be used to represent the conditions of the problem. x0 y0 147x  219y  1200 xy8

54. (m

46. 107  107 or

49.

22  3

3n)(mn2 )

cm.

an (a3 )

Negative Exponent Property

25

1 100,000 1 10,000

1 10,000,000 1 1 109  109 or 1,000,000,000 1 The range of x-rays is 107 1 cm. 1,000,000,000

1

 25(5)  210 53. Since each number is obtained by dividing the previous number by 3, 31  3 and 30  1.

The range of visible light is to

1

1 10

 25

 2n 1

1 2

1

52. A; 22  23  22  (3)

1 2n

45. 10 5  105 or

1101 2

89

22  23

12

 2n

1101 2

 10 Thus, a pH of 8 is ten times more acidic than a pH of 9.

1 1000 1 1000

n

 

10  9

1

44.

9

 10  6

 103 or

1 n

8

131 21b1 2 n

3 bn

y xy8

147x  219y  1200 O

Chapter 8

370

x

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68. To find the x-intercept, let y  0. 2x  7  3y 2x  7  3(0) 2x  7

61. Sample answers: 3 oz of mozzarella, 4 oz of Swiss; 4 oz of mozzarella, 3 oz of Swiss; 5 oz of mozzarella, 3 oz of Swiss 62. y  mx  b 63. y  mx  b y  1x  (4) y  2x  3 yx4 64. y  mx  b 65. y  mx  b 1

y

7



1

To find the y-intercept, let x  0. 2x  7  3y 2(0)  7  3y 0  7  3y 0  3y  7  3y  3y 3y  7

y  2x  2

1

66. To find the x-intercept, let y  0. 2y  x  10 2102  x  10 0  x  10 0  10  x  10  10 10  x To find the y-intercept, let x  0. 2y  x  10 2y  0  10 2y  10 2y 2

7

2

x  2 or 32

3

y  3x  (1) 1 3x

2x 2

3y 3

7

3 7

1

y  3 or 23 y

(0, 2 13) 2x  7  3y

10 2

y5

(3 12 , 0)

O

x

y

69. 1121  11 since both 112  121 and (11) 2  121.

(0, 5)

2y  x  10

O

(10, 0)

70. 13.24  1.8 since 1.82  3.24. 71. 152  7.21 since (7.21) 2  52. 72. 102 103  1023  105 73. 108 105  10(8)  (5)  1013 6 9 74. 10 10  10(6)  9  103 8 1 75. 10 10  108  (1)  107 4 104  104  (4) 76. 10  100 or 1 77. 1012 10  10(12)  1  1011

x

67. To find the x-intercept, let y  0. 4x  y  12 4x  0  12 4x  12 4x 4



12 4

x3 To find the y-intercept, let x  0. 4x  y  12 4102  y  12  y  12 (1)(y)  (1)(12) y  12 y

Page 424

(3, 0) 8

4

O

4

8x

4 8 12

Reading Mathematics

1. Sample answer: triangle; a three-sided polygon. 2. See students’ work. 3a. precisely half of 3b. six 3c. eight

4x  y  12 (0, 12)

371

Chapter 8

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8-3

Pages 428–430

Scientific Notation

Page 428

18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

Check for Understanding

1. When numbers between 0 and 1 are written in scientific notation, the exponent is negative. If the number is not between 0 and 1, use a positive exponent. 2. 65.2 103 is not written in scientific notation. The number 65.2 is greater than 10. 3. Sample answer: 6.5 million; 6,500,000; 6.5 106 4. 2 108  0.00000002 5. 4.59 103  4590 6. 7.183 1014  718,300,000,000,000 7. 3.6 105  0.000036 8. 56,700,000  5.67 107 9. 0.00567  5.67 103 10. 0.00000000004  4 1011 11. 3,002,000,000,000,000  3.002 1015 12. 15.3 102 2 14.1 105 2  15.3 4.121102 105 2  21.73 107  12.173 101 2 1107 2  2.173 1101 107 2  2.173 108 or 217,300,000 5 3 13. 12 10 2 19.4 10 2  12 9.421105 103 2  18.8 108  11.88 101 2 1108 2  1.88 1101 108 2  1.88 107 or 0.000000188 14.

1.5 102 2.5 1012



29. 30. 31. 32. 33. 34. 35. 36.

37.

38.

10 11.5 2.5 21 10 2

39.

2

12

 0.6 1010  16 101 2 1010  6 1101 1010 2  6 1011 or 0.00000000006 15.

1.25 104 2.5 106



40.

10 11.25 2.5 21 10 2

41.

4

6

 0.5 1010  15 101 2 1010  5 1101 1010 2  5 109 or 5,000,000,000 16. 1,650,000,000; 1.65 109; 1,540,000,000,000; 1.54 1012 17. Average  

42. 43. 44.

1.54 1012 1.65 109 1.54 1012 1.65 109

1 21 2

 0.933 103  19.33 101 2 103  9.33 1101 103 2  9.33 102  933. 33 An average of $933.33 is charged per credit card.

Chapter 8

45.

372

106

Practice and Apply

5  0.000005 6.1 109  0.0000000061 7.9 104  79,000 8 107  80,000,000 1.243 107  0.0000001243 2.99 101  0.299 4.782 1013  47,820,000,000,000 6.89 100  6.89 2 1011  200,000,000,000 2.389 105  238,900 1.67265 1027  0.00000000000000000000000000167265 9.1095 1031  0.00000000000000000000000000000091095 50,400,000,000  5.04 1010 34,402,000  3.4402 107 0.000002  2 106 0.00090465  9.0465 104 25.8  2.58 10 380.7  3.807 102 622 106  16.22 102 2 106  6.22 1102 106 2  6.22 108 11 87.3 10  18.73 102 1011  8.73 110 1011 2  8.73 1012 4 0.5 10  15 101 2 104  5 1101 10 4 2  5 105 0.0081 103  (8.1 103 ) 103  8.1 (103 103 )  8.1 106 7 94 10  19.4 102 107  9.4 110 107 2  9.4 106 12 0.001 10  (1 103 ) 1012  1 1103 1012 2  1 109 10 billion tons  10,000,000,000 tons  1 1010 tons 602,214,299,000,000,000,000,000  6.02214299 1023 18.9 104 214 103 2  18.9 421104 103 2  35.6 107  13.56 102 107  3.56 110 107 2  3.56 108 or 356,000,000 (3 106 )(5.7 102 )  (3 5.7)(106 102 )  17.1 108  (1.71 10) 108  1.71 (10 108 )  1.71 109 or 1,710,000,000

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46. (5 102 )(8.6 103 )  (5 8.6)(102 10 3 )  43 105  (4.3 10) 105  4.3 (10 105 )  4.3 104 or 0.00043 47. (1.2 105 ) (1.2 103 )  (1.2 1.2) (105 10 3 )  1.44 108 or 0.0000000144 48. (3.5 107 ) (6.1 108 )  (3.5 6.1) (107 108 )  21.35 101  (2.135 10) 101  2.135 (10 101 )  2.135 100 or 2.135 49. (2.8 102 ) (9.1 106 )  (2.8 9.1) (102 106 )  25.48 104  (2.548 10) 104  2.548 (10 104 )  2.548 105 or 254,800 50.

7.2 109 4.8 104

1 21 2 7.2 4.8



109 104

51.

7.2 103 1.8 107

 1.5 or 150,000 52.



1

21

5.745 trillion 283.9 million

3.162 104 5.1 102

2

58.

53.



1

21

55.

2.795 108 4.3 104

4.65 101 5 105



1 21 2

6

Rodriguez’s salary Foster’s salary

25.2 million

 2.04 million



 4 10 or 0.0004

10 125.2 2.04 21 10 2 6 6

 12.35 Rodriguez’s salary in 2000 was about 12 times Foster’s salary in 1982. 59. There are 365 24 60 60 or 31,536,000 or 3.1536 107 seconds per year. (4.4 106 )(3.1536 107 )  (4.4 3.1536) (106 107 )

2

 13.87584 1013  1.387584 1014 The sun burns about 1.4 1014 or 140 trillion tons of hydrogen per year. 60a. always (a 10n ) p  ap (10n ) p Product of Powers  ap 10np Power of a Product p np 60b. Sometimes; a 10 is only in scientific notation if 1  ap  10. Counterexample: (5 103)2  52 106 or 25 106, but 25 106 is not in scientific notation since 25 is greater than 10.

105

10 12.795 4.3 21 10 2 8 4

 0.65 104  (6.5 10 1 ) 10 4  6.5 (101 10 4 )  6.5 105 or 0.000065



12

25.2 106

 0.23  (2.3 101 ) 105  2.3 (101 105 )  2.3 106 or 0.0000023 54.

10 15.745 283.9 21 10 2

 2.04 106

7.2 103 1.8 107

106

1.035 102 4.5 103

5.745 1012 283.9 106

 0.020 106 or 20,236 In April 2001, each person’s share of the debt was about $20,236.

 0.62  (6.2 101 ) 106  6.2 (101 106 )  6.2 107 or 0.00000062 1.035 102 4.5 103

 

4

105

3.162 104 5.1 102



56. Multiply the daily growth rate by the number of days. Since we are considering 10 years, there are 365 10 days. 13.3 104 21365 102  13.3 36521104 102  1204.5 103  11.2045 103 2 103  1.2045 (103 103 )  1.2045 100 or 1.2045 After 10 years, the hair would be about 1.2 meters long. 57. Divide the national debt by the number of people.

10 14.65 5 21 10 2 1 5

 0.93 106  19.3 101 2 106  9.3 (101 106 )  9.3 107 or 0.00000093

373

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61. Astronomers work with very large numbers such as the masses of planets. Scientific notation allows them to more easily perform calculations with these numbers. Answers should include the following. •

Planet

69.

4,870,000,000,000,000,000,000,000

Earth

5,970,000,000,000,000,000,000,000

Mars

 70.

1,900,000,000,000,000,000,000,000,000

Saturn

569,000,000,000,000,000,000,000,000

Uranus

86,800,000,000,000,000,000,000,000

Neptune

102,000,000,000,000,000,000,000,000

Pluto

4.5 ⫻ 10

73. Yes; 74.

(

75.

KEYSTROKES:

ENTER

(

5.5 ⫻ 10

 1.09

Page 430 68.

49a4b7c2 7ab4c3

4 ⫼

6 ⫻ 10 7

)

ENTER

103

Maintain Your Skills



1 21 21 21 2 49 a4 b7 c2 7 a b4 c3

 7(a41 )(b74 )(c23 )  7a3b3c1  7a3b3  Chapter 8

7a3b3 c

3

16

and v2.

17 17  3 14 14} 14

7 7 7 7

14

1 3

12

10

9 99 18 186 18

20

22

36

34

11c 2

374

2

32

30

77. 5b  5122 78. c2  9  32  9  5142 99  20 0 79. b3  3ac  122 3  3152132  8  45  37 80. a2  2a  1  52  2(5)  1  25  10  1  34 81. 2b4  5b3  b  2(2) 4  5(2) 3  (2)  2(16)  5(8)  2  32  40  2  10 82. 3.2c3  0.5c2  5.2c  3.2(3) 3  0.5(3) 2  5.2(3)  3.21272  0.5192  5.2132  86.4  4.5  15.6  75.3

5 ⫼

)

131 2

x  11  23 x  11  11  23  11 x  34 (1)(x)  (1)34 x  34 {x|x  34}

2

8

6 6 6 6

9  d 9  d  9 d 5d|d

38

3.15 ⫻ 10

)

is the product of

m3 m33 m {m|m

12

76.

1.03 103 67.

v2 3

18

11 ⫻ 1.2 ⫻ 10

4.095 ⫻ 10

3

64(n14(6)

6

106

KEYSTROKES:

14

6

72. No; n shows division, not multiplication of variables.

5 ENTER 66.

1641 21nn 2131 2

 64(n20 )(27)  1728n20 71. No; 3a  4b shows addition, not multiplication of variables.

7.83 107

8.52

82 (n7 ) 2



9 ⫻ 1.74 ⫻ 10

7.1 ⫻ 10

4n5 p5

1 1p 2

 33 (n2 ) 3



2 ENTER

KEYSTROKES:

5

64n14

12,700,000,000,000,000,000,000

KEYSTROKES:

3

2

 33n6

• Scientific notation allows you to fit numbers such as these into a smaller table. It allows you to compare large values quickly by comparing the powers of 10 instead of counting zeros to find place value. For computation, scientific notation allows you to work with fewer place values and to express your answers in a compact form. 62. C; 360 104  3.6 102 104  3.6 102 63. (25 billion) (270 million)  (25 109)(270 106)  (25 270)(109 106)  6750 1015  6.75 103 1015  6.75 1018 There are about 6.75 1018 hemoglobin molecules in the human body.

65.

(8n7 ) 2 (3n2 ) 3

642,000,000,000,000,000,000,000

Jupiter

141 21nn 21p1 2

 4n5

330,000,000,000,000,000,000,000

Venus



 (4)(n3(2 ) (p5 )

Mass (kg)

Mercury

64.

4n3p5 n2

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Page 430

2.

Practice Quiz 1

1. n 1n 2 1n2  n  n8 3 2. 4ad(3a d)  (4  3)(a  a3 )(d  d)  12(a13 )(d11 )  12a4d2 3. (2w3z4 ) 3 (4wz3 ) 2 (2) 3 (w3 ) 3 (z4 ) 3 (4) 2 (w2 )(z3 ) 2  8(w9 )(z12 ) (16) (w2 ) (z6 )  128(w92 )(z126 )  128w11z18 3

4.

4

25p10 15p3

341

1 21 2 5  1 3 2 1 p103 2 5  1 3 2 (p7 ) 

 6.

p10 p3

25 15

5.

1

2

6k3 2 7np4



x

5. 6. 7. 8. 9.

(6k3 ) 2 (7np4 ) 2

62 1k3 2 2

 72n2 1p4 2 2 36k6

 49n2p8

2

6

 36z10 36z10 y4



10

2

x

2

x

x

x

x 1

1

2

3x  2x x2  x  4 2x2  3x  1 x2  2x  3 Sample answer: x2, x, and 1 represent the areas of tiles.



Polynomials

1y1 2 4

5.

6.

10 19.2 2.3 21 10 2 3 5

 4 102 or 0.04 

7.

10 13.6 1.2 21 10 2 7

8.

2

 3 10 or 3,000,000,000

9.

Algebra Activity (Preview of Lesson 8-4)

10.

1. x

Check for Understanding

1. Sample answer: 8 2. A negative exponent indicates division, and an expression involving division by a variable is not a monomial. If one of the terms of an expression is not a monomial, then the expression is not a polynomial. 3a. True; a binomial is a polynomial with two terms. 3b. False; 3x  5 is a polynomial but since it has two terms it is not a monomial. 3c. True; a monomial is a polynomial with one term. 4. 5x  3xy  2x  7x  13xy2 This expression is the sum of two monomials. Yes, it is a polynomial; it is a binomial.

9

2

x

Page 434

10

7. 16.4 103 217 102 2  16.4 721103 102 2  44.8 105  14.48 102 105  4.48 110 105 2  4.48 106 or 4,480,000 8. (4 102 ) (15 106 )  (4 15)(102 106 )  60 104  (6.0 10) 104  6.0 (10 104 )  6.0 103 or 0.006

x

2

8-4

 41y4 21921z10 2

Page 431

x

x

2

2

3.6 107 1.2 102

2

4(1) (y2 )

141 21yy 2131 21z 1 2 3 z  41y26 2 1 1 21 1 2

10.

x

1

 32 (y3 ) 2 (z5 ) 2



9.2 10 2.3 105

x

4.

4y2

9.

x

3.

 32y6z10

3

x

1 1 1 1

5p7 3

4x0y2 (3y3z5 ) 2

x

2

11.

12.

375

2z 5

2

 5z

This expression is a monomial. Yes, it is a polynomial. 9a2  7a  5  9a2  7a  152 This expression is the sum of three monomials. Yes, it is a polynomial; it is a trinomial. The polynomial 1 has only one term, whose degree is 0. Thus, the degree of 1 is 0. The polynomial 3x  2 has two terms, 3x and 2, whose degrees are 1 and 0, respectively. Thus, the degree of 3x  2 is 1, the greater of 1 and 0. The polynomial 2x2y3  6x4 has two terms, 2x2y3 and 6x4, whose degrees are 5 and 4, respectively. Thus, the degree of 2x2y3  6x4 is 5, the greater of 5 and 4. 6x3  12  5x  6x3  12x0  5x1  12  5x  6x3 2 3 2 7a x  4x  2ax5  2a  7a2x3  4x2  2ax5  2ax0  2a  4x2  7a2x3  2ax5 2c5  9cx2  3x  2c5x0  9cx2  3x1  9cx2  3x  2c5 Chapter 8

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13. y3  x3  3x2y  3xy2  y3x0  x3  3x2y  3xy2  x3  3x2y  3xy2  y3 14. Since the diameter of each semicircle is 2d, the radius is half of 2d or d. Area of shaded region  area of rectangle  area of semicircles

27. The polynomial 4ab has only one term, whose degree is 2. Thus, the degree of 4ab is 2. 28. The polynomial 13 has only one term, whose degree is 0. Thus, the degree of 13 is 0. 29. The polynomial c4  7c2 has two terms, c4 and 7c2, whose degrees are 4 and 2, respectively. Thus, the degree of c4  7c2 is 4, the greater of 4 and 2. 30. The polynomial 6n3  n2p2 has two terms, 6n3 and n2p2, whose degrees are 3 and 4, respectively. Thus, the degree of 6n3  n2p2 is 4, the greater of 3 and 4. 31. The polynomial 15  8ag has two terms, 15 and 8ag, whose degrees are 0 and 2, respectively. Thus, the degree of 15  8ag is 2, the greater of 0 and 2. 32. The polynomial 3a2b3c4  18a5c has two terms, 3a2b3c4 and 18a5c, whose degrees are 9 and 6, respectively. Thus, the degree of 3a2b3c4  18a5c is 9, the greater of 9 and 6. 33. The polynomial 2x3  4y  7xy has three terms— 2x3, 4y, and 7xy—whose degrees are 3, 1, and 2, respectively. Thus, the degree of 2x3  4y  7xy is 3, the greatest of 3, 1, and 2. 34. The polynomial 3z5  2x2y3z  4x2z has three terms—3z5, 2x2y3z, and 4x2z—whose degrees are 5, 6, and 3, respectively. Thus, the degree of 3z5  2x2y3z  4x2z is 6, the greatest of 5, 6, and 3. 35. The polynomial 7  d5  b2c2d3  b6 has four terms—7, d5, b2c2d3, and b6—whose degrees are 0, 5, 7, and 6, respectively. Thus, the degree of 7  d5  b2c2d3  b6 is 7, the greatest of 0, 5, 7, and 6. 36. The polynomial 11r2t4  2s4t5  24 has three terms—11r2t4, 2s4t5, and 24—whose degrees are 6, 9, and 0, respectively. Thus, the degree of 11r2t4  2s4t5  24 is 9, the greatest of 6, 9, and 0.

11 2 1  c12d2  2 1 2d2 2  lw  2 2r2  2cd  d2

Pages 434–436

Practice and Apply

15. 14 This expression is a monomial. Yes, it is a polynomial. 16.

6m2 p

 p3

The expression

6m2 p

is not a monomial. No, this is

not a polynomial. 17. 7b  3.2c  8b  15b  13.2c2 This expression is the sum of two monomials. Yes, it is a polynomial; it is a binomial. 18. 3x2  x  2  3x2  x  122 1

1

This expression is the sum of three monomials. Yes, it is a polynomial; it is a trinomial. 19. 6gh2  4g2h  g  6gh2  14g2h2  g This expression is the sum of three monomials. Yes, it is a polynomial; it is a trinomial. 5

20. 4  2a  a2 5

The expression a2 is not a monomial. No, this is not a polynomial. 21. Area of shaded region  area of rectangle  area of triangle 1

 bh  2bh 1

 2bh or 0.5bh

37. 2x  3x2  1  2x1  3x2  1x0  1  2x  3x2 3 5 38. 9x  7  3x  9x3  7x0  3x5  7  9x3  3x5 2 3 3 2 39. c x  c x  8c  c2x3  c3x2  8cx0  8c  c3x2  c2x3 40. x3  4a  5a2x6  x3  4ax0  5a2x6  4a  x3  5a2x6 5 2 41. 4  3ax  2ax  5a7  4x0  3ax5  2ax2  5a7x0  4  5a7  2ax2  3ax5 42. 10x3y2  3x9y  5y4  2x2  10x3y2  3x9y  5x0y4  2x2  5y4  2x2  10x3y2  3x9y 43. 3xy2  4x3  x2y  6y  3x1y2  4x3  x2y  6x0y  6y  3xy2  x2y  4x3 44. 8a5x  2ax4  5  a2x2  8a5x1  2ax4  5x0  a2x2  5  8a5x  a2x2  2ax4

22. Area of shaded region  area of rectangle  area of squares  ab  41x2 2  ab  4x2 23. Area of shaded region  area of triangle  area of circle 1

 2xy  r2 24. Since the radius of the circle is r, the diameter is 2r. Since the base of the triangle is a diameter, the base is 2r. Area of shaded region  area of circle  area of triangle 1

 r2  2 (2r)(r)  r2  r2 25. The polynomial 5x3 has only one term, whose degree is 3. Thus, the degree of 5x3 is 3. 26. The polynomial 9y has only one term, whose degree is 1. Thus, the degree of 9y is 1.

Chapter 8

376

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45. 5  x5  3x3  5x0  x5  3x3  x5  3x3  5 2 46. 2x  1  6x  2x1  1x0  6x2  6x2  2x  1 3x2  5a  2a2x3  4a3x2  5ax0  2a2x3 47. 4a  2a2x3  4a3x2  5a 2 2 2 48. b  x  2xb  b x0  x2  2x1b  x2  2xb  b2 49. c2  cx3  5c3x2  11x  c2x0  cx3  5c3x2  11x1  cx3  5c3x2  11x  c2 50. 9x2  3  4ax3  2a2x  9x2  3x0  4ax3  2a2x1  4ax3  9x2  2a2x  3 51. 8x  9x2y  7y2  2x4  8x1  9x2y  7x0y2  2x4  2x4  9x2y  8x  7y2 52. 4x3y  3xy4  x2y3  y4  4x3y  3x1y4  x2y3  x0y4  4x3y  x2y3  3xy4  y4 53. Multiply the number of each type of coin by its value, and add. 0.25q  0.10d  0.05n 54. t 23; For t 23, the model predicts a negative number of quadruplet births. 55. Volume

58. A polynomial model of a set of data can be used to predict future trends in data. Answers should include the following.

1

1 4

1

2

 r2h  3r3 56. Volume  r2h 

2

123 2r3 2

 (4)(6)  3(8) 88  3

19 19 22 23.5 25 34

Actual Data Values 19 19 22 24 26 36

Maintain Your Skills

61. 12,300,000  1.23 107 62. 0.00345  3.45 103 63. 12 106  1.2 10 106  1.2 107 64. 0.77 1010  7.7 101 1010  7.7 1011

2



0 1 2 3 4 5

Page 436

 (2) 2 (6)  3r(2) 3  24 

H

The polynomial function models the data exactly for the first 3 values of t, and then closely for the next 3 values. • Someone might point to this model as evidence that the time people spend playing video games is on the rise. This model may assist video game manufacturers in predicting production needs. 59. B; 3x3  2x2  x  1  3(1)3 + 2(1)2  (1)  1  3(1)  2(1)  0  3  2  1 60. C; The degree of 5x2y3 is 5. The degree of 3x3y2 is 5. So the two quantities are equal.

 volume of cylinder  2 (volume of sphere)  r2h  2 3 r3

t

65. a0b2c1  1

16  3

1b1 2 11c 2 2

1

 b2c

or about 92.15

66.

5n5 n8



1nn 2 5 8

 5(n58 )  5n3

The volume of the container is about 92.15 in3. 57. True; for the degree of a binomial to be zero, the highest degree of both terms would need to be zero. The terms would be like terms. With these like terms combined, the expression is not a binomial, but a monomial. Therefore, the degree of a binomial can never be zero. Only a monomial can have a degree of zero.



67.

68.

1

4x3y2 3z

5

2

 2



(4x3y2 ) 2 (3z) 2 42 (x3 ) 2 (y2 ) 2 32z2 16x6y4 9z2



y5m8 y3m7

 

8

(y) m y3m7



1 2

5 1 1 n3 5 n3

1yy 2 1mm 2 5

8

3

7

 (y53 ) (m8(7) )  y2m15 69. The graph represents a relation that is not a function. The element 1 in the domain is paired with both 3 and 4 in the range. 70. The table represents a function since, for each element in the domain, there is only one corresponding element in the range.

377

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3. 4x2  x 5x  2

71. There are 26 black cards and 52 total cards. P(black) 

26 52

or

1 2

72. 3n  5n  (3  5)n  8n 2 73. 9a  3a  2a2  9a2  2a2  3a  (9  2)a2  3a  7a2  3a 74. The expression 12x2  8x  6 is simplified since it has no like terms or parentheses. 75. 3a  5b  4a  7b  (3a  4a)  (5b  7b)  (3  4)a  (5  7)b  a  2b 76. 4x  3y  6  7x  8  10y  (4x  7x)  (3y  10y)  (6  8)  (4  7)x  (3  10)y  2  11x  7y  2

Page 438

x 2

x 2

x

x

x

x 2

x 2

x

x

x 1 1

4x 2



 2

6x

(4x2  x)  (5x  2)  4x2  6x  2 4. 3x2  4x  2 remove x2  5x  5

Algebra Activity (Preview of Lesson 8-5)

1. 5x2  3x  4 2x2  4x  1

x

x2 x2

x

x

1

1

1

1

1

1

1

1

1

1

1

x2

x2

x2

x2

x2

x

x

x2

x

x

1

1

x

x

x

1



x



x

x2

x

x

x

x

x

x





9x

x 2

x2

x2

x2

x2

5x 2

x x

 2x



1

1

1

1

x 2

1

1

1

1

1

x 2

11

x

x

x

x

x

x

x2

3x 2

(2x2  5)  (3x2  2x  6)  5x2  2x  11

x

x2

1

1

7

(3x2  4x  2)  (x2  5x  5)  2x2  9x  7 5. x2  7x remove 2x2  3x

3

(5x2  3x  4)  (2x2  4x  1)  7x2  x  3 2. 2x2  5 3x2  2x  6

x2

x

x

1 2x 2

7x 2

x

x2

1

Chapter 8

x

1



4x

(x2  7x)  (2x2  3x)  3x2  4x

378

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6. 8x  4 remove 6x2  x  3

1

x2

x 2

x2

x 2

x

x2

x 2

x2

x 2

x2

x 2

x

x

x 2

x2

1

1

1

1

1

1

1

1

x



7x

7

(8x  4)  (6x2  x  3)  6x2  7x  7 7. Method from Activity 2: You need to add zero pairs so that you can remove 2 x tiles and 3 1 tiles.

2

x

2

x

x

x

x

1

1

1

1

1

x x x

2x 2  5x  2

Method from Activity 3: You remove all zero pairs to find the difference in simplest form.

x

2

x

2

x x x

Pages 441–443

1

Opposite of 2x  3

Check for Understanding

1. The powers of x and y are not the same. 2. Sample answer: 6x2  4x  7 and 4x2  3x  4 3. Kendra; Esteban added the additive inverses of both polynomials when he should have added the opposite of the polynomial being subtracted. 4. (4p2  5p)  (2p2  p)  [4p2  (2p2 ) ]  (5p  p)  2p2  6p 5. (5y2  3y  8)  (4y2  9)  (5y2  4y2 )  (3y)  [ 8  (9) ]  9y2  3y  1 6. (8cd  3d  4c)  (6  2cd)  (8cd  2cd)  (3d)  4c  (6)  10cd  3d  4c  6 7. (6a2  7a  9)  (5a2  a  10)  (6a2  7a  9)  (5a2  a  10)  (6a2  5a2 )  [ 7a  (a) ] (9  10)  11a2  6a  1 8. (g3  2g2  5g  6)  (g2  2g)  (g3  2g2  5g  6)  (g2  2g)  g3  [2g2  (g2 ) ]  [ 5g  (2g) ]  6  g3  3g2  3g  6 9. (3ax2  5x  3a)  (6a  8a2x  4x)  (3ax2  5x  3a)  (6a  8a2x  4x)  3ax2  [ 5x  (4x) ]  [ 3a  (6a) ]  8a2x  3ax2  9x  9a  8a2x 10. You can find a model for T by adding the polynomials for F and M. T  (1247n  126,971)  (1252n  120,741)  (1247n  1252n)  (126,971  120,741)  2499n  247,712 11. The year 2010 is 2010 – 1990 or 20 years after 1990. T  2499n  247,712  2499(20)  247,712  297,692 If this trend continues, the population in 2010 would be about 297,692 thousand or 297,692,000.

x

x



Page 441

1

x

x

6x 2

x

Adding and Subtracting Polynomials

8-5

Practice and Apply

2

12. (6n  4)  (2n2  9)  [6n2  (2n2 ) ]  (4  9)  4n2  5 13. (9z  3z2 )  (4z  7z2 )  (9z  4z)  [3z2  (7z2 ) ]  13z  10z2 14. (3  a2  2a)  (a2  8a  5)  (a2  a2 )  [ 2a  (8a) ]  (3  5)  2a2  6a  8

x x 1 1 1

2x 2  5x  2

379

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15. (3n2  8  2n)  (5n  13  n2 )  (3n2  n2 )  (2n  5n)  (8  13)  2n2  7n  5 16. (x  5)  (2y  4x  2)  (x  4x)  2y  [5  (2) ]  5x  2y  3 17. (2b3  4b  b2 )  (9b2  3b3 )  (2b3  3b3 )  [b2  (9b2 ) ]  (4b)  5b3  8b2  4b 18. (11  4d2 )  (3  6d2 )  (11  4d2 )  (3  6d2 )  (4d2  6d2 )  [11  (3) ]  10d2  8 19. (4g3  5g)  (2g3  4g)  (4g3  5g)  (2g3  4g)  [4g3  (2g3 ) ]  [5g  (4g) ]  2g3  9g 20. (4y3  y  10)  (4y3  3y2  7)  (4y3  y  10)  (4y3  3y2  7)  [4y3  (4y3 ) ]  (3y2 )  (y)  (10  7)  8y3  3y2  y  17 21. (4x  5xy  3y)  (3y  6x  8xy)  (4x  5xy  3y)  (3y  6x  8xy)  [4x  (6x) ]  [ 5xy  (8xy) ]  [3y  (3y) ]  2x  3xy 22. (3x2  8x  4)  (5x2  4)  (3x2  8x  4)  (5x2  4)  [3x2  (5x2 ) ]  8x  (4  4)  2x2  8x  8 23. (5ab2  3ab)  (2ab2  4  8ab)  (5ab2  3ab)  (2ab2  4  8ab)  [5ab2  (2ab2 ) ]  (3ab  8ab)  (4)  3ab2  11ab  4 3 24. (x  7x  4x2  2)  (2x2  9x  4)  (x3  7x  4x2  2)  (2x2  9x  4)

29. (9x3  3x  13)  (6x2  5x)  (2x3  x2  8x  4)  (9x3  3x  13)  (6x2  5x)  (2x3  x2  8x  4)  (9x3  2x3 )  [6x2  (x2 ) ]  [3x  5x  (8x) ]  (13  4)  11x3  7x2  9 30. The measure of the third side is the perimeter minus the measures of the other two sides. (7x  3y)  (x  2y)  (2x  3y)  (7x  3y)  (x  2y)  (2x  3y)  [7x  (x)  (2x) ]  [ 3y  2y  (3y) ]  4x  2y The measure of the third side is 4x  2y. 31. The measure of the third side is the perimeter minus the measures of the other two sides. (10x2  5x  16)  (4x3  3)  (10x  7)  (10x2  5x  16)  (4x2  3)  (10x  7)  [10x2  (4x2 ) ]  [ 5x  (10x) ]  [16  3  (7) ]  6x2  15x  12 The measure of the third side is 6x2  15x  12. 32. You can find a model for D by subtracting the polynomial for I from the polynomial for T. D  (160.3n2  26n  24,226)  (161.6n2  20n  23,326)  (160.3n2  26n  24,226)  (161.6n2  20n  23,326)  [160.3n2  (161.6n2 ) ]  (26n  20n)  [24,226  (23,326) ]  1.3n2  6n  900 33. The year 2010 is 2010  1990 or 20 years after 1990. D  1.3n2  6n  900  1.3(20) 2  6(20)  900  260 If this trend continues, there will be 260 outdoor movie screens in 2010. 34. The result is always the original number with its digits swapped. 35. Original number  10x  y; show that the new number will always be represented by 10y  x. new number  9(y  x)  (10x  y)  9y  9x  10x  y  10y  x 36. The length is 60  x  x or 60  2x inches. 37. The width is 40  x  x or 40  2x inches. 38. Girth  2(width)  2(height)  2(40  2x)  2(x)  80  4x  2x  80  2x The girth is 80  2x inches.

 x3  [4x2  (2x2)] (7x  9x)  [2  (4)]

25.

26.

27.

28.

 x3  2x2  2x  6 (5x2  3a2  5x)  (2x2  5ax  7x)  (5x2  3a2  5x)  (2x2  5ax  7x)  [5x2  (2x2 ) ]  [5x  (7x) ]  5ax  3a2  3x2  12x  5ax  3a2 (3a  2b  7c)  (6b  4a  9c)  (7c  3a  2b)  [3a  (4a)  (3a) ]  [2b  6b  (2b) ]  [7c  9c  (7c) ]  4a  6b  5c (5x2  3)  (x2  x  11)  (2x2  5x  7)  (5x2  x2  2x2 )  [x  (5x) ]  (3  11  7)  8x2  6x  15 (3y2  8)  (5y  9)  ( y2  6y  4)  [3y2  8)  (5y  9)  (y2  6y  4)  [ 3y2  (y2 ) ]  [ 5y  (6y) ]  (8  9  4)

 2y2  y  5

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39. The sum of the length and the girth must not exceed 108 inches. (60  2x)  (80  2x)  108 (60  80)  [ (2x)  (2x) ]  108 140  4x  108 140  4x  140  108  140 4x  32 4x 4



Page 443

32 4

x8 The least possible value of x is 8 inches. 40. 19 inches; For integral values of x greater than 19, the width of the box, 40  2x, would be negative or zero. 41. The next integer greater than x is 1 greater, so x + 1. 42. The sum of two consecutive integers x and x  1 is x  (x  1) or 2x  1. When 2x  1 is divided by 2x  1 2x 1 1 2; the quotient is 2  2  2 or x  2.



6a 2



4b 2

60 50 40 30 20 10

wpm

The quotient is not an integer, so 2x  1 is not divisible by 2, and hence, not even. Thus, the sum of two consecutive integers is odd. 43. 2, 3, and 4 are three consecutive integers whose sum, 2  3  4 or 9, is not even. Let x, x  1, x  2, and x  3 be four consecutive integers. Their sum is x  (x  1)  (x  2)  (x  3) or 4x  6. When 4x  6 is divided by 2, the 4x  6 4x 6 quotient is 2  2  2 or 2x  3, which is an integer. Thus, 4x  6 is divisible by 2, so that the sum of four consecutive integers is always even. 4 is the least number. 44. In order to find the sum of the video games sales and the traditional toy sales, you must add the two polynomial models V and R, which represent each of these sales from 1996 to 1999. • T  0.45t3  1.85t2  4.4t  22.6 • If a person was looking to invest in a toy company, they might want to look at the trend in toy sales over the last several years and try to predict toy sales for the future. 45. A; subtract the width from the perimeter twice. (16a  2b)  (5a  b)  (5a  b)  (16a  2b)  (5a  b)  (5a  b)  [16a  (5a)  (5a) ]  (2b  b  b)  6a  4b This is twice the length, so divide by 2. 6a  4b 2

Maintain Your Skills

47. The polynomial 15t3y2 has only one term, whose degree is 5. Thus, the degree of 15t3y2 is 5. 48. The polynomial 24 has only one term, whose degree is 0. Thus, the degree of 24 is 0. 49. The polynomial m2  n3 has two terms, m2 and n3, whose degrees are 2 and 3, respectively. Thus, the degree of m2  n3 is 3, the greater of 2 and 3. 50. The polynomial 4x2y3z  5x3z has two terms, 4x2y3z and 5x3z, whose degrees are 6 and 4, respectively. Thus, the degree of 4x2y3z  5x3z is 6, the greater of 6 and 4. 51. 8 106  8,000,000 52. 2.9 105  290,000 53. 5 104  0.0005 54. 4.8 107  0.00000048 55–56.

O

1 2 3 4 5 6 7 8 9 10 Weeks

57. Sample answer: We use the points (4, 33) and (7, 45). y2  y1

mx

2

 

58.

59. 60.

or 3a  2b

61.

The length is 3a  2b. 62.

46. D; (a2  2ab  b2 )  (a2  3ab  b2 )  (a2  2ab  b2 )  (a2  3ab  b2 )  [a2  (a2 ) ]  [ (2ab)  3ab]  [b2  (b2 ) ]  ab This expression is also equivalent to 36 – 22 or 14. Thus, ab  14.

 x1

45  33 7  4 12 or 4 3

Use the point-slope form. y  y1  m(x  x1 ) y  33  4(x  4) y  33  4x  16 y  33  33  4x  16  33 y  4x  17 y  4x  17  4(12)  17  48  17  65 After 12 weeks, a student’s keyboarding speed should be about 65 wpm. No; there is a limit to how fast one can keyboard. The domain is {2, 0, 6}. The range is {5, 2, 3}. The domain is {4, 1, 5}. The range is {2,3, 0, 1}. Let x represent the length. 1 87



8 x

1(x)  87(8) x  696 The real locomotive is 692 inches or 58 feet long. 63. 6(3x  8)  6(3x)  6(8)  18x  48 64. 2(b  9)  2(b)  (2) (9)  2b  18

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65. 7(5p  4q)  7(5p)  (7)(4q)  35p  28q 66. 9(3a  5b  c)  9(3a)  9(5b)  9(c)  27a  45b  9c 67. 8(x2  3x  4)  8(x2 )  8(3x)  8(4)  8x2  24x  32 68. 3(2a2  5a  7)  3(2a2 )  (3)(5a)  (3)(7)  6a2  15a  21

8-6

2(w  1)  w  7  4w 2(w)  (2)(1)  w  7  4w 2w  2  w  7  4w w  2  7  4w 3w  2  7 3w  9 w3 x(x  2)  3x  x(x  4)  5 11. x(x)  x(2)  3x  x(x)  x(4)  5 x2  2x  3x  x2  4x  5 x2  x  x2  4x  5 x  4x  5 3x  5 5 x3

10.

Multiplying a Polynomial by a Monomial

Page 446

12. Subtract x from 10,000 to find the amount in the CD. The expression is 10,000  x. 13. The total is the sum of the original investment and the interest earned. T  10,000  0.04x  0.07(10,000  x) T  10,000  0.04x  0.07(10,000)  0.07(x) T  10,000  0.04x  700  0.07x T  10,700  0.03x 14. T  10,700  0.03x  10,700  0.03(3000)  10,700  90  10,610 If she puts $3000 in savings, she will have $10,610.

Check for Understanding

1. Distributive Property; Product of Powers Property 2. The three monomials that make up the trinomial are similar to the three digits that make up the 3-digit number. The single monomial is similar to a 1-digit number. With each procedure you are performing multiplications. The difference is that polynomial multiplication involves variables and the resulting product is often the sum of two or more monomials while numerical multiplication results in a single number. 3. Sample answer: 4x and x2  2x  3; 4x(x2  2x  3)  4x(x2 )  4x(2x)  4x(3)  4x3  8x2  12x 4. 3y(5y  2)  3y(5y)  (3y)(2)  15y2  (6y)  15y2  6y 2 3 2 5. 9b (2b  3b  b  8)  9b2 (2b3 )  9b2 (3b2 )  9b2 (b)  9b2 (8)  18b5  27b4  9b3  72b2 6. 2x(4a4  3ax  6x2 )  2x(4a4 )  2x(3ax)  2x(6x2 )  8a4x  6ax2  12x3 7. 4xy(5x2  12xy  7y2 )  4xy(5x2 )  (4xy)(12xy)  (4xy)(7y2 )  20x3y  (48x2y2 )  (28xy3 )  20x3y  48x2y2  28xy3 8. t(5t  9)  2t  t(5t)  t(9)  2t  5t2  9t  2t  5t2  11t 3 2 9. 5n(4n  6n  2n  3)  4(n2  7n)  5n(4n3 )  5n(6n2 )  5n(2n)  5n(3)  (4)(n2 )  (4)(7n)  20n4  30n3  10n2  15n  (4n2 )  (28n)  20n4  30n3  [ (10n2 )  (4n2 ) ]  [15n  (28n) ]  20n4  30n3  (14n2 )  (13n)  20n4  30n3  14n2  13n

Chapter 8

Pages 446–448

Practice and Apply

15. r(5r  r2 )  r(5r)  r(r2 )  5r2  r3 3 16. w(2w  9w2 )  w(2w3 )  w(9w2 )  2w4  9w3 17. 4x(8  3x)  4x(8)  (4x) (3x)  32x  12x2 2 18. 5y(2y  7y)  5y (2y2 )  5y(7y)  10y3  35y2 3 19. 7ag (g  2ag)  7ag (g3 )  7ag (2ag)  7ag4  14a2g2 2 20. 3np(n  2p)  3np(n2 )  (3np) (2p)  3n3p  (6np2 )  3n3p  6np2 2 2 21. 2b (3b  4b  9)  2b2 (3b2 )  (2b2 ) (4b)  (2b2 ) (9)  6b4  (8b3 )  (18b2 )  6b4  8b3  18b2 22. 6x3 (5  3x  11x2 )  6x3 (5)  6x3 (3x)  6x3 (11x2 )  30x3  18x4  66x5 23. 8x2y(5x  2y2  3)  8x2y(5x)  8x2y(2y2 )  8x2y(3)  40x3y  16x2y3  24x2y 24. cd2 (3d  2c2d  4c)  cd2 (3d)  (cd2 ) (2c2d)  (cd2 ) (4c)  3cd3  (2c3d3 )  (4c2d2 )  3cd3  2c3d3  4c2d2

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3

25. 4 hk2 (20k2  5h  8) 3

1

3

2

1

3

35. 3c2 (2c  7)  4c(3c2  c  5)  2(c2  4)

2

 3c2 (2c)  (3c2 ) (7)  4c(3c2 )  4c(c)

 4 hk2 (20k2 )  4 hk2 (5h)  4 hk2 (8)

1

2

15  15hk4   4 h2k2  (6hk2 )

 15hk4 

15 2 2 h k 4

 4c(5)  2(c2 )  2(4)  6c3  (21c2 )  12c3  4c2  20c  2c2  8  (6c3  12c3 )  (21c2  4c2  2c2 )  20c  8

 6hk2

 6c3  23c2  20c  8

2

26. 3 a2b(6a3  4ab  9b2 )  

2 2 2 a b(6a3 )  3 a2b(4ab) 3 8 4a5b  3 a3b2  6a2b3



36. 4x2 (x  2)  3x(5x2  2x  6)  5(3x2  4x)  4x2 (x)  4x2 (2)  3x(5x2 )  3x(2x)  3x(6)  (5) (3x2 )  (5) (4x)  4x3  8x2  15x3  6x2  18x  (15x2 )  (20x)  (4x3  15x3 )  (8x2  6x2  15x2 )  (18x  20x)  19x3  x2  2x 37. area of shaded region  area of large rectangle  area of small rectangle  4x(3x  2)  2x(3x)  4x(3x)  4x(2)  2x(3x)  12x2  8x  6x2  (12x2  6x2 )  8x  6x2  8x 38. area of shaded region  area of large rectangle  area of small rectangle  5p(3p  4)6(2p  1)  5p(3p)  5p(4)  (6) (2p)  (6) (1)  15p2  20p  (12p)  (6)  15p2  (20p  12p)  6  15p2  8p  6 39. 2(4x  7)  5(2x  9)  5 2(4x)  2(7)  5(2x)  5(9)  5 8x  14  10x  45  5 8x  14  10x  50 18x  14  50 18x  36 x  2 40. 2(5a  12)  6(2a  3)  2 2(5a)  2(12)  6(2a)  (6) (3)  2 10a  24  12a  (18)  2 10a  24  12a  18  2 10a  24  12a  20 22a  24  20 22a  44 a2 41. 4(3p  9)  5  3(12p  5) 4(3p)  4(9)  5  3(12p)  (3) (5) 12p  36  5  36p  (15) 12p  31  36p  15 48p  31  15 48p  16

2 2 a b(9b2 ) 3

27. 5a3b(2b  5ab  b2  a3 )  5a3b(2b)  (5a3b)(5ab)  (5a3b)(b2 )  (5a3b) (a3 )  10a3b2  (25a4b2 )  (5a3b3 )  (5a6b)  10a3b2  25a4b2  5a3b3  5a6b 28. 4p2q2 (2p2  q2  9p3  3q)  4p2q2 (2p2 )  4p2q2 (q2 )  4p2q2 (9p3 )  4p2q2 (3q)  8p4q2  4p2q4  36p5q2  12p2q3

29. d(2d  4)  15d  d(2d)  d(4)  15d  2d2  4d  15d  2d2  (4d  15d)  2d2  19d 30. x(4x2  2x)  5x3  x(4x2 )  (x)(2x)  5x3  4x3  (2x2 )  5x3  4x3  2x2  5x3  (4x3  5x3 )  2x2  9x3  2x2 2 31. 3w(6w  4)  2(w  3w  5)  3w(6w)  3w(4)  2(w2 )  2(3w)  2(5)  18w2  12w  2w2  6w  10  (18w2  2w2 )  (12w  6w)  10  20w2  18w  10 32. 5n(2n3  n2  8)  n(4  n)  5n(2n3 )  5n(n2 )  5n(8)  n(4)  n(n)  10n4  5n3  40n  4n  n2  10n4  5n3  n2  (40n  4n)  10n4  5n3  n2  44n 33. 10(4m3  3m  2)  2m(3m2  7m  1)  10(4m3 )  10(3m)  10(2)  (2m)(3m2 )  (2m)(7m)  (2m)(1)  40m3  30m  20  6m3  (14m2 )  (2m)  40m3  30m  20  6m3  14m2  2m  (40m3  6m3 )  14m2  (30m  2m)  20  46m3  14m2  32m  20 34. 4y(y2  8y  6)  3(2y3  5y2  2)  4y(y2 )  4y(8y)  4y(6)  (3)(2y3 )  (3)(5y2 )  (3)(2)  4y3  32y2  24y  (6y3 )  (15y2 )  (6)  4y3  32y2  24y  6y3  15y2  6  (4y3  6y3 )  (32y2  15y2 )  24y  6  2y3  17y2  24y  6

p 42.

16 48

1

or 3

7(8w  3)  13  2(6w  7) 7(8w)  7(3)  13  2(6w)  2(7) 56w  21  13  12w  14 56w  (8)  12w  14 44w  8  14 44w  22 22

1

w  44 or 2

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d(d  1)  4d  d(d  8) d(d)  d(1)  4d  d(d)  d(8) d2  d  4d  d2  8d d2  3d  d2  8d 3d  8d 11d  0 d0 c(c  3)  c(c  4)  9c  16 44. c(c)  c(3)  (c)(c)  (c) (4)  9c  16 c2  3c  c2  4c  9c  16 (c2  c2 )  (3c  4c)  9c  16 7c  9c  16 2c  16 c8 y(y  12)  8y  14  y(y  4) 45. y(y)  y(12)  8y  14  y(y)  y(4) y2  12y  8y  14  y2  4y y2  4y  14  y2  4y 4y  14  4y 8y  14

51. The increased length is 5x  12. Area  4x(5x  12)  4x(5x)  4x(12)  20x2  48x 52. Since 1 mile is at the $2.75 rate, m  1 miles are at the $1.25 rate for each taxi. So the cost for each taxi is 2.75(1)  1.25(m  1) or 2.75  1.25(m  1). Since there are t taxis, we multiply the cost of each taxi by t. t [ 2.75  1.25(m  1) ]  t(2.75  1.25m  1.25)  t(1.50  1.25m)  1.50t  1.25mt The total cost to transport the group is 1.50t  1.25mt dollars. 53. The next odd integer is 2 greater than x, so x  2 is the next odd integer. 54. x(x  2)  x(x)  x(2)  x2  2x 55. Let x and y be integers. Then 2x and 2y are even numbers, and (2x)(2y)  4xy. 4xy is divisible by 2 since one of its factors, 4, is divisible by 2. Therefore 4xy is an even number. 56. 2x  1 or 2x  1 57. Let x and y be integers. Then 2x is an even number and 2y  1 is an odd number. Their product, 2x(2y  1), is always even since one of its factors is 2. 58. Since a represents the number of apples, 10-a represents the number of oranges. T  0.25a  0.20(10  a) T  0.25a  0.20(10)  0.20(a) T  0.25a  2  0.20a T  2  0.05a 59. T  2  0.05a  2  0.05(4)  2  0.20  2.20 With 4 apples, the total cost is $2.20. 60. T  p  0.30p  0.01n(p  0.30p) T  p  0.30p  0.01n(p)  (0.01n)(0.30p) T  p  0.30p  0.01np  0.003np T  0.7p  0.007np 61. Replace p with 200 and n with 10. T  0.7(200)  0.007(10)(200)  140  14  126 The discounted cost is $126. 62. Distance apart at start  circumference of outer semicircle  circumference of inner semicircle

43.

y 46.

14 8

7

or 4

k(k  7)  10  2k  k(k  6) k(k)  k(7)  10  2k  k(k)  k(6) k2  7k  10  2k  k2  6k k2  7k  10  k2  8k 7k  10  8k 15k  10  0 15k  10 10

2

k  15 or 3 47. 2n(n  4)  18  n(n  5)  n(n  2)  7 2n(n)  2n(4)  18  n(n)  n(5)  n(n)  n(2)  7 2n2  8n  18  n2  5n  n2  2n  7 2n2  8n  18  2n2  3n  7 8n  18  3n  7 5n  18  7 5n  25 n  5 48. 3g(g  4)  2g(g  7)  g(g  6)  28 3g(g)  3g(4)  (2g)(g)  (2g)(7)  g(g)  g(6)  28 3g2  12g  2g2  14g  g2  6g  28 g2  2g  g2  6g  28 2g  6g  28 4g  28 g7 49. Since x represents the amount put into a savings account, 6000  x represents the amount used to buy a certificate of deposit. The total amount is the sum of the original investment and the interest earned. T  6000  0.03x  0.06(6000  x) T  6000  0.03x  0.06(6000)  .06(x) T  6000  0.03x  360  0.06x T  6360  0.03x or T  0.03x  6360 50. T  6360  0.03x 6315  6360  0.03x 45  0.03x 1500  x 6000  x  6000  1500 or 4500 Savings account: $1500; certificate of deposit: $4500 Chapter 8

1

1

 2 [2(x  2.5) ]  2 (2x)  (x  2.5)  x  x  2.5  x  2.5 The runners should be 2.5 or about 7.9 feet apart.

384

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70. 4x2  10ab  6  4x2  (10ab)  6 This expression is the sum of three monomials. Yes, it is a polynomial; it is a trinomial. 71. 4c  ab  c  3c  ab This expression is the sum of two monomials. Yes, it is a polynomial; it is a binomial.

63. Answers should include the following. • The product of a monomial and a polynomial can be modeled using an area model. The area of the figure shown at the beginning of the lesson is the product of its length 2x and width (x  3). This product is 2x(x  3), which when the Distributive Property is applied, becomes 2x(x)  2x(3) or 2x2  6x. This is the same result obtained when the areas of the algebra tiles are added together. • Sample answer: (3x)(2x  1) (3x)(2x  1)  (3x)(2x)  (3x)(1)  6x2  3x 1 x x

x

x

2

x

2

x

x

x

2

x

2

x

x

x

2

x

2

x

72.

 y2 7

The expression y is not a monomial. No, this is not a polynomial. 73.

n2 3

1

 3 n2

This expression is a monomial. Yes, it is a polynomial. 74. Let n be the number. 6  10n  9n 6  10n 10n  9n  10n 6  n (1)6 (1)(n) 6 n The solution set is {n|n  6}. 75. Let n be the number. 9n  4  7  13n 9n  4  13n  7  13n  13n 22n  4  7 22n  4  4  7  4 22n  3

64. B; [ (3x2  2x  4)  (x2  5x  2) ] (x  2)  [3x2  2x  4  (x2 )  (5x)  2] (x  2)  [2x2  7x  6](x  2)  2x2 (x  2)  (7x)(x  2)  6(x  2)  2x2 (x)  2x2 (2)  (7x) (x)  (7x)(2)  6(x)  6(2)  2x3  4x2  7x2  14x  6x  12  2x3  3x2  8x  12 65. A; Let m represent the number of minutes over 30 minutes and T represent the total charges. T  70  4m 122  70  4m 52  4m 13  m The total time is 13  30 or 43 minutes.

Page 449

7 y

22n 22

3

 22 3

n  22

5

3

6

The solution set is n|n  22 . 76. Find the slope. y2  y1

mx

2

 

 x1

4  (8) 1  (3) 12 or 3 4

Find the y-intercept. y  mx  b 8  3(3)  b 8  9  b 1b Write the slope-intercept form. y  mx  b y  3x  1 77. Find the slope.

Maintain Your Skills

66. (4x2  5x)  (7x2  x)  [4x2  (7x2 ) ]  (5x  x)  3x2  6x 67. (3y2  5y  6)  (7y2  9)  (3y2  5y  6)  (7y2  9)  (3y2  7y2 )  5y  (6  9)  4y2  5y  3 68. (5b  7ab  8a)  (5ab  4a)  (5b  7ab  8a)  (5ab  4a)  5b  (7ab  5ab)  (8a  4a)  5b  12ab  12a 69. (6p3  3p2  7)  (p3  6p2  2p)  (6p3  p3 )  (3p2  6p2 )  2p  7  7p3  3p2  2p  7

y2  y1

mx

2

 

 x1

7  5 2  (4) 12 or 2 6

Find the y-intercept. y  mx  b 5  2(4)  b 58b 3  b Write the slope-intercept form. y  mx  b y  2x  3

385

Chapter 8

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78. Find the slope.

Page 449

y2  y1

mx

2

 

 x1

2  (1) 3  3 3 1 or  2 6

Find the y-intercept. y  mx  b 1 1  2 (3)  b 3

1  2  b 1 2

b

Write the slope-intercept form. y  mx  b 1

1

y  2 x  2

5. 4x2  9x  12  5x3  4x2  9x1  12x0  5x3  5x3  4x2  9x  12  12  9x  4x2  5x3 6. 2xy4  x3y5  5x5y  13x2  2x1y4  x3y5  5x5y  13x2  5x5y  x3y5  13x2  2xy4  2xy4  13x2  x3y5  5x5y 7. (7n2  4n  10)  (3n2  8)  (7n2  3n2 )  4n  (10  8)  10n2  4n  2 8. (3g3  5g)  (2g3  5g2  3g  1)  (3g3  5g)  (2g3  5g2  3g  1)  [3g3  (2g3 ) ]  (5g2 )  (5g  3g)  1  g3  5g2  2g  1 9. 5a2 (3a3b  2a2b2  6ab3 )  5a2 (3a3b)  5a2 (2a2b2 )  5a2 (6ab3 )  15a5b  10a4b2  30a3b3 10. 7x2y(5x2  3xy  y)  7x2y(5x2 )  7x2y(3xy)  7x2y(y)  35x4y  21x3y2  7x2y2

79. Let x be the amount of money Kristen had originally. 1 x 5

gasoline: haircut: lunch: 7 1

1 2

1x  15 x2 or 21 145 x2 or 25 x

2

x  5 x  5 x  7  13 2 x 5

 7  13

2 x 5 5 2 x 2 5

 20

1 2  52 (20)

x  50 Kristen originally had $50. 80. Stem 3 4 5 6 81. Stem 1 2 3 4

0 5 1 2

Leaf 2 3 7 7 8 9 3 5 6 7 8 6|2  62

0 0 0 3

Leaf 4 5 8 8 8 0 1 1 2 4 4 3|4  34

Practice Quiz 2

1. The polynomial 5x4 has only one term, whose degree is 4. Thus, the degree of the polynomial is 4. 2. The polynomial 9n3p4 has only one term, whose degree is 7. Thus, the degree of the polynomial is 7. 3. The polynomial 7a2  2ab2 has two terms, 7a2 and 2ab2, whose degrees are 2 and 3, respectively. Thus, the degree of 7a2  2ab2 is 3, the greater of 2 and 3. 4. The polynomial 6  8x2y2  5y3 has three terms, 6, 8x2y2, and 5y3, whose degrees are 0, 4, and 3, respectively. Thus, the degree of 6  8x2y2  5y3 is 4, the greatest of 0, 4, and 3.

Page 451 Algebra Activity (Preview of Lesson 8-7) 1.

82. (a)(a)  a2

x2

83. 2x(3x2 )  (2  3)(x  x2 )  6x3 2 84. 3y (8y2 )  (3  8) (y2  y2 )  24y4 85. 4y(3y)  4y(6)  (4  3)(y  y)  (4  6)y  12y2  24y 86. 5n(2n2 )  (5n) (8n)  (5n)(4)  (5  2)(n  n2 )  (5  8)(n  n)  (5  4)n  10n3  40n2  20n 87. 3p2 (6p2 )  3p2 (8p)  3p2 (12)  (3  6)(p2  p2 )  (3  8)(p2  p)  (3  12)p2  18p4  24p3  36p2

Chapter 8

x

x3

1

1

x

x2

x x

1 1 1

x x x

1 1 1

1 1 1

(x  2)(x  3)  x2  5x  6

386

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2.

5.

x2

x

x2

x

x  3 1

x x

1 1 1

1 1

x

1 1

x

x1 x 1

x

x2

x x

x

x2

x x

1 1 1

x

1

1

x

1

1

x

1

1

2x  3

(x  2)(2x  3)  2x2  7x  6

(x  1)(x  3)  x2  4x  3 6.

3.

x3

x x1 x 1

x x2

1 1

x2 x x

x 2x  4

1 1

4.

x1 x 1

2x  1

x2

x

x

x2

x

1

x

1

(x  1) (2x  1) 

2x2

1

1

x

x2

x x x

x

x2

x x x

1 1 1

x

1 1 1

x

1 1 1

x

1 1 1

1

x

1 1 1

(x  3)(2x  4)  2x2  2x  12 7. By the Distributive Property, (x  3)(x  4)  x(x  4)  3(x  4). The top row represents x(x  4) or x2  4x. The bottom row represents 3(x  4) or 3x  12.

(x  1)(x  2)  x2  x  2

x

1

8-7 Page 455

Polynomials Check for Understanding x3

1.

 3x  1

x

2

x x x

x

2

x x x

2x  1

x

1 1 1

2a. (3x  4)(2x  5)  3x(2x  5)  4(2x  5)  6x2  15x  8x  20  6x2  7x  20 2b. (3x  4)(2x  5)  3x(2x)  3x(5)  4(2x)  4(5)  6x2  15x  8x  20  6x2  7x  20

387

Chapter 8

3x  4 ()2x  5 15x  20

2c.

6x2  6x2  2d.

8. (9p  1) (3p  2)  (9p) (3p)  (9p) (2)  (1) (3p)  (1) (2)  27p2  18p  3p  2  27p2  21p  2

8x

9. (2g  7)(5g  8)  (2g)(5g)  (2g)(8)  (7)(5g)  (7)(8)  10g2  16g  35g  56  10g2  19g  56 10. (3b  2c)(6b  5c)  (3b)(6b)  (3b)(5c)  (2c) (6b)  (2c)(5c)  18b2  15bc  12bc  10c2  18b2  3bc  10c2 11. (3k  5) (2k2  4k  3)  3k(2k2  4k  3)  5(2k2  4k  3)  6k3  12k2  9k  10k2  20k  15  6k3  2k2  29k  15

7x  20 3x  4

2x  5

x

2

x

2

x

2

x x x x

x

2

x

2

x

2

x x x x

x x x x x

x x x x x

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

x x x x x

1

12. A  2 bh 1

 2 (2x  3)(3x  1) 1

 2 [ (2x)(3x)  (2x)(1)  3(3x)  3(1) ] 1

 2 [6x2  2x  9x  3) ] 1

x

2

x

2

x

2

x

2

x

2

x

2

x x x x x x x

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

x x x x x x x x

 2 [6x2  7x  3]

x x x x x x x x



Pages 455–457

7

3

or 3x2  2 x  2

Practice and Apply

13. (b  8)(b  2)  (b)(b)  (b)(2)  (8) (b)  (8)(2)  b2  2b  8b  16  b2  10b  16 14. (n  6)(n  7)  (n)(n)  (n)(7)  (6)(n)  (6) (7)  n2  7n  6n  42  n2  13n  42 15. (x  4)(x  9)  (x)(x)  (x)(9)  (4)(x)  (4)(9)  x2  9x  4x  36  x2  13x  36 16. (a  3)(a  5)  (a)(a)  (a)(5)  (3)(a)  (3)(5)

3. See students’ work. 4. ( y  4) ( y  3)  ( y) ( y)  ( y)(3)  (4)( y)  (4)(3)  y2  3y  4y  12  y2  7y  12 5. (x  2) (x  6)  (x) (x)  (x)(6)  (2)(x)  (2)(6)  x2  6x  2x  12  x2  4x  12 6. (a  8) (a  5)  (a) (a)  (a)(5)  (8)(a)  (8)(5)  a2  5a  8a  40  a2  3a  40 7. (4h  5) (h  7)  (4h) (h)  (4h)(7)  (5)(h)  5(7)  4h2  28h  5h  35  4h2  33h  35

Chapter 8

6x2  7x  3 2

 a2  5a  3a  15  a2  8a  15 17. (y  4)(y  8)  (y)(y)  (y)(8)  (4)(y)  (4)(8)  y2  8y  4y  32  y2  4y  32 18. (p  2)(p  10)  (p)(p)  (p)(10)  (2)(p)  (2)(10)  p2  10p  2p  20  p2  8p  20 19. (2w  5)(w  7)  (2w)(w)  (2w)(7)  (5)(w)  (5)(7)  2w2  14w  5w  35  2w2  9w  35

388

20. (k  12) (3k  2)  (k)(3k)  (k)(2)  (12)(3k)  (12)(2)  3k2  2k  36k  24  3k2  34k  24 21. (8d  3) (5d  2)  (8d)(5d)  (8d)(2)  (3)(5d)  (3)(2)  40d2  16d  15d  6  40d2  31d  6 22. (4g  3) (9g  6)  (4g)(9g)  (4g)(6)  (3)(9g)  (3)(6)  36g2  24g  27g  18  36g2  51g  18 23. (7x  4) (5x  1)  (7x)(5x)  (7x)(1)  (4)(5x)  (4)(1)  35x2  7x  20x  4  35x2  27x  4 24. (6a  5)(3a  8)  (6a)(3a)  (6a)(8)  (5) (3a)  (5)(8)  18a2  48a  15a  40  18a2  63a  40 25. (2n  3)(2n  3)  (2n)(2n)  (2n)(3)  (3)(2n)  (3)(3)  4n2  6n  6n  9  4n2  12n  9 26. (5m  6) (5m  6)

33. (2x  5)(3x2  4x  1)  2x(3x2  4x  1)  5(3x2  4x  1)  6x3  8x2  2x  15x2  20x  5  6x3  23x2  22x  5 34. (3k  4) (7k2  2k  9)  3k(7k2  2k  9)  4(7k2  2k  9)  21k3  6k2  27k  28k2  8k  36  21k3  34k2  19k  36 35. 1n2  3n  221n2  5n  42  n2 (n2  5n  4)  3n(n2  5n  4)  21n2  5n  42  n4  5n3  4n2  3n3  15n2  12n  2n2  10n  8  n4  2n3  17n2  22n  8 36. (y2  7y  1)(y2  6y  5)  y2 (y2  6y  5)  7y(y2  6y  5)  1(y2  6y  5)  y4  6y3  5y2  7y3  42y2  35y  y2  6y  5  y4  y3  38y2  41y  5 37. (4a2  3a  7) (2a2  a  8)  4a2 (2a2  a  8)  3a(2a2  a  8)  7(2a2  a  8)  8a4  4a3  32a2  6a3  3a2  24a  14a2  7a  56  8a4  2a3  15a2  31a  56 38. 16x2  5x  2213x2  2x  42  6x2 (3x2  2x  4)  5x(3x2  2x  4)  2(3x2  2x  4)  18x4  12x3  24x2  15x3  10x2  20x  6x2  4x  8  18x4  3x3  20x2  16x  8 39. A  /w  1x  4212x  52  1x212x2  1x2152  412x2  4152  2x2  5x  8x  20  2x2  3x  20 The area is 2x2  3x  20 square units.

 (5m) (5m)  (5m) (6)  (6) (5m)  (6) (6)  25m2  30m  30m  3b  25m2  60m  36

27. (10r  4) (10r  4)  (10r)(10r)  (10r)(4)  (4)(10r)  (4)(4)  100r2  40r  40r  16  100r2  16 28. (7t  5) (7t  5)  (7t)(7t)  (7t)(5)  (5) (7t)  (5)(5)  49t2  35t  35t  25  49t2  25 29. (8x  2y)(5x  4y)  (8x)(5x)  (8x)(4y)  (2y)(5x)  (2y)(4y)  40x2  32xy  10xy  8y2  40x2  22xy  8y2 30. (11a  6b) (2a  3b)

1

40. A  2bh

 2 14x  3213x  22 1

 2 3 14x213x2  14x2122  13213x2  132122 4

 (11a) (2a)  (11a) (3b)  (6b) (2a)  (6b) (3b)

1

 22a2  33ab  12ab  18b2  22a2  21ab  18b2

 2 312x2  8x  9x  64 1

31. ( p  4)( p2  2p  7)  p( p2  2p  7)  4(p2  2p  7)

1

 2 [12x2  17x  6]  6x2 

 p3  2p2 7p  4p2  8p  28

 p3  6p2  p  28

17 x 2

The area is

3

6x2



17 x 2

 3 square units.

2

32. (a  3)(a  8a  5)  a(a2  8a  5)  3(a2  8a  5)  a3  8a2  5a  3a2  24a  15  a311a2  29a  15

389

Chapter 8

50. A  /w  (x  2)(x  4)  1x21x2  1x2142  1221x2  122142  x2  4x  2x  8  x2  2x  8 The area of the office is x2  2x  8 square feet. 51. Find the area of the square office. A  s2  92 or 81 Find the area of the new office. A  x2  2x  8  (9) 2  2(9)  8  81  18  8  91 The area of the new office is 91  81 or 10 square feet bigger. 52a. Sample answer: (30  5)(10  9)  13021102  30192  51102  5192  300  270  50  45  665 52b. Sample answer: (60  7)(100  2)  6011002  60122  711002  7122  6000  120  700  14  6834 52c. Sample answer:

1

41. A  2h(b1  b2 ) 1

 2 (5x  8) [ (x  7)  (2x  1) ] 1

 2 (5x  8) (3x  6)

 2 3 15x213x2  15x2162  182 13x2  182 162 4 1

 2 3 15x2  30x  24x  484 1

 2 3 15x2  6x  484 1



15 2 x 2

 3x  24

The area is

15 2 x 2

 3x  24 square units.

2

42. A  r  13x  42 2  13x  42 13x  42  3 13x2 13x2  13x2 142  14213x2  142 142 4   39x2  12x  12x  16 4  19x2  24x  162 The area is (9x2  24x  16) square units. 43. V  Bh  3 12a  22 1a  52 4 1a  12  3 12a21a2  12a2152  1221a2  122152 4 1a  12  32a2  10a  2a  104 1a  12  12a2  8a  1021a  12  a12a2  8a  102  112a2  8a  102  2a3  8a2  10a  2a2  8a  10  2a3  10a2  2a  10 The volume is 2a3  10a2  2a  10 cubic units. 44. V  Bh  3 13y2 13y212y2 162 4 17y  32  19y2  12y217y  32  19y2 2 17y2  19y2 2132  112y2 17y2  112y2132  63y3  27y2  84y2  36y  63y3  57y2  36y The volume is 63y3  57y2  36y cubic units. 45. Since a is the first integer, the second and third integers are a  1 and a  2. a(a  1) (a  2)  a [ (a) (a)  (a)(2)  (1) (a)  (1) (2) ]  a [ a2  2a  a  2]  a [a2  3a  2]  a3  3a2  2a 46. Sample answer: a  1; 1(2)(3)  6 47. Sample answer: 6; the result is the same as the product in Exercise 46. a3  3a2  2a  13  3(1) 2  2(1) 132 6 48. A  /w  (5y  6) (2y  10)  (5y) (2y)  (5y)(10)  (6)(2y)  (6)(10)  10y2  50y  12y  60  10y2  38y  60 The area is 10y2  38y  60 square feet. 49. Let x represent the length of a side of the square office. Then the new office has dimensions x  2 and x  4.

Chapter 8

18  12 216  34 2 3 1 1 3  8(6)  8 1 4 2  2 (6)  2 1 4 2 3

 48  6  3  8 3

 578 52d. Sample answer:

112  35 2110  23 2

 121102  12

123 2  35 1102  35 123 2 2

 120  8  6  5 2

 1345 53. Outer length: w  10 outer width: w  6 Area of concrete  total area  area of pool.  (w  10)(w  6)  w(w  4)  1w21w2  1w2162  11021w2  1102162  w(w)  w(4)  w2  6w  10w  60  w2  4w  12w  60 The area of the concrete is 300 square feet. Find w. 12w  60  300 12w  240 w  20 The pool should be 20 ft by 24 ft. 54. Sometimes; the product of x  1 and x2  2x  3 is x3  3x2  5x  3, which has 4 terms; the product of y  1 and x3  2x2  3x is x3y  2x2y  3xy  x3  2x2  3x, which has 6 terms.

390

f(x)  2x  5 f(4)  2(4)  5  8  5  13 67. g1x2  x2  3x g122  7  3 122 2  3122 4  7  14  62  7 5 v at 69. 68. f(x)  2x  5 v f(a  3)  2(a  3)  5 (t)a  (t) t  2a  6  5 at  v  2a  1 at v a a

55. Multiplying binomials and two-digit numbers each involve the use of the Distributive Property twice. Each procedure involves four multiplications and the addition of like terms. Answers should include the following. • 24  36  (4  20) (6  30)  (4  20)6  (4  20)30  (24  120)  (120  600)  144  720  864 • The like terms in vertical two-digit multiplication are digits with the same place value. 56. C; (x  2)(x  4)  (x  4)(x  2)

66.

v

ta

 [ (x) (x)  (x) (4)  (2) (x)  (2) (4) ]  [ (x) (x) 

(x) (2)  (4) (x)  (4)(2) ]  [x2  4x  2x  8][x2  2x  4x  8]  [ x2  2x  8][x2  2x  8 ]  [x2  2x  8]  [x2  2x  8]  4x

70.

by b



y

57. B; (x  y)(x2  xy  y2 )  x(x2  xy  y2 )  y(x2  xy  y2 )  x3  x2y  xy2  x2y  xy2  y3  x3  y3

Page 457

ax  by  2cz ax  by  ax  2cz  ax by  2cz  ax

y 71.

4x  3y  7 4x  3y  4x  7  4x 3y  4x  7 3y 3

Maintain Your Skills



y

58. 3d(4d2  8d  15)  (3d)(4d2 )  (3d)(8d)  (3d)(15)  12d3  24d2  45d

72.

2cz  ax b 2cz  ax b ax  2cz b

(6a) 2

4x  7 3 4 7 x 3 3

62 

 a2 2  36a

73. (7x) 2  72  x2  49x2

74. (9b) 2  92  b2  81b2

59. 4y(7y2  4y  3)  (4y)(7y2 )  (4y)(4y)  (4y)(3)  28y3  16y2  12y

75. (4y2 ) 2  42 (y2 ) 2  16y4

76. 12v3 2 2  22 1v3 2 2  4v6

2 (5m2

 7m  8) 60. 2m  (2m2 )(5m2 )  (2m2 )(7m)  (2m2 )(8)  10m4  14m3  16m2 61. 3x(2x  4)  6(5x2  2x  7)  6x2  12x  30x2  12x  42  36x2  42 62. 4a(5a2  2a  7) 3(2a2  6a  9)  20a3  8a2  28a  6a2  18a  27  20a3  2a2  10a  27 63. The measure of the third angle is 180º minus the measures of the other two angles. 180  (2x  1)  (5x  2)  180  (2x  1)  (5x  2)  181  7x The measure of the third angle is (181  7x)º. 64. 1st angle  2x  1  2(15)  1 or 31 2nd angle  5x  2  5(15)  2 or 73 3rd angle  181  7x  181  7(15) or 76 The angles measure 31º, 73º, and 76º. 65. The lines intersect at the point (6, 3). Thus, the system has one solution, (6, 3).

8-8 Page 461

77. 13g4 2 2  32 1g4 2 2  9g8

Special Products Check for Understanding

1. The patterns are the same except for their middle terms. The middle terms have different signs. 2. The square of a difference is (a  b)2, which equals a2  2ab  b2. The difference of squares is the product of a  b and a  b or a2  b2. 3. x3

x x3

2

x x x

x x x 1 1 1

1 1 1

1 1 1

4. Sample answer: x  1 and x  1 (x  1)(x  1)  x2  12  x2  1

391

Chapter 8

5. 1a  62 2  a2  21a2 162  62  a2  12a  36 6. 14n  3214n  32  14n  32 2  (4n) 2  2(4n)(3)  32  16n2  24n  9 7. (8x  5) (8x  5)  (8x) 2  52  64x2  25 8. (3a  7b) (3a  7b)  (3a) 2  (7b) 2  9a2  49b2 2 2 2 2 9. (x  6y)  (x )  2(x2 )(6y)  (6y) 2  x4  12x2y  36y2 10. (9  p) 2  92  2(9)(p)  p2  81  18p  p2 11. The genetic makeup of the purebred golden can be modeled by 1.0 G. The genetic makeup of the purebred cinnamon can be modeled by 1.0 g. Their offspring can be modeled by the product of 1.0 G and 1.0 g or 1.0 Gg. 12. 0%; All pups will be golden since only Gg combinations are possible and the golden gene G is dominant.

27. (2x  9y) 2  (2x) 2  2(2x)(9y)  (9y) 2  4x2  36xy  81y2 2 28. 13n  10p2  13n2 2  213n2110p2  110p2 2  9n2  60np  100p2 29. 15w  14215w  142  15w2 2  142  25w2  196 30. (4d  13)(4d  13)  (4d) 2  132  16d2  169 3 2 3 2 31. 1x  4y2  1x 2  21x3 214y2  14y2 2  x6  8x3y  16y2 2 2 2 32. 13a  b 2  13a2 2 2  213a2 21b2 2  1b2 2 2  9a4  6a2b2  b4 2 3 33. 18a  9b 218a2  9b3 2  18a2 2 2  19b3 2 2  64a4  81b6 4 4 34. (5x  y)(5x  y)  (5x4 ) 2  y2  25x8  y2 35.

123x  622  123x22  2123x2 162  62 4

 9x2  8x  36

36.

145x  1022  145x22  2145x2 1102  102 16

 25x2  16x  100

Page 462

37. (2n  1)(2n  1)(n  5)  [ (2n) 2  12 ] (n  5)  (4n2  1)(n  5)  (4n2 ) (n)  (4n2 )(5)  (1) (n)  (1)(5)  4n3  20n2  n  5 38. ( p  3)( p  4)( p  3)( p  4)  [ ( p  3)( p  3) ] [ ( p  4) ( p  4) ]  [p2  32 ] [p2  42 ]  [p2  9] [ p2  16]  (p2 )(p2 )  (p2 )(16)  (9)(p2 )  (9) (16)  p4  16p2  9p2  144  p4  25p2  144 39. Since Pam’s genes are Bb, her makeup can be modeled by 0.5 B  0.5 b. Since Bob has blue eyes, his genes are bb and can be modeled by 1.0 b. Their children’s genes can be modeled by the product of 0.5 B  0.5 b and 1.0 b.

Practice and Apply

13. 1y  42 2  y2  21y2142  42  y2  8y  16 14. 1k  821k  82  1k  82 2  k2  21k2 182  82  k2  16k  64 15. 1a  52 1a  52  1a  52 2  a2  21a2 152  52  a2  10a  25 2 2 16. 1n  122  n  21n2 1122  122  n2  24n  144 17. 1b  721b  72  b2  72  b2  49 18. (c  2) (c  2)  c2  22  c2  4 2 19. 12g  52  12g2 2  212g2152  52  4g2  20g  25 20. 19x  32 2  19x2 2  219x2 132  32  81x2  54x  9

(0.5 B  0.5 b) (1.0 b)  (0.5 B) (1.0 b)  (0.5 b) (1.0 b)  0.5 Bb  0.5 bb

40. Since only bb will be blue eyes, we see that 0.5 or 1 of their children should have blue eyes. The 2 1 probability is 2. 41. Sample answer: We pick 2. Square the number: 22  4 Add twice the original: 4  2(2)  8 Add 1: 8  1  9 Take square root: 19  3 Subtract original number: 3  2  1 Yes, the result is 1. 42. Square the number: a2 Add twice the original: a2  2a Add 1: a2  2a  1

21. 17  4y2 2  72  2172 14y2  14y2 2  49  56y  16y2 22. 14  6h2 2  42  214216h2  16h2 2  16  48h  36h2 23. (11r  8) (11r  8)  (11r) 2  (8) 2  121r2  64 24. (12p  3) (12p  3)  (12p) 2  32  144p2  9 25. 1a  5b2 2  a2  21a2 15b2  15b2 2  a2  10ab  25b2 2 26. 1m  7n2  m2  21m2 17n2  17n2 2  m2  14mn  49n2

Chapter 8

392

• Sample answer: (10x  11)(10x  11) (10x  11)(10x  11)  (10x  11) 2  (10x) 2  2(10x) (11)  112  100x2  220x  121 49. C; (a  b) 2  a2  2ab  b2  (a2  b2 )  2(ab)  40  2(12)  40  24  16 50. B; Since (x  y)(x  y)  x2  y2, we have x2  y2  (x  y)(x  y)  (10)(20)  200. 51a. (a  b)(a  b) (a  b)  (a  b)(a2  2ab  b2 )  b(a2  2ab  b2 )  a(a2  2ab  b2 )  b(a2  2ab2  b2 )  a3  2a2b  ab2  ba2  2ab2  b3  a3  3a2b  3ab2  b3 51b. (a  b) 3  a3  3a2b  3ab2  b3 (x  2) 3  x3  3x2 (2)  3x(2) 2  23  x3  6x2  12x  8 51c.

43. (a  1) 2  a2  2(a)(1)  12  a2  2a  1 2  2a  1 is the square of a  1. Thus, a 44. Take square root: 21a  12 2  a  1 Subtract original number: (a  1)  a  1 The result is 1. 45. Since each seating level is about 1 meter wide, the second and third levels have radii s  2 and s  3 meters, respectively. 46. A  (s  3) 2  (s  2) 2  1s2  2132 1s2  32 2  1s2  21221s2  22 2  (s2  6s  9)  (s2  4s  4)  1s2  6s  9  s2  4s  42  12s  52  6.3s  15.7 The area is about 6.3s  15.7 square meters. 1

47. Area of a trapezoid  2 (height) (base 1  base 2) A1  A2 

1 1a 2 1 1a 2

 b2 1a  b2  b2 1a  b2 a

a–b a

a

b b

b b a

a–b

3 12 1a  b2 1a  b2 4  3 12 1a  b21a  b2 4

a

b

Total area of shaded region 

 1a  b2 1a  b2 a

ab

Page 463

b

ab

ab

b

a ab

 20y2  29y  6 55. (3n  5)(8n  5)  3n(8n)  3n(5)  (5)(8n)  (5) (5)  24n2  15n  40n  25  24n2  25n  25

Area of rectangle  (a  b)(a  b) a

A1

a–b

a b

Maintain Your Skills

52. 1x  221x  72  1x21x2  1x2172  1221x2  122172  x2  7x  2x  14  x2  9x  14 53. 1c  921c  32  1c21c2  1c2132  1921c2  192132  c2  3c  9c  27  c2  6c  27 54. (4y  1)(5y  6)  (4y)(5y)  (4y)(6)  (1)(5y)  (1) (6)  20y2  24y  5y  6

A2

56. (x  2)(3x2  5x  4)  x(3x2  5x  4)  2(3x2  5x  4)  3x3  5x2  4x  6x2  10x  8  3x3  11x2  14x  8

b a–b

48. The product of two binomials is also a binomial when the two binomials are the sum and the difference of the same two terms. Answers should include the following. • Sample answer: 12x  132 12x  132 12x  13212x  132  12x2 2  132  4x2  169

57. (2k  5) (2k2  8k  7)  2k(2k2  8k  7)  5(2k2  8k  7)  4k3  16k2  14k  10k2  40k  35  4k3  6k2  26k  35

393

Chapter 8

58. 6(x  2)  4  5(3x  4) 6x  12  4  15x  20 6x  16  15x  20 9x  16  20 9x  36 x4 59. 3(3a  8)  2a  4(2a  1) 9a  24  2a  8a  4 7a  24  8a  4 15a  24  4 15a  20 20 4 a  15 or 3

Find y. 2x  3y  4 2152  3y  4 10  3y  4 3y  6 y  2 The solution is (5, 2). 65. Solve for y. 5x  5y  35 5y  5x  35 y  x  7 The slope of the given line is 1. The slope of any line perpendicular to it is 1. y  y1  m1x  x1 2

60. p( p  2)  3p  p( p  3) p2  2p  3p  p2  3p p2  5p  p2  3p 5p  3p 8p  0 p0 61. y(y  4)  2y  y(y  12)  7 y2  4y  2y  y2  12y  7 y2  2y  y2  12y  7 2y  12y  7 14y  7 y

7 14

or

y  2  11x  1322 y2x3 yx5 66. Solve for y. 2x  5y  3 5y  2x  3 2

2

The slope of the given line is 5. The slope of any 5 line perpendicular to it is 2.

1 2

y  y1  m1x  x1 2 5 y  7  2 (x  (2))

62. Add the equations. 3 x 4 3 x 4 6 x 4



1 y 5 1 y 5

3

y  5x  5

5

5

 5

y  7  2 (x  2)

0

y  7  2 x  5

x0 Find y.

y  2 x  2



3 x 4

3 102 4

 

1 y 5 1 y 5 1 y 5

5 5

67. Solve for y. 5x  y  2 y  5x  2 The slope of the given line is 5. The slope of any 1 line perpendicular to it is 5. The y-intercept is 6. y  mx  b

5 5 5

y  25 The solution is (0, 25). 63. Multiply the first equation by 3. Then add. 6x  3y  30 5x  3y  3 11x  33 x3 Find y. 2x  y  10 2132  y  10 6  y  10 y  4 y  4 The solution is (3, 4). 64. Rewrite the equations. 2x  3y  4 x  3y  11 Multiply the second equation by 1. Then add. 2x  3y  4 x  3y  11 3x  15 x5 Chapter 8

1

y  5x  6

68. a  a  1n  12d n 1 a18  3  118  12142  3  17142  3  68  71 69. a  a  (n  1)d n 1 a  5  112  12162 12  5  1112162  5  66  61 70. b

394

Chapter 8 Study Guide and Review Page 464 1. 3. 5. 7. 9.

25.

16a3b2x4y 48a4bxy3

2. 4. 6. 8. 10.

26.

5 8

(a) b a5b2

 

Pages 464–468 3

11. y  y  y  y  y7 12. (3ab) (4a2b3 )  [ 3  (4) ] (a  a2 )(b  b3 )  12a3b4 2 3 4 13. (4a x)(5a x )  [ (4)  (5) ] (a2  a3 )(x  x4 )  20a5x5 2 3 3 2 3 14. (4a b)  4 (a ) b3  64a6b3 2 15. 13xy2 14x2 3  132 2x2y2 142 3x 3  9  x2  y2  64  x3  (9  64) (x2  x3 )(y2 )  576x5y2 2 4 16. 12c d2 13c2 2 3  122 4 1c2 2 4d4 132 3 1c2 2 3  16c8d4 1272c6  [16  (27) ] (c8  c6 )(d4 )  432c14d4

27.

13bc 4d 2

2 3

28.

(3bc ) (4d) 3



(3) 3b3 (c2 ) 3 (4) 3 (d) 3

36.



23.

27b2 14b3

z3 x2

 

27b3c6 64d3 27 b  2 14 b3 27 2132 1b 2 14

1 21 2

 14 1b2 27

 24.

(3a3bc2 ) 2 18a2b3c4

      

42 (a1 ) 2 22 (a4 ) 2



42a2 4a8

144 21aa 2 2

2 6

141 21a1 2 3

6

1 64a6

135x5xy y 20  1 2

29. 2.4  105  240,000

2 6

8.4  106 1.4  109



10 18.4 1.4 21 10 2 6 9

 6  10 6192  6  103 or 6000 37. (3  102 )(5.6  10 8 )  (3  5.6)(102  10 8 )  16.8  10 6  1.68  10  10 6  1.68  10 5 or 0.0000168 38. The polynomial n  2p2 has two terms, n and 2p2, whose degrees are 1 and 2, respectively. Thus, the degree of n  2p2 is 2, the greater of 1 and 2. 39. The polynomial 29n2  17n2t2 has two terms, 29n2 and 17n2t2, whose degrees are 2 and 4, respectively. Thus, the degree of 29n2  17n2t2 is 4, the greater of 2 and 4.

 1 z3 22. x 2y0z3  x2  1  1



3.14  10 4  0.000314 4.88  109  4,880,000,000 0.00000187  1.87  10 6 796  103  7.96  102  103  7.96  105 34. 0.0343  102  3.43  102  102  3.43  10 4 5 35. 12  10 213  106 2  12  321105  106 2  6  1011 or 600,000,000,000 30. 31. 32. 33.

2 3



(4a1 ) 2 (2a4 ) 2



18. 15a2 2 3  71a6 2  152 3 1a2 2 3  7a6  125a6  7a6  132a6 2 2 3 4 3 19. 3 13 2 4  33 4  312 or 531,441 21.

1 21 21 21 2

1 21 2



1

1



bx3 3ay2

  4 3a 6

1

 6a

3

(4 21 )(a28 )

 2m4n8

13y2 0 6a

4

a5b8 a5b2 a5 b8  a5 b2



17. 2 1m2n4 2 2  2 1m2 2 2 1n4 2 2

20.

2

4

 (a55 )(b82 )  (a0 )(b6 )  (1) (b6 )  b6

Lesson-by-Lesson Review 331

1

3





Power of a Power monomial zero exponent FOIL method Product of Powers

a b x y 116 48 21 a 21 b 21 x 21 y 2

1 34 1a 21b21 21x41 21y13 2 3 1 1 1 3 2 a b x y 3 1 1 b x3 1 3 a 1 1 y2



Vocabulary and Concept Check

negative exponent Quotient of Powers trinomial polynomial binomial

3



27b 14

32 (a3 ) 2b2 (c2 ) 2 18a2b3c4 9a6b2c4 18a2b3c4 9 a6 b2 c4 18 a2 b3 c4 1 (a62 )(b23 )(c44 ) 2 1 4 1 0 a b c 2 1 a4 1 (1) 2 1 b a4 2b

1 21 21 21 2 1 21 2

395

Chapter 8

55. 2x (x  y2  5)  5y2 (3x  2)  2x(x)  2x(y2 )  2x(5)  5y2 (3x)  5y2 (2)  2x2  2xy2  10x  15xy2  10y2  2x2  17xy2  10x  10y2 56. m(2m  5)  m  2m(m  6)  16 2m2  5m  m  2m2  12m  16 2m2  4m  2m2  12m  16 4m  12m  16 8m  16 m2 57. 213w  w2 2  6  2w1w  42  10 6w  2w2  6  2w2  8w  10 6w  6  8w  10 14w  6  10 14w  16

40. The polynomial 4xy  9x3z2  17rs3 has three terms, 4xy, 9x3z2, and 17rs3, whose degrees are 2, 5, and 4, respectively. Thus, the degree of 4xy  9x3z2  17rs3 is 5, the greatest of 2, 5, and 4. 41. The polynomial 6x5y  2y4  4  8y2 has four terms, 6x5y, 2y4, 4, and 8y2, whose degrees are 6, 4, 0, and 2, respectively. Thus, the degree of 6x5y  2y4  4  8y2 is 6, the greatest of 6, 4, 0, and 2. 42. The polynomial 3ab3  5a2b2  4ab has three terms, 3ab3, 5a2b2, and 4ab, whose degrees are 4, 4, and 2, respectively. Thus, the degree of 3ab3  5a2b2  4ab is 4, the greatest of 4, 4, and 2. 43. The polynomial 19m3n4  21m5n has two terms, 19m3n4 and 21m5n, whose degrees are 7 and 6, respectively. Thus, the degree of 19m3n4  21m5n is 7, the greater of 7 and 6. 44. 3x4  x  x2  5  3x4  x1  x2  5x0  3x4  x2  x  5 2 3 45. 2x y  27  4x4  xy  5x3y2  2x2y3  27x0  4x4  x1y  5x3y2  4x4  5x3y2  2x2y3  xy  27 46. (2x2  5x  7)  (3x3  x2  2)  12x2  5x  72  13x3  x2  22  3x3  12x2  x2 2  5x  17  22  3x3  x2  5x  5 2 47. (x  6xy  7y2 )  (3x2  xy  y2 )  1x2  3x2 2  16xy  xy2  17y2  y2 2  4x2  5xy  6y2 48. (7z2  4)  (3z2  2z  6)  17z2  42  13z2  2z  62  (7z2  3z2 )  2z  (4  6)  4z2  2z  10 49. (13m4  7m  10)  (8m4  3m  9)  113m4  8m4 2  17m  3m2  110  92  21m4  10m  1 50. (11m2n2  4mn  6)  (5m2n2  6mn  17)  (11m2n2  5m2n2 )  (4mn  6mn)  (6  17)  16m2n2  10mn  11 51. 15p2  3p  492  12p2  5p  242  15p2  3p  492  12p2  5p  242  15p2  2p2 2  13p  5p2  149  242  7p2  2p  25 52. b(4b  1)  10b  b(4b)  b(1)  10b  4b2  b  10b  4b2  9b 53. x13x  52  71x2  2x  92  x13x2  x152  71x2 2  712x2  7192  3x2  5x  7x2  14x  63  10x2  19x  63 54. 8y(11y2  2y  13)  9(3y3  7y  2)  8y(11y2 )  8y(2y)  8y(13)  9(3y3 )  917y2  9122  88y3  16y2  104y  27y3  63y  18  61y3  16y2  167y  18

Chapter 8

16

8

1

w  14 or 7 or 17

58. 1r  321r  72  1r21r2  1r2172  1321r2  132172  r2  7r  3r  21  r2  4r  21 59. (4a  3)(a  4)  (4a)(a)  (4a)(4)  (3) (a)  (3)(4)  4a2  16a  3a  12  4a2  13a  12 60. (3x  0.25) (6x  0.5)  (3x) (6x)  (3x) (0.5)  0.25(6x)  (0.25) (0.5)  18x2  1.5x 1.5x  0.125  18x2  0.125

61. (5r  7s)(4r  3s)  (5r)(4r)  (5r)(3s)  (7s)(4r)  (7s) (3s)  20r2  15rs  28rs  21s2  20r2  13rs  21s2 62. (2k  1) (k2  7k  9)  2k(k2  7k  9)  1(k2  7k  9)  2k3  14k2  18k  k2  7k  9  2k3  15k2  11k  9 63. (4p  3)(3p2  p  2)  4p(3p2  p  2)  3(3p2  p  2)  12p3  4p2  8p  9p2  3p  6  12p3  13p2  11p  6 64. 1x  621x  62  x2  62  x2  36 2 65. 14x  72  14x2 2  214x2172  72  16x2  56x  49 2 66. 18x  52  18x2 2  218x2152  52  64x2  80x  25 67. 15x  3y215x  3y2  15x2 2  13y2 2  25x2  9y2 2 2 68. (6a  5b)  (6a)  2(6a) (5b)  (5b) 2  36a2  60ab  25b2 2 69. (3m  4n)  (3m) 2  2(3m) (4n)  (4n) 2  9m2  24mn  16n2

396

16. 13  103 212  104 2  13  221103  104 2  6  107 or 60,000,000

Chapter 8 Practice Test Page 469 1. (42 ) (43 )  (4  4)(4  4  4)  44444  45 and 165  45. 2.

1 5

17.

 51

5. (12abc)(4a2b4 )  (12  4)(a  a2 )(b  b4 )c  48a3b5c

1

135m22  135 22m2

7. 13a2 4 1a5b2 2  132 4 1a2 4 1a5 2 2 1b2 2  81a4  a10  b2  81(a4  a10 )b2  81a14b2

1 21 2 1m1 21n1 2 2

2

n2

 m2 9a2bc2 63a4bc

1639 21aa 21bb 21cc 2 1  1 7 2 (a24 )(b11 )(c21 ) 2



11.

48a2bc5 (3ab3c2 ) 2

117 21a1 2 (1) 11c 2 2

     

9 5

23. (x3  3x2y  4xy2  y3 )  (7x3  x2y  9xy2  y3 )  (x3  3x2y  4xy2  y3 )  (7x3  x2y  9xy2  y3 )  (x3  7x3 )  (3x2y  x2y)  (4xy2  9xy2 ) (y3  y3 )  6x3  4x2y  13xy2 24. To find the measure of the third side, subtract the measures of the two sides given from the perimeter. (11x2  29x  10)  (x2  7x  9)  (5x2  13x  24)  (11x2  29x  10)  (x2  7x  9)  (5x2  13x  24)  (11x2  x2  5x2 )  (29x  7x  13x)  (10  9  24)  5x2  23x  23 25. 1h  52 2  h2  21h2152  52  h2  10h  25

c 7a2



10 12.85 1.86 21 10 2

22. 15a  3a2  7a3 2  12a  8a2  42  17a3 2  13a2  8a2 2  15a  2a2  4  7a3  5a2  7a  4

1





5xy  7  2y4  x2y3  5xy1  7y0  2y4  x2y3  2y4  x2y3  5xy  7

2

4

 7 a 2b0c 

)

2y2  8y4  9y  2y2  8y4  9y1  8y4  2y2  9y 21. The polynomial 5xy  7  2y4  x2y3 has four terms, 5xy, 7, 2y4, and x2y3, whose degrees are 2, 0, 4, and 5, respectively. Thus, the degree of 5xy  7  2y4  x2y3 is 5, the greatest of 2, 0, 4, and 5.

m n4 m3 n2

 (m13 )(n42 )  m 2n2

10.

3

4(3)

 (1.53)(109 5 )  1.53  104 It will take about 1.53  104 seconds or 4.25 hours. 20. The polynomial 2y2  8y4  9y has three terms, 2y2, 8y4, and 9y, whose degrees are 2, 4, and 1, respectively. Thus, the degree of 2y2  8y4  9y is 4, the greatest of 2, 4, and 1.

8. (5a2 ) (6b3 ) 2  (5)(a2 )(6) 2 (b3 ) 2  (5)(a2 )(36)(b6 )  (5  36) (a2 )(b6 )  180a2b6



2

2.85  109 1.86  105

9



4

19. To find the time, divide the distance by the rate d tr .

 25m2

9.

10 114.72 3.2 21 10 2

18. 115  10 7 213.1  104 2  115  3.12110 7  104 2  46.5  10 3  4.65  10  10 3  4.65  10 2 or 0.0465

4. (a2b4 ) (a3b5 )  (a2  a3 )(b4  b5 )  a5b9

mn4 m3n2



 (4.6)(10  4.6  10 1 or 0.46

3. Sample answer: A monomial is a number, variable, or product of numbers and variables; 6x2.

6.

14.72  10  4 3.2  10  3

48a2bc5 (3) 2 (a) 2 (b3 ) 2 (c2 ) 2 48a2bc5 9a2b6c4 48 a2 b c5 9 a2 b6 c4 16 22 16 54 a b c 3 16 0 5 a b c 3 16 1 c (1) b5 1 3 16c 3b5

1 21 21 21 2 1 2 1 21 2

12. 46,300  4.63  104 13. 0.003892  3.892  10 3 14. 284  103  2.84  102  103  2.84  105 9 15. 52.8  10  5.28  10  10 9  5.28  10 8

397

Chapter 8

26. (4x  y) (4x  y)  (4x) 2  y2  16x2  y2

6. A; The cost of 4 sets of table and chairs plus the cost of b bookcases must be less than or equal to $7500. 415502  125b  7500 7. B; Let s  amount Sophia spent, and a  amount Allie spent. Write a system of equations. s  a  122 s  2a  25 Solve the system by substitution. 2a  25  a  122 3a  25  122 3a  147 a  49 Allie spent $49. 8. D; (2x3 )(4x4 )  (2  4)(x3  x4 )  8x7 9. B; 0.00037  3.7  104 10. A; 13x2  4x  52  1x2  2x  12  13x2  4x  52  1x2  2x  12  13x2  1x2 22  14x  2x2  15  1122  2x2  2x  4 11. a  a  (n  1)d n 1 a15  20  (15  1)(9)  20  (14)(9)  20  126  106 12. Each y-value is 4 times the corresponding x-value. f (x)  4x or y  4x 13. To find the y-intercept, let x  0. 3x  2y  8  0 3(0)  2y  8  0 2y  8  0 2y  8 y4 14. Graph 3x  y  2. Since the boundary is included in the solution set, draw a solid line. Test (0, 0). 3x  y  2 3(0)  0  2 0  2 true Shade the half-plane that contains (0, 0).

27. 3x2y3 (2x  xy2 )  3x2y3 (2x)  3x2y3 (xy2 )  6x3y3  3x3y5 28. 12a2b  b2 2 2  12a2b2 2  212a2b2 1b2 2  1b2 2 2  4a4b2  4a2b3  b4 29. (4m  3n)(2m  5n)

 (4m)(2m)  (4m) (5n)  (3n) (2m)  (3n) (5n)  8m2  20mn  6mn  15n2  8m2  14mn  15n2

30. (2c  5) (3c2  4c  2)  2c(3c2  4c  2)  5(3c2  4c  2)  6c3  8c2  4c  15c2  20c  10  6c3  7c2  16c  10 31. 2x (x  3)  2(x2  7)  2 2x2  6x  2x2  14  2 2x2  6x  2x2  12 6x  12 x2 32. 3a1a2  52  11  a13a2  42 3a3  15a  11  3a3  4a 15a  11  4a 11a  11  0 11a  11 a1 33. C; 31x  y2 2  31x2  2xy  y2 2  3182  24

Chapter 8 Standardized Test Practice Pages 470–471 1. A; Mean of first 5 games: 72 Median of first 5 games: 70 Mode of first 5 games: 70 Mean of 6 games: 65 Median of 6 games: 70 Mode of 6 games: 70 Mean changes the most. 2. D;

2100 60



5600 x

3. D; r 

20,750  20,542 4

2100x  336,000 x  160 4. A; From the graph, we see that the y-intercept is 1 1 and the slope is 5. Thus, the equation is 1 y  5x  1.

y

3x  y  2

5. B; Find the y-intercept. y  mx  b 4  2112  b 42b 2b Write the equation. y  mx  b y  2x  2

Chapter 8

O

x

15. P  Q  (3x2  2x  1)  (x2  2x  2)  (3x2  (x2 ))  (2x  2x)  (1  (2))  2x2  3

398

24a. A  /w  (3m  3)(m  1)  (3m)(m)  (3m)(1)  (3)(m)  (3) (1)  3m2  3m  3m  3  3m2  3 24b. A  /w  (3m  3)(m  4)  (3m)(m)  (3m)(4)  (3)(m)  (3) (4)  3m2  12m  3m  12  3m2  9m  12 24c. V  /wh  (3m  3)(m  1) (m  4)  [ (3m)(m)  (3m)(1)  (3)m  (3)(1) ] (m  4)  (3m2  3m  3m  3)(m  4)  (3m2  3)(m  4)

16. (x2  1)(x  3)  (x2 )(x)  (x2 ) (3)  (1)(x)  (1) (3)  x3  3x2  x  3 17. A; The x-coordinate of A is 6. The y-coordinate of B is 5. The quantity in column A is greater. 6(x  1) 3 18. A; 4x  10  20 8 4x  30 6(x  1)  24 x

30 4

The quantity in 19. B; x  3y  2 x  3y  0 2x 2 x1 The quantity in 20. C;

2b3c2 4bc

15 2

6x  6  24 6x  18 x  3 column A is greater. 3x  8y  6 x  8y  2 4x 8 x2 column B is greater.

or

124 21bb 21cc 2 1  1 2 2 (b31 )(c21 ) 

3

1  2 b4c

2

 (3m 2 ) (m )  (3m 2 ) (4)  (3) (m )  (3) (4)

 3m3  12m2  3m  12 24d. V  3m3  12m2  3m  12  3(2) 3  12(2) 2  3(2)  12  3(8)  12(4)  6  12  24  48  18  54 The volume is 54 cm3.

10b4 20b8c1

11020 21bb 21c1 2 1  1 2 2 (b48 )c



4 8

1

1  2 b4c

The two quantities are equal. 21. A; 5.01  102 50.1  104  0.0501  0.00501 The quantity in column A is greater. 22. C; The degree of The degree of x2  5  6x  13x3 10  y  2y2  4y3 is 3. is 3. The two quantities are equal. 23. B; (m  n) 2  m2  2mn  n2  (m2  n2 )  2(mn)  10  2(6)  10  12  2 (m  n) 2  m2  2mn  n2  (m2  n2 )  2(mn)  10  2(3)  10  6  16 The quantity in column B is greater.

399

Chapter 8

Chapter 9 Page 473

Factoring 3. Sample answer: 5x2 and 10x3 5x2  5  x  x 10x3  2  5  x  x  x GCF: 5  x  x or 5x2 4. List all pairs of numbers whose product is 8. 18 24 factors: 1, 2, 4, 8 composite 5. List all pairs of numbers whose product is 17. 1  17 factors: 1, 17 prime 6. List all pairs of numbers whose product is 112. 2  56 4  28 1  112 7  16 8  14 factors: 1, 2, 4, 7, 8, 14, 16, 28, 56, 112 composite 7. 45  3  15

Getting Started

1. 3(4  x)  3  4  3  x  12  3x 2. a(a  5)  a  a  a  5  a2  5a 2 3. 7(n  3n  1)  7(n2 )  (7)(3n)  (7)(1)  7n2  21n  7 4. 6y(3y  5y2  y3 )  6y(3y)  6y(5y2 )  6y(y3 )  18y2  30y3  6y4 5. (x  4) (x  7)  (x)(x)  (x)(7)  (4)(x)  (4)(7)  x2  7x  4x  28  x2  11x  28 6. (3n  4)(n  5)  (3n)(n)  (3n)(5)  (4)(n)  (4)(5)  3n2  15n  4n  20  3n2  11n  20 7. (6a  2b)(9a  b)  (6a)(9a)  (6a)(b)  (2b)(9a)  (2b)(b)  54a2  6ab  18ab  2b2  54a2  12ab  2b2 8. (x  8y)(2x  12y)

8.

 (x)(2x)  (x) (12y)  (8y) (2x)  (8y) (12y)  2x2  12xy  16xy  96y2  2x2  4xy  96y2

9.

9. (a  b) 2  a2  2ab  b2 (y  9) 2  y2  2(y) (9)  92  y2  18y  81 10. (a  b) 2  a2  2ab  b2 (3a  2) 2  (3a) 2  2(3a)(2)  22  9a2  12a  4 11. (a  b) (a  b)  a2  b2 (n  5) (n  5)  n2  52  n2  25 12. (a  b) (a  b)  a2  b2 (6p  7q) (6p  7q)  (6p) 2(7q) 2  36p2  49q2 13. 1121 is the positive square root of 121. 112  121 S 1121  11

10. 11. 12.

13.

14.

14. 10.0064 is the positive square root of 0.0064. (0.08) 2  0.0064 S 10.0064  0.08 15.

16.

25

3 36 is the positive square root of

156 22  2536 S 3 2536  56

3

8 98

is the positive square root of

127 22  494  988 S 3 988  27

9-1

25 . 36

15. 8 98

or

4 . 49

Factors and Greatest Common Factors

Page 477

16.

Check for Understanding

1. False; 2 is a prime number that is even. 2. Two numbers are relatively prime if their GCF is 1. Chapter 9

400

 3  3  5 or 32  5 32  1  32  1  2  16  1  2  2  8  1  2  2  2  4  1  2  2  2  2  2 or 1  25 150  1  150  1  2  75  1  2  3  25  1  2  3  5  5 or 1  2  3  52 2  22pp 4p 39b3c2  3  13  b  b  b  c  c 100x3yz2  1  100  x  x  x  y  z  z  1  2  50  x  x  x  y  z  z  1  2  2  25  x  x  x  y  z  z  1  2  2  5  5  x  x  x  y  z  z Factor each monomial and circle the common prime factors. 10  2  5 15  3  5 GCF: 5 Factor each monomial and circle the common prime factors. 18xy  2  3  3  x  y 36y2  2  2  3  3  y  y GCF: 2  3  3  y or 18y Factor each monomial and circle the common prime factors. 54  2  3  3  3 63  3  3  7 180  2  2  3  3  5 GCF: 3  3 or 9 Factor each monomial and circle the common prime factors. 25n  5  5  n 21m  3  7  m GCF: 1

factors: 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126 composite 27. List all pairs of numbers whose product is 304. 2  152 4  76 1  304 16  19 8  38 factors: 1, 2, 4, 8, 16, 19, 38, 76, 152, 304 composite 28. List all pairs of whole numbers whose product is 96. These are the possible dimensions for the rectangle. Find the perimeter of the rectangle associated with each pair and identify the least perimeter among them.

17. Factor each monomial and circle the common prime factors. 12a2b  2  2  3  a  a  b 90a2b2c  2  3  3  5  a  a  b  b  c GCF: 2  3  a  a  b or 6a2b 18. Factor each monomial and circle the common prime factors. 15r2  3  5  r  r 35s2  5  7  s  s 70rs  2  5  7  r  s GCF: 5 19. List all pairs of numbers whose product is 120 and both numbers are at least 5. 6  20 5  24 8  15 10  12 Since either number in each pair can be the number of rows and the other number the number of plants, Ashley can arrange the plants in any of the following ways. 5 rows of 24 plants 6 rows of 20 plants 8 rows of 15 plants 10 rows of 12 plants 12 rows of 10 plants 15 rows of 8 plants 20 rows of 6 plants 24 rows of 5 plants

Pages 477–479

A  lw 96  96  1 96  48  2 96  32  3 96  24  4 96  16  6 96  12  8

The minimum perimeter is 40 mm. 29. Identify the greatest perimeter in the chart in Exercise 28. The maximum perimeter is 194 mm. 30. Find the GCF of 18 and 24. Factor each number and circle the common prime factors. 18  2  3  3 24  2  2  2  3 GCF: 2  3 or 6 Each cellophane package should contain 6 cookies. 31. Since 18  3  6 and 24  4  6, 3 cellophane packages will go in each box of 18 cookies, and 4 cellophane packages will go in each box of 24 cookies. 32. 39  3  13 33. 98  1  98  1  2  49  1  2  7  7 or 1  2  72 34. 117  3  39  3  3  13 or 32  13 35. 102  2  51  2  3  17 36. 115  1  115  1  5  23 37. 180  2  90  2  2  45  2  2  3  15  2  2  3  3  5 or 22  32  5 38. 360  2  180  2  2  90  2  2  2  45  2  2  2  3  15  2  2  2  3  3  5 or 23  32  5 39. 462  1  462  1  2  231  1  2  3  77  1  2  3  7  11

Practice and Apply

20. List all pairs of numbers whose product 1  19 factors: 1, 19 prime 21. List all pairs of numbers whose product 1  25 55 factors: 1, 5, 25 composite 22. List all pairs of numbers whose product 2  40 4  20 1  80 8  10 5  16 factors: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80 composite 23. List all pairs of numbers whose product 1  61 factors: 1, 61 prime 24. List all pairs of numbers whose product 1  91 7  13 factors: 1, 7, 13, 91 composite 25. List all pairs of numbers whose product 1  119 7  17 factors: 1, 7, 17, 119 composite 26. List all pairs of numbers whose product 2  63 3  42 1  126 7  18 9  14 6  21

2l  2w  P 2(96)  2(1)  194 2(48)  2(2)  100 2(32)  2(3)  70 2(24)  2(4)  56 2(16)  2(6)  44 2(12)  2(8)  40

is 19.

is 25.

is 80.

is 61.

is 91.

is 119.

is 126.

401

Chapter 9

40. 66d4  2  33  d  d  d  d  2  3  11  d  d  d  d 2 2 41. 85x y  5  17  x  x  y  y 42. 49a3b2  7  7  a  a  a  b  b 43. 50gh  2  25  g  h  255gh 44. 128pq2  2  64  p  q  q  2  2  32  p  q  q  2  2  2  16  p  q  q  22228pqq  222224pqq  2222222pqq 45. 243n3m  3  81  n  n  n  m  3  3  27  n  n  n  m  3339nnnm  33333nnnm 46. 183xyz3  1  183  x  y  z  z  z  1  3  61  x  y  z  z  z 2bc2  1  169  a  a  b  c  c 47. 169a  1  13  13  a  a  b  c  c 48. Factor each monomial and circle the common prime factors. 27  3  3  3 72  2  2  2  3  3 GCF: 3  3 or 9 49. Factor each monomial and circle the common prime factors. 18  2  3  3 35  5  7 GCF: 1 50. Factor each monomial and circle the common prime factors. 32  2  2  2  2  2 48  2  2  2  2  3 GCF: 2  2  2  2 or 16 51. Factor each monomial and circle the common prime factors. 84  2  2  3  7 70  2  5  7 GCF: 2  7 or 14 52. Factor each monomial and circle the common prime factors. 16  2  2  2  2 20  2  2  5 64  2  2  2  2  2  2 GCF: 2  2 or 4 53. Factor each monomial and circle the common prime factors. 42  2  3  7 63  3  3  7 105  3  5  7 GCF: 3  7 or 21 54. Factor each monomial and circle the common prime factors. 15a  3  5  a 28b2  2  2  7  b  b GCF: 1

Chapter 9

55. Factor each monomial and circle the common prime factors. 24d2  2  2  2  3  d  d 30c2d  2  3  5  c  c  d GCF: 2  3  d or 6d 56. Factor each monomial and circle the common prime factors. 20gh  2  2  5  g  h 36g2h2  2  2  3  3  g  g  h  h GCF: 2  2  g  h or 4gh 57. Factor each monomial and circle the common prime factors. 21p2q  3  7  p  p  q 32r2t  2  2  2  2  2  r  r  t GCF: 1 58. Factor each monomial and circle the common prime factors. 18x  2  3  3  x 30xy  2  3  5  x  y 54y  2  3  3  3  y GCF: 2  3 or 6 59. Factor each monomial and circle the common prime factors. 28a2  2  2  7  a  a 63a3b2  3  3  7  a  a  a  b  b 91b3  7  13  b  b  b GCF: 7 60. Factor each monomial and circle the common prime factors. 14m2n2  2  7  m  m  n  n 18mn  2  3  3  m  n 2m2n3  2  m  m  n  n  n GCF: 2  m  n or 2mn 61. Factor each monomial and circle the common prime factors. 80a2b  2  2  2  2  5  a  a  b 96a2b3  2  2  2  2  2  3  a  a  b bb 128a2b2  2  2  2  2  2  2  2  a  a  b b GCF: 2  2  2  2  a  a  b or 16a2b 62. The next five pairs of twin primes are: 5 and 7, 11 and 13, 17 and 19, 29 and 31, and 41 and 43. 63. Find the GCF of 75 and 90. Factor each number and circle the common prime factors. 75  3  5  5 90  2  3  3  5 GCF: 3  5 or 15 The maximum number of rows is 15. 64. Since there is a total of 75  90 or 165 members in 15 rows and 165  15  11, there will be 11 members in each row.

402

65. The next prime number after 2 is 3, so let p  3. 2p  1  23  1 81 7 Since 7 is prime, the second Mersenne prime is 7. The next prime number after 3 is 5, so let p  5. 2p  1  25  1  32  1  31 Since 31 is prime, the third Mersenne prime is 31. 66. No; the first Mersenne prime is 3, so the formula does not generate the first prime number, 2. 67. The area of the triangle is 20 sq cm.

70. D; List all pairs of numbers whose product is 120. 2  60 3  40 1  120 5  24 6  20 4  30 10  12 8  15 factors: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 Thus, it is true that 120 has at least eight factors. Any number that has at least eight factors will have at least four factorizations into two numbers. 71. A; Find the prime factorization of each number. 53  53 74  2  37 99  3  3  11 117  3  3  13 Thus,   53,   37,   11, 53 74 99

1

A  2bh 1

20  2bh 2(20)  2

1

1 bh 2

and   13. 53 has the greatest value. 117

2

40  bh List all pairs of whole numbers whose product is 40. These are the possible dimensions for the triangle. 2  20 1  40 58 4  10 Since either number in each pair can be the base and the other number the height, the triangle can have any of the following whole-number dimensions. base: 1 cm, height: 40 cm base: 2 cm, height: 20 cm base: 4 cm, height: 10 cm base: 5 cm, height: 8 cm base: 8 cm, height: 5 cm base: 10 cm, height: 4 cm base: 20 cm, height: 2 cm base: 40 cm, height: 1 cm 68a. False; 6 is a factor of 3  4 or 12, but 6 is not a factor of 3 or 4. 68b. True; since 6 is a factor of ab, the prime factorization of ab must contain a factor of 3 since 3 is a factor of 6. Thus, 3 must be a factor of either a or b. 68c. False; 6 is a factor of 3  1082 or 3246, but 3 is not a factor of 1082. 69. Scientists listening to radio signals would suspect that a modulated signal beginning with prime numbers would indicate a message from an extraterrestrial. Answer should include the following. • 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 79, 83, 89, 97, 101, 103, 107, 109, 113; See students’ work. • Sample answer: It is unlikely that any natural phenomenon would produce such an artificial and specifically mathematical pattern.

Page 479

Maintain Your Skills

72. (a  b) 2  a2  2ab  b2 (2x  1) 2  (2x) 2  2(2x)(1)  12  4x2  4x  1 73. (a  b)(a  b)  a2  b2 (3a  5)(3a  5)  (3a) 2  52  9a2  25 74. (7p2  4)(7p2  4)  (7p2  4) 2 (a  b) 2  a2  2ab  b2 (7p2  4) 2  (7p2 ) 2  2(7p2 )(4)  42  49p4  56p2  16 75. (6r  7)(2r  5)  (6r)(2r)  (6r)(5)  (7)(2r)  7(5)  12r2  30r  14r  35  12r2  16r  35 76. (10h  k)(2h  5k)  (10h)(2h)  (10h)(5k)  (k)(2h)  (k) (5k)  20h2  50hk  2hk  5k2  20h2  52hk  5k2 77. (b  4)(b2  3b  18)  b(b2  3b  18)  4(b2  3b  18)  b3  3b2  18b  4b2  12b  72  b3  7b2  6b  72 78. Use the slope formula with (x1, y1)  (1, 2), (x2, y2)  (2, r) and m  3. y2  y 1

mx 3 3 1



2

 x1

r  2 2  1 r  2 3

3(3)  1(r  2) 9  r  2 9  2  r  2  2 7  r

403

Chapter 9

2. Model 6x  8.

79. Use the slope formula with (x1, y1)  (5, 9), 3 (x2, y2)  (r, 6) and m  5. y2  y1

mx

2

3 5 3 5

 

 x1

1 1 1 1

6  9 r  (5) 3 r  5

x

7900 125

84. 85. 86.

x x

2

x x

1 1 1 1 1 1 1 1

x x

25

 100



The rectangle has a width of 2 and a length of 3x  4. So, 6x  8  2(3x  4). 3. Model 5x2  2x.

125x 125

63.2  x The wholesale price was $63.20. 5(2x  8)  5(2x)  5(8)  10x  40 a(3a  1)  a(3a)  a(1)  3a2  a 2g(3g  4)  2g(3g)  (2g)(4)  6g2  8g 4y(3y  6)  (4y) (3y)  (4y)(6)  12y2  24y 7b  7c  7(b  c) 2x  3x  (2  3)x

Page 480

x

3x  4

(79  x) (100)  (x)(25) 7900  100x  25x 7900  100x  100x  25x  100x 7900  125x

83.

x

0

 3

79  x x

82.

x

Arrange the tiles into a rectangle.

r0 80. Let x  the wholesale price. Then 79  x is the amount of change. The percent of increase is 25%.

81.

x

1 1 1 1

3(r  5)  5(3) 3r  15  15 3r  15  15  15  15 3r  0 3r 3

x

x2

x2

x2

x2

x2

x

Arrange the tiles into a rectangle. 5x  2

x

x2

x2

x2

x2

x2

x x

The rectangle has a width of x and a length of 5x  2. So, 5x2  2x  x(5x  2). 4. Model 9  3x.

Algebra Activity (Preview of Lesson 9-2)

1. Model 2x  10.

x

1

1

1

1

1

1

1

1

1

1

x

x

1

1

1

x

1

1

1

x

1

1

1

Arrange the tiles into a rectangle.

Arrange the tiles into a rectangle.

x5 2

x x

1 1

1 1

3x 1 1

1 1

1 1

3

The rectangle has a width of 2 and a length of x  5. So, 2x  10  2(x  5).

Chapter 9

x x x

1 1 1

1 1 1

1 1 1

The rectangle has a width of 3 and a length of 3  x. So, 9  3x  3(3  x).

404

x

5. Yes;

4. The GCF of 9x2 and 36x is 9x. 9x2  36x  9x(x)  9x(4)  9x(x  4) 5. The GCF of 16xz and 40xz2 is 8xz. 16xz  40xz2  8xz(2)  8xz(5z)  8xz(2  5z) 6. The GCF of 24m2np2 and 36m2n2p is 12m2np. 24m2np2  36m2n2p  12m2np(2p)  12m2np(3n)  12m2np(2p  3n) 7. The GCF of 2a3b2, 8ab, and 16a2b3 is 2ab.

2x  5 2

x x

x x

1 1 1 1 1 1 1 1 1 1

6. No;

2a3b2  8ab  16a2b3  2ab(a2b)  2ab(4)  2ab(8ab2 )  2ab(a2b  4  8ab2 )

1 1 1 1 1 1 1

x x x

8. 5y2  15y  4y  12  (5y2  15y)  (4y  12)  5y(y  3)  4(y  3)  (y  3)(5y  4) 9. 5c  10c2  2d  4cd  (5c  10c2 )  (2d  4cd)  5c(1  2c)  2d(1  2c)  (1  2c)(5c  2d) 10. h(h  5)  0 h  0 or h  5  0 h  5 The solution set is {0, 5}. Check: h(h  5)  0 h(h  5)  0 ? ? 5(5  5)  0 0(0  5)  0 ? ? 5(0)  0 0(5)  0 00✓ 00✓ 11. (n  4)(n  2)  0 n  4  0 or n  2  0 n4 n  2 The solution set is {4, 2}. Check: (n  4)(n  2)  0 (n  4)(n  2)  0 ? ? (2  4)(2  2)  0 (4  4)(4  2)  0 ? ? 6(0)  0 0(6)  0 00✓ 00✓ 12. 5m  3m2 5m  3m2  0 m(5  3m)  0 m  0 or 5  3m  0 3m  5

7. Yes;

x2 x

x

2

x

2

x x

8. No;

1

x

1

1

2

9. Sample answer: Binomials can be factored if they can be represented by a rectangle. Examples: 2x  2 can be factored and 2x  1 cannot be factored.

9-2 Page 484

Factoring Using the Distributive Property

5 56

Check for Understanding

5

m3

The solution set is 0, 3 .

1. Sample answers: 4x2  12x  4(x2 )  4(3x)  4(x2  3x) 2  12x  x(4x)  x(12) 4x  x(4x  12) 4x2  12x  4x(x)  4x(3)  4x(x  3) 4x(x  3) is the completely factored form because 4x is the GCF of 4x2 and 12x. 2. Sample answer: The equation x2  6x  7  0 can be solved using the Zero Product Property because x2  6x  7 can be factored to (x  1)(x  7). 3. The equation has two solutions, 4 and 2. Dividing each side by x  2 would eliminate 2 as a solution.

Check:

5m  3m2 ?

5(0)  3(0) 2 00✓

5m  3m2 5

153 2  3153 22 25 ? 25  31 9 2 3 ?

25 3



25 3



13. Since the height is expressed in units of feet above sea level, the flare is at height 0 ft when it returns to the sea. 14. h  100t  16t2 0  100t  16t2 0  4t(25  4t) 4t  0 or 25  4t  0 t0 4t  25 25 t 4 t  6.25

405

Chapter 9

15. The flare returns to the sea in 6.25 s. The answer 0 is not reasonable since it represents the time at which the flare is launched.

37. 2my  7x  7m  2xy  2my  7m  7x  2xy

Pages 484–486

38. 8ax  6x  12a  9  (8ax  6x)  (12a  9)  2x(4a  3)  3(4a  3)  (4a  3)(2x  3)

   

Practice and Apply

16. 5x  30y  5(x)  5(6y)  5(x  6y) 17. 16a  4b  4(4a)  4(b)  4(4a  b) 5 18. a b  a  a(a4b)  a(1)  a(a4b  1) 3 2 19. x y  x  x(x2y2 )  x(1)  x(x2y2  1) 20. 21cd  3d  3d(7c)  3d(1)  3d(7c  1) 21. 14gh  18h  2h(7g)  2h(9)  2h(7g  9) 22. 15a2y  30ay  15ay(a)  15ay(2)  15ay(a  2) 23. 8bc2  24bc  8bc(c)  8bc(3)  8bc(c  3) 24. 12x2y2z  40xy3z2  4xy2z(3x)  4xy2z(10yz)  4xy2z(3x  10yz) 25. 18a2bc2  48abc3  6abc2 (3a)  6abc2 (8c)  6abc2 (3a  8c) 2b2  a3b3  a(1)  a(ab2 )  a(a2b3 ) 26. a  a  a(1  ab2  a2b3 ) 2y2  25xy  x  x(15xy2 )  x(25y)  x(1) 27. 15x  x(15xy2  25y  1) 28. 12ax3  20bx2  32cx  4x(3ax2 )  4x(5bx)  4x(8c)

39. 10x2  14xy  15x  21y  (10x2  14xy)  (15x  21y)  2x(5x  7y)  3(5x  7y)  (5x  7y)(2x  3) 1

29. 3p



9pq2

1

1

1

 2n(n  3) 41. Replace n with 10 in the polynomial. 1 n(n 2

1

 3)  2 (10)(5  3)

 5(7) or 35 A decagon has 35 diagonals. 1

1

42. g  2n2  2n 1

1

 2n(n)  2n(1) 1

 2n(n  1) 43. First find the number of games needed for 7 teams to play each other once. 1

g  2 (7)(7  1) 1

 2 (7)(6) 1

 2 (42)  21 Thus, the number of games needed for 7 teams to play each other 3 times is 3(21) or 63 games. 44. area of shaded  area of outer  area of inner region rectangle rectangle  (a  4)(b  4)  ab  ab  a(4)  4b  16  ab  4a  4b  16  4(a)  4(b)  4(4)  4(a  b  4) 45. area of shaded  area of the  2  area of a region rectangle circle  2r(4r)  2(r2 )  8r2  2r2  2r2 (4)  2r2 ()  2r2 (4  ) 46. The length of one side of the square is 1 (12x  20y) or 3x  5y in. 4

 36pq  3pq(p2 )  3pq(3q)  3pq(12)  3pq(p2  3q  12)

30. x2  2x  3x  6  (x2  2x)  (3x  6)  x(x  2)  3(x  2)  (x  2)(x  3) 31. x2  5x  7x  35  (x2  5x)  (7x  35)  x(x  5)  7(x  5)  (x  5)(x  7) 32. 4x2  14x  6x  21  (4x2  14x)  (6x  21)  2x(2x  7)  3(2x  7)  (2x  7) (2x  3) 33. 12y2  9y  8y  6  (12y2  9y)  (8y  6)  3y(4y  3)  2(4y  3)  (4y  3)(3y  2) 34. 6a2  15a  8a  20  (6a2  15a)  (8a  20)  3a(2a  5)  4(2a  5)  (2a  5) (3a  4) 2  30x  3x  5  (18x2  30x)  (3x  5) 35. 18x  6x(3x  5)  (3x  5)  (3x  5)(6x  1) 36. 4ax  3ay  4bx  3by  (4ax  3ay)  (4bx  3by)

A  (3x  5y) 2  (3x) 2  2(3x)(5y)  (5y) 2  9x2  30xy  25y2 The area of the square is 9x2  30xy  25y2 in2.

 a(4x  3y)  b(4x  3y)  (4x  3y)(a  b)

Chapter 9

3

40. 2n2  2n  2n(n)  2n(3)

 4x(3ax2  5bx  8c) 3q

(2my  7m)  (7x  2xy) m(2y  7)  x(7  2y) m(2y  7)  x(2y  7) (2y  7) (m  x)

406

47. The length of one side of the square is 1 (36a  16b) or 9a  4b cm. 4

57.

A  (9a  4b) 2  (9a) 2  2(9a)(4b)  (4b) 2  81a2  72ab  16b2 The area of the square is 81a2  72ab  16b2 cm2. Exercises 48–59 For checks, see students’ work. 48. x(x  24)  0 x  0 or x  24  0 x  24 {0, 24} 49. a(a  16)  0 a  0 or a  16  0 a  16 {16, 0} 50. (q  4)(3q  15)  0 or 3q  15  0 q40 q  4 3q  15 q5 {4, 5} 51. (3y  9)( y  7)  0 or y  7  0 3y  9  0 3y  9 y7 y  3 {3, 7} 52. (2b  3) (3b  8)  0 2b  3  0 or 3b  8  0 2b  3 3b  8

58.

59.

532, 83 6

6

x7

50, 67 6

6x2  4x 6x  4x  0 2x(3x  2)  0 2x  0 x0

or

3x  2  0 3x  2 2

x  3

523, 06

20x2  15x 20x  15x  0 5x(4x  3)  0 5x  0 x0

˛

or

4x  3  0 4x  3 3

x  4

60. Let h  0 and solve for t. h  20t  16t2 0  20t  16t2 0  4t(5  4t) 4t  0 or 5  4t  0 t0 4t  5

8

5

t4 ˛

or

t  1.25 The dolphin leaves the water at 0 seconds and returns to the water at 1.25 seconds. The dolphin is in the air for 1.25 s. 61. Let h  2 and solve for t. h  2  45t  16t2 2  2  45t  16t2 0  45t  16t2 0  16t(2.8125  t) 16t  0 or 2.8125  t  0 t  2.8125 t0 t  2.8125 Malik hits the ball at 0 seconds, and the catcher catches it at about 2.8 seconds. The ball is in the air about 2.8 s before it is caught.

3n  7  0 3n  7

5

7

n  4

n3

54. 3z2  12z  0 3z(z  4)  0 3z  0 or z  4  0 z0 z  4 {4, 0} 55. 7d2  35d  0 7d(d  5)  0 7d  0 or d  5  0 d0 d5 {0, 5} 56. 2x2  5x 2  5x  0 2x x(2x  5)  0 or 2x  5  0 x0 2x  5

62. axy  axby  aybx  bxy  (axy  axby )  (aybx  bxy )  (axay  axby )  (aybx  bxby )  ax (ay  by )  bx (ay  by )  (ay  by )(ax  bx )

˛

50, 52 6

7x  6  0 7x  6

534, 06

b3

53. (4n  5)(3n  7)  0 4n  5  0 4n  5

554, 73 6

or

2

˛

3

˛

2

˛

b2

7x2  6x 7x  6x  0 x(7x  6)  0 x0 2

5

x2

407

Chapter 9

63. Answers should include the following. • Let h  0 in the equation h  151t  16t2. To solve 0  151t  16t2, factor the right-hand side as t(151  16t). Then, since t(151  16t)  0, either t  0 or 151  16t  0. Solving each equation for t, we find that t  0 or t  9.44. • The solution t  0 represents the point at which the ball was initially thrown into the air. The solution t  9.44 represents how long it took after the ball was thrown for it to return to the same height at which it was thrown. 64. A; Since there are 3 feet in one yard, the number of feet in x yards is 3x. The number of feet in y feet is y. Since there are 12 inches per foot, the z number of feet in z inches is 12. 65. A; First determine the negative solution of the equation in Column A. (a  2) (a  5)  0 a  2  0 or a  5  0 a2 a  5 The solution set is {5, 2}. The negative solution is 5. Next determine the negative solution of the equation in Column B. (b  6) (b  1)  0 b60 or b  1  0 b  6 b1 The solution set is {6, 1}. The negative solution is 6. 5  6, so the quantity in Column A is greater.

Page 486

73.

s4 s7

   74.

11812 21xx 21yy 2 3

1

2

4

3 32 (x )(y14 ) 2 3 (x)(y5 ) 2 3 1 (x) y5 2 3x 2y5

34p7q2r5 17(p3qr1 ) 2

1 2

34p7q2r5

 17(p3 ) 2 (q) 2 (r1 ) 2 34p7q2r5

 17p6q2r2 

13417 21pp 21qq 21rr 2 7

2

6

2

5 2

)(q )(r5(2) )  2(p 0 3  2(p)(q )(r )  2(p)(1)(r3 ) 76



22

2p r3

75. Let x represent the number of shares that Michael can purchase. The total purchase price cannot exceed 60% of last year’s dividend. 14x  0.60(885) 14x  531 x

531 14

or about 37.9

Michael can purchase up to 37 shares. 76. (n  8)(n  3)  (n)(n)  (n)(3)  (8)(n)  8(3)  n2  3n  8n  24  n2  11n  24 77. (x  4)(x  5)  (x)(x)  (x)(5)  (4)(x)  (4)(5)  x2  5x  4x  20  x2  9x  20 78. (b  10) (b  7)  (b)(b)  (b)(7)  (10)(b)  (10)(7)  b2  7b  10b  70  b2  3b  70 79. (3a  1)(6a  4)  (3a)(6a)  (3a)(4)  (1)(6a)  (1)(4)

Maintain Your Skills

 18a2  12a  6a  4  18a2  6a  4 80. (5p  2)(9p  3)  (5p)(9p)  (5p)(3)  (2) (9p)  (2)(3)  45p2  15p  18p  6  45p2  33p  6 81. (2y  5)(4y  3)  (2y)(4y)  (2y)(3)  (5)(4y)  (5)(3)  8y2  6y  20y  15  8y2  14y  15

Page 486

Practice Quiz 1

1. List all pairs of numbers whose product is 225. 1  225 3  75 5  45 9  25 15  15 factors: 1, 3, 5, 9, 15, 25, 45, 75, 225 composite

 s4(7)  s11

Chapter 9

 

66. List all pairs of numbers whose product is 123. 1  123 3  41 factors: 1, 3, 41, 123 composite 67. List all pairs of numbers whose product is 300. 1  300 2  150 3  100 4  75 5  60 6  50 10  30 12  25 15  20 factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300 composite 68. List all pairs of numbers whose product is 67. 1  67 factors: 1, 67 prime 69. (4s3  3) 2  (4s3 ) 2  2(4s3 )(3)  (3) 2  16s6  24s3  9 70. (2p  5q) (2p  5q)  (2p) 2  (5q) 2  4p2  25q2 71. (3k  8) (3k  8)  (3k  8) 2  (3k) 2  2(3k)(8)  82  9k2  48k  64 72.

18x3y1 12x2y4

408

2. 320  1  320  1  2  160  1  2  2  80  1  2  2  2  40  1  2  2  2  2  20  1  2  2  2  2  2  10  1  2  2  2  2  2  2  5 or 1  26  5 2 3 3. 78a bc  2  39  a  a  b  c  c  c  2  3  13  a  a  b  c  c  c 4. 54x3  2  3  3  3  x  x  x 42x2y  2  3  7  x  x  y 30xy2  2  3  5  x  y  y GCF: 2  3  x or 6x 5. 4xy2  xy  xy(4x)  xy(1)  xy(4x  1) 2 6. 32a b  40b3  8a2b2  8b(4a2 )  8b(5b2 )  8b(a2b)

Page 488

Algebra Activity (Preview of Lesson 9-3)

1. Model x2  4x  3.

1

x2

x

x

x

x

1

1

Place the x2 tile at the corner. Arrange the 1 tiles into a rectangular array.

 8b(4a2  5b2  a2b)

7. 6py  16p  15y  40  (6py  16p)  (15y  40)

x2

 2p(3y  8)  5(3y  8)  (3y  8) (2p  5)

8. (8n  5)(n  4)  0 8n  5  0 8n  5

1

or

1

1

n40 n4

5

n  8

558, 46

Complete the rectangle with the x tiles.

x3

Check:

(8n  5) (n  4)  0

3 8158 2  5 4158  42  0 ? 5 (5  5) 1 8  4 2  0 37 ? (0) 1  8 2  0 ?

(8n  5) (n  4)  0 ?

[ 8(4)  5] (4  4)  0

x1

?

(37)(0)  0 00✓

x2

x x x

x

1

1

1

00✓

2

9. 9x  27x  0 9x(x  3)  0 9x  0 or x0 {0, 3} Check: 9x(x  3)  0

9x(x  3)  0

?

?

9(0) (0  3)  0

9(3)(3  3)  0

?

0(3)  0 00✓ 10. 10x2  3x 10x2  3x  0 x(10x  3)  0 x0

The rectangle has a width of x  1 and a length of x  3. Therefore, x2  4x  3  (x  1)(x  3). 2. Model x2  5x  4.

x30 x3

?

x2

27(0)  0 00✓

˛

or

10x  3  0 10x  3 3

Check:

10x2  3x

10x2  3x

?

10(0) 2  3(0)

1103 22  31103 2 ? 9 9 10 1 100 2  10

10

9 10



9 10



x

x

x

1

1

1

1

x

Place the x2 tile at the corner. Arrange the 1 tiles into a rectangular array.

x  10

5103 , 06

x

x2

?

1

1

1

1

?

10(0)  0 00✓

409

Chapter 9

Complete the rectangle with the x tiles.

4. Model x2  3x  2.

x4

x x x x

x2

x1

1

x2 x

1

1

1

x x x 1

1

The rectangle has a width of x  1 and a length of x  4. Therefore, x2  5x  4  (x  1)(x  4). 3. Model x2  x  6.

Place the x2 tile at the corner. Arrange the 1 tiles into a rectangular array.

x2 x

x2

1 1 1 1

1

1 1 1

Complete the rectangle with the x tiles.

x2

Place the x2 tile at the corner. Arrange the 1 tiles into a rectangular array.

x1 x2

x2

x x

x

1 1

1 1 1 1 1 1 The rectangle has a width of x  1 and a length of x  2. Therefore, x2  3x  2  (x  1)(x  2). 5. Model x2  7x  12.

Complete the rectangle with the x tile and two zero-pairs of x tiles

x2

x3

x2

x x x

x x

1 1 1 1 1 1

x

x

x

x

x

x

x

1

1

1

1

1

1

1

1

1

1

1

1

x2 Place the x2 tile at the corner. Arrange the 1 tiles into a rectangular array.

The rectangle has a width of x  2 and a length of x  3. Therefore, x2  x  6  (x  2)(x  3).

Chapter 9

x2

410

1 1

1 1

1 1

1 1

1

1

1

1

Complete the rectangle with the x tiles.

7. Model x2  x  2.

x4

x x x x

x2

x

x2

x3

x x x

1 1

1 1 1 1 1 1 1 1 1 1 1 1

Place the x2 tile at the corner. Arrange the 1 tiles into a rectangular array.

The rectangle has a width of x  3 and a length of x  4. Therefore, x2  7x  12  (x  3)(x  4). 6. Model x2  4x  4.

x2

1

1

1

1

1 1

x x x x

x2

Complete the rectangle with the x tile and a zero pair of x tiles.

x1 x2

Place the tile at the corner. Arrange the 1 tiles into a rectangular array.

x2

x

x x

1 1

x2

x2 1 1

1 1

The rectangle has a width of x  2 and a length of x  1. Therefore, x2  x  2  (x  2)(x  1).

Complete the rectangle with the x tiles.

x2

x2

x x

x x

1 1 1 1

x2

The rectangle has a width of x  2 and a length of x  2. Therefore, x2  4x  4  (x  2)(x  2).

411

Chapter 9

8. Model x2  6x  8.

x2

x x x x x x

1

1

1

1

1

1

1

1

3. 4.

Place the x2 tile at the corner. Arrange the 1 tiles into a rectangular array.

x2 5. 1 1 1 1 1 1 1 1

Complete the rectangle with the x tiles. 6.

x4

x2

x x x x

x x

1 1 1 1 1 1 1 1

x2

The rectangle has a width of x  2 and a length of x  4. Therefore, x2  6x  8  (x  2)(x  4).

9-3

Factoring Trinomials x 2  bx  c

Pages 492–493

7.

Check for Understanding

1. In this trinomial, b  6 and c  9. This means that m  n is positive and mn is positive. Only two positive numbers have both a positive sum and product. Therefore, negative factors of 9 need not be considered. 2. Sample answer: Factor x2  14x  40  0. Since b  14 and c  40, make a list of negative factors of 40, and look for the pair whose sum is 14. Factors of 40 Sum of Factors 1, 40 41 2, 20 22 4, 10 14 5, 8 13

Chapter 9

412

The correct factors are 4 and 10. x2  14x  40  0 (x  4)(x  10)  0 x  4  0 or x  10  0 x4 x  10 The solution set is {4, 10}. Aleta; to use the Zero Product Property, one side of the equation must equal zero. Since b  11 and c  24, make a list of positive factors of 24, and find the pair whose sum is 11. Factors of 24 Sum of Factors 1, 24 25 2, 12 14 3, 8 11 4, 6 10 The correct factors are 3 and 8. x2  11x  24  (x  m)(x  n)  (x  3)(x  8) Since b  3 and c  2, make a list of negative factors of 2, and find the pair whose sum is 3. Factors of 2 Sum of Factors 1, 2 3 The correct factors are 1 and 2. c2  3c  2  (c  m)(c  n)  (c  1)(c  2) Make a list of factors of 48, and find the pair whose sum is 13. Factors of 48 Sum of Factors 1, 48 47 1, 48 47 2, 24 22 2, 24 22 3, 16 13 3, 16 13 4, 12 8 4, 12 8 6, 8 2 6, 8 2 The correct factors are 3 and 16. n2  13n  48  (n  m)(n  p)  (n  3)(n  16) Make a list of factors of 35, and find the pair whose sum is 2. Factors of 35 Sum of Factors 1, 35 34 1, 35 34 5, 7 2 5, 7 2 The correct factors are 5 and 7. p2  2p  35  ( p  m)( p  n)  ( p  5)( p  7)

8. 72  27a  a2  a2  27a  72 Make a list of positive factors of 72, and find the pair whose sum is 27. Factors of 72 Sum of Factors 1, 72 73 2, 36 38 3, 24 27 4, 18 22 6, 12 18 8, 9 17 The correct factors are 3 and 24. 72  27a  a2  a2  27a  72  (a  m)(a  n)  (a  3)(a  24) 9. x2  4xy  3y2  x2  (4y)x  3y2 Make a list of negative factors of 3y2, and find the pair whose sum is 4y. Factors of 3y2 Sum of Factors y, 3y 4y y2, 3 y2  3 1, 3y2 1  3y2 The correct factors are y and 3y. x2  4xy  3y2  x2  (4y)x  3y2  (x  m)(x  n)  (x  y)(x  3y) 10. n2  7n  6  0 (n  1)(n  6)  0 or n  6  0 n10 n  1 n  6 The solution set is {6, 1}. Check: n2  7n  6  0 ? 2  7(1)  6  0 (1) ? 1760 00✓ n2  7n  6  0 ? (6) 2  7(6)  6  0

The solution set is {2, 21}. Check: p2  19p  42  0 2

?

(2)  19(2)  42  0 ?

4  38  42  0 00✓

p2  19p  42  0 2

?

(21)  19(21)  42  0 ?

441  399  42  0 00✓

y2  9  10y y  9  10y  0 y2  10y  9  0 ( y  1)( y  9)  0 or y  9  0 y10 y  1 y  9 The solution set is {9, 1}. Check: y2  9  10y y2  9  10y ? ? 2 (9)  9  10(9) (1) 2  9  10(1) ? ? 81  9  90 1  9  10 90  90 ✓ 10  10 ✓ 14. 9x  x2  22 9x  x2  22  0 x2  9x  22  0 (x  2)(x  11)  0 x  2  0 or x  11  0 x2 x  11 The solution set is {11, 2}. Check: 9x  x2  22 9x  x2  22 ? ? 2 9(11)  (11)  22 9(2)  (2) 2  22 ? ? 99  121  22 18  4  22 22  22 ✓ 22  22 ✓ 15. d2  3d  70 d2  3d  70  0 (d  7)(d  10)  0 or d  10  0 d70 d  7 d  10 The solution set is {7, 10}. Check: d2  3d  70 d2  3d  70 ? ? 2 2 (7)  3(7)  70 (10)  3(10)  70 ? ? 49  21  70 100  30  70 70  70 ✓ 70  70 ✓ 16. Let n be the first integer. Then n  1 is the second integer. n(n  1)  156 n2  n  156 2  n  156  0 n (n  12)(n  13)  0 n  12  0 or n  13  0 n  12 n  13 When n  12, n  1  13. When n  13, n  1  12. The two consecutive integers are either 12 and 13 or 13 and 12.

13.

2

?

36  42  6  0 00✓ 11. a2  5a  36  0 (a  4)(a  9)  0 a  4  0 or a  9  0 a4 a  9 The solution set is {9, 4}. Check: a2  5a  36  0 a2  5a  36  0 ? ? 2 2 (9)  5(9)  36  0 (4)  5(4)  36  0 ? ? 81  45  36  0 16  20  36  0 00✓ 00✓ 12. p2  19p  42  0 ( p  2)( p  21)  0 or p  21  0 p20 p  2 p  21

413

Chapter 9

Pages 493–494

31. 72  6w  w2  w2  6w  72 Among all pairs of factors of 72, choose 6 and 12, the pair of factors whose sum is 6. 72  6w  w2  w2  6w  72  (w  m)(w  n)  (w  6)(w  12) 32. 30  13x  x2  x2  13x  30 Among all pairs of factors of 3, choose 2 and 15, the pair of factors whose sum is 13. 30  13x  x2  x2  13x  30  (x  2)(x  15) 33. a2  5ab  4b2  a2  (5b)a  4b2 Among all pairs of positive factors of 4b2, choose b and 4b, the pair of factors whose sum is 5b. a2  5ab  4b2  a2  (5b)a  4b2  (a  m)(a  n)  (a  b)(a  4b) 34. x2  13xy  36y2  x2  (13y)x  36y2 Among all pairs of negative factors of 36y2, choose 4y and 9y, the pair of factors whose sum is 13y. x2  13xy  36y2  x2  (13y)x  36y2  (x  m)(x  n)  (x  4y) (x  9y) 35. Factor the expression given for the area. area  x2  24x  81  (x  27)(x  3) Assuming that the factors represent the length and width, find the perimeter of the rectangle. perimeter  2(x  27)  2(x  3)  2x  54  2x  6  4x  48 36. Factor the expression given for the area. area  x2  13x  90  (x  18)(x  5) Assuming that the factors represent the length and width, find the perimeter of the rectangle. perimeter  2(x  18)  2(x  5)  2x  36  2x  10  4x  26 Exercises 37–53 For checks, see students’ work. 37. x2  16x  28  0 (x  2)(x  14)  0 x20 or x  14  0 x  2 x  14 {14, 2} 38. b2  20b  36  0 (b  2)(b  18)  0 b20 or b  18  0 b  2 b  18 {18, 2} 39. y2  4y  12  0 ( y  2)( y  6)  0 y  2  0 or y  6  0 y2 y  6 {6, 2}

Practice and Apply

17. Among all pairs of positive factors of 15, choose 3 and 5, the pair of factors whose sum is 8. a2  8a  15  (a  m)(a  n)  (a  3)(a  5) 18. Among all pairs of positive factors of 27, choose 3 and 9, the pair of factors whose sum is 12. x2  12x  27  (x  m)(x  n)  (x  3)(x  9) 19. Among all pairs of positive factors of 35, choose 5 and 7, the pair of factors whose sum is 12. c2  12c  35  (c  m)(c  n)  (c  5) (c  7) 20. Among all pairs of positive factors of 30, choose 3 and 10, the pair of factors whose sum is 13. y2  13y  30  ( y  m)( y  n)  ( y  3)( y  10) 21. Among all pairs of negative factors of 21, choose 1 and 21, the pair of factors whose sum is 22. m2  22m  21  (m  p)(m  n)  (m  1)(m  21) 22. Among all pairs of negative factors of 10, choose 2 and 5, the pair of factors whose sum is 7. d2  7d  10  (d  m)(d  n)  (d  2)(d  5) 23. Among all pairs of negative factors of 72, choose 8 and 9, the pair of factors whose sum is 17. p2  17p  72  ( p  m)( p  n)  ( p  8)( p  9) 24. Among all pairs of negative factors of 60, choose 4 and 15, the pair of factors whose sum is 19. g2  19g  60  ( g  m)( g  n)  ( g  4)( g  15) 25. Among all pairs of factors of 7, choose 1 and 7, the pair of factors whose sum is 6. x2  6x  7  (x  m)(x  n)  (x  1)(x  7) 26. Among all pairs of factors of 20, choose 4 and 5, the pair of factors whose sum is 1. b2  b  20  (b  m)(b  n)  (b  4)(b  5) 27. Among all pairs of factors of 40, choose 5 and 8, the pair of factors whose sum is 3. h2  3h  40  (h  m)(h  n)  (h  5)(h  8) 28. Among all pairs of factors of 54, choose 6 and 9, the pair of factors whose sum is 3. n2  3n  54  (n  m)(n  p)  (n  6)(n  9) 29. Among all pairs of factors of 42, choose 6 and 7, the pair of factors whose sum is 1. y2  y  42  ( y  m)( y  n)  ( y  6)( y  7) 30. Among all pairs of factors of 40, choose 2 and 20, the pair of factors whose sum is 18. z2  18z  40  (z  m)(z  n)  (z  2)(z  20)

Chapter 9

414

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.

d2  2d  8  0 (d  2)(d  4)  0 d  2  0 or d  4  0 d2 d  4 {4, 2} a2  3a  28  0 (a  7)(a  4)  0 a  7  0 or a  4  0 a7 a  4 {4, 7} g2  4g  45  0 ( g  9)( g  5)  0 g  9  0 or g  5  0 g9 g  5 {5, 9} m2  19m  48  0 (m  3)(m  16)  0 m  3  0 or m  16  0 m3 m  16 {3, 16} n2  22n  72  0 (n  4)(n  18)  0 n  4  0 or n  18  0 n4 n  18 {4, 18} z2  18  7z 2 z  7z  18  0 (z  2)(z  9)  0 z  2  0 or z  9  0 z2 z  9 {9, 2} h2  15  16h 2 h  16h  15  0 (h  15) (h  1)  0 h  15  0 or h  1  0 h  15 h  1 {15,  1} 24  k2  10k 2 k  10k  24  0 (k  4)(k  6)  0 k  4  0 or k  6  0 k4 k6 {4, 6} x2  20  x 2 x  x  20  0 (x  4)(x  5)  0 x40 or x  5  0 x  4 x5 {4, 5} c2  50  23c 2  23c  50  0 c (c  2)(c  25)  0 c  2  0 or c  25  0 c2 c  25 {25, 2}

50.

y2  29y  54  29y  54  0 ( y  2)( y  27)  0 y  2  0 or y  27  0 y2 y  27 {2, 27} 14p  p2  51 2 p  14p  51  0 ( p  3)( p  17)  0 p  3  0 or p  17  0 p3 p  17 {17, 3} x2  2x  6  74 2 x  2x  6  74  0 x2  2x  80  0 (x  8)(x  10)  0 or x  10  0 x80 x  8 x  10 {8, 10} x2  x  56  17x 2  x  56  17x  0 x x2  18x  56  0 (x  4)(x  14)  0 x  4  0 or x  14  0 x4 x  14 {4, 14} To find the number of Justices on the Supreme Court, replace h with 36 in the equation and solve for n. y2

51.

52.

53.

54.

h 36 

n2  n 2 n2  n 2

72  n2  n 0  n2  n  72 0  (n  9)(n  8) n  9  0 or n  8  0 n9 n  8 Only 9 is a valid solution since the number of Justices cannot be negative. Thus there are 9 Justices on the Supreme Court. 55. Let n be the first integer. Let n  2 be the second integer. n(n  2)  168 n2  2n  168 2  2n  168  0 n (n  12)(n  14)  0 n  12  0 or n  14  0 n  12 n  14 When n  12, n  2  14. When n  14, n  2  12. The two consecutive even integers are either 12 and 14 or 14 and 12.

415

Chapter 9

62. Replace the area with 8160 and solve. area  w(w  52) 8160  w(w  52) 8160  w2  52w 0  w2  52w  8160 0  (w  68)(w  120) or w  68  0 w  120  0 w  120 w  68 Only 68 is a valid solution since dimensions cannot be negative. Now find the length of the field. length  w  52  68  52  120 The dimensions of the field are 120 m by 68 m. 63. Answers should include the following. • You would use a guess-and-check process, listing the factors of 54, checking to see which pairs added to 15. • To factor a trinomial of the form x2  ax  c, you also use a guess-and-check process, list the factors of c, and check to see which ones add to a. 64. C; Of the pairs of factors of 42 shown in the choices, only 3 and 14 have a sum of 17. So, the answer must be C. 65. p2  13p  30  0 ( p  2)( p  15)  0 or p  15  0 p20 p  2 p  15 The positive solution is 15. 66. No; the graphs of y1  x2  14x  48 and y2  (x  6)(x  8) do not coincide. The correct factorization is x2  14x  48  (x  6)(x  8). 67. Yes; the graphs of y1  x2  16x  105 and y2  (x  5)(x  21) coincide. 68. No; the graphs of y1  x2  25x  66 and y2  (x  33)(x  2) do not coincide. The correct factorization is x2  25x  66  (x  22)(x  3). 69. No; the graphs of y1  x2  11x  210 and y2  (x  10)(x  21) do not coincide. The correct factorization is x2  11x  210  (x  10)(x  21).

56. Write an equation for the area of the triangle and solve for h. 1

area  2 base  height 1

40  2 (2h  6)(h)

57.

58.

59.

60.

40  h2  3h 0  h2  3h  40 0  (h  5) (h  8) h  5  0 or h  8  0 h5 h  8 Only 5 is a valid solution since the height of the triangle cannot be negative. Thus, the height of the triangle is 5 cm. List all pairs of factors of 19 and the sums of the pairs. Factors of 19 Sum of Factors 1, 19 18 1, 19 18 k can be 18 or 18. List all pairs of factors of 14 and the sums of the pairs. Factors of 14 Sum of Factors 1, 14 15 1, 14 15 2, 7 9 2, 7 9 k can be 15, 9, 9, or 15. Because k is positive, m and n must have the same sign. Because b  8, m and n must be negative. List all pairs of negative integers whose sum is 8, and list the products of the pairs. Addends of 8 Product of Addends 1, 7 7 2, 6 12 3, 5 15 4, 4 16 k can be 7, 12, 15, or 16. Because k is positive, m and n must have the same sign. Because b  5, m and n must be negative. List all pairs of negative integers whose sum is 5, and list the products of the pairs.

Addends of 5 Product of Addends 1, 4 4 2, 3 6 k can be either 4 or 6. 61. area  length  width  (w  52) (w)  w(w  52) The area of the rugby field is [w(w  52)] m2.

Page 494

Maintain Your Skills

Exercises 70–72 For checks, see students’ work. 70. (x  3)(2x  5)  0 or 2x  5  0 x30 x  3 2x  5 5

x2

53, 52 6

71. b(7b  4)  0 b  0 or 7b  4  0 7b  4

50, 47 6 Chapter 9

416

4

b7

72.

5y2  9y  9y  0 y(5y  9)  0 y  0 or 5y  9  0 5y  9

83. 4g2  2g  6g  3  (4g2  2g)  (6g  3)  2g(2g  1)  3(2g  1)  2g(2g  1)  3(1) (2g  1)  2g(2g  1)  (3) (2g  1)  (2g  1)(2g  3)

5y2

9

y  5

595, 06

9-4

73. Factor each monomial and circle the common prime factors. 24  2  2  2  3 36  2  2  3  3 72  2  2  2  3  3 GCF: 2  2  3 or 12 74. Factor each monomial and circle the common prime factors. 9p2q5  3  3  p  p  q  q  q  q  q 21p3q3  3  7  p  p  p  q  q  q GCF: 3  p  p  q  q  q or 3p2q3 75. Factor each monomial and circle the common prime factors. 30x4y5  2  3  5  x  x  x  x  y  y  y  y  y 20x2y7  2  2  5  x  x  y  y  y  y  y  y  y 75x3y4  3  5  5  x  x  x  y  y  y  y GCF: 5  x  x  y  y  y  y or 5x2y4 76. Find the change. 28.2  1.54  26.66 Find the percent using the original number, 1.54, as the base. 26.66 1.54

Pages 498–499

r

 100

77.

78.

79.

80.

81.



Check for Understanding

1. m and n are the factors of ac whose sum is b. 2. Sample answer: 2x2  9x  7. ac  2(7)  14 and b  9. 2x2  9x  7  2x2  2x  7x  7  2x(x  1)  7(x  1)  (x  1)(2x  7) 3. Craig; when factoring a trinomial of the form ax2  bx  c, where a 1, you must find the factors of ac, not of c. 4. 3a2  8a  4 Since ac  12 and b  8, make a list of positive factors of 12 and find the pair whose sum is 8. Factors of 12 Sum of Factors 1, 12 13 2, 6 8 3, 4 7 The correct factors are 2 and 6. 3a2  8a  4  3a2  ma  na  4  3a2  2a  6a  4  (3a2  2a)  (6a  4)  a(3a  2)  2(3a  2)  (3a  2)(a  2) 5. 2a2  11a  7 Since ac  14 and b  11, make a list of negative factors of 14 and find the pair whose sum is 11. Factors of 14 Sum of Factors 1, 14 15 2, 7 9 There are no factors whose sum is 11. So, 2a2  11a  7 is prime.

26.66(100)  1.54(r) 2666  1.54r 2666 1.54

Factoring Trinomials: ax 2  bx  c

1.54r 1.54

1731  r The percent increase is about 1731%. 1.54  17.31(1.54)  1(1.54)  17.31(1.54)  (1  17.31) (1.54) or 18.31(1.54) 3y2  2y  9y  6  (3y2  2y)  (9y  6)  y(3y  2)  3(3y  2)  (3y  2)(y  3) 3a2  2a  12a  8  (3a2  2a)  (12a  8)  a(3a  2)  4(3a  2)  (3a  2)(a  4) 4x2  3x  8x  6  (4x2  3x)  (8x  6)  x(4x  3)  2(4x  3)  (4x  3)(x  2) 2p2  6p  7p  21  (2p2  6p)  (7p  21)  2p(p  3)  7(p  3)  (p  3) (2p  7)

6. 2p2  14p  24 First factor out the GCF, 2. 2p2  14p  24  2( p2  7p  12) Now factor p2  7p  12. Since the lead coefficient is 1, find two factors of 12 whose sum is 7. The correct factors are 3 and 4. So, p2  7p  12  ( p  3)( p  4). Thus, 2p2  14p  24  2( p  3)( p  4).

82. 3b2  7b  12b  28  (3b 2  7b)  (12b  28)    

b(3b  7)  4(3b  7) b(3b  7)  4(1) (3b  7) b(3b  7)  (4) (3b  7) (3b  7) (b  4)

417

Chapter 9

7. 2x2  13x  20 Since ac  40 and b  13, make a list of positive factors of 40 find the pair whose sum is 13. Factors of 40 Sum of Factors 1, 40 41 2, 20 22 4, 10 14 5, 8 13 The correct factors are 5 and 8. 2x2  13x  20  2x2  mx  nx  20  2x2  5x  8x  20  (2x2  5x)  (8x  20)  x(2x  5)  4(2x  5)  (2x  5)(x  4) 8. 6x2  15x  9 First factor out the GCF, 3. 6x2  15x  9  3(2x2  5x  3) Now factor 2x2  5x  3. Since ac  6 and b  5, make a list of the factors of 6 and find the pair whose sum is 5. Factors of 6 Sum of Factors 1, 6 5 1, 6 5 2, 3 1 2, 3 1 The correct factors are 1 and 6. 2x2  5x  3  2x2  mx  nx  3  2x2  x  6x  3  (2x2  x)  (6x  3)  x(2x  1)  3(2x  1)  (2x  1)(x  3) Thus, 6x2  15x  9  3(2x  1)(x  3).

10.

2

x  3 Check:

2

6

3x2  11x  6  0

1 2

1 2

3x2  11x  6  0

?

?

2 2 3 3 2  11 3 #  6  0

3

3(3) 2  11(3)  6  0

149 2  223  6  0 4 3



22 3

?



18

3 

18 ?  3 18 ? 3

?

3(9)  33  6  0 ?

0

27  33  6  0

0

6  6  0

?

00✓

11.

00✓

2

10p  19p  7  0 (5p  7)(2p  1)  0 5p  7  0 or 2p  1  0 5p  7 2p  1 7 1 p5 p2 The solution set is

512, 75 6.

Check: 10p2  19p  7  0 10



19 2



14

Sum of Factors 139 139 68 68 31 31 23 23 13 13 4 4

175 22  19175 2  7  0 ? 49 133 10 1 25 2  5  7  0

?

2  12.

10p2  19p  7  0

112 22  19112 2  7  0 ? 1 19 10 1 4 2  2  7  0 5 2

9. Factor  4n  35. Since ac  140 and b  4, make a list of the pairs of factors of 140 and find the pair whose sum is 4.

14 ?  2 14 ? 2

?

10

98 5

0



133 5



35

0

5 

00✓ 6n2  7n  20 6n2  7n  20  0 (3n  4)(2n  5)  0 3n  4  0 or 2n  5  0 3n  4 2n  5 4

n3

5

35 ?  5 35 ? 5

00✓

5

n  2 5 4

6

Check: 6n2  7n  20

1 2

1 2

6

20 20

20

1254 2  352  20 75 2



?

35 ?  2 40 ?  2

20  20 ✓

418

6n2  7n  20

143 22  7143 2  20 16 28 ? 6 1 9 2  3  20

5 5 ? 6 2 2  7 2  20

6

?

32 3

0

0

The solution set is 2, 3 .

The correct factors are 10 and 14. 4n2  4n  35  4n2  mn  pn  35  4n2  10n  14n  35  (4n2  10n)  (14n  35)  2n(2n  5)  7(2n  5)  (2n  5) (2n  7)

Chapter 9

5

The solution set is 3, 3 .

4n2

Factors of 140 1, 140 1, 140 2, 70 2, 70 4, 35 4, 35 5, 28 5, 28 7, 20 7, 20 10, 14 10, 14

3x2  11x  6  0 (3x  2)(x  3)  0 or x  3  0 3x  2  0 3x  2 x  3



28 ?  3 60 ?  3

20

20  20 ✓

21. Among all pairs of factors of 40, choose 5 and 8, the pair of factors whose sum is 3. 2x2  3x  20  2x2  5x  8x  20  (2x2  5x)  (8x  20)  x(2x  5)  4(2x  5)  (2x  5)(x  4) 22. Among all pairs of negative factors of 70, choose 10 and 7, the pair of factors whose sum is 17. 5c2  17c  14  5c2  10c  7c  14  (5c2  10c)  (7c  14)  5c(c  2)  7(c  2)  (c  2)(5c  7) 23. Among all negative factors of 48, there are no pairs whose sum is 25. Thus, 3p2  25p  16 is prime. 24. Among all pairs of factors of 72, choose 12 and 6, the pair of factors whose sum is 6. 8y2  6y  9  8y2  12y  6y  9  (8y2  12y)  (6y  9)  4y(2y  3)  3(2y  3)  (2y  3)(4y  3) 25. Among all pairs of factors of 60, choose 15 and 4, the pair of factors whose sum is 11. 10n2  11n  6  10n2  15n  4n  6  (10n2  15n)  (4n  6)  5n(2n  3)  2(2n  3)  (2n  3)(5n  2) 26. Among all pairs of factors of 270, choose 10 and 27, the pair whose sum is 17. 15z2  17z  18  15z2  10z  27z  18  (15z2  10z)  (27z  18)  5z(3z  2)  9(3z  2)  (3z  2) (5z  9) 27. Among all pairs of factors of 168, choose 8 and 21, the pair whose sum is 13. 14x2  13x  12  14x2  8x  21x  12  (14x2  8x)  (21x  12)  2x(7x  4)  3(7x  4)  (7x  4) (2x  3) 28. First factor out the GCF, 2. 6r2  14r  12  2(3r2  7r  6) Now factor 3r2  7r  6. Among all pairs of factors of 18, choose 9 and 2, the pair whose sum is 7. 3r2  7r  6  3r2  9r  2r  6  3r(r  3)  2(r  3)  (r  3)(3r  2) Thus, 6r2  14r  12  2(r  3)(3r  2) . 29. First factor out the GCF, 5. 30x2  25x  30  5(6x2  5x  6) Now factor 6x2  5x  6. Among all pairs of factors of 36, choose 9 and 4, the pair whose sum is 5. 6x2  5x  6  6x2  9x  4x  6  3x(2x  3)  2(2x  3)  (2x  3)(3x  2) Thus, 30x2  25x  30  5(2x  3)(3x  2).

13. Use the model for vertical motion with h  0, v  8, and s  8. h  16t2  vt  s 0  16t2  8t  8 0  8(2t2  t  1) 0  8(2t  1) (t  1) 2t  1  0 or t  1  0 2t  1 t1 1

t  2 1

Only 1 is a valid solution since 2 represents a time before the vault was made. Thus the gymnast’s feet reach the mat in 1 s.

Pages 499–500

Practice and Apply

14. Among all positive pairs of factors of 10, choose 2 and 5, the pair of factors whose sum is 7. 2x2  7x  5  2x2  2x  5x  5  (2x2  2x)  (5x  5)  2x(x  1)  5(x  1)  (x  1)(2x  5) 15. Among all pairs of positive factors of 6, choose 2 and 3, the pair of factors whose sum is 5. 3x2  5x  2  3x2  2x  3x  2  (3x2  2x)  (3x  2)  x(3x  2)  1(3x  2)  (3x  2)(x  1) 16. Among all pairs of factors of 36, choose 4 and 9, the pair of factors whose sum is 5. 6p2  5p  6  6p2  4p  9p  6  2p(3p  2)  3(3p  2)  (3p  2)(2p  3) 17. Among all pairs of factors of 40, choose 4 and 10, the pair of factors whose sum is 6. 5d2  6d  8  5d2  4d  10d  8  d(5d  4)  2(5d  4)  (5d  4)(d  2) 18. Among all pairs of negative factors of 72, there are no pairs whose sum is 19. Thus, 8k2  19k  9 is prime. 19. Among all pairs of negative factors of 36, choose 6 and 6, the pair of factors whose sum is 12. 9g2  12g  4  9g2  6g  6g  4  (9g2  6g)  (6g  4)  3g(3g  2)  2(3g  2)  3g(3g  2)  2(1)(3g  2)  3g(3g  2)  2(3g  2)  (3g  2) (3g  2) or (3g  2) 2 20. Among all pairs of factors of 36, choose 3 and 12, the pair of factors whose sum is 9. 2a2  9a  18  2a2  12a  3a  18  (2a2  12a)  (3a  18)  2a(a  6)  3(a  6)  (a  6)(2a  3)

419

Chapter 9

30. 9x2  30xy  25y2  9x2  (30y)x  (25y2 ) Among all pairs of positive factors of 225y2, choose 15y and 15y, the pair of factors whose sum is 30y. 9x2  30xy  25y2  9x2  (30y)x  25y2  9x2  15yx  15yx  25y2  (9x2  15yx)  (15yx  25y2 )  3x(3x  5y)  5y(3x  5y)  (3x  5y)(3x  5y) or (3x  5y) 2 2 2 31. 36a  9ab  10b  36a2  (9b)a  (10b2 ) Among all pairs of positive factors of 360b2, choose 15b and 24b, the pair of factors whose sum is 9b. 36a2  9ab  10b2  36a2  9ba  10b2  36a2  15ba  24ba  10b2  (36a2  15ba)  (24ba  10b2 )  3a(12a  5b)  2b(12a  5b)  (12a  5b) (3a  2b) 32. List all factors of 24 and the sums of the pairs of factors. Factors of 24 Sum of Factors 1, 24 25 1, 24 25 2, 12 14 2, 12 14 3, 8 11 3, 8 11 4, 6 10 4, 6 10 The possible values for k are 25, 14, 11, and 10. 33. List all factors of 30 and the sums of the pairs of factors. Factors of 30 Sum of Factors 1, 30 31 1, 30 31 2, 15 17 2, 15 17 3, 10 13 3, 10 13 5, 6 11 5, 6 11 The possible values for k are 31, 17, 13, and 11. 34. Because ac  2k is positive and b  12 is positive, m and n must both be positive and mn must be even. List all pairs of positive numbers whose sum is 12 and list the products of the pairs. Possible values for 2k will be the even products.

Exercises 35–48 For checks, see students’ work. 35. 5x2  27x  10  0 (5x  2)(x  5)  0 or x  5  0 5x  2  0 5x  2 x  5

36.

55, 25 6

3x2  5x  12  0 (3x  4)(x  3)  0 or x  3  0 3x  4  0 3x  4 x3 4

x  3

37.

543, 36

24x2  11x  3  3x 24x2  14x  3  0 (4x  3)(6x  1)  0 4x  3  0 or 6x  1  0 4x  3 6x  1 3

1

x4

38.

516, 34 6

x  6

17x2  11x  2  2x2 15x2  11x  2  0 (5x  2)(3x  1)  0 5x  2  0 or 3x  1  0 5x  2 3x  1

513, 25 6

2

1

x5

x3

39. 14n2  25n  25 14n2  25n  25  0 (7n  5)(2n  5)  0 or 7n  5  0 7n  5 5 n  7

40.

2n  5  0 2n  5 5 n2

557, 52 6

12a2  13a  35 12a  13a  35  0 (4a  5)(3a  7)  0 or 3a  7  0 4a  5  0 4a  5 3a  7 2

5

a  4

41.

554, 73 6

2

x  3

523, 36 420

7

a3

6x2  14x  12 6x  14x  12  0 2(3x2  7x  6)  0 3x2  7x  6  0 (3x  2)(x  3)  0 or x  3  0 3x  2  0 3x  2 x3 2

Addends of 12 Product of Addends 1, 11 11 2, 10 20 (even) 3, 9 27 4, 8 32 (even) 5, 7 35 6, 6 36 (even) The possible values of 2k are 20, 32, and 36. Thus, the possible values of k are 10, 16, and 18.

Chapter 9

2

x  5

42.

21x2  6  15x 21x  15x  6  0 3(7x2  5x  2)  0 7x2  5x  2  0 (7x  2)(x  1)  0 or x  1  0 7x  2  0 7x  2 x1

48.

2

2

x  7

43.

527, 16

2

44.

512, 23 6

1

24x2  46x  18 24x  46x  18  0 2(12x2  23x  9)  0 12x2  23x  9  0 (3x  1) (4x  9)  0 3x  1  0 or 4x  9  0 3x  1 4x  9 2

1

45.

513, 94 6 12

1

x2 12

x2 12





2x 3

2x 3

x2

9

x4

40

2

 4  12(0)

5

t2

 8x  48  0 (x  12) (x  4)  0 x  12  0 or x  4  0 x  12 x  4 {4, 12} 46.

1

t

t2  6 

2

The negative value represents a time before the diver leaps. The diver will enter the water after 5 or 2.5 seconds. 2 52. Use the model for vertical motion with h  30, v  56, and s  6. h  16t2  vt  s 30  16t2  56t  6 0  16t2  56t  24 0  8(2t2  7t  3) 0  8(2t  1)(t  3) 2t  1  0 or t  3  0 2t  1 t3 1 t2

35 6

t 6 t2  6  6

1356 2

6t2  t  35 6t  t  35  0 (3t  7) (2t  5)  0 or 2t  5  0 3t  7  0 3t  7 2t  5 7 5 t  3 t2 2

47.

573, 52 6

1

The first time, t  2, represents how long it takes the hook to reach a height of 30 feet on its way up. The second time, t  3, represents the time the hook anchors on the ledge on its way down. Thus, the hook is in the air for 3 seconds.

(3y  2) (y  3)  y  14 3y2  9y  2y  6  y  14 3y2  11y  6  y  14 3y2  10y  8  0 (3y  2) ( y  4)  0 3y  2  0 or y  4  0 3y  2 y  4

54, 23 6

1

a2

49. Write an equation for the area of the smaller rectangle and solve for x. (9  2x)(7  2x)  35 63  18x  14x  4x2  35 4x2  32x  63  35 4x2  32x  28  0 4(x2  8x  7)  0 4(x  7)(x  1)  0 x  7  0 or x  1  0 x7 x1 If x  7, the dimensions of the smaller rectangle would be negative. Thus, the width of each strip is 1 inch. 50. The dimensions of the new rectangle are (7  2) in. by (9  2) in. or 5 in. by 7 in. 51. Use the model for vertical motion with h  0, v  8, and s  80. h  16t2  vt  s 0  16t2  8t  80 0  8(2t2  t  10) 0  8(2t  5) (t  2) 2t  5  0 or t  2  0 2t  5 t  2

x2

x  3

7

a2

512, 72 6

24x2  30x  8  2x 24x2  28x  8  0 4(6x2  7x  2)  0 6x2  7x  2  0 (3x  2)(2x  1)  0 3x  2  0 or 2x  1  0 3x  2 2x  1 x3

(4a  1)(a  2)  7a  5 4a2  8a  a  2  7a  5 4a2  9a  2  7a  5 4a2  16a  7  0 (2a  7)(2a  1)  0 2a  7  0 or 2a  1  0 2a  7 2a  1

2

y3

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60. (2k  9) (3k  2)  0 or 2k  9  0 2k  9

53. Answers should include the following. • 2x  3 by x  2 • With algebra tiles, you can try various ways to make a rectangle with the necessary tiles. Once you make the rectangle, however, the dimensions of the rectangle are the factors of the polynomial. In a way, you have to go through the guess-and-check process whether you are factoring algebraically or geometrically (using algebra tiles.) Guess: (2x  1)(x  3) The guess is incorrect because one more x tile is needed to complete the rectangle.

9

k  2

592, 23 6

3k  2  0 3k  2 2

k  3

61. 12u  u2 0  u2  12u 0  u(u  12) u  0 or u  12  0 u  12 {0, 12} 62. Two points are (2, 83) and (5, 185). Find the slope. y2  y1

x

2

x x x

x

2

mx

x x

2



x

2

x

x x x

x

1 1

x x x

1 1

1 1

2



x x 1 1 1

1 1 1

54. D; 2p2  p  3  0 (2p  3)(p  1)  0 2p  3  0 2p  3

or

p10 p  1

63. 64. 65. 66. 67. 68. 69. 70.

3

p2 55. B; Solve the equation for h  338. h  16t2  48t  402 338  16t2  48t  402 0  16t2  48t  64 0  16(t2  3t  4) 0  16(t  1)(t  4) t10 or t  4  0 t  1 t4 Choose the positive solution. The ball will be 338 feet from the ground after 4 seconds.

Page 500

Chapter 9

Find the y-intercept. y  mx  b 83  34(2)  b 83  68  b 15  b Write the equation. y  mx  b y  34x  15 42  16 S 116  4 72  49 S 149  7 62  36 S 136  6 52  25 S 125  5 102  100 S 1100  10 112  121 S 1121  11 132  169 S 1169  13 152  225 S 1225  15

Page 500

Practice Quiz 2

1. Among all pairs of factors of 72, choose 4 and 18, the pair of factors whose sum is 14. x2  14x  72  (x  4)(x  18) 2. Among all pairs of factors of 280, choose 20 and 14, the pair whose sum is 6. 8p2  6p  35  8p2  20p  14p  35  (8p2  20p)  (14p  35)  4p(2p  5)  7(2p  5)  (2p  5) (4p  7) 3. Among all pairs of negative factors of 80, choose 4 and 20, the pair of factors whose sum is 24. 16a2  24a  5  16a2  4a  20a  5  (16a2  4a)  (20a  5)  4a(4a  1)  5(4a  1)  (4a  1) (4a  5) 4. Among all pairs of negative factors of 52, choose 4 and 13, the pair of factors whose sum is 17. n2  17n  52  (n  4)(n  13)

Maintain Your Skills

56. Among all pairs of factors of 21, choose 3 and 7, the pair of factors whose sum is 4. a2  4a  21  (a  m)(a  n)  (a  3)(a  7) 57. Among all pairs of positive factors of 2, there is no pair of factors whose sum is 2. Thus, t2  2t  2 is prime. 58. Among all pairs of factors of 44, choose 4 and 11, the pair of factors whose sum is 15. d2  15d  44  (d  m)(d  n)  (d  4)(d  11) Exercises 59–61 For checks, see students’ work. 59. (y  4)(5y  7)  0 y  4  0 or 5y  7  0 y4 5y  7

575, 46

 x1

185  83 5  2 102 or 34 3

7

y  5

422

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5. First factor out the GCF, 2. 24c2  62c  18  2(12c2  31c  9) Now factor 12c2  31c  9. Among all pairs of positive factors of 108, choose 4 and 27, the pair whose sum is 31. 12c2  31c  9  12c2  4c  27c  9  4c(3c  1)  9(3c  1)  (3c  1)(4c  9) Thus, 24c2  62c  18  2(3c  1)(4c  9). 6. First factor out the GCF, 3. 3y2  33y  54  3(y2  11y  18) Now factor y2  11y 18. Since the lead coefficient is 1, find two factors of 18 whose sum is 11. The correct factors are 2 and 9. So, y2  11y  18  (y  2)( y  9). Thus, 3y2  33y  54  3(y  2)( y  9). 7. b2  14b  32  0 (b  2)(b  16)  0 b  2  0 or b  16  0 b2 b  16 {16, 2} Check: b2  14b  32  0 ?

2

256  224  32  0

4  28  32  0

32  32  0

32  32  0

Check: 6a2  25a  14 ? 2 2 6 3 2  25 3  14

12 12 4 ? 50 42 619 2  3  3 8 3

Page 501

3

4

y  4

y3

Check:

1 2

12

1169 2  214  484  0 27 4



21 4

?



48 ?  4

0

00 ✓



147 2



Algebra Activity

12y2  7y  12  0

143 22  7143 2  12  0 16 28 36 ? 12 1 9 2  3  3  0 ?

12

64 3



28 3



36 ?  3

Check for Understanding

1. The binomial is the difference of two terms, each of which is a perfect square. 2. Sample answer: x2  25  x2  52  (x  5)(x  5) 3. Yes, you can use the differences of squares pattern to factor 3n2  48 after you factor out the GCF. 3n2  48  3(n2  16)  3(n2  42 )  3(n  4) (n  4) 4. Manuel; 4x2  y2 is not the difference of squares. 5. n2  81  n2  92  (n  9) (n  9) 6. 4  9a2  22  (3a) 2  (2  3a) (2  3a) 7. 2x5  98x3  2x3 (x2  49)  2x3 (x2  72 )  2x3 (x  7) (x  7) 4  2y4  2(16x4  y4 ) 8. 32x  2[ (4x2 ) 2  (y2 ) 2 ]  2(4x2  y2 ) (4x2  y2 )  2(4x2  y2 ) [ (2x) 2  y2 ]  2(4x2  y2 ) (2x  y) (2x  y) 2  27 is not the difference of perfect squares, 9. 4t and 4t2 and 27 have no common factors. Thus 4t2  27 is prime. 10. x3  3x2  9x  27  (x3  3x2 )  (9x  27)  x2 (x  3)  9(x  3)  (x  3) (x2  9)  (x  3) (x2  32 )  (x  3) (x  3) (x  3)  (x  3) (x  3) (x  3) or (x  3) (x  3) 2

x2  45  18x x  45  18x  0 x2  18x  45  0 (x  3)(x  15)  0 x  3  0 or x  15  0 x3 x  15 {3, 15} Check: x2  45  18x x2  45  18x ? ? 2 (15) 2  45  18(15) (3)  45  18(3) ? ? 225  45  270 9  45  54 54  54 ✓ 270  270 ✓ 9. 12y2  7y  12  0 (4y  3)(3y  4)  0 or 3y  4  0 4y  3  0 4y  3 3y  4

1 2

147 2

Factoring Differences of Squares

Pages 504–505

00 ✓

2

3 3 12 4 2  7 4  12  0

12 12 49 ? 175 28 61 4 2  2  2

6a2  25a  14 ? 7 7 6 2 2  25 2  14

1. (a  b)(a  b) 2. Since a2  b2 and (a  b)(a  b) describe the same area, a2  b2  (a  b)(a  b).

8.

?

8

3 ✓

9-5

?

00 ✓

12y2  7y  12  0

7

a2

?

?

534, 43 6

2

a3

523, 72 6

?

(2)  14(2)  32  0

6a2  25a  14 6a  25a  14  0 (3a  2)(2a  7)  0 3a  2  0 or 2a  7  0 3a  2 2a  7 2

b2  14b  32  0 2

(16)  (14)(16)  32  0 ?

10.

0

00 ✓

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11.

4y2  25 4y  25  0 (2y) 2  52  0 (2y  5)(2y  5)  0 or 2y  5  0 2y  5

Check: 121a  49a3

2

5

2y  5  0 2y  5

6

?

00 ✓ 121a 

4

121

4y2  25

1 2 25 4 1 4 2  25

152 22  25 25 ? 4 1 4 2  25

? 5 2 2  25 ?

1331 7

?

4

1

5

1 1

17  68k2  0

1 2 1 ? 17  68 1 4 2  0

112 22  0 1 ? 17  68 1 4 2  0

13.

Pages 505–506

17  17  0 00 ✓

1

x2 

1

x60

1

x60

or 1

x  6

1

5

1 1

x6

6

The solution set is 6, 6 . Check: 1

1

x2  36  0

x2  36  0

116 22  361  0

116 22  361  0

?

1 36

1

?

?

1 36

 36  0 00 ✓

14.

1

?

 36  0

00 ✓ 121a  49a3

49a3  121a  0 a(49a2  121)  0 2

5

11

a  7 11

The solution set is  7 , 0,

Chapter 9

11 7

6

 72

1 2  2(72) 1 2 x 2

Practice and Apply

26. 8d2  18  2(4d2  9)  2[ (2d) 2  32 ]  2(2d  3) (2d  3)

2

a[ (7a)  11 ]  0 a(7a  11) (7a  11)  0 or a  0 or 7a  11  0 a0 7a  11



16. x2  49  x2  72  (x  7)(x  7) 17. n2  36  n2  62  (n  6)(n  6) 18. 81  16k2 is prime. 19. 25  4p2  52  (2p) 2  (5  2p) (5  2p) 20. 16  49h2  49h2  16  (7h) 2  42  (7h  4) (7h  4) 21. 9r2  121  121  9r2  112  (3r) 2  (11  3r) (11  3r) 22. 100c2  d2  (10c) 2  d2  (10c  d) (10c  d) 23. 9x2  10y2 prime 24. 144a2  49b2  (12a) 2  (7b) 2  (12a  7b) (12a  7b) 2 2 25. 169y  36z  (13y) 2  (6z) 2  (13y  6z) (13y  6z)

x2  36  0

116 22  0 1x  16 21x  16 2  0

1331 7

?

?

17  17  0 00 ✓



x2  144 x  144  0 (x  12)(x  12)  0 or x  12  0 x  12  0 x  12 x  12 Since length cannot be negative, the only reasonable solution is 12. The square is 12 in. by 12 in.

1

17  68

?



2

17  68k2  0

? 1 17  68 2 2  0

1331 7



?

2

The solution set is 2, 2 . Check:

7

49a3

1 2 x 2

k2

6



15. The area of the square is x  x or x2 square inches, 1 1 and the area of the triangle is 2  x  x or 2x2 square inches. Thus, the area of the shaded 1 1 region is x2  2x2 or 2x2 square inches.

25  25 ✓ 25  25 ✓ 20 17  68k 12. 17(1  4k2 )  0 17 [12  (2k) 2 ]  0 17(1  2k)(1  2k)  0 or 1  2k  0 1  2k  0 2k  1 2k  1 k  2

1 2 1331 49 1 49  7 2

1117 2  491117 23 1331 ? 1331  49 1 49  7 2 7

The solution set is 2, 2 . Check: 4y2  25

1331 ?  7  1331 ?

0  49(0)

y2

2

11 ? 11 121  7  49  7 3

?

5

5 5

1

?

121(0)  49(0) 3

5

y  2

121a  49a3

7a  11  0 7a  11 a

27. 3x2  75  3(x2  25)  3(x2  52 )  3(x  5) (x  5)

11 7

28. 8z2  64  8(z2  8) 29. 4g2  50  2(2g2  25)

424

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30. 18a4  72a2  18a2 (a2  4)  18a2 (a2  22 )  18a2 (a  2) (a  2)

39.

31. 20x3  45xy2  5x(4x2  9y2 )  5x[ (2x) 2  (3y) 2 ]  5x(2x  3y)(2x  3y) 32. n3  5n2  4n  20  (n3  5n2 )  (4n  20)  n2 (n  5)  (4)(n  5)  (n  5)(n2  4)  (n  5)(n2  22 )  (n  5)(n  2)(n  2) 33. (a  b) 2  c2  [ (a  b)  c ] [ (a  b)  c ]  (a  b  c) (a  b  c) Exercises 34–45 For checks, see students’ work. 25x2  36 34. 2  36  0 25x (5x) 2  62  0 (5x  6)(5x  6)  0 or 5x  6  0 5x  6  0 5x  6 5x  6 x

5 6

6 5

x

36.

37.

5 83 6

8

40.

41.

6 5

2  3n  0 3n  2

43.

2

n3

44.

5  2a  0 2a  5

5 27 6

2

5 186

r

630

or

r

 6

3  6

r  18

r  18

1 2 x  25 4 1 2 x  52 2

0

0 1 2 112x  52112x  52  0 50

1 x 2

or

50 1 x 2

 5

5

x  10

12d3  147d  0 3d(4d2  49)  0 3d(2d  7) (2d  7)  0 3d  0 or 2d  7  0 d0 2d  7

572, 0, 72 6

7

5

n  3

2d  7  0 2d  7 7

d  2

18n3  50n  0 2n(9n2  25)  0 2n(3n  5) (3n  5)  0 2n  0 or 3n  5  0 n0 3n  5

553, 0, 53 6

or

d2

or

3n  5  0 3n  5 5

n3

x3  4x  12  3x2 x  4x  12  3x2  0 x3  3x2  4x  12  0 3 (x  3x2 )  (4x  12)  0 x2 (x  3)  4(x  3)  0 (x  3) (x2  4)  0 (x  3) (x  2) (x  2)  0 or x  2  0 or x  2  0 x30 x  2 x2 x  3 {3, 2, 2} 3

5

a2

w2 

w  7

13r 22  0 16  3r 2 16  3r 2  0

{10}

4

or

1

36  9r2  0

r 3

w2  49  0

2

5 109 6

r

42.

5

w70

9

p  10

x  10

2

127 22  0 1w  27 21w  27 2  0

p0 9

1 x 2

a  2 38.

9 10

p  10

1 x 2

n  3

50  8a2  0 2(25  4a2 )  0 2 [52  (2a) 2 ]  0 2(5  2a)(5  2a)  0 or 5  2a  0 2a  5

or

9 10

630

8

5 23 6

p0

62 

y3

12  27n2  0 3(4  9n2 )  0 3 [22  (3n) 2 ]  0 3(2  3n)(2  3n)  0 or 2  3n  0 3n  2

1 2  p2  0 1109  p21109  p2  0 p

9y2  64 9y2  64  0 (3y) 2  82  0 (3y  8)(3y  8)  0 or 3y  8  0 3y  8  0 3y  8 3y  8 y  3

 p2  0

9 10

6 5

35.

81 100 9 2 10

2

w70 2

w7

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45.

36x  16x3  9x2  4x4  9x2  36x  0 4x  3 2 x(4x  16x  9x  36)  0 x [ (4x3  16x2 )  (9x  36) ]  0 x [4x2 (x  4)  9(x  4) ]  0 x [ (x  4)(4x2  9) ]  0 x(x  4) (2x  3)(2x  3)  0 x  0 or x  4  0 or 2x  3  0 or 2x  3  0 x4 2x  3 2x  3 4

Since length cannot be negative, the only reasonable solution is 3. The box is 3 in. by 12 in. by 2 in. 51. The flaw is in line 5. Since a  b, a  b  0. Therefore, dividing by a  b is dividing by zero, which is undefined. 52. Answers should include the following: • 1 foot • To find the hang time of a student athlete who attains a maximum height of 1 foot, solve the equation 4t2  1  0. You can factor the left side using the difference of squares pattern since 4t2 is the square of 2t and 1 is the square of 1. Thus the equation becomes (2t  1)(2t  1)  0. Using the Zero Product Property, each factor can be set equal to zero, 1 1 resulting in two solutions, t  2 and t  2. Since time cannot be negative, the hang time 1 is 2 second.

16x3

3

x  2

532, 0, 32, 46

3

x2

46. Use factoring by grouping. a2  b2  a2  ab  ab  b2  (a2  ab)  (ab  b2 )  a(a  b)  b(a  b)  (a  b)(a  b) 47. Replace b with 3600 in the equation and solve for c. 900c2  b 900c2  3600 2 900c  3600  0 900(c2  4)  0 900(c  2) (c  2)  0 or c  2  0 c20 c  2 c2 Since circumference cannot be negative, the only reasonable solution is 2 in. 1

1

53. A; 25b2  1  (5b) 2  1  (5b  1) (5b  1)  (5b  1) (5b  1) 54. Let x represent the length of a side of the larger square. Then the perimeter of the larger square is 4x in., and the perimeter of the smaller square is 68  4x in. Thus, the length of a side of the 1 smaller square is 4 (68  4x) or 17  x in. Write an equation for the area between the two squares. x2  (17  x) 2  17 2 2 x  (17  2(17)(x)  x2 )  17 x2  (289  34x  x2 )  17 x2  289  34x  x2  17 34x  289  17 34x  306 x9 The length of a side of the larger square is 9 in.

48. p  2dv2  2dv2 1

1

1

2

2

 2d v21  v22 1  2d(v  v )(v  v ) 1 2 1 2 49. Replace d with 54 in the equation and solve for s.

24

1

1 2 s 24 1 2 s 24 1 2 s 24

d  54

2  24(54)

2

s  1296 s2  1296  0 s2  362  0 (s  36) (s  36)  0 or s  36  0 s  36  0 s  36 s  36 Since speed cannot be negative, the only reasonable answer is 36. The car that left the 54-foot skid marks was traveling approximately 36 mph when the brakes were applied. 50. The volume of the box is x(x  9)(x  1). x(x  9) (x  1)  72 x(x2  x  9x  9)  72 x(x2  8x  9)  72 3  8x2  9x  72  0 x x2 (x  8)  9(x  8)  0 (x  8)(x2  9)  0 (x  8)(x  3) (x  3)  0 or x  3  0 or x  3  0 x80 x  8 x  3 x3

Chapter 9

Page 506

Maintain Your Skills

55. Among all pairs of positive factors of 14, there are no pairs whose sum is 5. Thus, 2n2  5n  7 is prime. 56. Among all pairs of negative factors of 24, choose 3 and 8, the pair whose sum is 11. 6x2  11x  4  6x2  3x  8x  4  3x(2x  1)  4(2x  1)  (2x  1) (3x  4) 57. Among all pairs of factors of 210, choose 6 and 35, the pair whose sum is 29. 21p2  29p  10  21p2  6p  35p  10  3p(7p  2)  5(7p  2)  (7p  2) (3p  5) Exercises 58–60 For checks, see students’ work. 58. y2  18y  32  0 ( y  2)( y  16)  0 or y  16  0 y20 y  2 y  16 {16, 2}

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k2  8k  15 k  8k  15  0 (k  3)(k  5)  0 k  3  0 or k  5  0 k3 k5 {3, 5} 60. b2  8  2b 2  8  2b  0 b b2  2b  8  0 (b  2)(b  4)  0 or b  4  0 b20 b4 b  2 {2, 4} 61. Let x represent Amy’s score on the fourth biology test. Her average must be between 88 and 92, inclusive.

Page 507

59.

2

88  88  88  4(88) 

88  90  91  x  92 4 269  x  92 4 269  x 269  x and 4 4 269  x 269  x 4 4 4 4

1

2

1

9-6  92

Page 512

2  4(92)

63. 5  10r 7 2 10r 7 7 7 10

6 10

8 10

12 10

1

5. 6. 7. 8.

14 10

64. 13x  3 6 23 13x 6 26 x 6 2 6

4

2

0

2

Check for Understanding

x3  5x2  4x  20  x2 (x  5)  4(x  5)  (x  5) (x2  4)  (x  5) (x  2) (x  2) 4. Yes; The first term is a perfect square: y2. The last term is a perfect square: 42. The middle term is 2(y)(4) or 8y.

1 2 3 4 5 6 7 8 9 10 11 12

4 10

Perfect Squares and Factoring

1. Determine if the first term is a perfect square. Then determine if the last term is a perfect square. Finally, check to see if the middle term is equal to twice the product of the square roots of the first and last terms. 2. Never; (a  b)2  a2  2ab  b2 3. Sample answer: x3  5x2  4x  20

269  x  368 352  269  x x  99 83  x 83  x  99 Her test score must be between 83 and 99, inclusive. Exercises 62–64 For checks, see students’ work. 62. 6  3d  12 18  3d 6d

r 7

Reading Mathematics

1. (1) explains how to factor a perfect square trinomial; (2) summarizes methods used to factor polynomials; (3) explains how to solve equations involving perfect squares using the Square Root Property 2. GCF, perfect square trinomial; x2  bx  c, ax2  bx  c 3. a greatest common factor 4. For any number n, where n is positive, the square of x equals n, then x equals plus or minus the square root of n. This property can be applied to the equation (a  4)2  49 since the variable x  a  4 and n  49 in the equation x2  n.

9.

4

65. (x  1)(x  1)  (x  1) 2  x2  2x(1)  12  x2  2x  1 66. (x  6)(x  6)  (x  6) 2  x2  2x(6)  62  x2  12x  36 67. (x  8) 2  x2  2x(8)  82  x2  16x  64

10.

y2  8y  16  y2  2(y) (4)  42  (y  4) 2 No; the last term, 10, is not a perfect square. 2x2  18  2(x2  9) c2  5c  6  (c  3) (c  2) 5a3  80a  5a(a2  16)  5a(a  4) (a  4) 2 8x  18x  35  8x2  28x  10x  35  4x(2x  7)  5(2x  7)  (2x  7) (4x  5) There is no pair of factors of 36 whose sum is 12. 9g2  12g  4 is prime.

11. 3m3  2m2n  12m  8n  m2 (3m  2n)  4(3m  2n)  (3m  2n) (m2  4)  (3m  2n) (m  2) (m  2)

68. (3x  4) (3x  4)  (3x  4) 2  (3x) 2  2(3x)(4)  42  9x2  24x  16 2 2 69. (5x  2)  (5x)  2(5x)(2)  22  25x2  20x  4 70. (7x  3) 2  (7x) 2  2(7x)(3)  32  49x2  42x  9

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12.

4y2  24y  36  0 4(y2  6y  9)  0 2  2(y)(3)  32 ]  0 4[ y 4(y  3) 2  0 y30 y  3 {3}

Ignore the negative solution. It took about 3.35 seconds for the objects to hit the ground.

Pages 512–514

4y2  24y  36  0

Check:

?

4(3) 2  24(3)  36  0 ?

36  72  36  0 00 ✓ 13. 3n2  48 3n2  48  0 3(n2  16)  0 3(n  4)(n  4)  0 or n  4  0 n40 n4 n  4 {4, 4} Check: 3n2  48

3n2  48

? 2

?

48 3(4) 2  48 3(4) 48  48 ✓ 48  48 ✓ 14. a2  6a  9  16 a2  2(a) (3)  32  16 (a  3) 2  16 a  3  116 a  3  4 a3  4 a  3  4 or a  3  4  1 7 {1, 7} Check: a2  6a  9  16 ?

(1) 2  6(1)  9  16 16  16 ✓ a2  6a  9  16 ? 72  6(7)  9  16 16  16 ✓ 2 15. (m  5)  13 m  5  113 m  5  113 {5  113} or {1.4, 8.6} (m  5) 2  13 ? (5  113  5) 2  13 ? ( 113) 2  13 13  13 ✓ (m  5) 2  13 ? (5  113  5) 2  13 ? (113) 2  13 13  13 ✓ 16. Let h  0 and h0  180 and solve for t. h  16t2  h0

x 1 2 3 4 5

Check:

Area 16 1 36 121 256

Perimeter 16 4 24 44 64

If the pattern in the table continues, the perimeter will continue to get larger for increasing positive integer values of x. The least perimeter is 4 meters. 25. 4k2  100  4(k2  25)  4(k  5) (k  5) 26. 9x2  3x  20  9x2  15x  12x  20  3x(3x  5)  4(3x  5)  (3x  5) (3x  4) 27. x2 + 6x  9 is prime.

0  16t2  180 16t2  180 t2  11.25 t  111.25 t  3.35

Chapter 9

Practice and Apply

17. No; the middle term is not 2(x)(9). 18. Yes; The first term is a perfect square: a2. The third term is a perfect square: 122. The middle term is 2(a)(12). a2  24a  144  a2  2(a) (12)  122  (a  12) 2 19. Yes; The first term is a perfect square: (2y)2. The third term is a perfect square: 112. The middle term is 2(2y)(11). 4y2  44y  121  (2y) 2  2(2y)(11)  112  (2y  11) 2 20. No; the first term, 2c2, is not a perfect square. 21. 9n2  49  42n  9n2  42n  49 Yes; The first term is a perfect square: (3n)2. The third term is a perfect square: 72. The middle term is 2(3n)(7). 9n2  42n  49  (3n) 2  2(3n) (7)  72  (3n  7) 2 22. Yes; The first term is a perfect square: (5a)2. The third term is a perfect square: (12b)2. The middle term is 2(5a)(12b). 25a2  120ab  144b2  (5a) 2  2(5a)(12b)  (12b) 2  (5a  12b) 2 23. To find the radius of the circle, write the expression for the area in the form r2. (16x2  80x  100)   [ (4x) 2  2(4x) (10)  102 ]  (4x  10) 2 The radius is 4x  10 inches, so the diameter is 2(4x  10) or 8x  20 inches. 24. Make a table.

428

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28. 50g2  40g  8  2(25g2  20g  4)  2[ (5g) 2  2(5g)(2)  22 ]  2(5g  2) 2 3 2 29. 9t  66t  48t  3t(3t2  22t  16)  3t(3t2  2t  24t  16)  3t [t(3t  2)  8(3t  2) ]  3t(3t  2)(t  8) 30. 4a2  36b2  4(a2  9b2 )  4[ a2  (3b) 2 ]  4(a  3b) (a  3b) 31.

32. 33.

34.

35.

36. 37.

38. 39.

40.

20n2

The length s must be positive. The expression 7 4x  7 is positive when x 7 4 and 7  4x is 7 positive when x 6 4. Write two expressions for the area of the rectangle. 7

If x 7 4, s  4x  7.

11 2

A  (s  3) 2 s

31

 [ (4x  7)  3] 2 (4x  7)  (4x  4)

3

1 (4x 2

 7)

4

4

 (2x  2) (4x  7)  8x2  14x  8x  14  8x2  22x  14 in2

2(10n2

 34n  6   17n  3)  2(10n2  15n  2n  3)  2[5n(2n  3)  (2n  3) ]  2(2n  3)(5n  1) 5y2  90  5(y2  18) 24x3  78x2  45x  3x(8x2  26x  15)  3x(8x2  6x  20x  15)  3x [2x(4x  3)  5(4x  3) ]  3x(4x  3)(2x  5) 18y2  48y  32  2(9y2  24y  16)  2[ (3y) 2  2(3y)(4)  42 ]  2(3y  4) 2 2  75  27g2  90g  75 90g  27g  3(9g2  30g  25)  3[ (3g) 2  2(3g)(5)  52 ]  3(3g  5) 2 2 45c  32cd  c(45c  32d) 4a3  3a2b2  8a  6b2  a2 (4a  3b2 )  2(4a  3b2 )  (4a  3b2 )(a2  2) 5a2  7a  6b2  4b is prime. x2y2  y2  z2  x2z2  y2 (x2  1)  z2 (1  x2 )  y2 (x2  1)  z2 (x2  1)  (x2  1)(y2  z2 )  (x  1)(x  1)(y2  z2 ) 4m4n  6m3n  16m2n2  24mn2  2mn(2m3  3m2  8mn  12n)  2mn [m2 (2m  3)  4n(2m  3) ]

7

If x 6 4, s  7  4x.

11 2

A  (s  3) 2 s

31

 [ (7  4x)  3] 2 (7  4x)  (10  4x)

3

1 (7 2

 4x)

4

4

 (5  2x) (7  4x)  35  20x  14x  8x2  8x2  34x  35 in2 Exercises 43–54 For checks, see students’ work. 43. 3x2  24x  48  0 3(x2  8x  16)  0 2 3[x  2(x) (4)  42 ]  0 3(x  4) 2  0 x40 x  4 {4} 44. 7r2  70r  175 2  70r  175  0 7r 7(r2  10r  25)  0 7[r2  2(r) (5)  52 ]  0 7(r  5) 2  0 r50 r5 {5} 45. 49a2  16  56a 2 49a  56a  16  0 (7a) 2  2(7a) (4)  42  0 (7a  4) 2  0 7a  4  0 7a  4

 2mn(2m  3) (m2  4n) 41. Factor the polynomial. x3y  63y2  7x2  9xy3  x3y  9xy3  7x2  63y2  xy(x2  9y2 )  7(x2  9y2 )  (x2  9y2 )(xy  7)  (x  3y)(x  3y)(xy  7) The dimensions are x  3y m, x  3y m, and xy  7 m. 42. Express the polynomial for the area of the square in the form s2. 16x2  56x  49  (4x) 2  2(4x)(7)  72  (4x  7) 2 or 16x2  56x  49  49  56x  16x2  72  2(7)(4x)  (4x) 2  (7  4x) 2

46.

547 6

18y2  24y  8  0 2(9y2  12y  4)  0 2  2(3y) (2)  22 ]  0 2[ (3y) 2(3y  2) 2  0 3y  2  0 3y  2

523 6 429

4

a7

2

y  3

Chapter 9

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2

1

L

y2  3y  9  0

47.

y2  2(y)

55. B  16 (D2  8D  16)

113 2  113 22  0 1y  13 22  0

L

B  16 [D2  2(D)(4)  42 ] L

B  16 (D  4) 2 56. Replace B with 256 and L with 16 and solve for D.

1

y30

56

1

y3

16

256  16 (D  4) 2

1 3

48.

4

4

a2  5a  25  0

a2  2(a)

125 2  125 22  0 1a  25 22  0 2

a50

525 6

2

a  5

z2  2z  1  16 z  2(z)(1)  12  16 (z  1) 2  16 z  1  116 z  1  4 z  1  4 z  1  4 or 3 {5, 3} 50. x2  10x  25  81 2  2(x)(5)  52  81 x (x  5) 2  81 x  5  181 x  5  9 x  5  9 x  5  9 or 4 {14, 4} 51. (y  8) 2  7 y  8  17 49.

57.

2

z  1  4  5

58.

x  5  9  14

59.

y  8  17 {8  17} or {5.4, 10.6} 52. (w  3) 2  2 w  3  12 w  3  12 {3  12} or {4.4, 1.6} 53. p2  2p  1  6 2  2(p) (1)  12  6 p (p  1) 2  6 p  1  16 p  1  16 {1  16} or {3.4, 1.4} 54. x2  12x  36  11 2  2(x)(6)  62  11 x (x  6) 2  11 x  6  111 x  6  111 {6  111} or {2.7, 9.3}

Chapter 9

60.

256  (D  4) 2 1256  D  4 16  D  4 4  16  D D  4  16 or D  4  16  20  12 Since diameter must be positive, 20 is the only reasonable solution. The logs should be 20 inches in diameter. Let h  0 and t  3. Solve for s. h  16t2  s 0  16(3) 2  s 0  16(9)  s 0  144  s 144  s The starting height of the car should be at least 144 feet. Let h  0 and s  160. Solve for t. h  16t2  s 0  16t2  160 16t2  160 t2  10 t  110 or 3.16 Ignore the negative solution. The riders will be in free fall about 3.16 seconds. h  16t2  vt  s 70  16t2  64t  6 0  16t2  64t  64 0  16(t2  4t  4) 0  16[ t2  2(t)(2)  22 ] 0  16(t  2) 2 0t2 2t Yes; the acrobat will reach a height of 70 feet after 2 seconds. x2  kx  64  x2  kx  82 This will be a perfect square trinomial if kx  2(x)(8) or k  16x. Thus, k can be 16 or 16.

61. 4x2  kx  1  (2x) 2  kx  12 This will be a perfect square trinomial if kx  2(2x)(1) or kx  4x. Thus, k can be 4 or 4. 62. 25x2  kx  49  (5x) 2  kx  72 This will be a perfect square trinomial if kx  2(5x)(7) or kx  70x. Thus, k can be 70 or 70. 63. x2  8x  k  x2  2(x) (4)  k This would be a perfect square trinomial if k  42 or 16.

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64. x2  18x  k  x2  2(x)(9)  k This would be a perfect square trinomial if k  92 or 81. 65. x2  20x  k  x2  2(x)(10)  k This would be a perfect square trinomial if k  102 or 100. 66. Answers should include the following. • The length of each side of the pavilion is 8  x  x or 8  2x feet. Thus, the area of the pavilion is (8  2x)2 square feet. This area includes the 80 square feet of bricks and the 82 or 64-square-foot piece of art, for a total area of 144 square feet. These two representations of the area of the pavilion must be equal, so we can write the equation (8  2x)2  144. • (8  2x) 2  144 Original equation Square Root 8  2x  12 Property 8  2x  12 or 8  2x  12 Separate into two equations. 2x  4 2x  20 Solve each equation. x2 x  10 Since length cannot be negative, the border should be 2 feet wide. 67. C; Solve the equation. 205  16t2 205 16



71.

72.

73.

74.

597, 97 6

8k2  22k  6  0 2(4k2  11k  3)  0 2(4k2  12k  k  3)  0 2[4k(k  3)  (k  3) ]  0 2(k  3) (4k  1)  0 or 4k  1  0 k30 k  3 4k  1 1

k4

53, 14 6

12w2  23w  5 12w  23w  5  0 12w2  20w  3w  5  0 4w(3w  5)  (3w  5)  0 (3w  5) (4w  1)  0 or 4w  1  0 3w  5  0 3w  5 4w  1 5

553, 14 6

1

w  4

6z2  7  17z 6z  17z  7  0 6z2  14z  3z  7  0 2z(3z  7)  (3z  7)  0 (3z  7) (2z  1)  0 3z  7  0 or 2z  1  0 3z  7 2z  1

512, 73 6

68. D; Since represents the  2ab  principal square root of a2  2ab  b2, its value cannot be negative. Thus, a  b  0 or a  b.

4 3

1

1

1

1

1

9

y  4  2x  2 y  2x  2  4 y  2x  2 2

2

76. The slope of y  3x  7 is 3. The slope of any 3 line perpendicular to it is 2. y  y1  m(x  x1 ) 3 y  7  2 [ x  (4) ]

3x  4  0 3x  4 x

1

z2

y  y1  m(x  x1 ) 1 y  4  2 (x  1)

Maintain Your Skills

9x2  16  0 (3x  4)(3x  4)  0 or 3x  4  0 3x  4

7

z3

75. The slope of y  2x  1 is 2. The slope of any line 1 perpendicular to it is 2.

Exercises 69–74 For checks, see students’ work. 69. s2  25 2  25  0 s (s  5)(s  5)  0 or s  5  0 s50 s  5 s5 { 5, 5}

543, 43 6

9

m7

2

b2

x

9

m  7

w  3

t2

2a2

70.

7m  9  0 7m  9

2

12.8125  t2 112.8125  t 3.6  t Ignore the negative solution; it will take about 3.6 seconds.

Page 514

49m2  81 49m  81  0 (7m  9)(7m  9)  0 or 7m  9  0 7m  9 2

3

y  7  2 (x  4)

4 3

3

y  7  2x  6 3

y  2x  6  7 3

y  2 x  13

431

Chapter 9

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77. Let x represent the horizontal change. slope  1 2



21. 20n2  2  2  5  n  n 25np5  5  5  n  p  p  p  p  p GCF: 5n 22. 60x2y2  2  2  3  5  x  x  y  y 35xz3  5  7  x  z  z  z GCF: 5x 23. 13x  26y  13(x)  13(2y)  13(x  2y) 2 2 24. 24a b  18ab  6ab(4ab)  6ab(3)  6ab(4ab  3) 25. 26ab  18ac  32a2  2a(13b)  2a(9c)  2a(16a)

vertical change horizontal change 1015 x

1(x)  2(1015) x  2030 The horizontal change is 2030 feet. 78. 17 13 9 5

   4 4 4

Add 4 three more times. The next three terms are 1, 3, and 7. 79. 5 4.5 4 3.5

 2a(13b  9c  16a)

   0.5 0.5 0.5

2

26. a  4ac  ab  4bc  (a2  4ac)  (ab  4bc)  a(a  4c)  b(a  4c)  (a  4c) (a  b) 27. 4rs  12ps  2mr  6mp  2(2rs  6ps  mr  3mp)  2[ (2rs  6ps)  (mr  3mp) ]  2[2s(r  3p)  m(r  3p) ]  2(r  3p)(2s  m) 28. 24am  9an  40bm  15bn  (24am  9an)  (40bm  15bn)  3a(8m  3n)  5b(8m  3n)  (8m  3n)(3a  5b) Exercises 29–31 For checks, see students’ work. 29. x(2x  5)  0 x  0 or 2x  5  0 2x  5

Add 0.5 three more times. The next three terms are 3, 2.5, and 2. 80. 45 54 63 72

   9 9 9

Add 9 three more times. The next three terms are 81, 90 and 99.

Chapter 9 Study Guide and Review Page 515 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

false; true false; true false; false; true false; true true

Vocabulary and Concept Check composite sample answer: 64 24  3 difference of squares

50, 52 6

30. (3n  8)(2n  6)  0 3n  8  0 3n  8

sample answer: x2  2x  2

Pages 515–518

Lesson-by-Lesson Review

11. 28  2  14 12. 33  3  11  2  2  7 or 22  7 13. 150  2  75 14. 301  7  43  2  3  25  2  3  5  5 or 2  3  52 15. 83  1  83 16. 378  1  378  1  2  189  1  2  3  63  1  2  3  3  21  1  2  3  3  3  7 or 1  2  33  7 17. 35  5  7 18. 12  2  2  3 18  2  3  3 30  2  3  5 GCF: 5 40  2  2  2  5 GCF: 2 19. 12ab  2  2  3  a  b 4a2b2  2  2  a  a  b  b GCF: 2  2  a  b or 4ab 20. 16mrt  2  2  2  2  m  r  t 30mr2  2  3  5  m  r  r GCF  2mr Chapter 9

5

x2

31.

5

8 3,

3

6

or

˛

2n  6  0 2n  6

8

n  3

4x2  7x  7x  0 x(4x  7)  0 x  0 or

n3

4x2

4x  7  0 4x  7 7

574, 06

x  4

32. Find a pair of factors of 12 whose sum is 7: 3 and 4. y2  7y  12  (y  3)(y  4) 33. Find a pair of factors of 36 whose sum is 9: 12 and 3. x2  9x  36  (x  12)(x  3) 34. Find a pair of factors of 6 whose sum is 5: 6 and 1. b2  5b  6  (b  6)(b  1) 35. 18  9r  r2  r2  9r  18 Find a pair of factors of 18 whose sum is 9: 3 and 6. 18  9r  r2  (r  3)(r  6)

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36. Find a pair of factors of 40x2 whose sum is 6x: 10x and 4x. a2  6ax  40x2  (a  10x)(a  4x) 37. Find a pair of factors of 32n2 whose sum is 4n: 4n and 8n. m2  4mn  32n2  (m  4n)(m  8n) Exercises 38–40 For checks, see students’ work. 38. y2  13y  40  0 ( y  8)( y  5)  0 or y  5  0 y80 y  8 y  5 {8, 5} 39. x2  5x  66  0 (x  11) (x  6)  0 or x  6  0 x  11  0 x  11 x  6 {6, 11} 40. m2  m  12  0 (m  4)(m  3)  0 m  4  0 or m  3  0 m4 m  3 {3, 4} 41. There are no pairs of factors of 6 whose sum is 9. 2a2  9a  3 is prime. 42. Find a pair of factors of 48 whose sum is 13: 3 and 16.

48.

49.

7

a3

52, 73 6

40x2  2x  24 40x  2x  24  0 2(20x2  x  12)  0 2(20x2  16x  15x  12)  0 2[4x(5x  4)  3(5x  4) ]  0 2(5x  4) (4x  3)  0 or 4x  3  0 5x  4  0 5x  4 4x  3 2

˛

˛

4

3

x  5

545, 34 6

x4

50. 2y3  128y  2y( y2  64)  2y( y  8) ( y  8) 51. 9b2  20 is prime. 1

9

112n22  134r22 1 3 1 3  1 2n  4r 21 2n  4r 2

52. 4n2  16r2 

Exercises 53–55 For checks, see students’ work. 53. b2  16  0 (b  4)(b  4)  0 b40 or b  4  0 b  4 b4 {4, 4} 54. 25  9y2  0 (5  3y)(5  3y)  0 5  3y  0 or 5  3y  0 3y  5 5  3y

2m2  13m  24  2m2  3m  16m  24  (2m2  3m)  (16m  24)  m(2m  3)  8(2m  3)  (2m  3)(m  8) 43. Find a pair of factors of 100 whose sum is 20: 10 and 10. 25r2  20r  4  25r2  10r  10r  4  (25r2  10r)  (10r  4)  5r(5r  2)  2(5r  2)  (5r  2)(5r  2) 44. There are no pairs of factors of 18 whose sum is 7. 6z2  7z  3 is prime. 45. Find a pair of factors of 72 whose sum is 17: 8 and 9. 12b2  17b  6  12b2  8b  9b  6  (12b2  8b)  (9b  6)  4b(3b  2)  3(3b  2)  (3b  2)(4b  3) 46. 3n2  6n  45  3(n2  2n  15) Now factor n2  2n  15. Find a pair of factors of 15 whose sum is 2: 5 and 3. n2  2n  15  (n  5)(n  3) Thus, 3n2  6n  45  3(n  5)(n  3). Exercises 47–49 For checks, see students’ work. 47. 2r2  3r  20  0 2  8r  5r  20  0 2r 2r(r  4)  5(r  4)  0 (r  4)(2r  5)  0 r  4  0 or 2r  5  0 r4 2r  5

552, 46

3a2  13a  14  0 3a  6a  7a  14  0 3a(a  2)  7(a  2)  0 (a  2) (3a  7)  0 a  2  0 or 3a  7  0 a2 3a  7 2

˛

˛

55.

5

5

5 3

y  3 5 5 3, 3

6

y

16a2  81  0 (4a  9)(4a  9)  0 or 4a  9  0 4a  9  0 4a  9 4a  9 ˛

9

a  4

594, 94 6

9

a4

56. a2  18a  81  a2  2(a) (9)  92  (a  9) 2 2 57. 9k  12k  4  (3k) 2  2(3k) (2)  22  (3k  2) 2 2 58. 4  28r  49r  22  2(2) (7r)  (7r) 2  (2  7r) 2 2 59. 32n  80n  50  2(16n2  40n  25)  2[ (4n) 2  2(4n) (5)  52 ]  2(4n  5) 2

5

r  2

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10. 25y2  49w2  (5y) 2  (7w) 2  (5y  7w) (5y  7w) 2 11. t  16t  64  t2  2(t) (8)  82  (t  8) 2 2 12. x  14x  24 Find two numbers whose product is 24 and whose sum is 14: 12 and 2. x2  14x  24  (x  12)(x  2) 13. 28m2  18m  2m(14m  9) 14. a2  11ab  18b2 Find two numbers whose product is 18b2 and whose sum is 11b: 2b and 9b. a2  11ab  18b2  (a  2b)(a  9b) 15. 12x2  23x  24  12x2  9x  32x  24  3x(4x  3)  8(4x  3)  (4x  3) (3x  8) 16. 2h2  3h  18 is prime. 17. 6x3  15x2  9x  3x(2x2  5x  3)  3x(2x2  6x  x  3)  3x[ 2x(x  3)  (x  3) ]  3x(x  3) (2x  1) 18. 64p2  63p  16 is prime. 19. 2d2  d  1  2d2  2d  d  1  2d(d  1)  (d  1)  (d  1) (2d  1) 20. 36a2b3  45ab4  9ab3(4a  5b) 21. 36m2  60mn  25n2  (6m) 2  2(6m) (5n)  (5n) 2

Exercises 60–63 For checks, see students’ work. 60. 6b3  24b2  24b  0 6b(b2  4b  4)  0 2 6b(b  2(b) (2)  22 )  0 6b(b  2) 2  0 6b  0 or b  2  0 b0 b2 {0, 2} 61. 49m2  126m  81  0 (7m) 2  2(7m)(9)  92  0 (7m  9) 2  0 7m  9  0 7m  9 9

m7

597 6

62. (c  9) 2  144 c  9  1144 c  9  12 c  9  12 c  9  12 or c  9  12 c  21 c  3 {3, 21} 63. 144b2  36 36

b2  144 1

b2  4 1

b  2

512 6

 (6m  5n) 2

22. a2  4  a2  22  (a  2)(a  2) 23. 4my  20m  3py  15p  4m( y  5)  3p( y  5)

Chapter 9 Practice Test

 (y  5)(4m  3p) 2

24. 15a b  5a  10a  5a(3ab  a  2) 25. 6y2  5y  6  6y2  9y  4y  6  3y(2y  3)  2(2y  3)  (2y  3) (3y  2) 26. 4s2  100t2  4(s2  25t2 )  4(s  5t) (s  5t) 3  4x2  9x  36  x2 (x  4)  9(x  4) 27. x  (x  4) (x2  9)  (x  4) (x  3) (x  3) 28. area of shaded  area of large  area of small region rectangle rectangle  (x  6) (y  6)  xy  xy  6x  6y  36  xy  6x  6y  36  6(x  y  6) 29. area of shaded  area of square  area of circles region  (4r) 2  4(r2 )  16r2  4r2  4r2 (4  )

Page 519 1. Sample answer: 7; Its only factors are 1 and itself. 2. Sample answer: n2  100 n2  100  n2  102  (n  10)(n  10) 3. Check for a GCF other than 1 and factor it out. 4. 63  3  21  3  3  7 or 32  7 5. 81  3  27 6. 210  1  210  339  1  2  105  3  3  3  3 or 34  1  2  3  35  1  2  3  5  7 7. 48  2  2  2  2  3 64  2  2  2  2  2  2 GCF: 2  2  2  2 or 16 8. 28  2  2  7 75  3  5  5 GCF: 1; 28 and 75 are relatively prime. 9. 18a2b2  2  3  3  a  a  b  b 28a3b2  2  2  7  a  a  a  b  b GCF: 2  a  a  b  b or 2a2b2

Chapter 9

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38. Let x represent the amount of the increase. Write an equation for the area of the new rectangle. (4  x) (7  x)  28  26 28  4x  7x  x2  54 x2  11x  28  54 x2  11x  26  0 (x  13) (x  2)  0 or x  2  0 x  13  0 x  13 x2 Ignore the negative solution. Thus, 2 inches are added to each dimension of the rectangle. The new rectangle is 6 in. by 9 in. 39. Let x represent the width of the sidewalk. Write an equation for the area of the remaining lawn. (24  2x) (32  2x)  425 768  48x  64x  4x2  425 4x2  112x  768  425 4x2  112x  343  0 4x2  14x  98x  343  0 2x(2x  7)  49(2x  7)  0 (2x  7) (2x  49)  0 or 2x  49  0 2x  7  0 2x  7 2x  49

Exercises 30–37 For checks, see students’ work. 30. (4x  3)(3x  2)  0 4x  3  0 or 3x  2  0 4x  3 3x  2 3

2

x4

523, 34 6

x  3

31. 18s2  72s  0 18s(s  4)  0 18s  0 or s  4  0 s0 s  4 {4, 0} 4x2  36 32. 2  36  0 4x 4(x2  9)  0 4(x  3)(x  3)  0 or x  3  0 x30 x  3 x3 {3, 3} t2  25  10t 33. 2  10t  25  0 t t2  2(t)(5)  52  0 (t  5) 2  0 t50 t5 {5} 34. a2  9a  52  0 (a  13) (a  4)  0 or a  4  0 (a  13)  0 a  13 a  4 {4, 13} 35. x3  5x2  66x  0 x(x2  5x  66)  0 x(x  11)(x  6)  0 or x  6  0 x  0 or x  11  0 x  11 x  6 {6, 0, 11}

˛

˛

˛

7

x  2 or 3.5

1

2

2x2  9x  5 2x  9x  5  0 2x2  10x  x  5  0 2x(x  5)  (x  5)  0 (x  5)(2x  1)  0 x  5  0 or 2x  1  0 x5 2x  1

37.

5

6

 98

1 2  2(98)

Chapter 9 Standardized Test Practice Pages 520–521

1

x  2

1. A; The line passes through the points (5, 0) and (0, 3). 3  0 (5)

m0

3b2  6  11b 2  11b  6  0 3b 3b2  9b  2b  6  0 3b(b  3)  2(b  3)  0 (b  3)(3b  2)  0 b  3  0 or 3b  2  0 b3 3b  2

523, 36

1 2 x 2 1 2 x 2

x2  196 x  1196 x  14 Ignore the negative solution. Thus, the square is 14 meters by 14 meters.

2

5

or 24.5

x2  2x2  98

˛

1 2,

49 2

The walk cannot be 24.5 feet wide since this is wider than the original lawn. Thus, the sidewalk is 3.5 feet wide. 40. Write an equation for the shaded part of the square.

˛

36.

x

3

or 5 3

The equation is y  5x  3. 2. D; Use (12, 176) and (18, 224) to find the slope (price per ticket). m 

2

b3

224  176 18  12 48 or 8 6

Since the band initially had $80, the intercept is 80. The equation is a  8t  80.

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1

Shade the half plane that contains (0, 0).

3. A; The slope of the boundary line is 3; this eliminates answer choices C and D. Test (0, 0) in answer choice B.

y

1

y  3x  1 0

1 (0) 3

xy3

1 O

0  1 false Since (0, 0) is part of the shaded region but not a solution of B, we can eliminate B. This leaves only answer choice A. 4. D; The total was 258. n  f  258 Twice as many bags were sold today as last Friday. n  2f 5. B; 5.387 103  0.005387 6. C;

16x8 8x4



12. Let a represent the number of adult tickets sold and c represent the number of child tickets. Write an equation for the total number of tickets. a  c  145 Write an equation for the total receipts. 7.50a  4c  790 Solve the first equation for c. a  c  145 c  145  a Substitute 145  a for c in the second equation. 7.50a  4c  790 7.50a  4(145  a)  790 7.50a  580  4a  790 3.5a  580  790 3.5a  210 a  60 The theater sold 60 adult tickets. 13. 3x  y  8 4x  2y  14 Multiply the first equation by 2. Then add. 6x  2y  16 () 4x  2y  14 10x  30 x3 Use 3x  y  8 to find y. 3x  y  8 3(3)  y  8 9y8 y  1 The solution is (3, 1). 14. (x  t)x  (x  t)y  (x  t) (x  y) 15. Let n be the first odd integer. Then n  2 is the next odd integer. n(n  2)  195 n2  2n  195 2  2n  195  0 n (n  15)(n  13)  0 or n  13  0 n  15  0 n  15 n  13 When n  15, n  2  15  2 or 13. When n  13, n  2  13  2 or 15. The integers are 15 and 13 or 13 and 15.

1168 21xx 2 8 4

84

7.

 2x  2x4 3x2  48  0 A; 3(x2  16)  0 3(x  4)(x  4)  0 or x  4  0 x40 x  4 x4 2  3x  8  6x  6 D; x x2  9x  14  0 (x  2)(x  7)  0 x  2  0 or x  7  0 x2 x7 B; Factor the expression for the area. 12x2  21x  6  3(4x2  7x  2)  3(4x2  8x  x  2)  3 [4x(x  2)  (x  2) ]  3(x  2)(4x  1) The width is 3x  6 or 3(x  2). Thus, the length is 4x  1. 6 0x  2 0  18 ˛

8.

9.

10.

1 [ 6 0x 6

 2 0 ]  6 (18) 1

0x  2 0  3 x  2  3 or x  2  3 x5 x  1 Both 5 and 1 make the equation true. 11. x  y  3 Graph the boundary as a solid line since the inequality includes equal. Test the point (0, 0). xy3 003 0  3 true

Chapter 9

x

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16.

21. B; 3x  y  5 x  3y  6

2x2  5x  12  0 2x  8x  3x  12  0 2x(x  4)  3(x  4)  0 (x  4)(2x  3)  0 x40 or 2x  3  0 x  4 2x  3 2

Solve the second equation for x. x  3y  6 x  3y  6 Substitute 3y  6 for x in the first equation. 3x  y  5 3(3y  6)  y  5 9y  18  y  5 8y  18  5 8y  13

3

x2

54, 32 6

17. 2x2  7x  3  2x2  6x  x  3  2x(x  3)  (x  3)  (x  3)(2x  1) 18. C; 0x 0  0y 0  015 0  07 0  15  7 8 0x  y 0  015  (7) 0  015  7 0  08 0 8 The two quantities are equal.

13

y  8

2a  3b  3 a  4b  24 Solve the second equation for a. a  4b  24 a  24  4b Substitute 24  4b for a in the first equation. 2a  3b  3 2(24  4b)  3b  3 48  8b  3b  3 48  11b  3 11b  51

2

19. B; 3x  27  39 2 x 3 3 2 x 2 3

 66

1 2  32 (66)

51

b  11

x  99

3 y 4

51

 55  20 3 y 4 4 3 y 3 4

 75

13

22. C; The GCF of 2x3, 6x2, and 8x is 2x. The GCF of 18x3, 14x2, and 4x is 2x. The two quantities are equal. 23a. Since the area is 28 square yards, the width is 3 yards less than the length, so the width is L  3. Thus, L(L  3)  28. 23b. Write an equation for the area and solve for L. L(L  3)  28 L2  3L  28  0 (L  7)(L  4)  0 L  7  0 or L  4  0 L7 L  4 Ignore the negative solution. The length is 7 yards. 23c. Find the perimeter. Since the length is 7 yd, the width is 7  3 or 4 yd. P  2(7)  2(4) or 22

y  100 The quantity in Column B is greater. 20. A; The line appears to pass through (0, 2) and (2, 1). 1  (2) 2  0

51 11

Since 11 7  8 , the quantity in Column B is greater.

1 2  43 (75)

m

or

3

or 2 2

A line perpendicular to it has slope 3. y  y1  m(x  x1 ) 2 y  (4)  3 (x  6) 2

y  4  3x  4 2

y  3x This line passes through (0, 0). Its x-intercept is 0. Since the x-intercept of the given line is between 1 and 2, the quantity in Column A is greater.

He will need 22 yd of fencing.

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Chapter 10 Page 523

Quadratic and Exponential Functions 6. Sample answer:

Getting Started

1. Sample answer: x 6 4 2 0 2

y 1 1 3 5 7

x 5 0

y yx5

y

y 0 2

5y  10  2x

x

O

O

x

7. Sample answer: 2. Sample answer: x 0 1 2 3

x 6 4 2 0 2

y

y 3 1 1 3

x

O

y

y 0 1 2 3 4

O

x

x  2y  6

y  2x  3

8. Sample answer: 3. Sample answer: x 4 2 0 2 4

y 1 0 1 2 3

x 1 1 3

y y  0.5x  1

O

4. Sample answer:

49  72 14a  2(a) (7) a2  14a  49  (a  7) 2 11. No; 81 is not a perfect square. 12. No; 12 is not a perfect square. 13. Yes; 9b2  (3b) 2

x

y  3x  2

5. Sample answer: y 4 2 0

1  12 6b  2(3b) (1) 9b2  6b  1  (3b  1) 2 14. No; 6x2 is not a perfect square. 15. Yes; 4p2  (2p) 2

y

x

O

9  32 12p  2(2p) (3) 4p2  12p  9  (2p  3) 2

2x  3y  12

Chapter 10

t2  (t) 2

36  62 12  2(t) (6) t2  12t  36  (t  6) 2 10. Yes; a2  (a) 2

y

y 4 1 2

O

x 0 3 6

3x  2y  9

x

O

9. Yes; x 2 1 0

y

y 6 3 0

438

x

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16. Yes;

The lowest point on the graph is (3, 1). The vertex is (3, 1). 2. The fold line is a vertical line on which each point has x-coordinate 3. The equation is x  3. 3. The vertex, (3, 1), lies on the fold line. 4. Sample answer: A parabola is symmetrical. The vertex of the parabola lies on the line that divides the parabola into two matching halves.

16s2  (4s) 2 2

93 24s  2(4s)(3) 16s2  24s  9  (4s  3) 2 17. 5 9 13 17

 4 4 4

Add 4 three more times. The next three terms are 21, 25, and 29. 18. 12 5 2 9

 7 7 7

Page 528

Add 7 three more times. The next three terms are 16, 23, and 30. 19. 4 1 2 5

Check for Understanding

1. Both types of parabolas are U shaped. A parabola with a maximum opens downward, and its corresponding equation has a negative coefficient for the x2 term. A parabola with a minimum opens upward, and its corresponding equation has a positive coefficient for the x2 term. 2. Sample answer:

 3 3 3

Add 3 three more times. The next three terms are 8, 11, and 14. 20. 24 32 40 48

 8 8 8

y

Add 8 three more times. The next three terms are 56, 64, and 72. 21. 1 6 11 16

 5 5 5

O

x

Add 5 three more times. The next three terms are 21, 26, and 31. 22. 27 20 13 6

 7 7 7

3. Sample answer: If you locate several points of the graph on one side of the axis of symmetry, you can locate the corresponding points on the other side of the axis of symmetry to help graph the equation. 4. y  x2  5 Sample answer:

Add 7 three more times. The next three terms are 1, 8, and 15. 23. 5.3 6.0 6.7 7.4

 0.7 0.7 0.7

Add 0.7 three more times. The next three terms are 8.1, 8.8, and 9.5. 24. 9.1 8.8 8.5 8.2

 0.3 0.3 0.3

x 3 2 1 0 1 2 3

Add 0.3 three more times. The next three terms are 7.9, 7.6, and 7.3.

10-1 Graphing Quadratic Functions Page 525

y

x

O

y  x2  5

Algebra Activity

1. y  x2  6x  8 Sample answer: x 5 4 3 2 1

y 4 1 4 5 4 1 4

y 3 0 1 0 3

5. y  x2  4x  5 Sample answer: y

x 1 0 1 2 3 4 5

y  x 2  6x  8 O

x

439

y 0 5 8 9 8 5 0

y y  x 2  4x  5

O

x

Chapter 10

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The equation of the axis of symmetry is x  2. y  x2  4x  3 y  (2) 2  4(2)  3 y  4  8  3 y1 The vertex is at (2, 1). Since the coefficient of the x2 term is negative, the vertex is a maximum.

6. In y  x2  4x  9, a  1 and b  4. x x

b 2a 4 2(1)

or 2

The equation of the axis of symmetry is x  2. y  x2  4x  9 y  (2) 2  4(2)  9 y489 y  13 The vertex is at (2, 13). Since the coefficient of the x2 term is positive, the vertex is a minimum. 2

y

9. B; Since all four graphs have the same axis of symmetry, x  0, you cannot eliminate any graphs using the axis of symmetry. Since the coefficient of the x2 term is negative, the graph opens downward. Eliminate choices A and C. Let x  0.

7. In y  x2  5x  6, a  1 and b  5. b

x  2a

1

5 2(1)

or

5 2

y  2 x2  1

or 2.5

1

y  2 (0) 2  1 or 1

The equation of the axis of symmetry is x  2.5. y  x2  5x  6 y  (2.5) 2  5(2.5)  6 y  6.25  12.5  6 y  12.25 The vertex is at (2.5, 12.25). Since the coefficient of the x2 term is negative, the vertex is a maximum. 14 12 10 8 6 4 2 21 2

x

O

O x 654321 2 1 2 4 6 8 10 12 14 y  x 2  4x  9

x

y  (x  2)2  1

y

y

The graph passes through the point (0, 1). Eliminate choice D. The answer is B.

Pages 528–530

Practice and Apply

10. y  x2  3 Sample answer: x 2 1 0 1 2

y  x 2  5x  6

y 1 2 3 2 1

y

O 1 2 3 4 5 6x

y  x2  3

11. y  x2  7 Sample answer:

8. y  (x  2) 2  1  (x2  4x  4)  1  x2  4x  4  1  x2  4x  3 In y  x2  4x  3, a  1 and b  4.

x 2 1 0 1 2

b

x  2a 4

x  2(1) or 2

y 3 6 7 6 3

y  x 2  7

y

O

Chapter 10

x

O

440

x

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12. y  x2  2x  8 Sample answer: x 1 0 1 2 3

17. In y  4x2  5x  16, a  4 and b  5. b

x  2a y

y 5 8 9 8 5

(5)

5

x  8 is the equation of the axis of symmetry. 18. In y  4x2, a  4 and b  0. b

x  2a 0

x  2(4) or 0 x  0 is the equation of the axis of symmetry. y  4x2 y  4(0) 2 y0 The vertex is at (0, 0). Since the coefficient of the x2 term is positive, the vertex is a minimum.

y  x 2  2x  8

13. y  x2  4x  3 Sample answer: x 0 1 2 3 4

5 8

x   2(4) or

x

O

y

y 3 0 1 0 3

y x

O

y  x 2  4x  3 y  4x 2

14. y  3x2  6x  4 Sample answer: x 3 2 1 0 1

y 5 4 7 4 5

y

y  3x 2  6x  4

19. In y  2x2, a  2 and b  0. b

x  2a 0

x  2(2) or 0

O

x  0 is the equation of the axis of symmetry. y  2x2 y  2(0) 2 y0 The vertex is at (0, 0). Since the coefficient of the x2 term is negative, the vertex is a maximum.

x

15. y  3x2  6x  1 Sample answer: x 1 0 1 2 3

x

O

y

y 8 1 4 1 8

y O

x y  2x O

2

x

y  3x 2  6x  1

16. In y  3x2  2x  5, a  3 and b  2. b

x  2a 2

x  2(3) or

1 3

1

x  3 is the equation of the axis of symmetry.

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20. In y  x2  2, a  1 and b  0. x x

b 2a 0 2(1)

y y  x 2  2x  3

or 0

x  0 is the equation of the axis of symmetry. y  x2  2 y  (0) 2  2 y2 The vertex is at (0, 2). Since the coefficient of the x2 term is positive, the vertex is a minimum.

23. In y  x2  6x  15, a  1 and b  6. b

x  2a

y

(6)

x  2(1) or 3 x  3 is the equation of the axis of symmetry. y  x2  6x  15 y  (3) 2  6(3)  15 y  9  18  15 y  24 The vertex is at (3, 24). Since the coefficient of the x2 term is negative, the vertex is a maximum.

y  x2 2 x

O

21. In y  x2  5, a  1 and b  0. b

x  2a

y  x 2  6x  15 0

x  2(1) or 0 x  0 is the equation of the axis of symmetry. y  x2  5 y  (0) 2  5 y5 The vertex is at (0, 5). Since the coefficient of the x2 term is negative, the vertex is a maximum. y  x 2  5

12 8

4 4

y

O 4x

b

x  2a x

(14) 2(1)

or 7

x  7 is the equation of the axis of symmetry. y  x2  14x  13 y  (7) 2  14(7)  13 y  49  98  13 y  36 The vertex is at (7, 36). Since the coefficient of the x2 term is positive, the vertex is a minimum.

x

22. In y  x2  2x  3, a  1 and b  2. b

x  2a

y 5

2

x  2(1) or 1

O 25 10 15 20 25 30 35

x  1 is the equation of the axis of symmetry. y  x2  2x  3 y  (1) 2  2(1)  3 y  1  2  3 y4 The vertex is at (1, 4). Since the coefficient of the x2 term is negative, the vertex is a maximum.

Chapter 10

28 24 20 16 12 8 4

24. In y  x2  14x  13, a  1 and b  14.

y

O

x

O

442

2 4 6 8 10 12 14 x

y  x 2  14x  13

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25. In y  x2  2x  18, a  1 and b  2. x x

b 2a 2 2(1)

Since the coefficient of the x2 term is positive, the vertex is a minimum. y

or 1

x  1 is the equation of the axis of symmetry. y  x2  2x  18 y  (1) 2  2(1)  18 y  1  2  18 y  17 The vertex is at (1, 17). Since the coefficient of the x2 term is positive, the vertex is a minimum.

y  x 2  2x  18

35 30 25 20 15 10 5

654321 5

y  3x 2  6x  4

x

O

28. In y  5  16x  2x2, a  2 and b  16.

y

b

x  2a 16

x  2(2) or 4 x  4 is the equation of the axis of symmetry. y  5  16x  2x2 y  5  16(4)  2(4) 2 y  5  64  32 y  37 The vertex is at (4, 37). Since the coefficient of the x2 term is negative, the vertex is a maximum.

O 1 2x

26. In y  2x2  12x 11, a  2 and b  12. b

x  2a 12

x  2(2) or 3

35 30 25 20 15 10 5

x  3 is the equation of the axis of symmetry. y  2x2  12x  11 y  2(3) 2  12(3)  11 y  18  36  11 y  29 The vertex is at (3, 29). Since the coefficient of the x2 term is positive, the vertex is a minimum.

1 5

y

y  5  16x  2x 2

O 1 2 3 4 5 6 7 8 9x

29. In y  9  8x  2x2, a  2 and b  8.

y 3 O x 7654321 3 1 6 9 12 15 18 21 24 27 30 2 y  2x  12x  11

b

x  2a (8)

x   2(2) or 2 x  2 is the equation of the axis of symmetry. y  9  8x  2x2 y  9  8(2)  2(22 ) y  9  16  8 y1 The vertex is at (2, 1). Since the coefficient of the x2 term is positive, the vertex is a minimum. y

27. In y  3x2  6x  4, a  3 and b  6. b

x  2a (6)

x   2(3) or 1 x  1 is the equation of the axis of symmetry. y  3x2  6x  4 y  3(1) 2  6(1)  4 y364 y1 The vertex is at (1, 1).

y  9  8x  2x2 O

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30. y  3(x  1) 2  20  3(x2  2x  1)  20  3x2  6x  3  20  3x2  6x  17 In y  3x2  6x  17, a  3 and b  6. x x

b 2a 6 2(3)

32.

y  2  x2  10x  25 y  2  2  x2  10x  25  2 y  x2  10x  23 2 In y  x  10x  23, a  1 and b  10. b

x  2a x

or 1

(10) 2(1)

or 5

x  5 is the equation of the axis of symmetry. y  x2  10x  23 y  52  10(5)  23 y  25  50  23 y  2 The vertex is at (5, 2). Since the coefficient of x2 is positive, the vertex is a minimum.

x  1 is the equation of the axis of symmetry. y  3x2  6x  17 y  3(1) 2  6(1)  17 y  3  6  17 y  20 The vertex is at (1, 20). Since the coefficient of the x2 term is positive, the vertex is a minimum.

y

y 4 O x 654321 4 1 2 8 12 16 20 y  3(x  1)2 20 24 28

x

O

y  2  x 2  10x  25

33.

31. y  2(x  4) 2  3 y  2(x2  8x  16)  3 y  2x2  16x  32  3 y  2x2  16x  35 In y  2x2  16x  35, a  2 and b  16.

y  1  3x2  12x  12 y  1  1  3x2  12x  12  1 y  3x2  12x  11 In y  3x2  12x  11, a  3 and b  12. b

x  2a 12

x  2(3) or 2

b

x  2a

x  2 is the equation of the axis of symmetry. y  3x2  12x  11 y  3(2) 2  12(2)  11 y  12  24  11 y  1 The vertex is at (2, 1). Since the coefficient of the x2 term is positive, the vertex is a minimum.

16

x  2(2) or 4 x  4 is the equation of the axis of symmetry. y  2x2  16x  35 y  2(4) 2  16(4)  35 y  32  64  35 y  3 The vertex is at (4, 3). Since the coefficient of the x2 term is negative, the vertex is a maximum.

y

y x

O 2

y  2(x  4)  3 O

y  1  3x 2 12x  12

Chapter 10

444

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1

Since the coefficient of the x2 term is positive, the vertex is a minimum.

y  5  3 (x  2) 2

34.

1

y  5  5  3 (x2  4x  4)  5 1

4

4

1

4

4

19 , 3

19 3 1 3

y

y  3x2  3x  3  5 y  3x2  3x  1

In y  3x2  3x 

a

4

and b  3.

b

x  2a x

4 3  1 23

12

O

or

4 3

3 2

or 2

y  1  23 (x  1)2

36. If the vertex of the parabola is (4, 3), the axis of symmetry is x  4. The point (11, 0) is 7 units from the axis of symmetry. The other x-intercept must be (3, 0), which is 7 units on the other side of the axis of symmetry. 37. The equation of the axis of symmetry is:

x  2 is the equation of the axis of symmetry. y y y y

1 2 4 19 x  3x  3 3 1 4 (2) 2  3 (2) 3 4 8 19 3 3 3 15 or 5 3



x

19 3

6  4 2 2 2

x

The vertex is at (2, 5). Since the coefficient of the x2 term is positive, the vertex is a minimum.

x

x  1 38. 2 m; parabolas are symmetric.

y

39. In h  16t2  32t  3, a  16 and b  32. b

t  2a 32

t  2(16)

y  5  13 (x  2)2

O

t  1 is the equation of the axis of symmetry h  16(1) 2  32  3 h  19 The vertex is at (1, 19). The maximum height of the weight is 19 feet. 40. No, the prize will not be won since the maximum height of 19 feet is less than the 20 feet required to win. 41. A  x(20  x) A  20x  x2 A  x2  20x 42. In A  x2  20x, a  1 and b  20.

x

2

y  1  3 (x  1) 2

35.

2

y  1  1  3 (x2  2x  1)  1 2

4

2

2

4

1

y  3x2  3x  3  1 y  3x2  3x  3 2

4

1

2

4

In y  3x2  3x  3, a  3 and b  3. b

x  2a x

b

4 3  2 23

12

x  2a 4

3

 3  4 or 1

20

x  2(1) or 10 The x-coordinate of the vertex is 10. Since the coefficient of the x2 term is negative, the vertex is a maximum. The greatest area will result when x is 10 m. 43. A  x2  20x A  102  20(10) A  100  200 A  100 The second coordinate of the vertex, which is a maximum, is 100. Thus, the maximum possible area for the pen is 100 m2. 44. In h  0.00635x2  4.0005x  0.07875,

x  1 is the equation of the axis of symmetry. 2

4

1

y  3x2  3x  3 2

4

1

y  3 (1) 2  3 (1)  3 2

4

1

y333 3

y  3 or 1 The vertex is at (1, 1).

a  0.00635 and b  4.0005. b

x   2a 4.0005

x   2 (0.00635 ) or 315 x  315 is the equation of the axis of symmetry.

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51. In order to coordinate a firework with recorded music, you must know when and how high it will explode. Answers should include following. • The rocket will explode when it reaches the 39.2 vertex or when t  2(4.9) , which is 4 seconds. • The height of the rocket when it explodes is the height when t  4. Therefore, h  4.9(42)  39.2(4)  1.6 or 80 meters. 52. A; Since the parabola opens upward, the coefficient of the x2 term is positive. Eliminate choices B and D. The y-intercept of choice C is y  02  4(0)  5 or 5. The y-intercept of the graph is 5. Eliminate choice C. The answer is A. 53. D; y  x  5 is not a quadratic equation because there is no x2 term. 54. y  x2  10x  25 Step 1 Graph the equation on the screen.

45. The ends of the arch are the x-intercepts of the graph of the equation. Since the axis of symmetry is an equal distance from both x-intercepts, the axis is 315 feet from each end. Thus, the ends are 315  315 or 630 feet apart. 46. To find the maximum height of the arch, find the second coordinate of the vertex by substituting 315 for x in the equation for the height. h  0.00635x2  4.0005x  0.07875 h  0.00635(315) 2  4.0005(315)  0.07875 h  630 The maximum height of the arch is 630 ft. 47. To avoid confusion, we replace a with A in the equation given. In A  0.003x2  0.115x  21.3, a  0.003 and b  0.115. b

x  2a (0.115)

x   2(0.003) or about 19

KEYSTROKES:

The x-coordinate of the vertex is about 19. Since the coefficient of the x2 term is positive, the vertex is a minimum. The minimum age occurs about 19 years after 1940 or in 1959. 48. Substitute 19 for x in the equation. A  0.003x2  0.115x  21.3 A  0.003(19) 2  0.115(19)  21.3 A  20 In 1959, the average age of brides was about 20 years old. 49. Step 1 Graph the equation on the screen.

X,T,␪,n

X,T,␪,n

10

25 GRAPH

Step 2

Approximate the minimum. KEYSTROKES: 2nd [CALC] 3 Use the and keys to set the left and right bounds and to guess the minimum. The vertex is a minimum with ordered pair (5, 0).

0.003 X,T,␪,n

KEYSTROKES:

0.115 X,T,␪,n 21.3 GRAPH Step 2 Approximate the minimum. KEYSTROKES: 2nd [CALC] 3

55. y  x2  4x  3 Step 1 Graph the equation on the screen.

Use the and keys to set the left and right bounds and to guess the minimum. The minimum is about (19.2, 20.2).

KEYSTROKES:

X,T,␪,n

( )

X,T,␪,n

4

3 GRAPH

Step 2

Approximate the maximum. KEYSTROKES: 2nd [CALC] 4 Use the and keys to set the left and right bounds and to guess the maximum. The vertex is a maximum with ordered pair (2, 7). 50. Sample answer: y  4x2  3x  5. A quadratic equation y  ax2  bx  c with 3 x  8 as its axis of symmetry must satisfy b

3

b

3

2a  8. 3

2a  8 or 2(4) Choose b  3 and a  4. Then c can be any number, such as 5.

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56. y  2x2  8x  1 Step 1 Graph the equation on the screen. KEYSTROKES:

X,T,␪,n Step 2

59. y  0.5x2  2x  3 Step 1 Graph the equation on the screen.

( ) 2 X,T,␪,n 1 GRAPH

8

KEYSTROKES:

X,T,␪,n

Approximate the maximum. KEYSTROKES: 2nd [CALC] 4

2nd [CALC] 4

Use the and keys to set the left and right bounds and to guess the maximum. The vertex is a maximum with ordered pair (2, 5).

Use the and keys to set the left and right bounds and to guess the maximum. The vertex is a maximum with ordered pair (2, 7).

57. y  2x2  40x  214 Step 1 Graph the equation on the screen. KEYSTROKES:

2 X,T,␪,n

Page 530

40

214 GRAPH Step 2 Approximate the minimum. KEYSTROKES: 2nd [CALC] 3 Use the and keys to sset the left and right bounds and to guess the minimum. The vertex is a minimum with ordered pair (10, 14).

65. 1  16g2  12  (4g) 2  (1  4g) (1  4g) 66. (13x  9y)  11y  13x  (9y  11y)  13x  20y 67. (7p2  p  7)  (p2  11)  (7p2  p  7)  (p2  11)  (7p2  p2 )  p  (7  11)  6p2  p  18 68. Let m represent the cost for a member and n the cost for a nonmember. Solve the following system of equations. 3m  3n  180 5m  3n  210 Multiply the first equation by 1. Then add. 3m  3n  180 (  ) 5m  3n  210 2m  30 m  15 Substitute 15 for m in the first equation and solve for n. 3m  3n  180 3(15)  3n  180 45  3n  180 3n  135 n  45 An aerobics class costs $15 for members and $45 for nonmembers.

58. y  0.25x2  4x  2 Step 1 Graph the equation on the screen. 0.25 X,T,␪,n

Maintain Your Skills

60. There are no factors of 9 whose sum is 6. Thus, x2  6x  9 is prime. 61. a2  22a  121  a2  2(a) (11)  112  (a  11) 2 2 62. 4m  4m  1  (2m) 2  2(2m) (1)  12  (2m  1) 2 2 2 63. 4q  9  (2q)  32  (2q  3) (2q  3) 2 64. 2a  25  2a2  0a  25 There are no factors of 25 whose sum is 0. Thus, 2a2  25 is prime.

X,T,␪,n

KEYSTROKES:

2

Step 2

Approximate the maximum.

KEYSTROKES:

( ) 0.5 X,T,␪,n 3 GRAPH

4

X,T,␪,n

2 GRAPH Step 2 Approximate the minimum. KEYSTROKES: 2nd [CALC] 3 Use the and keys to set the left and right bounds and to guess the minimum. The vertex is a minimum with ordered pair (8, 18).

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Exercises 69–71 For checks, see students’ work. 69. 12b 7 144 70. 5w 7 125 12b 12

7

144 12

5w 5

b 7 12 {b|b 7 12} 71.

r

125 5

72. y  y  m(x  x ) 1 1 y  13  4(x  2)

12

y  13  4x  8

8 9 8 9

5r|r  6

y  4x  5

y  y1  m(x  x1 ) y  (7)  0[x  (2) ] y70 y  7 74. y  y  m(x  x ) 1 1 73.

Graphing Calculator Investigation (Follow-Up of Lesson 10-1)

1. y  x2 y  3x2 y  6x2

w 6 25 {w|w 6 25}

3r 2 3 4 4 3r 4 2 33 3 4

1 2

6

Page 532

All the graphs open downward from the origin. y  3x2 is narrower than y  x2, and y  6x2 is the narrowest. 2. y  x2 y  0.6x2 y  0.4x2

3

y  6  2 [x  (4) ] 3

y  6  2x  6 3

y  2x  12 75. To find the x-intercept, let y  0. 3x  4y  24 3x  4(0)  24 3x  24 x8 76. To find the x-intercept, let y  0. 2x  5y  14 2x  5(0)  14 2x  14 x7 77. To find the x-intercept, let y  0. 2x  4y  7 2x  4(0)  7 2x  7

All the graphs open downward from the origin. y  0.6x2 is wider than y  x2, and y  0.4x2 is the widest. 3. y  x2 y  (x  5) 2 y  (x  4) 2

7

x  2 or 3.5 78. To find the x-intercept, let y  0. 7y  6x  42 7(0)  6x  42 6x  42 x7 79. To find the x-intercept, let y  0. 2y  4x  10 2(0)  4x  10 4x  10 10

All the graphs open downward, have the same shape, and have vertices along the x-axis. However, each vertex is different. 4. y  x2 y  x2  7 y  x2  5

5

x   4 or 2 or 2.5 80. To find the x-intercept, let y  0. 3x  7y  9  0 3x  7(0)  9  0 3x  9  0 3x  9 x  3

Chapter 10

All the graphs open downward, have the same shape, and have vertices along the y-axis. However, each vertex is different.

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5. y  0.1x2 Sample answer: The graph will have a vertex at the origin, open downward, and be wider than y  x2. x 2 1 0 1 2

y 0.4 0.1 0 0.1 0.4

9. If |a| 7 1, the graph is narrower than the graph of y  x2. If 0 6 |a| 6 1, the graph is wider than the graph of y  x2. If a  0, it opens downward. If a  0, it opens upward. 10. The graph has the same shape as y  x2, but is shifted h units (left if h  0, right if h  0). 11. The graph has the same shape as y  x2, but is shifted k units (up if k  0, down if k  0). 12. The graph has the same shape as y  x2, but is shifted h units left or right and k units up or down as prescribed in Exercises 10 and 11.

y x

O y   0.1 x

2

6. y  (x  Sample answer: The graph will open upward and have the same shape as y  x2, but the vertex will be at (1, 0). x 3 2 1 0 1

Pages 535–536

7x2  2x  8 7x  2x  8  8  8 7x2  2x  8  0 Replace zero with f(x). f(x)  7x2  2x  8 3. Sample answer: 2

y  (x  1)2 x

O

7. y  4x2 Sample answer: The graph will open upward, have a vertex at the origin, and be narrower than y  x2. x 2 1 0 1 2

y

y

y 16 4 0 4 16

The graph has only one x-intercept. Thus, the related quadratic equation has only one distinct solution. 4. Graph f(x)  x2  7x  6.

y  4x 2 x

Sample answer:

8. y  x2  6

x 1 2 3 4 5 6

Sample answer: The graph will open upward and have the same shape as y  x2, but its vertex will be at (0, 6). y 2 5 6 5 2

x

O

O

x 2 1 0 1 2

Check for Understanding

1. The x-intercepts of the graph are 3 and 1. Thus, the roots of the equation are 3 and 1. 2. First rewrite the equation so one side is equal to zero.

y

y 4 1 0 1 4

Solving Quadratic Equations by Graphing

10-2

1) 2

y

O

x

f(x) 0 4 6 6 4 0

f (x ) x

O

f (x)  x 2  7x  6

The x-intercepts of the graph are 1 and 6. Thus, the solutions of the equation are 1 and 6.

y  x2 6

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5. Graph f(a)  a2  10a  25. Sample answer: a 3 4 5 6 7

9. First rewrite the equation so that one side is equal to zero. w2  3w  5 2  3w  5  5  5 w w2  3w  5  0 Graph f(w)  w2  3w  5. Sample answer:

f (a )

f(a) 4 1 0 1 4

w 2 1 1 2 4 5

a O

f (a)  a 2  10a  25

The graph has one a-intercept, 5. Thus, the solution of the equation is 5. 6. Graph f(c)  c2  3. Sample answer: c 2 1 0 1 2

The w-intercepts of the graph are between 2 and 1 and between 4 and 5. So one root is between 2 and 1, and the other root is between 4 and 5. 10. Let m and n represent the two numbers. The sum of the numbers is 4. mn4 The product of the numbers is 12. mn  12 Solve the first equation for m. mn4 m4n Replace m with 4  n in the second equation. mn  12 (4  n)n  12 4n  n2  12 2 n  4n  12  0

f (c)  c 2  3 c

7. Graph f(t)  t2  9t  5. Sample answer: f (t )  t 2  9t  5

6 4 2

98 7654 321 2 4 6 8 10 12 14 16

f (t )

O

t

Graph y  x2  4x  12.

The t-intercepts of the graph are between 9 and 8 and between 1 and 0. So one root is between 9 and 8, and the other is between 1 and 0. 8. Graph f(x)  x2  16. Sample answer: x 4 2 0 2 4

f(x) 0 12 16 12 0

2

[10, 10] scl: 1 by [10, 20] scl: 2

The x-intercepts are 2 and 6. Let n  2. Then m  4  (2) or 6. Let n  6. Then m  4  6 or 2. The numbers are 2 and 6.

f (x ) O 1 2 3 4x

4 321 2 4 f (x)6  x 2  16 8 10 12 14 16

The x-intercepts of the graph are 4 and 4. Thus, the roots of the equation are 4 and 4. Chapter 10

w

O

f (w)  w 2  3w  5

The graph has no c-intercept. Thus, the equation has no real number solutions: .

f(t) 5 3 15 15 3 5

f (w )

f (c )

f(c) 7 4 3 4 7 O

t 9 8 5 4 1 0

f(w) 5 1 7 7 1 5

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Pages 536–538

15. Graph f(x)  x2  2x  5  0. Sample answer:

Practice and Apply 2

11. Graph f(c)  c  5c  24. Sample answer: c 3 0 3 6 7 8

x 3 2 1 0 1

f (c ) 5 O 4 2 2 4 6 8 10 12 c 5 10 15 20 25 30 2 35 f (c)  c  5c  24

f(c) 0 24 30 18 10 0

f(n) 22 9 6 13 30

35 30 25 20 15 10 5 2 1 5

r 6 4 2 0 2

1 2 3 4 5 6n

2

f (r )

7654 321 O 1 2 3 r 2 4 6 8 10 12 14 16 f (r)  r 2  4r  12 18

The r-intercepts of the graph are 6 and 2. Thus, the solutions of the equation are 6 and 2.

2

13. Graph f(x)  x  6x  9. Sample answer:

17.

y

f (x )

f(x) 4 1 0 1 4

(2, 0)

x

O

(6, 0) (4, 2)

f (x )  x 2  6x  9

O

x

18.

The graph has one x-intercept, 3. Thus, the solution of the equation is 3. 14. Graph f(b)  b2  12b  36. Sample answer: b 4 5 6 7 8

f(r) 0 12 16 12 0

f (n)  5n 2  2n  6

The graph has no n-intercepts. Thus, the equation has no real number solutions: .

x 1 2 3 4 5

x

The graph has no x-intercept. Thus, the equation has no real number solutions: . 16. Graph f(r)  r2  4r  12. Sample answer:

f (n )

O

f (x)  x 2  2x  5

O

The c-intercepts of the graph are 3 and 8. Thus, the solutions of the equation are 3 and 8. 12. Graph f(n)  5n2  2n  6. Sample answer: n 2 1 0 1 2

f (x )

f(x) 8 5 4 5 8

(3, 4)

f (b )

f(b) 4 1 0 1 4

y

(6, 0) O

O

f (b )  b 2  12b  36

(0, 0) x

b

The graph has one b-intercept, 6. Thus, the solution of the equation is 6.

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21. Graph f(a)  a2  12. Sample answer:

19. Let m and n represent the numbers. mn9 mn  20 Solve the first equation for m. mn9 m9n Substitute 9  n for m in the second equation. mn  20 (9  n)n  20 9n  n2  20 2  9n  20  0 n Graph y  x2  9x  20.

a 4 3 2 0 2 3 4

f(a) 4 3 8 12 8 3 4

f (a ) 4

2

4a

2

O 4 8

12 f (a)  a 2  12

The a-intercepts are between 4 and 3 and between 3 and 4. So, one root is between 4 and 3, and the other root is between 3 and 4. 22. Graph f(n)  n2  7. Sample answer: n 3 2 1 0 1 2 3

[2, 8] scl: 1 by [2, 2] scl: 1

The x-intercepts are 4 and 5. Let n  4. Then m  9  4 or 5. Let n  5. Then m  9  5 or 4. The numbers are 4 and 5. 20. Let m and n represent the numbers. mn5 mn  24 Solve the first equation for m. mn5 m5n Substitute 5  n for m in the second equation. mn  24 (5  n)n  24 5n  n2  24 2 n  5n  24  0 Graph y  x2  5x  24.

f (n )

f(n) 2 3 6 7 6 3 2

n

O

f (n)  n 2  7

The n-intercepts of the graph are between 3 and 2 and between 2 and 3. So, one root is between 3 and 2, and the other root is between 2 and 3. 23. Graph f(c)  2c2  20c  32. Sample answer: c 8 6 4 2

f(c) 0 16 16 0

3 12 8

4

3 6 9 12 15 18 f (c)  2c 2  20c  3221

1 f (c ) O c

The c-intercepts of the graph are 8 and 2. So, the roots of the equation are 8 and 2.

[10, 10] scl: 1 by [10, 40] scl: 10

24. Graph f(s)  3s2  9s  12. Sample answer:

The x-intercepts are 3 and 8. Let n  3. Then m  5  (3) or 8. Let n  8. Then m  5  8 or 3. The numbers are 3 and 8.

s 4 3 2 1 0 1

f(s) 0 12 18 18 12 0

3 6

4

f (s ) O

2

3 6 9 12 15 18 f (s)  3s2  9s  1221

The s-intercepts of the graph are 4 and 1. So, the roots of the equation are 4 and 1.

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25. Graph f(x)  x2  6x  6. Sample answer: x 5 4 3 2 1

28. Rewrite the equation. x2  6x  7 2 x  6x  7  7  7 x2  6x  7  0 Graph f(x)  x2  6x  7. Sample answer:

f (x )

f(x) 1 2 3 2 1

x 5 4 3 2 1

x

O

f (x)  x 2  6x  6

The x-intercepts of the graph are between 5 and 4 and between 2 and 1. So, one root is between 5 and 4, and the other root is between 2 and 1. 26. Graph f(y)  y2  4y  1. Sample answer: y 0 1 2 3 4

f(y) 1 2 3 2 1

f (x )

f(x) 2 1 2 1 2

x

O

f (x)  x 2  6x  7

The x-intercepts are between 5 and 4 and between 2 and 1. So, one root of the equation is between 5 and 4, and the other root is between 2 and 1. 29. Rewrite the equation.

f (y )

m2  10m  21 m  10m  21  21  21 m2  10m  21  0 Graph f(m)  m2  10m  21. Sample answer: 2

O

y

f (y)  y 2  4y  1

m 3 4 5 6 7

The y-intercepts of the graph are between 0 and 1 and between 3 and 4. So, one root is between 0 and 1, and the other root is between 3 and 4. 27. Rewrite the equation. a2  8a  4 a  8a  4  4  4 a2  8a  4  0

f(m) 0 3 4 3 0

f (m )

O

m

2

f (m)  m 2  10m  21

Graph f(a)  a2  8a  4. Sample answer: a 1 0 2 4 6 8 9

f(a) 5 4 16 20 16 4 5

The m-intercepts of the graph are 3 and 7. So, the roots of the equation are 3 and 7. 30. Rewrite the equation. p2  16  8p 2  16  8p  8p  8p p p2  8p  16  0 Graph f(p)  p2  8p  16. Sample answer:

( ) 3 f a O 42 2 4 6 8 10 12a 3 6 9 12 15 18 21 f (a)  a 2  8a  4

p 2 3 4 5 6

The a-intercepts of the graph are between 1 and 0 and between 8 and 9. So, one root of the equation is between 1 and 0, and the other root is between 8 and 9.

f(p) 4 1 0 1 4

f (p )

O

f (p)  p 2  8p  16

p

The graph has only one p-intercept, 4. So, the root of the equation is 4.

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34. Sample answer:

31. Rewrite the equation. 12n2  26n  30 2 12n  26n  30  30  30 12n2  26n  30  0 Graph f(n)  12n2  26n  30. Sample answer: n 1 0 1 2 3

y

( ) 20 f n 10 O 21 1 2 3 4 5 6n 10 20 30 40 50 f (n)  12n 2  26n  30 60

f(n) 8 30 44 34 0

(3, 5)

35. f(x)  x2  4x  12 Sample answer: x 6 4 2 0 2

One n-intercept of the graph is 3, and the other n-intercept is between 1 and 0. So, one root of the equation is 3, and the other root is between 1 and 0. 32. Rewrite the equation. 4x2  35  4x 2  35  4x  4x  4x 4x 4x2  4x  35  0 Graph f(x)  4x2  4x  35. Sample answer: x 4 3 1 1 2 3

( ) f (x)  x 2  4x  12 f x 15 12 9 6 3 O 765 4 321 1 2 3x 3 6

2

38. A  3 bh 2

 3 (8) (16) 

256 3

1

or 85 3 1

The area to be painted is about 85 3 square feet. 39. Multiply the area under one arch by 12 to find the area under 12 arches.

1 12

The x-intercepts of the graph are between 4 and 3 and between 2 and 3. So, one root of the equation is between 4 and 3, and the other root is between 2 and 3. 33. Sample answer:

1256 2

12 853  12 3 or 1024 ft2 Since there are two coats, multiply the area under the 12 arches by 2. 2(1024)  2048 ft2 Each gallon covers 200 ft2, so

y

2048 200

(1, 6)

O

f(x) 0 12 16 12 0

The x-intercepts are 6 and 2. 36. The length of the segment is the distance between the x-intercepts, which is |2  (6)| on 8 feet. 37. The highest point on the arch is at the vertex of the parabola, which is at (2, 16). Thus, the height of the arch is 16 feet.

( ) 5 f x O 4321 1 2 3 4x 5 10 15 20 25 30 35 f (x)  4x 2  4x  35

f(x) 13 11 35 27 11 13

x

O

256

6

 25 or 10 25 gallons. Round up to 11 gallons and find the total cost. 11(27)  297 dollars The paint for the walls would cost $297. 40. Graph y  0.005x2  0.22x and use the Zero feature to find the relevant x-intercept.

x

[0, 50] scl: 10 by [0, 5] scl: 1

The x-intercept is about 44. The ball travels a horizontal distance of 44 yd.

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41. Graph y  16x2  30x  1000 and use the Zero feature to find the relevant x-intercept.

48. Since quadratic functions can be used to model a golf ball after it is hit, solving the related quadratic equation will determine where the ball hits the ground. Answers should include the following. • In the golf problem, one intercept represents the ball’s original location and the other intercept represents where the ball hits the ground. • Using the quadratic function y  0.0015x2  0.3x, the ball will hit the ground 200 yd from the starting point. 49. C; The graph in answer choice C has no x-intercepts. 50. A; The x-intercepts of the graph are 2 and 2. Thus, 2 and 2 are the roots of the equation. 51. Graph y  x3  x2  4x  4.

[0, 10] scl: 1 by [0, 1100] scl: 100

42. 43.

44.

45.

The x-intercept is about 9. It took about 9 seconds for the ball to hit the ground. Yes; there are almost 9  3  1 or 5 seconds to warn them. The total area to be mowed is 500  400 or 200,000 ft2. Since each will mow half, each mows 1  200,000 or 100,000 ft2. 2 A  w 100,000  (500  2x)(400  2x) 100,000  200,000  1000x  800x  4x2 100,000  200,000  1800x  4x2 0  100,000  1800x  4x2 or 4x2  1800x  100,000  0 Graph y  4x2  1800x  100,000 and find the relevant x-intercept.

[5, 5] scl: 1 by [10, 10] scl: 1

The x-intercepts of the graph are 2, 1, and 2. Thus, the solutions of the equation are 2, 1, and 2. 52. Graph y  2x3  11x2  13x  4.

[0, 70] scl: 5 by [0, 100,000] scl: 10,000

The x-intercept is about 65. Kirk should mow a width of about 65 ft.

[1, 5] scl: 1 by [10, 1] scl: 1

46. Kirk must mow a width of 65 feet. Since the mower cuts a width of 5 feet each time, Kirk 65 should go around the field 5 or 13 times.

1

The x-intercepts of the graph are 2, 1, and 4. 1 Thus, the solutions of the equation are 2 , 1, and 4.

x3  2x2  3x

47. The x-intercepts of the graph of f(x)  x  5 are the values of x that satisfy the following equation. x3  2x2  3x x  5

0

The value of a fraction is 0 only when the numerator is 0, so we can solve x3  2x2  3x  0 to find the x-intercepts. x3  2x2  3x  0 x(x2  2x  3)  0 x(x  3)(x  1)  0 or x  1  0 x  0 or x  3  0 x  3 x1 The x-intercepts are 3, 0, and 1.

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Page 538

Since the coefficient of the x2 term is positive, the vertex is a minimum.

Maintain Your Skills 2

53. In y  x  6x  9, a  1 and b  6. b

x  2a

y

6

x  2(1) or 3

O

x  3 is the equation of the axis of symmetry. y  x2  6x  9 y  (3) 2  6(3)  9 y  9  18  9 or 0 The vertex is at (3, 0). Since the coefficient of the x2 term is positive, the vertex is a minimum.

4

8

x

12

4 8 12

y  0.5x 2  6x  5

Exercises 56–58 For checks, see students’ work.

y

m2  24m  144 m  24m  144  0 m2  2(m)(12)  122  0 (m  12) 2  0 m  12  0 m  12 2 57. 7r  70r  175 56.

2

x

O

y  x 2  6x  9

1 (7r2 ) 7

r2  10r  25 r  10r  25  0 r2  2(r)(5)  52  0 (r  5) 2  0 r50 r5 58. 4d2  9  12d 4d2  12d  9  0 2  2(2d)(3)  32  0 (2d) (2d  3) 2  0 2d  3  0 2d  3

54. In y  x2  4x  3, a  1 and b  4.

2

b

x  2a 4

x  2(1) or 2 x  2 is the equation of the axis of symmetry. y  x2  4x  3 y  (2) 2  4(2)  3 y  4  8  3 or 1 The vertex is at (2, 1). Since the coefficient of the x2 term is negative, the vertex is a maximum.

3

y y  x 2  4x  3 O

1

 7 (70r  175)

59. x

10m4 30m



d  2 or 1.5

11030 21mm 2 4

1

 3m41

m3 3

1

 3m3 or 60.

22a2b5c7 11abc2



22 a b c 111 21 a 21 b 21c 2 2

5

7 2

21 51 72

55. In y  x x

0.5x2

b 2a (6) 2(0.5)

 2a b  2ab4c5

 6x  5, a  0.5 and b  6. 61.

9m3n5 27m2n5y4

 

or 6



x  6 is the equation of the axis of symmetry. y  0.5(6) 2  6(6)  5 y  18  36  5 or 13 The vertex is at (6, 13).



c

m n 1 19 27 21 m 21 n 21 y 2 3

5

2

5

4

1 3m3(2)n55y4 1 3m5n0y4 m5y4  3

62. Let n be the number of books. 30  1.5n 55 and 30  1.5n  60 1.5n 25 1.5n  30 3n 50 n  20 n

50 3

2

or 163

Since the number of books must be a whole number, there must be at least 17 but no more than 20 books in the crate.

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Square the result. (6)2  36 Thus, c  36. 7. t2  5t  c

63. Yes; a2  14a  49  a2  2(a)(7)  72  (a  7) 2 64. Yes; m2  10m  25  m2  2(m)(5)  52  (m  5) 2 65. No; 64 is not a perfect square. 66. Yes; 4y2  12y  9  (2y) 2  2(2y)(3)  32  (2y  3) 2 67. No; 4 is not a perfect square. 68. Yes; 25x2  10x  1  (5x) 2  2(5x)(1)  12  (5x  1) 2

Find 1 (5) 2

152 2

2.

3. 4.

5.

x2

x x

1 (12) 2

25 . 4 1 2

of

1 2

of

Add 9 to each side. c2  6c  9  7  9 (c  3) 2  16 c  3  4 c  3  3  4  3 c3 4 c  3  4 or c  3  4 7  1 The solutions are 1, 7. 9. x2  7x  12 The x2 and x terms are already isolated. Find 7 and square the result.

1 1

172 22  494

Add

49 4

to each side. 49 4 7 2 2

2

x  7x 

 12 

1x  2

7

1

7

7

1

49 4

1

4

x  2  2 7

x  2  2  2  2 7

x  2 7

1

x  2  2 6

 2 or 3 10.

or

1 2 7

1

x  2  2 8

 2 or 4

The solutions are 4, 3. v2  14v  9  6 2  14v  9  9  6  9 v v2  14v  15 Find

1 2

m  7  120 m  7  7  120  7 m  7 120 m  7  120 or m  7  120 m  11.5 m  2.5 The solutions are 11.5, 2.5. 6. a2  12a  c 1 2

25 4



162 22  (3)2 or 9

The area is x2  4x  4. Graphing f(x)  x2  5x  7 would not result in an exact answer, and x2  5x  7 cannot be factored. Divide each side by 5. b2  6b  9  25 (b  3) 2  25 2(b  3) 2  125 0b  3 0  5 b  3  5 b  3  3  5  3 b3 5 b  3  5 or b  3  5 b  2 b8 The solutions are 2, 8. m2  14m  49  20 (m  7) 2  20 2(m  7) 2  120 0m  7 0  120

Find

2

8. c2  6c  7 The c2 and c terms are already isolated. Find 6 and square the result.

Check for Understanding 1 1

5

2

Thus, c 

1. Sample answer: x x

of 5.

Square the result.

10-3 Solving Quadratic Equations by Completing the Square Page 542

1 2

14 2 2

1 2

of 14 and square the result.

 72 or 49

Add 49 to each side. v2  14v  49  15  49 (v  7) 2  64 v  7  8 v  7  7  8  7 v  7 8 v  7  8 or v  7  8 1  15 The solutions are 15, 1.

of 12.  6

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11. r2  4r  2 The r2 and r terms are already isolated. Find 4 and square the result.

142 22  (2)2 or 4

1 2

14. Let s represent the length of a side of the square. The area of the square is then s2. When the length is increased by 6 inches and the width by 4 inches, the area of the resulting rectangle is (s  6)(s  4). The area of the rectangle is twice the area of the square.

of

Add 4 to each side. r2  4r  4  2  4 (r  2) 2  6 r  2  16 r  2  2  16  2 r  2  16 r  2  16 or r  2  16  4.4  0.4 The solutions are 0.4, 4.4. 12. a2  24a  9  0 2  24a  9  9  0  9 a a2  24a  9 Find

1 2

1 2

24 2 2

(s  6)(s  4)  2s2  4s  6s  24  2s2 s2  10s  24  2s2 2  10s  24  s2  10s  2s2  s2  10s s 24  s2  10s s2

Find

Add 25 to each side and reverse the sides. s2  10s  25  24  25 (s  5) 2  49 s  5  7 s  5  5  7  5 s5  7 s  5  7 or s  5  7  12  2 Since we are looking for a length, ignore the negative number. The length of a side of the square is 12 in.

 (12) 2 or 144

Add 144 to each side. a2  24a  144  9  144 (a  12) 2  135 a  12  1135 a  12  12  1135  12 a  12  1135 a  12  1135 or a  12  1135  23.6  0.4 The solutions are 0.4, 23.6. 13. 2p2  5p  8  7 Since the coefficient of the p2 term is not 1, first divide each side by 2, the coefficient of the p2 term.

3 1 24

1 2

Pages 542–543 2

1

2

5

17

result.

5

17

p  4   3 16 5

5

p  4  4   3 16  4 5

p4  5

p4  2.3

Chapter 10

17

3 16

or

17

3 16 5

p4

Practice and Apply

15. b  4b  4  16 (b  2) 2  16 2(b  2) 2  116 0b  2 0  4 b  2  4 b  2  2  4  2 b2  4 b  2  4 or b  2  4  2 6 The solutions are 2, 6. 16. t2  2t  1  25 (t  1) 2  25 2(t  1) 2  125 0t  1 0  5 t  1  5 t  1  1  5  1 t  1  5 t  1  5 t  1  5  6 4 The solutions are 6, 4. 17. g2  8g  16  2 (g  4) 2  2 2(g  4) 2  12 0g  4 0  12 g  4  12 g  4  4  12  4 g  4  12 g  4  12 or g  4  12  2.6  5.4 The solutions are 2.6, 5.4.

5 2 5 2 25 2  4 or 16 25 Add 16 to each side. 5 25 1 25 p2  2p  16  2  16 5 2 17 p  4  16 1 2

of 10 and square the result.

 (5) 2 or 25 110 2 2

of 24 and square the result.

2p2  5p  8 7 2 2 5 7 p2  2p  4  2 5 7 p2  2p  4  4  2  4 5 1 p2  2p  2 1 5 Find 2 of 2 and square the

1 2

17

3 16

 0.2

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18. y2  12y  36  5 (y  6) 2  5 2(y  6) 2  15 0y  6 0  15 y  6  15 y  6  6  15  6 y  6  15 y  6  15 or y  6  15  3.8  8.2 The solutions are 3.8, 8.2. 19. w2  16w  64  18 (w  8) 2  18 2(w  8) 2  118 0w  8 0  118 w  8  118 w  8  8   118  8 w  8  118 w  8  118 or w  8  118  12.2  3.8 The solutions are 12.2, 3.8. 20. a2  18a  81  90 (a  9) 2  90 2(a  9) 2  190 0a  9 0  190 a  9  190 a  9  9  190  9

26. k2  11k  c Find c

c

1 of 16 2 16 2 2

1 2

Square

12

c 2 2

c

1 of 10 2 10 2 2

1 2

2

c 4

4

c 2

c

1 of 2 22 2 2

1 2

c 25.

1 of 2 34 2 2

1 2

and set it equal to 81. Then solve for c.

 81

1c4 2  4(81) 2

Square

12

c 2 2 2

c 4

4

c 2

and set it equal to 144. Then solve for c.

 144  144

1 2  4(144) c2 4

c2  576 c  24 29. s2  4s  12  0 2 s  4s  12  12  0  12 s2  4s  12 2 s  4s  4  12  4 (s  2) 2  16 s  2  4 s  2  2  4  2 s2  4 s  2  4 or s  2  4  2 6 The solutions are 2, 6. d2  3d  10  0 30. d2  3d  10  10  0  10 d2  3d  10

and square the result.

9

9

d2  3d  4  10  4

1d  32 22  494

and square the result.

3

7

3

7

d  2  2 3

3

d  2  2  2  2

22 and square the result.

3

d  2  3

7

d  2  2

 112 or 121 24. a2  34a  c Find

121 4

 81

 (5) 2 or 25 23. w2  22w  c Find

or

c2  324 c  18 28. x2  cx  144

 82 or 64 22. y2  10y  c Find

1 2

11 and square the result.

27. x2  cx  81

a  9  190 a  9  190 or a  9  190  18.5  0.5 The solutions are 18.5, 0.5. 21. s2  16s  c Find

1 of 2 11 2 2

10

  2 or 5

˛˛

or

7 2

3

7

d  2  2 4

 2 or 2

The solutions are 5, 2.

34 and square the result.

 172 or 289  7p  c

p2

Find c

1 of 2 7 2 2

1 2

7 and square the result. or

49 4

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31.

y2  19y  4  70 y  19y  4  4  70  4 y2  19y  66

d2  8d  7  0 d  8d  7  7  0  7 d2  8d  7 2  8d  16  7  16 d (d  4) 2  9 d  4  3 d  4  4  3  4 d4  3 d  4  3 or d  4  3 1 7 The solutions are 1, 7. 37. s2  10s  23 2  10s  25  23  25 s (s  5) 2  48 s  5   148 s  5  5   148  5 s  5  148 s  5  148 or s  5  148  1.9  11.9 The solutions are 1.9, 11.9. 38. m2  8m  4 2  8m  16  4  16 m (m  4) 2  20 m  4  120 m  4  4  120  4 m  4  120 m  4  120 or m  4  120  0.5  8.5 The solutions are 0.5, 8.5. 39. 9r2  49  42r 9r2  49  42r  42r  42r 9r2  42r  49  0 (3r  7) 2  0 3r  7  0 3r  7  7  0  7 3r  7 36.

2

361 4 19 2 y 2 19 y 2 19 19  2  2

y2  19y 

1

y

y 

2

 66 

2

361 4

625 4 25  2 25 19  2  2 19 25 y 2  2 19 25 or y  2  2 44  2 or 22

19 25  2 2 6 2 or 3



˛˛˛˛˛˛

The solutions are 3, 22. d2  20d  11  200 2  20d  11  11  200  11 d d2  20d  189 2  20d  100  189  100 d (d  10) 2  289 d  10  17 d  10  10  17  10 d  10  17 d  10  17 or d  10  17  27 7 The solutions are 27, 7. 33. a2  5a  4

32.

25 4 5 a22 5 a2 5 5 22

a2  5a 

1

a

2

25 4

 4  9

4 3

 2 3

5

 2  2 5

3 2 5 2 8 2

a2  5

3

a22 2

 2 or 1

˛˛˛˛˛˛

or

a

3

2 or 4

7

Chapter 10

1

r  3 or 23

The solutions are 1, 4. 34. p2  4p  21 2 p  4p  4  21  4 (p  2) 2  25 p  2  5 p  2  2  5  2 p2  5 p  2  5 or p  2  5  3 7 The solutions are 3, 7. 35. x2  4x  3  0 2 x  4x  3  3  0  3 x2  4x  3 2 x  4x  4  3  4 (x  2) 2  1 x  2  1 x  2  2  1  2 x  2  1 x  2  1 or x  2  1  3  1 The solutions are 3, 1.

The solution is 40.

7 3

1

or 23.

4h2  25  20h  25  20h  20h  20h 4h2  20h  25  0 (2h  5) 2  0 2h  5  0 2h  5  5  0  5 2h  5 4h2

5

h2 The solution is

460

5 2

1

or 22.

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41. 0.3t2  0.1t  0.2 2

0.3t  0.1t 0.3 1 t2  3t 1 1 t2  3t  36 1 t62

 0.3

1

5

1

5

1

2

44. 9w2  12w  1  0 9w2  12w  1 9 4 1 w2  3 w  9 4 1 1 w2  3 w  9  9 4 w2  3w 4 4 w2  3 w  9 2 w32

0.2 2

3  

2 1  36 3 25 36

t  6  6 1

1

1

t  6  6  6  6 1

t  6  1

5

t  6  6 

6 6

˛˛˛˛˛

5 6 1

or 1

4 6



or

2

v

v

5 2



5 2 5 2

2

w3

 5v

45.

 5v  5v

43.

2

0 0

0 5 2

1 22

5 2

1 22.

or



5

39 2

w3

or

5

39

1

0

2  2(0)

5

d2  2d  6  0

5

11

5

5

11

5

d  4  4  4  4 5

11 4 5 11 4 4 16  4 or 4

d4  5

11 4

d4

7 5



6 4

or

or

d

3 2 3

7  5 12 5

The solutions are 2, 4.

1

46. 3

12 5

x  1  1  3

12 5

1

x  1 

35

or

25

d  4  4

0 5

0



25

1d  54 22  121 16

x  1  3

12

5

39

5

0

 2x  1 

35

2

d2  2d  16  6  16

7

x  1 

5

5

x2  2x  5 (x 

2

d2  2 d  6

5 2

1) 2

5

1 2 5 d  4d  3 2 1 2 5 d  4d  3 2

 10x  7  0

x2

2

5

0

5x2  10x  7 5 7 x2  2x  5 7 7 x2  2x  5  5

5

9

d2  2 d  6  6  0  6

v  or

5x2

4

 0.1  1.4 The solutions are 0.1, 1.4.

2v

 0.4

The solution is

1

99

2

42. 0.4v2  2.5  2v

2

1

9

w3 

2

1

1

09

w  3  3  3 9  3

2 3

The solutions are 1, 3. 0.4v2  2.5 0.4 25 v2  4 25 v2  4  5v 25 v2  5v  4 5 v22

0

w  3  3 9

5

t  6  6

or

2

0

9

0

2  3(0)

7

3

f 2  2f  2  0 7

3

3

3

f 2  2f  2  2  0  2

12

x  1 

1

1 2 7 1 f  6f  2 3 1 2 7 1 f  6f  2 3

3 2 7 49 3 49 f 2  2f  16  2  16 7 25 f  4 2  16 7 5 f  4  4 7 7 5 7 f  4  4  4  4 7 5 f4  4 7 5 7 5 f  4  4 or f  4  4 2 1 12  4 or 2  4 or 3 1 The solutions are 2, 3. 7

f 2  2f 

12

35

1

 2.5  0.5 The solutions are 2.5, 0.5.

2

˛˛˛

461

Chapter 10

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x2  4x  c  0 x  4x  c  c  0  c x2  4x  c 2  4x  4  c  4 x (x  2) 2  4  c x  2   14  c x  2  2   14  c  2 x  2  14  c 2 48. x  6x  c  0 x2  6x  c  c  0  c x2  6x  c 2 x  6x  9  c  9 (x  3) 2  9  c x  3   19  c x  3  3   19  c  3 x  3  19  c 49. area of garden  9  6 or 54 area of path  area of large rectangle  area of garden  (2x  9)(2x  6)  54  4x2  12x  18x  54  54  4x2  30x Since these areas are equal, set the expressions equal to each other and solve for x. 4x2  30x  54

x2  4x  12  0 x  4x  12  12  0  12 x2  4x  12 2  4x  4  12  4 x (x  2) 2  8 There are no real solutions since the square of a number cannot be negative. 52. The dimensions of the photograph are 12  4x and 12  2x. The area of the photograph is 54 square inches. (12  4x)(12  2x)  54 144  24x  48x  8x2  54 8x2  72x  144  54 51.

47.

2

4x2  30x 4 15 x2  2 x 15 225 x2  2 x  16 15 x 4 2 15 x 4 15 15 x 4  4

1

2

x 



21 4

8x2  72x  144 8

    

˛˛

or

x2  9x  18  18  x2  9x 

1x

x 9

x23

˛˛

or

3

x  2 or 1.5

15

3 3 2

or

y  0.059x2  7.423x  362.1 300  0.059x2  7.423x  362.1 0.059x2  7.423x  362.1 0.059

9  3 9

 3  2

9

x23 x

15 2

x2  8x  35

5084.75  x2  125.81x  6137.29 1052.54  x2  125.81x 2904.50  x2  125.81x  3957.04 2904.50  (x  62.91) 2

x2

or 7.5

Original equation

 8x  16  35  16

Since

2

182 22  16, add 16 to each side.

Factor x2  8x  16.

(x  4)  51 x  4   151

Take the square root of each side.

x  4  4  151  4

 12904.50  x  62.91 62.91  12904.50  x x  62.91  12904.50 or x  62.91  12904.50 9  117 or in 1909 or in 2017

Subtract 4 from each side.

x  4  151

Simplify.

x  4  151 or x  4  151 x  11.14 x  3.14 The solutions are 11.14, 3.14. 54. C; 225 is not the square of any real number. 55. A; x2  5x  14

In the year 2017, an average American will consume 300 pounds of bread and cereal per year.

x2  5x 

1

2

25 4

 14 

5 81 A; x  2 2  4

Chapter 10

2

Use x  1.5 because the solution must satisfy 12  2x  0. (The width must be positive.) Then the width of the photograph is 12  2(1.5) or 9 inches, and the height is 12  4(1.5) or 6 inches. 53. Al-Khwarizmi used squares to geometrically represent quadratic equations. Answers should include the following. • Al-Khwarizmi represented x2 by a square whose sides were each x units long. To this square, he added 4 rectangles with length 8 x units long and width 4 or 2 units long. This area represents 35. To make this a square, four 4  4 squares must by added. • To solve x2  8x  35 by completing the square, use the following steps.

21 4

Ignore the negative number. The path is 1.5 m wide. 50. Replace y with 300 and solve for x.





9

 4 or 2

300 0.059

81 4 9 2 2 9 x2 9 9 2 2

x2  9x 

54 8 27 4 27  18 4 45 4 45 81  4 4

x2  3

x  4  6

or 9



x2  9x  18 

54 4 27 2 27 225  16 2 441 16 21 4 21 15 4  4 15 21 4  4



x 15 4 36 4

2

462

25 4

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Page 544

60. y  x2  3x  10 Sample answer:

Maintain Your Skills 2

56. Graph f(x)  x  7x  12. Sample answer: x 5 4 3 2 1

f(x) 2 0 0 2 6

x 2 0 1 2 3 5

f (x )

x

O

y

y 0 10 12 12 10 0

2

O

x 4 2 0 2 4

x 0 1 2 3 4

 16. f (x )

f(x) 0 12 16 12 0

4

2

O

2

4x

4 8

8 12

y  x 2  3x  10

x2

O

x y  x 2  3x  4

f(x) 9 6 5 6 9

GCF: a  b  b or ab2 63. 32m2n3  2  2  2  2  2  m  m  n nn m  m  n 8m2n  2  2  2  m  m  m n n 56m3n2  2  2  2  7 

 2x  6. f (x )

GCF: 2  2  2  m  m  n or 8m2n 64. y  2x xy9 Substitute 2x for y in the second equation. xy9 x  2x  9 3x  9

f (x )  x 2  2x  6

x

O

3x 3

The graph has no x-intercepts. Thus, the equation has no real number solutions: .

y 32 20 16 20 32

32

y

24 16 8 4

2

O

y  4x 2  16

2

9

3

x3 Use y  2x to find the value of y. y  2x y  2(3) or 6 The solution is (3, 6). 65. x  y  3 2x  3y  5 Substitute y  3 for x in the second equation. 2x  3y  5 2(y  3)  3y  5 2y  6  3y  5 y  6  5 y  6  6  5  6 y  1 (1)(y)  (1) (1) y1 Use x  y  3 to find the value of x. xy3 x  1  3 or 4 The solution is (4, 1).

59. y  4x2  16 Sample answer: x 2 1 0 1 2

y

62. 14a2b3  2  7  a a b  b b 20a3b2c  2  2  5  a  a  a  b  b  c 35ab3c2  5  7  a  b  b bcc

f (x )  x 2  16

The x-intercepts of the graph are 4 and 4. Thus, the solutions of the equation are 4 and 4. x 1 0 1 2 3

y 4 2 2 4 8

12 16

58. Graph f(x) 

x

61. y  x2  3x  4 Sample answer:

The x-intercepts of the graph are 4 and 3. Thus, the solutions of the equation are 4 and 3. 57. Graph f(x) 

4

4

f (x )  x 2  7x  12

x2

2

4x

463

Chapter 10

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66. x  2y  3 3x  y  23 Solve the first equation for x. x  2y  3 x  2y  2y  3  2y x  2y  3 Substitute 2y  3 for x in the second equation. 3x  y  23 3(2y  3)  y  23 6y  9  y  23 7y  9  23 7y  9  9  23  9 7y  14 7y 7



72. 2b2  4ac  2(2) 2  4(1) (15)  14  60  164 8 73. 2b2  4ac  272  4(2) (3)  149  24  125 5 74. 2b2  4ac  252  4(1) (2)  125  8  133  5.7 75. 2b2  4ac  272  4(2) (5)

14 7

 149  40  189  9.4

y2 Use x  2y  3 to find the value of x. x  2y  3 x  2(2)  3 or 7 The solution is (7, 2). 67. 3 6 x 6 1 68. x 2 or x 7 1 69. 5x  3y  7 5x  3y  5x  7  5x 3y  5x  7 3y 3



y

Page 544 b

x  2a (1)

x   2(1) 1

x  2 or 0.5

5x  7 3 5 7 x 3 3

x  0.5 is the equation of the axis of symmetry. y  x2  x  6 y  (0.5) 2  (0.5)  6 y  0.25  0.5  6 y  6.25 The vertex is at (0.5, 6.25). The vertex is a minimum since the coefficient of the x2 term is positive.

The slope of any line perpendicular to this line is 3 5 5, the opposite of the reciprocal of 3. Use the point-slope form. y  y  m(x  x ) 1

1

3

y  (2)  5 (x  8) 3

y  2  5 x  3

y  2  2  5 x  3

y  5 x 

24 5 24 5 14 5

Practice Quiz 1

1. In y  x2  x  6, a  1 and b  1.

y

2

O

x

70. We use the points (2, 0) and (0, 2). Find the slope. y2  y1

mx

2

 

 x1

2  0 0  2 2 or 1 2

y  x2  x  6

Use the slope-intercept form. y  mx  b y  x  (2) yx2 71. We use the points (1, 2) and (0, 0). Find the slope. y2  y1

mx

2

 x1

0  2 (1) 2 or 2 1

0 

Use the slope-intercept form. y  mx  b y  2x  0 y  2x

Chapter 10

464

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2. In y  2x2  3, a  2 and b  0.

5. Graph f(x)  x2  2x  1. Sample answer:

b

x  2a

x 1 0 1 2 3

0

x  2(2) or 0 x  0 is the equation of the axis of symmetry. y  2x2  3 y  2(0) 2  3 y  0  3 or 3 The vertex is at (0, 3). The vertex is a minimum since the coefficient of the x2 term is positive. 21 18 15 12 9 6 3

x

O

The x-intercepts lie between 1 and 0 and between 2 and 3. So, one root is between 1 and 0, and the other is between 2 and 3. 6. Graph f(x)  x2  5x  6. Sample answer:

y

x 1 0 2 4 6

3. In y  3x2  6x  5, a  3 and b  6. b

x  2a (6)

x  2(3) 6

x  6 or 1 x  1 is the equation of the axis of symmetry. y  3x2  6x  5 y  3(1) 2  6(1)  5  3  6  5 8 The vertex is at (1, 8). The vertex is a maximum since the coefficient of the x2 term is negative. y  3x 2  6x  5

x

4. Graph f(x)  x2  6x  10. Sample answer: f(x) 5 2 1 2 5

f (x )

f (x )  x 2  6x  10

O

f(x) 0 6 12 10 0

( ) 6 f x 4 2 O 422 2 4 6 8 10 12 x 4 6 8 10 12 14

f (x )  x 2  5x  6

The x-intercepts are 1 and 6. Thus, the solutions of the equation are 1 and 6. s2  8s  15 7. 2  8s  16  15  16 s (s  4) 2  1 s  4  1 s  4  4  1  4 s  4  1 s  4  1 or s  4  1  5  3 The solutions are 5, 3. 8. a2  10a  24 2  10a  25  24  25 a (a  5) 2  1 a  5  1 a  5  5  1  5 a5  1 a  5  1 or a  5  1 4 6 The solutions are 4, 6. 9. y2  14y  49  5 (y  7) 2  5

y

x 5 4 3 2 1

f (x )

f (x )  x 2 2x  1

y  2x 2  3 O x 4321 3 1 2 3 4

O

f(x) 2 1 2 1 2

y  7  15 y  7  7  15  7 y  715 y  7  15 or y  7  15  4.8  9.2 The solutions are 4.8, 9.2.

x

The graph has no x-intercepts. The equation has no real number solutions: .

465

Chapter 10

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2b2  b  7  14

10.

2b2  b  7 2 1 7 2 b  2b  2 1 7 7 b2  2b  2  2 1 b2  2b 1 1 b2  2b  16 1 2 b4 1 b4

1

2

1



5. y  x2  6x  1 y  (x2  6x  9)  1  9 y  (x  3) 2  10 The vertex is (3, 10) .

14 2

7 7

72    

y

21 2 21 1  16 2 169 16 13 4

1

13

y  x 2  6x  1 4 4

6

2

4 8

1

b  4  4  4  4 1

b4  1

13 4

12 4

or 3

b4 

or

b 

13 4 1 13  4 4 14 7 or 2 4

Solving Quadratic Equations by Using the Quadratic Formula

10-4

The solutions are 3, 3.5.

Page 550 Page 545

Check for Understanding

1. Sample answer: (1) Factor x2  2x  15 as (x  3)(x  5). Then according to the Zero Product Property, either x  3  0 or x  5  0. Solving these equations, x  3 or x  5. (2) Rewrite the equation as x2  2x  15. Then add 1 to each side of the equation to complete the square on the left side. Then (x  1)2  16. Taking the square root of each side, x  1  4. Therefore, x  1  4 and x  3 or x  5. (3) Use the Quadratic Formula.

Graphing Calculator Investigation (Follow-Up of Lesson 10-3)

1. y  a(x  h) 2  k The vertex is (h, k) . 2. y  x2  2x  3 y  (x2  2x  1 )  3  1 y  (x  1 ) 2  4 3. y  x2  2x  7 y  (x2  2x  1)  7  1 y  (x  1) 2  8 The vertex is (1, 8) .

Therefore, x 

2  2(2) 2  4(1) (15) 2(1)

2  164 . 2

x

y  x 2  2x  7

x

2

4. y  x  4x  8 y  (x2  4x  4)  8  4 y  (x  2) 2  4 The vertex is (2, 4). y

b  2b2  4ac 2a



7  272  4(1) (6) 2(1)



7  149  24 2



7  125 2



7  5 2 7  5 2

x

or

x

7  5 2

 6  1 The solutions are 6, 1. y  x 2  4x  8 O

Chapter 10

or

x Simplifying the expression, x  3 or x  5. See students’ preferences. 2. Sample answer: x2  x  5  0. See students’ work. Any quadratic equation for which the discriminant is negative is correct. 3. Juanita; you must first write the equation in the form ax2  bx  c  0 to determine the values of a, b, and c. Therefore, the value of c is 2, not 2. 4. x2  7x  6  0

y O

2x

O

x

466

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5.

t2  11t  12 t  11t  12  12  12 t2  11t  12  0

2

t     t

2

1

w

11  13 2

10  2102  4(1) (12) 2(1)



10  1100  48 2



10  152 2

r

10  152 2

or

r

5  252  4(3) (11) 2(3)



5  125  132 6



5  1107 6

10  152 2

4x2  2x  17 4x  2x  17  17  17 4x2  2x  17  0 b  2b2  4ac 2a



2  222  4(4) (17) 2(4)



2  14  272 8



2  1276 8

x

2  1276 8

or

x

2

 15w  2  0

b  2b2  4ac 2a

15  1225  200 50



15  125 50 15  5 50 15  5 or 50 10 1 or 5 50

15  5 50 20 2 or 5 50

w 

The solutions are

2

x

2

1 2 , . 5 5

10. m2  5m  6  0 b2  4ac  52  4(1) (6)  25  24  49 Since the discriminant is positive, the equation has 2 real roots. 11. s2  8s  16  0 b2  4ac  82  4(1) (16)  64  64 0 Since the discriminant is 0, the equation has 1 real root. 12. 2z2  z  50 2 2z  z  50  50  50 2z2  z  50  0 b2  4ac  12  4(2) (50)  1  400  399 Since the discriminant is negative, the equation has no real roots. 13. Write an expression for the volume of the pan. V  /wh  (x  4) (x  4) (2)  (x2  8x  16)(2)  2x2  16x  32 The volume is to be 441 cubic centimeters.

; There are no real solutions because the discriminant is negative. 8.

3





b  2b2  4ac 2a



2

(15)  2(15) 2  4(25) (2) 2(25)

w

 8.6  1.4 The solutions are 8.6, 1.4. 7. 3v2  5v  11  0 v

3





b  2b2  4ac 2a



3

w2  5w  25  0

25w2

 12 1 The solutions are 12, 1. 6. r2  10r  12  0 r

3

25 w2  5w  25  25  0

11  2112  4(1) (12) 2(1) 11  1121  48 2

t

3

w2  25  5w  5w  5w

b  2b2  4ac 2a

11  1169 2 11  13 2 11  13 or 2

3

w2  25  5w

9.

2

2  1276 8

2x2  16x  32  441 2x2  16x  409  0

 2.3  1.8 The solutions are 2.3, 1.8.

x

(16)  2(16) 2  4(2) (409) 2(2)

16  13528 4 16  13528  4

 x

or

x

16  13528 4

 10.8  18.8 Ignore the negative number. The original sheet should be about 18.8 cm by 18.8 cm.

467

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Pages 550–552

19. r2  25  0

Practice and Apply

14. x2  3x  18  0 x

r

b  2b2  4ac 2a



3  232  4(1) (18) 2(1)



3  181 2 3  9 2 3  9 or 2

 x

x

b  2b2  4ac 2a



0  10  4(1) (25) 2(1)



 1100 2

; There are no real solutions because the discriminant is negative. 20. 2x2  98  28x 2  98  28x  28x  28x 2x 2x2  28x  98  0

3  9 2

 6 3 The solutions are 6, 3. 15. v2  12v  20  0

x

b  2b2  4ac 2a

b  2b2  4ac 2a



(28)  2(28) 2  4(2) (98) 2(2)



12  2122  4(1) (20) 2(1)





12  164 2 12  8 2 12  8 or 2



28  10 4 28 or 7 4

v

 v

v

21.

12  8 2

 10  2 The solutions are 10, 2. 16. 3t2  7t  20  0 t

(7)  2(7) 2  4(3) (20) 2(3)



7  1289 6 7  17 6 7  17 6 10 2  6 or 13

 t 

s

b  2b2  4ac 2a



or t  

7  17 6 24 or 4 6

17. 5y  y  4  0

 

1  181 10 1  9 10 1  9 or 10 4 5

1  9 10

1





1  1113 4 1  1113 4

r

or

1  1113 4

b  2b2  4ac 2a



(7)  2(7) 2  4(2) (3) 2(2)



7  173 4 7  173 4

or

n

7  173 4

 0.4  3.9 The solutions are 0.4, 3.9. 24. 5v2  7v  1 2  7v  1  1  1 5v 5v2  7v  1  0

18. x2  25  0



1  212  4(2) (14) 2(2)

n

4





n y

b  2b2  4ac 2a

 2.9  2.4 The solutions are 2.9, 2.4. 23. 2n2  7n  3  0

The solutions are 5, 1. x

40  10 8 40 or 5 8

r

b  2b2  4ac 2a (1)  2(1) 2  4(5) (4) 2(5)





r

2

y

(40)  2(40) 2  4(4) (100) 2(4)

The solution set is {5}. 22. 2r2  r  14  0

2



b  2b2  4ac 2a

 

The solutions are 13, 4. y

The solution set is {7}. 4s2  100  40s 2 4s  100  40s  40s  40s 4s2  40s  100  0

b  2b2  4ac 2a 0  202  4(1) (25) 2(1)  1100 2 10 2

v

 5 The solutions are 5, 5.

b  2b2  4ac 2a



(7)  2(7) 2  4(5) (1) 2(5)



7  169 10

v

7  169 10

or v 

7  169 10

 0.1  1.5 The solutions are 0.1, 1.5. Chapter 10

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25.

11z2  z  3 11z  z  3  3  3 11z2  z  3  0

29.

2

z

b  2b2  4ac 2a

x



(1)  2(1) 2  4(11) (3) 2(11)



1  1133 22

z

1  1133 22

z

or

  x

1  1133 22

 

7  125 4 7  5 4 7  5 or 4

 w

27.

7  5 4 1  2 1 3, 2.

5

1

1

1

2y2  4y  2  0 5

1

2

8y2  5y  2  0 b  2b2  4ac 2a



5  189 16

y

5  189 16

y

or

5  189 16

 0.3  0.9 The solutions are 0.3, 0.9. 1 2 v 2

31.

4

1

3

v4

1 2 3 v v4 2 1 2 3 v v4 2 1 2 3 v v4 2

2

3

3

44 0

2  4(0)

2v  4v  3  0 g

or

The solutions are



v

2  38 48 40 5 or 6 48

3 5 , . 4 6

1.34d2  1.1d  1.02 1.34d  1.1d  1.02  1.02  1.02 1.34d2  1.1d  1.02  0

b  2b2  4ac 2a



(4)  2(4) 2  4(2) (3) 2(2)



4  140 4

v

2



1

(5)  2(5) 2  4(8) (2) 2(8)

b  2b  4ac 2a



d

5



2

2  11444 48 2  38 48 2  38 48 36 3 or 4 48

 28.

1

(2)  2(2) 2  4(24) (15) 2(24)

g

1

4 2y2  4y  2  4(0)





0.7  1.7 4 1 4

2y2  4y  2  2  2

2(12g2  g)  15 24g2  2g  15 2  2g  15  15  15 24g 24g2  2g  15  0 g

5

w

The solutions are

x

2y2  4y  2

y

 3

0.7  12.89 4 0.7  1.7 4 0.7  1.7 or 4 2.4 4

30.

b  2b2  4ac 2a 7  272  4(2) (3) 2(2)

0.7  2(0.7) 2  4(2) (0.3) 2(2)

   0.6  0.3 The solutions are 0.3, 0.6.

 0.5  0.6 The solutions are 0.5, 0.6. 26. 2w2  (7w  3) 2  (7w  3)  (7w  3)  (7w  3) 2w 2w2  7w  3  0 w

2x2  0.7x  0.3 2x  0.7x  0.3  0.3  0.3 2x2  0.7x  0.3  0 2

4  140 4

v

4  140 4

 0.6  2.6 The solutions are 0.6, 2.6.

2

b  2b  4ac 2a (1.1) ; 2(1.1) 2  4(1.34) (1.02) 2(1.34) 1.1  14.2572

 2.68 ; There are no real solutions because the discriminant is negative.

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34. Let n be the first odd integer. Then n  2 is the next odd integer. Write an equation for the product and solve. n(n  2)  255 n2  2n  255 2 n  2n  255  0

32. Write an equation for the perimeter. 2/  2w  60 Solve for . 2/  2w  60 2/  60  2w 2/ 2



60  2w 2

n

/  30  w Thus, the length and width can be expressed as 30  w and w, respectively. Write an equation for the area and solve. (30  w)(w)  221 30w  w2  221 2 w  30w  221  0 w   w

  n

w

30  4 2





n   n

 w

4  11024 4 4  32 4 4  32 or 4

n

4  32 4

36. f(x)  4x2  9x  4 To find the x-intercepts, set f(x)  0 and solve for x. 4x2  9x  4  0 x

21  2212  4(1) (80) 2(1)

w

4  242  4(2) (126) 2(2)

 9 7 When n  9, n  2  9  2 or 7. When n  7, n  2  7  2 or 9. The numbers are 9 and 7 or 7 and 9.

42  2w 2

21  1121 2 21  11 2 21  11 2

2  32 2

2

/  21  w Thus, the length and width can be expressed as 21  w and w, respectively. Write an equation for the area and solve. (21  w)(w)  80 21w  w2  80 2 w  21w  80  0 w

n

n2  (n  2) 2  130 n  (n2  4n  4)  130 2n2  4n  4  130 2n2  4n  126  0

 17  13 When w  17,   30  17 or 13. When w  13,   30  13 or 17. The rectangle is 13 in. by 17 in. 33. Write an equation for the perimeter. 2/  2w  42 Solve for . 2/  2w  42 2/  42  2w 2/ 2

2  11024 2 2  32 2 2  32 or 2

 17  15 When n  17, n  2  17  2 or 15. When n  15, n  2  15  2 or 17. The numbers are 17 and 15 or 15 and 17. 35. Let n be the first odd integer. Then n  2 is the next odd integer. Write an equation for the sum of their squares.

30  2302  4(1) (221) 2(1) 30  116 2 30  4 2 30  4 or 2

2  222  4(1) (255) 2(1)



(9)  2(9) 2  4(4) (4) 2(4)



9  117 8

x

21  11 2

b  2b2  4ac 2a

9  117 8

or x 

9  117 8

 0.6  1.6 The x-intercepts are about 0.6 and about 1.6.

 16 5 When w  16,   21  16 or 5. When w  5,   21  5 or 16. The rectangle is 5 cm by 16 cm.

37. f(x)  13x2  16x  4 To find the x-intercepts, set f(x)  0 and solve for x. 13x2  16x  4  0 x

b  2b2  4ac 2a



(16)  2(16) 2  4(13) (4) 2(13)



16  1464 26

x

16  1464 26

or

x

16  1464 26

 0.2  1.4 The x-intercepts are about 0.2 and about 1.4.

Chapter 10

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38. x2  3x  4  0 b2  4ac  (3) 2  4(1) (4)  9  16  25 Since the discriminant is positive, the equation has 2 real roots. 39. y2  3y  1  0 b2  4ac  32  4(1)(1) 94 5 Since the discriminant is positive, the equation has 2 real roots.

45. f(x)  x2  4x  7 The number of x-intercepts is equal to the number of roots of the equation x2  4x  7  0. b2  4ac  42  4(1) (7)  16  28  12 Since the discriminant is negative, the graph of the function has 0 x-intercepts. 46. h(t)  16t2  35t  5 To find how many chances Jorge will have, set h(t)  25 and solve for t. 25  16t2  35t  5 25  25  16t2  35t  5  25 0  16t2  35t  20 Check the discriminant.

4p2  10p  6.25 4p2  10p  6.25  6.25  6.25 4p2  10p  6.25  0 b2  4ac  102  4(4)(6.25)  100  100 0 Since the discriminant is 0, the equation has 1 real root. 41. 1.5m2  m  3.5 2 1.5m  m  3.5  3.5  3.5 1.5m2  m  3.5  0 b2  4ac  12  4(1.5)(3.5)  1  21  20 Since the discriminant is negative, the equation has no real roots. 40.

2r2 

42. 1

2

1 r 2

2 3

2r2  2r  3  2

2r 



b2  4ac    

1 r 2 1 r 2

 

2 3 2 3

1 2

0  16t2  35t  5 t 

1

2

 2r  3

4(2)

4 2 n 3 4 2 n 3 4 2 n 3

b2

123 2

 t

t

35  11545 32

96  2962  4(16) (96) 2(16) 96  13072 32 96  13072 32

or

t

96  13072 32

 4.7  1.3 The distance, s, is 96 feet when t is about 1.3 seconds and again when t is about 4.7 seconds.

 4n  3  3  3

 4ac  42  4

35  11545 32

t

 4n  3

 4n  3  0

35  2352  4(16) (5) 2(16)

 2.3  0.1 Ignore the negative number. The camera will hit the ground after about 2.3 seconds. 48. Replace s with 96 and solve for t. s  96t  16t2 96  96t  16t2 0  16t2  96t  96

Since the discriminant is negative, the equation has no real roots. 43.

b  2b2  4ac 2a

35  11545 32

 t

0

1 2  2 1 16  3 4 3 64  12 12 61 12

b2  4ac  (35) 2  4(16)(20)  1225  1280  55 Since the discriminant is negative, there are no real solutions to the equation. Jorge has no chances to catch the camera. 47. Set h(t)  0 and solve for t.

143 2 (3)

 16  16 0 Since the discriminant is 0, the equation has 1 real root. 44. f(x)  7x2  3x  1 The number of x-intercepts is equal to the number of roots of 7x2  3x  1  0. b2  4ac  (3) 2  4(7)(1)  9  28  37 Since the discriminant is positive, the graph of the function has 2 x-intercepts.

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49. Let L  20, D  6, and H  10. Solve for v. 4v2

 5v  2 

4v2

 5v  2 

54. If a population trend can be modeled by a quadratic function, the Quadratic Formula can be used to solve when the function equals a particular value. Answers should include the following. • 15  0.0055t2  0.0796t  5.2810 Original equation 15  15  0.0055t2  0.0796t  5.2810  15 Subtract 15 from each side. 0  0.0055t2  0.0796t  9.7190 Simplify.

1200 HD L 1200(10) (6) 20

2

4v  5v  2  3600 4v2  5v  2  3600  3600  3600 4v2  5v  3602  0 v

v

5  157,657 8

b  2b2  4ac 2a



5  252  4(4) (3602) 2(4)



5  157,657 8

v

t

5  157,657 8

t

 30.6  29.4 Ignore the negative number. The water is flowing at a rate of about 29.4 feet per second. 50. The equation 0  ax2  10x  3 will have 2 real solutions if the discriminant is positive. b2  4ac 7 0 2 10  4a(3) 7 0 100  12a 7 0 100  12a  100 7 0  100 12a 7 100 12a 12

6

a 6

100 12 25 or 3

x



1.87  2(1.87) 2  4(0.048) (4) 2(0.048)



1.87  14.264 0.096

or

x

56. C;

x

1.87  14.264 0.096

1.87  2(1.87) 2  4(0.048) (154) 2(0.048) 1.87  133.0649 0.096 1.87  133.0649 0.096

or

x

1.87  133.0649 0.096

 79.4  40.4 We would expect the death rate to be 0 per 100,000 79 years after 1970 or about 2049. Sample answer: No; the death rate from cancer will never be 0 unless a cure is found. If and when a cure will be found cannot be predicted.

Chapter 10

t

0.0796  10.22015416 0.011

or t 

b  2b2  4ac 2a

Simplify. 0.0796  10.22015416 0.011



5  252  4(2) (1) 2(2)



5  117 4

Maintain Your Skills 2

x  8x  7 x2  8x  16  7  16 (x  4) 2  9 x  4  3 x  4  4  3  4 x4  3 x  4  3 or x  4  3 1 7 The solutions are 1, 7. a2  2a  5  20 58. a2  2a  5  5  20  5 a2  2a  15 2  2a  1  15  1 a (a  1) 2  16 a  1  4 a  1  1  4  1 a  1  4 a  1  4 or a  1  4  5 3 The solutions are 5, 3.

57.

0  0.048x2  1.87x  154



0.0796  10.22015416 0.011

Page 552

 41.0  2.0 When y  150, x  2.0 or x  41.0. 52. Ignoring the negative solution, we would expect the death rate to be 150 per 100,000 in 1970  41 or 2011. 53. Solve for x when y  0. x

t

b2  4ac  (5) 2  4(8) (1)  7

1

83

b  2b2  4ac 2a

1.87  14.264 0.096

(0.0796)  2(0.0796)  4(0.0055) (9.7190) 2(0.0055)

t  35.42 t  49.89 • Graphing the related function would not give precise solutions. The quadratic equation cannot be factored and completing the square would involve difficult computations. 55. A; the discriminant is negative.

150  0.048x2  1.87x  154 150  150  0.048x2  1.87x  154  150 0  0.048x2  1.87x  4 x

Quadratic Formula 2

a  0.0055, b  0.0796, and c  9.7190.

The graph of the function has two x-intercepts 1 when a 6 83. 51.

b  2b2  4ac 2a

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59.

n2  12n  5 n  12n  36  5  36 (n  6) 2  41 n  6  141 n  6  6   141  6 n  6  141 n  6  141 or n  6  141  0.4  12.4 The solutions are 0.4, 12.4.

63. 15xy3  y4  y3 (15x)  y3 (y)  y3 (15x  y) 64. 2ax  6xc  ba  3bc  (2ax  6xc)  (ba  3bc)  2x(a  3c)  b(a  3c)  (a  3c) (2x  b)

2

60.

65. 0.000000000000000000001672  1.672  1021 66. x 2 y4 5 y

x2  x  6 2 x x666 x2  x  6  0 Graph f(x)  x2  x  6. Sample answer: x 2 1 0 1 2 3

x2

y45

f(x) 0 4 6 6 4 0

x

O

f (x )

x

O

67. x  y 7 2 xy 2 y

f (x )  x 2  x  6

xy2

The x-intercepts of the graph are 2 and 3. The solutions of the equation are 2 and 3. 61. 2x2  x  2 2 2x  x  2  2  2 2x2  x  2  0 Graph f(x)  2x2  x  2. Sample answer: f (x ) x f(x) 2 4 1 1 0 2 1 1 2 8 x

x

O

xy2

68. y 7 x y x4 y yx4

x

O

O

yx

f (x )  2x 2  x  2

The x-intercepts of the graph lie between 2 and 1 and between 0 and 1. So, one root is between 2 and 1, and the other root is between 0 and 1.

Exercises 69–71 For checks, see students’ work. 69. 2m  7 7 17 2m  7  7 7 17  7 2m 7 10 2m 10 7 2 2 m 7 5 {m 0m 7 5} 70. 2  3x 2 2  3x  2 2  2 3x 4 3x 4

3 3

62. Graph f(x)  x2  3x  6. x 2 1 0 2 4 5

f(x) 4 2 6 8 2 4

f (x )

O

f (x )  x 2  3x  6

5x 0x 43 6

x

The x-intercepts of the graph lie between 2 and 1 and between 4 and 5. So, one root is between 2 and 1, and the other root is between 4 and 5.

473

4

x 3

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71.

72. c(ax )  1(24 ) 20 8  7k 20  8 8  7k  8  1(16) or 16 28 7k 28 7k 7 7 4 k {k 0k 4}

73. c(ax )  3(72 )  3(49) or 147

3. Solve the first equation for y. 1.8x  y  3.6 y  3.6  1.8x Enter y  3.6  1.8x as Y . 1

Enter y  x2  3x  1 as Y2. Graph. Approximate the first intersection point. KEYSTROKES: 2nd [CALC] 5 ENTER ENTER

74. c(ax )  2(53 )  2(125) or 250

ENTER

Page 553

Graphing Calculator Investigation (Follow-Up of Lesson 10-4)

1. Enter y  2(2x  3) as Y . 1

Enter y  x2  2x  3 as Y2. Then graph. Approximate the intersection point. KEYSTROKES: 2nd [CALC] 5 ENTER

[10, 10] scl: 1 by [10, 10] scl: 1

ENTER

Use the TRACE feature and right arrow to move the cursor near the other intersection point. Use the intersect feature again to approximate the other intersection point.

ENTER

[5, 1] scl: 1 by [1, 15] scl: 1

The solution is (3, 6). 2. Solve the first equation for y. y50 y5 Enter y  5 as Y1.

The solutions are approximately (1.6, 6.5) and (2.8, 1.5). 4. Enter y  1.4x  2.88 as Y . 1

Enter y  x2  0.4x  3.14 as Y . 2 Graph. Approximate the first intersection point. KEYSTROKES: 2nd [CALC] 5 ENTER ENTER

Enter y  x2 as Y . 2 Graph.

ENTER

[10, 10] scl: 1 by [10, 10] scl: 1

The graphs do not intersect. Thus, the system has no solution.

[10, 10] scl: 1 by [10, 10] scl: 1

Use the TRACE feature and left arrow to move the cursor near the other intersection point. Use the intersect feature again.

The solutions are approximately (1.9, 0.2) and (0.1, 3.1).

Chapter 10

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5. Enter y  x2  3.5x  2.2 as Y1.

Enter y  2x  5.3625 as Y2. Graph. Approximate the intersection point. KEYSTROKES: 2nd [CALC] 5 ENTER

The graphs are the same shape. The graph of y  2x  3 is the graph of y  2x translated 3 units up. The graph of y  2x  4 is the graph of y  2x translated 4 units down. 2. Enter y  2x as Y . 1

ENTER

Enter y  2x5 as Y . 2

ENTER

Enter y  2x4 as Y . 3 Then graph.

[1, 5] scl: 1 by [5, 5] scl: 1

The solution is approximately (2.8, 0.1). 6. Enter y  0.35x  1.648 as Y1.

The graphs are the same shape. The graph of y  2x  5 is the graph of y  2x translated 5 units to the left. The graph of y  2x  4 is the graph of y  2x translated 4 units to the right. 3. Enter y  2x as Y . 1 Enter y  3x as Y2. Enter y  5x as Y3. Then graph.

Enter y  0.2x2  0.28x  1.01 as Y . 2 Graph. Approximate the first intersection point. KEYSTROKES: 2nd [CALC] 5 ENTER ENTER

ENTER

[10, 10] scl: 1 by [10, 10] scl: 1

Use the TRACE feature and left arrow to move the cursor near the other intersection point. Use the intersect feature again.

All of the graphs cross the y-axis at 1. The graph of y  3x is steeper than the graph of y  2x, and the graph of y  5x is steeper yet. 4. Enter y  3  2x as Y . 1 Enter y  3(2x  1) as Y . 2 Enter y  3(2x  1) as Y3. Then graph.

The solutions are approximately (3.8, 3.0) and (3.5, 0.4).

10-5 Page 556

Exponential Functions The graphs are the same shape. The graph of y  3(2x  1) is the graph of y  3(2x) translated 3 units down. The graph of y  3(2x  1) is the graph of y  3(2x) translated 3 units up.

Graphing Calculator Investigation

1. Enter y  2x as Y . 1 Enter y  2x  3 as Y . 2 Enter y  2x  4 as Y . 3 Then graph.

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Pages 557–558

6. y  9x Sample answer:

Check for Understanding

1. Never; there is no value of x for which ax  0. 2. Sample answer: y  2x The y-intercept of the graph is 1. The graph increases quickly for x  0. y

x

9x

2

92

1 81

1

91

1 9

0 1 2

y

90 91 92

80 60 40

1 9 81 4

y  2x

3. Kiski; the graph of y  4. y  3x Sample answer: x

3x

2

32

1 9

1

31

1 3

30

0 1 2 3

31 32 33

113 2x decreases as x increases.

y

35 30 25 20 15 10 5

1 3 9 27

4 3 2 1O 5

y

12

1 2 3 4x

2 1 0 1 2

12 114 22 114 21 114 20 114 21 114 22

16 4

()

y  14

1 1 4

x

56 48 40 32 24 16 8

4 3 2 1O 8

y

2 9

1

231

2 3

O

2

4x

y

y

230 231 2 32

56 48 40 32 24 16 8

2 6 18

4 3 2 1O 8

4(5x  10)

y  2 · 3x 1 2 3 4x

y

2

4(52

1

4(51  10)

395

4(50  10) 4(51  10) 4(52  10)

36 20 60

0 1 2

40

21

 10) 3925 1

y

20

1 2 3 4x 4

1 16

2

O

2

4x

20

The y-intercept is 1.

114 21.7  0.1

Chapter 10

2

232

x

y

y  9x

The y-intercept is 2. 8. y  4(5x  10) Sample answer:

Sample answer: 1 x 4

2 3x

y  3x

1 x 4

x

x

0 1 2

The y-intercept is 1. 31.2  3.7 5. y 

2

20

The y-intercept is 1. 90.8  5.8 7. y  2  3x Sample answer:

x

O

y

40

y  4(5x  10)

The y-intercept is 36. 9. Yes; the domain values are at regular intervals, and the range values have a common ratio 6. 10. No; the domain values are at regular intervals, and the range values have a common difference 4. 11. 264  1.84  1019 grains.

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12. Divide the number of grains by the number of grains per pound to find the number of pounds. 1.84  1019 2.4  104



10 11.84 2.4 2 1 10 2

16. y 

Sample answer:

19 4

1015

x 2 1 0

1014

 0.767  or 7.67  Now divide the number of pounds by 2000 of 2  103 to find the number of tons. 7.67  1014 2  103



115 2x

10 17.67 2 2 1 10 2

y 25 5 1

40 30

14 3

 3.84  1011 The man will receive about 3.84  1011 tons of rice the last day.

y

1

1 5

2

1 25

20 x

(1)

y 5 4

2

10 O

4x

2

The y-intercept is 1.

Pages 558–560

115 20.5  0.4

Practice and Apply

13. y  5x Sample answer: x

17. y  6x Sample answer: y

y

2

1 25

1

1 5

0 1 2

x

40 30 20

1 5 25 4

2

10

y  5x

O

2

2 1

1 10

40

y

x

10 2

y  10x 2

O

2

y

2

1 64

y  6x

O

1

1 8

80

2

4x

y

60 40 20 4

2

y  8x 2

O

4x

The y-intercept is 1. 80.8  5.3 19. y  5(2x ) Sample answer:

1101 2x

y 100 10 1

40

y

30 20

1 10 x

(1)

1 100

20

4x

Sample answer:

2

40

1 6 36

0 1 1 8 2 64

The y-intercept is 1. 100.3  2.0

1

y

60

4

20

4

x 2 1 0

1 6

4x

30

0 1 1 10 2 100

15. y 

1

80

The y-intercept is 1. 60.3  1.7 18. y  8x Sample answer:

y 1 100

2

1 36

0 1 2

The y-intercept is 1. 51.1  5.9 14. y  10x Sample answer: x

y

y  10 4

2

y

2

1 14

1

22

0 1 2

10 O

x

2

4x

40 30

1

20

5 10 20

10 4

The y-intercept is 1.

1101 21.3  20.0

y

2

O

y  5(2x) 2

4x

The y-intercept is 5.

477

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20. y  3(5x ) Sample answer: x

y

40

3 25

2

x

y

30

3 5

1 0 1 2

24. y  5(2x )  4 Sample answer:

20

3 15 75 4

2

O

y

2

8 69

30

2

y

x

10

y

2

1 44

2

O

4x

2

10

16

5 6 8

y

4 2

2

4x

1

3

0 1 2

1

1 5 17

y  2(3x)  1 O

x

The y-intercept is 1.

Chapter 10

O

2

4x

23

y

30

2

20

4 8 20

10

y  2(3x  1) 2

O

2

4x

2

4x

y

x

y

2

1 144

1

132

20

1

12 9 3

10 4

2

O 10

y  3(2x  5)

The y-intercept is 12. 27. No; the domain values are at regular intervals, and the range values have a common difference 3. 28. Yes; the domain values are at regular intervals, and the range values have a common ratio 0.5. 29. Yes; the domain values are at regular intervals, and the range values have a common ratio 0.75. 30. No; the domain values are at regular intervals, but the range values do not have a positive common ratio. 31. No; the domain values are at regular intervals, but the range values do not change. 32. Yes; the domain values are at regular intervals, and the range values have a common ratio 0.5.

y

2

1

0 1 2

y  2x  4

O

y 7 9

2

40

4

The y-intercept is 5. 23. y  2(3x )  1 Sample answer: x

2

2 29

y3 7

8

4

y  5(2x)  4

x

12

42

0 1 2

10

The y-intercept is 4. 26. y  3(2x  5) Sample answer:

1

1

20

9 14 24

y

0 1 2

The y-intercept is 6. 22. y  2x  4 Sample answer: x

30

1

4

20

4

62

y

The y-intercept is 9. 25. y  2(3x  1) Sample answer:

1 63 0 6 1 4 2 2 3 20

1

40

4x

2

The y-intercept is 3. 21. y  3x  7 Sample answer: x

2

1 54

0 1 2

y  3(5x )

10

y

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33. T(x)  12(1.12)x In 2005, x  10. T(10)  12(1.12)10  37.27 In 2005, the sales are expected to be about $37.27 million. In 2006, x 11. T(11)  12(1.12)11  41.74 In 2006, the sales are expected to be about $41.74 million. In 2007, x  12. T(12)  12(1.12)12  46.75 In 2007, the sales are expected to be about $46.75 million. 34. T(x)  12(1.12)x Sample answer: x 0 1 2 3

T(x) 12 13.44 15.05 16.86

16

40. D(x)  20(1.1)x Week 1 2 3 4

41. Continue to find D(x) for integers greater than 4. A partial table is shown. Week 9 10

42. Graph y  5x and y 

2

O

graph of y 

T (x )

4x

The y-intercept is 12. 35. The y-intercept represents the sales 0 years after 1995. There were $12 million in sales in 1995. 36. There are four 15-minute time periods in one hour. Thus, the process will be completed four times. Each time the number of bacteria doubles. Thus, there are 100 (24) or 1600 bacteria after one hour. 1

37. Since 3 of the schools remain after each round, we 1 multiply by 3 for each round. y  729

115 2x is the graph of y  5x reflected

around the y-axis. 43. Graph y  5x and y  5x  2 in the same coordinate plane or viewing rectangle. You will see that the graph of y  5x  2 is the graph of y  5x translated 2 units up. 44. Graph y  5x and y  5x  4 in the same coordinate plane or viewing rectangle. You will see that the graph of y  5x  4 is the graph of y  5x translated 4 units down. 45. If the number of items on each level of a piece of art is a given number times the number of items on the previous level, an exponential function can be used to describe the situation. Answers should include the following. • For the carving of the pliers, y  2x. • For this situation, x is an integer between 0 and 8, inclusive. The values of y are 1, 2, 4, 8, 16, 32, 64, 128, and 256.

T (x )  12(1.12)x

2

115 2x in the same coordinate

plane or viewing rectangle. You will see that the

4 4

Distance(miles) about 47.2 about 51.9

The runner will exceed 50 miles for the first time in the 10th week.

12 8

Distance (miles) 22 24.2 26.62 29.282

y

113 2x

38. Let x  3.

113 23 1  729 1 27 2

y  729

y  2x

 27 There are 27 schools after 3 rounds. 39. Consider y for various values of x. In particular, let x  6.

O

x

46. B; f(x)  6x is an exponential function since the variable is the exponent. 47. A; graph y  2x and y  6x in the same coordinate plane or viewing rectangle. You will see that the graph of y  6x is steeper than the graph of y  2x.

113 26 1  729 1 729 2

y  729

1 After 6 rounds, there will be only one school remaining. This school is the winner.

479

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Page 560

Maintain Your Skills

x

2

b  2b2  4ac 2a



(9)  2(9) 2  4(1) (36) 2(1)



9  1225 2 9  15 2 9  15 or 2

 x

x

9  15 2

 3  12 The solutions are 3, 12. 49. 2t2  3t  1  0 t

54.

b  2b2  4ac 2a



3  232  4(2) (1) 2(2)



3  117 4

t

3  117 4

t

or

55.

56.

3  117 4

 1.8  0.3 The solutions are 1.8, 0.3. 50.

57.

5y2  3  y 23yyy 5y 5y2  y  3  0 y

b  2b2  4ac 2a



(1)  2(1) 2  4(5) (3) 2(5)



1  159 10

; Since the discriminant is negative, there are no real solutions. x2  7x  10

51.

49 4 7 2 2 7 x2 7 7 2 2

x2  7x 

1x

x

2

 10  9 3

 2 3

3

7

 2  2 7

7

49 4

4

x2  x22

or

x

3 2 7 2

3

2

58.

2 5 The solutions are 2, 5. 52.

t2  6t  3  0

53.

48. x2  9x  36  0

a2  12a  3 a  12a  36  3  36 (a  6) 2  39 a  6  ; 139 2

60.

t  6t  3  3  0  3 t2  6t  3 t2  6t  9  3  9 (t  3) 2  6 t  3  ; 16 t  3  3  ; 16  3 t  3  16 t  3  16 or t  3  16  5.4  0.6 The solutions are 5.4, 0.6. Two numbers whose product is 40 and whose sum is 14 are 10 and 4. m2  14m  40  (m  10)(m  4) Among all pairs of integers whose product is 35, there are no pairs whose sum is 2. Thus, t2  2t  35 is prime. A pair of numbers whose product is 24 and whose sum is 5 and 8 and 3. z2  5z  24  (z  8)(z  3) Let n be the first number. Let m be the second number. Solve the following system. 3n  2m 2n  m  3 Solve the second equation for m. 2n  m  3 2n  3  m Substitute 2n  3 for m in the first equation and solve for m. 3n  2m 3n  2(2n  3) 3n  4n  6 3n  4n  6 n  6 n6 Use m  2n 3 to find m. m  2n  3  2(6)  3 or 9 The numbers are 6 and 9. 59. x7 7 2 10 x  8 x77 7 27 10  8 x  8  8 x 7 5 2 x {x|x 7 5} {x|x 2} y  7 6 12 y  7  7 6 12  7 y 6 5 {y|y 6 5}

1

a  6  6  ; 139  6

2

1 61. p(1  r) t  5 1  2 2

a  6  139 a  6  139 or a  6  139  0.2  12.2 The solutions are 0.2, 12.2.

2

 5(1.5)  5(2.25)  11.25

1

2

1 62. p(1  r) t  300 1  4 3

300(1.25) 3

  300(1.953125)  585.9375

Chapter 10

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63. p(1  r) t  100(1  0.2) 2  100(1.2) 2  100(1.44)  144

5. y  5x  4 Sample answer:

64. p(1  r) t  6(1  0.5) 3  6(1.5) 3  6(3.375)  20.25

Page 560 1.

y

x

y

2

24 325

1

35

4

3 1 21

0 1 2

x

O

y  5x  4

Practice Quiz 2

x2  2x  35 2  2x  35  35  35 x x2  2x  35  0 x

The y-intercept is 3.

b  2b2  4ac 2a



2  222  4(1) (35) 2(1)



2  1144 2 2  12 2 2  12 x 2

 2  12

10-6 Growth and Decay Page 563

or x 2  7 5 The solutions are 7, 5. 2. 2n2  3n  5  0 n

b  2b2  4ac 2a



(3)  2(3) 2  4(2) (5) 2(2)



3  131 4

Check for Understanding

1. Exponential growth is an increase by the same percent over a period of time, while exponential decay is a decrease by the same percent over a period of time. 2. Determine the amount of the investment if $500 is invested at an interest rate of 7% compounded quarterly for 6 years. 3. Sample answer: y

; Since the discriminant is negative, there are no real solutions. 3.

2v2  4v  1  4v  1  1  1 2v2  4v  1  0 2v2

v

v

4  124 4

b  2b2  4ac 2a

t



(4)  2(4) 2  4(2) (1) 2(2)



4  124 4

or

v

r) t

4. I  C(1  I  37,060(1  0.005) t I  37,060(1.005) t 5. First find t. t  2009  1979 or 30 I  37,060 (1.005) 30  43,041 The median household income in 2009 is expected to be $43,041.

4  124 4

 0.2  2.2 The solutions are 0.2, 2.2. 4. y  0.5(4x ) Sample answer: x 2 1 0 1 2

1

y

y 0.01325 0.125 0.5 2 8

2

r 6. A  P 1  n nt

1

A  400 1 

A  400(1.018125) 28 A  661.44 The investment will be about $661.44.

y  0.5(4x) O

2

0.0725 47 4

x

The y-intercept is 0.5.

481

Chapter 10

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16. y  C(1  r) t y  2,405,000(1  0.011) t y  2,405,000(0.989) t In 2015, t  2015  2000 or 15. y  2,405,000(0.989) 15 y  2,037,321 In 2015, the population of Latvia will be about 2,037,321.

7. Write an equation. y  C(1  r) t y  1,821,000(1  0.002) t y  1,821,000(0.998) t Find t. t  2010  1995 or 15 y  1,821,000(0.998) 15 y  1,767,128 The population of West Virginia will be about 1,767,128 in 2010. 8. y  C(1  r) t y  16,000(1  0.18) 8

17. y  C(1  r) t y  2,976,400,000(1  0.186) t y  2,976,400,000(0.814) t In 2009, t  2009  1994 or 15. y  2,976,400,000(0.814) 15 y  135,849,289 In 2009, the sales of cassettes will be about $135,849,289. 18. y  C(1  r) t y  71,601(1  0.0563) t y  71,601(1.0563) t In 2020, t  2020  1920 or 100. y  71,601(1.0563) 100 y  17,125,650 There will be about 17,125,650 visitors in 2020. 19. y  C(1  r) t y  25,000(1  0.10) 8 y  25,000(0.9) 8 y  10,761.68 After 8 years, the equipment will be worth about $10,761.68. 20. y  C(1  r) t y  23,000(1  0.12) 5 y  23,000(0.88) 5 y  12,137.83 In 5 years, the value of the car will be about $12,137.83. 21. In 2010, t  2010  1900 or 110. P  3.86(1.013) 110 P  15.98 In 2010, about 15.98% of the population will be 65 or older.

y  16,000(0.82) 8 y  3270.63 After 8 years, the value of the car will be about $3270.63.

Pages 563–565

Practice and Apply

9. y  C(1  r) t y  18.9(1  0.19) t y  18.9(1.19) t 10. In 2015, t  2015  1980 or 35. y  18.9(1.19) 35 y  8329.24 In 2015, there will be about 8329.24 computers. 11. W  C(1  r) t W  43.2(1  0.06) t W  43.2(1.06) t 12. In 2007, t  2007  1997 or 10. W  43.2(1.06) 10 W  77.36 In 2007, about 77.36 million people will be using free weights. 13. Write an equation. y  C(1  r) t y  100,350,000(1  0.017) t y  100,350,000(1.017) t In 2012, t  2012  2000 or 12. y  100,350,000(1.017) 12 y  122,848,204 In 2012, the population of Mexico will be about 122,848,204.

1

14. A  P 1 

1

2

r nt t

A  500 1 

22.

1

1

A  250 1

2

2

r nt t 0.103  4 440

2

A  14,607.78 The amount will be about $14,607.78.

Chapter 10

P 19.91 20.16

The percentage of the population 65 or older will first be over 20% when t  128 or about 2028. 23. This equation represents growth since 1.026 7 1. 1  r  1.026 r  0.026 The annual rate of change is 2.6%. 24. This equation represents decay, since 0.761 6 1. 1  r  0.761 r  0.239 r  0.239 The annual rate of change is 2.39%.

0.0575 1225 12

A  2097.86 The amount will be about $2097.86. 15. A  P 1 

t 127 128

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5730

34. y  2x  5 Sample answer: x y

25. y  256(0.5) 5730 y  256(0.5) 1 y  128 There are 128 grams of Carbon-14 remaining. 1000

26. y  256(0.5) 5730 y  226.83 There are about 226.83 grams of Carbon-14 remaining.

2

3 44

1

1 42

0 1 2

10,000

27. y  256(0.5) 5730 y  76.36 There will about 76.36 grams of Carbon-14 remaining. 28.

t 17,190

The organism lived 17,190 years ago. 29. See students’ work. 30. If the sales are growing by the same percent each year, an exponential equation can be used to model sales and predict future sales. Answers should include the following. • The equation states that the new value equals the amount in the year 1994, or $1698 times the sum of 1 plus 4.6% raised to the power, that is equal to the number of years since 1994. • According to the equation, the average family will spend about $3486.94 for restaurant meals in 2010. 31. C; In the equation y  35(1.05x ), the exponent is the variable and the base is greater than 1.

1

2

1

33. y 

12

2 1

223

2

2

20

y  4(3x  6)

b  2b2  4ac 2a (9)  2(9) 2  4(1) (10) 2(1)

9  11 2

 1  10 The solutions are 1, 10. 2t2  4t  3 37. 2t2  4t  3  3  3 2t2  4t  3  0

16

y

12

t

8 4 2

4x

2

9  1121 2 9  11  2 9  11 or m  2

t

4

O



Maintain Your Skills

1 64

2

10



m

1 8

4

20 12 12

Sample answer:

1

10

m

2

y 64 8 1

y

The y-intercept is 20. 36. m2  9m  10  0

1 x 8

x 2 1 0

x

O

5 239

0 1 2

0.0825 4.4 4

A  6931.53

Page 565

4 3 1

r nt

A  5000 1 

y  2x  5

The y-intercept is 4. 35. y  4(3x  6) Sample answer: x y

y 32

32. D; A  P 1  n

y

O

2



(4)  2(4) 2  4(2) (3) 2(2)



4  140 4

or

t

4  140 4

 0.6  2.6 The solutions are 0.6, 2.6. 38. 7x2  3x  1  0

x

(1)

y 8

4  140 4

b  2b2  4ac 2a

x

4x



The y-intercept is 1.

3  232  4(7) (1) 2(7) 3  119 14

; There are no real solutions since the discriminant is negative. 39. m7 (m3b2 )  (m7m3 )b2  m10b2

483

Chapter 10

40. 3(ax3y) 2  3a2 (x3 ) 2y2  3a2x6y2

4a. Angela bought a car for $18,500. If the rate of depreciation is 11%, find the value of the car in 4 years.

41. (0.3x3y2 ) 2  0.32 (x3 ) 2 (y2 ) 2  0.09x6y4 42. |7x  2|  2 ; There are no solutions since absolute value is never negative. 43. |3x  3|  0 3x  3  0 3x  3 x1 The solution is 1. 44. ƒt  4ƒ  3 or t  4  3 t43 t  4  4  3  4 t4434 t  1 t  7 {t|t  7 or t  1} 45. slope  

4b. y  C(1  r) t; 18,500(1  0.11) 4 or about $11,607.31 5a. The population of Centerville is increasing at an average annual rate of 3.5%. If its current population is 12,500, predict its population in 5 years. 5b. y  C(1  r) t; y  12,500(1  0.035) 5 or about 14,846 people

Geometric Sequences

10-7 Page 569

2 1

 0.24 Yes, the hill meets the requirements since 0.24 6 0.33. 46. 8 11 14 17 ?

2

The common ratio is 2. 16

 3 3 3 3

O

2 1

Add 1.1 three more times. The next three terms are 5.9, 7.0, and 8.1.

Reading Mathematics

 2

O 8

2

16

r nt

2. A  P 1  n ; since the final amount is greater than the initial amount, P is multiplied by a number greater than 1. Also, the annual rate is divided by the number of times it is compounded per year. Time is equal to nt because this is the total number of times that the interest is compounded over the course of t years. 3a. Suppose that $2500 is invested at an annual rate of 6%. If the interest is compounded quarterly, find the value of the account after 5 years. $3367.14

Chapter 10

1

; A  2500 1 

2

0.06 4(5) 4

an

8

1. y  C(1  r) t; Since the final amount is less than the initial amount, the initial amount is multiplied by a number less than 1. If an amount is decreased by r percent, then 1  r percent will remain.

2

n

The common ratio is 2. 16

r nt

6

2. 1, 2, 4, 8, 16, . . . Divide the second term by the first.

 1.1 1.1 1.1 1.1

1

4

16

Add 3 three more times. The next three terms are 5, 8, and 11. 48. 1.5 2.6 3.7 4.8 ?

3b. A  P 1  n

2

8

 3 3 3 3

1

an

8

Add 3 three more times. The next three terms are 20, 23, and 26. 47. 7 4 1 2 ?

Page 566

Algebra Activity

1. 1, 2, 4, 8, 16, . . . Divide the second term by the first.

rise run 60 250

or about

484

4

6

n

7. Graphs 1, 3, and 5—those with positive common factors—appear to be similar to an exponential function. 8. Both types of graphs show rapid change. If r 7 0, the graph of the geometric sequence is similar to an exponential function. If r 6 0, the graph of the geometric sequence has points above and below the n-axis and is not similar to an exponential function.

3. 81, 27, 9, 3, 1, . . . Divide the second term by the first. 27 81

1

3 1

The common ratio is 3. 80

an

40 O

2

4

9. The two values c(ax ) and a rn1 are very similar. 1 Both have a number multiplied by another number that is raised to a power.

n

6

40

Page 570

80

4. 81, 27, 9, 3, 1, . . . Divide the second term by the first. 27 81

1

 3 1

The common ratio is 3. 80

an

40 O

2

4

6

n

40 80

5. 0.2, 1, 5, 25, 125, . . . Divide the second term by the first. 1 0.2

5

The common ratio is 5. 100

 3 3 3

an

Yes; the common ratio is 3. 5. 56 28 14 7

50 O

2

4

6

 1 1 1  1 2 2  1 2 2  1 2 2

n

6. 25

15

10

No; the difference between successive terms is constant. This sequence is arithmetic, not geometric. 7. 5, 20, 80, 320, . . . Divide the second term by the first.

6. 0.2, 1,5, 25,125, . . . Divide the second term by the first.  5

The common ratio is 5.

20 5

an

2

4

6

4

The common ratio is 4. Multiply by 4 three more times. The next three terms are 1280, 5120, and 20,480. 8. 176, 88, 44, 22, . . . Divide the second term by the first.

50 O

20

 5 5 5

100

100

1

Yes, the common ratio is 2.

50

1 0.2

Check for Understanding

1. Both arithmetic sequences and geometric sequences are lists of related numbers. In an arithmetic sequence, each term is found by adding the previous term to a constant called the common difference. In a geometric sequence, each term is found by multiplying the previous term by a constant called the common ratio. 2. If a common ratio equals 0, all of the terms except possibly the first term will equal 0, since any number times 0 equals 0. If a common ratio equals 1, all of the terms will equal the first term, since 1 is the multiplicative identity. 3. Sample answer: 1, 4, 9, 16, 25, 36, . . . The difference between the first and second terms is 4  1 or 3, and the difference between the second and third terms is 9  4 or 5. Since these are not equal, the sequence is not arithmetic. The 4 9 corresponding ratios are 1 or 4 and 4. Since these are not equal, the sequence is not geometric. 4. 5 15 45 135

n

50

88 176

100

1

 2 or 0.5

The common ratio is 0.5. Multiply by 0.5 three more times. The next three terms are 11, 5.5, and 2.75.

485

Chapter 10

9. 8, 12, 18, 27, . . . Divide the second term by the first. 12 8

18. 7

3

The common ratio is 1.5. Multiply by 1.5 three more times. The next three terms are 40.5, 60.75, and 91.25.

No; the difference between consecutive terms is constant. This sequence is arithmetic, not geometric. 20. 640 160 40 10

a6  (1) 25 a6  (1) 32 a6  32

 1 1 1 40

12. a  a rn1 n 1 a7  4 (3) 71 a  4 7

21. 36

3

Yes; the common ratio is 4.5. 25. 1, 4, 16, 64, . . . Divide the second term by the first. 4 1

r

114 2  12 or

6 1

4r2 4

25  r2 5  r The geometric mean is (4)(5)  20 or (4) (5)  20.

512 1024

112 224

 0.5

Multiply by 0.5 three more times. The next three terms are 14, 7, and 3.5. 29. 80, 20, 5, 1.25, . . . Divide the second term by the first.

The perimeter of the smallest triangle is 7.5 cm.

20 80

 0.25

Multiply by 0.25 three more times. The next three terms are 0.3125, 0.078125, and 0.01953125.

Practice and Apply 24

 3 3 3

Yes; the common ratio is 3.

Chapter 10

 0.5

Multiply by 0.5 three more times. The next three terms are 64, 32, and 16. 28. 224, 112, 56, 28, . . . Divide the second term by the first.

Perimeter 120cm 120(0.5) or 60 cm 60(0.5) or 30 cm 30(0.5) or 15 cm 15(0.5) or 7.5 cm

Pages 571–572

6

Multiply by 6 three more times. The next three terms are 1296, 7776 and 46,656. 27. 1024, 512, 256, 128, . . . Divide the second term by the first.

100  (4) r2

Triangle Largest Next largest Next largest Next largest Smallest

 4

Multiply by 4 three more times. The next three terms are 256, 1024, and 4096. 26. 1, 6, 36, 216, . . . Divide the second term by the first.

a3  a1 r31

18

21

3

90 405 1822.5  (4.5) (4.5) (4.5)

 r2

6

3

23. 20

48r2 48

1 2

17. 2

63

  1 1

1

The geometric mean is 48 1 48 4  12.

16.

16

Yes; the common ratio is 3.

7r2 7



9

189

 1

3  48 r2

100 4

25

22. 567

4  r2 2  r The geometric mean is 7(2)  14 or 7(2)  14. 14. a3  a1 r31

15.

16

1 . 40

No; the ratios are not the same.

7



25

36

13. a3  a1 r31 28  7 r2

3 48 1 16 1 4

40

 25 16 9

(3) 6

a  2916



40

Yes; the common ratio is

a  4 729 7

28 7

37

 3 3 3

11. a  a rn1 n 1 a6  (1) 261

a5  3 44 a5  3 256 a5  768

27

No; the difference between consecutive terms is constant. This sequence is arithmetic, not geometric. 19. 19 16 13 10

 2 or 1.5

10. a  a rn1 n 1 a5  3 451

17

 10 10 10

486

30. 10,000, 200, 4, 0.08, . . . Divide the second term by the first. 200 10,000

41. a  a rn1 n 1 a10  300 (0.5) 101

 0.02

a10  300 (0.5) 9 a10  300 0.001953125 or 0.5859375

Multiply by 0.02 three more times. The next three terms are 0.0016, 0.000032, and 0.00000064. 1 1 2

42. a  a rn1 n 1 a6  14 (1.5) 61

4

31. 2, 3, 9, 27, . . .

a6  14 (1.5) 5 a  14 7.59375 or 106.3125

Divide the second term by the first. 1 3 1 2

6

43. a  a r31 3 1

2

3

20  5 r2

2 3

Multiply by terms are

20 5

three more times. The next three

8 16 , , 81 243

and

5r2 5

4  r2 2  r The geometric mean is 5(2)  10 or 5(2)  10.

32 . 729

3 1 1 2

32. 4, 2, 3, 9, . . . Divide the second term by the first. 1 2 3 4



44. a  a r31 3 1 54  6 r2

2

3

54 6

Multiply by

2 3

4

16

terms are 27, 81, and 243. 33. The area of the original rectangle is 6 8 or 48 square inches. Multiply by 0.5 four times to find the areas of the next four rectangles. The areas of the five rectangles are 48 in2, 24 in2, 12 in2, 6 in2, and 3 in2. 34. Multiply the measurement of the first angle by 0.5 three times to find the measurements of the other three angles. The measurements are 160 , 80 , 40 , and 20 . 35. a  a rn1 n 1 a7  5 271

45.

a3  a1 r31 225  (9) r2 225 9



9r2 9

25  r2 5  r The geometric mean is (9) 5  45 or (9) (5)  45. 46.

36. a  a rn1 n 1 a5  4 351

a7  5 26 a7  5 64 or 320

6r2 6

9  r2 3  r The geometric mean is 6(3)  18 or 6(3)  18.

three more times. The next three 8



a3  a1 r31 80  (5) r2 80 5

a5  4 34 a5  4 81 or 324



5r2 5

16  r2 4  r The geometric mean is (5)4  20 or (5)(4)  20.

37. a  a rn1 n 1 a4  (2) (5) 41

a4  (2) (5) 3 a4  (2) (125) or 250

47. a  a r31 3 1 8  128 r2

38. a  a rn1 n 1 a6  3 (4) 61

8 128 1 16 1 4

(4) 5

a6  3 a6  3 (1024) or 3072

39. a  a rn1 n 1 a3  (8) 631



128r2 128

 r2 r

1 2  32.

The geometric mean is 128 128

1 4

114 2  32 or

48. a  a r31 3 1 5  180 r2

a3  (8) 62 a3  (8) 36 or 288

5 180 1 36 1 6

40. a  a rn1 n 1 a8  (10) 281 27

a8  (10) a8  (10) 128 or 1280



180r2 180

 r2 r

1 12

The geometric mean is 180 180 6  30.

487

116 2  30 or Chapter 10

49.

a3  a1 r31 98  (2) r2 98 2



58. Multiply 848 by 0.75 six times to find the level on the next six days. The levels during the first week are 848, 636, 477, 357.75, 268.3125, 201.234375, and 150.9257813 parts per million.

2r2 2

49  r2 7  r The geometric mean is (2)(7)  14 or (2) (7)  14. 50.

59. a  848(0.75) 14 or about 15.1 15 a16  848(0.75) 15 or about 11.3 The lake will be safe in 16 days. 60. Sample answer: Although the amount of pesticide will become very small, 75% of something will always be something. Therefore, there will always be pesticides according to this model.

a3  a1 r31 384  (6) r2 384 6



6r2 6

61. Always; if each term of a geometric sequence whose nth term is a rn1 is multiplied by a

64  r2 8  r The geometric mean is (6)(8)  48 or (6) (8)  48. 51.

1

nonzero real number c, the nth term of the new sequence is c(a rn1 ) or (c a )rn1, which is a 1 1 geometric sequence.

a3  a1 r31

62. Never; if the same nonzero real number is added to each term of a geometric sequence, the resulting sequence will have an nth term of the a1rn1  b, which is not in the form of a geometric sequence. 3 63. Since the distance of each bounce is 4 times the distance of the last bounce, the list of the distances from the stopping place is a geometric sequence. Answers should include the following.

1.75  7 r2 1.75 7



7r2 7

0.25  r2 0.5  r The geometric mean is 7(0.5)  3.5 or 7(0.5)  3.5. 52.

a3  a1 r31 0.75  3 r2 0.75 3



• To find the 10th term, multiply the first term 3 80 by 4 to the 9th power. • The 17th bounce will be the first bounce less than 1 ft from the resting place.

3r2 3

0.25  r2 0.5  r The geometric mean is 3(0.5)  1.5 or 3(0.5)  1.5. 53.

100

64. C; The common ratio is 40 or 2.5. To find the next term, multiply 625 by 2.5. The next term is 1562.5.

a3  a1 r31 3 20 5 3 3 20 1 4 1 2

3

 5 r2

49

1 2 1 2

The geometric mean is 54.

a3  a1 r31 2 45 5 2 2 45 1 9 1 3

1 2

66. The terms are getting closer to 0. 67. The limit is 0.

 r2 r

2

1 2  103 or 35 112 2  103 .

68. If 0 6 r 6 1, the nth term will approach 0. If r 7 1, the nth term will approach infinity.

3 1 5 2

 5 r2

1 2

Page 572

5 2 2 r 5

2 

1

Maintain Your Skills

2

r 69. A  P 1  n n  t

r2

r

1

65. The common ratio is 343 or 7. To find the next 1 1 term, multiply 1 by 7. The next term is 7.

5 3 2 r 3 5

1

A  1500 1 

1 2  152 or 25 113 2  152 .

2

0.065 12  3 12

55. Multiply 10 by 0.6 three times. The heights of three rebounds are 6 m, 3.6 m, and 2.16 m. 56. Multiply 10 by 2 five times to find the next five scores. The first six scores are 10, 20, 40, 80, 160, and 320.

 1822.01 The value of the investment will be about $1822.01. 70. No; the domain values are at regular intervals, but the range values have a common difference 2. 71. Yes; the domain values are at regular intervals, and the range values have a common factor 3.

57. a  10 216 or 655,360 17 a18  10 217 or 1,310,720

72. 7a2  22a  3  7a2  a  21a  3  a(7a  1)  3(7a  1)  (7a  1) (a  3)

The geometric mean is

2 1 5 3

The score will be greater than a million when 18 questions are answered correctly.

Chapter 10

488

73. 2x2  5x  12  2x2  3x  8x  12  x(2x  3)  4(2x  3)  (2x  3)(x  4)

Richter Number (x)

74. 3c2  3c  5 is prime.

Page 573

Algebra Activity: (Follow-Up of Lesson 10-7)

1. The graph begins by increasing slowly and then increases rapidly for the last few values. 1,000,000 900,000 800,000 700,000 600,000 500,000 400,000 300,000 200,000 100,000 0

1

0.00017

2

0.006



3

0.179

0.173

4

5

4.821

0.00583

5

179

174

6

5643

5464

7

179,100

173,357

The regression equation is y  5.54  10 6 (31.7x ) .

Chapter 10 Study Guide and Review Page 574 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

0 1 2 3 4 5 6 7x

y 200,000 180,000 160,000 140,000 120,000

Vocabulary and Concept Check

d; exponential growth equation g; quadratic function i; symmetry a; equation of axis of symmetry c; exponential function i; vertex b; exponential decay equation h; roots f; Quadratic Formula e; parabola

Pages 574–578

Lesson-by-Lesson Review

11. In y  x2  2x, a  1 and b  2.

100,000 80,000

x x

60,000 40,000

0 1 2 3 4 5 6 7 x

The graph begins increasing slowly and then increases rapidly for the last few values. Use the slope formula to find the values for the table. 0.006  0.00017 2  1 0.179  0.006 3  2 5  0.179 4  3 179  5 5  4 5643  179 6  5 179000  5643 7  6

b 2a 2 2(1)

or 1

x  1 is the equation of the axis of symmetry. y  (1) 2  2(1) 12  1 The vertex is (1, 1). Since the coefficient of the x2 term is positive, the vertex is a minimum.

20,000 0

Rate of Change (slope)

There is no constant value that can be multiplied by each rate of change to obtain the next value.

y

2. No. 3. No; the rate of change between any two points is always a different value. 4. To move from one rate of change to the next, you multiply by 10. 5. The regression equation is y  0.1(10x ) . 6.

Energy Released (y)

 0.00583

y

 0.173  4.821  174  5464

O

 173,357

x

2

y  x  2x

489

Chapter 10

12. In y  3x2  4, a  3 and b  0. x x

b 2a 0 2(3)

Since the coefficient of the x2 term is positive, the vertex is a minimum.

or 0

8 4

x  0 is the equation of the axis of symmetry. y  3(0)  4 4 The vertex is (0, 4). Since the coefficient of the x2 term is negative, the vertex is a maximum.

6 5 4 3 2 1O 4 8 12 16 20 24 y  3x 2  6x  17

y

x x

13. In y  x2  3x  4, a  1 and b  3. x

The

12

x

O

x  2 is the equation of the axis of symmetry. y

or 0

y  2x 2  1 y

3

or 2

3

b 2a 0 2(2)

x  0 is the equation of the axis of symmetry. y  2(0) 2  1 1 The vertex is (0, 1). Since the coefficient of the x2 term is negative, the vertex is a maximum.

x

O

b 2a (3) 2(1)

1 2x

15. In y  2x2  1, a  2 and b  0.

y  3x 2  4

x

y

12

3 2 3  2 32 4 9 9 424 25  4 3 25 vertex is 2, 4 .

1

2

Since the coefficient of the x2 term is positive, the vertex is a minimum.

16. In y  x2  3x, a  1 and b  3. x

y

x

x

O

b 2a (3) 2(1)

3

or 2

3

x  2 is the equation of the axis of symmetry. y  

y  x 2  3x  4

132 22  3132 2

9 4 9 4

9

2

The vertex is 14. In y 

3x2

 6x  17, a  3 and b  6.

x

b 2a 6 2(3)

or 1

x

132, 94 2.

Since the coefficient of the x2 term is negative, the vertex is a maximum. y

y  x 2 3x

x  1 is the equation of the axis of symmetry. y  3(1) 2  6(1)  17  3  6  17  20 The vertex is (1, 20).

Chapter 10

O

490

x

17. Graph f(x)  x2  x  12. Sample answer: x 3 1 1 3 4

f(x) 0 10 12 6 0

21.

2 4 3 2 1O 2 4 6 8 10 12 14

f (x ) 1 2 3 4x

x 3 4 5 6 7

f (x )  x 2  x  12

The solutions are 3 and 4, the x-intercepts of the graph.

22.

6x2  13x  15  13x  15  15  15 6x2  13x  15  0 Graph f(x)  6x2  13x  15. Sample answer:

O

x 1 0 1 2 3

x

The solution is 3, the x-intercept of the graph. 19. Graph f(x)  x2  4x  3. Sample answer: f(x) 2 3 7 3 2

f(x) 4 15 22 17 0

4

O

f (x )

1O 1 2 3 4 5 6 7 x 4 8 12 16 20 24 2 28 f (x )  6x  13x  15

f (x ) x

One root is 3, and the other root is between 1 and 0. 23.

3x2  4  0 3x2  4  4  0  4 3x2  4 4

x2  3

f (x )  x 2  4x  3

4

One root is between 5 and 4, and the other root is between 0 and 1.

x  3 3 x  1.2 The solution set is {1.2, 1.2}.

20. Graph f(x)  2x2  5x  4. Sample answer: x 1 0 1 2 3

x

O

6x2

f (x )  x 2  6x  9

x 5 4 2 0 1

f (x )

The solutions are 3 and 7, the x-intercepts of the graph.

f (x )

f(x) 4 1 0 1 4

f(x) 0 3 4 3 0

f (x )  x 2  10x  21

18. Graph f(x)  x2  6x  9. Sample answer: x 5 4 3 2 1

x2  10x  21 x  10x  21  21  21 x2  10x  21  0 Graph f(x)  x2  10x  21. Sample answer: 2

f(x) 11 4 1 2 7

24.

f (x )  2x 2  5x  4 O

x2  16x  32  0 x  16x  32  32  0  32 x2  16x  32 2  16x  64  32  64 x (x  8) 2  32 x  8  132 x  8  8  132  8 x  8  132 x  8  132 or x  8  132 x  2.3 x  13.7 The solution set is {2.3, 13.7}. 2

f (x )

x

; The graph has no x-intercepts. Therefore, the equation has no real solutions.

491

Chapter 10

m2  7m  5

25.

m2 

49 7m  4 7 2 m2 7 m2

1

2

7

29.

5 

69 4

x

69

 3 4

7



69

7

m  2  2  3 4  2 m m

7 2

69 34



7 2

or



69 34



7 2

m

 69 34



x

4a2  16a  15  0 1 (4a2 4

 16a  15)  a2  4a 

a2  4a 

15 4



15 4 15 4

30. r2  10r  9  0

1 (0) 4

r

0



0

15 4



15

a2  4a   4 a2

 4a  4  2) 2

(a 



a2 a22 a a  2 

1 2



15 4  4 1 4 1 2 1 2  2 1 2  2

r

31.

5



3

 2 or 1.5

2

1

1 2 y  2y  1 2 1 2 y  2y  1 2



0



2  2(0)

p

y2  4y  2  0 2  4y  2  2  0  2 y y2  4y  2 2  4y  4  2  4 y (y  2) 2  6 y  2  16 y  2  2  16  2 y  2  16

32.

5

5 4

1n  32 22  1

9

4

3

n  2  1 3

3

3

n  2  2  1  2 n  1.5  1 n  1.5  1 or n  1.5  1  0.5  2.5 The solution set is {0.5, 2.5}. Chapter 10

8  282  4(2) (3) 2(2)



8  140 4 8  140 4

or

y

8  140 4

 3.6  0.4 The solution set is {3.6, 0.4}.

5

n2  3n  4 9

b  2b2  4ac 2a



y

n2  3n  4  4  0  4 n2  3n  4 

2y2  3  8y  3  8y  8y  8y 2y2  8y  3  0 y

5 5

4  242  4(4) (15) 2(4) 4  1256 8 4  16 8 4  16 4  16 or p  8 8

2y2

n2  3n  4  0 5

b  2b2  4ac 2a

 2.5  1.5 The solution set is {2.5, 1.5}.

y  2  16 or y  2  16  4.4  0.4 The solution set is {4.4, 0.4}. 28.

4p2  4p  15 4p  4p  15  15  15 4p2  4p  15  0 p

The solution set is {2.5, 1.5}. 27.

10  2102  4(1) (9) 2(1) 10  164 2 10  8 2 10  8 10  8 or r  2 2

2

1

 2 or 2.5

b  2b2  4ac 2a

 9  1 The solution set is {9, 1}.

a  2  2

or

b  2b2  4ac 2a (8)  2(8) 2  4(1) (20) 2(1) 8  1144 2 8  12 2 8  12 8  12 or x  2 2

 2  10 The solution set is {2, 10}.

 0.7  7.7 The solution set is {0.7, 7.7}. 26.

x2  8x  20 x  8x  20  20  20 x2  8x  20  0 2

49 4

492

33.

1

2d2  8d  3  3 2d  8d  3  3  3  3 2d2  8d  0 d

8  282  4 (2) (0) 2(2)



8  164 4 8  8 4 8  8 or 4

 d

1

d



5  1613 42

or

 0.7

 5500(1.004375) 180  12,067.68 The final amount is $12,067.68.

1

a

1

5  1613 42

14 12 10 8 6 4 2

3 2 1O 2

y

43. an  a1 rn1 2 3 a4  7 3

12 8  7 1 27 2

1 2 3 4 5x

y  3x  2

4 2 1

2

1 2



243 81

a3  a1 r31

45.

20  5r2 1 (20) 5

 r2

4  r2 2  r The geometric mean is 5(2)  10 or 5(2)  10.

1 2x

a3  a1 r31

46.

48  12r2 1 (48) 12

Sample answer:

1 0 1

1 2 1  243 1 81 2 3

y 14 12 10 8 6 4 2

12

8

44. an  a1 rn1 1 a3  243 3 4

56

1 x 2

2

2

0.0975 365 40 365

 27

1 3 9 27

y

1

42. an  a1 rn1 a5  2(24 )  2(16)  32

y  3x  6

1 3

x

2

0.075 12  25 12

 500(1.000267123) 14,600  $24,688.36 The final amount is $24,688.36.

O The y-intercept is 9. 6 5 4 3 2 12

37. y  2

2

A  500 1 

36. y  3x2 Sample answer: x y 2 1 0 1

1

r 41. A  P 1  n nt

 0.5

1 69 1 63

3

2

r nt

 15,000(1.00625) 300  97,243.21 The final amount is $97,243.21.

The solution set is {0.7, 0.5}. 35. y  3x  6 Sample answer: x y

0 7 1 9 2 15 The y-intercept is 7.

2

0.0525 12 15 12

A  15,000 1 



1

1

40. A  P 1  n

5  252  4(21) (7) 2(21)

2

2

A  5500 1  8  8 4

b  2b2  4ac 2a

5  1613 42

32

r 39. A  P 1  n nt

34. 21a2  5a  7  0

a

2

0.08 4 8 4

 2000(1.02)  3769.08 The final amount is $3769.08.

 4 0 The solution set is {4, 0}. a

1

A  2000 1 

b  2b2  4ac 2a



2

r 38. A  P 1  n nt

2



1 (12r2 ) 12

4  r2 2  r The geometric mean is 12(2)  24 or 12(2)  24.

y

47. a  a r31 3 1 1 4 1 2

( 12 )x

y2

O

x

 1r2 r

The geometric mean is 1

The y-intercept is 2.

493

112 2  12 or 1112 2  12. Chapter 10

6. In y  2x2  3, a  2 and b  0.

Chapter 10 Practice Test

x

Page 579 1. 2. 3. 4.



c; Quadratic Formula b; exponential growth equation a; exponential decay equation In y  x2  4x  13, a  1 and b  4. x 

b 2a (4) 2(1)

b 2a 0 2(2)

or 0

x  0 is the equation of the axis of symmetry. y  2(0) 2  3 3 The vertex is (0, 3). Since the coefficient of the x2 term is positive, the vertex is a minimum.

or 1

x  2 is the equation of the axis of symmetry. y  22  4(2)  13  4  8  13 9 The vertex is (2, 9). Since the coefficient of the x2 term is positive, the vertex is a minimum.

y

y  2x 2  3 O

y

14 12 10 8 6 4 2

2 7. y  1(x  2)  1 y  (x2  4x  4)  1 y  x2  4x  4  1 y  x2  4x  3 a  1 and b  4

y  x 2  4x  13

2 1O 2

1 2 3 4 5 6x

x 

5. In y  3x2  6x  4, a  3 and b  6. x 

b 2a (6) 2(3)

x  1 is the equation of the axis of symmetry. y  3(1) 2  6(1)  4  3  6  4 7 The vertex is (1, 7). Since the coefficient of the x2 term is negative, the vertex is a maximum. y  3x  6x  4

Chapter 10

or 2

2 y y  1(x  2)  1

O

y

O

b 2a 4 2(1)

x  2 is the equation of the axis of symmetry. y  1(x  2) 2  1  1(0) 2  1 1 The vertex is (2, 1). Since the coefficient of the x2 term is negative, the vertex is a maximum.

or 1

2

x

x

494

x

12. x2  7x  6  0 Sample answer: Solve using the Quadratic Formula.

8. Graph f(x)  x2  2x  2. Sample answer: x

f(x)

1

5

x

0 1 2 3

2 1 2 5



7  272  4(1) (6) 2(1)



7  125 2 7  5 2 7  5 or 2

f (x )

f (x )  x 2 2x  2

 x

O

13. 2x2  5x  12  0 Sample answer: Solve using the Quadratic Formula.

x2  6x  7 2 x  6x  7  7  7 x2  6x  7  0 Graph f(x)  x2  6x  7. Sample answer: x

x

f(x)

f (x )

(5)  2(5) 2  4(2) (12) 2(2)



5  1121 4 5  11 4 5  11 or 4 1 12

2

4

1



3

2

x

2

1

1

2

 x

14

4

13 12 11 10

1 0 1 4

8

4

12



7  2(7) 2  4(6) (20) 2(6)

8



7  1529 12 7  23 12 7  23 or 12



f(x)

4

22

3 2 7 8

0 50 0 22

O x

The solutions are 2.5, 1.3. 15. 3k2  2k  5 Sample answer: Solve using the Quadratic Formula. 3k2  2k  5  0 k

f (x ) 20 4

O

4

8

b  2b2  4ac 2a



2  222  4(3) (5) 2(3)



2  164 6 2  8 6 2  8 or 6 2 13



x

k

20 40

7  23

n  12  1.3

n  2.5

The solution is 12, the x-intercept of the graph. 2x2  8x  42 11. 2  8x  42  42  42 2x 2x2  8x  42  0 Graph f(x)  2x2  8x  42. Sample answer: x

b  2b2  4ac 2a

n

4 f (x )  x  24x  144 12

4

f (x )

2

16

5  11 4

14. 6n2  7n  20 Sample answer: Solve using the Quadratic Formula. 6n2  7n  20  20  20 6n2  7n  20  0

One root is between 5 and 4, and the other root is between 2 and 1. 10. Graph f(x)  x2  24x  144. Sample answer: f(x)

x

The solutions are 112, 4.

f (x )  x 2  6x  7

x

b  2b2  4ac 2a



5

O

7  5

x 2 x  6  1 The solutions are 6, 1.

; The graph has no x-intercepts. Thus, the equation has no real number solutions. 9.

b  2b2  4ac 2a



k

2  8 6

1 2

f (x)  2x 2  8x  42

The solutions are 13, 1.

The solutions are 3 and 7, the x-intercepts of the graph.

495

Chapter 10

20. 7m2  m  5 Sample answer: Solve using the Quadratic Formula.

3 2 16. y2  5y  25  0

Sample answer: Solve by completing the square. 3

2

2

2

y2  5y  25  25  0  25

7m2  m  5  0

2  25 2 9  25  100 1  100 1  10 1 3  10  10 3 1 y  10  10 3 1 3 1 y  10  10 or y  10  10 1 2 5 5 1 2 The solutions are 5, 5. 3 y  5y 3 9 y2  5y  100 3 y  10 2 3 y  10 3 3 y  10  10

2

1

2

1  1141 14

21. y 

1  1141 14

x

y

2

4

1

2

0

1

1 2

14  1256 6 14  16 6 14  16 or 6 1 3

m

y

1 2 1 4

x

14  16 6

x

 5

2 1 0 1

14 12 10 8 6 4 2

1 2

1 2 4 8

b  2b  4ac 2a (13)  2(13) 2  4(1) (32) 2(1)



13  1297 2

23. y 

or

3x2

 4x  8  0

x

b  2b2  4ac 2a

z



4  242  4(3) (8) 2(3)



4  2112 6

1 2 3 4x

113 2x  3

or

x

y

x y 2 6 1 0 0 2 2

1 2 3

O

8

2 2 9

x y 1 3

( 3)

The y-intercept is 2. 24. an  a1 # rn1 25. an  a1 # rn1 5 a6  12(2 ) a4  20(33 )  12(32)  20(27)  384  540

4  2112 6

 1.1

The solutions are 2.4, 1.1. Chapter 10

y  4 · 2x

Sample answer: 13  1297 2

19. 3x2  4x  8 Sample answer: Solve using the Quadratic Formula.

 2.4

1 2 3x

The y-intercept is 4.

 2.1  15.1 The solutions are 2.1, 15.1.

4  2112 6

y

y

4 3 2 1O 2

2



x

1 2

5 4 3 2 1O 2

z2  13z  32  0

z

( )x

y

3

The solutions are 5, 18. 72  13z  32 Sample answer: Solve using the Quadratic Formula.

13  1297 2

14 12 10 8 6 4 2

The y-intercept is 1. 22. y  4 2x Sample answer:

1 . 3

z

1  1141 14

112 2x

2



or

Sample answer:

b  2b  4ac 2a (14)  2(14)  4(3) (5) 2(3)





2



x

(1)  2(1) 2  4(7) (5) 2(7)

 0.8  0.9 The solutions are 0.8, 0.9.

17. 3x  5  14x Sample answer: Solve using the Quadratic Formula. 3x2  5  14x  14x  14x 3x2  14x  5  0





m

2

x

b  2b2  4ac 2a

m

496

x

a3  a1 r31

26.

4. D; Let x represent the number of grapefruit. Then 2x represents the number of apples. 0.20(2x)  0.25x  1.95 0.4x  0.25x  1.95 0.65x  1.95

63  7r2 1 (63) 7

1

 7 (7r2 )

9  r2 3  r The geometric mean is 7(3)  21 or 7(3)  21.

0.65x 0.65

x3 2x  6 The shopper bought 6 apples. 5. D; Area  (2x  3) (2x  6)  4x2  12x  6x  18  4x2  6x  18 2 6. B; x  x  12  0 (x  4)(x  3)  0 x  4  0 or x  3  0 x  4 or x  3 7. B; Substitution of the x-values into y  x2  9 yields the corresponding y-values.

a3  a1 r31

27.

1

12  3 r2 3(12)  3 36  r2 6  r

113 r22 1

The geometric mean is 3 (6)  2 or 1 3 (6)  2. 28. y  C(1  r) t  17,369(1  0.16) 2  12,255.57 14,458  12,255.57  2202.43 The car will be worth $2202.43 less than the buyout price of the lease.

1

2

r 29. A  P 1  n nt

1

 1500 1 

8. B; y  x2  2x  3 x 

b 2a (2) 2(1)

or 1

2

y  1  2(1)3  4 The vertex is at (1, 4). This is the only equation with the correct vertex.

2

0.06 4.10 4

 1500(1.015) 40  2721.03 The total amount will be $2721.03.

9. B; 2x2  8x  6  0 2(x2  4x  3)  0 2(x  3)(x  1)  0 or x10 x30 x3303 x1101 x  3 x  1 The graph intersects the x-axis at (3, 0) and (1, 0). 10. Let x be the amount earned during the 4th week.

12

30. A; The common ratio is 4 or 3. Multiply 108 by 3 to find the next term. The next term is 324.

Chapter 10 Standardized Test Practice Pages 580–581 1. B; Translating the line down two units does not change the slope of the line, but it does change the y-intercept by 2 units.

x  18.50  23.00  15.00 4 x  56.5 4 x  56.5 4 4

1

2. C; a  kb Find k. 21  k(6) 21 6 7 2

1.95

 0.65

 18  18

2  4(18)

x  56.5  72 x  56.5  56.5  72  56.5 x  15.50 She should earn at least $15.50 during the 4th week. 11. 8x  4y  9  0 4y  8x  9

k k 7

a  2b Now find a when b  28.

y

7

a  2 (28) or 98

y

3. A; Since the slope of the graphed line is negative, the answer must be A or B. From inspection of the graph, the y-intercept must be less than 5. So, by elimination, A is the answer.

8x  9 4 9 2x  4

The slope of this line is 2. The slope of a line 1 perpendicular to it is 2. y  y1  m(x  x1 ) 1 y  3  2 (x  2) 1

y  3  2x  1 1

y  2x  4

497

Chapter 10

12. 5a  4b  25 3a  8b  41 Multiply the first equation by 2. Then add. 10a  8b  50 3a  8b  41 13a  91 a7 Use 5a  4b  25 to find b. 5(7)  4b  25 35  4b  25 4b  10 5 1 b  2 or 22

b

or

1

4(x  1) 2  4  0 1

1

1

4(x  1) 2  4 1

4(x  1) 2 4



4 4 1

(x  1) 2  16 1

x  1  4 1

x  1  1  4  1 x1  1

x14

1 4

1

x14

or

3

4

The solution set is e 4, 14 f . 3

20e. y  4x2  8x 

14  1388 2

 4  8 

1

15 4

 4(12 )  8(1) 

 16.85  2.85 Ignore the negative value. Each side was increased by 2.85 inches. 16. B; By inspection, the median is 40. The values greater than 40 are clumped near 40, whereas the values less than 40 are spread out further from the median. These low values will bring the average below 40.

15 4

15 4

1

4

1 12

The minimum point is located at 1, 4 . y

20f. O

(1, 14 ) x

3 0, 4

(

17. A; The solution of 6p  12 is 2. The solution of 1 1 10q 5 is 2. 2 7 2

y  4x 2  8x  15 4

18. B; 5.3  103  5300 6 53,000 19. C; All even-numbered terms of the first sequence are the same as the even-numbered terms in the second sequence.

Chapter 10

1

4(x  1) 2  4  4  0  4

1

m

15 4

a  4 and b  8

 14

14  1388 2

1

4x2  8x  4  4

The equation of the axis of symmetry is x  1. 20c. Since the coefficient of the x2 term is negative, the parabola opens downward. 20d. Solve the equation, using the right-side expression.

2



1

4(x2  2x  1)  4

Equation for the axis of symmetry of a parabola

8

b  2b  4ac 2a 14  214  4(1) (48) 2(1)

1

4(x  1) 2  4

 4x2  8x 

x  2(4) or 1

2



m

15 4

20b. x  2a

b2  4ac  (11) 2  4(6)(4)  25 Since the discriminant is positive, the equation has two real solutions. 15. Let m be the amount by which each side is increased. (m  8) (m  6)  2(8.6) m2  6m  8m  48  96 m2  14m  48  96 m2  14m  48  0

14  1388 2

4x2  8x 

4x2  8x 

13. x2  4x  5  (x2  4x  4)  5  4  (x  2) 2  9 h  2 and k  9 14. y  6x2  11x  4. Find the number of real solutions to 6x2  11x  4  0. Find the value of the discriminant.

m

4x2  8x 

15 ?  4 15 ?  4 15 ?  4

20a. 4x2  8x 

498

)

1 0, 1 4

(

)

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Chapter 11 Page 585

Radical Expressions and Triangles 3. Sample answer: 212  313 and 212  313; (212  313) (212  313)  (212) 2  (313) 2  8  27  19 4. 120  12  2  5 5. 12  18  116 4  222  15  215

Getting Started

1. 125  5 2. 180  8.94 3. 156  7.48 4. 1324  18 5. 3a  7b  2a  (3a  2a)  7b  a  7b 6. 14x  6y  2y  14x  (6y  2y)  14x  4y 7. (10c  5d)  (6c  5d)  (10c  6c)  (5d  5d)  16c

6. 3110  4110  (3  4) ( 2102 )  (12) (10)  120 2 2 7. 254a b  22  33  a2  b2  12  232  13  2a2  2b2  12  3  13  0a 0  0b 0  3 0ab 0 16

8. (21m  15n)  (9n  4m)  (21m  15n)  (9n  4m)  (21m  4m)  (15n  9n)  25m  6n 9. x(x  5)  0 x  0 or x  5  0 x5 {0, 5}

8. 260x5y6  222  3  5  x5  y6  222  13  15  2x4  1x  2y6  2  13  15  x2  1x  0y3 0  2x2 0y3 0 115x

10. x2  10x  24  0 (x  4)(x  6)  0 x  4  0 or x  6  0 x  4 or x  6 {6, 4} 11.

12.

9.

x2  6x  27  0 (x  9)(x  3)  0 x  9  0 or x  3  0 x9 x  3 {3, 9}

11.

2x2  x  1  2 2x2  x  1  0 (2x  1)(x  1)  0 2x  1  0 or x  1  0 2x  1 x  1

13.

12.

14.

8  3  2  12 24  24 yes 8 ? 12  16 10 ?

15.

10  12  8  16 120  128 no



4 16 6



13 110



2 16 3



130 10

8 3  12

 16

10.

8

3

12

3

3 10  110 110

 110

3  12 12

3 



24  8 12 9  2



24  8 12 7

2 15 4  18

4 ? 16  25 5 ?

5  16  4  25 80  100 no 16.

13

4 16

8(3  12) ( 12) 2

1

2 ? 8  12 3 ?

16



 32 

x2

512, 16

4 16

6 ? 3  15 30 ?

6  15  30  3 90  90 yes

2 15 18

 4 

4  18 18

 4 



2 15(4  18) (4) 2  ( 18) 2



8 15  2 140 16  8



8 15  2 222  2  5 8



8 15  4 110 8



2 15  110 2

13. A  s2  (217 ) 2  22 ( 17 ) 2  4(7)  28 ft2

11-1 Simplifying Radical Expressions

14. P  2 3 32 /

8

Pages 589–590

 2 3 32

Check for Understanding

1

 2 3 4

1. Both x4 and x2 are positive even if x is a negative number. 2.

1 1a



1a 1a



1 11 14 2 1  2 1 2 2  2

1a a

  3.14 s

499

Chapter 11

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Pages 590–592

Practice and Apply 2

29.

3

15. 118  22  3  12  232  12  3  312

16. 124  22  3  222  12  13  216

17. 180  224  5  224  15  22 15  415

18. 175  252  3  252  13  513

19. 15  16  130 20. 13  18  124  223  3  222  12  13  2 16 21. 7130  216  (7  2)( 130  16)  141180  14222  32  5  14222  232  15  14  2  3  15  8415

31.

1t

t

3 8  18 1t 18

 18





1t  222  2 8









1t  222  12 8 2 12t 8



12t 4

5c5

25c5



12 13



12 13



16 3

28.

13

 13

34.



15  2c4  1c 14  1d4  1d



23x3 14y5

c2 15c



23  x2  x 14y4  y

1d

c2 15cd 2 0 d3 0 18 18  6  12 6  12

x 13x x 13x

 6  12 12

6 

18(6  12) 62  ( 12) 2



18(6  12) 36  2



18(6  12) 34



9(6  12) 17



54  9 12 17

10 17  12

2 15  18

 4

4  18 18

 4 



2 15 (4  18) (4) 2  ( 18) 2



8 15  2 140 16  8



8 15  2 22 3  5 8



8 15  2 22 2  12  15 8



8 15  4 110 8



2 15  110 2 17  12 12

3 5  3 4  3 20



10( 17  12) 7  2

9

18

 17 



3 10



10( 17  12) 5



19 110



2( 17  12) 1



19 110

110

 217  212

 110 36.

500

2 13  16

1y

 2y2 1y  1y



2 15 4  18

3x3

 2y2 1y

10( 17  12) ( 17) 2  ( 12) 2

3 110 10

3 12x2y6 3 4y5



6

9x5y

 

10 17  12

35.

3 13 0p 0

25  c4  c 14  d4  d



33.

232  13 0p 0





3

29x5y 112x2y6

c2 15cd



Chapter 11

32.

 2d2 1d2

26. 272x3y4z5  223  32  x3  y4  z5  222  12  232  2x2  1x  2y4  2z4  1z  2  12  3  0x 0  1x  y2  z2  1z  6 0x 0 y2z2 12xz 2

18

3 4d5  14d5

25. 2147x6y7  272  3  x6  y7  272  13  2x6  2y6  1y  7  13  0x3 0  0y3 0  1y  7 0x3y3 0 13y

7

233 0p 0



c2 15c

24. 250m3n5  252  2  m3  n5  252  12  2m2  1m  2n4  1n  5  12  0m 0  1m  n2  1n  5n2 0m 0 12mn

2

127 1p2

27

 2d2 1d  1d

23. 240a4  223  5  a4  222  12  15  2a4  2  12  15  a2  2a2 110

37  33  33

3 p2 

 2d2 1d

22. 213  5127  (2  5)( 13  127)  10( 181 )  10(9)  90

27.

30.

13  16 16



2 13  16



2( 13  16) ( 13) 2  ( 16) 2



2( 13  16) 3  6



2 13  2 16 3



2 16  2 13 3

 13 

x 13xy 2 0 y3 0

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37.

4 4  3 13

4

4  3 13

46. s  130fd  130(0.8)d  124d  223  3  d  222  12  13  1d  216d 47. wet road: s  312d  312(110)  31220  44.5 mph dry road: s  216d  216(110)  21660  about 51.4 mph

4  3 13  3 13

4

4(4  3 13) (3 13) 2

 42 

38.



4(4  3 13) 16  27



16  12 13 11



16  12 13 11

3 17 5 13  3 15

3 17

5 13  3 15

 5 13  3 15  5 13  3 15 

3 17(5 13  3 15) (5 13) 2  (3 15) 2



15 121  9 135 75  45



15 121  9 135 30



5 121  3 135 10

1

48. s  2 (a  b  c)

39. A  /w  (4110)(315)  12150  12252  2  12252  12  6012 or about 84.9 cm2

40. A  /w

41. A  s2 s  1A s  172  223  32  222  22  232  2  12  3

42.

1

 2 (13  10  7)

a

1

 2 (30)

a



38  32



3 16

 15 49. A  1s(s  a) (s  b) (s  c)  115(15  13)(15  10)(15  7)  115(2)(5)(8)  11200  224  3  52  224  13  252  22  13  5  2013 or about 34.6 ft2

a2

2a2 116 a  4 m2 1 E  2mv2



2E  mv2 2E m

 v2

v

2E

3m

50.

 a(a  1

 612 in. 43. v 

3

1 (a  1  1a)

(a  1  1a) 1a)

 (a  1 

a  1  1a  1a)  1(a  1  1a )  1a(a  1  1a) a  1  1a  1  1a  a 1a  1a  a

 a2  a  a 1a  a

2(54) 0.6

 1180  222  32  5  222  232  15  2  3  15  615 or about 13.4 m/s 44. Escape Velocity 

3

2GM R



3

2(6.7  1020 ) (7.4  1022 ) (1.7  103 )



3

2(6.7  7.4) (1020  1022 ) (1.7  103 )



3

(99.16) (102 ) (1.7) (103 )



31



a  1  1a a2  3a  1

51. A lot of formulas and calculations that are used in space exploration contain radical expressions. Answers should include the following. • To determine the escape velocity of a planet, you would need to know its mass and radius. It would be very important to know the escape velocity of a planet before you landed on it so you would know if you had enough fuel and velocity to launch from it to get back into space. • The astronomical body with the smaller radius would have a greater escape velocity. As the radius decreases, the escape velocity increases. 52. C; surface area  96a2

21 2

99.16 102 1.7 103

 258.3  101  15.83  2.4 km/s The Moon has a much lower escape velocity than Earth. 45. s  130fd  130(0.6)d  118d  22  32  d  12  232  1d  312d

area of one face 

96a2 6

 16a2 length of each edge  4a Volume  s3  (4a) 3  64a3 53. B; x  81b2 1x  281b2  181  2b2  9b

501

Chapter 11

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54.

63. 1, 2, 4, 8

y  91.4  (91.4  t) [0.478  0.301( 1x  0.02) ] 9  91.4  (91.4  12) [0.478  0.301( 1x  0.02) ] 9  91.4  79.4[0.478  0.3011x  0.00602] 9  91.4  79.4[0.47198  0.3011x] 9  91.4  37.475212  23.8994 1x 9  53.924788  23.8994 1x 62.924788  23.89941x 2.632902416  1x 7 mph  x

2 1

The common factor is 2. 1, 2, 4, 8 16 32

The next three terms are 16, 32, and 64. 64. 384, 192, 96, 48 192 384

 91.4  (91.4  12) [ 0.478  0.301( 14  0.02) ]  91.4  (91.4  12) [ 0.478  0.301(2  0.02) ]  91.4  (79.4) [ 0.478  0.301(1.98) ]  91.4  (79.4) [ 0.478  0.59598]  91.4  (79.4) [ 1.07398]  91.4  85.274012  6.125988  6F 1

1

1

1

5 2

x 58. x  x

5 1 2

57.

59.

3

 x2  2x3  2x2  1x  x 1x

The next three terms are 24, 12, and 6. 1 2

65. 9, 3, 4, 24 2 3

1

96

The common factor is 6. 1 2

1 2 , , 9 3

1a a  1 3 a1 a a a3

1 2

4, 24

3

3

1

3 4

4

The common factor is 4.

34 1

a3

3

4

3, 4,

3 3 , 16 64

0y3 0 

4

1 3 13

10 50

0y3 0  3  2

1 2

y  3

23

1

13

3 4096

.

1

 5  0.2

69. y  75(0.875) t  75(0.875) 15 y  10.1 So the coffee is 95  10.1  84.9C 6  (5)  30 70. 6x2  7x  5

y  13  13 13 3

1 61. s2t 2 8 2s5t4  (s16t4 ) 2s4  s  t4

 (s16t4 ) 2s4  1s  2t4  s16  t4  s2  1s  t2  (s16  s2 )(t4  t2 ) 1s  s18t6 1s

Factors of 30 10, 3

Sum of Factors 7

6x2  7x  5  6x2  mx  nx  (5)  6x2  10x  (3)x  5  2x(3x  5)  (1) (3x  5)  (3x  5) (2x  1)

62. 2, 6, 18, 54 3

The common factor is 3. 2, 6, 18, 54 162 486 1458

   3 3 3

The next three terms are 162, 486, and 1458.

Chapter 11

and

The next three terms are 0.08, 0.016, and 0.032.

1

6 2

3 3 , , 256 1024

68. y  art y  1000(212 )  4,096,000

1

2

y  13

2

4

    0.2  0.2  0.2

3

1

4

The common factor is 0.2. 50, 10, 2, 0.4 0.08 0.016 0.032

3

32

y

3 4096

67. 50, 10, 2, 0.4 1

313 2 1

1

3 1024

The next three terms are

1

0y3 0  3

3 256

   1 1 1

5

0y3 0 

5184

3

1

a2

 a6 

864

66. 3, 4, 16, 64

1

a1 3

1

60.

144

   6 6 6

The next three terms are 144, 864, and 5184.

1

a2

 a2  3 0y3 0

6

    0.5  0.5  0.5

1 2



 0.5

The common factor is 0.5. 384, 192, 96, 48 24 12

1x 24  x2



64

 

 (2)  (2)  (2)

55. y  91.4  (91.4  t) [ 0.478  0.301( 1x  0.02) ]

56. x 2  x 2  x 2  2 x

 2

502

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71. 35x2  43x  12

78. 3a  2b  11 (3, 10); 3(3)  2(10)  11 9  20  11 11  11, true (4, 1); 3(4)  2(1)  11 12  2  11 14  11, false (2, 2.5); 3(2)  2(2.5)  11 6  5  11 11  11, true (3, 2); 3(3)  2(2)  11 9  4  11 5  11, false {(3, 10), (2, 2.5)}

35  12  420

Factors of 420 28, 15

Sum of Factors (43) 43

35x2  43x  12  35x2  mx  nx  12  35x2  (28)x  (15)x  12  (35x2  28x)  (15x  12)  7x(5x  4)  3(5x  4)  (5x  4) (7x  3) 72. 5x2  3x  31 prime 73. 3x2  6x  105 3  (105)  315 Factors of 315 21, 15

Sum of Factors (6) 6

3

3x2  6x  105  3x2  mx  nx  105  3x2  (21)x  15x  105  (3x2  21x)  (15x  105)  3x(x  7)  15(x  7)  (x  7)(3x  15)  3(x  7)(x  5) 74. 4x2  12x  15 prime 75. 8x2  10x  3 8  3  24 Factors of 24 6, 4 2

79. 5  2 x  2y 3

(0, 1); 5  2 (0)  2(1) 502 5  2, false 3

(8, 2); 5  2 (8)  2(2)

14, 12 2;

Sum of Factors (10) 10

5  12  4 7  4, false 5  6  1 1  1, true

8x2

 mx  nx  3 8x  12x  15   8x2  (6)x  (4x)  3  (8x2  6x)  (4x  3)  2x(4x  3)  1(4x  3)  (4x  3)(2x  1) 76. y  3x  2 (1, 5); 5  3(1)  2 532 5  5, true (2, 6); 6  3(2)  2 662 6  8, false (2, 2); 2  3(2)  2 2  6  2 2  4, false (4, 10); 10  3(4)  2 10  12  2 10  10, true {(1, 5), (4, 10)} 77. 5x  2y  10 (3, 5); 5(3)  2(5)  10 15  10  10 25  10, false (2, 0); 5(2)  2(0)  10 10  0  10 10  10, true (4, 2); 5(4)  2(2)  10 20  4  10 24  10, false (1, 2.5); 5(1)  2(2.5)  10 5  5  10 10  10, true {(2, 0), (1, 2.5)}

1 12

3

5  2 (4)  2 2

3

(2, 1); 5  2 (2)  2(1) 532 2  2, true

514, 12 2, (2, 1) 6

80. 40  5d 40 5



5d 5

20.4 3.4

8  d 82. (11)

81. 20.4  3.4y 

3.4y 3.4

6y

h 11 h 11

 25

1 2  (11)(25) h  275

83.

r

65  29 29(65)  29

129r 2

1885  r 84. (x  3)(x  2)  x(x)  x(2)  3(x)  3(2)  x2  2x  3x  6  x2  x  6 85. (a  2)(a  5)  a(a)  a(5)  2(a)  2(5)  a2  5a  2a  10  a2  7a  10 86. (2t  1)(t  6)  2t(t)  2t(6)  1(t)  1(6)  2t2  12t  t  6  2t2  11t  6 87. (4x  3)(x  1)  4x(x)  4x(1)  3(x)  3(1)  4x2  4x  3x  3  4x2  x  3

503

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12. P  4s  4(4  316)  4(4)  4(316)  16  1216 ft

88. (5x  3y) (3x  y)  5x(3x)  5x(y)  3y(3x)  3y(y)  15x2  5xy  9xy  3y2  15x2  4xy  3y2 89. (3a  2b)(4a  7b)  3a(4a)  3a(7b)  2b(4a)  2b(7b)  12a2  21ab  8ab  14b2

The perimeter is 16  1216 ft. A  s2  (4  316)(4  316)  4(4)  4(316)  (316) (4)  (316) (316)  16  1216  1216  9136  16  2416  9(6)  16  2416  54  70  2416 ft2 The area is 70  2416 ft2. V  1PR 13. 100-watt: V  1100(110)  1100  1110  101110  104.88 volts 75-watt: V  175(110)  125  3  110  252  13  1110  51330  90.83 The 100-watt bulb takes 101110  51330 or about 14.05 volts more than a 75-watt bulb.

 12a2  13ab  14b2

11-2

Operations with Radical Expressions

Page 595

Check for Understanding

1. to determine if there are any like radicands 2. The Distributive Property allows you to add like terms. Radicals with like radicands can be added or subtracted. 3. Sample answer: ( 12  13) 2  2  216  3 or 5  2 16 4. 413  713  (4  7) 13  1113 5. 216  716  (2  7) 16  516 6. 515  3120  515  3222  5  515  3 ( 222  15)  515  3 (215)  515  615  (5  6) 15  15

Pages 595–597

7. 213  112  213  222  3  213  222  13  213  213  (2  2) 13  413 8. 315  516  3120  315  516  3222  5  315  516  3 ( 222  15)  315  516  3 (215)  315  516  615  (3  6) 15  516  915  516 9. 813  13  19  813  13  3  (8  1) 13  3  913  3 10. 12( 18  413)  ( 12 )( 18)  ( 12) (413)  116  416  4  416

20. 813  212  312  513  813  513  212  312  (8  5) 13  (2  3) 12  1313  12 21. 416  117  612  4117  416  612  117  4117  416  612  (1  4) 117  416  612  5117 22. 118  112  18  19  2  14  3  24  2  232  2  222  3  222  2  312  213  212  312  212  213  (3  2) 12  213  512  213

11. (4  15)(3  15)  4(3)  4 15  ( 15)(3)  ( 15) ( 15)  12  415  315  125  12  (4  3) 15  5  17  715

Chapter 11

Practice and Apply

14. 815  315  (8  3) 15  1115 15. 316  1016  (3  10) 16  1316 16. 2115  6115  3115  (2  6  3) 115  7115 17. 5119  6119  11119  (5  6  11) 119  0119 0 18. 161x  21x  (16  2) 1x  181x 19. 315b  415b  1115b  (3  4  11) 15b  1015b

504

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23. 16  213  112  16  213  14  3  16  213  222  3  16  213  213  16  (2  2) 13  16  413 24. 317  2128  317  214  7  317  2222  7  317  2(217)  317  417  (3  4) 17  17 25. 2150  3132  2125  2  3116  2  2252  2  3242  2  2(512)  3(412)  1012  1212  (10  12) 12  212 26. 12 

30. 16( 13  512)  ( 16) ( 13)  ( 16) (512)  118  5112  19  2  514  3  232  2  5222  3  312  5(213)  312  1013 31. 15(2110  312)  ( 15) (2110)  ( 15) (312)  2150  3110  2125  2  3110  2252  2  3110  2(512)  3110  1012  3110 32. (3  15)(3  15)  32  ( 15) 2 95 4 33. (7  110) 2  (7) 2  2(7) ( 110)  (110) 2  49  14110  10  59  14110 34. ( 16  18)  ( 124  12)

11

1

3 2  12  12  12 

1 12

 12 

1 12



12 1



2 12 2 3 12 2

 27. 110 





 ( 16)( 124)  ( 16)( 12)  ( 18) ( 124)  ( 18)( 12)  1144  112  1192  116  12  14  3  164  3  4

12 12

 16  222  3  282  3  16  213  813  16  (2  8) 13  16  1013

12 2



12 2

35.

3 5  110 

22 15

 110 

12 15

 110 

110 5



5 110 5

110 5



4 110 5

2



1

15 15



 252  7  222  7  517  217  (5  2) 17  317

36. (2110  3115) (313  212)  (2110) (313)  (2110) (212)  (3115) (313)  (3115) (212)  6130  4120  9145  6130  414  5  919  5  4222  5  9232  5  4(215)  9(315)  815  2715  1915 37. (512  315) (2110  3)  (512) (2110)  (512) (3)  (315) (2110)  (315) (3)  10120  1512  6150  915  1014  5  1512  6125  2  915  10222  5  1512  6252  2  915  10(215)  1512  6(512)  915  2015  1512  3012  915  (2015  915)  (1512  3012)  1115  1512

1 11 13 2 1 2  313  23  5  3 1 13 2 1 13  313  315  3 1 13  13 2 13  313  315  3 1 3 2

28. 313  145  3 3 3  313  19  5  3

 313  315  13  313  13  315  (3  1) 13  315  413  315

11  3 14  7  10 1 17 2 29. 6 3 74  3128  10 3 17  6 1 17 14 2

1 172 2  3 222  7  101 171 2 6 17 1 17  2  3(217)  10 1 17  17 2 6 17 17  2  6 17  10 1 7 2 6



6 17 2



42 17 14



(42  84  20) 17 14



106 17 14



53 17 7



6 17 1



( 15  12) ( 114  135)  ( 15)( 114)  ( 15)( 135)  ( 12)( 114)  ( 12)( 135)  170  1175  128  170  125  7  14  7



84 17 14

10 17 7



20 17 14

505

Chapter 11

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38. P  2/  2w  2(817  415)  2(517  315)  1617  815  1017  615  (1617  1017)  (815  615)  (16  10) 17  (8  6) 15  2617  215

45. V  12gd from 25 feet: V  12(32)(25)  11600  40 ft/s from 100 feet: V  12(32)(100)  16400  80 ft/s 46. The velocity doubled. 47. The velocity should be 19 or 3 times the velocity of an object falling 25 feet; 3  40  120 ft/s, 12(32)(225)  120 ft/s. 48. 10201P (1  0.01 1P)  1020155 (1  0.01 155)  1020(7.42) [1  0.01(7.42) ]  7564.52[ 1  0.074]  7564.52[ 0.93]  7003.5 The pumping station must supply about 7003.5 gal/min. 49. Sample answer: a  4, b  9; 14  9  14  19 50. a  0 or b  0 or both 51. The distance a person can see is related to the

The perimeter is 2617  215 in. 39. P  2/  2w 213  4111  6  2(2111  1)  2w 213  4111  6  4111  2  2w 213  4  2w 13  2  w The width is 13  2 cm. 1

40. A  2 d d 1 2 1

 2 (3 16)(5 14) 1

 2 (3 16) (5) (2)  1516 The area is 15 16 cm2. 41. d 

3h

32

Sears Tower: d 

3

3(1450) 2



3

4350 2

height of the person using d 

500

should include the following. • You can find how far each lifeguard can see from the height of the lifeguard tower. Each tower should have some overlap to cover the entire beach area. • On early ships, a lookout position (crow’s nest) was situated high on the foremast. Sailors could see farther from this position than from the ship’s deck. 52. C; 917  2128  917  214  7  917  2222  7  917  2(217)  917  417  (9  4) 17  517 53. D; 13(4  112) 2  13(42  2(4)( 112)  ( 112) 2 )  13(16  8112  12)  13(28  8112)  ( 13)(28)  ( 13) (8112)  2813  8136  2813  8(6)  2813  48

100 

Page 597

 12175  125  87  252  87  5187 Empire State Building: d 

3

3(1250) 2



3

3750 2

 11875  1625  3  2252  3  2513 5287  25 23  5(9.33)  25(1.73)  46.65  43.30  3.34 mi A person can see about 3.34 mi farther. 42. Approximately 1000 feet; Solve 3(1250) 2

3h

 3 2  4.57; may use guess and test, graphical, or analytical methods.

3

F

43. r 

3 5

r

3 5



3



1100 1 10 1



3h

3 2 . Answers

 5.64  6 in. The radius of the pipe is about 6 in. 44. No, each pipe would need to carry 500 gallons per minute, so the pipes would need a radius greater than 5.6 in. Chapter 11

Maintain Your Skills

54. 140  14  10  222  110  2110

55. 1128  164  2  282  12  812

56. 2196x2y3  2142  x2  y2  y  2142  2x2  2y2  1y  14|x|y1y

506

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57.

150 18



150 18



1400 8 20 8 5 2

 

18

 18

3

58.

225c4d 18c2



2225c4d 118c2



1225  2c4  1d 118  1c2

    

59.

3

63a 128a3b2

3



163 1128a2b2



19  7 164  2  a2  b2



232  17 182  12  1a2  1b2

15 0 c2 0 1d 19  2  c

1

15 0 c2 0 1d 3c 12 5 0 c 0 1d 12 5 0 c 0 1d 12



5 0 c 0 12d 2

65.

12 12

 4(4) 5  4(1024)  4096

?

48(0) 3  75(0)  0 000✓ or

4

or

154 23  751 54 2  0 125 5 ? 48 1 64 2  75 1 4 2  0 3 125 5 ? 48 1 64 2  75 1 4 2  0 4

375 4

2

63. 81  49y 0  49y2  81 0  (7y  9)(7y  9) 7y  9  0 or 7y  9  0 7y  9 7y  9 9

66.

y  7

197 22 ? 81 81  49 1 49 2 ?

81  49

or

1 2 81 81  49 1 49 2

? 9 81  49 7 2

1

?

81 49  49

81 

1

Check:

1

81 49  49 1

81  81 ✓

81  81 ✓

36

q2  121  0

6

q  11

5 116 6

or

or

6

q  11  0 q

6 11

507

375 4

0✓

5x  80x  240  15x2 5x  15x  80x  240  0 (5x3  15x2 )  (80x  240)  0 5x2 (x  3)  80(x  3)  0 (x  3) (5x2  80)  0 or 5x2  80  0 x30 5x2  80 x  3 or x2  16 x  4

54,3, 46

?



3

3

9

y7

6

?

48

 2(0.8) 7  2(0.2097152)  0.4194304

q  11  0

5

n  4

1 2 ? 48 1 2  75154 2  0 3 125 5 ? 48 1  64 2  75 1 4 2  0 375 375  4  1 4 2  0 ✓

 7(93 )  7(729)  5103

1q  116 21q  116 2  0

4n  5  0 4n  5

or

5

62. an  a1 (rn1 ) a8  2(0.8) 81

64.

36

 121  0 ✓

125  64

61. a  a (rn1 ) n 1 a4  7(9) 41

81 

36 121

1 2

3 114

?

?

36

 121  0

5 5 ? 48 4 3  75 4  0

12

 16|ab|

Check:

2

n4

Check:

 8|ab| 12  12

5 97 6

or

6 11 2

554, 0, 54 6

3 17

60. a  a (rn1 ) n 1 a6  4(4) 61

36

 121  0 ✓

48n3  75n  0 3n(16n2  25)  0 3n(4n  5) (4n  5)  0 3n  0 or 4n  5  0 n0 4n  5

 8|ab| 12 3 17

?

36 121

63 128a2b2



36 0 1116 22  121

Check:

2

?

5(4) 3  80(4)  240  15(4) 2 ? 320  (320)  240  (240) 00✓ or ? 5(3) 3  80(3)  240  15(3) 2 ? 135  (240)  240  (135) 105  105 ✓ or ? 5(4) 3  80(4)  240  15(4) 2 ? 320  320  240  240 00✓

Chapter 11

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67. 8n 5 8n 8

3.

5 8 5 8

n

5

4

8

158 2 5 ?

8

5 5✓

68.

6 14

?

8

45

5

5 8

168 2 5 ?

6 5✓

6.

6 14  9

w 6 126 Let w  126 Check: 126 9

Let w  125

?

125 9

6 14

14 14

69.

Let n  8 .

148 2 5

The solution set is n|n w 9 w 9 9

6

Let n  8 .

Let n  8 .

Check:

7k 2 2 7k  7 2

7 7

k 7 Check:

Let w  127

?

127 9

6 14

13.88 6 14 ✓

Page 600

?

6 14

14.11 14

Let k 

Let k 

3

75

? 21 10



75 2 3

? 21 5

75 7 21 5



21

 10

21 5

2 1



75

4

? 2 1

7

2 2

21

 10

2 1

? 21 5

75 7 14 5

? 21 10

7

2

2

? 2 1

7

3. Sample answer: 1x  1  8 ( 1x  1) 2  82 x  1  64 x  63 4. Alex; the square of 1x  5 is x  5.

4 . 5

4

75

? 21 10

7

2

3

2 1

Let k 

2

75

7

2

2 . 5



21 5



75

? 2 1

7

2

21

 10

7

21 5

71. (x  2) 2  (x 2 )  2(x)(2)  (2) 2  x2  4x  4 2 72. (x  5)  (x) 2  2(x)(5)  (5) 2  x2  10x  25 2 73. (x  6)  (x) 2  2(x)(6)  (6) 2  x2  12x  36 2  (3x) 2  2(3x)(1)  (1) 2 74. (3x  1)  9x2  6x  1 75. (2x  3) 2  (2x) 2  2(2x)(3)  (3) 2  4x2  12x  9 2  (4x) 2  2(4x)(7)  (7) 2 76. (4x  7)  16x2  56x  49

12b  8 ( 12b) 2  (8) 2 2b  64

12  32  8 ? 164  8 8  8 ✕



Check:

?

64 2

b  32 Since 32 does not satisfy the original equation, there is no solution. 7.

17x  7 ( 17x) 2  72 7x  49 x7 The solution is 7.

8.

13a  6 Check: ( 13a) 2  (6) 2 3a  36 a  12 The solution is 12.

9. 18s  1  5 18s  4 ( 18s ) 2  (4) 2 8s  16 s2 The solution is 2.

Graphing Calculator Investigation

1.

2. (10, 5)

Chapter 11

6.

2b 2

Radical Equations

Page 600

125  5 55✓



70. There are 6  6  6  216 possible outcomes in the sample space. Only one outcome has each roll a 1. 1 Therefore, the probability is 216.

11-3

1x  5 Check: ( 1x) 2  52 x  25 The solution is 25.

75 7 28 5

?

5.

? 21 5

4

Check for Understanding

1. Isolate the radical on one side of the equation. Square each side of the equation and simplify. Then check for extraneous solutions. 2. The solution may not satisfy the original equation.

21 10 2 21  7 10 3 5 3 . 5

13x  5  x  5 ( 13x  5) 2  (x  5) 2 3x  5  (x) 2  2(x) (5)  (5) 2 3x  5  x2  10x  25 0  x2  13x  30 0  (x  3) (x  10) x  3  0 or x  10  0 x  3 or x  10 x  10; the solution is the same as the solution from the graph. However, when solving algebraically, you have to check that x  3 is an extraneous solution.

508

Check:

Check:

?

17  7  7 ? 149  7 77✓

?

1(3) (12)  6 ? 236  6 66✓

?

18(2)  1  5 ? 116  1  5 ? 415 55✓

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Check:

s  3.11d 240  3.11d 77.42  1d (77.42) 2  ( 1d ) 2 5,993.76  d The depth of the water is about 5994 m. 16. s  3.11d 10,000 m: s  3.1110,000  3.1(100)  310 20 m: s  3.1120  13.9 310  13.9  296 The wave loses about 296 m/s.

?

17x  18  9 Check: ( 17x  18) 2  (9) 2 7x  18  81 7x  63 x9 The solution is 9. 11. 15x  1  2  6 15x  1  4 ( 15x  1) 2  (4) 2 5x  1  16 5x  15 x3

15.

17(9)  18  9 ? 163  18  9 ? 281  9 99✓

10.

?

15(3)  1  2  6 ?

115  1  2  6 ?

116  2  6

Pages 601–603

?

426 66✓ 12.

17.

The solution is 3. 16x  8  x  4 ( 16x  8) 2  (x  4) 2 6x  8  (x) 2  2(x) (4)  (4) 2 6x  8  x2  8x  16 0  x2  14x  24 0  (x  12)(x  2) x  12  0 or x  2  0 x  12 x  2 Check: 16x  8  x  4

?

1100  10 10  10 ✓ 1k  4 18. ( 1k) 2  (4) 2 k  16 k  16 Check:

Check:

?

1(16)  4 ?

116  4

16x  8  x  4

?

44✓

16(2)  8  2  4

16(12)  8  12  4 ?

172  8  12  4

19.

112  8  6

?

164  8 220  6 ✕ 88✓ Since 2 does not satisfy the original equation, 12 is the only solution. 13. 4  1x  2  x 1x  2  x  4 ( 1x  2) 2  (x  4) 2 x  2  x2  2(x)(4)  (4) 2 x  2  x2  8x  16 0  x2  9x  18 0  (x  3)(x  6) x  3  0 or x  6  0 x  3 or x6 Check:

Practice and Apply

1a  10 # ( 1a) 2  102 a  100

?

4  13  2  3

4  16  2  6

4  11  3

4  14  6

?

512  1x (512) 2  ( 1x) 2 25(2)  x 50  x Check:

?

512  150 ?

512  125  2 512  512 ✓ 317  1y 20. (317 ) 2  ( 1y ) 2 9(7)  y 63  y 63  y Check:

?

317  1(63) ?

317  163 ?

317  19  7 317  317 ✓ 21. 314a  2  10 314a  12 14a  4 ( 14a ) 2  (4) 2 4a  16 a4

?

426 413 53✕ 66✓ Since 3 does not satisfy the original equation, 6 is the only solution. 14. s  3.11d s  3.1110  3.1(3.16)  9.8 The speed of the tsunami is about 9.8 m/s.

Check:

?

314(4)  2  10 ?

3116  2  10 ?

3(4)  2  10 ?

12  2  10 10  10 ✓

509

Chapter 11

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22. 3  51n  18 51n  15 1n  3 ( 1n ) 2  (3) 2 n9 Check:

?

18  1  3  0 ?

19  3  0 ?

330 00✓

?

3  519  18

28. 13r  5  7  3 13r  5  4 ( 13r  5) 2  (4) 2 3r  5  16 3r  21 r7

?

3  5(3)  18 ?

3  15  18 18  18 ✓ 23. 1x  3  5 ( 1x  3 ) 2  (5) 2 x  3  25 x  22 Check:

?

14(2)  1  3  0

Check:

?

13(7)  5  7  3

Check:

?

?

121  5  7  3

?

116  7  3

122  3  5

?

125  5 5  5 ✕

?

473 11  3 ✕

No solution 24. 1x  5  2 16 ( 1x  5 ) 2  (216) 2 x  5  4(6) x  5  24 x  29 Check:

No solution 29.

4x

35

93 4x

1 3 4x5 22  (12)2

?

129  5  216

4x 5

?

124  216

93

720 5

93

? ?

1144  9  3 ? 12  9  3 33✓

?

13(5)  12  313

4t

30. 5 3 3  2  0 4t

53 3  2

?

127  313 ?

19  3  313 313  313 ✓

4t

33

2

5

1 3 4t3 22  125 22

12c  4  8 ( 12c  4) 2  (8) 2 2c  4  64 2c  68 c  34

4t 3

4

 25 12

4t  25 3

t  25

?

12(34)  4  8 ?

168  4  8

Check:

?

164  8 88✓

3 425 ? 20 5B 3 12 25

?

4

?

5B  2  0 3

27. 14b  1  3  0 14b  1  3 ( 14b  1) 2  (3) 2 4b  1  9 4b  8 b2

Chapter 11

?

4(180) 5

3

?

Check:

3

Check:

115  12  313

26.

 144

4x  720 x  180

?

14  6  216 216  216 ✓ 25. 13x  12  313 ( 13x  12 ) 2  (313) 2 3x  12  9(3) 3x  12  27 3x  15 x5 Check:

 12

35

5 3 25  2  0 5

125 2  2  0 ? ?

220 00✓

510

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31.

2x2  9x  14  x  4 ( 2x2  9x  14 ) 2  (x  4) 2 x2  9x  14  x2  2(x)(4)  42 x2  9x  14  x2  8x  16 9x  14  8x  16 x  14  16 x2 Check:

36.

?

222  9(2)  14  2  4

Check:

?

y  2  2y2  5y  4 (y  2) 2  ( 2y2  5y  4 ) 2 2 y  2(y)(2)  22  y2  5y  4 y2  4y  4  y2  5y  4 4y  4  5y  4 y  4  4 y  0 y0

Check:

?

0  2  202  5(0)  4 2  10  0  4

38.

?

157  7  8 ?

 81

9

181  9 99✓ x  16  x 35. (x) 2  ( 16  x) 2 x2  6  x 2 x x60 (x  3)(x  2)  0 or x  2  0 x30 x  3 or x2 ? ?

128  3(4)  4

?

128  12  4

?

x  486

3  16  (3)

?

19  3 33✓

116  4 149  7 7  7 ✕ 44✓ Since 7 does not satisfy the original equation, 4 is the only solution. 39. 1x  1  x  1 ( 1x  1 ) 2  (x  1) 2 x  1  x2  2(x) (1)  (1) 2 x  1  x2  2x  1 0  x2  3x 0  x(x  3) x  0 or x  3  0 x3 ? ? Check: 10  1  0  1 13  1  3  1

1 3 6x 22  92

Check:

?

?

128  3(7)  7 128  21  7

x

486 ?  6 ?

115  6  3

128  3x  x ( 128  3x ) 2  (x) 2 28  3x  x2 0  x2  3x  28 0  (x  7) (x  4) x  7  0 or x  4  0 x  7 x4 Check:

36  9

3

?

?

164  8 88✓ 34. Let x  the number.

Check:

15(3)  6  3

?

14  2 22✓

?

2  14 22✓ 33. Let x  the number. 1x  7  8 ( 1x  7 ) 2  82 x  7  64 x  57

x 6

?

15(2)  6  2 110  6  2

?

Check:

?

4  14  20

4  115 ✕ 5  125 55✓ Since 4 does not satisfy the original equation, 5 is the only solution. 15x  6  x 37. ( 15x  6 ) 2  (x) 2 5x  6  x2 0  x2  5x  6 0  (x  2) (x  3) x  2  0 or x  3  0 x3 x  2 or

136  6 66✓

Check:

?

5  15  20 ?

?

14  18  14  6

32.

x  1x  20 (x) 2  ( 1x  20) 2 x2  x  20 2  x  20  0 x (x  5)(x  4)  0 x  5  0 or x  4  0 x5 x  4

?

?

11  1 14  2 1  1 ✕ 22✓ Since 0 does not satisfy the original equation, 3 is the only solution.

?

2  16  2 ?

2  14 3  19 3  3 ✕ 22✓ Since 3 does not satisfy the original equation, 2 is the only solution.

511

Chapter 11

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40.

11  2b  1  b ( 11  2b ) 2  (1  b) 2 1  2b  (1) 2  2(1)(b)  (b) 2 1  2b  1  2b  b2 0  b2  4b 0  b(b  4) b  0 or b  4  0 b  4 Check:

Check:

?

11  8  3

? ?

19  3 11  1 11✓ 3  3 ✕ Since 4 does not satisfy the original equation, 0 is the only solution. 41. 4  1m  2  m 1m  2  m  4 ( 1m  2 ) 2  (m  4) 2 m  2  m2  2(m)(4)  (4) 2 m  2  m2  8m  16 0  m2  9m  18 0  (m  3)(m  6) m  3  0 or m  6  0 m3 m6 Check:

?

4  16  2  6

?

4  14  6

4  13  2  3 4  11  3

?

16  3(25)  25  16 ?

16  75  9

?

19  8  1 ?

?

16  3(10)  10  16 ?

16  30  6 ?

136  6 66✓

Since 25 does not satisfy the original equation, 10 is the only solution. 45.

? ?

22r2  121  r ( 22r2  121 ) 2  r2 2r2  121  r2 r2  121  0 (r  11)(r  11)  0 or r  11  0 r  11  0 r  11 or r  11 Check: ?

22(11) 2  121  11

?

12(121)  121  11

22(112 )  121  11 12(121)  121  11

? ?

?

?

1121  11 1121  11 11  11 ✓ 11  11 ✕ Since 11 does not satisfy the original equation, 11 is the only solution. 46.

?

13(4)  8  4  2 ?

112  8  2 ?

14  2 11  1 11✓ 22✓ 43. x  16  x  4 16  x  4  x ( 16  x ) 2  (4  x) 2 6  x  42  2(4)(x)  (x) 2 6  x  16  8x  x2 0  x2  7x  10 0  (x  2)(x  5) x  2  0 or x  5  0 x5 x2

Chapter 11

?

181  9 9  9 ✕

?

?

?

?

?

426 413 53✕ 66✓ Since 3 does not satisfy the original equation, 6 is the only solution. 42. 13d  8  d  2 ( 13d  8 ) 2  (d  2) 2 3d  8  d2  2(d)(2)  (2) 2 3d  8  d2  4d  4 0  d2  7d  12 0  (d  3)(d  4) d  3  0 or d  4  0 d3 d4 Check: 13(3)  8  3  2

5  11  4

514 224 44✓ 64✕ Since 5 does not satisfy the original equation, 2 is the only solution. 44. 16  3x  x  16 ( 16  3x ) 2  (x  16) 2 6  3x  x2  2(x)(16)  162 6  3x  x2  32x  256 0  x2  35x  250 0  (x  25)(x  10) or x  10  0 x  25  0 x  25 x  10 Check:

?

?

?

?

11  2(4)  1  4

11  0  1

5  16  5  4

?

2  14  4

?

11  2(0)  1  0

?

2  16  2  4

25p2  7  2p ( 25p2  7 ) 2  (2p) 2 5p2  7  4p2 p2  7  0 p2  7 p  17 Check: ?

25( 17) 2  7  2 17

?

25(17) 2  7  2 (17)

?

15(7)  7  217

?

135  7  217

15(7)  7  2 17 135  7  2 17 ?

128  2 17 2 17  217 ✓

?

? ?

128  217 217  217 ✕

Since 17 does not satisfy the original equation, 17 is the only solution. 47.

512

2(x  5) 2  x  5 ( 2(x  5) 2 ) 2  (x  5) 2 (x  5) 2  (x  5) 2 Sometimes; only numbers that make (x  5) non-negative.

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L  1kP 42  10.1669P (42) 2  ( 10.1669P ) 2 1764  0.1669P 10,569  P The maximum take off weight is 10,569 lb. 49. L  1kP 232  1k(870,000) (232) 2  ( 1870,000k ) 2 53,824  870,000k 0.0619  k 50. A  r2 48.

3

A 

 r2

A 

 2r2

56. P  2 3 32 /

t  2 3 32 /

t 2

t2 42

r

52. r 

3

96 

r

32t2 2

r  196 r  116  6 r  416 or about 9.8 m

A 48 

r  148 r  116  3 r  413 or about 6.9 m

53. It increases by a factor of 12. 54.

/

1 3

 2 3 32 /



3 32 /

131 22  1 3 32/ 22 1 92 32 92

 32 /

/

/  0.36 ft 55. P  2 3 32 /

1  2 3 32 /

121 22  1 3 32/ 22 1 42

 32

32 42

/

/

/  0.81 ft

2  2 3 32 /

2 2



1 2 32 2

 32

2t 2



t2 2

 32

3 32 /

1t 22  1 3 32/ 22 /

/

32t2 2

8t2 2

8t2 2



24t2 2

58.

V  201t  273 340  201t  273 17  1t  273 (17) 2  ( 1t  273 ) 2 289  t  273 t  16C

1h  9  1h  13 ( 1h  9  1h ) 2  ( 13 ) 2 ( 1h  9 ) 2  2( 1h  9 ) (1h )  (1h ) 2  3 h  9  21h(h  9)  h  3 2h  9  2 1h(h  9)  3 21h(h  9)  6  2h 1h(h  9)  3  h ( 1h(h  9) ) 2  (3  h) 2 h(h  9)  32  2(3) (h)  (h) 2 h2  9h  9  6h  h2 3h  9 h3 Check: 13  9  13  13 112  13  13 213  13  13 13  13 ✓ 61. You can determine the time it takes an object to fall from a given height using a radical equation. Answers should include the following. • It would take a skydiver approximately 42 seconds to fall 10,000 feet. Using the equation, it would take 25 seconds. The time is different in the two calculations because air resistance slows the skydiver. • A skydiver can increase the speed of his fall by lowering air resistance. This can be done by pulling his arms and legs close to his body. A skydiver can decrease his speed by holding his arms and legs out, which increases the air resistance. 62. A; 1x  3  6 1y  3  6 ( 1x  3) 2  (6) 2 1y  3 ( 1y ) 2  (3) 2 x  3  36 x  33 y9

P  2 3 32 2 3



32t2 42

/

V  201t  273 356  201t  273 17.8  1t  273 (17.8) 2  ( 1t  273 ) 2 316.84  t  273 t  43.84C 59. V  201t  273 V  2010  273  201273  330.45 V 6 330.45 m/s 60.

3 3

/



A

A

/

 32

/

57.

3

3 32

12t 22  1 3 32/ 22

3  r 51. r 



2t  2 3 32

3 32 /

11 22  1 3 32/ 22 /

/

/  3.24 ft 3.24  0.81  2.43 ft longer

513

Chapter 11

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76. d2 50d  225 No; 50d  2(d)(15) 77. 4n2  28n  49 Yes; 28n  2(2n)(7) (2n 7)2 78. 16b2  56bc  49c2 Yes; 56bc  2(4b)(7c) (4b  7c)2 79. (r  3)(r  4)  r(r)  r(4)  3(r)  3(4)  r2  4r  3r  12  r2  r  12 80. (3z  7)(2z  10)  3z(2z)  3z(10)  7(2z)  7(10)  6z2  30z  14z  70  6z2  44z  70

2 2(a  1) 2 63. C; ( 1a  1) a1 a1 64. Clear the Y  list. Enter the left side of the equation as Y1. Enter the right side of the equation as Y2. Use the intersect feature on the CALC menu to find the x value of the point of intersection is 8. 65. Clear the Y  list. Enter the left side of the equation as Y1. Enter the right side of the equation as Y2. Use the intersect feature on the CALC menu to find the x value of the point of intersection is 11. 66. Clear the Y  list. Enter the left side of the equation as Y1. Enter the right side of the equation as Y2. Use the intersect feature on the CALC menu to find the x value of the point of intersection is 2. 67. Clear the Y  list. Enter the left side of the equation as Y1. Enter the right side of the equation as Y2. Use the intersect feature on the CALC menu to find the x value of the point of intersection is 15.08. 68. Clear the Y  list. Enter the left side of the equation as Y1. Enter the right side of the equation as Y2. Use the intersect feature on the CALC menu to find the x value of the point of intersection is 1.70. 69. Clear the Y  list. Enter the left side of the equation as Y1. Enter the right side of the equation as Y2. Use the intersect feature on the CALC menu, or the fact that the graph has no points of intersection, to determine that the equation has no solution.

Page 603

81.

 2p(3p2 )  2p(4p)  2p(9)  5(3p2 )  5(4p)  5(9)  6p3  8p2  18p  15p2  20p  45  6p3  7p2  2p  45 9

82. F  5C  32 9

83.

Maintain Your Skills

21 110  13



21( 110  13) ( 110) 2  ( 13) 2



21( 110  13) 10  3



21( 110  13) 7

 110 

3

y  2x  7

89. 2a2  b2  282  122  164  144  1208  116  13  4113

 3( 110  13 )

Chapter 11

 72  32  104F

88. 2a2  b2  212  12  11  1  12

110  13 13

21 110  13

 63  32  95F 95  F  104

87. 2a2  b2  2(24) 2  (7) 2  1576  49  1625  25

74. 16  110  160  14  15  2115



F  5 (40)  32

86. 2a2  b2  2(3) 2  (4) 2  19  16  125 5

 3 12  512  3(412 )  3 12  512  1212  (3  5  12) 12  4 12

75.

9

F  5 (35)  32

7y  14x  3 14x  7y  3 14x  7y  3 84. y  3  2(x  6) y  3  2x  12 y  2x  15 2x  y  15 85. y  2  7.5(x  3) y  2  7.5x  22.5 y  7.5x  24.5 7.5x  y  24.5 15x  2y  49

70. 516  1216  (5  12) 16  1716 71. 112  6127  14  3  619  3  213  6(313 )  213  1813  2013 72. 118  512  3132  19  2  512  3116  2

73. 1192  164  3  813

(2p  5)(3p2  4p  9)

514

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Page 603

Practice Quiz 1

1. 148  116  3  413 3.

3 2  110

2

3  110

10.

2. 13  16  118  19  2  312 2  110  110

2



3(2  110) (2) 2  ( 110) 2



3(2  110) 4  10



3(2  110) 6



2  110 2

5

x2 Check:

5

?

5

? ?

14  2 2  2 ✕

? ?

110  1  10  7 ?

19  3 33✓

5

Since 2 does not satisfy the original equation, 5 is the only solution.

Page 604

Graphing Calculator Investigation (Follow-Up of Lesson 11-3)

1. y  1x  1

[10, 10] scl: 1 by [10, 10] scl: 1

The domain is {x|x 0}; the graph is shifted up 1 unit. 2. y  1x  3

23x2  32  x ( 23x2  32) 2  x2 3x2  32  x2 2x2  32  0 2x2  32 x2  16 x  4 Check:

3 2 1 2 2  1  2 1 2 2  7 12(5)  1  2(5)  7 15  1  5  7

4. 615  3111  515  615  515  3111  (6  5) 15  3111  1115  3111 5. 213  9112  213  9( 14.3)  213  9(213)  213  1813  (2  18) 13  2013 2 6. (3  16 )  (3) 2  2(3)(16 )  (16 ) 2  9  616  6  15  616 7. A  s2  (2  17) 2  22  2(2)( 17)  ( 17) 2  4  417  7  11  417 or 21.6 cm2 115  x  4 8. ( 115  x) 2  42 15  x  16 x  1 x  1 115  (1)  4 Check: 116  4 4  4✓ 9.

12x  1  2x  7 ( 12x  1) 2  (2x  7) 2 (2x  1)  (2x) 2  2(2x) (7)  (7) 2 2x  1  4x2  28x  49 0  4x2  30x  50 0  2(2x2  15x  25) 0  2(2x  5) (x  5) 2x  5  0 or x  5  0 x5 2x  5

[10, 10] scl: 1 by [10, 10] scl: 1

The domain is {x|x 0} ; the graph is shifted down 3 units. 3. y  1x  2 ?

23(4) 2  32  4

?

13(16)  32  4

?

148  32  4

23(4) 2  32  4 13(16)  32  4 148  32  4 ?

116  4 44✓

? ? ? ?

116  4 4  4 ✕

Since 4 does not satisfy the original equation, 4 is the only solution.

[10, 10] scl: 1 by [10, 10] scl: 1

The domain is {x|x 2} ; the graph is shifted left 2 units.

515

Chapter 11

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4. y  1x  5

9. y  12x  5  4

[5, 15] scl: 1 by [10, 10] scl: 1

[10, 10] scl: 1 by [10, 10] scl: 1

The domain is {x|x 5} ; the graph is shifted right 5 units.

The domain is {x|x 2.5}; the graph is shifted left 2.5 units, down 4 units. 10. No, you must consider the graph of y  1x and the graph of y  1x. This graph fails the vertical line test. For every value of x 7 0, there are two values for y.

5. y  1x

11. No; the equation y  21  x2 is not a function since there are both positive and negative values for y for each value of x. 12. Enter y  |x|  21  x2 as Y1 and y  |x|  21  x2 as Y2.

[10, 10] scl: 1 by [10, 10] scl: 1

The domain is {x|x  0} ; the graph is reflected across y-axis.

1 X,T,␪,n

KEYSTROKES:

6. y  13x

1 X,T,␪,n

ENTER [1 ] 1

)

X,T,␪,n

2nd [ 1 ] 1

X,T,␪,n

)

)

)

2nd

GRAPH

The graph is shaped like a heart.

11-4 The Pythagorian Theorem

[10, 10] scl: 1 by [10, 10] scl: 1

The domain is {x|x 0} ; the graph is expanded.

Pages 607–608

7. y  1x

Check for Understanding

1. hypotenuse leg

leg

2. Compare the lengths of the sides. The hypotenuse is the longest side, which is always the side opposite the right angle. 3. s2  s2  d2 2s2  d2 22s2  2d2 22s2  d or d  s 12 2 4. c  a2  b2 c2  142  122 c2  196  144 c2  340 c  1340 c  18.44 The length of the hypotenuse is 18.44 units. c2  a2  b2 5. 412  a2  402 1681  a2  1600 81  a2

181  a 9a The length of the leg is 9 units.

[10, 10] scl: 1 by [10, 10] scl: 1

The domain is {x|x 0} ; the graph is reflected across x-axis. 8. y  11  x  6

[10, 10] scl: 1 by [5, 15] scl: 1

The domain is {x|x  1} ; the graph is reflected across y-axis, shifted right 1 unit, up 6 units.

Chapter 11

516

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6. c2  a2  b2 c2  102  242 c2  100  576 c2  676 c2  1676 c  26 The length of the hypotenuse is 26 units. c2  a2  b2 7. 612  112  b2 3721  121  b2 3600  b2

13600  b 60  b The length of the leg is 60 units. c2  a2  b2 8. ( 1233) 2  a2  132 233  a2  169 64  a2

164  a 8a The length of the leg is 8 units. 9. c2  a2  b2 c2  72  42 c2  49  16 c2  65 c  165 c  8.06 The length of the hypotenuse is about 8.06 units. 10. No; 42  62  92. 11. Yes; 162  302  342. 12. A; c2  a2  b2 82  a2  62 64  a2  36 28  a2

128  a 217  a The length of the leg is 217 units. A A

14. c2  a2  b2 c2  72  92 c2  49  81 c2  130 c  1130 c  11.40 The length of the hypotenuse is about 11.40 units. 15. c2  a2  b2 c2  282  452 c2  784  2025 c2  2809 c  12809 c  53 The length of the hypotenuse is 53 units. c2  a2  b2 16. 142  52  b2 196  25  b2 171  b2

1171  b 13.08  b The length of the leg is about 13.08 units. c2  a2  b2 17. 1802  a2  1752 32,400  a2  30,625 1775  a2

11775  a 42.13  a The length of the leg is about 42.13 units. 18. c2  a2  b2 1012  992  b2 10,201  9,801  b2 400  b2

1400  b 20  b The length of the leg is 20 units. 19. c2  a2  b2 c2  162  632 c2  256  3,969 c2  4,225 c  14225 c  65 The length of the hypotenuse is about 65 units. 20. c2  a2  b2 342  162  b2 1156  256  b2 900  b2

1900  b 30  b The length of the leg is 30 units. 21. c2  a2  b2 c2  ( 1112 ) 2  32 c2  112  9 c2  121 c  1121 c  11 The length of the hypotenuse is 11 units.

bh 2 6  2 17 2

 617 units2

Pages 608–610 13.

Practice and Apply

c2  a2  b2 152  a2  52 225  a2  25 200  a2

1200  a 14.14  a The length of the leg is about 14.14 units.

517

Chapter 11

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22. c2  a2  b2 c2  ( 115) 2  ( 110) 2 c2  15  10 c2  25 c  125 c5 The length of the hypotenuse is 5 units. c2  a2  b2 23. 142  92  b2 196  81  b2 115  b2

1115  b 10.72  b The length of the leg is about 10.72 units. 24. c2  a2  b2 c2  62  32 c2  36  9 c2  45 c  145 c  6.71 The length of the hypotenuse is about 6.71 units. c2  a2  b2 25. 122  a2  ( 177) 2 144  a2  77 67  a2

167  a 8.19  a The length of the leg is about 8.19 units. 26. c2  a2  b2 c2  42  ( 111) 2 c2  16  11 c2  27 c  127 c  5.20 The length of the hypotenuse is about 5.20 units. 27. c2  a2  b2 c2  ( 1225) 2  ( 128) 2 c2  225  28 c2  253 c  1253 c  15.91 The length of the hypotenuse is about 15.91 units. 28. c2  a2  b2 ( 1155) 2  ( 131) 2  b2 155  31  b2 124  b2

1124  b 11.14  b The length of the leg is about 11.14 units. 29. c2  a2  b2 c2  (8x) 2  (15x) 2 c2  64x2  225x2 c2  289x2 c  2289x2 c  17x The length of the hypotenuse is about 17x units.

Chapter 11

30.

31. 32. 33. 34. 35. 36. 37.

38.

39.

518

c2  a2  b2 (7x) 2  a2  (3x) 2 49x2  a2  9x2 40x2  a2

240x2  a

140x  a 6.32x  a The length of the leg is about 6.32x units. Yes; 302  402  502. No; 62  122  182. No; 242  302  362. Yes; 452  602  752. Yes; 152  ( 131) 2  162. Yes; 42  72  ( 165) 2. Use the formula for the area of a square to find the length of a side. A  s2 162  s2

1162  s 1162  s Use the Pythagorean Theorem to find the diagonal. d 2  s2  s2 d2  ( 1162) 2  ( 1162) 2 d2  162  162 d2  324 d  1324 d  18 The length of the diagonal is 18 ft. Let x  length of one leg. x  5  length of second leg. c2  a2  b2 252  x2  (x  5) 2 625  x2  x2  2(x) (5)  52 625  2x2  10x  25 0  2x2  10x  600 0  2(x2  5x  300) 0  2(x  20)(x  15) x  20  0 or x  15  0 x  20 ✕ x  15 The first leg is 15 cm; the second leg is 20 cm. The diagonal of the cube is the hypotenuse of a triangle with one leg being the diagonal of a face and the other leg being a side. Diagonal of face: d2  s2  s2 d2  42  42 d2  16  16 d2  32 d  132 Diagonal of cube: d2  ( 132) 2  42 d2  32  16 d2  48 d  148 d  116.3 d  413 or 6.93 in.

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40. Let x the length of the shorter leg. 144 x

41.

42.

43.

44. 45.

46.

47.



49. Use the Pythagorean Theorem to calculate the length of the vertical leg of the right triangle whose hypotenuse is 180 ft and horizontal leg is 130 ft. c2  a2  b2 1802  a2  1302 32400  a2  16900 15500  a2

115500  a 124.5  a The maximum height reached is the sum of the three vertical sections. 80 ft  124.5 ft  120 ft  324.5 ft This value of 324.5 ft is the vertical leg of the right triangle whose hypotenuse is 381.2 ft. The Pythagorean Theorem can be used to determine the horizontal leg is about 200 ft. The plateau at the top can be calculated by considering the corresponding segment of equal length at the bottom. 750 ft  (100 ft  130 ft  50 ft  200 ft)  270 ft The Pythagorean Theorem can be used to show that the right triangle with legs of 50 ft and 80 ft has a hypotenuse of about 94.3 ft. The Pythagorean Theorem can also be used to show that the right triangle with legs of 100 ft and 120 ft has a hypotenuse of about 156.2 ft. The total distance traveled is about 381.2 ft  270 ft  94.3 ft  180 ft  156.2 ft  1081.7 ft The maximum height is about 324.5 ft. 50. Engineers can use the Pythagorean Theorem to find the total length of the track, which determines how much material and land area they need to build the attraction. Answers should include the following. • A tall hill requires more track length both going uphill and downhill, which will add to the total length of the tracks. Tall, steep hills will increase the speed of the roller coaster. So a coaster with a tall, steep first hill will have more speed and a longer track length. • The steepness of the hill and speed are limited for safety and to keep the cars on the track. 51. C; Use the Pythagorean Theorem to find x. c2  a2  b2 152  x2  (2x) 2 225  x2  4x2 225  5x2 45  x2 145  x Find the area of the triangle with b  145 and h  2145.

8 5

8x  720 x  90 The shorter leg is 90 m. c2  a2  b2 1442  902  b2 20,736  8100  b2 12,636  b2

112,636  b 112.41  b The longer leg is about 112.41 m. c2  a2  b2 c2  2082  3602 c2  43,264  129,600 c2  172,864 c  1172,864 c  415.8 It will travel about 415.8 ft. c2  a2  b2 c2  442  2082 c2  1936  43,264 c2  45,200 c  145,200 c  212.6 It will travel about 212.6 ft. The roller coaster makes a total horizontal advance of 404 feet, reaches a vertical height of 208 feet, and travels a total track length of about 628.4 feet. See students’ work. c2  a2  b2 c2  1002  602 c2  10,000  3600 c2  13,600 c  113,600 c  116.6 The longest edge is about 116.6 ft. c2  a2  b2 c2  52  122 c2  25  144 c2  169 c  1169 c  13 The missing length is 13 ft. The garage roof is made up of two 30 by 15 rectangles. The 15 ft dimension was obtained by adding the 2 foot overhang to the 13 ft calculated in the previous problem. A  /w A  30  15 A  450 ft2 The total area is 900 ft2.

48. The area of the largest semicircle is The sum of the other two areas is

 4

c2 4



bh 2 ( 145) (2 145) 2

 4 c2.



2

 ( 145 ) ( 145 )  45 units2

1a2  b22.

2

A

2

Using the Pythagorean Theorem, c  a  b , we can show that the sum of the two small areas is equal to the area of the largest semicircle.

519

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52. B; Find the length of the side by using the Pythagorean Theorem. d2  s2  s2 102  s2  s2 100  2s2 50  s2 150  s

61.

26a4b7c5 13a2b4c3



a b c 126 13 21 a 21 b 21 c 2 4

7

5

2

4

3

 2(a42 )(b74 ) (c53 )  2a2b3c8 a2 1 2a2b3 c8 2

 1 



b3 1

1

 c8

62. Let p  air speed of the plane and w  wind speed.

Find the perimeter of the square with s  150. p  4s  4150  4125  2  4(512)  2012 cm

Rate

Time (hr)

With

pw

Against

pw

2 3 3 4

Distance (m) 300 300

Write two equations based on the fact that the product of the rate and the time is the distance.

Page 610 53.

Maintain Your Skills

1y  12 ( 1y) 2  122 y  144

31s  126 1s  42 ( 1s) 2  422 s  1764 55. 412v  1  3  17 412v  1  20 12v  1  5 ( 12v  1) 2  52 2v  1  25 2v  24 v  12 Check:

1144  12 12  12 ✓

Check:

54.

2 (p 3 3 (p 4

?

311764  126 ? 3(42)  126 126  126 ✓

63. 2(6  3) 2  (8  4) 2  232  42  19  16  125 5 64. 2(10  4) 2  (13  5) 2  262  82  136  64  1100  10

?

412(12)  1  3  17 ?

4 124  1  3  17

65. 2(5  3) 2  (2  9) 2  222  (7) 2  14  49  153

?

4125  3  17 ? 20  3  17 17  17 ✓

66. 2(9  5) 2  (7  3) 2  2(14) 2  42  1196  16  1212  14  53  2153

56. 172  136  2  262  12  612 57. 71z  101z  (7  10) 1z  3 1z 58.

59.

3

58 53

3 7

 121 

13 17



13 17



121 7



121 1



121 7



7 121 7



(1  7) 121 7



8 121 7

67. 2(4  5) 2  (4  3) 2  2(9) 2  (7) 2  181  49  1130

 121 17

 17 

121 1

68. 2(20  5) 2  (2  6) 2  2152  (8) 2  1225  64  1289  17

11-5

 583  55 or 3125

The Distance Formula

Pages 612–613

1 60. d7  d7

Chapter 11

 w)  300

Simplify each equation and solve the system. p  w  450 p  w  400 2p  850 p  425 mph The air speed of the plane is 425 mph.

?

Check:

 w)  300

Check for Understanding

1. The values that are subtracted are squared before being added and the square of a negative number is always positive. The sum of two positive numbers is positive, so the distance will never be negative

520

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2. See students’ graph; the distance from A to B equals the distance from B to A. Using the Distance Formula, the solution is the same no matter which ordered pair is used first. 3. See students’ diagrams; there are exactly two points that lie on the line y  3 that are 10 units from the point (7, 5). 4. d  2(x  x 2 1

)2

 (y2  y1

10.

)2

2

 2(11  5)  [7  (1) ] 2  262  82  136  64  1100  10  2(2  3) 2  (5  7) 2  2(5) 2  (12) 2  125  144  1169  13 6. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(5  2) 2  (1  2) 2  232  (3) 2  19  9  118  312 or about 4.24 7. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

8.

2(6  (3)) 2  [ 4  (5) ] 2 2(3) 2  (1) 2 19  1 110 or about 3.16

d  2(x2  x1 ) 2  (y2  y1 ) 2 10  2(a  3) 2  [7  (1) ] 2 10  2a2  6a  9  82 10  2a2  6a  9  64 102  ( 2a2  6a  73 ) 2 100  a2  6a  73 0  a2  6a  27 0  (a  3)(a  9) a30 or a  9  0 a  3 a9

9.

AB    

2(5  (3) ) 2  (2  4) 2 282  (2) 2 164  4 168

BC    

2(1  5) 2  (5  2) 2 2(6) 2  (7) 2 136  49 185

AC  2(1  (3) ) 2  (5  4) 2  222  (9) 2  14  81  185 Yes, BC  AC. 11. The quarterback is located at (40, 10). The top receiver is located at (20, 25). The bottom receiver is located at (15, 5). Distance from quarterback to top receiver: d  2(40  20) 2  (10  25) 2 d  2202  (15) 2 d  1400  225 d  1625 d  25 yd Distance from quarterback to bottom receiver: d  2(40  15) 2  (10  5) 2 d  2252  52 d  1625  25 d  1650 d  125  26 d  5126 or 25.5 yd 12. The receivers are located at (20, 25) and (15, 5). d  2(20  15) 2  (25  5) 2 d  252  202 d  125  400 d  1425 d  125  17 d  5117 or 20.6 yd

5. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

   

d  2(x2  x1 ) 2  (y2  y1 ) 2

Pages 613–615

Practice and Apply

13. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

d  2(x2  x1 ) 2  (y2  y1 ) 2

 2(8  12) 2  (3  3) 2  2(20) 2  02  1400  20

1145  2(1  10) 2  (6  a) 2 1145  2(9) 2  36  12a  a2 ( 1145) 2  ( 281  36  12a  a2 ) 2 145  117  12a  a2 0  a2  12a  28 0  (a  14)(a  2) or a  2  0 a  14  0 a  14 or a2

14. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(5  0) 2  (12  0) 2  252  122  125  144  1169  13

521

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15. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

22. d  2(x2  x1 ) 2  (y2  y1 ) 2

1 15 2  3 (2) 2  1 4 2

 2(3  6) 2  (4  8) 2  2(3) 2  (4) 2  19  16  125 5 16. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(4  (4)) 2  (17  2) 2  282  152  164  225  1289  17 17. d  2(x  x 2 1

)2

 (y2  y1



34 



3



10 3

2



183 22

289

3 12  5 2

4 2

3 15 2

6 2



36

1

3 25  4



112 2

2

3 12  (1) 4

2

169

3 100

 1.30 24. d  2(x2  x1 ) 2  (y2  y1 ) 2

2



3 (4  3)



31



3 1  49



3 49



174 7

2



2

157 2

 2

127  37 2

2

25

74

 1.23 25. d  2(x2  x1 ) 2  (y2  y1 ) 2  2(615  415) 2  (1  7) 2  2(215) 2  (6) 2  120  36  156  2114  7.48 26. d  2(x2  x1 ) 2  (y2  y1 ) 2  2(712  512) 2  (10  8) 2  2(212) 2  (2) 2  18  4  112  213  3.46

123  222

27.

d  2(x2  x1 ) 2  (y2  y1 ) 2 5  2(a  4) 2  (3  7) 2 5  2a2  8a  16  (4) 2 5  2a2  8a  32 (5) 2  ( 2a2  8a  32 ) 2 25  a2  8a  32 0  a2  8a  7 0  (a  1)(a  7) a  1  0 or a  7  0 a1 a7

64 9

100 9

 3.33

Chapter 11

225 16

13

21. d  2(x2  x1 ) 2  (y2  y1 ) 2

32

17 4

2

1 2

 44

 10

 2(10  2) 2  (4  7) 2  282  (11) 2  164  121  1185  13.60







20. d  2(x2  x1 ) 2  (y2  y1 ) 2



3 16



 2(3  (8)) 2  [ 8  (4) ] 2  252  (4) 2  125  16  141  6.40

2





 2(3  9) 2  [ 6  (2) ] 2  2(6) 2  (4) 2  136  16  152  14  13  2113  7.21

3 (6  4)

34 



18. d  2(x2  x1 ) 2  (y2  y1 ) 2





2

23. d  2(x2  x1 ) 2  (y2  y1 ) 2

)2

19. d  2(x2  x1 )  (y2  y1 )

3 (3  5)

 4.25

 2(5  (3)) 2  (4  8) 2  282  (4) 2  164  16  180  116  5  415  8.94

2



522

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28.

d  2(x  x ) 2  (y  y ) 2 2 1 2 1

AC  2(5  7) 2  [ 6  (4) ] 2  2(2) 2  (2) 2  14  4  18 The triangle has two sides that are equal in length—AB  BC  10.

17  2(4  (4)) 2  (2  a) 2 17  282  4  4a  a2 17  268  4a  a2 2 (17) 2  ( 268  4a  a2 ) 2 289  68  4a  a 0  221  4a  a2 0  a2  4a  221 0  (a  17)(a  13) a  17 or a  13 29.

34.

d  2(x2  x1 ) 2  (y2  y1 ) 2 110  2(6  5) 2  (1  a) 2 110  212  1  2a  a2 ( 110 ) 2  ( 21  1  2a  a2 ) 2 10  a2  2a  2 0  a2  2a  8 0  (a  4)(a  2) a  4 or a  2

30.

35.

BD    

2(0  10) 2  (5  6) 2 2(10) 2  (1) 2 1100  1 1101

d  2(x2  x1 ) 2  (y2  y1 ) 2

PM  2(x  2) 2  (2  5) 2  2x2  4x  4  (7) 2  2x2  4x  53 PL  PM 2x2  8x  17  2x2  4x  53 ( 2x2  8x  17 ) 2  ( 2x2  4x  53 ) 2 x2  8x  17  x2  4x  53 8x  17  4x  53 12x  36 x3

d  2(x2  x1 ) 2  (y2  y1 ) 2

36.

d  2(x2  x1 ) 2  (y2  y1 ) 2 QR    

d  2(x2  x1 ) 2  (y2  y1 ) 2 1340  2(a  20) 2  (9  5) 2 1340  2a2  40a  400  42 ( 1340 ) 2  ( 2a2  40a  400  16 ) 2 340  a2  40a  416 0  a2  40a  76 0  (a  2)(a  38) a20 or a  38  0 a2 a  38

33.

2[9  (2) ] 2  (8  2) 2 2112  62 1121  36 1157

PL  2(x  (4) ) 2  [ 2  (3) ] 2  2x2  8x  16  1  2x2  8x  17

d  2(x2  x1 ) 2  (y2  y1 ) 2

1130  2(3  6) 2  [ a  (3) ] 2 1130  2(9) 2  a2  6a  9 ( 1130 ) 2  ( 281  a2  6a  9 ) 2 130  a2  6a  90 0  a2  6a  40 0  (a  10)(a  4) a  10 or a  4 32.

AC    

1157  1101; The trapezoid is not isosceles.

129  2(7  a) 2  (3  5) 2 129  249  14a  a2  (2) 2 ( 129 ) 2  ( 249  14a  a2  4 ) 2 29  a2  14a  53 0  a2  14a  24 0  (a  2)(a  12) a  2 or a  12 31.

d  2(x2  x1 ) 2  (y2  y1 ) 2

2(3  1) 2  (1  7) 2 222  (6) 2 14  36 140

ST  2(7  9) 2  (d  3) 2  2(2) 2  d2  6d  9  2d2  6d  13 QR  ST 140  2d2  6d  13 ( 140) 2  ( 2d2  6d  13 ) 2 40  d2  6d  13 0  d2  6d  27 0  (d  9) (d  3) d  9  0 or d  3  0 d9 d  3 The distance formula demonstrates that d  3 does not make TQ  140. Therefore, d  9.

d  2(x2  x1 ) 2  (y2  y1 ) 2 AB  2(1  7) 2  [2  (4) ] 2  2(8) 2  (6) 2  164  36  1100  10 BC  2[5  (1) ] 2  (6  2) 2  2(6) 2  (8) 2  136  64  1100  10

523

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43. Compare the slopes of the two potential legs to determine whether the slopes are negative reciprocals of each other. You can also compute the lengths of the three sides and determine whether the square of the longest side length is equal to the sum of the squares of the other two side lengths. Neither test holds true in this case because the triangle is not a right triangle. 44. You can determine the distance between two points by forming a right triangle. Drawing a line through each point parallel to the axes forms the legs of the triangle. The hypotenuse of this triangle is the distance between the two points. You can find the lengths of each leg by subtracting the corresponding x- and y-coordinates, then use the Pythagorean Theorem. Answers should include the following. • You can draw lines parallel to the axes through the two points that will intersect at another point forming a right triangle. The length of a leg of a triangle is the difference in the x- or y-coordinates. The length of the hypotenuse is the distance between the points. Using the Pythagorean Theorem to solve for the hypotenuse, you have the Distance Formula. • The points are on a vertical line so you can calculate distance by determining the absolute difference between the y-coordinates.

37. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(254  132) 2  (105  428) 2  2(122) 2  (323) 2  114,884  104,329  1119,213  345.27 units Since each unit is equal to 0.316 mile, the distance is 345.27(0.316)  109 mi. 38.

c2  a2  b2

114  18 2 9 3 2 c2  64  1 8 2 c2 

138 2 9

2



2

9

c2  64  64 18

c2  64 18

2c2   3 64 c  0.53 mi 39.

0.53 3

 0.18 hr

(0.18)(60)  10.6 minutes Yes; it will take her about 10.6 minutes to walk between the two buildings. 40. Duluth, (44, 116); St. Cloud, (46, 39); Eau Claire, (71, 8); Rochester, (27, 58) 41. d  2(x2  x1 ) 2  (y2  y1 ) 2 Minneapolis – St. Cloud:

45. B; d  2(x  x ) 2  (y  y ) 2 2 1 2 1

2(46  (7)) 2  (39  3) 2 2(39) 2  (3) 2 11521  1296 12817  53 mi

 2(2  6) 2  (4  11) 2  2(8) 2  (15) 2  164  225  1289  17 units

St. Paul – Rochester: 2(27  0) 2  (58  0) 2 2272  (58) 2 1729  3364 14093  64 mi Minneapolis – Eau Claire:

46. B; d  2(x  x ) 2  (y  y ) 2 2 1 2 1 AB  2(3  3) 2  (4  7) 2  2(6) 2  (3) 2  136  9  145 P  4s  4145  419  5  1215 units

2(71  (7)) 2  (8  3) 2 2(78) 2  (11) 2 16084  121 16205  79 mi Duluth – St. Cloud:

Page 615

2(46  44) 2  (39  116) 2 2(90) 2  (77) 2 18100  5929 114,029  118 mi 42. all cities except Duluth

Chapter 11

47.

524

Maintain Your Skills

c2  a2  b2 c2  72  242 c2  49  576 c2  625 2c2  1625 c  25 The length of the hypotenuse is 25 units.

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48.

c2  a2  b2 342  a2  302 1156  a2  900 256  a2 1256  2a2 16  a The length of the leg is 16 units.

49.

c2  a2  b2 ( 116 ) 2  ( 17 ) 2  b2 16  7  b2 9  b2 19  2b2 3  b The length of the leg is 3 units.

50.

53.

? ?

16  2  8  6 ?

14  8  6

1500  29  20  3

?

?

149  7 77✓

1529  23 23  23 ✓

The solution set is {2, 10}. 54. Asia, 1,113,000,000,000; Europe, 1,016,000,000,000; U.S./Canada, 884,000,000,000; Latin America, 241,000,000,000; Middle East, 101,200,000,000; Africa, 56,100,000,000. 55. Asia, 1.113  1012; Europe, 1.016  1012; U.S./Canada, 8.84  1011; Latin America, 2.41  1011; Middle East, 1.012  1011; Africa, 5.61  1010. 56. $8.72  1011 or $872 billion 57. 8 m1 9 m {m 0m 9} 2

58.

3

4

5

6

7

8

9

10

3 7 10  k 7 7 k {k 0k 6 7} ⫺8 ⫺7 ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 0

59.

?

111  2  8  11 ?

19  8  11 ?

?

286 3  8  11 10  6 ✕ 11  11 ✓ Since 6 does not satisfy the original equation, 11 is the only solution. 1r  5  r  1 52. ( 1r  5) 2  (r  1) 2 r  5  r2  2r  1 0  r2  3r  4 0  (r  4)(r  1) r  4  0 or r  1  0 r4 r  1 Check:

?

120  29  4  3

51. 1p  2  8  p 1p  2  p  8 ( 1p  2 ) 2  (p  8) 2 p  2  p2  16p  64 0  p2  17p  66 0  ( p  6) ( p  11) p  6  0 or p  11  0 p6 p  11 Check:

?

25(2) 2  29  2(2)  3 25(10) 2  29  2(10)  3

c2  a2  b2 c2  ( 113 ) 2  ( 150 ) 2 c2  13  50 c2  63 2c2  163 c  163 c  19  7 c  317 The length of the hypotenuse is 317 or about 7.94 units.

?

25t2  29  2t  3 ( 25t2  29 ) 2  (2t  3) 2 5t2  29  4t2  12t  9 2  12t  20  0 t (t  2)(t  10)  0 t  2  0 or t  10  0 t2 t  10 Check:

?

?

?

?

3x 2x  3 x 3 {x 0x 3} ⫺8 ⫺7 ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 0

60. v  (4) 7 6 v4 7 6 v 7 2 {v 0v 7 2} ⫺4 ⫺3 ⫺2 ⫺1 0

1

2

3

4

7

8

9

10

61. r  5.2 3.9 r 9.1 {r 0r 9.1}

14  5  4  1 11  5  1  1

19  3 14  2 33✓ 2  2 ✕ Since 1 does not satisfy the original equation, 4 is the only solution.

2

62. s 

3 1 6



s s

4

5

6

2 3 3 6 1 2

5s 0s 12 6

⫺4 ⫺3 ⫺2 ⫺1 0

525

1

2

3

4

Chapter 11

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63.

x 4

3

2

64.

2x  4(3) 2x  12 x6 65.

6 9



8 x

x  2 7



5 2

8.

5x  2(20) 5x  40 x  8 66.

6x  9(8) 6x  72 x  12 67.

20 x

10 12

3

2 3

7(x  2)  3(7) 7x  14  21 7x  7 x1

x

9.

5

7

c

x

 18

68.

BC

 EF

7c  30 AC DF b 5

12x  10(18) 12x  180 x  15

7

AB DE c 6

 

b

2(x  4)  3(6) 2x  8  18 2x  10 x5

AC

 DF 

15 e

10e  90

30 7

e9

BC EF 5 7

AB DE 10 6

BC

 EF 

17 d

10d  102 d  10.2

7b  25

6  4

AB DE 10 6

25 7

10. Let x  the height of the school building. 10 26



25 x

10x  650 x  65 feet The school building is 65 feet high.

11-6 Similar Triangles Pages 618–619

Pages 619–620

Check for Understanding

1. If the measures of the angles of one triangle equal the measures of the corresponding angles of another triangle, and the lengths of the sides are proportional, then the two triangles are similar. 2. Sample answer: ^ ABC  ^DEF

B

C

A

AB DE 15 5

AC

 DF b

9

7.

AB DE 9 6

KL NO m 4

19.

5a  105 a  21

Chapter 11

KL NO 11 6

KL NO 11 6

b

18.



 

LM OP K 5

 

55 6 KM NP / 4

/ 

BC

18 d

9d  108 d  12

526

44 6 22 3

LM OP 24 16

 

KM NP 30 o

24o  480 o  20

LM OP 9 6

6/  44

 EF 



K

 10

AB DE 9 6

a 7



KM NP / 8

6K  55

AC

AB DE 15 5





KL NO 15 p

6m  36 m6

 DF

6b  90 b  15

BC

LM OP 9 6

F corresponding sides AB and DE BC and EF

5b  135 b  27  EF

17.

6/  72 /  12

C and  F AC and DF 3. Consuela; the arcs indicate which angles correspond. The vertices of the triangles are written in order to show the corresponding parts. 4. No; the angle measures are not equal. 5. Yes; the angle measures are equal. 6.

Yes; the angle measures are equal. No; the angle measures are not equal. No; the angle measures are not equal. Yes; the angle measures are equal. No; the angle measures are not equal. Yes; the angle measures are equal.

E

D corresponding angles  A and  D  B and  E

Practice and Apply

11. 12. 13. 14. 15. 16.

 

LM OP 24 16

24p  240 p  10 20.

KL NO 12 p

 

KM NP 13 7

13p  84 84

p  13 LM OP 16 n

 

KM NP 13 7

13n  112 n

112 13

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21.

KL NO



LM OP

1.25 2.5



K 6

22.

2.5K  7.5 K3 KL NO 1.25 2.5

 

KL NO m 2.7

 

KM NP 4 o

 

LM OP

7.5 5



10.5 n

KL NO 7.5 5

33. Yes; all circles are similar because they have the same shape. 34. 2:1; Let the first circle have radius r and the larger have radius 2r. The circumference of the first is 2r and the other has circumference 2(2r)  4r. 35. 4:1; The area of the first is r2 and the area of the other is (2r)2  4r2. 36. The size of an object on the film of a camera can be related to its actual size using similar triangles. Answers should include the following. • Moving the lens closer to the object (and farther from the film) makes the object appear larger. • Taking a picture of a building; you would need to be a great distance away to fit the entire building in the picture. 37. D; We do not have enough information to determine if the triangles are similar. 38. A; AB AC and AC DC. Therefore, AB DC.

 

KM NP 15 o

7.5o  75 o  10

KM NP 4.4 3.3

24.

3.3m  11.88 m  3.6 LM QP K 2.1



7.5n  52.5 n7

1.25o  10 o8 23.

KL NO

KL NO 5 2.5

 

LM QP 12.6 n

5n  31.5 n  6.3

KM NP 4.4 3.3

KL NO 5 2.5

 

KM NP / 8.1

3.3K  9.24 2.5/  40.5 K  2.8 /  16.2 25. Always; if the measures of the sides form equal ratios, the triangles are similar, and the measures of their corresponding angles are equal. 26. Let x  the distance the man is from the camera. x 3

Page 621

2

 1.5

 2(2  1) 2  (4  8) 2  2(3) 2  (4) 2  19  16  125 5

1.5x  6 x4m 27. Let x  height on the model. 1 12

x

 40

40. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

12x  40 1

 2(12  6) 2  [ 5  (3) ] 2  2(6) 2  (8) 2  136  64  1100  10

x  33 in. 28. Let x  distance from pocket B. 84 x 84 x

 42 

Maintain Your Skills

39. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

28 (10  x)

28

 32  x

28x  2688  84x 112x  2688 x  24 in.

41. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(3  4) 2  (12  7) 2  2(1) 2  (5) 2  11  25  126  5.1

29.

4

42. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(6  1) 2  (716  516) 2  252  (216) 2  125  24  149 7 43. Yes; 252  602  652.

4

8 pieces 30. twice as big; 4 by 4 by 5 in. 31. Let x  height of the building. x 6



80 9

44. No; 202  252  352.

9x  480

45. Yes; 492  1682  1752.

1

x  533

46. No; 72  92  122.

The building is about 53 feet. 32. Viho’s eyes are 6 feet off the ground, Viho and the building each create right angles with the ground, and the two angles with the ground at P have equal measure.

47. 3x2  7x  1 48. 5x3  2x2  4x  7 49. 3x2  6x  3 50. x7  abx2  bcx  34

527

Chapter 11

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2x  y  4 () x  y  5 3x 9 x3 Substitute 3 for x into either equation. 3y5 y  2 y  2 The solution is (3, 2). 52. 3x  2y  13  2 6x  4y  26 () 2x  5y  5  3 () 6x  15y  15 11y  11 y  1 Substitute 1 for y into either equation. 3x  2(1)  13 3x  2  13 3x  15 x  5 The solution is (5, 1). 53. 0.6m  0.2n  0.9 0.3m  0.45  0.1n 51.

58.

 60.

()

1

1

1

()

 2y  8  2  4y  0

1 x 6 1 x 2 4 x 6



1.

4

6

5 6

  b2 2 c  14  482 c2  196  2304 c2  2500 2c2  12500 c  50 The length of the hypotenuse is 50 units.

4.

c2  a2  b2 c2  ( 15 ) 2  ( 18 ) 2 c2  5  8 c2  13 2c2  113 c  113 The length of the leg is 113 or about 3.61 units.

 2(5  1) 2  (11  3) 2  2(6) 2  (8) 2  136  64  1100  10

or 0.83

 4

Chapter 11



Practice Quiz 2

6. d  2(x2  x1 ) 2  (y2  y1 ) 2

32,500 ft 739,200 ft b

1.5 (1.5) 1.5 4.5 1 3 or 0.3

6

a2

 2(3  6) 2  (3  (12)) 2  2(9) 2  (15) 2  181  225  1306  17.49

32,500 ft 140 mi

57. a 

c2

or 1.1

5. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

 0.044 

c a  c

c2  a2  b2 ( 184 ) 2  a2  82 84  a2  64 20  a2 120  2a2 215  a The length of the leg is 215 or about 4.47 units.

1

 232 ft/mi. Alternatively, since there are 5280 ft per mile, there is a horizontal gain of (140) (5280)  739,200 ft.

56.

61.

3.

 4y  0

y  12 The solution is (6, 12). 55. The plane has a vertical decent of 32,500 ft and a horizontal gain of 140 mi giving a slope of

6 1.5



c2  a2  b2 412  402  b2 1681  1600  b2 81  b2 181  2b2 9  b The length of the leg is 9 units.

1

1

a c



6(1.5) 5 9 5 9 or 1.8 5

2.

 4y  4

2  2y  8

Therefore m 



2

1

m 

ac b



10 9

Page 621

 2y  8 1 y 2

5 (1.5)

6

59.

5

x6 Substitute 6 for x into either equation. 1 (6) 3

b a  c

6  5 1.5 1 1.5 2 3 or 0.6

 4.5

Substitute 1.5 for m into either equation. 0.6(1.5)  0.2n  0.9 0.9  0.2n  0.9 0.2n  0 n0 The solution is (1.5, 0). 1 x 3 1 x 2

 

0.6m  0.2n  0.9 0.6m  0.2n  0.9 () 0.3m  0.1n  0.45  2 () 0.6m  0.2n  0.9 1.2m  1.8 m  1.5

54.

a  b c

528

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Pages 627–628

7. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(4  2) 2  (7  5) 2  2(2) 2  (2) 2  14  4  18  212  2.83 8. d  2(x  x 2 1

)2

 (y2  y1

sin 50 

)2

BC EF a 2

CA

 FD 

10.

10 1

a  20 BA ED c 1.5

BA

 ED 

CA

BC EF 12 8

10 1

c  15

cos 50 

BC  6.43

C

9 f

50˚

B

3. They are equal. leg 4. sin Y  opposite hypotenuse



CA

 FD

cos Y 

36 45



 0.8

b

 12

8b  144 b  18



Algebra Activity (Preview of Lesson 11-7)

6.

cos Y 

10 26

Side Lengths

Ratios

7.



5

6.1

0.7

0.574

35º

24 26

 0.9231

36 27

 1.3333 tan Y  

opposite leg adjacent leg 10 24

 0.4167

SIN 60 ENTER .8660254038

55º

90º

7

10

12.2

0.7

0.574

35º

55º

90º

20

24.4

0.7

0.574

35º

55º

90º

28

40

48.8

0.7

0.574

35º

55º

90º

35

50

61

0.7

0.574

35º

55º

90º

42

60

73.2

0.7

0.574

35º

55º

90º

KEYSTROKES:

COS 75 ENTER .2588190451

TAN 10 ENTER .1763269807 tan 10º  0.1763 9. sin W  0.9848 1 KEYSTROKES: 2nd [SIN ] .9848 ENTER 79.99744219 sin W  80º 10. cos X  0.6157 1 KEYSTROKES: 2nd [COS ] .6157 ENTER 8.

angle angle B C

14

KEYSTROKES:

51.99719884 cos X  52º 11. tan C  0.3249 1 KEYSTROKES: 2nd [TAN ] .3249 ENTER 17.99897925 tan C  18º 12. Let A  the angle to find.

1. All ratios and angle measures are the same for any 7:10 right triangle. 2. 7:10 3. 55º

sin A 

11-7 Trigonometric Ratios Page 626

adjacent leg hypotenuse



opposite leg adjacent leg

cos 75º  0.2588

Angle Measures

3.5

27 45

tan Y 

sin 60º  0.8660

Steps 1–4. Sample answers are given in the table. side side side angle BC:AC BC:AB BC AC AB A

KEYSTROKES:

adjacent leg hypotenuse

 0.6

leg 5. sin Y  opposite hypotenuse

 0.3846

Page 622

BC 10

10

12f  72 f6

 FD 

BC EF 12 8

AC 10

AC  7.66 A

 2(5  (2)) 2  [ 4  (9) ] 2  2(3) 2  (13) 2  19  169  1178  13.34 9.

Check for Understanding

1. If you know the measure of the hypotenuse, use sine or cosine, depending on whether you know the measure of the adjacent side or the opposite side. If you know the measures of the two legs, use tangent. 2. Sample answer: A  180  (90  50 ) or 40

opposite leg hypotenuse

7

 13

2nd [SIN1] 7 ⫼ 13 ENTER 32.57897039 sin A  33º

KEYSTROKES:

13. Let A  the angle to find.

Algebra Activity

opposite leg

1. See students’ work. 2. The angle measured by the hypsometer is not the angle of elevation. It is the other acute angle formed in the triangle. So, to find the measure of the angle of elevation, subtract the reading on the hypsometer from 90 since the sum of the measures of the two acute angles in a right triangle is 90°. 3. See students’ work.

6

tan A  adjacent leg  15

2nd [TAN1] 6 ⫼ 15 ENTER 21.80140949 tan A  22º 14. Let A  the angle to find. KEYSTROKES:

cos A 

adjacent leg hypotenuse

9.3

 9.7

1 KEYSTROKES: 2nd [COS ] 9.3 ⫼ 9.7 ENTER 16.5114644 cos A  17º

529

Chapter 11

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15. Find the measures of  A, AC, and BC. Find the measure of  A. The sum of the measures of the angles in a triangle is 180. 180  90  30  60  A  60 Find the measure of AC, which is the side opposite B. Use the sine ratio. sin 30  0.50 

18. Draw a diagram.

0.8660 

AC 42 AC 42

opposite leg

tan A  adjacent leg 40

tan A  1000 Use a calculator. 1 KEYSTROKES: 2nd [TAN ] 40 ⫼ 1000 ENTER 2.290610043  A  2.3

BC 42 BC 42

Pages 628–630

36.4  BC or 36.4 in. BC  36.4 in. 16. Find the measures of  A, AC, and AB. Find the measure of  A. 180  90  35  55  A  55 Find the measure of AC, which is the side opposite B. tan 35  0.7002 

19. sin R  



cos R 

6 10



12 37

21. sin R 

AC 18 AC 18



opposite leg hypotenuse

 cos R  

 0.7241

22. sin R  

cos R 

16 34



leg 23. sin R  opposite hypotenuse

18

0.8192  AB



21.97  AB AB  22 m 17. Find the measures of B, AB, and BC. Find the measure of B. 180  90  55  35  A  35 Find the measure of AB, which is the hypotenuse.

7 1170



 0.5369

24. sin R  

4 4

0.5736  AB 6.973  AB AB  7.0 in. Find the measure of BC, which is the side opposite  A. BC 4 BC 4

5.71  BC BC  5.7 in.

530

35 37

adjacent leg hypotenuse 20 29

18 22



6 8

adjacent leg hypotenuse 30 34

tan R 

11 1170

tan R 

4 110 22

 0.5750

26. 28. 30. 32.

21 20

opposite leg adjacent leg 16 30

opposite leg adjacent leg 7 11

 0.6364 tan R  

 0.8182

opposite leg adjacent leg

 0.5333 tan R  

adjacent leg hypotenuse

12 35

 1.05



adjacent leg hypotenuse

opposite leg adjacent leg

 0.3429



 0.8437 cos R 

tan R  

sin 30  0.5 cos 45  0.7071 tan 32  0.6249 tan 67  2.3559 cos 12  0.9781 cos V  0.5000 V  cos1 0.5000  60 36. sin K  0.9781 K  sin1 0.9781  78 38. tan S  1.2401 S  tan1 1.2401  51 40. sin V  0.3832 V  sin1 0.3832  23 25. 27. 29. 31. 33. 34.

cos 55  AB

opposite leg hypotenuse

opposite leg adjacent leg

 0.75

adjacent leg hypotenuse

 0.8824 cos R 

tan R  

 0.6897

 0.4706

18

8 10

 0.9459

21 29

opposite leg hypotenuse

adjacent leg hypotenuse

 0.8 cos R 

 0.3243

cos 35  AB

Chapter 11

Practice and Apply

 0.6

Find the measure of AB, which is the hypotenuse.

1.4281 

opposite leg hypotenuse

leg 20. sin R  opposite hypotenuse

12.6  AC AC  12.6 m

tan 55 

A

 0.04 1000 ft  4% To find the angle of elevation, note that the 40 ft vertical rise is opposite the angle and the 1000 ft horizontal change is adjacent the angle. This suggests we use tangent.

21  AC AC  21 in. Find the measure of BC, which is the side adjacent B. Use the cosine ratio. cos 30 

40 ft

40

percent grade  1000

opposite leg adjacent leg 18 4 110

 1.4230

sin 80  0.9848 cos 48  0.6691 tan 15  0.2679 sin 53  0.7986

35. cos Q  0.7658 Q  cos1 0.7658  40 37. sin A  0.8827 A  sin1 0.8827  62 39. tan H  0.6473 H  tan1 0.6473  33 41. cos M  0.9793 M  cos1 0.9793  12

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42. tan L  3.6541 L  tan1 3.6541  75 43. Let A  the angle that we need to find. sin A  sin A  A

Now find the measure of C. cos C  cos C  C

opposite leg hypotenuse 10 16 10 sin1 16

1 2

opposite leg

tan A  adjacent leg 2

A  tan1

1142 2

 8 45. Let A  the angle that we need to find. cos A  cos A  A

adjacent leg hypotenuse 10 16 10 cos1 16

1 2

 51 46. Let A  the angle that we need to find. sin A  sin A  A

opposite leg

tan A  adjacent leg

opposite leg hypotenuse 21 25 21 sin1 25

24

tan A  16 A  tan1

1 2

cos A  A

adjacent leg hypotenuse 17 21 17 cos1 21

b

tan 45  8

1 2

b

18 8b

 36 48. Let A  angle that we need to find.

cos 45 

opposite leg

0.7071 

tan A  adjacent leg tan A 

5 8

A  tan1

sin A  A

8 c 8 c

0.7071c  8 c  11.3 ft 53. 180  90  27  63 The measure of A is 63 .

158 2

 32 49. Let A  the angle that we need to find. sin A 

12416 2

 56 The measure of A is 56 . 52. 180  90  45  45 The measure of A is 45 .

 57 47. Let A  the angle that we need to find. cos A 

1 2

 69 Now the measure of A is the difference between the measures of C and B. 69  62  7 The measure of A is 7 . 51. Let A  the angle that we need to find. We can use the Pythagorean Theorem to find the length of the lower leg of the smaller right triangle. a2  b2  c2 a2  242  252 a2  576  625 a2  49 a  149 a7 Now the length of the lower leg of the larger right triangle is 23  7 or 16 units. Now find the measure of A.

 39 44. Let A  the angle that we need to find. tan A  14

adjacent leg hypotenuse 16 45 16 cos1 45

b

sin 27  20

opposite leg hypotenuse 9 15 9 sin1 15

b

0.4540  20

1 2

9.1 in  b a

cos 27  20

 37 50. Let A  the angle that we need to find, B  the upper angle in the smaller right triangle, and C  the upper angle in the larger right triangle. First find the measure of B.

a

0.8910  20 17.8 in  a

opposite leg

tan B  adjacent leg 30

tan B  16 B  tan1  62

13016 2 531

Chapter 11

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54. 180  90  21  69 The measure of B is 69 .

58.

13 c 13 c

cos 21  0.9336 

0.9336c  13 c  13.9 AB is about 13.9 cm long.

8

tan A  6

a

tan 21  13

A  tan1

5.0  a CB is about 5.0 cm long. 55. 180  90  40  50 The measure of B is 50 . a a

0.6428  16 10.3  a BC is about 10.3 ft long. b

cos 40  16 b

0.7660  16 12.3  b AC is about 12.3 ft long. 56. 180  90  20  70 The measure of A is 70 . 0.3640 

5

tan A  12 A  tan1

33  b AC is about 3.3 m long. 0.9397 

9 c 9 c

0.9397c  9 c  9.6 AB is about 9.6 m long. 57. 180  90  38  52 The measure of B is 52 . tan 38  0.7813 

3

sin A  6 A  sin1

24 b 24 b

0.6157 

626

61. sin A  4420 A  sin1

24 c 24 c

 8.1 62.

0.6157c  24 c  39 AB is about 39 in. long.

Chapter 11

136 2

 30 The measure of A is 30 . 180  90  30  60 The measure of B is 60 .

0.7813b  24 b  30.7 AC is about 30.7 in. long. sin 38 

1125 2

 23 The measure of A is 23 . 180  90  23  67 The measure of B is 67 . c2  a2  b2 60. 62  32  b2 36  9  b2 27  b2 127  2b2 5.2  b AC is about 5.2 cm long.

b 9 b 9

cos 20 

186 2

 53 The measure of A is 53 . 180  90  53  37 The measure of B is 37 . 59. c2  a2  b2 c2  122  52 c2  144  25 c2  169 2c2  1169 c  13 AB is 13 ft long.

sin 40  16

tan 20 

c2  a2  b2 c2  62  82 c2  36  64 c2  100 2c2  1100 c  10 AB is 10 ft long.

532

626 14420 2

c2  a2  b2 (4420) 2  a2  (626) 2 19,536,400  a2  391,876 19,144,524  a2 119,144,524  2a2 4375  a The submarine traveled a horizontal distance of about 4375 m.

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3000

63. tan A  8000 A  tan1  20.6 64.

c2

a2

Page 630

13000 8000 2

70.

b2

  c2  80002  30002 c2  64,000,000  9,000,000 c2  73,000,000 2c2  173,000,000 c  8544 The distance is 8544 ft. x

72.

0.7  x x

sin 20  8 x

0.3420  8 2.74  x from about 0.7 m to about 2.74 m b

66. Let sin A  c and let cos A  c , where a and b are legs of a right triangle and c is the hypotenuse. a2 c2



b2 c2





a2  b2 . c2

71.

KL NO 3 4.5

 

KM NP 9 o

5p  60 p  12

3o  40.5 o  13.5

LM OP 5 10

KL NO 3 4.5

 

KM NP 3 o

 

LM OP k 12

4.5k  36 k8

d  2(x2  x1 ) 2  (y2  y1 ) 2 39  2(a  9) 2  (8  28) 2 39  2a2  18a  81  (36) 2 39  2a2  18a  81  1296 (39) 2  ( 2a2  18a  1377) 2 1521  a2  18a  1377 0  a2  18a  144 0  (a  24)(a  6) or a  6  0 a  24  0 a  24 a  6

x

0.0872  8

Then sin2A  cos2 A 



Maintain Your Skills KL NO 6 p

5o  30 o6

65. sin 5  8

a

LM OP 5 10

73.

Since the

d  2(x2  x1 ) 2  (y2  y1 ) 2 165  2(10  3) 2  (1  a) 2 165  272  1  2a  a2 ( 165 ) 2  ( 250  2a  a2 ) 2 65  a2  2a  50 0  a2  2a  15 0  (a  5) (a  3) a  5 a  3

Pythagorean Theorem states that a2  b2  c2, the c2 expression becomes c2 or 1. Thus sin2A  cos2A  1. 67. If you know the distance between two points and the angles from these two points to a third point, you can determine the distance to the third point by forming a triangle and using trigonometric ratios. Answers should include the following. • If you measure your distance from the mountain and the angle of elevation to the peak of the mountain from two different points, you can write an equation using trigonometric ratios to determine its height, similar to Example 5. • You need to know the altitude of the two points you are measuring. 68. A; Use the Pythagorean Theorem to find RT. c2  a2  b2 42  a2  22 16  a2  4 12  a2 112  2a2 213  a

74. c2 (c2  3c)  c4  3c3 75. s(4s2  9s  12)  4s3  9s2  12s 76. xy2 (2x2  5xy  7y2 )  2x3y2  5x2y3  7xy4 77. a  3b  2 4a  7b  23 Substitute 3b  2 for a in the other equation. 4(3b  2)  7b  23 12b  8  7b  23 5b  8  23 5b  15 b3 Substitute 3 for b into either equation. a  3(3)  2 92  11 Checking values in both equations confirms the solution is (11, 3).

RT is 213 units so TS is 213 units. Use the Pythagorean Theorem to find RS. c2  (213 ) 2  (213 ) 2 c2  12  12 c2  24 2c2  124 c  216 RS is 216 units. 2

69. D; cos Q  4 Q  cos1  60

124 2 533

Chapter 11

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p  q  10 3p  2q  5 Solve the first equation for p. p  10  q Substitute 10  q for p into the other equation. 3(10  q)  2q  5 30  3q  2q  5 30  5q  5 5q  35 q7 Substitute 7 for q into either equation. p  10  7 3 Checking values in both equations confirms the solution is (3, 7). 79. 3r  6s  0 4r  10s  2 Solve the first equation for r. 3r  6s  0 3r  6s r  2s Substitute 2s for r into the other equation. 4(2s)  10s  2 8s  10s  2 2s  2 s1 Substitute 1 for s into either equation. r  2(1)  2 Checking values in both equations confirms the solution is (2, 1).

3a. Sample answer: dome, domestic, domicile 3b. Sample answer: eradicate, radicand, radius 3c. Sample answer: simile, similarity, similitude

78.

Chapter 11 Study Guide and Review Page 632

2. true 4. true

6. true

7. false,

Pages 632–636 9.

3

60 y2

Lesson-by-Lesson Review



160 1y2



14  115 0y 0



2 115 0y 0

10. 244a2b5  24  11  a2  b4  b  14  111  2a2  2b4  1b  2  111  0a 0  b2  1b  2 0a 0b2 111b

11. (3  2112 ) 2  32  2(3) (2112 )  (2112 ) 2  9  12112  48  57  12(213 )  57  2413 9 3  12

Reading Mathematics

1a. Sample answer: A circumstance that brings about a result; any of two or more quantities that form a product when multiplied together; the quantities bring about a result, a product. 1b. Sample answer: The legs of an animal support the animal; one of the two shorter sides of a right triangle; the legs of a triangle support the hypotenuse. 1c. Sample answer: To devise rational explanations for one’s acts without being aware that these are not the real motives; to remove the radical signs from an expression without changing its value; to justify an action without changing its intent. 2a. Sample answer: Rank as determined by the sum of a term’s exponents; the degree of x2y2 is 4. 1 th of a circle; a semicircle measures 180 . 360 2b. Sample answer: The difference between the greatest and least values in a set of data; the range of 2, 3, 6 is 6  2 or 4. The set of all y-values in a function; the range of {(2, 6), (1, 3)} is {6, 3}. 2c. Sample answer: Circular or spherical; circles are round. To abbreviate a number by replacing its ending digits with zeros; 235,611 rounded to the nearest hundred is 235,600.

Chapter 11

x 12xy y

8. true

12.

Page 631

Vocabulary and Concept Check

1. false, 3  17 3. true 5. false, 3x  19  x2  6x  9

13.

14.

9

3

3  12 12

3 



9(3  12) 9  2



27  9 12 7

2 17 3 15  5 13

23a3b4 28ab10

12

2 17

3 15  5 13

 3 15  5 13  3 15  5 13 

2 17(3 15  5 13) 45  75



2 17(3 15  5 13) 30



 17(3 15  5 13) 15



5 121  3 135 15

3

a3

b4



3 8  a  b10



3 8b6



13  2a2 18  2b6

3a2

0 a 0 13

 2 0 b3 0 12 0 a 0 13

12

 2 0 b3 0 12  12 

0 a 0 16 4 0 b3 0

15. 213  815  315  313  (2  3) 13  (8  3) 15  513  515 16. 216  148  216  116  3  216  413

534

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17. 4127  6148  4(313)  6(413)  1213  2413  3613 18. 417k  717k  217k  (4  7  2) 17k  17k 19. 5118  31112  3198  5(312 )  3(417 )  3(712)  1512  1217  2112  (15  21) 12  1217  6 12  1217 20. 18 

1

29.

Check:

?

1

2 12 1



12 4



8 12 4



12 4



9 12 4

21. 12(3  313 )  312  (313 )( 12 )  312  316 22. 15(215  17 )  (215 )( 15 )  ( 17 )( 15 )  10  135 23. ( 13  12 ) (212  13 )

?

13(5)  14  5  6

 ( 13) (212)  ( 13) ( 13)  12(212)  12( 13)  2 16  3  4  16  1  16

?

24. (6 15  2) (3 12  15 )  (6 15) (3 12)  (6 15) ( 15)  2 (3 12)  2 15  18 110  30  6 12  2 15

25. 10  21b  0

Check:

21b  10 1b  5 ( 1b ) 2  (5) 2 b  25 No solution 26.

1a  4  6 ( 1a  4 ) 2  62 a  4  36 a  32

27.

Check: 17x  1  5 2 2 ( 17x  1 )  (5) 7x  1  25 7x  26 x

28.

4a

33

20 4a

33

2

1 3 4a3 22  22 4a 3

4 a3

Check:

Check:

31.

?

10  2125  0 ? 10  2(5)  0 ? 10  10  0 20  0 ✕

?

32.

132  4  6 ? 136  6 66✓

37 1 7 2  1  5 ?

26

?

126  1  5 ? 125  5 55✓

26 7

3

4(3) 3

?

15  4  5  8

?

112  4  12  8 ?

19  3 116  4 3  3 ✕ 44✓ Since 5 does not satisfy the original equation, 12 is the only solution. 30. 13x  14  x  6 13x  14  6  x ( 13x  14 ) 2  (6  x) 2 3x  14  36  12x  x2 0  x2  15x  50 0  (x  5) (x  10) x  5  0 or x  10  0 x  10 x5 Check:

3 8  212  2 12 

1x  4  x  8 ( 1x  4 ) 2  (x  8) 2 x  4  x2  16x  64 0  x2  17x  60 0  (x  5) (x  12) x  5  0 or x  12  0 x5 x  12

33.

?

20 ?

14  2  0 ?

220 34.

00✓

535

?

13(10)  14  10  6 ?

116  10  6 11  5  6 ? ? 156 4  10  6 66✓ 14  6 ✕ Since 10 does not satisfy the original equation, 5 is the only solution. c2  a2  b2 c2  302  162 c2  900  256 c2  1156 2c2  11156 c  34 The hypotenuse is 34. c2  a2  b2 c2  62  102 c2  36  100 c2  136 2c2  1136 c  11.66 The hypotenuse is 11.66. c2  a2  b2 152  102  b2 225  100  b2 115  b2 1115  2b2 10.72  b The length of the leg is 10.72. c2  a2  b2 562  a2  42 3136  a2  16 3120  a2 13120  2a2 55.86  a The length of the leg is 55.86.

Chapter 11

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c2  a2  b2 302  182  b2 900  324  b2 576  b2 1576  2b2 24  b The length of the leg is 24. 36. c2  a2  b2 c2  (1.2) 2  (1.6) 2 c2  1.44  2.56 c2  4 2c2  14 c  2 The length of the hypotenuse is 2.

46. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

35.

 2[5  (2) ] 2  (11  6) 2  272  52  149  25  174  8.60 d  2(x2  x1 ) 2  (y2  y1 ) 2

47.

5  2 [1  (3) ] 2  (a  2) 2 5  242  a2  4a  4 5  2a2  4a  20 52  ( 2a2  4a  20 ) 2 25  a2  4a  20 0  a2  4a  5 0  (a  5)(a  1) a  5  0 or a  1  0 a5 a  1

37. no; 92  162  202 38. yes; 202  212  292 39. yes; 92  402  412

d  2(x2  x1 ) 2  (y2  y1 ) 2

48.

40. no; 182  ( 124 ) 2  302

5  2(4  1) 2  (a  1) 2 5  232  a2  2a  1 (5) 2  ( 2a2  2a  10 ) 2 25  a2  2a  10 0  a2  2a  15 0  (a  5)(a  3) a  5  0 or a  3  0 a5 a  3

41. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(1  9) 2  [ 13  (2) ] 2  2(8) 2  (15) 2  164  225  1289  17 42. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

d  2(x2  x1 ) 2  (y2  y1 ) 2

49.

 2(7  4) 2  (9  2) 2  232  72  19  49  158  7.62

1145  2(5  6) 2  [ a  (2) ] 2 1145  2(1) 2  a2  8a  4 ( 1145 ) 2  ( 2a2  8a  5 ) 2 145  a2  8a  5 0  a2  8a  140 0  (a  14)(a  10) a  14  0 or a  10  0 a  14 a  10

43. d  2(x2  x1 ) 2  (y2  y1 ) 2  2(2  4) 2  [7  (6) ] 2  2(6) 2  (13) 2  136  169  1205  14.32

d  2(x2  x1 ) 2  (y2  y1 ) 2

50.

1170  2(a  5) 2  [ 3  (2) ] 2 1170  2a2  10a  25  1 ( 1170) 2  ( 2a2  10a  26 ) 2 170  a2  10a  26 0  a2  10a  144 0  (a  18)(a  8) a  18  0 or a  8  0 a  18 a  8

44. d  2(x2  x1 ) 2  (y2  y1 ) 2  2(415  2 15) 2  (3  9) 2  2(215) 2  (6) 2  120  36  156  2114  7.48

51.

45. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(7  4) 2  (12  8) 2  2(11) 2  (4) 2  1121  16  1137  11.70

AB DE 16 9

AC

 DF 

12 e

16e  108 e AB DE 16 9

27 4 BC

 EF 

10 d

16d  90 d

Chapter 11

536

45 8

52.

AB DE 10 12

AC

 DF 

6 e

10e  72 e  7.2 AB DE 10 12

BC

 EF 8

d

10d  96 d  9.6

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53.

AB DE 12 9

AC

 DF b

 11

9b  132 b AB DE 12 9

AB DE 15 6

54.

BC

AB DE 15 6

8

d

12d  72 d6

BC

 EF 

28 53

 0.8491 28

60. sin A  53

 1.6071  0.5283 61. tan M  0.8043 62. sin I  0.1212 I  sin1 (0.1212) M  tan1 (0.8043)  39  7 63. cos B  0.9781 64. cos F  0.7443 B  cos1 (0.9781) F  cos1 (0.7443)  12  42 65. sin A  0.4540 66. tan Q  5.9080 A  sin1 (0.4540) Q  tan1 (5.9080)  27  80



16 1



3 16 3



4 16 3



3

10 3

4

 3 30  

3

4  4 ✕

?

?

7  15(7)  14 2  15(2)  14 7  135  14 ?

?

2  110  14 ?

7  149 2  14 77✓ 2  2 ✕ Since 2 does not satisfy the original equation, 7 is the only solution. 15. 14x  3  6  x ( 14x  3 ) 2  (6  x) 2 4x  3  36  12x  x2 0  x2  16x  39 0  (x  3) (x  13) x  3  0 or x  13  0 x  3 or x  13

13

 13

16 3

6. 2112x4y6  216  7  x4  y6  4  17  x2  |y3|  4x2|y3|17 7.

4  216

?

?

16 3



4  26(4)  8

?

x  15x  14 x2  ( 15x  14 ) 2 x2  5x  14 2  5x  14  0 x (x  7)(x  2)  0 x  7  0 or x  2  0 x7 x  2 Check:

12 12 13

?

2  26(2)  8

No solution 14.

3 3  16  13  16 

?

10  1  11 11  11 ✓

2  24

b; sine a; cosine c; tangent 2127  163  413  2(313 )  317  413  613  317  413  213  317 2

?

1100  1  11

2  2 ✕

Page 637

5. 16 

?

?

Check:

Chapter 11 Practice Test 1. 2. 3. 4.

14(25)  1  11

14x  1  5 Check: 14(6)  1  5 ? ( 14x  1 ) 2  52 124  1  5 ? 4x  1  25 125  5 4x  24 55✓ x6 x  16x  8 13. x2  ( 16x  8 ) 2 x2  6x  8 2 x  6x  8  0 (x  2)(x  4)  0 or x  4  0 x20 x  2 x  4

45

45 28

Check:

12.

28 45

58. cos A  53

 0.8491

110(40)  20 ? 1400  20 20  20 ✓

( 14s ) 2  102 4s  100 s  25

 0.6222

45 53

?

Check:

14s  10

a 7

56. tan A 

110x  20 ( 110x ) 2  202 10x  400 x  40

11. 14s  1  11

6a  105 a  17.5

 0.5283

59. tan B 

20 e

e8

 EF

57. sin B 



10.

15e  120

44 3

55. cos B 

AC

 DF

Check:

?

14(3)  3  6  3 ?

112  3  3 ?

19  3 33✓

40 90

?

14(13)  3  6  (13) ?

152  3  19 155  19 ✕

Since 13 does not satisfy the original equation, 3 is the only solution.

4

39 2

3 8. 16(4  112 )  416  172  416  612 9. (1  13 )(3  12 )  3  12  313  16

537

Chapter 11

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c2  a2  b2 c2  82  102 c2  64  100 c2  164 2c2  1164 c  1164  12.81 The hypotenuse is about 12.81 units. c2  a2  b2 17. 122  (612 ) 2  b2 144  72  b2 72  b2 172  2b2 172  b 8.49  b The length of the leg is about 8.49. c2  a2  b2 18. 172  a2  132 289  a2  169 120  a2 1120  2a2 1120  a 10.95  a The length of the leg is about 10.95. 16.

19. d  2(x2  x1

)2

 (y2  y1

24.

BC KH 4.5 j

25.

a

 12

26.

)2

15a  240 a  16 AB JK 20 15

AC

 JH b

 16

15b  320 b

64 3

Chapter 11

3 j 1

 42 AC

 JH 

b 1

24 1

 108

c2  a2  b2 292  a2  212 841  a2  441 400  a2 1400  2a2 20  a

12021 2

A  44 B  180  90  44  46 c2  a2  b2 27. c2  212  152 c2  441  225 c2  666 2c2  1666 c  25.8 15

tan A  21 A  tan1

28.

11521 2

A  36 B  180  90  36  54 A  180  90  42  48 b

sin 42  10

6 j

6.7  b a

cos 42  10 7.4  a c  a2  b2 29. c2  92  122 c2  81  144 c2  225 2c2  1225 c  15 miles

AC

 JH

2

13 k

12k  130 k



A  tan1

BC



BC

 KH

20

12j  60 j5 AB JK 12 10

7.5 5

tan A  21

 KH 



3

 2 [21  (9) ] 2  (7  2) 2  2302  52  1900  25  1925  5137  30.41 AB JK 12 10

AC

 JH

b64

21. d  2(x2  x1 ) 2  (y2  y1 ) 2

23.

7.5 5

j1

 2(1  (1)) 2  (5  1) 2  222  (6) 2  14  36  140  2110  6.32

BC

AB JK 1 42 1 12 1 42 j

AB JK 1 42 1 12 1 12 b

20. d  2(x2  x1 ) 2  (y2  y1 ) 2

 KH



7.5j  22.5 j3

 2(4   (2  7) 2 2  20  (9) 2  181 9

AB JK 20 15

AC

 JH

7.5h  32.5 h  4.3

4) 2

22.

AB JK 6.5 h

65 6

538

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30. B; A  /w  16 (2132  316 )  2 1192  3(6)  2(813 )  18  1613  18 units2

1

3

10. 2y  2x  4  0 m 

A B 3 2 1 2

 3 m  3 since slopes of parallel lines are equal.

Chapter 11 Standardized Test Practice 11.

y

Pages 638–639 1. D; y  2x  1 2. C; /w6 2/  2w  92 3. C; Let x  cost of highway resurfacing project y  cost of bridge repair project x  y  2,500,000 y  2x  200,000 x  2x  200,000  2,500,000 3x  2,700,000 x  $900,000

O

x

12. Let x  first integer y  second integer x  y  66 1

y  18  2x

4. C; 32,800,000  3.28  107 n7 2 5. A; x  7x  18  0 (x  2)(x  9)  0 x  2  0 or x  9  0 x2 x  9

1

Substitute 18  2x for y into the other equation. 1

x  18  2x  66 1

12x  48 x  32 Substitute 32 for x into either equation. 32  y  66 y  34 The two integers are 32 and 34.

g  t2  t 132  t2  t 0  t2  t  132 0  (t  12)(t  11) t  12  0 or t  11  0 t  12 t  11 12 teams are in the league.

6. B;

13. h(t)  16t2  v t  h 0 0 When the ball hits the ground h(t)  0. 0  16t2  60t  100 0  4t2  15t  25 0  (4t  5) (t  5) 4t  5  0 or t  5  0 4t  5

7. B; c2  a2  b2 Let x  length of shorter leg x  4  length of longer leg 202  x2  (x  4) 2 400  x2  x2  8x  16 0  2x2  8x  384 0  2(x2  4x  192) 0  2(x  16)(x  12) x  16  0 or x  12  0 x  16 x  12 The shorter leg is 12 in. 8. A; c2  a2  b2 c2  122  52 c2  144  25 c2  169 2c2  1169 c  13 The distance is 13 yards. 9. D; 4, one in each quadrant.

5

t5

t  4

It will take the ball 5 seconds. 14. x2  8x  6  0 x

b  2b2  4ac 2a



8  2(8) 2  4(1) (6) 2(1)



8  164  24 2



8  140 2

x

8  140 2

or

x

 7.16 3

8  140 2

 0.84 3

15. 23181  23(9) 3  127 3

539

Chapter 11

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1 2 1 2 3 4 2 3

16. x

2x x



1 2

20. C; c2  a2  b2 c2  102  112 c2  100  121 c2  221 2c2  1221 c  1221

1

(x2 ) x2 1

x2 x1

1 2

x  x 1

x22  x

21a.

3

17.

 x2  x 1x A  /w 64  x

x3  1



6 7 3

Value of Column B 12y  16  8y 4y  16 y  4

start c2

7

  b2 2 c  7  92 c2  49  81 c2  130 2c2  1130 c  11.4 mi 21c. The sketch shows that the distance she is from her starting point is the length of the hypotenuse of a right triangle with legs 7 mi and 9 mi long. 21b.

2

A

m  B 2

3 Value of Column B Substitute 0 for x to find the y-intercept. 7(0)  4y  4 4y  4 y1

Chapter 11

finish

x  1 x

64  x2 164  2x2 8  x 18. B; Value of Column A 13x  12  10x  3 3x  12  3 3x  15 x  5 19. B; Value of Column A 2x  3y  10

( 1390 ) 2  a2  132 390  a2  169 221  a2 1221  2a2 1221  a

540

a2

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Chapter 12 Page 641 y 9

1.

Rational Expressions and Equations



2.

16(y)  (9)(7) 16y  63 63

3 15

1

n

1.1 0.6



4.

9.

10.

11.

12.

2.7 3.6



x 8



8.47 n

6.

8.1 a

9 8

0.21 2

y

6

0.19 2

6m 6

x

 24

6c2d2

3x 3





10(5  x)  14(x  3) 50  10x  14x  42 50  10x  50  14x  42  50 10x  14x  92 10x  14x  14x  92  14x 4x  92 4x 4



92 4

x  23 23. 6

1

7n  1 6 7n  1 6

5

2  5(6)

7n  1  30 7n  1  1  30  1 7n  31 7n 7

3c2d(1

  2d) 3c  6nm  15m2  3m(2n  5m) x2  11x  24  (x  3)(x  8) x2  4x  45  (x  9)(x  5) 2x2  x  21  2x2  6x  7x  21  (2x2  6x)  (7x  21)  2x(x  3)  7(x  3)  (x  3)(2x  7) 18. 3x2  12x  9  3x2  3x  9x  9  (3x2  3x)  (9x  9)  3x(x  1)  9(x  1)  (x  1)(3x  9)  3(x  1)(x  3) 3x  2  5 19. 3x  2  2  5  2 3x  3 13. 14. 15. 16. 17.

m  10 11

149 6 149 m 6 5  x 14  10 x  3

22.

2.7(a)  3.6(8.1) 0.19(24)  2(x) 2.7a  29.16 4.56  2x a  10.8 2.28  x 30  2  3  5 42  2  3  7 GCF  2  3 or 6 60r2  2  2  3  5  r  r 45r3  3  3  5  r  r  r GCF  3  5  r  r or 15r2 32m2n3  2  2  2  2  2  m  m  n n  n 12m2n  2  2  3  m  m  n GCF  2  2  m  m  n or 4m2n 14a2b2  2  7  a  a  b  b 18a3b  2  3  3  a  a  a  b GCF  2  a  a  b or 2a2b 2d



11(m  9)  5(m  10) 11m  99  5m  50 11m  99  99  5m  50  99 11m  5m  149 11m  5m  5m  149  5m 6m  149

9(6)  8(y) 54  8y 6.75  y 8.

m  9 5

21.

2(x)  8(0.21) 2x  1.68 x  0.84

1.1(n)  0.6(8.47) 1.1(n)  5.082 n  4.62 7.



2 10

x  20

3(n)  15(1) 3n  15 n5 5.

y x

2(x)  4(10) 2x  40

y  16 3.

20. 5x  8  3x  (2x  3) 2x  8  2x  3 2x  8  2x  2x  3  2x 8  3 No Solution

Getting Started 7 16



n

24. 9

1

4t  5 9 4t  5 9

7

2  7(9)

4t  5  63 4t  5  5  63  5 4t  58

31 7 31 7

4t 4

58

 4

t  14.5

2

x  x  56  0 (x  8)(x  7)  0 x  8  0 or x  7  0 x8 x  7 x2  2x  8 26. x2  2x  8  0 (x  4)(x  2)  0 or x  2  0 x40 x  4 x2 25.

3 3

x  1

541

Chapter 12

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7.

12-1 Inverse Variation Page 645

24 6

Check for Understanding

1. Sample answer: xy  8 2. Sample answer: Direct variation equations are in the form y  kx and inverse variation equations are in the form xy  k. The graph of a direct variation is linear while the graph of an inverse variation is nonlinear. 3. b; Sample answer: As the price increases, the number purchased decreases. 4. Solve for k. xy  k (24) (8)  k 192  k xy  192 Choose values for x and y whose product is 192. Sample answer: x 48 24 12 8 8 12 24 48

y 4 8 16 24 24 16 8 4

100

21.87 5.4

x 12 6 4 1 4 6 12

y 1 2 3 12 3 2 1

100 x

5120 8

Pages 645–647 11.

y

4



8x

4 8

8y2 8

9  y2 Thus, y  9 when x  8.

Chapter 12

8y2 8

542

Practice and Apply

xy  k (8)(24)  k 192  k Choose values for x and y whose product is 192. Sample answer: x 24 12 8 8 12 24

x1y1  x2 y2 (6) (12)  (8)(y2 ) 72  8y2 72 8



640  y2 Thus, the frequency of an 8-inch string is 640 cycles per second.

4 O

 x2

10. (length of string)  (frequency of vibrations) x1y1  x2 y2 (10) (512)  (8)(y2 ) 5120  8y2

xy  12

4

32x2 32

1

100

8



Thus, x  4 when y  32.

50

8

5.4x2 5.4

8  32x2

8 32 1 4

50 50



112 2 (16)  (x2)(32)

y

O

6y2 6

4.05  x2 Thus x  4.05 when y  5.4. 9. x y x y 1 1 2 2

xy  192

100 50



4  y2 Thus, y  4 when x  6. 8. x1y1  x2 y2 (8.1)(2.7)  (x2 )(5.4) 21.87  5.4x2

5. Solve for k. xy  k (2)(6)  k 12  k xy  12 Choose values for x and y whose product is 12. Sample answer:

6.

x1y1  x2 y2 (3)(8)  (6)(y2 ) 24  6y2

y 8 16 24 24 16 8

100

y xy   192

50 100 50 O 50 100

50

100 x

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12.

xy  k (4)(3)  k 12  k Choose values for x and y whose product is 12. Sample answer: x 6 4 3 3 4 6

y 2 3 4 4 3 2

8

15.

y

x 8 4 2 2 4 8

xy  12 4 8

4

O

4

xy  k (8)(9)  k 72  k xy  72 Choose values for x and y whose product is 72. Sample answer:

8x

4 8

y 9 18 36 36 18 9

y xy  72

16 8

16 8 O

16 x

8

8 16

13.

xy  k (15) (5)  k 75  k xy  75 Choose values for x and y whose product is 75. Sample answer: x 25 15 5 5 15 25

y 3 5 15 15 5 3

xy  75

16.

y 40

x 8 4 2 2 4 8

20 20 10 O

10

20 x

20 40

14.

y 2 4 12 12 4 2

xy  48

17.

y 20 10 20 10 O 10

10

20x

xy  19.44

3y2 3



20  y2 Thus, xy  60 and y  20 when x  3. 18. x1y1  x2 y2 (2)(7)  (7) (y2 ) 14  7y2

10 10

x y x y 1 1 2 2 (5)(12)  (3) (y2 ) 60  3y2 60 3

y 20

20 10 O

y 2.43 4.86 9.72 9.72 4.86 2.43

20

xy  k (12) (4)  k 48  k xy  48 Choose values for x and y whose product is 48. Sample answer: x 24 12 4 4 12 24

xy  k (8.1)(2.4)  k 19.44  k xy  19.44 Choose values for x and y whose product is 19.44. Sample answer:

20 x

14 7

10



7y2 7

2  y2 Thus, xy  14 and y  2 when x  7. 19. x1y1  x2 y2 (1)(8.5)  (x2 ) (1) 8.5  x2

20

8.5 1



x2 1

8.5  x2 Thus, xy  8.5 and x  8.5 when y  1. 20. x y x y 1 1 2 2 (1.55)(8)  (x2 ) (0.62) 12.4  0.62x2 12.4 0.62



0.62x2 0.62

20  x 2 Thus, xy  12.4 and x  20 when y  0.62.

543

Chapter 12

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21.

x1y1  x2 y2 (4.4)(6.4)  (x2 )(3.2) 28.16  3.2x2 28.16 3.2



28.

3.2x2 3.2

1.25 20

8.8  x2 Thus, xy  28.16 and x  8.8 when y  3.2. 22. x1y1  x2 y2 (0.5)(1.6)  (x2 )(3.2) 0.8  3.2x2 0.8 3.2



3.2x2 3.2

16 7



 y2 12 5

3

14 3 14 3

when x  5.

108 15

 7y2

1 2  3(7y2) 

21y2 21

 y2

Thus, xy  26.

12 1 2

14 3

220 65 220 65

 32y2 

t2 

32y2 32

t  2

 y2

x1y1  x2 y2 (6.1)(4.4)  (x2 )(3.2) 26.84  3.2x2 26.84 3.2



3.2x2 3.2

8.3875  x2 Thus, xy  26.84 and x  8.3875 when y  3.2.

Chapter 12

15h2 15



65t2 65

 t2

220 h 65 (220) (60) 65

min

t  203 min 2 t2  3 h and 23 min It takes about 3 hours and 23 minutes if they drive 65 miles per hour. 33. 4 h 0 min S 3 h 60 min  3 h 23 min  3 h 23 min 37 min If they drive at 65 miles per hour, they can save about 37 min.

1

Thus, xy  8 and y  4 when x  32. 27.



or

8  32y2 8 32 1 4

1.6p2 1.6

(55) (4)  (65)(t2 ) 220  65t2

2

and y  3 when x  7.

x y x y 1 1 2 2 (16)  (32) (y2 ) 16 2



7.2  h2 It takes 7.2 hours, if 15 students hand out 1000 flyers. 32. Let r1  55, t1  4, and r2  65. Solve for t2. r t r t 1 1 2 2

14  21y2 14 21 2 3

90w2 90

660  p2 The pitch of a tone with a wavelength of 1.6 feet is 660 vibrations per second. 31. Let s1  12, h1  9, and s2  15. Solve for h2. s h s h 1 1 2 2 (12) (9)  (15)(h2 ) 108  15h2

5y2 5

x y x y 1 1 2 2 2 (7)  (7) (y2 ) 3

12



1056 1.6

Thus, xy  12 and y  25.

when x  7.

x y x y 1 1 2 2 (2)(6)  (5)(y2 ) 12  5y2 12 5 12 5

20x2 20

8  w2 The width of the second rectangle is 8 inches. 30. (pitch)  (wavelength) Let p1  440, w1  2.4, and w2  1.6. Solve for p2. p1w1  p2w2 (440)(2.4)  (p2 )(1.6) 105.6  1.6p2

 y2

Thus, xy  16 and y  24.

720 90

7y2 7





0.0625  x2 Thus, xy  1.25 and x  0.0625 when y  20. 29. A    w Let 1  20, w1  36, and 2  90. Solve for w2. 1w1  2w2 (20) (36)  (90)(w2 ) 720  90w2

0.25  x2 Thus, xy  0.8 and x  0.25 when y  3.2. 23. x1y1  x2 y2 (4)(4)  (7) (y2 ) 16  7y2 16 7 16 7

x1y1  x2 y2 (0.5)(2.5)  (x2 )(20) 1.25  20x2

544

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42. A; Graphs B and D can be ignored since they are linear, showing direct variation. The point at (2, 4) is one solution to the equation, so it must be on the graph. Graph A contains this point and Graph C does not. Therefore, the correct choice is A.

34. If V is the volume of a gas at pressure P, then PV  k. or If V1 is the volume at pressure P1, and V2 is the volume at pressure P2, then P1V1  P2V2. 35. P1V1  P2V2 (1) (60)  (3)(V2 ) 60  3V2 60 3



Page 647

3V2 3

tan A 

20  V2 The new volume of the gas is 20 cubic meters. 36. P1V1  P2V2 (1) (22)  (0.8)(V2 ) 22  0.8V2 22 0.8





0.8V2 0.8

sin A 

The volume of the balloon at 0.8 atmosphere is 27.5 cubic meters. 37. Let w1  36, d1  8, and d2  20  d1. Solve for w2. w d w d 1 1 2 2 (36) (8)  (w2 )(20  8) 288  12w2 





1

y1 2

Use a calculator. 2nd [SIN1] 10 ⫼ 12 ENTER 56.44269024 The measure of the missing angle is 56. 45. Let A represent the missing angle.

12w2 12

cos A  

46.

1

1 y 2 1

 y2

Thus, if the value of x1 is doubled, the value of y 1 will be 2 of what it was. 39. Let y2  3y1 x1y1  x2 y2 x1y1  x23y1 x1y1 3y 1

x1 3



adjacent leg hypotenuse 3 10

Use a calculator. 1 KEYSTROKES: 2nd [COS ] 3 ⫼ 10 ENTER 72.54239688 The measure of the missing angle is 73.

2x1y2 2x

 y2 or

opposite leg hypotenuse 10 12

KEYSTROKES:

24  w2 The second piece of art must weigh 24 kg. 38. Let x2  2x1 x1y1  x2 y2 x1y1  2x1y2 x1y1 2x

opposite side adjacent leg 7 8

Use a calculator. 1 KEYSTROKES: 2nd [TAN ] 7 ⫼ 8 ENTER 41.18592517 The measure of the missing angle is 41. 44. Let A represent the missing angle.

27.5  V2

288 12

Maintain Your Skills

43. Let A represent the missing angle.

a d 3 12

f



b e 10 e

3 12



10 e



c



9 f

3(e)  (12) (10) 3e  120 3e 3

x23y1 3y 1



120 3

47.

a d a 21

 

a 21

Thus, if the value of y is tripled, the value of x will be one third of what it was. 40. Sample answer: When the gear ratio is lower, the pedaling revolutions increase to keep a constant speed. Answers should include the following. • Shifting gears will require that the rider increase pedaling revolutions. • Lower gears at a constant rate will cause a decrease in speed, while higher gears at a constant rate will cause an increase in speed. 41. B; xy  k (1.3)(4.25)  k 5.525  k

3 12



9 f

3(f )  (12) (9) 3f  108 3f 3

48.

545



108 3

c

f 

4 f

8

 28

28(a)  (8) (21) 28a  168 28a 28

e  40

1

 x2 or 3 x1  x2

b e 8 28



168 28

a6 8 28



4 f

8(f )  (4) (28) 8f  112 8f 8



112 8

f  36 f  14 A is 220  2  2  5  11 C is 264  2  2  2  3  11 C sharp is 275  5  5  11 GCF of A and C  2  2  11  44 GCF of A and C sharp  5  11  55 GCF of C and C sharp  11 Therefore, A and C sharp have the closest harmony. Chapter 12

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49.

7(2y  7)  5(4y  1) 7(2y)  7(7)  5(4y)  5(1) 14y  49  20y  5 14y  49  49  20y  5  49 14y  20y  5  49 14y  20y  20y  20y  54 6y  54 6y 6

50.

36  2  2  3  3 15  3  5 45  3  3  5 GCF  3 56. 48  2  2  2  2  3 60  2  2  3  5 84  2  2  3  7 GCF  2  2  3  12 58. 17a  17  a 57. 210  10  3  7

55.

54

 6

y  9 w(w  2)  2w(w  3)  16 w2  2w  2w2  6w  16 2  2w  w2  2w2  6w  16  w2 w 2w  w2  6w  16 2w  2w  w2  6w  16  2w 0  w2  8w  16 0  (w  4)(w  4) w40 w4404 w4

51.

330  10  3  11

34a2  2  17  a  a GCF  17a

150  10  3  5 GCF  10  3  30 59. 12xy2  2  2  3  x  y  y 18x2y3  2  3  3  x  x  y  y  y GCF  2  3  x  y  y  6xy2 60. 12pr2  2  2  3  p  r  r 40p4  2  2  2  5  p  p  p  p GCF  2  2  p  4p

y

12-2 Rational Expressions y  x  1

Page 651

x

O

Check for Understanding

1. Sample answer: Factor the denominator, set each factor equal to 0, and solve for x.

y  3x  5

2. Sample answer: 52.

y  2x  3 x

2y  5x  14

53.

2x 2

y

y0 x

xy1

2

54.

7. y 3x  2y  16

6 2

56x y 70x3y2

2



(14x y) (4) (14x2y) (5xy)



(14x y) (4) (14x2y) (5xy) 4 5xy

2

Exclude the values for which 70x3y2  0. 70x3  0 or y2  0 x3  0 y0 x0 The excluded values are x  0 and y  0.

x x  4y  4

Chapter 12





5x  8y  8 O

1 x2  11x  28

x  3 The excluded value is 3. 6. Exclude the values for which n2  n  20  0. (n  5)(n  4)  0 n50 or n  4  0 n  5 n4 The excluded values are 5 and 4.

xy3

O

or

3. Sample answer: You need to determine excluded values before simplifying. One or more factors may have been canceled in the denominator. 4. Exclude the values for which 3  a  0. 3a0 a  3 The excluded value is 3. 5. Exclude the values for which 2x  6  0. 2x  6  0 2x  6

y

O

1 (x  4) (x  7)

546

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8.

x2  49 x  7



(x  7) (x  7) x  7



(x  7 ) (x  7) (x  7 )

13.

b2  3b  4 b2  13b  36



(b  1) (b  4) (b  9) (b  4)



(b  1) (b  4 ) (b  9) (b  4 )



b  1 b  9

1

1

1

1

x7 Exclude the values for which x  7  0. x70 x  7 The excluded value is 7. 9.

x  4 x2  8x  16



x  4 (x  4) (x  4)



x  4 (x  4 ) (x  4)

Exclude the values for which b2  13b  36  0. b  9  0 or b  4  0 b9 b4 The excluded values are 9 and 4. 14. Let g  the number of guppies.

1

4 4  g

represents the fraction of neon fish in the aquarium. 15. Let g  the number of guppies. Double the guppy population, 2g, plus the four neon fish, plus five new fish can be represented 4 by 9  2g.

1

x

1  4

Exclude the values for which x2  8x  16  0. or x  4  0 x40 x  4 x  4 The excluded value is 4. 10.

x2  2x  3 x2  7x  12



(x  1) (x  3) (x  4) (x  3)



(x  1) (x  3 ) (x  4) (x  3 )

Pages 651–653

1

1

x  1  4

x

Exclude the values for which x2  7x  12  0. (x  4)(x  3)  0 x  4  0 or x  3  0 x4 x3 The excluded values are 4 and 3. 11.

a2  4a  12 a2  2a  8



(a  6) (a  2) (a  2) (a  4)



(a  6) (a  2 ) (a  2 ) (a  4)



a  6 a  4

1

1

Exclude the values for which a2  2a  8  0. a2  2a  8  0 (a  2)(a  4)  0 a  2  0 or a  4  0 a2 a  4 The excluded values are 2 and 4. 12.

2x2  x  21 2x2  15x  28



(2x  7) (x  3) (2x  7) (x  4)



(2x  7 ) (x  3) (2x  7 ) (x  4)

1

1

x  3  4

x

Exclude the values for which 2x2  15x  28  0. (2x  7)(x  4)  0 2x  7  0 or x  4  0 2x  7 x4 7

x2 The excluded values are

7 2

Practice and Apply

16. Exclude the values for which m  2  0. m20 m2 The excluded value is 2. 17. Exclude the values for which b  5  0. b50 b  5 The excluded value is 5. 18. Exclude the values for which n2  36  0. (n  6)(n  6)  0 n  6  0 or n  6  0 n6 n  6 The excluded values are 6 and 6. 19. Exclude the values for which x2  25  0. (x  5)(x  5)  0 x  5  0 or x  5  0 x5 x  5 The excluded values are 5 and 5. 20. Exclude the values for which a2  2a  3  0. (a  3)(a  1)  0 or a  1  0 a30 a  3 a1 The excluded values are 3 and 1. 21. Exclude the values for which x2  2x  15  0. (x  5)(x  3)  0 or x  3  0 x50 x  5 x3 The excluded values are 5 and 3. 22. Exclude the values for which n2  n  30  0. (n  6)(n  5)  0 or n  5  0 n60 n  6 n5 The excluded values are 6 and 5. 23. Exclude the values for which x2  12x  35  0. (x  5)(x  7)  0 or x  7  0 x50 x  5 x  7 The excluded values are 5 and 7.

and 4.

547

Chapter 12

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24.

35yz2 14y2z



(7yz) (5z) (7yz) (2y)



(7yz ) (5z) (7yz ) (2y)

Exclude the values for which 36mn3  12m2n2  0. 36mn3  12m2n2  12mn2 (3n  m) 12m  0 or n2  0 or 3n  m  0 m0 n0 3n  m The excluded values are m  0, n  0, and 3n  m.

1

1

5z

 2y Exclude the values for which 14y2z  0. 14y2  0 or z  0 y2  0 y0 The excluded values are y  0 and z  0. 25.

14a3b2 42ab3



(14ab2 ) (a2 ) (14ab2 ) (3b)



(14ab ) (a2 ) (14ab2 ) (3b)



a2 3b

1

30.

2

31.

(16qrs) (4r) (16qrs) (q)



(16qrs ) (4r) (16qrs ) (q)



4r q

32.



(3xyz) (3x) (3xyz) (8z)



(3xyz ) (3x) (3xyz ) (8z)



3x 8z



(z  2 ) (z  8) z  2

4x  8 x2  6x  8



4(x  2) (x  4) (x  2)



4(x  2 ) (x  4) (x  2 ) 1

x

33.



(7ab) (a2b) (7ab) (3a  7b2 )



(7ab ) (a2b) (7ab ) (3a  7b2 )

2(y  2) (y  5) (y  2) 1 2(y  2 ) (y  5) (y  2 ) 1

2  5

Exclude the values for which y2  3y  0. (y  5)(y  2) or y  2  0 y50 y  5 y2 The excluded values are 5 and 2.

1

a2b

 3a  7b2

34.

Exclude values for which 21a2b  49ab3  0. 7a  0 or b  0 or 3a  7b2  0 a0 3a  7b2

m2  36 m2  5m  6

 

(m  6) (m  6) (m  1) (m  6) 1 (m  6) (m  6 ) (m  1) (m  6 ) m  6

1

m1

7

a  3b2

Exclude the values for which m2  5m  6  0. (m  1)(m  6) m10 or m  6  0 m  1 m6 The excluded values are 1 and 6.

7

The excluded values are a  0, b  0, and a  3b2. (3mn2 )mn  4m(mn2 )

 12n(3mn2 )

(3mn2 )mn  4m(3mn2 )

 12n

1

(3mn2 )mn  4m(3mn2 )

 12n



y

1

3m2n3 36mn3  12m2n2

2y  4 y2  3y  10



Exclude the values for which 24xyz2  0. 24x  0 or y  0 or z2  0 x0 z0 The excluded values are x  0, y  0, and z  0. 7a3b2 21a2b  49ab3

4  4

Exclude the values for which x2  6x  8  0. (x  4)(x  2)  0 x40 or x  2  0 x  4 x  2 The excluded values are 4 and 2.

1

1

mn  4m

 12n Chapter 12

(z  2) (z  8) z  2

1

1

29.



1

Exclude the values for which 16q2rs  0. 16q2  0 or r  0 or s  0 q2  0 q0 The excluded values are q  0, r  0, and s  0.

28.

z2  10z  16 z  2

z8 Exclude the values for which z  2  0. z20 z  2 The excluded value is 2.

1

9x2yz 24xyz2

(x  5 ) (x  4) x  5

1

1

27.



1

2



(x  5) (x  4) x  5

x4 Exclude the values for which x  5  0. x50 x  5 The excluded value is 5.

Exclude the values for which 42ab3  0. 42a  0 or b3  0 a0 b0 The excluded values are a  0 and b  0. 64qr s 16q2rs



1

1

26.

x2  x  20 x  5

548

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35.

a2  9 a2  6a  27



(a  3) (a  3) (a  9) (a  3)



1 (a  3) (a  3 ) (a  9) (a  3 )



2(2x2  3x  2) 2(x2  4x  4) 2(2x  1) (x  2) 2(x  2) (x  2)

a  3 a  9



1 1 2 (2x  1) (x  2 ) 2 (x  2) (x  2 )



2x  1 x  2

40.

4x2  6x  4 2x2  8x  8

1



1

Exclude the values for which a2  6a  27  0. (a  9)(a  3) a90 or a  3  0 a  9 a3 The excluded values are 9 and 3. 36.

x2  x  2 x2  3x  2





(x  2) (x  1) (x  2) (x  1)



1 (x  2) (x  1 ) (x  2) (x  1 )



x  2 x  2

Exclude the values for which 2x2  8x  8  0. 2(x  2)(x  2) x20 x2 The excluded value is 2. 41.

3m2  9m  6 4m2  12m  8

 

1

3

4 Exclude the values for which 4m2  12m  8  0. 4(m  2)(m  1) m20 or m  1  0 m  2 m  1 The excluded values are 2 and 1. 42. a  4500  1000  4.5 43. 3500  1000  3.5

(b  4) (b  2) (b  4) (b  16) (b  4) (b  2) (b  4) (b  16)

Exclude the values for which b2  20b  64. (b  4)(b  16) b  4  0 or b  16  0 b4 b  16 The excluded values are 4 and 16. 38.

x2  x  20 x3  10x2  24x

t  

1 (x  4 ) (x  5)  4 ) (x  6)

 x(x

t

1

x  5 x(x  6)

n2  8n  12 n3  12n2  36n

40(37.95)



(n  2) (n  6)  36) (n  2) (n  6) n(n  6) (n  6)



1 (n  2) (n  6 ) n(n  6) (n  6 )

 40.97 about 41 min to cook a potato. 45. The times are not doubled even though the altitude is. The difference between the times is 12 minutes.

 n(n2  12n

n  2  6)

40 [ 25  1.85(7) ] 50  1.85(7)

 50  12.95

Exclude the values for which x3  10x2  24x  0. x(x  4)(x  6) x  0 or x  4  0 or x  6  0 x  4 x  6 The excluded values are 0, 4, and 6. 39.

40 [ 25  1.85(3.5) ] 50  1.85(3.5) 40(25  6.475) 50  6.475 40(31.475) 43.525

 28.93 about 29 min 44. 7000  1000  7

(x  4) (x  5)  4) (x  6)

 x(x



1

3(m  3m  2 )

 4(m2  3m  2 )

Exclude the values for which x2  3x  2  0. (x  2)(x  1) x  2  0 or x  1  0 x2 x1 The excluded values are 2 and 1. b2  2b  8 b2  20b  64

3(m2  3m  2)

 4(m2  3m  2) 2

1

37.

1

46. MA 

s  r r

47. MA 

17.5  0.4 0.4

MA  42.75 48. Force on lid  MA  force applied  42.75  6  256.5 lb 450  4n 49. 450  4n 50. n

1

 n(n

Exclude the values for which n3  12n2  36n  0. n(n  6) (n  6) n  0 or n  6  0 n6 The excluded values are 0 and 6.

51. 15 

450  4n n

15n  450  4n 11n  450 n  40.9 41 students must attend. 52.

54.

450  4(n  2) n

 4

53.

Area Circle Area Square



x2 (2x) 2



x2 4x2



 4

 0.785  79%

549

Chapter 12

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61.

55a. Sample answer: The graphs appear to be identical because the second equation is the simplified form of the first equation. 55b. Sample answer: The first graph has a hole at x  4 because it is an excluded value of the equation. 56. Sample answer: Use the rational expression for light intensity to help determine the brightness of the picture on the screen for the distance between the projector and the screen. Answers should include the following. • Find the solutions for the expression in the denominator. • Use the light intensity expression to determine the brightness of a search light. 57. C; The other expressions can be factored. 58. B;

7.5 8

4.374606646 The measure of V is 4. 65. sin A  0.7011 1 KEYSTROKES: 2nd [SIN ] .7011 ENTER 44.51532391 The measure of A is 45.

Maintain Your Skills

xy  k (10) (6)  k 60  k xy  60 x1y1  x2 y2 (10) (6)  (12) (y2 ) 60  12y2 60 12



66.

12y2 12

112 2 (16)  k k

8k xy  8 x1y1  x2 y2 16 2

 (x2 )(32)  32x2

Check:

8  32x2 8 32 1 4



12z  2  z  3 ?

12(1)  2  1  3

32x2 32

?

14  2 2  2 ✕

 x2

The only solution is 7.

1

Thus, x  4 when y  32.

Chapter 12

?

1a  3  2 ? 11  3  2 ? 14  2 22 67. 12z  2  z  3 ( 12z  2) 2  (z  3) 2 2z  2  z2  6z  9 2z  2  2z  2  z2  6z  9  2z  2 0  z2  8z  7 0  (z  1) (z  7) z  1  0 or z  7  0 z1 z7

Thus, y  5 when x  12. 60. xy  k

1 (16) 2

1a  3  2 ( 1a  3) 2  22 a34 a1 Check:

5  y2

16 2

8y2 8

13.57393947 The measure of N is 14. 63. cos B  0.3218 1 KEYSTROKES: 2nd [COS ] .3218 ENTER 71.22818376 The measure of B is 71. 64. tan V  0.0765 1 KEYSTROKES: 2nd [TAN ] .0765 ENTER

x2  6x  5  0 (x  5)(x  1)  0 x  5 or x  1 5 and 1 are excluded values.

59.



0.9375  y2 Thus, y  0.9375 when x  8. 62. sin N  0.2347 1 KEYSTROKES: 2nd [SIN ] .2347 ENTER

x2  3x  2 x2  6x  5

Page 653

xy  k (3)(2.5)  k 7.5  k xy  7.5 x y x y 1 1 2 2 (3)(2.5)  (8) (y2 ) 7.5  8y2

550

12z  2  z  3 ?

12(7)  2  7  3 ?

116  4 44✓

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68.

74. A  /w A  (2x  y) (x  y)  2x2  2xy  xy  y2  2x2  3xy  y2 The area is 2x2  3xy  y2 square units.

113  4p  p  8 113  4p  p  p  8  p 113  4p  8  p 13  4p  (8  p) 2 13  4p  64  16p  p2 13  4p  13  64  16p  p2  13 4p  51  16p  p2 4p  4p  51  16p  p2  4p 0  p2  20p  51 0  ( p  3)( p  17) or p  17  0 p30 p  3 p  17 113  4p  p  8

Check:

?

113  4(3)  (3)  8

7

84 in. 1

76.

4.5 m 1

77.

113  4p  p  8

78.

?

181  17  8 ?

538

9  17  8

88✓

12 in. 1 ft



84 in. 1

1 ft

12 in.  7 ft

100 cm  m

1

4h 1

 450 cm

1



60 min h 15 min 1

60 s

 min  14,400 s 60 s

 min  900 s

14,400 s  900 s  15,300 s

?

125  3  8



1

113  4(17)  (17)  8

?

79.

26  8 ✕

80.

The only solution is 3. 69.

75.

18 mi 5280 ft  mi  95,040 ft 1 3 days 24 h  day  72 h 1 220 mL 1000 mL 220 mL   1 1 L

L

1000 mL 

220 L 1000

 0.22 L

23r2

 61  2r  1  61) 2  (2r  1) 2 3r2  61  4r2  4r  1 0  4r2  4r  1  3r2  61 0  r2  4r  60 0  (r  6)(r  10) r  6  0 or r  10  0 r6 r  10 ( 23r2

23r 2  61  2r  1

Check:

23(6) 2

?

 61  2(6)  1

Page 654

1

1.

?

1108  61  12  1

3x  6 x2  7x  10



3(x  2 ) (x  2 ) (x  5)



3 x  5

1

When x  5, x  5  0. When x  2, x  2  0. Therefore, x cannot equal 2 or 5 because you cannot divide by zero.

23r 2  61  2r  1 23(10) 2

Graphing Calculator Investigation (Follow-Up of Lesson 12-2)

?

 61  2(10)  1 ?

1300  61  20  1

?

?

1169  13

1361  21

13  13 ✓

19  21 ✕

The only solution is 13. 70. Common factor:

3 1

3

27 3  81 81 3  243 243 3  729 Thus, the next three terms are 81, 243, and 729. 71. Common factor:

24 6

[10, 10] scl: 1 by [10, 10] scl: 1

2.

4

1



(x  8 ) (x  1) (x  8 ) (x  8) 1

x  1  8

x

384 4  1536 1536 4  6144 6144 4  24,576 Thus, the next three terms are 1536, 6144, and 24,576. 1

x2  9x  8 x2  16x  64

When x  8, x  8  0. Therefore, x cannot equal 8 because you cannot divide by 0.

1

72. Common factor: 2  4  2 2 (2)  4 4 (2)  8 8 (2)  16 Thus, the next three terms are 4,8, and 16. 73. Common factor: 27 16 81 64 243 256

3

81

3

243

3

729

[5, 15] scl: 1 by [10, 10] scl: 1

3 4

4  64 4  256 4  1054

Thus, the next three terms are

81 243 , , 64 256

729

and 1054.

551

Chapter 12

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3.

1

5x2  10x  5 3x2  6x  3

5(x2  2x  1 )  1)

 3(x2  2x

10.

1 24 feet 1 second



60 seconds 1 minute

60 minutes 1 hour



1 mile

 5280 feet 220

1

5

1  60  60  1 mile

3

 1  1  1 hour  220

2

When x  1, x  2x  1  0. Therefore, x cannot equal 1 because you cannot divide by 0.

3600 miles 220 hour



 16.36 mph 11.

4,000 240,000 miles 1

1 hour

1 day

 60 minutes  24 hours 1

40 4,000  1  1  1 day 1  100  1  24 40 days 24 2 13 days

  

[10, 10] scl: 1 by [10, 10] scl: 1

1 minutes 100 miles



1

4.

2x  9 4x2  18x

2x  9  9)

 2x(2x

Pages 657–659

1

1

 2x

12.

a. Sample answer: Examine the values and verify that they are identical. b. It displays ERROR.

8 x2

2x

4

x

8x

 4x  4x3 1

 2x 13.

10r3 6n3

2 1 , 1 x

15.

2. Sample answer: When the negative sign in front of the first expression is distributed, the numerator is x  6. 3. Amiri; Sample answer: Amiri correctly divided by the GCF. 4.

64y 5y

5y

 8y 

8 y 2

64y 5y

16.

17. 15 s2 t 3 12 s t 3 11



7.

m  4 3m

x2  4 2

 (m

x

1 3

2 4t

n2  16 n  4

1

1

1 m  4 3m



4m 3(m  5)

1

1 (x  2) (x  2 ) 2

n  2  16



m 4m2

19.

 (m  4) (m  5)

x

2 4  2

20.

12 g

1

24 g2 h 5 1 b

12ag 5 (x  4) (x  3) (x  8)



(n  1) (n  1) (n  1)

 (n

(z  4) (z  6) (z  6) (z  1)



1 (x  8 ) (x  8) (x  3)



x  4 x  8

(z  1) (z  5)  4)



 (n

x  5  5

x





n  2  4) (n  4) 1

21.

1 x  5 (x  5) (x  2) 1

x2  25 9

1

y2  4 y2  1

y  1  2

y

1

1 (z  4 ) (z  6) (z  6) (z  1)



(z  6) (z  5) (z  6) (z  3)

1

(x  1) (x  7) (x  7) (x  4)



(x  1) (x  7) (x  7) (x  1)

1

x  5  5

x

1

(x  5) (x  5) 9 (x  5) 2 9

1 (y  2 ) (y  2) (y  1) (y  1)

1 y  1  2

y

1

y

(x  3) (x  2) 5

x  3 5

22.

1 x2  x  12

x  3  5

x

 

552

(n  4)  1) (n  4) 1

1 (x  4) (x  3) 1 1 (x  4) (x  5)

1 x  3  5

x

1 (z  1 ) (z  5)  4)

 (z  3) (z



1 (x  5) (x  5 )

1

 (n

n  4 n  4

y  2  1

1





1 (x  4) (x  3 ) (x  8) 1

1 1 (n  1 ) (n  1 ) (n  1)





(x  4) (x  10)  1) (x  10)

 (x



1

x  x  6 5

(x  1) (x  7) (x  7) (x  4)



1

(n  4)  1) (n  4)

 (z  3) (z

1

1 (n  4 ) (n  4) n  4



Chapter 12

1 z2

 15 a b2

1

n  2  4



(x  8) (x  8) (x  3)

1





1

 n2  8n

x  5 x2  7x  10

1

y

2 s 1

n 9.

1 x

2t s



2

5

3 a2 b 2gh

1

4 1 t

 2(x  2) 8.

1 a

24g2h

 15ab2 

w

16 s t2  10 s3 t 3

18.

4m2  4) (m  5)

4  2



4wy 5xz2

4

5y

1 1



120 w2 x2 y3 z2 150 w x3 y2 z 4

 8y 3 1 1

16st2  10s3t3

3a2b 2gh



1 1 1

 3  2s

6.

12w2x2 25y2z4



 8y 5.



n

1

1

15s2t2 12st

10y3z2 6wx3

1

1

2 n

Check for Understanding

1. Sample answer:

1

1

420 n2 r3

 14.

2

2

42n2

 35r3  210 n3 r3

12-3 Multiplying Rational Expressions Page 657

Practice and Apply

4

1

1 1 (x  4 ) (x  10 )  1) (x  10)

 (x

1

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23.

x  6 x2  4x  32

x  4  2

x



1

 24.

x  3 x  4

x

 x2  7x

 12

1 x  3  4

 n2

n  8n  15



2n  10 n2

1 x  4  2

x

36.

x  6 (x  8) (x  2)

x 

25.

x  6 (x  8) (x  4)

1

37.

x (x  4) (x  4) x (x  4) 2 n (n  5) (n  3)





1

1 2(n  5 ) n2

 26.

b2  12b  11 b2  9

b  9  99

 b2  20b

1 (b  11 ) (b  1) (b  3) (b  3) b  1 (b  3) (b  3)

 

27.

28.

a2  a  6 a2  16



a2  7a  12 a2  4a  4

2.54 centimeters 1 inch



12 inches  1 foot

1



(a  3) (a  2 ) (a  4) (a  4)



(a  3) (a  3) (a  4) (a  2)

1

1 b  9

 (b  11) (b  9) 1

1

40. D;



42.

165

43.

14 feet 1



1 yard 3 feet 1



1 yard 3 feet 1

6



18 dollars yard2

44.

 336 It will cost $336 to carpet the room. 33.

21.95 Canadian dollars 1 21.95 U.S. dollar  1.37

1 kilowatt

 1000 watts



8x2z2 2y3

41. A;

4a  4 a2  a

a2

 3a  3



104x3y2z3 8x2y4

 a(a  1)  3(a  1)

4(a  1)  a2



13xz3 y2

 3a(a

s  6 s2  36

4aa  1) 4a 3(a  1)

Maintain Your Skills 

s  6 (s  6) (s  6)

a2  25 a2  3a  10



(a  5) (a  5) (a  2) (a  5)

Exclude the values for which a2  3a  10  0. a  2  0 or a  5  0 a2 a  5 The excluded values are 2 and 5.

1

4

8

Exclude the values for which s2  36  0. or s  6  0 s60 s  6 s6 The excluded values are 6 and 6.





13xyz 4x2y

Page 659 1 mile

 5280 feet

18 feet 3 feet 1 yard3  1  27 feet3 1 20 540 feet3 1 yard3  27 feet3 1

12 feet 1

1 hour

 60 minutes  1



 20 yard3 32.

60 watts light 15 cents 1 dollar  1 kilowatt  hour  100 cents

1 minute

 21.82 miles/hour 31.

1

1

1 32 feet 60 seconds 60 minutes   1 hour 1 second 1 minute 1  60  60  1 mile  1  1  1 hour  165 3600 miles  165 hours

10 feet 1

1 vehicle 30 feet

 60 minutes  60 seconds

 16.67 m/s 30.



• Sample answer: converting units of measure

3 feet  1 yard

1 hour

176 5280 feet 1 mile

25 lights  h hours 

1

 91.44 cm/yd 29.



6864 vehicles 24 seconds 1 minute  1 vehicle  60 seconds 1 3 6864  24  1  1 hour 1  8 1  1  60  60 20,592  3600



1

(a  3) (a  4)

 (a  2) (a  2)

2.54  12  3 centimeters yard

1 60 kilometers 1000 meters  1 kilometer 1 hour 1  1000 meters  1  1  1  1  1  60 seconds

13 miles 1 lane

 5.72 It will take about 5.72 hours. 38. c and e; Sample answer: The expressions each have a GCF that can be used to simplify the expressions. 39. Sample answer: Multiply rational expressions to perform dimensional analysis. Answers should include the following.

2n  3) 2n n  n(n  3) 2 n(n  3)

 n2 (n 



 3  13  176  1 vehicle  6864 vehicles 6864 vehicles are involved in the backup.

x  4) (x  3)

 (x

3 lanes 1

x  3 x3  6x  9



x  3 (x  3) (x  3)

Exclude the values for which x2  6x  9  0. x30 x  3 The excluded value is 3.

1 U.S. dollar

 1.37 Canadian dollars

 16.02 Johanna spent about $16.02 in U.S. dollars. 34.

18 miles 3 hours



6 18 hours 1

 108 miles

108 miles of streets can be cleaned. 35. 3 lanes 

13 miles 1 lane



5280 feet 1 mile



1 vehicle 30 feet

553

Chapter 12

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45.

xy  k (8) (9)  k 72  k xy  72 x y x y 1 1 2 2 (8)(9)  (x2 )(6) 72  6x2 72 6



Check:



3.5r 3.5



712 79

54.

51.

24a3b4c7 6a6c2

g 8

52. g 8

6

8

1 6

24a3b4c7  2 6a6  3c2



4b4c5 a3

3.375 6 3.5 ✓

3.625 3.5

Let r  2.1.

Let r  2.2.

3.5r 7.35

3.5r 7.35

3.5r 7.35

3.5(2.1) 7.35

3.5(2.2) 7.35

3.5(2.0) 7.35

9k 7

3 5 3 4 5 1 12 5

9k 7

12 1 5 9

Let r  2.0.

7.7 7.35 ✓

7 7.35

127 12 7

20p6 8p8

3

Check:

4

3

5

Let k  15 .

Let k  15 .

Let k  15 .

? 12 7 5

? 12 7 5

? 12 7 5

9k

115 2 7 125

3 4

9

?

5

12 5



12 5

9k

115 2 7 125

3 3

9

?

5

9 5



12 5

9k

115 2 7 125

3 5

9

?

5

15 5

7

12 5



55. Let m  money earned and d  number of days. 935  85x m  kx 340  k(4) 11  x 85  k It will take 11 days to earn $935. 56. x2  3x  40  (x  8) (x  5) 57. n2  64  (n  8)(n  8) 58. x2  12x  36  (x  6) (x  6) or (x  6) 2 59. a2  2a  35  (a  7) (a  5) 60. 2x2  5x  3  (2x  1) (x  3) 61. 3x3  24x2  36x  3x(x2  8x  12)  3x(x  2) (x  6)

Page 659 1.

5

20p6

 8p8  6 2

5

 2p2

1

7 2 4 7 8  2 1 1

554

Practice Quiz 1

xy  k (7)(28)  k 196  k Choose values for x and y whose product is 96. x 2 4 7 7 4 2

{g|g 6 28}

Chapter 12

7 2

7.35 3.5

12 12 4 5k|k 7 15 6

4



29 ? 7 6 2 8

6

4

3.2x2 3.2

50.

9k 4 9k 4 4 1

1 9

4y2 4

 73 or 343

6

7.35 7.35 ✓

3.6y2 3.6

 7129

7 2

{r|r 2.1}

8.8  x2 Thus, x  8.8 when y  3.2. 49.



Check:

48  y2 Thus, y  48 when x  4. 48. xy  k (4.4)(6.4)  k 28.16  k xy  28.16 x1y1  x2 y2 (4.4)(6.4)  x2 (3.2) 28.16  3.2x2 28.16 3.2

g 8

6

3.5 3.5

5.4  y2 Thus, y  5.4 when x  3.6. 47. xy  k (8)(24)  k 192  k xy  192 x1y1  x2 y2 (8)(24)  (4)(y2 ) 192  4y2 192 4

Let g  27.

g 8

7 2

53. 3.5r 7.35

12  x2 Thus, x  12 when y  6. 46. xy  k (8.1)(2.4)  k 19.44  k xy  19.44 x1y1  x2 y2 (8.1)(2.4)  (3.6) (y2 ) 19.44  3.6y2 

Let g  29.

g 8

28 ? 7 6 2 8

6x2 6

19.44 3.6

Let g  28.

y 98 49 28 28 49 98

y 160 120 80 40 8642O 80 120 160

2 4 6 8x

xy  196

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2.

xy  k (9) (6)  k 54  k Choose values for x and y whose product is 54. x 2 4 6 6 4 2

3.

y 27 13.5 9 9 13.5 27



4.

1

y  3y2 3y  1

4a 7b

a  7

2a

a  3  7

1

2a

a7 6.

xy  54

3m  15 m  4

m  5



3m  15 6m  24  m5 m  4 3(m  5) 6(m  4)  m5 m  4



3(m  5 ) m  4

 6m  24 

x

1

2 4 6 8

1

7.

2x  6 x  5

1



y(1  3y ) (1  3y )

 

2(x  3 ) x  5

x 8.

1

k  3 k2  4k  4



2k  6 k  2

3m2 2m





k  3

5a  10 10x2

(3n  1 ) (n  2) (3n  1 ) (n  4)



k  3 (k  2) (k  2 )

n  2  4

 2(k  2)

4n  8 n2  25

18  m  m 9m



9.

2x  4 x2  11x  18

4x3

 

10.

x2  x  6 x2  9



1

5(a  2) 2x2  5

2x  9 4(n  2) (n  5) (n  5) 4 5(n  5)

x2  7x  12 x2  4x  4

 



x2  5x  6 x1 2(x  2) (x  2) (x  3)  (x  9) (x  2) x1



2(x  2) (x  9) (x  2 )



2(x  2) (x  3) (x  9) (x  1)

x1

2x  4

 x2  5x  6  x2  11x  18 

1

 a2  11a  18  n  5  10

k  2

 2(k  3 )

1

1

 5n

1

1

1

3mm 2m

k  2

 2(k  3)

1

1

2x2  2x

 (a  9) (a  2)



1

a

9.

k  2

 k2  4k  4  2k  6

 3m2 8.

1

k  3 (k  2) (k  2)

2

18m2 9m

1  3



n 7.

x

2  5

1



1

1

1

1

b  1  9

3n2  5n  2 3n2  13n  4

6(m  4 ) m  5

2x  6 1  x  5 x  3 2(x  3) 1 x  3 x  5

 (x  3) 

b 6.

1



 18

(b  4 ) (b  1) (b  9) (b  4 )



a  3  7 1

y

b2  3b  4 b2  13b  36

2a

 a  3  a  3a  a  3a

40 60 80

(7a ) (4a) (7a ) (7b)

2a a  3

y

8642O

1

28a2 49ab

 5.

80 60 40 20

5.

n  5  2)

 5(n

(x  3) (x  2) (x  3) (x  3) x  4 x  2

10.

85 kilometers 1 hour



(x  4) (x  3)  2) (x  2)

 (x

 



1000 meters 1 kilometer

1 hour

(x  2 ) (x  3) x  1

1 minute

 60 minutes  60 seconds

(85 kilometers) (1000 meters) (1 hour) (1minute) (1 hour) (1 kilometer) (60 minutes) (60 seconds) 85  1000 meters  1  1 1  1  60  60 seconds 85,000 meters 3600 seconds

 23.61 m/s 11.

12-4 Dividing Rational Expressions

32 pounds 1 square foot



Page 662

15z 4y2

32 pounds

3x

 4y

2

 9 pound/square inch

2. Sometimes; Sample answer: 0 has no reciprocal. 3. Sample answer: Divide the density by the given volume, then perform dimensional analysis. 4.

10n3 7



5n2 21



10n3 7



10n 7

2n

1

3

(32 pounds) (1 square foot) (1 square foot) (144 square inches)

 144 square inches

Check for Understanding

1. Sample answer:

1 square foot

 144 square inches

12. There are 32 ounces in a quart, The pan can hold 2.32  64 ounces.

21  5n2 3

21

 5n2

If the pan is candy in it.

2 3

128 3

3 

3

4

128 3

2

full, there are 64  3  4

512 9

128 3

ounces of

 57

Latisha will make about 57 pieces of candy.

1

 6n

555

Chapter 12

PQ249-6481F-12[541-568] 26/9/02 6:33 PM Page 556 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

Pages 662–664 13.

a2 b2

a b3



 

a2 b2 a 2

a b2

b3 a



14.

n4 p2



n2 p3

b 3

8x2 y2







x



3

4x y4 y2

n4 p2

y2

 8x2

16.

10m2 7n2



25m4 14n3

1 2

y

17.



 

2

a4bc3 g2h2

ab2c2 g3h2



19.

b2  9 4b

10m2 7n2 2

10m 7n2 1

x2y3z s2t2

2n

14n

27.

 25m4

a4bc3 g2h2



a bc g 2h2

1 mi

60 s

 5280 ft  1 min 

60 min 1h 1

1



(330 ft ) (1 mi) (60 s ) (60 min ) (1s ) (5280 ft ) (1 min ) (1 h)



1,188,000 mi 5280 h

1

1

1

 220 mi/h 1

28.

z2



330 ft 1s

cm3

1

s t  x2yz3

y2s z2

1m 1m 1m 1 (0.35 m3 ) (100 cm) (100 cm) (100 cm) 1 m3

5m2

s 3 2

x y z s2t2

100 cm 100 cm 100 cm m3   

 350,000

3

s3t2  x2yz3

y2 2 3

1730 plants 1 km2

km

km

 1000 m  1000 m 

a3c 4 3 1 3

g3h2

 ab2c2

 0.00173 plants/m2

g 3 2

29.

g h

 ab2c2

2x  6 x  5

x

2  5



2x  6 x  5



2 (x  3) x  5



1

 

(b  3) (b  3 ) 4b

30.

1 b  3

m  8 m  7

5d d  3

m  8 m  8 1

m  8

23.

3x  12 4x  18



2x  8 x  4



(m  4 ) (m  4) 5m

 31.

x2  2x  1 2

x  1  1

x

1

m  4

d

 

(x  1) (x  1 ) 2



(x  1(x  1) 2

1

1

32.

n2  3n  2 4



n  1 n  2

1  1

33.

1

3(x  4 )  9)

a2  8a  16 a2  6a  9

x  4  4)

 2(x



2a  4 a  4

4a  8 2



1

a  4  2 (a  2 )

34.

1



(n  2) 2 4

n  2  1

n

1



a2  8a  16 3a   a2  6a  9 2a  (a  4) (a  4) 3(a  (a  3) (a  3) 2(a



(a  4 ) (a  4) (a  3 ) (a  3)



3(a  4) 2(a  3)



9 8  3)  4) 1

b  2 b2  4b  4



2b  4 b  4

b  2

3(a  3 )  4)

 2(a

1

b  4

 b2  4b  4  2b  4 1

a  4  3

a

Chapter 12

(n  2) (n  1 ) 4

1

a  4

1



1

 2a  6  2a  4 4 (a  2 ) 2(a  3)



1

3(x  4)  9)

4a  8 2a  6

1

n2  3n  2 n  2 n  1 4 (n  2) (n  1) n  2 n  1 4



2a  8  9

 3a

 4(x 24.

x  1  1

x

1

3x  12 x  4   18 2x  8 3(x  4) x  4  2(x  9) 2(x  4)

 2(x

1 (m  7) (m  1)

x2  2x  1 x  1 x  1 2 (x  1) (x  1) x  1 x  1 2



 4x 

1

1

5d (d  3) (d  1)



1

 m  7  (m  8 ) (m  1)

1



5d  3

1

 m  7  (m  8) (m  1)

m2  16 1 m  4 5m (m  4) (m  4) 1 m  4 5m

 (d  1)  d

1

 m2  7m  8  m  7  m2  7m  8

1 b  3

m  4  5m 3k 1  k  1 k  2 3k (k  1) (k  2)

 (k  2) 

x  5 2

x3

b  3 4b

 (m  4) 

 22.



1

b2  9 1 b  3 4b (b  3) (b  3) 4b

1

3k k  1

x  5 2

1

b

a cg b



21.

1

1730 plants

 (b  3) 

m2  16 5m

1

(1730 plants) (1 km ) (1 km ) (1 km2 ) (1000 m) (1000 m)

 1,000,000 m2

1

20.

1

ft3

1

14n3

 25m4

4n







 5m2

1

18.

 

x

x2yz3 s3t2

26. 0.35

1

3

(24 yd ) (3 ft) (3 ft) (3 ft) 1 yd3

 648

2

 8x2

3 ft

p 3

p

 n2

n2p

 2y2 x2y3z s2t2

4

n p2

3 ft

25. 24 yd3  1 yd  1 yd  1 yd 

p3  n2

2

1



4x3 y4

3 ft

n

b a

 ab 4x3 y4



1

1

15.

1

Practice and Apply

556



b  2 (b  2) (b  2)



b  4 2(b  2) 2

b  4

 2(b  2 ) 1

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35.

x2  x  2 x2  5x  6

x2  2x  3  12

 x2  7x



x2  x  2 x2  7x  12   6 x2  2x  3 (x  2) (x  1) (x  4) (x  3)  (x  3) (x  2) (x  3) (x  1)



(x  2 ) (x  1 ) (x  3) (x  2 )



x  4 x  3

1

1

1

1

36.

x2  2x  15 x2  x  30

x2  3x  18 x2  2x  24



45. Sample answer: Divide the number of cans 5 recycled by 8 to find the total number of cans produced. Answers should include the following.

 x2  5x

(x  4) (x  3 )

 (x  3 ) (x  1 ) 1

46. B;



x2  2x  15 x2  2x  24  x2  x  30 x2  3x  18 (x  5) (x  3) (x  6) (x  4)  (x  6) (x  5) (x  6) (x  3)



(x  5 ) (x  3) (x  6 ) (x  5 )



1



60 minutes 1 hour

1

V 



18b 15c



3b 5c



3b 5c

 18b

3

1

1

(x  6 ) (x  4)

 (x  6) (x  3)

3

15c 6

A  /w x2  4 

x2  4 x  1  x2  x  2 1 (x  2) (x  2) x  1  (x  2) (x  1) 1 (x  2) (x  2) x  1  (x  2) (x  1) 1

60

 6.5  9.2

1 yd3

Page 664

39. An equation that represents the number of truck loads.

48.

15 ft (18 ft2  15 ft)  9 ft  271 ydft 2 742.5 ft 1 yd  20,000 yd3  1 1  27 ft 2 742.5 yd  20,000 yd3  1 27 2

x  2 1



3

3

49.

x2  3x  10 x2  8x  15



x  5 (x  5 ) (x  2)

x2  5x  6  4

1

rev

30 in. 1 rev

x  4 4y

1

2y

2y  x 2xy

(x  5) (x  2) (x  5) (x  3)



(x  5 ) (x  2) (x  5 ) (x  3 )

(x  3) (x  2)  2)

 (x  2) (x 1

1

1

(x  3 ) (x  2 )  2)

 (x  2) (x

x  4 4y 1

4



x  4 4y

16y (x  4 ) (x  3)

16y

1 hr  60 min



1

x

(12) (5280) (55) (26) (60)

1 ft

51.

1

x2  8x  15 x  y

7x  14y x  3



1 mi

 12 in.  5280 ft  63.5 mph

x

52.

c  6 c2  12c  36

2y  x 2xy

1

x  2y . x2  4y2 1

21

3

43. Volume of new block  x x  2 x  4

 (10  85 pounds) 



 3)

1

4  3



(x  5) (x  3) x  y



(x  5) (x  3 ) (x  y)



7(x  2y) x  3



7(x  2y) x  3



7(x  2y) (x  5)  x  y c  6 (c  6) (c  6)



c  6 (c  6 ) (c  6)

1

1

is not equivalent to

44. Weight of original block

16y

 (x  4) (x

1

 2y  2xy  2xy 



 x2  7x  12 

42. d; Sample answer: 1 x

1

1

55 mi  1 hr

60 min 1h

x  2 1

x  2  2

50.





x

 711 rpm 41. 1 revolution  circumference  d  30 711min 

x  2 1

1

27.5 yd3

5280 ft  1 mi



1

 x2  4x

It will take 727.27 truck loads. 40. 1 revolution  circumference  d  26. 12 in.  1 ft

w

x  5 (x  5) (x  2)

1

 727.27

1 rev 26 in.

w



 20,000 yd3  27.5 yd3 

w

1

3

3

20,000 yd3 1 20,000 27.5

w

Maintain Your Skills

x  5 x2  7x  10

3



x2  x  2 x  1

x2w

 9 ft  27 ft3

3

15c

 18b

47. C;

(x  3) (x  4) (x  6) (x  3)

n  20,000 yd3 

1 pound

 6 or 2

Jorge should ride at 9.2 miles per hour. 38. There are 27 cubic feet in 1 cubic yard. d(a  b) w 2 5 ft(18 ft  15 ft) 2

3b 5c

1

37. Convert minutes per mile into miles per hour. 1 mile 6.5 minutes

5

x  63,900,000 cans  8  33 cans

1

1





1

1

2

c 53.

25  x2 x2  x  30

1  6



1(x  5) (x  5) (x  6) (x  5)



1(x  5) (x  5 ) (x  6) (x  5 )



x  5 x  6

1

1x  12 21x  34 2 (x)

1

x3

557

Chapter 12

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54.

a  3 a2  4a  3



a  3 (a  3) (a  1)



a  3 (a  3 ) (a  1)

61. The degree of z3  2z2  3z  4 is 3, since 3 is the greatest sum of a term’s exponents. 62. The degree of a5b2c3  6a3b3c2 is 10, since 10 is the greatest sum of a term’s exponents. 6  0.8 g 63.

1

1

1

a1 55.

n2  16 n2  8n  16



(n  4) (n  4) (n  4) (n  4)



(n  4 ) (n  4) (n  4 ) (n  4)

6 0.8

1

n  4  4

n 56. 3y2  147 y2  49 y  7 {7} Check:

15b 15

3(7) 2  147 3(49)  147 147  147 ✓ 3(7) 2  147 3(49)  147 147  147 ✓ 57. 9x2  24x  16 9x2  24x  16  0 (3x  4)(3x  4)  0 3x  4  0 3x  4

56

x

Check:

9 1

5b|b 7 6

Check:

1 2 1

12

24 4 1 3

0.049 0.07

15b 6 28

15b 6 28

1 2 6 28 28 15

?

15

1 2 6 28 29 15

15

29 6 28 ✓

12715 2 6 28

27 28



0.07x 0.07

?

?

?

0.049  0.07(0.6) 0.049  0.042 ✓ 66.

3 h 7 7 3

3 49 3 7 49 3 1 7 1 7

5h|h 6 6

Check:

1

Let h  7. 3 h 7

6

3 49

1 2 6 493

3 1 7 7

3 49

2 ?

[ (6  114)  6]  14 ? (6  3.742  6) 2  14 14.003  14 ✓ ?

is 0, since there are no

558

0.049  0.07(0.8) 0.049  0.056

137h21 2 6 1 21 2 6

h 6

?

[ (6  114)  6] 2  14 ? (6  3.742  6) 2  14 14.003  14 ✓

Chapter 12

15b 6 28

0.049  0.07(0.7) 0.049  0.049 ✓ Let x  0.6. Let x  0.8. 0.049  0.07x 0.049  0.07x

 16

1

1 8

27

Let b  15 .

0.7  x {x|x 0.7} Let x  0.7. Check: 0.049  0.07x

(15) 2  225  30(15) ? 225  225  450 450  450 ✓

60. The degree of 13  exponents.

29

Let b  15 .

28 28

59. (n  6) 2  14 n  6  114 n  6  114 {6  114} Check:

28

65. 0.049  0.07x

16  32  16 16  16 ✓ 58. a2  225  30a a2  30a  225  0 (a  15) (a  15)  0 a  15  0 a  15 {15} Check:

Let g  7. 6  0.8 g ? 6  0.8(7) 6  5.6

Let b  15 .

15

4 3

8



Let g  8. 6  0.8 g ? 6  0.8(8) 6  6.4 ✓

28 15 28 15 28 15

7

b 7

143 22  24143 2  16

9 16 1 9

0.8 g 0.8

7.5  g {g|g 7.5} Let g  7.5. Check: 6  0.8 g ? 6  0.8(7.5) 66✓ 64. 15b 6 28

1

4 3



?

3

49

2

Let h  7. 3 7

h 6

3 49

1 2 6 493

3 2 7 7

6 49

?

3

49

1

Let h  8. 3 7

h 6

3 49

1 2 6 493

3 1 7 8

3 56

6

3 49



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67.

12r 4

7

3 20

70.

3

12r 4

7

3r 7 3r 3

7

3 20 3 20 3 20

6x2 x4

1

6  x2

 x 2  x2 1

 72.

3 1 r 6 20

18a3 45a5

1

1

1

12

4 1



Let

12r 4 1 20

1

4 1 12 20 4 12 1

4 1



2

74. 7 ?

7

2

3 20

1

29aaa 1

1

1

73.

1

b6c3 b3c6

7

3 20

1

4

2

?

7

3 20



4 1

1181 2 7 1220 ?

12

12

1

Let r  22. 12r 4 1 22

2

4 

1

1

12 22 4 12 1

3

2

4

1bb 21cc 2 6

3

3

6

75.

7x4z2 z3

b3 c3

 7(x4 )  7(x

1zz 2 2 3

4 ) (z23 )

 7(x4 ) (z 1 ) 

7x4 z

2

7 ?

7 ?

7

7.

3x  6 9

9.

5x2  25x 10x

3 20

 

3 20 3 20

Reading Mathematics



4 1

10.

1221 2 7 1220

3(x  2) 9 x  2 3

4n  12 8

 

4(n  3) 8 n  3 2

1

?

12

8.

5x(x  5)  10x x  5  2 x  3 x  3  (x  4) (x  3) x2  7x  12



x  3 (x  4) (x  3 ) 1

12

22 20



1  4 x  y (x  y) (x  y)



x  y (x  y) (x  y )



1 x  y

x

1

2

1 2 12  61

11.

x  y x2  2xy  y2

1

y 3 {y|y 3} Check: Let y  3. y 6

1

1. Sample answer: the quantity m plus two, divided by 4 2. Sample answer: three x divided by the quantity x minus 1, 3. Sample answer: the quantity a plus 2 divided by the quantity a squared plus 8 4. Sample answer: the quantity x squared minus 25 divided by the quantity x plus 5 5. Sample answer: the quantity x squared minus 3 x plus 18 divided by the quantity x minus 2 6. Sample answer: the quantity x squared plus 2 x minus 35 divided by the quantity x squared minus x minus 20

18 7 20 ✓

4 1

11228 21xx 21yy 2 3  1 7 2 (x34 ) (y21 ) 3  1 7 2 (x1 ) (y) 

Page 665

2

1

1





 7x

4  1



m3 5

?

12 18

12

12x3y2 28x4y

5mmmm 55m

 (b63 ) (c36 )  (b3 ) (c3 )

3y

?

12 12 20 20 1 r  18. 12r 3 7 20 4 1 12 18 ? 3 7 20 4

12 1

y 6 6 y 1 6

1

 59aaaaa

3 20

1201 2 7 1220

1

68.

1

1



1

6 x2

2

Let r  20.

1

5m4 25m

 5a2

5r|r 6 201 6

Check:

71.

1

2

3 ? 1 2 6 1 1 2✓ 2

1

Let y  4.

Let y  2.

y 6

y 6

1

2

4 ? 1 2 6 2 1 2✓ 3

12.

1

2

x2  16 x2  8x  16



(x  4) (x  4) (x  4) (x  4)



(x  4 ) (x  4) (x  4 ) (x  4)

1

2 ? 1 2 6 1 1 2 3

1

x  4  4

x

69. Multiply 3250 by 12. 3250 covers 1 month

 12 months  39,000 covers

559

Chapter 12

PQ249-6481F-12[541-568] 26/9/02 6:33 PM Page 560 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

12-5

Dividing Polynomials

Page 667

Algebra Activity

x2

1. (x2  3x  4)  (x  1) Step 1:

x2

x

Step 3:

x2

1 1

x

x 3

x

x x

x x x

1 1 Step 2:

1 1 1

1 1 1

x1 The width of the array, x  3, is the quotient. 3. (x2  16)  (x  4)

x2

Step 1:

1

1 1 1 1 1 1 1 1 1 1 1

x2

1 1 1 1 1 Step 2: Step 3:

x1 x4

x4

x2

x

x x x x

1 1 1 1

x2 1 1 1 1

The width of the array, x  4, is the quotient. 2. (x2  5x  6)  (x  2) Step 1:

x2

1

1

1

1

1

1

Step 3:

x4

x x x x x

x4

Step 2:

x x x x

x x

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

x x

x2

x2

x2

x x

The width of the array, x  4, is the quotient. 4. (2x2  4x  6)  (x  3) Step 1:

x2

x2

x x x x

1 1 1 1 1 1

Chapter 12

560

PQ249-6481F-12[541-568] 26/9/02 6:33 PM Page 561 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

Step 2:

5.

14a2b2  35ab2  2a2 7a2b2



14a2b2 7a2b2



14a2b2 7a2b2



35ab2 7a2b2



35ab2 7a2b2

2

x3

5

2

1 1 1

x3 x x x

x2

x x x

x x

1 1 1 1 1 1

The quotient is r  3  r 8. 2m

The width of the array, 2x  2, is the quotient. 5. You cannot do it. There is a remainder.

Page 669

Check for Understanding

2x  3 b. x  32x2  9x  9 ()2x2  6x 3x  9 ()3x  9 0

x3 c. 2x  32x2  9x  9 ()2x2  3x 6x  9 ()6x  9 0

x6 d. 2x  32x2  9x  9 ()2x2  3x 12x  9 ()12x  18 27

2m2  3m  7  0m2  5m  21  6m2 6m2  5m () 6m2  9m 14m  21 ()14m  21 0

 34m3 ()4m3

3

The quotient is b  2  2b  1. 10. Let p  0.75. C

120,000(0.75) 1  0.75

 360,000 The company will have to pay $360,000.

Pages 669–671

Practice and Apply

2

11. (x  9x  7)  3x  

x2  9x  7 3x x2 9x 7   3x 3x 3x x 7  3  3x 3

Therefore, b and c are the divisors that give a remainder of 0. 2. Sample answer: A remainder of zero means that the divisor is a factor of the dividend.

12. (a2  7a  28)  7a 

3. Sample answer: x3  2x2  8; x3  2x2  0x  8



4. (4x3  2x2  5)  2x   



4x3  2x2  5 2x 4x3 2x2 5   2x 2x 2x 2x2

4x 3 2x 1

 13.

x



2x2 2x 1

9 .  9

The quotient is 2m2  3m  7. b2 9. 2b  12b2  3b  5 ()2b2  b 4b  5 ()4b  2 3

1. Test each divisor to see which divisors give a remainder of 0. 2x  15 a. x  32x2  9x  9 ()2x2  6x 15x  9 ()15x  45 54



7b2

2 7b2

r3 7. r  9r2  12r  36 ()r2  9r 3r  36 () 3r  27 9

Step 3:

2x  2

5 a

2a2 7a 2b2

n4 6. n  3n2  7n  12 ()n2  3n 4n  12 () 4n  12 0 The quotient is n  4.

x2

x2

2



a

1

x2

2a2

 7a2b2

5

 2x

9s3t2  15s2t  24t3 3s2t2



15s2t 3s2t2

 3s2t2 

15s2t 3s2t2



9s3t2 3s2t2

a2  7a  28 7a a2 7a 28  7a  7a 7a a 4 1a 7

3

5

9s3t2

5

1

 2x2  x  2x

3

561

5 t

t



24t3

 3s2t2 8t

24t3

 3s2t2 s2

8t s2

Chapter 12

PQ249-6481F-12[541-568] 26/9/02 6:33 PM Page 562 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

14.

12a3b  16ab3  8ab 4ab



12a3b 4ab



12a3b 4ab



16ab3 4ab



16ab3 4ab

3a2

4b2

1

1

6n  3 24. 2n  512n2  36n  15 ()12n2  30n 6n  15 ()6n  15 0 The quotient is 6n  3.

8ab

 4ab 2

8ab

 4ab 1

 3a2  4b2  2 x8 x4 15. x  5x2  9x  20 16. x  2x2  6x  16 ()x2  5x ()x2  2x 4x  20 8x  16 ()4x  20 ()8x  16 0 0 The quotient is x  4. The quotient is x  8. s2 n7 17. n  5n2  2n  35 18. s  9s2  11s  18 ()n2  5n ()s2  9s 7n  35 2s  18 ()7n  35 ()2s  18 0 0 The quotient is n  7. The quotient is s  2.

3x2  2x  3 25. x  2  8x2  x  7 () 3x  6x2 2x2  x () 2x2  4x 3x  7 () 3x  6 1 3x3 3

The quotient is 3x2  2x  3  x 5b2  3b  1 26. 4b  27b2  13b  3  15b2 12b2  13b () 12b2  9b 4b  3 () 4b  3 0 The quotient is 5b2  3b  1.  3 20b3 () 20b3

z9 19. z  7z2  2z  30 ()z2  7z 9z  30 ()9z  63 33 33

3x2 27. 2x  36x3  9x2  0x  6 () 6x3  9x2 6

The quotient is z  9  z  7. a7 20. a  3a2  4a  22 ()a2  3a 7a  22 ()7a  21 1 The quotient is a  7  a

6

The quotient is 3x2  2x  3. 3g2  2g 28. 3g   0g2  5g () 9g  6g2 6g2  5g () 6g2  4g 9g () 9g 29g3 3

1 .  3

2r  7 21. r  52r2  3r  35 ()2r2  10r 7r  35 ()7r  35 0

3 8

8 6 2

The quotient is 3g2  2g  3  3g

The quotient is 2r  7. 3p  2 22. p  63p2  20p  11 ()3p2  18p 2p  11 ()2p  12 1

29. 2n

1

The quotient is 3p  2  p  6.

3n2  2n  3  5n2  0n  12  9n2 4n2  0n () 4n2  6n 6n  12 () 6n  9 3

t2  4t  1 30. 4t  17t2  0t  1  t2 16t2  0t () 16t2  4t 4t  1 () 4t  1 0 The quotient is t2  4t  1.  14t3 () 4t3

562

2 .  2

 36n3 () 6n3

The quotient is 3n2  2n  3  2n

t6 23. 3t  43t2  14t  24 ()3t2  4t 18t  24 ()18t  24 0 The quotient is t  6.

Chapter 12

1 .  2

3 .  3

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31. Let W  150 and L  60.

210(72  20) 20

 546 He could lift a 546-lb rock. 33. Perimeter of bedroom  14(12)  12(12)  14(12)  12(12)  34.5  34.5  42  42  471 inches Convert 471 inches into yards and divide by 5. 5 yd roll

64

10w3  23w2  5w  2 2w  1

64C d2

15w2

35. Size 10-inch Price $4.99 Number of slices 5 Cost per slice $1.02 $4.99

1

(10) 2 64

2

$12.99

1(18) 64 2 2

 $1.02

14-inch $8.99 10 $0.93

14-inch:

 $0.82

18-inch $12.99 16 $0.82

$8.99

1(14) 64 2 2

mass volume

Concrete

Lead

Brass

Blood

 B

 9w 2w  110w3  23w2  5w () 10w3  5w2 18w2  5w () 18w2  9w 4w () 4w

2 2

2 2 0 Area of the base: 5w2  9w  2

 $0.93

1

The area of a triangle is equal to 2bh. 1

5w2  9w  2  2 b(5w  1) 1

2(5w2  9w  2)  2  2 b(5w  1)

The 18-inch pizza offers the best price per slice. 36.

Iron

10w3  23w2  5w  2  B(2w  1) (10w3  23w2  5w  2)  B(2w  1)

s  C  d2

18-inch:

6 4

38. The densities are clustered around 9. 39. Find the area of the base using the formula V  Bh.

1d64 2

10-inch:

8

Aluminum

 2.62

Therefore, Anoki needs to buy 3 rolls of border. C 34. s  2

s

10

2 0

1 yd

471 in.  36 in.

12

Copper



Steel

W(L  x) x

Density (g/cm3)

32. Let W  210 pounds, L  6  12  72 inches, and x  20 inches.

20 18 16 14

Silver



37.

150(60  x) x

Gold

W(L  x) x

10w2  18w  4 5w  1

 density

b

2w 5w  110w2  18w () 10w2  2w 20w () 20w

4.15 Al  1.54  2.69 g/cm3 2.32 gold  0.12  19.33 g/cm3 6.30 silver  0.60  10.5 g/cm3 7.80 steel  1  7.8 g/cm3 15.20 iron  1.95  7.79 g/cm3 2.48 copper  0.28  8.86 g/cm3 4.35 blood  4.10  1.06 g/cm3 11.30 lead  1  11.3 g/cm3 17.90 brass  2.08  8.61 g/cm3 40 concrete  20  2 g/cm3

4 4 4 4 0

The base is 2w  4. 40.

x2  7x  12 x  k



(x  4) (x  3) (x  k)

In order for the quotient to have no remainder, k must be either 3 or 4. 41.

563

x5 x  2x2  7x  k () x2  2x 5x  k () 5x  10 k  10  2 In order for the quotient to have a remainder of 2, k  10  2. The value of k is 12.

Chapter 12

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x9 x  7x2  2x  k () x2  7x 9x  k () 9x  63 k  63  0 In order for the quotient to have a remainder of 0, k  63  0. The value of k is 63. 43. Sample answer: Division can be used to find the number of pieces of fabric available when you divide a large piece of fabric into smaller pieces. Answers should include the following. • The two expressions are equivalent. If you use the Distributive Property, you can separate the numerator into two expressions with the same denominator. • When you simplify the right side of the equation, the numerator is a  b and the denominator is c. This is the same as the expression on the left. 44. D; A    w m2  4m  32    (m  4) m8 m  4m2  4m  32 () m2  4m 8m  32 () 8m  32 0

51. 172  132  262  2  242  2  612  412  (6  4) 12  1012 52. 112  118  148  222  3  232  2  242  3

x2  3x  14 45. B; x   0x2  5x  20 () x  3x2 3x2  5x () 3x2  9x 14x  20 () 14x  42 22

58. (3x2  4xy  2y2 )  (x2  9xy  4y2 )  (3x2  x2 )  (4xy  9xy)  (2y2  4y2 )  4x2  13xy  2y2

42.

 213  312  413  (2  4) 13  312  613  312

53. d2  3d  40  d2  8d  5d  40  (d2  8d)  (5d  40)  d(d  8)  5(d  8)  (d  8) (d  5) 54. x2  8x  16  x2  4x  4x  16  (x2  4x)  (4x  16)  x(x  4)  4(x  4)  (x  4) 2 55. This polynomial is prime since t2  t  1 cannot be factored. 56. Solve for x. 6x  11  150 6x  139 x  23 boxes The greatest number of cards Mr. Martinez can have printed is 23  100 or 2300 cards. 57. (6n2  6n  10m3 )  (5n  6m3 )  [10m3  (6m3 ) ]  6n2  [ 6n  5n ]  4m3  6n2  n

3x3 3

59. (a3  b3 )  (3a3  2a2b  b2  2b3 )  [ a3  (3a3 ) ]  2a2b  b2  [b3  (2b3 ) ]  2a3  2a2b  b2  3b3

60. (2g3  6h)  (4g2  8h)  2g3  4g2  [ 6h  (8h) ]

The quotient is x2  3x  14 

Page 671 46.

2

x  5x  6 x2  x  12

 2g3  4g2  2h

22 . x  3

Maintain Your Skills x  2  20

 x2  x

2

x  5x  6  12

 x2  x 1



x  x  20 x  2

1

(x  3 ) (x  2 ) (x  4 ) (x  3 )



1

1



Page 674

(x  5) (x  4) x  2

m2  m  6 m 2  8m  15



m2  m  2 m 2  9m  20



m2  m  6 m 2  8m  15



(m  3 ) (m  2 ) (m  5 ) (m  3 )



m  4 m  1

1

b2  19b  84 b  3

1

1

(m  5 ) (m  4 )  2 ) (m  1 )

 (m

1

b2  9

 b2  15b  36 

1

1

(b  12 ) (b  7) b  3 1

1

(b  3 ) (b  3 )

 (b  12 ) (b  3 ) 1

1

b7

49.

z2  16z  39 z2  9z  18

1

z  5

 z2  18z  65 

1

(z  13 ) (z  3 ) (z  6) (z  3 ) 1

1

z  5

 (z  5 ) (z  13 ) 1

1

5.

1

z6

a  2 4



a  2 4

 

50. 317  17  (3  1) 17  217 Chapter 12

x  6 x  2

x  4  2

x

 1.

2. Sample answer: When you add rational expressions with like denominators, you combine the numerators and keep the common denominator. This is the same process as adding fractions with like denominators. 3. Sample answer: Two rational expressions whose sum is 0 are additive inverses, while two rational expressions whose difference is 0 are equivalent expressions. 4. Russell; sample answer: Ginger factored incorrectly in the next-to-last step of her work.

m 2  9m  20 m2  m  2

1

1

48.



Check for Understanding

1. Sample answer:

1

x5 47.

Rational Expressions with Like Denominators

12-6

2



564

a  2  a  2 4 2a 4 a 2

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6.

3x x  1

x

3  1

 

3x  3 x  1 3(x  1) x  1

23.

3 7.

8.

9.

11.

12.

2  n n  1

24.

2  n  1   n1 3  n n1 4t  1 2t  3 4t  1  2t  3  1  4t  1  4t 1  4t 6t  2  1  4t 5a 7a 5a  7a 7 4  12  12 10. n  3  n  3 12 2a  12 a  6 3m 6 3m 6  2  m  m  2  (m  2) m  2 3m 6 m2m2 3m  6  m2 2

x x  y

1 n  1

x

2

y  y

 

2

25.

26.  

7  4 n  3 3 n  3

27.

x  y x  y (x  y) (x  y) (x  y)

  

Pages 675–676 14.

m 3



2m 3

 

45  29  10  12 960 96 960 1 10

15.

12z 7



5z 7

  

m x  3 5



x  2 5

 

17.

n  7 2



n  5 2



a  5 6

18.



6 y  3

 

2y  6 y  3 2(y  3) y  3

19.

3r r  5

2 20.

k  5 k  1

4

k1 

32.

21.

22.

4x  5 x  2



1 n  3

 

x  3 x  2

 





a  3 6



2 x  7



5 x  7

33.

4 z  2

34.

5 3x  5

r

15  5

 



 

4n  2n 3 2n 3

(a  5)  (a  3) 6 a  5  a  3 6 2 6 1 3 2  2  x  7 x 

x 

7 5 7

x

5  7

7

6 z  2



4 6 z2z2 10 z2 3x 5  3x  3x  5 3x  5 (3x  5)  3x  5

 1

3r  15 r  5 3(r  5) r  5

35. 36.

3

k  5  4 k  1 k  1 k  1



3x 7



1 n  2 n  3





n1 2y y  3

11x  5  11x  12 2x  5 22x  7 2x  5

31.

5x 7



n  7  n  5 2 2n  2 2 2(n  1) 2





30.

x  3  x  2 5 2x  5 5





11x  12 2x  5



12z  (5z) 7 12z  5z 7 7z 7

z 16.

11x  5 2x  5

5x  3x 4n 2n 29. 3  3 7 2x  7 x  4 x  2 (x  4)  (x  2)  5  5 5 x  4  x  2  5 2 5

28.

Practice and Apply

m  2m 3 3m 3

2a  3  a  2 a  4 3a  1  a4 5s  1 3s  2 5s  1  3s  2  2s  1  2s  1 2s  1 8s  1  2s  1 9b  3 5b  4 9b  3  5b  4  2b  6  2b  6 2b  6 14b  7  2b  6 12x  7 9x  5 12x  7 9x  5  2  3x  3x  2  (3x  2) 3x  2 12x  7 9x  5  3x  2  3x  2 (12x  7)  (9x  5)  3x  2 12x  7  9x  5  3x  2 3x  2  3x  2





xy 13.

a  2  4

a

1

2

number of students absent total number of students

2a  3 a  4

37.

n  2  (1) n  3 n  3 n  3 4x  5  x  3 x  2 5x  2 x  2

565

4 7m 4  7m  7m  2  7m  2 7m  2 2x 2x 2x 2x  2  x  x  2  (x  2) x  2 2x 2x x2x2 4x x2 5y 5y 5y 5y  3  y  y  3  (y  3) y  3 5y 5y y3y3 5y  5y  y3 10y y3

Chapter 12

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38.

8 3t  4

 3t

6t  4

8  6t  4 2(3t  4) 3t  4

47. For figure a:

 3t 

area 

x2

 16

 2 39.

3  1

15x 5x  1

 5x

15x 3  5x  1  1 15x  3 5x  1 3(5x  1) 5x  1

perimeter  x

 5x  

40.

ratio 

41.

b  15 2b  12





For figure b:

10a 2a 10a 2a 10a

area 

x2

perimeter  x ratio 

x x  18

For figure c:





x2

perimeter  x

x x  24

180 3n 60 n

48. Figure a has the greatest ratio since it has the smallest number in the denominator. 49. Put the numbers in the same form. a.

1

b.

area of one wall  area of trapezoid  area of triangle area of one wall 1 1 (25  10)  2 (10)(20  5)  2 (15)(10)

1

25  10

250  125  75 250 50 1 or 5 250

2 1

2

45. There are 2 points in a quart and 4 quarts in a gallon. 1 ft3 7.48 gal

12(1 qt)  4(1 pt) 121 ptqt 22  14gal qt 3

1 ft

 7.48 gal (2 qt  2 qt)  3

1 ft

1

k  2

1 gal 4 qt

3  2 3 x  2



d. x

3  2



3  7



k  2  3 k  7 k  1 k  7

5r 52. B; perimeter of rectangle ABCD  2 1 2r 9r  2 1 2r  6s 2  6s 2

1gal

r

1

 7.48 ft3



1

 7.48 ft3  8.3 lb

Chapter 12

3

c. 2  x  x



1

62.4 lb ft3

3  2 3 x  2

 x

51. A; k  7  k

 7.48 gal (4 qt )  4 qt

46.

3 2  x 3 x  2

Number c is not equivalent to the others. 50. Sample answer: Since any rational number can be expressed as a fraction, values on graphs can be written as rational expressions for clarification. Answers should include the following. • The numbers in the graphic are percents that can be written as rational expressions with a denominator of 100. • To add the rational expressions, factor 1 out of either denominator so that it is like the other.

fraction of walls painted red



124x 2 2

44. The area of a trapezoid is equal to 2h(b  b ) 1 2 1 and the area of a triangle is 2bh.



1 x x 2 4 3

 24

The fraction of the trees that could be planted on 60 Monday, Tuesday, and Saturday is n .



1 21 2 1 21 2

1 3x 4x 12 12

area  2

ratio 





118x 2 2

8,695 269,817 140  20  20 n  2n

13x 216x 2

 18

42. 80 years or older: 8,634,000  61,000 total population: 77,525,000  79,112,000  68,699,000  35,786,000  8,634,000  61,000

43.

x

x

 16

 12 6a  (2a  6)  6  12 6a   2a  6  6  12  6a  2a  6 16a  12  2a  6 2(8a  6)  2(a  3) 8a  6  a3 3b  8 (b  15)  (3b  8)  2b  12 2b  12 b  15  3b  8  2b  12 4b  23  2b  12



6a 6  2a

116x 2 2

3 10a  12 2a  6

14x 22

r

9r  5r r  3s

r

566

9r  3s

14r  3s

5r  3s

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Page 677

66.

Maintain Your Skills

x2  2x  3 53. x  2x3  0x2  7x  6 () x3  2x2 2x2  7x () 2x2  4x 3x  6 () 3x  6 0 The quotient is x2  2x  3.

68.

69.

8x2 9 54. 7x  456x3  32x2  63x  36 () 56x3  32x2 63x  36 () 63x  36 0

70.

71.

The quotient is 8x2  9. 55.

b2  9 4b

 (b  3) 

b2  9 4b

1

b  3 1

56.

x x  2





(b  3) (b  3 ) 4b



b  3 4b

x2 x2  5x  6



b

x2  5x  6  x2

1



1

Page 677

x x  2

x

1  3

67. 45  3  3  5 55 10  2  5 6  23 6  23 15  3  5 LCM  2  3  3  5  90 LCM  2  3  5  30 8  222 9  33 12  2  2  3 LCM  2  2  2  3  3  72 16  2  2  2  2 20  2  2  5 25  5  5 LCM  2  2  2  2  5  5  400 36  2  2  3  3 48  2  2  2  2  3 60  2  2  3  5 LCM  2  2  2  2  3  3  5  720 9  33 16  2  2  2  2 24  2  2  2  3 LCM  2  2  2  2  3  3  144

1.

1

x  2 1



a a  3

Practice Quiz 2



a  11 a  3

a

a  3

a

a  3

 a  3  a  11 1

(x  3) (x  2 ) x2

 a  3  a  11

x

1

a

x  3 x

 a  11

57. a2  9a  14  a2  7a  2a  14  a(a  7)  2(a  7)  (a  2)(a  7)

2.

4z  8 z  3



4z  8 1  z  3 z  2 4(z  2) 1 z  2 z  3



4(z  2 ) z  3

 (z  2) 

1

58. p2  p  30  p2  6p  5p  30  p( p  6)  5( p  6)  ( p  5)( p  6) 3.

(2x  1) (x  2) (x  2) (x  3)



(2x  1) (x  5) (x  3) (x  1)



(2x  1 ) (x  2 ) (x  2 ) (x  3 )

(x  3) (x  1)  1) (x  5)

 (2x

1

1

1

1

(x  3 ) (x  1)  1 ) (x  5)

 (2x

1

x  1  5

 (4x  9x)  7

4. (9xy2  15xy  3)  3xy 

61. (5x2  6x  14)  (2x2  3x  8)  (5x2  2x2 )  (6x  3x)  (14  8)  7x2  3x  22 62. 1 foot  12 inches 8

(2x  1) (x  2) (x  2) (x  3)

x

 3x2  5x  7

riser tread



1

60. (3x2  4x)  (7  9x)  (3x2  4x)  (7  9x) 

1

4

z3

59. y2  11yz  28z2  y2  7yz  4yz  28z2  y(y  7z)  4z(y  7z)  (y  4z) (y  7z) 3x2

1

z  2

9xy2 3xy



15xy 3xy

3

 3xy 1

 3y  5  xy x5 5. 2x  32x2  7x  16 () 2x2  3x 10x  16 () 10x  15 1

2

 12  3

4  22 9  33 12  2  2  3 LCM  2  2  3  3  36 64. 77 21  7  3 55 LCM  3  5  7  105 65. 6  23 12  2  2  3 24  2  2  2  3 LCM  2  2  2  3  24 63.

1

The quotient is x  5  2x  3. y  15 6. y  4y2  19y  9 () y2  4y 15y  9 () 15y  60 51 The quotient is y  15  y

567

51 .  4

Chapter 12

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7.

8.

2  5 x  7 7 x7 6 2m 6 m3m3 m  3 2m  6  m3 2(m  3)  m3

2 x  7

x

2m m  3



5  7

9. The LCD is (y  5)(y  5).



2y y2  25

2 9.

5x  1 3x  2



2x  1 3x  2

  

1

3

2

1

3 34

in.

2

11. The LCD is 3z 6w2

16.5 minutes  1

z

 4a 2a  6



a

3  3

 

13. The LCD is (b  4)(b  4). b  8 b2  16

b

1  4



 

a a  4

a  4 a  4

b  4

1

b  8

  

b  4 (b  4) (b  4) b  8  (b  4) (b  4) (b  4) b  8  b  4 (b  4) (b  4) 4 (b  4) (b  4)

14. The LCD is (x  2)(x  5). 1)2(n

x x  2

 4)

3

 x2  3x

 10

x

x  2

x  5  5

x

3

 x2  3x



x2  5x (x  2) (x  5)



x2  5x  3 (x  2) (x  5)



 10

3 (x  2) (x  5)

15. C; The LCD is ( y  4)( y  3). 2y y2  7y  12



2

a  4a (a  4) (a  4)

y  2  4

y

2y

 y2  7y



 y2  7y

a  4 a  4

4a  16 (a  4) (a  4)

2



a  4a  4a  16 (a  4) (a  4)



a2  8a  16 (a  4) (a  4)

Pages 681–683

568

y  2  4

 12

y

 12



2y

4 a  4

16.

Chapter 12

b  8

 b2  16  b  4  b  4  b2  16 

6 2x 7   10x2 5x 2x 12x 7  10x2 10x2 12x  7 10x2

4 a  4

4a 3 2  a  32  6 4a 6  2(a  3) 2a  6 4a  6 2(a  3) 2(2a  3) 2(a  3) 2a  3 a  3

 2a 

8. The LCD is (a  4)(a  4). a a  4

3w

3wz



x 5 , 2x  6 x  3

LCM  (n  1)(n  1)(n  4) or (n  7. The LCD is 10x2. 

z

12. The LCD is 2(a  3).

6. n2  3n  4  (n  4)(n  1) (n  1) 2  (n  1)(n  1)



2

6z  3wz 12w2 3(2z  wz) 12w2 2z  wz 4w2

LCM  5  7  a  a  35a2 5. 2x  4  2(x  2) 3x  6  3(x  2) LCM  (2  3)(x  2)  6(x  2)



a  2 6(a  1)  (a  3) (a  1) (a  3) (a  1) a  2  6a  6 (a  3) (a  1) 7a  8 (a  3) (a  1)

12w2. 3z



Check for Understanding

7 10x2

y2  12y  25 (y  5) (y  5)

6z

5a2  5  a  a 7a  7  a

6 5x



 12w2  12w2

1. Sample answer: To find the LCD, determine the least common multiple of all of the factors of the denominators. 2. Sample answer: Multiply both the numerator and denominator by factors necessary to form the LCD.

4.

2y  y2  10y  25 (y  5) (y  5)



 4w  6w2  2  4w  3w

Rational Expressions with Unlike Denominators

3. Sample answer:







Page 681

y2  10y  25 (y  5) (y  5)

2y (y  5) (y  5)



 1 revolution 1 minute  4125 or about 12,959 in.

12-7

y  5  5

y





1

333 revolutions

y  5  5

2y

 y2  25  y

10. The LCD is (a  3)(a  1). a  2 6 a  2 6 a  1  a  3  a2  4a  3  a  3  a  1 a2  4a  3

(5x  1)  (2x  1) 3x  2 5x  1  2x  1 3x  2 3x 3x  2

10. Circumference of record  2r  2 2 34 in.

y  5  5

y

y2  5y  6 (y  4) (y  3)



2y  y2  5y  6 (y  4) (y  3)



y2  7y  6 (y  4) (y  3)

Practice and Apply

a2b  a  a  b ab3  a  b  b  b LCM  a  a  b  b  b  a2b3

y  3  3

y

7xy  7  x  y 21x2y  3  7  x  x  y LCM  3  7  x  x  y  21x2y 18. LCM  (x  4)(x  2) 19. LCM  (2n  5)(n  2)

30. The LCD is 3(x  3).

17.

5 3x  9

 

31. The LCD is 3(3m  2). m 3m  2

21. p2  5p  6  (p  6)(p  1) LCM  (p  6)(p  1) 23. The LCD is a3. 22. The LCD is x2. 

5 x

3

5

x

2 a3

 x2  x  x

7

2

7

 9m

2  6

5x x2 3  5x x2

7a a3 2  7a a3

 a3 







  

3 5  a



4 5m2

  

3 5m 4   5m2 7m 5m 15m 28  35m2 35m2 15m  28 35m2

x

4  4

3

5

 



n

3  3

18 y2  9

7 7

n  4

n  3  3

n

3  3

n



n2  3n  3n  12 (n  4) (n  3)



n2  12 (n  4) (n  3) a  2  2

a

 a  2  a  5a

x x2  2x  1

7a2  14a (a  5) (a  2)



7a2  14a  a2  5a (a  5) (a  2)



8a2  9a (a  5) (a  2)

x

x  1

x

6x  3

x  1  1

x

x

x  1



6x2  6x (x  3) (x  1)



6x2  6x  x2  3x (x  3) (x  1)



7x2  3x (x  3) (x  1)



x

1  1

x

 x2  2x



 1 x  x  1 (x  1) 2 2x  1 (x  1) 2

 

1 x  1   1 x  1 x  1 (x  1) 2

x

 1

x

35. The LCD is (x  4)(x  1)2.

n  4  4

n

2x  1 x  2  x2  3x  4 (x  1) 2 2x  1 x  4 x  2  (x  1) 2  x  4  (x  4) (x  1)

a  5



2x2  9x  4 (x  1) 2 (x  4)



2x2  9x  4  x2  3x  2 (x  1) 2 (x  4)



3x2  6x  6 (x  1) 2 (x  4)

x  1  1

x

x2  3x  2 (x  1) 2 (x  4)



36. The LCD is (2x  3)(2x  3)2.

a2  5a (a  5) (a  2)



7y  21 (y  3) (y  3)

 x2  2x

 a  2a  5



y  3  3

y

34. The LCD is (x  1)2.

3n  12 (n  4) (n  3)



7  3

7y  39 (y  3) (y  3)



x2 4x2  9



x (2x  3) 2

29. The LCD is (x  3)(x  1). 6x x  3

7  3)

18

18

n2  3n (n  4) (n  3)

7a

5

 y2  9  (y

 y2  9 



a

7  y

18

3 x  4 4 x  5   x  4x  5  5 x  4 3x  12 4x  20  (x  4) (x  5) (x  5) (x  4) 3x  12  4x  20 (x  5) (x  4) 7x  8 (x  5) (x  4)

n

3

 y2  9  y

28. The LCD is (a  5)(a  2). 7a a  5

a  5

33. The LCD is (y  3)(y  3).

27. The LCD is (n  4)(n  3). n n  4

5

3a  15  5 (a  5) (a  5) 3a  20 (a  5) (a  5)



x 

2

 a2  25  (a  5)  a2  25 3

26. The LCD is (x  5)(x  4). 3 x  5

 3(3m  2)

 a  5  a  5  a2  25

7 5 2a  3a  2a 6a2 7 10a  6a2 6a2 7  10a 6a2

25. The LCD is 35m2. 3 7m



32. The LCD is (a  5)(a  5).

24. The LCD is 6a2. 5 3a

2

 9m  6



2

 x2 

m 3   2 3 3m 3(3m  2) 3m  2 3(3m  2) 1 3

 3m 

a

 a2  a3  a2  a

3

7 6a2

5 3 3  x  33  9 5 9  3x  9 3x  9 5  9 3(x  3) 14 3(x  3)

 3x 

20. x2  5x  14  (x  7)(x  2) (x  2) 2  (x  2)(x  2) LCM  (x  7)(x  2)(x  2) or (x  7)(x  2)2

3 x2

3  3

x

x  3  3

x

x2  3x (x  3) (x  1)

x2  9

 4x2

2x  3  3

 2x



x (2x  3) 2



2x3  3x2 (4x2  9) (2x  3)



2x3  3x2  2x2  3x (2x  3) (2x  3) 2



2x3  5x2  3x (2x  3) (2x  3) 2



2x  3  3

 2x

2x2  3x (2x  3) 2 (2x  3)

37. The LCD is (a  b)(a  b)2. a2 a2  b2

569



a (a  b) 2

a2

a  b

 a2  b2  a  b 

a (a  b) 2



a3  a2b (a2  b2 ) (a  b)



a3  a2b  a2  ab (a  b) 2 (a  b)



a  b

a  b

a2  ab (a  b) 2 (a  b)

Chapter 12

46. The LCD is x(x  5).

38. The LCD is 6x2. 7 3x



3 6x2

2x x2  5x

7(2x) 3  6x2 3x(2x) 14x 3  6x2 6x2 14x  3 6x2

  

3x  5

x

5

4

 

5(5x)



4  25x 15x2

7x 6y



3 a  6

11x(2) 7x(y)  6y(y) 3y2 (2) 22x 7xy  6y2 6y2 22x  7xy 6y2

  

3a

 21x2     

5a(3x) 7x(3x)

6

 a2  6a  

n 5  n

3

n

3a

 21x2

15ax 3a  21x2 21x2 15ax  3a 21x2 3(5ax  a) 3(7x2 ) 5ax  a 7x2



x2  1 x  1



3  n2  25  5) n(n  5) 3  n2  25 (n  5) (n  5)



n2  5n  3 (n  5) (n  5)

 n2  25  1(n

49. The LCD is 3(a  2)(a  2). 3a  2 6  3a

a  2

3a  2 a  2  a2  4  2) 3a  2(a  2) (a  2) (3)  (a2  4) (3)  3(a  2) (a  2)

 a2  4  3(a 

3a2  8a  4  2) (a  2)



(x2  1) (x  1) (x  1) (x  1)



(x2  1) (x  1) (x  1) (x  1)



x3  x2  x  1 (x  1) (x  1)



x3  x2  x  1 (x  1) (x  1)



x3  x2  x  1  x3  x2  x  1 (x  1) (x  1)



2x2  2x (x  1) (x  1)



2x(x  1) (x  1) (x  1)



3a2  11a  10  2) (a  2) (3a  5) (a  2) 3(a  2) (a  2) 3a  5 3(a  2)

 

50. The LCD is (x  2)2(x  1). 3x x2  3x  2

3x  6

 x2  4x  4 

43. The LCD is (k  5)(k  3). 3

k(k  3) (k  5) (k  3)

3(k  5) (k  3) (k  5)





k2  3k  3k  15 (k  5) (k  3)



k2  6k  15 (k  5) (k  3)

k

2  2



k(k  2) (2k  1) (k  2)



2(2k  1) (k  2) (2k  1)



k2  2k (2k  1) (k  2)



4k  2 (k  2) (2k  1)



k2  2k  4k  2 (k  2) (2k  1)



k2  2k  2 (2k  1) (k  2)

4

 2m  5 

(m  1) (2m  5) (m  1) (2m  5)

3x  6 (x  2) 2



3x(x  2) (x  2) (x  1) (x  2)



3x2  6x (x  2) 2 (x  1)



3x2  6x  3x2  3x  6 (x  2) 2 (x  1)



9x  6 (x  2) 2 (x  1)



4(m  1) (2m  5) (m  1)



4m  4 (2m  5) (m  1)



5a2  5a (a  4) (a  1) (a  1)



5a2  5a  a2  3a  4 (a  4) (a  1) (a  1)



4a2  2a  4 (a  4) (a  1) (a  1)





(3x  6) (x  1) (x  2) 2 (x  1)

3x2  3x  6 (x  2) 2 (x  1)



a2  3a  4 (a  4) (a  1) (a  1)

52. The LCD is (x  3)(x  1). 

x2  4x  5 (x  3) (x  1)

2m2  3m  5  4m  4 (m  1) (2m  5)



x2  4x  5  2x  6 (x  3) (x  1)

2m2  m  9 (m  1) (2m  5)



x2  2x  1 (x  3) (x  1)



(x  1) 2 (x  3) (x  1) x  1 x  3

2



2m  3m  5 (m  1) (2m  5)

 

x2  4x  5 x2  2x  3

x

2  1



Chapter 12



5a a  1  a2  1 a2  3a  4 5a a  1  (a  4) (a  1)  (a  1) (a  1) 5a(a  1) (a  1) (a  4)  (a  4) (a  1) (a  1)  (a  1) (a  1) (a  4)

45. The LCD is (m  1)(2m  5). m  1 m  1

3x (x  2) (x  1)

51. The LCD is (a  1)(a  1)(a  4).

44. The LCD is (2k  1)(k  2). k 2k  1

 2)

 3(a

2x  1

k3

 3(a  2) (a

3a2  8a  4  3a  6 3(a  2) (a  2)

x k k  5

3a  6

 3(a

42. The LCD is (x  1)(x  1). x2  1 x  1

3(a) 6  a2  6a (a  6) (a) 3a  6 3a  6 or a(a  6) a2  6a

48. The LCD is (n  5)(n  5).

41. The LCD is 21x2. 5a 7x

2x  3x2 x2  5x x(2  3x) x(x  5) 2  3x x  5

47. The LCD is a(a  6).

40. The LCD is 6y2. 11x 3y2

3x2

2x

 3x  15x2  3x(5x) 

3x(x) (x  5) (x)

 x2  5x  x2  5x

39. The LCD is 15x2. 4 15x2

2x

 x2  5x 

570



2(x  3) (x  1) (x  3)

60. D; a2  2ab  b2  (a  b)(a  b) a2  b2  (a  b)(a  b) The LCD is (a  b)(a  b)(a  b) or (a  b)2 (a  b). 61. C; The LCD is (x  3)(x  2)2.

53. The LCD is (m  4)2(m  4). m  4 m2  8m  16

m  4 m  4





m  4(m  4) (m  4) 2 (m  4)



(m  4) (m  4) 2 (m  4) (m  4) 2



m2  8m  16 (m  4) 2 (m  4)



m3  12m2  48m  64 (m  4) 2 (m  4)



m2  8m  16  m3  12m2  48m  64 (m  4) 2 (m  4)



m3  11m2  56m  48 (m  4) 2 (m  4)

x  4 (2  x) 2

54. Find the LCM of 2, 3, and 4. 22 33 4  22 The LCM is 2  2  3 or 12. The least number of students is 12. 55. Find the number of miles each girl walks in an hour. 60 minutes 12 minutes 60 minutes 15 minutes 60 minutes 20 minutes

Maya : Makalla : Monya :

Page 683 62. 63.

 1 mile  5 miles

64.

1 bag 12 days

total 





b y

a

y

b





x2  x  12 (x  2) 2 (x  3)



x2  7x  10 (x  3) (x  2) 2



x2  x  12  x2  7x  10 (x  3) (x  2) 2



6x  22 (x  3) (x  2) 2

 3  1 5 3

2y 5  (y  3)  3 2y 5 y3 y  3 2y  5 y  3

y



(b  10) (b  2) b  2

 4t3 () t3

 

t2  4t  3  19t  9

0t2 4t2

The quotient is t2  4t  3  t

3 .  4

2m  3 67. 2m  74m2  8m  19 () 4m2  14m 6m  19 () 6m  21 2

x 7

 16  1.4875

The quotient is 2m  3  2m

2 .  7

68. 2x2  10x  8  (2x2  8x)  (2x  8)  2x(x  4)  2(x  4)  (2x  2)(x  4)  2(x  1)(x  4) 69. 5r2  7r  6  5r2  10r  3r  6  5r(r  2)  3(r  2)  (5r  3) (r  2)

x

 xy  yx 

(x  5) (x  2) (x  3) (x  2) (x  2)

4t2  19t () 4t2  16t 3t  9 () 3t  12 3

x  7 days

Therefore, she must buy 2 bags for one week. 57. Find the LCM of 3000, 6000, and 15,000. 3000  2  2  2  3  5  5  5 6000  2  2  2  2  3  5  5  5 15,000  2  2  2  3  5  5  5  5 The LCM is 2  2  2  2  3  5  5  5  5  30,000 Therefore, the car’s odometer would read 36,000  30,000 or 66,000 miles. 58. Sample answer: This method will always work. a x

b  8b  20 b  2

66. t

7  16x 7 15



 b  10

7  15x

7 12

2

x

x

7 16

5  y



7 x 15 1 bag x  16 days days

third dog :

3

 7 days

1 bag 15 days

second dog :

2y y  3

 1 mile  3 miles

7  12x 7 12

(x  4) (x  3) (x  2) 2 (x  3)

Maintain Your Skills



65.



3m 3 3m  2m  1  2m 2m  1 5 4x  4x  2x  3  2x  2x  3

 1 mile  4 miles

Total miles  5  4  3  12 miles The amount of money raised is 12  $2.50 or $30. 56. first dog :

x  5  6

 x2  x

70. 16p2  4pq  30q2  2(8p2  2pq  15q2 )  2(8p2  10pq  12pq  15q2 )  2[ (2p(4p  5q)  3q(4p  5q) ]  2(2p  3q)(4p  5q) 71. amount spent on entertainment  0.05(1782  525  120  40)  0.05(1097)  $54.85

ay bx  yx xy ay  bx xy

59. Sample answer: You can use rational expressions and their least common denominators to determine when elections will coincide. Answers should include the following. • Use each factor of the denominators the greatest number of times it appears. • 2012

571

Chapter 12

72.

x 2



3x 5

x

5

 2  3x

73.

a2 5b

a2

4a

 10b2  5b 

1

x

5

 2  3x 1

5

 5b 

a2

5b2 2a

a

b

 5b  1

 74.

x  7 x

2

a

6

x  7  3

x



x  7 x



x  7 x

10b2 4a

9.

75.

3n 2n  5

 12n2  5

 2n

 2n

5b 2a

ab 2

10. Average 



2n  5 12n2

1

77.

x2  7x  12 x  6

3x  2

x

3x (x  2) (x  1)

 (x  3) 

x

1  3

x

1  3

11. 8 

1



(x  4) (x  3 ) x  6

x

 15. 2m 

Mixed Expressions and Complex Fractions



2 3



2

6

16. 3a 

Check for Understanding

4  m m

a  1 2a

5 6

3(x) 4 3  x x 3x  4  x a  1 a  1  2a  3a 3a



7 2 19 4



7 2



7 4  2 19 14 19

Chapter 12

2z w

14. 6z 

 

2m(m) m



6z(w) 2z w w 6wz  2z w

4  m m



6a2  a  1 2a  b  b



b2a  b3  a  b a  b



r2 (r  3) r  3

or

b3  b2a  a  b a  b

r 18. r2  r

 4  3

r  4  3

19. 5n2 

r3  3r2  r  4 r  3 n  3 n  3 2  5n n2  9 (n  3) (n  3)

r



4 x











4(a) 5 a a 4a  5 a



3. Bolton; Lian omitted the factor (x  1).

7.

5

12. 4  a 

2m2  4  m 2m2  m  4 or m m 3a(2a) a  1  2a 2a

4

1 32 3 44

Practice and Apply

2x(y) x y y 2xy  x y

5

6.

204 20



5

a  b b2 (a  b) a 17. b2  a  b  a  b  a

 35

4.

165 20





1. Sample answer: Both mixed numbers and mixed expressions are made up by the sum of an integer or a monomial and a fraction or rational expression. 2. Sample answer: 2 3 5 6





8(n) 3 n n 8n  3 n

13. 2x  y 

x  4  6

Pages 686–687

3 n



1

x

12-8



Pages 687–688

(x  4) (x  3) x  6

130 20

729 1  20 5 29 7100 min



1  1

90 20

1

5



4n

1

729 20



1

1

5 140 20

 2n  5 12n2

1

7  42  62  84  105

1



1

a  b x  y



1



(a  b) (a  b) (x  y) (x  y) 1 (a  b) (a  b ) (x  y ) (x  y)

1

 4n  (x  1)  x

x  y 1 x  y

x  3 x  7

1

3x x  2

a2  b2

 a  b

1

x  3  7

3n  5

76.

x  y

 a  b  x2  y2  a  b

x

3n  5

 2n

x2  y2 a2  b2

2

1

x  3 x

x  y a  b

5. 7 

 





7(6y) 5  6y 6y 42y  5 6y

s  1

20. 3s2  s2  1  

2a(3a) 3a



a  1  6a2 3a



6a2  a  1 3a

n

1  3

5n3  15n2  1 n  3 s  1 3s2  (s  1) (s  1) 3s2 (s  1) s  1

s

1  1

3s3  3s2  1 s  1 x  2 (x  5) (x  3)  x  3 x  3



21. (x  5) 

8. 19 4

5 6y

5n2 (n  3) n  3

x3 y2 y3 x

 x3

 y2  

x3 y2



x4 y5

y3 x

x  2  3

x

x2  8x  15  x  2 x  3

x2  7x  17 x  3 p  1 (p  4) (p  4)  p  4 p  4



22. ( p  4) 

x  y3

572

p  1

p4



p2  16  p  1 p  4



p2  p  15 p  4

3

23.

54 2

73

2



23 4 23 3 23 4





23 4

 23



24.

87 4

45

23 3

3

25.

a2 b

a2 b







a b3

b  a2

1 a

1 b

b2

a

26.

n3 m2 n2 m2

 b3  a2

x2 y2





n3 m2 n n3 m2

m2  n2 1 m2  n2

1

1

27.



n3 m2



 ab2 x  4 y  2

145 84

x



x2

s t2 s  t s  t

x2  x  20 x  1 x2  2x  15 x  2

x2  x  20 x  1



 

(x  3) (x  1) (x  2) (x  4)

34.

1

35  12 63 n  2

nn n

n(n  12) n  12 n(n  2) n  2



35  12 63 n  2

n 

n2  12n  35 n  12



s3 t2

s

 x2

y2



s3 t2

s

s  t  t



n2  12n  35 n  12

y2 (x  4)  2)



s3 (s  t) t2 (s  t)

 

n2  12n  35 n  2  n2  2n  63 n  12 1 (n  7 ) (n  5) n  2  (n  9) (n  7 ) n  12



(n  5) (n  2) (n  12) (n  9)

x  4  2

 y2

x  4  2

y

y2  1 y2  3y  4

28.

y1

y2  1  4

 y2  3y



s  t  t



y  1 1

1

y a2  2a  3 a2  1

a3

1

35.

1

1

aa

a  2a  3 a2  1



a2  2a  3 1 a  3 a2  1 1 1 (a  3 ) (a  1 ) 1  (a  1) (a  1 ) a  3



a  3 1

1

n2  5n n2  n  30

1

36.

n  2n  18

 n2  9n  

2

n  5n  30

 n2  n

n2  2n n2  n  30  n2  5n n2  9n  18 1 1 1 n (n  2) (n  6 ) (n  5 )  (n  6 ) (n  3) n (n  5 ) 1

1

1

n  2  3

x2  3x  28 x2  10x  24



b(b) b a(a) a

1

b 1

a

b2  1 b a2  1 a



b2  1 b



a2  1 a



b2  1 b

 a2  1

a

2b2 5c



4b3 2c 7b3 8c2

¢ 

2b2 5c





2b2 5c





2b2 5c





32b2 35

14b2c

3

3



x2  3x  28 x2  10x  24

7b3 8c2

2

4c 2

3

1

7

1 16c 7

37. number of servings 

x2  4x  21 x2  9x  18



14b2c  7b8c 2 4

1

n x2  4x  21 x2  9x  18

n2  2n  63 n  2

a(b2  1)

1

2



 b(a2  1)

a1 n2  2n n2  9n  18









bb

1  4 2

n2  2n  63 n  2

1

y2  1 1   4 y  1 1 1 (y  1 ) (y  1 ) 1  (y  4) (y  1 ) y  1



32.

x2  2x  15 x  2

1

 y2  3y

31.

15  2 20  1)

x2  2x  15 x  1  x2  x  20 x  2 1 (x  5 ) (x  3) x  1  (x  5 ) (x  4) x  2



n2 m2

x(x  2) x x  2 x(x  1)  (x (x  1)



3

 x2 (y

30.

15  2 20 x  1

xx



n

y

29.

33.

12

3

a b3





4 a b3



58 7 24 5 58 24  5 7 29 58 5  7 24



5  66 ounces 1 52

ounces

 330  30

x2  4x  21 x2  10x  24   18 x2  3x  28 1 1 1 1 (x  7 ) (x  3 ) (x  6 ) (x  4 )  (x  6 ) (x  3 ) (x  7 ) (x  4 )



 30  2  60 servings

1

11 2

2

 330  11

1

330

11 2

 x2  9x 1



1

1

1

573

Chapter 12

38. h      

f

a

v s

1



f s s





v s

b.

a

1 3

b



a b

fs  v 370  760 760  65 281,200 695





4.799  1012 4,231,604,706





c.

4,651,208,508 4,231,604,706

1 2

 2a

b1 

4.794  1012

 4,231,604,706 

 1133 people per square mile 41. Let P  30 lb/in2 and V 

2 13



k



12



k

30 

5 3

153 2 k



 

(30)



50  k



3

Let k  50 and V  4 ft3. Solve for P. P

12



50 3 4

P  50  4

P

200 3 2 66 3

lb/in2

42. Simplify all expressions. a.

a 1

3 a



a 3 a

1

 

a a a

 a

3 a

 

a 3 a

a

a a

a  3 3  a a a a a  3 a  a  1 1

1 2 1  3 a a 2 a a a a  11  a  3 2  11  3  a 2 

Chapter 12

a

a2  3

3a

a

a2  3

 (a  3)

1

3 b

3a  1 3

 

3a 3

1

3 b

3a  1 3

1 2

b

3a  1 3

b

b 1

3a  1 3a  1  3b 3b 3a  1  3a  1 3b 2 0 3b

1





2a  2 1b

4a 2

b1

1  4a 2

b1

 

4a 2

1

2

1b

4a  1 2

1b

1  4a 2

1  4a 4a  1  2  2b 2b  2 1  4a 4a  1  (2b  2) 2b  2 1  4a 4a  1  2b  2 2b  2 1  4a  4a  1 2b  2 8a  2 2b  2 2(4a  1) 2(b  1) 4a  1 0 b  1

Therefore, expression a is equivalent to 0. 43. Sample answer: Most measurements used in baking are fractions or mixed numbers, which are examples of rational expressions. Answers should include the following. • You want to find the number of batches of cookies you can make using the 7 cups of flour you have on hand when a batch requires 1 12 cups of flour. • Divide the expression in the numerator of a complex fraction by the expression in the denominator.

3 4

P  50  3 P

3a 3

4a  1  b  12  1 2  1  b2 1 1  4a 1 4a  1 1  1 2  b  12  1 2  1  b2

ft3. Solve for k.

PV k 30  2 13

5 (30) 3

a2  a2 a  3 0 a  3

3a  1 b  2  1 3  12 1 3a  1 1 3a  1 1  1 3  b2  1 3  b2

population of New Jersey population of Alaska  land area of Alaska land area of New Jersey 8,414,350 626,932  7419  570,374







 404.60 cycles /s 40.

1 3



hs 

a2

a3

0

f s  v s s  v f s s fs  v fs s  v

39. Let s  760, v  65, and f  370. 

a2  3

a2

a2

574

44. C; Set up an equation. P  4BC  12  4BC  6n 12  n  8 12(n  8) 6n n8 n  8 12n  96 6n n8 n  8 12n  96  6n n  8 6n  96 n  8 6n  96 1 4 n  8 6n  96 4n  32

45. C;

6mn 5p 24n2 20mp

1

3n 2 n8 6n n  8

2

 4BC

7 x2

51.

x (x  3) 2

52.

2 t2  t  2

53.

2n n2  2n  24

 x2  

 4BC

 20mp



6mn 5p



20mp 24n2



6mn 5p



20mp 24n2



4m2 4n



m2 n

m

4m

4n

p2 





Check:



36x

8(2y) 6y(2y)

 

16y

36x  16y 12y2 4(9x  4y) 4(3y2 ) 9x  4y 3y2

57.

47. The LCD is (a  b)(2b  3a). a a  b

 2b

b  3a



a(2b  3a) (a  b) (2b  3a)



2ab  3a2  ab  b2 (a  b) (2b  3a)



3a2  3ab  b2 (a  b) (2b  3a)



b(a  b) (a  b) (2b  3a)

48. The LCD is (3a  2)(a  4)2.

a2  a  12 (3a  2) (a  4) 2



a2  a  12  6a2  4a (3a  2) (a  4) 2



7a2  3a  12 (3a  2) (a  4) 2



n  5  6

 n2  n

9p2  64

183 22  64 64 ? 9 1 9 2  64 ?

or

1 8 22  64 64 ? 9 1 9 2  64

9 3

?

64  64 ✓ 64  64 ✓ z3  9z  45  5z2 z3  5z2  9z  45  0 3  5z2 )  (9z  45)  0 (z z2 (z  5)  9(z  5)  0 (z  5)(z2  9)  0 or z2  9  0 z50  5 z2  9 z  3 The solution set is {5, 3, 3}. z3  9z  45  5z2 ?

3

5  9(5)  45  5(5) 2 ? 125  45  45  125 80  80 ✓ or 3 z  9z  45  5z2

6a2  4a (3a  2) (a  4) 2

?

33  9(3)  45  5(3) 2 ? 27  27  45  45 00✓ or

49. The LCD is (n  2)2(n  3). n  4 (n  2) 2

8

Check:

a  3 2a  a2  8a  16 3a2  10a  8 (a  3) (a  4) 2a(3a  2)  (3a2  10a  8) (a  4)  (a2  8a  16) (3a  2)



 t2  t

64 9

9

 12y2  12y2 

2  t  2 2  t  (t  2) (t  1) (t  2)  (t  2) (t  1) 1  t  1 8 2n  8  n2  2n  24 n2  2n  24 2(n  4)  (n  6) (n  4) 2 n6  2

64 9

8

46. The LCD is 12y2. 12x(3) 4y2 (3)

1  3

p  3 or 3

Maintain Your Skills 8 6y

x

55. s2  16 s  116 s  4 or 4 s2  16 Check: 42  16 ✓ 42  16 ✓ 56. 9p2  64

p  3

12x 4y2

x  3 (x  3) 2

54. The answer is 100 multiplied by 2 ten times. 1000  210  1,024,000 bacteria.

24n2

6mn 5p

Page 689



 BC  BC



t

 t2  t

 4BC  4BC

4

 x2

3 (x  3) 2

 4BC



1

7  3 x2

3

50.



(n  4) (n  3) (n  2) 2 (n  3)



(n  5) (n  2) (n2  n  6) (n  2)



n2  n  12 (n  2) 2 (n  3)



n2  7n  10 (n  2) 2 (n  3)



n2  n  12  n2  7n  10 (n  2) 2 (n  3)



2n2  8n  2 (n  2) 2 (n  3)

z3  9z  45  5z2 ? 3  9(3)  45  5(3) 2 ? 27  27  45  45 00✓ 3

575

Chapter 12

58. 160,140 53,310 27,990 22,980 18,120 15,750 11,190 10,800 59.

 16014  105  5.331  104  2.799  104  2.298  104  1.812  104  1.575  104  1.119  104  1.08  104

amount spent on food amount spent on clothing



12-9 Solving Rational Equations Page 694

2.

amount spent on housing total amount spent

Sample answer:



4.83  3.39 24  15 1.44 9

x

12  4

6. 10

4(12)  4 48  x 3 n 4

65. 3 n 4

14x 2

3 n 4 4 3 n 3 4

1 2

k(k  1)

3.2  (7) (3.2) 

3n  (4) 6 3n  4 (6) 6

2  x(6)

x  1  x  4  6x 2x  5  6x 5  4x x

1

k1

x  2 x  2

x

2  2

7

 3

1xx  22  x 2 2 2  3(x  2) (x  2) 173 2

3(x  2) 2  6(x  2)  (x2  4) (7)  4x  4)  6x  12  7x2  28 2 3x  12x  12  6x  12  7x2  28 3x2  6x  24  7x2  28 10x2  6x  4  0 2(5x2  3x  2)  0 (5x  2) (x  1)  0

x2  4

5x  2  0 5x  2

2

or

x10 x  1

2

x5 10.

32  b 128  b

 0.300

32  b  0.3(128  b) 32  b  38.4  0.3b 0.7b  6.4 b  9.14 Omar needs 10 more hits.

 9  (6)(9)

50

 3 50

n  3

Chapter 12

1

3(x2

3n  4  54 3n  4  4  54  4 3n  50 3n 3

x

6

2  k(k  1) 1k 1 1 2

3(x  2) (x  2)

22.4  8  n 22.4  8  8  n  8 14.4  n 68.

1

5 7 k k  1 5 7 k k  1

9.

x  2

1

7x 10

x  1 x  4  x x x  1 x  4  x x

7

4 (12) 3 8  n 7 8  n 7 7

7.

3  k

n  16

67.

7x

 10

5k  7(k  1)  k 5k  7k  7  k 2k  7  k 7  3k

66. 7x2  28

 12

5

a3

7(a  3)  5(a  1) 7a  21  5a  5 2a  26 a  13

2  101 2

8.

2.4  g

3393

1

3x 3 2 5 3x 3 2 5

7 a  1

5.

5 4

1.8  0.6  g  0.6  0.6

39

0

3 x  1

6x  15  7x 15  x

1.8  g  0.6

64.



x 4

2(x  1)  3x 2x  2  3x x  2 x2

 0.16 Find the equation of the line using (15, 3.39). y  mx  b 3.39  0.16(15)  b 3.39  2.4  b 0.99  b Let y  C and x  m. A linear equation is C  0.16m  0.99 62. Let m  9. C  0.16(9)  0.99 C  1.44  0.99 C  2.43 The cost of a 9-minute call is $2.43. 63.

2 x

4.

 0.3328 or 33.3% 61. Find the slope of the line containing points (15, 3.39) and (24, 4.83). m

1

4

5.331  10  105

 1.6014

n 3

3. The solution of a rational equation can never be zero.

2.799  104 1.08  104

 2.59  100 60.

Check for Understanding

1. Sample answer: When you solve the equation, n  1. But n  1, so the equation has no solution.

576

Pages 694–695 4 a

11.

Practice and Apply 3

a2

4(a  2)  3(a) 4a  8  3a a8 x  3 x

13.

3 x

12.

x

1  2

p5

3(x  2)  x 3x  6  x 2x  6 x3

6

1

x  3  6

x

21.

15(a  2)

x

6 x  4 3 x  2 2n 1 2n  3 2 6 3 2n 1 2n  3 2 6 6 3

2

1

4n  3  2n  3 2n  6

22.

16.

2

12

1

5 3y  2 4 5 3y  2 4

17. (a  1) (a  1)

1



(4  c)

7y 6

2  121 2 15

a  1 a  1

2a  1

2

x(x 

1

  

19. (2x  3) (2x  3)

1

4x 2x  3





2x 2x  3

2x 2x  3

6 4  c 6 4  c 6  c)

c c

2  (4  c)(c)

2b  5 b  2

2b

3  2

12bb  25  22  (b  2)(b  2) 1b 3 2 2

2b2  b  10  2b2  8  3b  6 b  2  3b  6 4b  4 b1 7 k  3

24.

4 x 4 x 4 x

2(k  3) (k  4)

1

2k

3  4

1k 7 3  12 2  2(k  3) (k  4) 1k 3 4 2

14(k  4)  (k  3) (k  4)  6(k  3) 14k  56  (k2  7k  12)  6k  18

7  3x  4x  20 7x  27 4x 2x  3

c  c  4 c  (4  c) c  (4 (4  c)

(b  2)(2b  5)  2(b  2)(b  2)  (b  2)(3)

2  x(x  5) 14x 2 4 7  3x  (x2  5x) 1 x 2 x

5a  3a  6(a  2) 2a  6a  12 4a  12 a  3

(b  2)(b  2)

 (a  1) (a  1) (1)

(a  1) (a  1)  2a(a  1)  (a2  1) (1) a2  2a  1  2a2  2a  a2  1 a2  4a  1  a2  1 4a  0 a0 7 3 5x x2  5x 7 3  (x  5) x(x  5) 7 3  (x  5) x(x  5) 7 3 5) x(x  5)  (x  5)

2

5

2  25 (15(a  2) )

23.

2a

 a  1  1

a

2

5

2

y  4

a  1 a  1

1

1

a a  5a  10 3a  6 a a  5(a  2) 3(a  2) a a  5(a  2) 3(a  2)

c  6  4c  c2 c  5c  6  0 (c  6)(c  1)  0 c  6  0 or c  1  0 c6 c1

7y 6

15  18y  14y 4y  15

n  3

18.

2

55  p2  8p  40 p  8p  15  0 p2  8p  15  0 Since p2  8p  15 does not factor, there are no solutions.

x(x  1)  (x  6)(x  1) x2  x  x2  5x  6 4x  6

15.

1 p 55 5  p p 5 2  ( p  5) (8)

x  6  1

x x  1

p2

 p  5  8

2

(x  3)(x  6)  (x  3)(x) x2  9x  18  x2  3x 6x  18  0 6x  18 x3 14.

55 p  5

20.

k50 k5

27 7

25.

1

2  (2x  3) (2x  3)

k2  21k  68  6k  18 k2  15k  50  0 k2  15k  50  0 (k  5) (k  10)  0

x2  4 x  2 (x  2) (x  2) x  2

or

k  10  0 k  10

 x2  4  x2  4

x  2  x2  4 x2  x  2  0 (x  2)(x  1)  0 or x  1  0 x20 x  2 x1

4x2

(2x  3) (4x)  (2x  3) (2x)  9 8x2  12x  4x2  6x  4x2  9 4x2  18x  4x2  9 18x  9 1

x2

577

Chapter 12

2n n  1

26. (n  1)(n  1)

n  5  1

 n2

31. Find the distance for Jim. d  rt

1

1n 2n 1  (n n1)(n5  1) 2  (n  1) (n  1)

d

(n  1) (2n)  n  5  (n  1) (n  1) 2n2  2n  n  5  n2  1 n2  3n  4  0 (n  4) (n  1)  0

Find the distance for Mateo. d  rt d

or n  1  0 n40 n  4 n1 The number 1 is an extraneous solution, since 1 is an excluded value for n. Thus, 4 is the solution of the equation. 27. (z  4) (z  1)

1

3z z2  5z  4 3z (z  4) (z  1)



2 z  4



1803 2  21.82  0.82 mi 1101 2  21.82  2.2 mi

They will be 0.82 mile from the nearest shore. 32. R  1

1 car 2 hr

112  12  13 2x  7 136  36  26 2x  7

3 z  1

8 x 6

3z  2(z  1)  3(z  4) 3z  2z  2  3z  12

x

14  2z 7z

(m  2) (m  6)

m

m  2

x

1

m6

1 (m  2)4(m  6) 2  (m  2)(m  6) 1m m 2  m 1 6 2

x

4  m2  6m  m  2 0  m2  5m  6 0  (m  6)(m  1) m10

or

m6

34. 600 ft3 

m  1

The number 6 is an extraneous solution, since 6 is an excluded value for m. Thus, 1 is the solution of the equation. 29. Let x  number of quizzes. 36  10x 5  x

35.

9

1

x2  x  2 x  5

x1 x

3

20

1  1

20

(x  3) (x  2) (x  2) (x  5)

20

(x  3) (x  2) (x  2) (x  5)

4

 2  (x  2) (x  5) (0)

(x  3) (x  2)  2(x  2) (x  5)  0 x2  5x  6  2x2  6x  20  0 3x2  11x  14  0 (3x  14) (x  1)  0 3x  14  0 or x  1  0 3x  14 x1 14

x  3

r

The number 1 is an extraneous solution. Thus, 14  3 is the only solution. 36. Sample answer: Rational equations are used in solving rate problems, so they can be used to determine traveling times, speeds, and distances related to subways. Answers should include the following. • Sample answer: Since both trains leave at the same time, their traveling time is the same. The sum of the distances of both trains is equal to the total distance between the two stations. So, add the two expressions to represent the distance each train travels and solve for time.

r r

3 1 t  10 t 80 3 1 t  10 t 80

3

2  80(3)

3t  8t  240 11t  240 t  21.82 They will meet in about 22 minutes.

Chapter 12





(x  2) (x  1) x  5

(x  2) (x  5)

Let t  time in minutes.

80

 4500 gal.

x  3 x  2 x  3 x  2

Find the rate for Mateo. d  rt 3 mi  r  30 min 3 mi 30 min 1 mi 10 min

7.5 gal. 1ft3

It will take 4500 gallons to fill the pool.

36  10x  9(5  x) 36  10x  45  9x x9 She must score 10 points on 9 quizzes to reach her goal. 30. Find the rate for Jim. d  rt 3 mi  r  80 min 3 mi 80 min

42 8 2 58 1 54

168 2

It will take them 5 hours and 15 minutes to clean 7 cars. 33. V  /  w  h V  15 ft  10 ft  4 ft  600 ft3

4  (m  6)(m)  (m  2)

m60

7

x  (7)

3z  5z  14

28.

1 car 3 hr

Let x  the number of hours.

2  (z  4)(z  1) 1z 2 4  z 3 1 2

4 m2  8m  12

and R2 

578

37. A;

a  2 a

a  3

1

a6a

a(a  6)

1a a 2  aa  36 2  a(a  6) 1a1 2

41.

2  5 6 x  1

x2x x6

1 n  2



n2  7n  8  8

 3n2  2n

x2  7x  12 x  5





x2  7x  12 (x  5)

 x2  7x

3 2m  3

m

 6  4m   

1, 2, 3, 4, 6, 8, 12, 24 1

y y2  2y  1



x2  8x  15 x2  x  6



(x  5) (x  3) (x  3) (x  2)



(x  5 ) (x  3 ) (x  3 ) (x  2)

1



y

1  1

a  2 a2  9

2a

 6a2  17a  3 

(x  3) (x  1)  5) (x  3) (x  3 ) (x  1)  5 ) (x  3 ) 1

1

x  1  2

x 40.

a2  6a  5 a2  13a  42 2

a  4a  3 a2  3a  18

2

a  6a  5

2

a  4a  3

 a2  13a  42  a2  3a  18 

(a  5) (a  1) (a  6) (a  7)



(a  5) (a  1 ) (a  6 ) (a  7)



(a  5) (a  7)

(a  6) (a  3)  3) (a  1)

 (a

1

1

1

1

(a  6 ) (a  3 )  3 ) (a  1 )

 (a

1

a  2 (a  3) (a  3)



2a (a  3) (6a  1)



(a  2) (6a  1) (a  3) (a  3) (6a  1)



2a(a  3) (a  3) (6a  1) (a  3)



6a2  13a  2 (a  3) (a  3) (6a  1)



2a2  6a (a  3) (a  3) (6a  1)



6a2  13a  2  2a2  6a (a  3) (a  3) (6a  1)



4a2  7a  2 (a  3) (a  3) (6a  1)

45. 20x  8y  4(5x  2y) 46. 14a2b  21ab2  7ab(2a  3b) 47. 10p2  12p  25p  30  2p(5p  6)  5(5p  6)  (2p  5)(5p  6) 48. Write two equations. 1. 0.5x  0.3y  0.45(100) 2. x  y  100 Solve for x in equation 1 and plug it into equation 2. 0.5(100  y)  0.3y  45 50  0.5y  0.3y  45 0.2y  5 y  25 x  25  100 x  75 Therefore, there are 25 gallons of the 30% glycol solution and 75 gallons of the 50% glycol solution.

1

1

y y  1  ( y  1) 2 ( y  1) 2 y  y  1 ( y  1) 2 1 1 or y2  2y  1 (y  1) 2

44. The LCD is (a  3)(a  3)(6a  1).

 (x  (x

 

x2  2x  15 x2  2x  3

1

(x  1)  12

3(2) m  2(2m  3) (2m  3) (2) 6 m  2(2m  3) 2(2m  3) 6  m 2(2m  3)



Maintain Your Skills

x2  2x  15 x2  2x  3

x2  7x  12 x  1

43. The LCD is (y  1)2.

Since n  2 is an extraneous solution, n  6 is the solution.

39.

x2  7x  6  6 x  1

42. The LCD is 2(2m  3).

Therefore, n3  2n2  20n  24  (n  2)(n2  4n  12)  (n  2)(n  6) (n  2) n20 or n  6  0 n  2 n6

x2  8x  15 x2  x  6

x  7x  10  2 x  5

x  1  5

Since (2)3  2(2)2  20(2)  24  0, 2 is a solution. By synthetic division, 2 1 2 20 24 2 8 24 1 4 12 0

Page 695





The possible factors of this polynomial must be 

2  5 6 x  1

x

x

1(3n2  2n  8)  (n  2) (n2  7n  8) 3n2  2n  8  n3  7n2  8n  2n2  14n  16 3n2  2n  8  n3  5n2  22n  16 3 n  2n2  20n  24  0 factors of 24 factors of 1

(x  2) (x  5) x  5 (x  6) (x  1) x  1 2

(a  2)(a  6)  a(a  3)  a  6 a2  8a  12  a2  3a  a  6 5a  12  a  6 6a  18 a3 38. D;



1

Chapter 12 Study Guide and Review Page 696

Vocabulary and Concept Check

1. false, rational 3. true 5. false, x2  144

579

2. true 4. false, 6. true

5a  20 a  3

Chapter 12

Pages 696–700

Lesson-by-Lesson Review

x1y1  xy 42  28  xy 1176  xy 1176  56y 21  y 9. x1y1  xy

8. x1y1  xy 5  15  xy 75  xy 75  3y 25  y 10. x1  y1  xy 175  35  xy 6125  xy 6125  75y 81.67  y

7.

8  18  xy 144  xy 144  x(3) 48  x

11.

x 1

3x2y 12xy3z

3x2y

 12x y3z

12.

4 y2

n2  3n n  3

x

13.

a  25 a2  3a  10

14.

x  10x  21 x3  x2  42x

15.

7b2 9



b

6a2 b

2

6a2 b



7b2 9



14a2b 3

3



16.

1

25.

3a  6 a2  9

a  3

 a2  2a  

5x2y 8ab

19.

x2  x  12 x  2

1

26.

x  4  6

 

20.

b2  19b  84 b  3

b2  9  b2  15b  36



12a2b 25x

xy



5x2y 8ab



3axy 10

3a



2

12a2b 25x 5

10  10(x  10 )

 (x

1

a  3  2)

 a(a

1

3 a2  3a

22.

p3 2q

p2

p3

 (x

x  4  3 ) (x  2)

29.

1

1

1



m  1 5



 

(x  4) 2 (x  2) 2 (b  7) (b  12 ) b  3

m  4 5

1

30.

1

(b  3 ) (b  3 ) (b  12 ) (b  3 ) 1

5 2n  5

2n  5

 2n

y2 y  4

 2q  p2



(a  b ) (a  b) a  b

 2p

ab

1

1

y

y2  4

3y



y2  16 3y



(y  4 ) (y  4) 3y

a2 a  b

b2 a  b



1

32.

7a b2



5a b2



7a  5a b2



2a b2

33.

2x x  3

x

6  3

3

2

b

3c2 b



2x  6 x  3



2(x  3 ) x  3

1

1

1

3

2

y(y  4) 3

 ( y2  6y  8) 

3y  12 y  4

34.

1

 y2  6y

 8

1



3(y  4 ) y  4 1



Chapter 12

6abc 2ab2

1

y2  4

3y  12 y  4



1

 y2  16  y



4a 3 2

8a b c 2ab2

5  2n 2n  5 2n  5 2n  5

a2  b2 a  b

2



1 31.

y

23.





4q

6abc2 2ab2

m  4  m  1 5 2m  3 5



1

4q

p



2

8

 4q  2q  p2 p3

4a2b2c2 2ab2

8a3b2c 2ab2

x2  4x  2 x  3x3  7x2  10x  6 () x3  3x2 4x2  10x () 4x2  12x 2x  6 () 2x  6 0 The quotient is x2  4x  2.

b7

21.



The quotient is 4b  1  12b  1.

(x  4) (x  3 ) x  2



4a2b2c2 2ab2

x2  2x  3  0x2  7x  6 x  2x2 2x2  7x () 2x2  4x 3x  6 3x  6 0 The quotient is x2  2x  3. 28. 4b  1 12b  148b2  8b  7 () 48b2  4b 12b  7 () 12b  1 8

1

 x2  x

1

 2ac2  4a2c 

1

3(a  2 ) (a  3) (a  3 )

or

(4a2b2c2  8a3b2c  6abc2 )  2ab2 

2a

30  10

3 a(a  3)

3m  2  2) (3m  2)

 2x3 () x3

1

18.

1

 (3m

27.

3(x  10 ) 1

x

3m  2  4

 9m2

(2m  3) (m  5 ) m  5



1

10

17. (3x  30)  x2  100 

2m2  7m  15 m  5

1

(x  3) (x  7)  42) (x  3) (x  7) x(x  7) (x  6) x  3 x(x  6)





2m  3  2

 x(x2  x 

9m2  4 3m  2

 3m

(a  5) (a  5) (a  5) (a  2) a  5 a  2

 2





n 

2m2  7m  15 m  5

1

n(n  3) n  3



 4y2z 2

24.

 (y

1  4) (y  2)

m2 m  n



2mn  n2 m  n



m2  2mn  n2 m  n



(m  n) 2 m  n

mn

3 (y  4) (y  2)

580

35. The LCD is 6cd2. 2c 3d2

3 2cd



2c(2c) 3d2 (2c)



4c



2

44.

3(3d) 2cd(3d)

9d

x2 y3

4



3x 9y2

 6cd2  6cd2



3r  3

r



r2  21r (r  3) (r  3)



r2  21r  3r2  9r (r  3) (r  3)



4r2  12r (r  3) (r  3)



3x 9y2

x

31

x2

9y2 3x

y

36. The LCD is r 2  9. r2  21r r2  9



 y3 

4c2  9d 6cd2



x2 y3

45.



3x y

   

46.

1

4r  3

y9y

6  4

y4y

2  1



37. The LCD is (a  2) (a  1 ). 3a a  2

a1

3a(a  1) (a  2) (a  1)



5a(a  2) (a  2) (a  1)



3a2  3a (a  2) (a  1)



5a2  10a (a  2) (a  1)

5a

2

2

8a  7a (a  2) (a  1)

2.

38. The LCD is 21. 

9n 7

   



39. The LCD is 6a

7n(7) 9n(3)  7(3) 3(7) 49n 27n  21 21 49n  27n 21 22n 21

7 3a

3

7(2a)



14a 3  6a2 6a2 14a  3 6a2

4x 3

47.

40. The LCD is 5(x  4). 2x 2x  8

 5x

4  20

   x  2

41. 4  x

  

12

2x 4  5x  20  4) x 4  5x  20 x  4 x(5) 4  5x  20 (x  4) (5) 5x 4  5x  20 5x  20 5x  4 5x  20

 2(x 

4

   43. 3 

2

2

x  y x2  y2



9(y  4) y  4

y

6  4

y(y  1) y  1



4(y  1) y  1

y

2  1

y2  4y  9y  36  6 y  4 y2  y  4y  4  2 y  1 y2  13y  30 y  4 y2  5y  6 y  1

48. 6x 3

1



49.



3x  3y  x  y x2  y2



4x2  2y2 x2  y2

 (y

y  1  3 ) (y  2)



(y  10) (y  1) (y  4) (y  2)

or

y2  11y  10 y2  6y  8

7

7x

1

1



1

12 (7) 2



1

3

12 (7x) 12



1

12 (1) 4 1

1

2  6x116 2

2



1

6

x

6x (2) 3x 1



6x (1) 6 1

2 3r r2 3r 2 3r (3r)(r  2) 3r  r  2 (3r) (r  2) (2) (3r) (r  2) (3r)  3r r  2

1

 3

2  (3r)(r  2) (3)  (3r)(r  2) (3)

2

2r  4  9r  9r2  18r 16r  4 4

r  16

x2  y2 x2  y2 2

(y  10) (y  3 ) y  4

33  4  x 29  x

1

2

y2  5y  6 y  1

 2  12  4

11 2  3x 2x 11 2  3x 2x

6x (11) 2x

1 2x2 2(x  2) 1 x2 x  2 2x  4  1 x  2 2x  5 x  2

2





16x  42  7x  3 9x  45 x  5

x  2 (x  2 ) (x  2)

3(x2  y2 ) x2  y2

y2  13y  30 y  4

14x3  72 2  121127x  14 2

1

1

x  2







6

12 (4x) 3

4(x  2) x x2 x  2 4x  8  x x  2 5x  8 x  2

42. 2  x2  4  2 

5a  4 4a  6  8 a 5a  4 8  4a  6 a 5a  4 4  2a  3 a 20a  16 2a2  3a

1

3

 6a2  3a(2a)  6a2 

5a 4 a a a(4) 3(2)  2(4) 2(4) 5a  4 a 4a  6 8

y(y  4) y  4

2

 7n 3



3a  3a  5a  10a (a  2) (a  1)





3

4

3r(r  3) (r  3) (r  3)

4r(r  3 ) (r  3) (r  3 )

r

a 2

11

1



5a

1

r  4

2

581

Chapter 12

50. (x) (x  6)

1

x  2 x  3 x6 x x  2 x  3 x6 x

1 x



8.

2  (x)(x  6) 11x 2

(x) (x  3) 3(x) (x  3) x2  3x

1



1

3 x  2 x3 x2  3x 3 x  2 x3 x2  3x



The excluded values occur when c  7  0 or 3 c  7 and when 2c  3  0 or c  2. 9.

1

1 x

1

(n  4 ) (n  1) (1) n  4



1

1

1 n  4 1 n  4





(x ) (x  3) (1) (x )

n

n

1  1

1  1

1

2

(n  4) (n  1 ) (1) n  1

 4

 (n  4) (n  1) 1



1

1

9 t

t2 81  t2 t2 t  9 t

 

t  9 t

1

2 n2  3n  4

t

1



 (t

t2  9) (t  9 ) 1

t t  9

The excluded values occur when t  0, t2  0, t  9  0, and t  9  0. or t  9  0 t  0 or t  9  0 t  9 t9

2

10.

(n  4 ) (n  1 ) (2) (n  4 ) (n  1 ) 1



1

2

 n2  3n

t t

t2  81 t2 t  9 t2  81  t2 t

1

3  x2  2x  x  3 x2  x  0 x(x  1)  0 x  0 or x  1  0 x  1 Since the value 0 gives a 0 in the denominator, 0 is an extraneous solution. (n  4) (n  1)

1

9 t 81 t2

1



1

52.

1

2  (x)(x  3) 11x 2

(x) (x  3 ) (x  2) x  3

(2c  3) 2 (2c  3) (c  7) 2c  3 c  7

 

(x  6) (x  2)  (x)(x  3)  x  6 x2  8x  12  x2  3x  x  6 5x  12  x  6 6x  18 x3 51.

4c2  12c  9 2c2  11c  21

5 u t 6 2u 3 t

1



(5) (t) 6u  6t 6t 2u 3t  t t 5t  6u 6t 2u  3t t 5t  6u 2u  3t  6t t



5t  6u 6t

 

(n  1)  (n  4)  2 5  2

There is no solution to this equation.

1

Chapter 12 Practice Test

6

5t  6u  18t

The excluded values occur when t  0 and 12u  18t  0. 12u  18t  0 12u  18t

1. a; complex fraction 2. c; mixed expression 3. b; rational expression 4. xy  (40) (21) 5. xy  (4)(22) xy  840 xy  88 84y  840 x(16)  88 y  10 x  5.5

u 11.

1

5  2m 6m  15

5  2m

 3(5  2m )

5  2 15 x  2

x4x x6

1

1

 3 The excluded value occurs when 6m  15  0 5 or m  2. 3  x 2x2  5x  3



3  x (2x  1) (x  3 )



(x  4) (x  2)  5 x  2 (x  6) (x  2)  15 x  2



(x  4) (x  2)  5 x  2



x2  2x  8  5 x  2

 x2  4x



x2  2x  3 x  2

x  2  3



x2  2x  3 x2  4x  3

1  1

The excluded values occur when 2x  1  0 and x  3  0. 2x  1  0 or x  3  0 2x  1 x  3

12.

1

x2

2x x  7

x

14  7

 

2x  14 x  7 2(x  7) x  7

2

Chapter 12



(x  6) (x  2)  15 x  2 x  2  12  15

1

 x2  4x

The excluded values occur when x  2  0 and x2  4x  3  0. x  2  0 and x2  4x  3  0 x2 (x  3)(x  1)  0 or x  1  0 x30 x  3 x  1

1

 2x

3t 2

1

1

7.

t  3t

 12u

Page 701

6.

 2u

582

1

13.

n  3 2n  8



6n  24 2n  1

n  3  4)

 2(n   

14.

15.

1

(n  4)(n  5)

6n  18 4n  2 2(3n  9) 2(2n  1) 3n  9 2n  1

(n  4) (n  5) (2n) n  4

x  3  7x  12

 x2



z  2z  15 z2  9z  20

4n  56 n  14 3 x2  5x  6 3 (x  3) (x  2)

23.

(x  3)(x  2)

 (x

x  3  4 ) (x  3 ) 1



z  2z  15 1

x70 x7

1

z  3

1

1

4x2  11x  6 x2  x  6

x2  x  12  16

18.

(10z4



5z3

z2 )





5z3

1

(4x  3) (x  2 ) (x  3 ) (x  2 ) 1



1



10z4 5z3



10z4 5z3 1



5z3 5z3



6 6  3y

(x  4 ) (x  3 )  4 ) (x  4)

 (x



1

z2 5z

 

y2  2y  14y  28 7( y  2) ( y  2)

 

15

1 5z

y 6  3(y  2) 7(y  2) y 2 y2 7(y  2) y(y  2) 2(7) (y  2)  7(y  2) (y  2) 7(y  2) (y  2)

 21.

x2  1 x  1



1

A2

1x 36 y 21x

2

A

 y2 12

2

1x 36 y 21 (x  y12) (x  y) 2 3

1

1

1

3(x  y) 2

1

3

or 2 (x  y)

Chapter 12 Standardized Test Practice





(x2  1) (x  1) (x  1) (x  1)



x3  x2  x  1 (x  1) (x  1)



x3  x2  x  1 (x  1) (x  1)



x3  x2  x  1  x3  x2  x  1 (x  1) (x  1)



2x2  2x (x  1) (x  1) 2x(x  1) (x  1) (x  1) 2x x  1



7

or 18

1

A2

(x2  1) (x  1) (x  1) (x  1)



15 8

25. B; Area of triangle  2bh

x  5 6(x  2)  x2  2 x  5  6x  12 x  2 7x  17 x  2

x2  1 x  1

2  15(1)

3t  5t  15 8t  15

7

6x 

1

It will take them 18 hours to rake a lawn.

y2  16y  28  2) (y  2)

x  5 x  2

1

1 1 t  3t 5 1 1 t  3t 5

t

 7(y 20.

1 lawn . 3 hr

Let t  time for 1 lawn.

z2 5z3

1

1 lawn . 5 hr

The rate Kalyn can rake a lawn is

1

5z3 1

or

24. The rate Scott can rake a lawn is

 5z3  5z3

 2z  1  19.

 5x  14  0 (x  7) (x  2)  0 x20 x  2

1

4x  3 x  4

2z

y 7y  14

(1  x) (x  3) (x  2) x  2

Since the value 2 gives a 0 in the denominator, 2 is an extraneous solution.

 x2  8x

1





x2

1

(z  5 ) (z  3 ) (z  5 ) (z  4)



1  x

3  7x  14  x  3  x2  3x

1

x2  8x  16 x2  x  12

 x  2

3  7(x  2)  (1  x)(x  3)

 (z  3)  z2  9z  20  z  3





7(x  3) (x  2) x  3

1

z4 4x2  11x  6 x2  x  6

x  1

 x  2

7x  11  x2  2x  3

2

1

17.

7  3 7 x  3

x

1 (x  3)3(x  2)  x 7 3 2  (x  3)(x  2) 11x  2x 2

3(x  3) (x  2) (x  3) (x  2)

1

(x  8) (x  4 ) x  5



4(n  4) (n  5) n  5

8n  40  4n  16

x  8  5

16.

1n 2n 4  22  (n  4)(n  5) 1n 4 5 2

 2(n  4)(n  5) 

x 2

4  5

2n2  10n  2n2  2n  40  4n  16

1

x  4x  32 x  5

2n

(n  5)(2n)  2(n  4)(n  5)  4(n  4)

5m  12 2m  310m2  9m  36 () 10m2  15m 24m  36 ()24m  36 0 The quotient is 5m  12. 2

2n n  4

22.

6(n  4) 2n  1



Pages 702–703 1. D; Volume of a cylinder is r2h. V  (2.5) 2 (8)  157 2. B; Find the equation of the line. Find the slope using points (3, 10) and (5, 14). m

14  10 5  (3)



24 8

 3

y  mx  b 10  (3)(3)  b 1b y  3x  1

583

Chapter 12

10. Let w  width. length  5  w Area of playground    w 750  (5  w)w 750  5w  w2 2 w  5w  750  0 (w  30)(w  25)  0 or w  25  0 w  30  0 w  30 w  25 The width is 25 meters since measurements must be positive. Therefore, the length is 5  25 or 30 meters. 11. Find the discriminant. b2  4ac  (24) 2  4(16)(9)  576  576 0 Since the discriminant is 0, the equation has one real root.

3. B; The graph contains the points (0,8) and (10, 0). Find the slope. 0  (8) 10  0

8

4

 10  5

y  mx  b 4

8  5 (0)  b 8  b 4 x 5

y

5 4. A;

1

4 y  5x 4 y  5x

8

 8

2  5(8)

5y  4x  40 or 4x  5y  40 y  mx  b

1 12

5  4 (12)  b 53b 2b

12.

1

y  4x  2

1

1

x2  9 x3  x

x

3x  3

2

4y  x  8 or x  4y  8 5. D; Find the equation of the line. Find the slope using points (1, 0) and (0, 2) m



2 1

13.

 2



3(x  3) x2  1



3x  9 x2  1

x x  4

x

3x  3

x

3x  3

3

 x2  16  x

4x

x  4

x

x  4

 14. B; Plug x  1 x2  2x

1 y2  2y 1 8



 7

12

1 2 4

1 4

1



x2  16 4x



(x  4 ) (x  4) 4x

1

1

x  4 4

4

or

1 x 4

1

and y  4 into both expressions.

1 2

12 1 4

1 (4) 2  2(4) 16 7



1 1 16

1

2



1 1 16

1

8

 16



1 7 16



16 7

1

 16  8  8

15. C; Simplify both expressions. 1500  120  1180  1720  15  100  14  5  136  5  1144  5  1015  215  615  1215  215 1125  145  125  5  19  5  515  315  215 215  215 16. A; Find the excluded value of both expressions. 32a  0 b60 a0 b  6 0 7 6

100

 144(1  0.5) 20  144(0.5) 5  4.5 9. Let x  miles driven on the third day.  350

630  x  1050 x  420 They drove 420 miles on the third day.

Chapter 12

(x  3 ) (x  3) x (x2  1)

1

y  mx  b 0  2(1)  b 2b y  2x  2 Since the origin is not in the shaded area, the inequality is y 2x  2. 6. B; Use elimination to solve the system of equations. 3x  y  2 () 2x  y  8 5x  10 x  2 Substitute 2 for x in either equation. 3(2)  y  2 y4 The solution is (2, 4). 7. B; Let w  width. length  /  2.5 w Area of rectangle  /  w 9750  (2.5 w)w 9750  2.5 w2 8. C; y  C(1  r) t

360  270  x 3



1

4 y  4x  4(2)

2  0 0  1

(x  3) (x  3) x(x2  1) 1

y  4x  2 1



584

18a. The height increases. 18b. Sample answer: As the distance between the bottom of the ladder and the base of the building decreases, the height that the ladder reaches increases. 18c. No; Sample answer: When the bottom of the ladder is 6 feet from the base of the building, it reaches a height of about 10.4 feet. When the bottom of the ladder is 4 feet from the base of the building, it reaches a height of about 11.3 feet. In order to form an inverse variation 6  10.4 must approximately equal 4  11.3. However, 6  10.4  63.6 and 4  11.3  45.2. Because the products are not equal, this relationship does not form an inverse variation.

17. D; Simplify both expressions. 5

3x x  1



5(x  1) 3(x) x1 x  1 5x  5  3x x  1 8x  5 x  1



24y  15 3



3 (8y  5) 3



8y  5 y  1

 

24y  15 3 6y  6 6



6y  6 6

1

1

1

 6( y 1

6  1)

Since x and y can be any number, the relationship cannot be determined.

585

Chapter 12

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Chapter 13 Page 707

Statistics not be representative of the voting population. Since readers voluntarily write letters, this is a voluntary response sample. 5. The sample is the work from 4 students. The population is work from all students in the 1st period math class. The sample is a biased sample because students who raise their hands first may be expected to do better work. Since the students volunteer to have their work shown at the open house, this is a voluntary sample. 6. The sample is 25 nails. The population is all nails on the store shelves. This is a stratified random sample. Each of the nails was randomly selected from each of the 25 boxes. 7. The sample is 12 pencils. The population is all pencils in the school store. This sample is a biased sample, since all of the pencils came from the same box. Since Namid grabbed the closest box of pencils and 12 pencils from the top of the box, it is a convenience sample.

Getting Started

1. Sample answer: If a  5 and b  2, then c  3. However, 5  3. 2. Sample answer: It could be a yellow rose. 3. Sample answer: The speed limit could be 55 mph, and Tara could be driving 50 mph. 4. Sample answer: 6 is even, but not divisible by 4. 5. Order the set: 1, 7, 9, 15, 25, 59, 63 Find the middle value: 1, 7, 9, 15 , 25, 59, 63 The median is 15. 6. Order the set: 0, 2, 2, 2, 3, 3, 4, 5, 7, 8, 8, 9, 10 Find the middle value: 0, 2, 2, 2, 3, 3, 4 , 5, 7, 8, 8, 9, 10 The median is 4. 7. Order the set: 211, 218, 235, 317, 355, 395, 407, 407, 411, 726 Find the two middle values: 211, 218, 235, 317, 355 , 395 , 407, 407, 411, 726 Average the two middle values:

355  395 2

 375

The median is 375.

Pages 711–713

8. 7 8 9 10 11 12 13 14 15

9. 15 16 17 18 19 20 21 22 23

10. 3

4

5

6

7

1

2

3

4

5

11.

13-1 Sampling and Bias Page 710

Check for Understanding

1. All three are unbiased samples. However, the methods for selecting each type of sample are different. In a simple random sample, a sample is as likely to be chosen as any other from the population. In a stratified random sample, the population is first divided into similar, nonoverlapping groups. Then a simple random sample is selected from each group. In a systematic random sample, the items are selected according to a specified time or item interval. 2. A convenience sample is a biased sample that is determined based on the ease with which it is possible to gather the sample. A voluntary sample is a biased sample composed of voluntary responses. 3. Sample answer: Ask the members of the school’s football team to name their favorite sport. 4. The sample is a group of readers of the newspaper. The population is all readers of the newspaper. The sample is a biased sample because readers of a particular newspaper may

Chapter 13

Practice and Apply

8. The sample is 3 sophomores. The population is all sophomores in the school. The sample was randomly selected from all sophomores in the school, so the sample is unbiased. This is a simple random sample, since each name was equally likely to be chosen. 9. The sample is 20 shoppers. The population is all shoppers. Since shoppers at a fast-food restaurant are more likely to name the cola sold in the restaurant as their preference, this sample is biased. Since only shoppers outside the fast-food restaurant were selected for the sample, it is a convenience sample. 10. The sample is people who are home between 9 A.M. and 4 P.M. The population is all people in the neighborhood. Since people who are home during the weekdays may tend to behave similarly, the sample is biased. Since people were interviewed only during the typical work day, it is a convenience sample. 11. The sample is 860 people from a state. The population is all people in a state. The people are chosen at random, so the sample is unbiased. Since the people were selected by county, which is a sub-division of a state, it is a stratified random sample. 12. The sample is 10 scooters. The population is all scooters manufactured on the particular production line during one day. The sample is biased because samples were collected only on one day, and it is possible that more manufacturing mistakes occur early in the morning and late in the afternoon. Since the scooters were selected early at the beginning of the day and at the end of the day, the sample is a convenience sample.

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22. We know that the results are from a national survey conducted by Yankelovich Partners for Microsoft Corporation. 23. Additional information needed includes how the survey was conducted, how the survey respondents were selected, and the number of respondents. 24. Sample answer: Get a copy of the school’s list of students and call every 10th person on the list. 25. Sample answer: Get a copy of the list of registered voters in the city and call every 100th person. 26. The graph shows four phrases with a percent associated with each phrase. We can assume that the percents indicate the percent of respondents who said the indicated topic was discussed during family dinners. Based on the sum of the percents, respondents must have been able to choose or state more than one topic. We do not know how many respondents there were, whether the respondents selected topics from a list of choices or stated their own topics, whether there were any restrictions that may have existed about the topics, and the time period of the family dinners considered in this survey (a night, a week, a month, or more). 27. Sample answer: Randomly pick 5 rows from each field of tomatoes and then pick a tomato every 50 ft along each row. 28. Sample answer: Every hour pull one infant seat from the end of the assembly line for testing. 29. It is a good idea to divide the school population into groups and to take a simple random sample from each group. The problem that prevents this from being a legitimate stratified random sample is the way the three groups are formed. The three groups probably do not represent all students. The students who do not participate in any of these three activities will not be represented in the survey. Other students may be involved in two or three of these activities. These students will be more likely to be chosen for the survey. 30. Usually it is impossible for a company to test every item coming off its production lines. Therefore, testing a sample of these items is helpful in determining quality control. Answers should include the following. • Sample answer: An unbiased way to pick the CDs to be checked is to take every 25th CD off the production line. • Sample answer: A biased way to pick the CDs to be checked is to take the first 5 CDs coming off the production line in the morning. 31. B; The most accurate result would be achieved through an unbiased, random sample of the population in question. Choices A, C, and D result in biased samples. Therefore, the correct answer is B, which samples the appropriate population. 32. D; The most accurate result would be achieved through an unbiased, random sample. Choices A, B, and C result in biased samples. Therefore, the correct answer is D.

13. The sample is 3 students. The population is all students in Ms. Finche’s class. The sample is unbiased since each student’s number was equally likely to be selected. For the same reason, the sample is a simple random sample. 14. The sample is an 8-oz jar of corn. The population is all corn in the storage silo. The sample is biased because it was selected from corn near or at the surface of all corn in the silo, which is unlikely to represent all the corn in the storage silo. Since only corn that was easily accessible was sampled, the sample is a convenience sample. 15. The sample is a group of U.S. district court judges. The population is all U.S. district court judges. Since every 20th number is selected, the sample is unbiased. Since the population was divided into 11 federal districts before the sample was selected, the sample is a stratified random sample. 16. The sample is a group of people who watch a television station. The population is all people who watch a television station. The sample is biased because people who watch a particular television station may be more likely to have similar viewpoints about building the golf course. Since viewers decide whether to call the 1–900 number, the sample is a voluntary sample. 17. The sample is 4 U.S. Senators. The population is all U.S. Senators. The sample is biased because the President’s four closest colleagues are more likely to be members of his political party and share his political viewpoints. Since the President discussed issues with people who are easily accessible, the sample is a convenience sample. 18. The sample is a handful of Bing cherries. The population is all Bing cherries in the produce department. The sample is biased since the cherries were not randomly selected. Since the manager selected cherries at the edge of one case, this sample is a convenience sample. 19. The sample is a group of high-definition television sets. The population is all high-definition television sets manufactured on one line on one shift. The sample is unbiased since the sets were selected randomly. Since every 15th set was selected, the sample is a systematic random sample. 20. The sample is a group of employees. The population is all employees of the company. The sample is unbiased, because employees were selected at random. Since one employee was selected from each department, the sample is a stratified random sample. 21. The sample is a group of readers of a magazine. The population is all readers of the magazine. The sample is biased because the readers of a particular magazine may be more likely to select a particular actor as their favorite. Since the readers mailed their inputs to the magazine’s office, the sample is a voluntary sample.

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Page 713 33.

Maintain Your Skills 

41. 6b2

1 4

2  12y114 2 4 6 3 1 10 5 (12y ) 1 3y 2  (12y ) 1 2y 2  12y 1 4 2 12y

1

10 5  2y 3y 10 5  2y 3y

1

1

6b2  15  19b  19b  15  0 b  

1

40  30  3y 10  3y y

34. r(r  4) 1

r(r  4) (3) r  4

1

3 1 r r  4 3 1 r r  4





1 r

42.



1

r (r  4) r



1

4m(m  3) 1

4m (m  3) 4m



1

1

1 2m m3 4m 1 2m m3 4m

44. (c  3)(c  7)  c2  7c  3c  21  c2  10c  21 45. (x  4)(x  8)  x2  8x  4x  32  x2  4x  32 46. 4.5  3.8  8.3 47. 16.9  7.21  24.11 48. 3.6  18.5  22.1 49. 7.6  3.8  3.8 50. 18  4.7  13.3 51. 13.2  0.75  12.45

4m(m  3)(2)

1

(m  3)  8m2  8m(m  3) 8m2  m  3  8m2  24m 25m  3 3 m  25 36.

2x  5 x 2x  5 6



2x  5 x





2x  5 x 6 x

 2x

 37.

a(a  12)  35 a  12 a  7

38.

t2



a2  12a  35 a  12



(a  5) (a  7) (a  12) (a  7) a  5 a  12

t  2  3

t

1

a  7

1  2

1  3

39. 516  3154  4124  516  319  6  414  6  516  916  816  2216 cm 40. x2  6x  40  0 x  

Reading Mathematics

1a. This question will bias people toward answering “yes” because it gives them a reason to think that recycling will help alleviate a shortage in resources. 1b. This question will bias people toward answering “no” because most citizens are against the government making laws that require certain behaviors. 2a. This question is not biased. It does not influence a person to answer one way or the other. 2b. This question will bias people toward answering “no” because most people do not want taxes to be raised. 3a. Sample answer: Since we had hamburgers at the last party, would you prefer pizza for the next party? 3b. Sample answer: Would you prefer hamburgers or pizza for the class party?

6  5

t t

Page 714

2x  5 6

 (t  2) (t  2) (t  3) (t  2)

9  1105 4

 4.8, 0.3

2

4m(m  3) (2m)  m  3

(9)  2(9) 2  4(2) (3) 2(2)

43. (y  5)(y  7)  y2  7y  5y  35  y2  12y  35

2  4m(m  3)(2)

1

2

d

3r  (r  4)  r  4 2r  4  r  4 r8 35.

19  11 12 19  1 12 20 18 , 12 12 2 1 13, 12

2d  9d  3 2d2  9d  3  0

1

r (r  4) r

1

 1 33

2  r(r  4) 11r 2

1





10 3

(19)  2(19) 2  4(6) (15) 2(6)

(6)  2(6) 2  4(1) (40) 2(1)

13-2 Introduction to Matrices

6  1196 2 6  14 2

Page 717–718

 10, 4

Check for Understanding

1. A 2-by-4 matrix has 2 rows and 4 columns, and a 4-by-2 matrix has 4 rows and 2 columns. 2. Sample answer: 2 3 1 3 2 1 6 6 c d  c d 5 6 2 5 4 7 6 14

Chapter 13

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3. Estrella; Hiroshi did not multiply each element of the matrix by 5. 4. 3 by 3; first row, second column 5. 1 by 4; first row, first column 6. 4 by 1; third row, first column 7. 3 by 2; first row, second column 8. A is a 2 2 matrix, while C is a 1 2 matrix. To add matrices, the matrices must have the same dimension. Therefore, it is impossible to perform A  C. 9. B  A  c

15 14 20 10 d  c d 10 6 12 19

 c

15  20 14  (10) d 10  12 6  19

 c

5 24 d 22 13

10. 2A  2 c  c

0 0 26. £ 0 8 § 0 4 1 5 9 12 7 16 27. A  B  £ 0 4 2 §  £ 5 10 13 § 3 7 6 20 11 8

2(10) d 2(19)

40 20 d 24 38 11. 4C  4[5 7]  [4(5) 4(7) ]  [20 28] 12 10 F  £ 11 8 14 8

3 13 8§, R  £ 1 10 8

12 5 11

11 11 8 6 10 § , N  £ 1 8 11 § 2 10 15 11

 £

3 13 12 11 11 8§  £ 1 5 10 §  £ 1 10 8 11 2 10

12  13  11 11  1  1 14  8  10

36 30  £ 13 21 32 34

10  12  8 858 8  11  15

8 6 8 11 § 15 11

3  11  6 8  10  11 § 10  2  11

17. 18. 19. 20. 21. 22. 23. 24.

2 3 3 2 3 4 2 2

25. c

by by by by by by by by

2; 2; 1; 3; 3; 2; 3; 2;

34 91 63 52 9 70 d  c d 81 79 60 49 8 45

 c

34  (52) 81  (49)

91  9 63  70 d 79  (8) 60  45

 c

18 100 133 d 32 71 105 34 91 63 52 9 70 d  c d 81 79 60 49 8 45

 c

34  (52) 81  (49)

91  9 63  70 d 79  (8) 60  45

 c

86 82 7 d 130 87 15 12 7 16 1 5 9 5 10 13 §  £ 0 4 2 § 20 11 8 3 7 6

 £

12  (1) 75 16  9 5  0 10  (4) 13  (2) § 20  3 11  7 86

 £

11 2 5 14 17 4

25 15 § 2

1 5 9 31. 5A  5 £ 0 4 2 § 3 2 6

20 29 § 23

5(1)  £ 5(0) 5(3)

15. The total sales for the weekend 16. The greatest number in the matrix is 36, which represents small, thin crust pizzas. Therefore, small, thin crust pizzas had the most sales over the entire weekend.

Pages 718–721

13 12 7 5 6 11 § 23 18 14

30. B  A  £

13. No; the corresponding elements are not equal. 14. T  F  R  N 12 10  £ 11 8 14 8

 £

29. C  D  c

 c

12.

1  (12) 5  7 9  (16) 0  5 4  10 2  13 § 3  20 7  11 68

28. C  D  c

20 10 d 12 19

2(20) 2(12)

 £

5(5) 5(4) 5(2)

5(9) 5(2) § 5(6)

5 25 45  £ 0 20 10 § 15 10 30 32. 2C  2 c

Practice and Apply

first row, first column second row, first column third row, first column second row, second column second row, third column fourth row, first column second row, third column second row, first column

34 91 63 d 81 79 60

 c

2(34) 2(81)

 c

68 182 126 d 162 158 120

2(91) 2(79)

2(63) d 2(60)

2 1 1 d 1 5 1

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41. 2F  V  S  C

1 5 9 34 91 63 d 33. A  C  £ 0 4 2 §  c 81 79 60 3 7 6

 2[70 6 3 0.8]  [70 2 2 0.3]  [160 0 0 0]  [185 2 11 3.9]  [2(70)  2(6)  2(3)  2(0.8)]  [70 2 2 0.3]  [160 0 0 0]  [185 2 11 3.9]  [140  70  160  185 12  2  0  2 6  2  0  11 1.6  0.3  0  3.9]  [555 16 19 5.8]

Since the matrices do not have the same dimension, it is impossible to add them. 34. B  D  £

12 7 16 52 9 70 5 10 13 §  c d 49 8 45 20 11 8

18 42. N  £ 24 17

Since the matrices do not have the same dimension, it is impossible to add them. 35. 2B  A  2 £  £

12 7 16 1 5 9 5 10 13 §  £ 0 4 2 § 20 11 8 3 7 6

2(12) 2(5) 2(20)

2(16) 1 2(13) §  £ 0 2(8) 3

2(7) 2(10) 2(11)

43. 1.20 44.

5 9 4 2 § 7 6

5

9

3

7

6

4(1)  £ 4(0) 4(3)

12 7 5 10 20 11

45.

16 13 § 8

4  (12) 20  7 36  (16) 0  5 16  10 8  13 § 12  20 28  11 24  8

8  £ 5 8

37.

38.

2C  3D  2 c

34 81

13 52 26 21 § 17 16

2(34) 2(81)

 c

68 162

 c

68  (156) 162  (147)

 c

224 155 84 d 309 182 15

5D  2C  5 c

2(91) 2(79) 182 158

2(63) 3(52) d  c 2(60) 3(49)

3(9) 3(8)

3(70) d 3(45)

126 156 27 210 d  c d 120 147 24 135 182  27 126  210 d 158  (24) 120  135

52 9 70 34 91 d  2c 49 8 45 81 79

 c

5(52) 5(49)

 c

260 45 350 68 d  c 225 40 225 162

 c

260  68 45  182 225  162 40  158

5(9) 5(8)

63 d 60

5(70) 2(34) d  c 5(45) 2(81) 182 158

2(91) 2(79)

2(63) d 2(60)

126 d 120

350  126 d 225  120

331 304 325 343

1.20(32) 1.20(45) 1.20(26)

1.20(24) 1.20(47) 1.20(30)

1.20(21) 1.20(25) § 1.20(28)

38 29 25 54 56 30 § 31 36 34 4135 3840 4353 4436

26 24 41 36

15 571 14 473 T, B  D 13 347 15 533

4135 3840 4353 4436

26 24 41 36

15 571 14 473 TD 13 347 15 533

533  571 515  473 D 499  347 571  533

331  357 304  284 325  235 343  324

1104 988 D 846 1104

8548 7270 7782 8166

51. C; Since c

192 227 476  c d 63 118 345

39. V  [70 2 2 0.3], S  [160 0 0 0], C  [185 2 11 3.9] 40. 2F  2[ 70 6 3 0.8]  [2(70) 2(6) 2(3) 2(0.8) ]  [ 140 12 6 1.6] Chapter 13

331 304 325 343

1.20(28) 1.20(30) 1.20(19)

357 284 235 324

4413 3430 3429 3730

33 28 21 19

15 11 T 18 18

688 588 560 667

59 52 62 55

357 284 235 324

4413 3430 3429 3730

4135  4413 26  33 3840  3430 24  28 4353  3429 41  21 4436  3730 36  19

33 28 21 19

15 11 T 18 18

15  15 14  11 T 13  18 15  18

30 25 T 31 33

48. the total of the passing statistics for the 1999 and 2000 seasons 49a. sometimes 49b. always 49c. sometimes 49d. sometimes 49e. sometimes 49f. sometimes 50. Matrices can be used to organize data that can be displayed in a rectangular array of numbers. Answers should include the following: • A table is a rectangular array of numbers with headings to indicate what each row and column represents. A matrix is just the rectangular array of numbers. • Sample answer: The grades from each of five different tests for each student in a math class can be organized in a matrix.

91 63 52 9 70 d d  3c 79 60 49 8 45

 c

533 515 499 571

533 515 D 499 571

4 20 36 12 7 16  £ 0 16 8 §  £ 5 10 13 § 12 28 24 20 11 8  £

AD

21 25 § 28

46. 4 by 5, 4 by 5 47. T  A  B

4(9) 12 7 16 4(2) §  £ 5 10 13 § 4(6) 20 11 8

4(5) 4(4) 4(7)

28 32 24 30 45 47 19 26 30

22 34  £ 29 36 20 23

25 19 23  £ 10 16 24 § 43 29 22 1

18 rN  1.20 £ 24 17 1.20(18)  £ 1.20(24) 1.20(17)

24 14 32 1 5 9  £ 10 20 26 §  £ 0 4 2 § 40 22 16 3 7 6

36. 4A  B  4 £ 0 4 2 §  £

28 32 24 21 30 45 47 25 § 19 26 30 28

590

1 3 5 2 7 0 d  c d 7 2 0 13 1 8

 c

1  2 3  7 5  0 d 7  (13) 2  1 0  8

 c

3 6

4 5 d 1 8

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65. a1  4, n  5, r  2

52. B; When adding two matrices, one must add each of the corresponding elements. If M  N  M, the 0 0 d. only way the equation is true is if N  c 0 0 53. c

0.7 1.6

17 54. c 12.1 55. c

5.3 2.4

0.4 4

2.3 d 2.4

66. 67. 68.

4.6 3.9 d 13.5 14.3 12.4 7.7

21.1 d 4

69.

14.22 9.72 12.24 56. c d 10.62 7.92 13.86 57. c

3.92 3.12

0.48 2.04

Page 721

2.08 d 3.6

70.

Maintain Your Skills

58. The sample is 10 calendars. 59. Since the calendars examined each day are not selected randomly, the sample is biased. The sample is a convenience sample because the last 10 calendars printed each day are examined. 60. a  (a  1)

Page 721

1a 4 1  a3 2  a  (a  1)(1)

13x  x 4x 3 2  x  (x  3)(4)

3(x  3)  x(4x)  4x(x  3) 3x  9  4x2  4x2  12x 15x  9 9

3

x  15 or 5

62.

2(d  5) (d  9)

1dd  35  d 2 9 2  2(d  5) (d  9) 12d 5 10 2

2(d  3) (d  9)  2  2(d  5)  5(d  9) 2(d2  6d  27)  4(d  5)  5d  45 2d2  12d  54  4d  20  5d  45 2d2  13d  11  0 (2d  11) (d  1)  0 d

11 , 2

Practice Quiz 1

1. The sample is half of the households in a neighborhood. The population is all households in a neighborhood. Since each household is selected randomly, the sample is unbiased. The sample is a systematic random sample since every second household is surveyed. 2. The sample is half of the households in a neighborhood. Sample answer: voters in the state. Since the sample was not randomly drawn from all voters in the state, it is a biased sample. The sample is a convenience sample because only households in a particular neighborhood are surveyed. 5 7 8 3 d  c d 3. c 4 9 1 0

4a  3(a  1)  a  (a  1) 4a  3a  3  a2  a a  3  a2  a 0  a2  2a  3 0  (a  3)(a  1) a  3, 1 61. x  (x  3)

Since an  a1  r(n1) a  4  (2) (51) 5  4  (16)  64 b2  7b  12  (b  3)(b  4) a2  2ab 3b2  (a  b)(a  3b) Since the trinomial d2  8d  15 cannot be factored, it is prime. Sample answer: Megan saved steadily from January to June. In July, she withdrew money to go on vacation. She started saving again in September. Then in November, she withdrew money for holiday presents. In this graph, the domain is the months of the year. The range is the amount of money in Megan’s bank account.

4. c

1

d  5.5, 1

63. a1  4, n  5, r  3 Since an  a1  r(n1)

 c

8  5 3  (7) d 4  (1) 9  0

 c

3 4 d 5 9

7 2 8 9 6 4 d  c d 1 3 2 5 3 1  c

9  7 6  (2) 1  5 3  (3)

 c

16 8 4 d 6 6 1

5. 3 c

(3) (51)

a5  (4)   4  81  324 64. a1  2, n  3, r  7 Since an  a1  r(n1)

48 d 21

8 3 4 5 d 6 1 2 10

 c

3(8) 3(6)

 c

24 9 12 15 d 18 3 6 30

3(3) 3(1)

3(4) 3(2)

3(5) d 3(10)

a3  (2)  (7) (31)  (2)  49  98

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12. For the men, the graph represents 5  6  4  2  17 values. So, the median of the data is the 9th value, which occurs in the range 220–240 rebounds. For the women, the graph represents 1  2  5  5  1  1  15 values, so the median of the data is the 8th value, which falls in the range 240–260 rebounds. The men’s rebounding data show a vast majority of values are between 200 and 240 rebounds. The women’s rebounding data are somewhat symmetrical. The top two women have more rebounds than any of the men. 13. The graph representing age at inauguration has 8  24  9  41 values. The median of the data is the 21st value, which occurs in the range 50–60 years. The graph representing age at death has 2  5  12  11  5  2  37 values. The median of the data is the 19th value, which occurs in the range 60–70 years. Both distributions show a symmetrical shape. The two distributions differ in their spread. The inauguration ages are not spread out as much as the death ages data. 14. Sample answer:

Histograms

13-3 Page 725

Check for Understanding

1. First identify the greatest and least values in the data set. Use this information to determine appropriate measurement classes. Using these measurement classes, create a frequency table. Then draw the histogram. Always remember to label the axes and give the histogram a title. 2. 50 v 6 60 3. Sample answer: 1, 1, 2, 4, 5, 5, 8, 9, 10, 11, 12, 13, 22, 24, 41 4. $200–250 5. There are no gaps. The data are somewhat symmetrical. 6. In Group A, there are 31 values, so the median value is the 16th. It falls in the 40–45 range. In Group B, there are 26 values, so the median value is the average between the 13th and 14th values. It also falls in the 40–45 range. 7. The Group A test scores are somewhat more symmetrical in appearance than the Group B test scores. There are 25 of 31 scores in Group A that are 40 or greater, while only 14 of 26 scores in Group B are 40 or greater. Also, Group B has 5 scores that are less than 30. Therefore, we can conclude that Group A performed better overall on the test. 8. Sample answer:

Semester Scoring Average in Mathematics Class Frequency

8 6 4 2 0 7075

8 6 4 2 0

75- 80- 85- 90- 9580 85 90 95 100 Weighted Average

15. Sample answer:

Raisins Counted in Snack-Size Boxes 30- 40- 50- 60- 70- 8040 50 60 70 80 90 Passengers (millions)

Frequency

Frequency

Passenger Traffic at U.S. Airports, 2000

9. B; The graph represents 32 employees, which means that the median is the average of the 16th and 17th values. Those values both fall in the $30–40 thousand range, so the median of the data falls between $30,000 and $40,000. The statement that is not correct is answer choice B.

14 12 10 8 6 4 2 0

50- 6060 70

70- 80- 90- 10080 90 100 110

Number of Raisins in a Box

Pages 726–728

Practice and Apply

16. Sample answer:

10. The median occurs at the average of the 10th and 11th values. So it falls in the range 350–600 thousand. There is a gap between 1100 and 1600 thousand. The histogram is highly skewed to the left, with a vast majority of the data values in the lowest measurement class. Over half of the top 20 newspapers have circulations between 350 and 600 thousand. 11. The graph represents 8  5  3  4  3  1  24 values. So, the median of the data is the average of the 12th and 13th values, so it is 3400–3800 championship points. There are no gaps. The data appear to be skewed to the left. Chapter 13

Frequency

Payrolls for Major League Baseball Teams in 2000 12 10 8 6 4 2 0

median $54,000,000

10- 20- 30- 40- 50- 60- 70- 80- 90- 100- 11020 30 40 50 60 70 80 90 100 110 120 Team Payroll (millions of dollars)

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27. Sample answer:

17. Sample answer: —See graph above. 18. The data has 50 values, so the median is the average between the 25th and 26th value in the ordered set. The median, then, is 54.07  53.66 2

 53.865. 19. Sample answer:

Percent of Eligible Voters Who Voted in the 2000 Presential Election

28. Sample answer:

Frequency

15 12 9 6 3 0 40- 45- 50- 55- 60- 6545 50 55 60 65 70

29. Sample answer:

Percent Eligible Voters Who Voted

Frequency

20. Sample answer: There are no gaps. The data are somewhat symmetrical. 21. See students’ work. 22. Sample answer: 4

30. A is a 3 3 matrix. B is a 2 3 matrix. It is impossible to add these matrices.

2

7 5 2 2 3 7 31. C  A  £ 0 0 3 §  £ 0 4 6 § 1 4 6 1 5 4

0 15- 20- 25- 30- 3520 25 30 35 40

4045

4550

5055

5560

6065

 £

23. Histograms can be used to show how many states have a median within various intervals. Answers should include the following. • A histogram is more visual than a frequency table and can show trends easily.

9 8 5  £ 0 4 3 § 2 9 2

Frequency

Year 2000 State Mean SAT Mathematics Scores 20 15 10 5 0

7  (2) 5  3 2  7 0  0 0  (4) 3  6 § 1  1 4  (5) 6  4

32. 2B  2 c

480- 500- 520- 540- 560- 580- 600500 520 540 560 580 600 620 Score

8 1 1 d 2 3 7

 c

2(8) 2(2)

 c

16 2 4 6

2(1) 2(3)

2(1) d 2(7)

2 d 14

2 3 7 33. 5A  5 £ 0 4 6 § 1 5 4

24. C; According to the graph, the number of employees represented in the graph is 9  12  11  6  6  2  46 employees. 25. B; The data has 46 values. The median is the average of the 23rd and 24th values. Therefore, the median falls in the range 4–6 days absent. 26. Sample answer:

5(2)  £ 5(0) 5(1)

5(3) 5(4) 5(5)

5(7) 5(6) § 5(4)

10 15 35  £ 0 20 30 § 5 25 20 34. Since each CD player has an equal chance of being selected, the sample is unbiased. The CD players are selected at a regular time interval, so the sample is a systematic random sample.

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1

35.

36.

s s  7

s  5  7

s

s

s  7

s

s  5

1

2m2  7m  15 m  2

Find the middle values: 1.6, 2.1, 3.901 , 5.21 , 7.4, 13.9 Average the values:

2m  3

 m2  5m  6

1



43. Order the set: 1.6, 2.1, 3.901, 5.21, 7.4, 13.9

s  7  5

s

1



(m  2) (m  3) 2m  3 1

 (m  5)(m  3)

Page 730

37. 1y  3  5  9 1y  3  4 y  3  16 y  13 Check:

 4.56

The median is about 4.56.

1

(2m  3) (m  5) m  2

3.901  5.21 2

Graphing Calculator Investigation (Follow-Up of Lesson 13-3)

1. To make a scatterplot, enter the x values in L1 and the y values in L2. Use STAT PLOT to graph the scatter plot. Use ZOOM

?

1(13)  3  5  9

9 to graph.

?

116  5  9 ?

459 99 38. 1x  2  x  4 x  2  x2  8x  16 0  x2  9x  18 0  (x  6) (x  3) x  6, 3 Check:

The graph appears to be an exponential regression. To find the coefficient of determination, select

?

1(6)  2  (6)  4

ExpReg on the

?

14  6  4 22

STAT

KEYSTROKES: STAT

CALC menu. 0 ENTER

R2

 0.9969724389 2. To make a scatterplot, enter the x values in L1 and the y values in L2. Use STAT PLOT to graph

?

1(3)  2  (3)  4 ?

11  1 1 1 x  3 is an extraneous solution.

the scatter plot. Use ZOOM

9 to graph.

39. 13  12w  5 169  2w  5 174  2w 87  w Check:

?

13  12(87)  5 ?

13  1174  5

The graph appears to be a linear regression. To find the coefficient of determination, select

?

13  1169 13  13 40. Order the set: 2, 4, 7, 9, 12, 15 Find the middle values: 2, 4, 7 , 9 , 12, 15 Average the values:

7  9 2



16 2

LinReg(ax  b) on the KEYSTROKES: STAT

8  10 2

CALC menu.

4 ENTER

R2  0.9389164209 3. To make a scatterplot, enter the x values in L1 and the y values in L2. Use STAT PLOT to graph

or 8

The median is 8. 41. Order the set: 1, 3, 6, 8, 10, 12, 15, 7 Find the middle values: 1, 3, 6, 8 , 10 , 12, 15, 17 Average the values:

STAT

the scatter plot. Use ZOOM 9 to graph.

9

The median is 9. 42. Order the set: 2, 4, 7, 7, 9, 19 Find the middle values: 2, 4, 7 , 7 , 9, 19 Average the values:

7  7 2

7

The median is 7.

The graph appears to be a linear regression. To find the coefficient of determination, select LinReg(ax  b) on the KEYSTROKES: STAT

R2  0.9974802029

Chapter 13

594

STAT

CALC menu.

4 ENTER

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4. To make a scatterplot, enter the x values in L1 and the y values in L2. Use STAT PLOT to graph

13-4 Measures of Variation

the scatter plot. Use ZOOM 9 to graph.

Pages 733–734

The graph appears to be a quadratic regression. To find the coefficient of determination, select QuadReg on the

STAT

KEYSTROKES: STAT

Check for Understanding

1. Sample answer: 1, 4, 5, 6, 7, 8, 15 and 1, 2, 4, 5, 9, 9, 10 2. Outliers may make the mean much greater or much less than the mean of the data excluding the outliers. 3. Alonso; the range is the difference between the greatest and the least values of the set. 4. Range: 85 – 25  60 median

CALC menu.

678

5 ENTER

25 58 59 62 67 69 69 73 75 76 77 77 82 85 64748 c c 6973 Q2  2  71 Q3 Q1

R2  0.97716799 5. Enter the year in L1 and the cost in L2. Use STAT PLOT to graph the scatter plot. Use

ƒ d Q  Q  IQR  77  62  15 S ƒ 3

ZOOM 9 to graph.

1

To find the outliers: Q1  1.5(IQR)  62  1.5(15)  39.5 Q3  1.5(IQR)  77  1.5(15)  99.5 The data set consists of one value which is less than 39.5 or greater than 99.5. Therefore, 25 is the only outlier. 5. Range: 11.9  7.3  4.6

6. Find the coefficient of determination for linear, quadratic, and exponential regression. Choose the equation whose coefficient of determination is closest to 1. Linear: Select LinReg(ax  b) on the STAT CALC menu. R2  0.9312567132 Quadratic: Select QuadReg on the STAT CALC menu. R2  0.9880773362 Exponential: Select ExpReg on the STAT CALC menu. R2  0.9624472328 The value of R2 is closest to 1 for the quadratic regression. The regression equation is

8.7  9.4  9.05 2 10.0  10.1 Q :  10.05 2 3 8.0  8.0 Q1:  8.0 2

Median:

IQR  Q  Q 3 1

 10.05  8.0  2.05 To find the outliers: Q1  1.5(IQR)  8  1.5(2.05)  4.925 Q3  1.5(IQR)  10.05  1.5(2.05)  13.125 The data set has no values less than 4.925 or greater than 13.125, so it has no outliers. 6. Range: 21 – 1  20 runs 7. There are 54 values in the data set, so the median is the average of the 27th and 28th values. The

y  0.4107142857x2  1645.696429x  1,648,561. The coefficient of determination is R2  0.9880773362. 7. Use the quadratic regression equation with x  2004.

5  5

median, then, is 2  5 runs. 8. The lower quartile is the 14th value, 4 runs. The upper quartile is the 41st value, 10 runs. 9. IQR  Q  Q 3 1

y  0.4107142857x2  1645.696429x  1,648,561 y  0.4107142857(2004) 2  1645.696429(2004)  1,648,561 y  20.5 The cost in 2004 is expected to be about 20.5¢. 8. No; the price will start to increase using the quadratic model. 9. An exponential model may be more appropriate for predicting cost beyond 2003, since it will continue the pattern of decreasing prices where the annual decrease is getting smaller.

 10  4  4 runs.

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14. The greatest number is 5.3. The least number is 1. The range is 5.3  1 4.3. Order the set:

10. To find the outliers: Q1  1.5(IQR)  4  1.5(4)  2 Q  1.5(IQR)  10  1.5(4) 3  16 runs The data set has no values less than 2, but it does have two values greater than 16 runs. The two values greater than 16 are 17 runs and 21 runs, which are outliers.

Pages 734–736

Q1 

Q2 

 108 S ƒ

ƒ d 41  1.5(11.5)  58.25 S ƒ

 30.6 ƒd IQR  30.9  30.05  0.85 S ƒ ƒd 30.9  1.5(0.85)  32.175 Sƒ

65  68 2 84  88 2

 66.5  86

 8.65  6.35  2.3 To find the outliers: 6.35  1.5(2.3)  2.9 8.65  1.5(2.3)  12.1 Since the data set has no values less than 2.9 and no values greater than 12.1, it has no outliers.

Since the data set has no values less than 28.775 or greater than 32.175, the set has no outliers.

Chapter 13

 7.15 S ƒ

 221  202  19 To find the outliers: 202  1.5(19)  173.5 221  1.5(19)  249.5 Since the data set has no values less than 173.5 and no values greater than 249.5, it has no outliers. 17. The greatest value in the set is 11.7. The least value in the set is 5.0. The range is 11.7  5.0  6.7. The data set has 25 values, so the median is the 13th value, or 7.6. The lower quartile is the average of the 6th and 7th values, which is 6.3  6.4  6.35. The upper quartile is the average 2 8.5  8.8 of the 19th and 20th values, or  8.65. 2 IQR  Q  Q 3 1

30.8  31.0

 28.775 Sƒ

ƒ d 4.15  1.5(2)

 86  66.5  19.5 To find the outliers: 66.5  1.5(19.5)  37.25 86  1.5(19.5)  115.25 Since the data set has no values less than 37.25 or greater than 115.25, it has no outliers. 16. The greatest value in the set is 232. The least value in the set is 193. The range is 232  193  39. The data set has 19 values, so the median is the 10th value, or 218. The lower quartile is the 5th value, or 202. The upper quartile is the 15th value, or 221. IQR  Q  Q 3 1

 30.05 median Q 3   30.9 Q1  2 2 144 4442444443 T 144442444443 29.9 30.0 30.1 30.5 30.7 30.8 31.0 31.0 6447448 30.5  30.7 Q  2 2

ƒd 30.05  1.5(0.85)

 3.1

IQR  Q  Q 3 1

Since the data set has no values less than 12.25 or greater than 58.25, the set has no outliers. 13. The greatest number is 31.0. The least number is 29.9. The range is 31.0  29.9  1.1. Order the set: 30.0  30.1

3  3.2 2

 0.85 S ƒ

Q3 

ƒ d IQR  41  29.5  11.5 S ƒ  12.25 S ƒ

 4.15

ƒ d 2.15  1.5(2)

Q  1

Since the data set has no values less than 32 and no values greater than 108, the data set has no outliers. 12. The greatest number is 42. The least number is 28. The range is 44  28  16. Order the set: median T 28 28 29 30 31 34 37 38 39 40 42 42 44 6447448 6447448 c 29  30 40  42 Q1   29.5 Q2 Q3   41 2 2 ƒ d 29.5  1.5(11.5)

3.4  4.9 2

Since the data set has no values less than 0.85 and no values greater than 7.15, it has no outliers. 15. The greatest value is 99. The least value is 53. The range is 99  53  46. The data set has 16 values. The median is the 77  77 average of the 8th and 9th values, or  77. 2

ƒ d IQR  79.5  60.5  19 S ƒ

 32 S ƒ

Q  3

ƒ d IQR  4.15  2.15  2 S ƒ

11. The greatest number is 92. The least number is 55, the range is 92  55  37. Order the set: median T 55 58 59 62 67 69 73 75 77 77 82 448 85 92 644474 4448 6444744 c 62  59 77  82  60.5 Q2 Q3   79.5 Q1  2 2 ƒ d 79.5  1.5(19)

 2.15 median

6447448 T 144424443 1 2 2.3 3 3.2 3.4 4.9 5.3 6474 8

Practice and Apply

ƒ d 60.5  1.5(19)

2  2.3 2

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29. The greatest value for a cable-stayed bridge is 1630 ft. The least value for a cable-stayed bridge is 630 ft. The range for cable-stayed bridges is 1630  630  1,000 ft. The greatest value for a steel-arched bridge is 1700 ft. The least value for a steel-arched bridge is 730 ft. The range for steel-arched bridges is 1700  730  970 ft. 30. For cable-stayed bridges: 1000  1000 Median:  1000 ft 2

18. The greatest value in the set is 6.8. The least value in the set is 0.0. The range is 6.8  0.0  6.8. The data set has 20 values, so the median is the average of the 10th and 11th values, or 2.6  3.3  2.95. The lower quartile is the average 2 1.7  1.9  1.8. The of the 5th and 6th values, or 2 upper quartile is the average of the 15th and 16th 3.9  4.0 values, or  3.95. 2 IQR  Q3  Q1

 3.95  1.8  2.15 To find the outliers: 1.8  1.5(2.15)  1.425

3.95  1.5(2.15)  7.175

Since the data set has no values less than 1.425 and no values greater than 7.175, it has no outliers. 19. The greatest value on the chart is 588,641 visitors. The least value on the chart is 117,080. The range is 588,641  117,080  471,561 visitors. 20. Order the data: 117,080

123,947 138,600 155,533 167,552 242,938

343,149

395,604 465,978 474,082 571,775 588,641

31.

Average the middle two values to find the median:

21.

242,938  343,149  293,043.5 visitors 2 138,600  155,533 Q   147,066.5 visitors 2 1 465,978  474,082 Q   470,030 visitors 2 3

32.

22. IQR  Q3  Q1  470,030  147,066.5  322,963.5 visitors 23. To find any outliers: Q1  1.5(IQR)  147,066.5  1.5(322,963.5)  337,378.75 Q3  1.5(IQR)  470,030  1.5(322,963.5)  954,475.25 Since the set has no data less than 337,378.75 or greater than 954,475.25, there are no outliers. 24. The greatest value is 304. The least value is 9. The range of the data is 304  9  295 Calories. 25. Order the data: 9 9 10 13 14 17 17 17 20 25 28 30 35 60 60 66 89 304. The median is the average of the middle two 20  25 values:  22.5 Calories. 2

33.

26. Q  14 Calories, Q  60 Calories 1 3 27. IQR  Q  Q 3 1  60  14  46 Calories 28. To find any outliers: Q1  1.5(IQR)  14  1.5(46)  55 Q3  1.5(IQR)  60  1.5(46)  129 The set has no values less than 55 Calories, but one value, 304, that is greater than 129 Calories. The outlier is 304 Calories (avocado).

34.

597

Q  760 ft 1 Q3  1280 ft For steel-arch bridges: Median: 980 ft Q1  820 ft Q3  1100 ft For cable-stayed bridges: IQR  Q3  Q1  1280  760  520 ft For steel-arch bridges: IQR  Q3  Q1  1100  820  280 ft To find outliers for cable-stayed bridges: Q1  1.5(IQR)  760  1.5(520)  20 ft Q3  1.5(IQR)  1280  1.5(520)  2060 ft Since the set has no values less than 20 ft and no values greater than 2060 ft, there are no outliers among the cable-stayed bridges. To find the outliers for steel-arch bridges: Q1  1.5(IQR)  820  1.5(280)  400 ft Q  1.5(IQR)  1100  1.5(280) 3  1520 ft The set has no values less than 400 ft, but it does have two values—1650 and 1700—greater than 1520 ft. There are two outliers—1650 ft and 1700 ft—among the steel-arch bridges. Although the range of the cable-stayed bridges is only somewhat greater than the range of the steel-arch bridges, the interquartile range of the cable-stayed bridges is much greater than the interquartile range of the steel-arch bridges. The outliers of the steel-arch bridges make the ranges of the two types of bridges similar, but in general, the data for steel-arch bridges are more clustered than the data for the cable-stayed bridges. The median, lower quartile, and upper quartile will each decrease by 2 in., since all values will decrease by 2 in. The range and interquartile range will remain the same, since the difference between values will not increase.

Chapter 13

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35. Measures of variation can be used to discuss how much the weather changes during the year. Answers should include the following. • The range of temperatures is used to discuss the change in temperatures for a certain area during the year, and the interquartile range is used to discuss the change in temperature during the moderate 50% of the year. • The monthly temperatures of the local area listed with the range and interquartile range of the data. 36. B; The greatest value in the set is 65. The least value is 40. The range is 65  40  25. 37. A; Order the set: 1 2 3 7 8 9 14 15 18 24 Average the two middle values to find the median: 8  9 2

45. 3 4 5 6 7 8 9 10 11 12 13

46. 10 12 14 16 18 20 22 24 26 28 30

47. 20 25 30 35 40 45 50 55 60 65 70

Page 736

 8.5

Page 736

Maintain Your Skills

38. Sample answer: Data 30–35 35–40 40–45 45–50 50–55 55–60 60–65

The median is

Frequency 2 6 2 5 0 0 3

4 2 0 3540

4045

4550

5055

5560

6065

39. 1 by 3; first row, first column 40. 3 by 2; third row, second column 41. 2 by 4; second row, second column 42.

15a 39a2



13-5

1 3a  5 3a  13a

5

t  3 t2  7t  12

 1

t

m  3 m2  9



1 t  3 (t  3) (t  4)

1 ;  4

1 m  3 (m  3) (m  3 ) 1

1

 m  3;

m2  9  (m  3)(m  3) or m  3  0 m30 m  3 m3 The excluded values are 3 and 3.

Chapter 13

Check for Understanding

1. The extreme values are 10 and 50. The quartiles are 15, 30, and 40. There are no outliers. 2. The scale must include the least and greatest values. 3. Sample answer: 2, 8, 10, 11, 11, 12, 13, 13, 14, 15, 16

t2  7t  12  (t  3)(t  4) t  3  0 or t  4  0 t3 t4 The excluded values are 3 and 4. 44.

 1075.

Box-and-Whisker Plots

Pages 739–740

1

 13a; excluded value is 0 43.

1075  1075 2

Q1  1025 Q3  1125 IQR  Q3  Q1  1125  1025  100 5. To find the outliers: Q  1.5(IQR)  1025  1.5(100) 1  875 Q  1.5(IQR)  1125  1.5(100) 3  1275 The set has one value, 835, less than 875 and no values greater than 1275. The set has one outlier—835.

6

3035

Practice Quiz 2

1. There are 4  7  6  2  19 values in the data set. As a result, the median occurs at the 10th value. The median occurs in the $10–20 range. 2. There is a gap in the $30–$40 measurement class. Most of the books cost less than $30. The distribution is skewed to the left. 3. The greatest value is 1175. The least value is 835. The range of the data is 1175  835  340. 4. Order the set: 835 975 1005 1025 1050 1055 1075 1075 1095 1100 1125 1125 1145 1175

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4. Step 1

6. Step 1 Determine the quartiles and outliers for each set of data. A: 17 18 22 22 24 26 28 31 32 c c c Q1  20 Q2  24 Q3  29.5 B: 23 24 27 27 28 30 30 32 45 c c c Q1  25.5 Q2  28 Q3  31 Set B has one outlier, 45. Step 2 Draw the box-and-whisker plots using the same number line.

Determine the quartiles and any outliers. Order the data from least to greatest. Use this list to determine the quartiles. 15, 16, 17, 21, 22, 22, 24, 24, 28, 30 Q  2

22  22 2

 22

Q1  17 Q3  24 Determine the interquartile range. IQR  Q3  Q1  24  17 7 Check to see if there are any outliers. 17  1.5(7)  6.5 24  1.5(7)  34.5 There are no values less than 6.5 or greater than 34.5. There are no outliers. Step 2 Draw a number line. 15 17

22 24

A B

30 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46

The B data have a greater range than the A data. However, if you exclude the one outlier in the B data, the B data are less diverse than the A data. 7. Step 1 Determine the quartiles and outliers for each set. A:

14 16 18 20 22 24 26 28 30

Step 3

Complete the box-and-whisker plot.

14 16 18 20 22 24 26 28 30

5. Step 1

Determine the quartiles and any outliers. Order the data from least to greatest. Use this list to determine the quartiles. 64 65 66 66 67 67 68 69 71 74 Q  2

67  67 2

8 11.4

14 14 c Q1  14

 67

66 67

69

71

64

66

68

70

Step 2 Draw the box-and-whisker plots using the same number line.

A B 8

10 12 14 16 18 20 22 24

The A data are more diverse than the B data. 8. Step 1 Determine the quartiles and outliers for the data.

74

72

452 467 472 524 559 573 620 678 693 1397 Q1  472

74

66

68

70

72

c Q2  566

c Q3  678

The data has one outlier, 1397. Step 2 Draw the box-and-whisker plot for the data.

Complete the box-and-whisker plot.

64

16.7 19 24 c Q3  16.7

13 14 15 15.8 16 16 18 20 c c c Q1  13 Q2  15.4 Q3  16

c

Step 3

15.5 16 c Q2  15.25

B: 9 12

Q1  66 Q3  69 Determine the interquartile range. IQR  Q3  Q1  69  66 3 Check to see if there are any outliers. 66  1.5(3)  61.5 69  1.5(3)  73.5 The set has no values less than 61.5 and one value, 74, greater than 73.5. It is the only outlier. Step 2 Draw a number line. 64

15

74 400 500 600 700 800 900 1000 1100 1200 1300 1400 Million Dollars

599

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Complete the box-and-whisker plot.

9. Most of the data are spread fairly evenly from about $450 million to $700 million. The one outlier ($1397 million) is far removed from the rest of the data.

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28

Pages 740–742

17. Order the data. Use this list to determine the quartiles.

Practice and Apply

10. The greatest value is 130. The least value is 85. The range of the data is 130  85  45.

23

11. From the graph, Q3  120 and Q1  90. As a result, the interquartile range is 120  90  30. 12. Since Q  90, 1 13. Since Q  95, 2

1 4 1 2

5

of the data is less than 90. of the data is greater than 95.

10

15

20

25

2

4

5

5

5

6

7

12

16

16

17

5.8

5

10

15

20

25

30

53 54 c Q2  52

55

60

69

81

c Q3  57.5

40

50

70

60

80

6.2

7.6 8.5 c Q1

8.5

8.8 9.0 c Q2  8.65

10.5 11.5 c Q3

15.1

5 6 7 8 9 10 11 12 13 14 15 16

19. Order the data. Use the list to determine the quartiles. 1.2 1.3 1.3 1.3 1.3 1.6 1.8 2 2.2 3.3 3.7 5.7 7.7 8.5 14 c c c Q1 Q2 Q3

Determine the interquartile range. IQR  5.7  1.3  4.4 Check to see if there are any outliers. 1.3  1.5(4.4)  5.3 5.7  1.5(4.4)  12.3 There is one value, 14, greater than 12.3. It is the only outlier. Complete the box-and-whisker plot.

30

16. Order the data. Use this list to determine the quartiles. 1 1 1 2 2 4 4 5 5 5 6 8 10 10 14 15 27 c c c Q1  2 Q2 Q3  10

Determine the interquartile range. IQR  10  2  8 Check to see if there are any outliers. 2  1.5(8)  10 10  1.5(8)  22 There is one value, 27, greater than 22. It is the only outlier.

Chapter 13

51

Determine the interquartile range. IQR  10.5  7.6  2.9 Check to see if there are any outliers. 7.6  1.5(2.9)  3.25 10.5  1.5(2.9)  14.85 There is one value, 15.1, greater than 14.85. It is the only outlier. Complete the box-and-whisker plot.

c c c Q1 Q2  6.5 Q3 Determine the interquartile range. IQR  16  5  11 Check to see if there are any outliers. 5  1.5(11)  11.5 16  1.5(11)  32.5 There are no values less than 11.5 or greater than 32.5, so there are no outliers. Complete the box-and-whisker plot.

0

30

20

30

20

46 46 c Q1  42.5

18. Order the data. Use the list to determine the quartiles.

15. Order the data. Use this list to determine the quartiles. 0

39

Determine the interquartile range. IQR  57.5  42.5  15 Check to see if there are any outliers. 42.5  1.5(15)  20 57.5  1.5(15)  80 There is one value, 81, greater than 80. It is the only outlier. Complete the box-and-whisker plot.

14. Order the data. Use this list to determine the quartiles. 1 1 2 3 4 5 6 8 10 15 27 c c c Q2 Q3 Q1 Determine the interquartile range. IQR  10  2  8 Check to see if there are any outliers. 2  1.5(8)  10 10  1.5(8)  22 Since 27 is greater than 22, it is the only outlier. Complete the box-and-whisker plot.

0

27

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

20. The least value in both sets is 20. It is part of data set B. 21. The greatest value in both sets is 70. It is part of data set B. 22. The IQR for data set A is 60  30  30. The IQR for data set B is 60  40  20. Data set A has the greater IQR.

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23. The range for data set A is 65  25  40. The range for data set B is 70  20  50. Data set B has the greater range. 24. Determine the quartiles and outliers for each data set. A: 15 16 17 19 22 24 26 28 32 38 40 c c c Q1 Q2 Q3 B: 24 25 27 27 28 29 30 30 31 32 37 c c c Q1 Q2 Q3 A: IQRA  32  17  15 17  1.5(15)  5.5; 32  1.5(15)  54.5; no outliers B: IQRB  31  27  4

26. Determine the quartiles and outliers for each data set. A: 1.5 3.5 3.8 3.9 4.0 4.0 4.1 4.1 4.2 4.4 { c c c outlier Q1 Q2  4.0 Q3 B: 4.2 4.8 5.5 6.7 6.8 7.1 7.6 12.2 { c c c Q1  5.15 Q2  6.75 Q3  7.35 outlier A: IQRA  4.1  3.8  0.3 3.8  0.3  3.5; 4.1  0.3  4.4; 3.1 is an outlier. B: IQRB  7.35  5.15  2.20 5.15  2.20  2.05; 7.35  2.20  9.55; 12.2 is an outlier. Draw the box-and-whisker plots using the same number line.

27  1.5(4)  21; 31  1.5(4)  37; no outliers Draw the box-and-whisker plots using the same number line.

A B

A 0

B

1

2

3

4

5

6

7

8

10 11 12

9

Each set of data has an outlier. In general, the B data are more diverse than the A data. 27. Determine the quartiles and outliers for each data set. A: 2.9 4.1 4.4 4.4 4.4 4.5 4.5 4.6 4.9 c c c Q1  4.25 Q2 Q3  4.55 B: 3.9 4.1 4.2 4.3 4.5 4.5 4.9 5.1 5.2 c c c Q1  4.15 Q2 Q3  5.0 A: IQRA  4.55  4.25  0.30 4.25  1.5(0.30)  3.80; 4.55  1.5(0.30)  5.00; 2.9 is an outlier. B: IQRB  5.0  4.15  0.85 4.15  0.85  3.30; 5.0  0.85  5.85; no outliers Draw the box-and-whisker plots using the same number line.

14 16 18 20 22 24 26 28 30 32 34 36 38 40

The A data are much more diverse than the B data. 25. Determine the quartiles and outliers for each data set. A: 45 47 47 48 49 50 51 51 51 55 { 58 c c c Q1 Q2 Q3 outlier B: 35 37 37 38 39 41 41 41 45 { 48 c c c Q1 Q2  40 Q3 outlier A: IQRA  51  47  4 47  1.5(4)  41; 51  1.5(4)  57; 58 is an outlier. B: IQRB  41  37  4 37  1.5(4)  31; 41  1.5(4)  47; 48 is an outlier. Draw the box-and-whisker plots using the same number line.

A B

A 2

B

3

4

5

The A data have an outlier. Excluding the outlier, the B data are more diverse than the A data. 34 36 38 40 42 44 46 48 50 52 54 56 58

The distribution of both sets of data are similar. In general, the A data are greater than the B data.

601

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IQR: 128  71  57 71  1.5(57)  14.5, 128  1.5(57)  213.5; 223 and 253 are outliers Draw the box-and-whisker plot.

28. Determine the quartiles and outliers for the data set. 3.5 3.5 4 4.5 4.5 5.5 10 17 c c c Q1  3.75 Q2  4.5 Q3  7.75 A: IQR  7.75  3.75  4 3.75  1.5(4)  2.25; 7.75  1.5(4)  13.75; no outliers. Draw the box-and-whisker plot.

80

60

37.

100 120 140 160 180 200 220 240 260

Life-Time Scores for Top 50 U.S. Soccer Players 20

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

18 Frequency

29. The upper half of the data is very dispersed. The range of the lower half of the data is only 1. 30. Determine the quartiles and outliers for the data set. 35 35 35 36 36 36 36 38 38 38 39 39 40 43 44 c Q1

2

123

56 56 57 58 58 60 64 66 76 89 98 100 181 c outliers Q3

60- 80- 100- 120- 140- 160- 180- 200- 220- 24080 100 120 140 160 180 200 220 240 260 Number of Scores

IQR  56  38  18 38  1.5(18)  11; 56  1.5(18)  83; outliers are 89, 98, 100, and 181 Draw the box-and-whisker plot of the data.

40

60

80

38. Both the box-and-whisker plot and the histogram give a visual summary of the data. The box-andwhisker plot shows the intervals by quartile, while the histogram shows the number of data values for intervals of width 20. 39. Sample answer: 40, 45, 50, 55, 55, 60, 70, 80, 90, 90, 90 40. Box-and-whisker plots use a number line to show the least value in the data, the greatest value in the data, and the quartiles of the data. They also indicate outliers. Answers should include the following.

100 120 140 160 180

31. Top half; the top half of the data goes from $48,000 to $181,000, while the bottom half goes from $35,000 to $48,000. 32. The range appears to be 80  39  41 years. The IQR appears to be 74  54  20 years. 33. Bottom half; the top half of the data goes from 70 yr to 80 yr, while the bottom half goes from 39 yr to 70 yr. 34. From the graph, the least value is 39, Q1 is 54, Q2 is 70, Q3 is 74, and the greatest value is 80. Therefore, three intervals of ages that contain half the data are 39 to 70 years, 54 to 74 years, and 70 to 80 years. 35. No; although the interval from 54 yr to 70 yr is wider than the interval from 70 yr to 74 yr, both intervals represent 25% of the data values. 36. Determine the quartiles and outliers for the data set. 61

61

61

62

63

63

63

64

65

67

68

69

71 72 c Q1

73

73

74

74

76

78

78

80

81

82

83

87

92

Least value that is not Greatest value that is not an outlier an outlier Lower Upper Quartile Quartile Outlier Outlier Median

Interquartile range

Range

• The students should draw a box-and-whisker plot representing data they found in a newspaper or magazine. 41 C; From the box-and-whisker plot. The median appears to be 25. 42. D; According to the box-and-whisker plot, the least value is 0, Q1 is 10, Q2 is 25, Q3 is 45, and the greatest value is 50. The interval 0–45 represents the first three quartiles, or 75% of the data.

96

c Q2  85 100

101

102

107

108

118

119

124

126 128 129 c Q3

131 131 135 137 138 150 152 189 193 223 253 14243 outliers

Chapter 13

12 10 8 6 4

45 46 46 47 47 48 48 49 50 51 52 54 55 55 55 c Q2

20

16 14

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Page 742

51. Find the measure of B. 180  90  39  51 The measure of B is 51 .

Maintain Your Skills

43. Range: 93  13  80 54  55 2

Median:

 54.5

Find the length of BC , which is the side opposite A.

Frequency

Lower quartile: 45 Upper quartile: 67 Interquartile range: 67  45  22 To determine if there are outliers: 45  1.5(22)  12 67  1.5(22)  100 The data set has no values less than 12 and no values greater than 100. It has no outliers. 44. Sample answer:

45.

tan 39  0.8097 

9.7  BC BC is about 9.7 meters long. Find the length of AB, the hypotenuse. 12

cos 39  AB

4 2

ABcos 39  12 12

AB  cos 39

0 020

3 y  3

y

20- 40- 60- 8040 60 80 100

y  4

 

12

AB  0.7771

3(y  4)  y(y  3) (y  3) (y  4)

AB  15.4 AB is about 15.4 meters long. 52. Find the measure of A. 180  90  46  44 The measure of A is 51 .

2

3y  12  y  3y (y  3) (y  4)

y2  6y  12 (y  3) (y  4) 2 3 2(r  2)  3(r  3)   r  3 r  2 (r  3) (r  2) 2r  4  3r  9  (r  3) (r  2) 5r  5  (r  3) (r  2) w 4 3w  4  15w  6  3(5w  2) 5w  2



46.

47. 48.

7a 5

2

15

 14a 

6r  3 r  6



Find the length of BC , which is the side opposite A. tan 44  0.9657 

1

 49.

BC 12 BC 12

14.5  BC

3  5  7  a  a 2  5  7  a

BC is about 14.5 feet long. Find the length of AB, the hypotenuse.

1

3a 2

r2  9r  18 2r  1



3(2r1  1) r  6



15

cos 44  AB

(r1  6) (r  3) 2r  1

ABcos 44  15

1

 3(r  3) 50. Find the measure of B. 180  90  42  48 The measure of B is 48 . Find the length of BC , which is the side opposite A. sin 42  0.6691 

15

AB  cos 44 15

AB  0.7193 AB  20.9 AB is about 20.9 feet long. 53. a2  7a  6  0 a2  7a  6

BC 22 BC 22

 6 

7

25 4

1a  2

BC is about 14.7 inches long.



49 4

25 4

a  2  3

Find the length of AC , which is the side adjacent to A. 0.7431 

49 4 7 2 2

a2  7a 

14.7  BC

cos 42 

BC 15 BC 15

7

5

a  2  2

AC 22 AC 22

7

a2  7

5

a22

16.3  AC



AC is about 16.3 inches long.

12 2

6 {1, 6}

603

or

a 

5 2 7 2 2 2

5

2

1

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54. x2  6x  2  0 x2  6x  2 2  6x  9  2  9 x (x  3) 2  7 x  3  17 x  3  17 x  3  17 or x  3  17  5.6  0.4 {0.4, 5.6}

8. Sample answer: Weight Interval (pounds) 3–4 4–5 5–6 6–7 7–8 8–9 9–10 10–11 11–12

55. t2  8t  18  0 t2  8t  18 t2  8t  16  18  16 (t  4) 2  34 t  4  134 t  4  134 t  4  134 or t  4  134  1.8  9.8 {9.8, 1.8} 2

Weight of Babies 50 Cumulative Number of Babies

56. (7p  p  7)   11)  (7p2  p  7)  (p2  11)  (7p2  p2 )  p  (7  11)  6p2  p  18 57. (3a2  8)  (5a2  2a  7)  (3a2  5a2 )  2a  (8  7)  8a2  2a  1

s 160

90

 100

40 30 20 10 0

Algebra Activity (Follow-Up of Lesson 13-5)

3-4 4-5 5-6 6-7 7-8 8-9 9-10 10-1111-12 Weight (pounds)

1. To find the number of students in Column 3, add the number from the previous cell to the number of students for the new interval. 2. The horizontal axes of both histograms represent the test score intervals, and the vertical axes of both histograms represent the number of students. The bars for the second histogram are taller than the bars for the first histogram because each bar in the second histogram indicates the cumulative number of students. 3. Sample answer: I prefer the first histogram because you can see the number of students that scored in each interval. 4. 40; 80; 120 5. Sample answer: greater than 600 6.

Cumulative Number of Babies 1 2 6 13 29 40 43 44 45

9. Sample answer:

(p2

Pages 743–744

Number of Babies 1 1 4 7 16 11 3 1 1

10. 8–9 lb

Chapter 13 Study Guide and Review Page 745 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

7. 600  700

100s  14,400 s  144

Vocabulary and Concept Check

simple random sample measures of variation quartile systematic random sample biased sample stratified random sample interquartile range voluntary response sample outlier range

Pages 745–748

Lesson-by-Lesson Review

11. The sample is 8 test tubes with results of chemical reactions. The population is all test tubes in the laboratory. The sample is biased because not all test tubes are equally likely to be selected. Since the first 8 test tubes from Tuesday were selected, the sample is a convenience sample.

Chapter 13

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12. The sample is every 50th chocolate bar. The population is all chocolate bars on the conveyor belt. Since every chocolate bar is equally likely to be selected, the sample is unbiased. The sample is systematic, as every 50th chocolate bar is selected for testing.

21. C  3D  c

1 3 1 1 1 3 13. A  B  C 2 0 4§  C 2 3 1 § 1 1 3 1 2 0 11 3  1 1  (3) C 22 03 4  (1) § 1  (1) 1  (2) 30

3(1)  C 3(2) 3(1)

3(1) 3(3) 3(2)

2(2) 2(2)

3(3) 3(1) § 3(0)

 c

9 1 d 5 4

3(1) d 3(0)

3 1 1 1 3 4§  £ 2 3 1 § 0 3 1 2 0 1 2(3) 2(0) 2(1)

2(1) 1 1 2(4) §  £ 2 3 2(3) 1 2

2 1 8§  £ 2 6 1

3 1 § 0

1 3 3 1 § 2 0

21 6  1 2  (3) 42 03 8  (1) § 2  (1) 2  (2) 60

1 5  £ 2 3 1 0

1 9§ 6

23. Sample answer:

Cellular Minute Usage

2(1) d 2(0)

3 2 2 16. C  D  c d  c 2 1 4

3  2 2  1 d 1  (2) 40

 c

1 3 d 3 4

6 4 2 0

1 d 0

 c

0- 50- 100- 150- 200- 250- 30050 100 150 200 250 300 350 Cellular Minutes

24. Sample answer:

Coffee Consumption 16 14

3 2 2 1 d  c d 1 4 2 0 Frequency

12

3  2 2  1  c d 1  (2) 40 5 1  c d 1 4

10 8 6 4

18. impossible

2

1 3 1 19. 5A  5 £ 2 0 4§ 1 1 3 5(1)  £ 5(2) 5(1)

3  6 2  3 d 1  (6) 40

 £

4 2 d  c 4 0

17. C  D  c

 c

2 6  £ 4 0 2 2

Frequency

 c

3 2 6 3 d  c d 1 4 6 0

2(1)  £ 2(2) 2(1)

1 d 0

2 2

 c

1

3 3 9 9 3 § C 6 3 6 0 15. 2D  2 c

3 2 3(2) d  c 1 4 3(2)

1

3 1 § 0

1 3 2

 c

22. 2A  B  2 £ 2

2 4 4 C 4 3 3§ 2 3 3 1 14. 3B  3C 2 1

3 2 2 1 d  3c d 1 4 2 0

5(3) 5(0) 5(1)

0 0-1 1-2 2-3 3-4 Cups

5(1) 5(4) § 5(3)

5 15 5  £ 10 0 20 § 5 5 15 20. impossible

605

Chapter 13

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25. Order the set of data from least to greatest. 30 40 50 60 70 80 90 100 c c c Q2 Q3 Q1 The range is 100 – 30 or 70. The median is

60  70 2

The lower quartile is

The interquartile range is 262 – 206 or 56. Check to see if there are any outliers. 206  1.5(56)  122 262  1.5(56)  346 There is one outlier, 348.

or 65. 40  50 or 45. 2 80  90 or 85. 2

The upper quartile is The interquartile range is 85 – 45 or 40. The outliers would be less than 45 – 1.5(40) or 15 or greater than 85  1.5(40) or 145. There are no outliers. 26. Order the set of data from least to greatest. 1 2 3 3.2 3.4 4 5 5 5.3 7 8 21 45 78 c c c Q2 Q3 Q1 The range is 78  1 or 77. The median is

5  5 2

60

60

Q2 77.1  82.3 2

50

c

 58.1 Q2 

c

67.0  69.8 2

 68.4 Q3 

78.1  79.1 2

 78.6

78.6  1.5(20.5)  109.35

60

70

80

90

Chapter 13 Practice Test Page 749 1. 2. 3. 4. 5. 6.

b; column a; element c; row e; scalar d; dimensions The sample is the first five dogs to run from the pen. The population is all dogs in the pen. Since each dog did not have an equal chance of being selected for the sample, the sample is biased. The first five dogs leaving the pen were selected; therefore, it is a convenience sample. 7. The sample is a book for each hour the library is open. The population is all book titles checked out Wednesday. Since any book title checked out on Wednesday has an equal chance of being selected, the sample is unbiased. A book title is selected every hour, so the sample is systematic.

125 199 200 212 220 230 239 240 240 250 274 327 348 c c c 250  274 2

100

There are no outliers.

64.3  68.6 or 66.45. 2 91.7  110.5 or 101.1. 2

Q3 

55.1  61.1 2

58.1  1.5(20.5)  27.35

The interquartile range is 101.1  66.45 or 34.65. The outliers would be less than 66.45  1.5(34.65) or 14.475 or greater than 101.1  1.5(34.65) or 153.075. There is one outlier, 254.8. 29. Order the set of data from least to greatest.

or 206 Q2

90

The interquartile range is 78.6  58.1 or 20.5. Check to see if there are any outliers.

or 79.7.

The upper quartile range is

80

c

Q1 

Q3

The lower quartile range is

70

52.4 55.2 55.1 61.1 61.9 67.0 69.8 73.4 78.1 79.1 81.2 81.6

The range is 254.8  59.8 or 195.

Chapter 13

70

31. Order the data from least to greatest.

59.8 63.8 64.3 68.6 70.7 77.1 82.3 88.9 91.7 110.5 111.5 254.8 c c c

200  212 2

325

or 5.

The range is 92  55 or 37. The median is the middle value or 73. The lower quartile is 62. The upper quartile is 77. The interquartile range is 77 – 62 or 15. The outliers would be less than 62  1.5(15) or 39.5 or greater than 77  1.5(15) or 99.5. There are no outliers. 28. Order the set of data from least to greatest.

Q1 

275

70 75 80 85 85 90 95 100 c c c 80  85 Q Q1 Q   82.5 3 2 2 The interquartile range is 90  70 or 20. Check to see if there are any outliers. 70  1.5(20)  40 90  1.5(20)  120 There are no outliers.

55 58 59 62 67 69 69 73 75 76 77 77 82 85 92 c c c Q1 Q2 Q3

The median is

225

30. Order the data from least to greatest.

The lower quartile is 3.2. The upper quartile is 8. The interquartile range is 8 – 3.2 or 4.8. The outliers would be less than 3.2  1.5(4.8) or 4 or greater than 8  1.5(4.8) or 15.2. There are three outliers: 21, 45, and 78. 27. Order the set from least to greatest.

Q1

175

125

or 262

606

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14. Sample answer:

2 3 1 4 2 1 8. W  X  £ 1 0 1 §  £ 2 2 0§ 2 2 0 0 1 2 3  2 1  (1) 0  (2) 1  0 § 2  1 02

14 12 Frequency

24  £ 1  (2) 20

Ages of Men in Billiards Tournament

6 5 0  £ 3 2 1 § 2 1 2

 £  £

3 2 1 3 1 6 §  £ § 1 2 4 4 1 1 33 1  4

2  1 2  (1)

16 § 4  (1)

8

 c

2

Number of Cards

16. Order the set of data from least to greatest. 975 1005 1025 1055 1075 1075 1095 1100 1125 1125 1145 The range is 1145  975 or 170. The median is the middle value or 1075. The lower quartile is 1025. The upper quartile is 1125. The interquartile is 1125  1025 or 100. The outliers would be less than 1025  1.5(100) or 875 or greater than 1125  1.5(100) or 1275. There are no outliers. 17. Order the set of data from least to greatest. 0.1 0.2 0.2 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.7 0.8 0.9 1.9 The range is 1.9  0.1 or 1.8. The median is the middle value or 0.5. The lower quartile is 0.3. The upper quartile is 0.7. The interquartile range is 0.7  0.3 or 0.4. The outliers would be less than 0.3  1.5(0.4) or 0.3 or greater than 0.7  1.5(0.4) or 1.3. There is one outlier, 1.9.

2(1) 2(1)

2(6) d 2(1)

6 2 12 d 8 2 2 12. impossible  c

13. Y  2Z  c

4

0- 20- 40- 60- 8020 40 60 80 100

3(1) 3(0) § 3(2)

3 1 6 d 4 1 1

2(3) 2(4)

6

0

12 6 3 0§  £ 6 6 3 6 0 11. 2Z  2 c

4 2 0

Number of Trading Cards to Share

4 2 1 10. 3X  3C 2 2 0§ 0 1 2 3(2) 3(2) 3(1)

6

15. Sample answer:

0 3 5 § 5 1 5

3(4)  £ 3(2) 3(0)

8

65- 70- 75- 80- 85- 9070 75 80 85 90 95 Ages

Frequency

9. Y  Z  £

10

3 2 1 3 1 6 d  2c d 1 2 4 4 1 1

 c

3 2 1 2(3) d  c 1 2 4 2(4)

 c

3 2 1 6 d  c 1 2 4 8

 c

36 1  8

 c

3 4 11 d 9 0 6

2  2 2  (2)

2(1) 2(1)

2(6) d 2(1)

2 12 d 2 2 1  12 d 4  (2)

607

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7. D; Choices A, B, and C do not result in a sample of students who represent the entire student body. 8. A; Consecutive entries in the Earth row are all 10 years apart. Consecutive entries in the Mars row are all 5.3 years apart. Since the difference between consecutive entries is constant in both rows, the relationship is linear. 9. C; According to the graph, the least variation occurs with Car C, which has the least range. 10. B; For Cars A and D, the median is 17. For Car C, the median is 24. For Car B, the median is 26.

18. Order the data from least to greatest. 1 1 1 1 1 2 2 3 3 4 4 5 6 6 9 10 10 c c c 1  1 6  6 Q  2 6 Q1  2  1 Q2 3 The interquartile range is 6  1 or 5. Check to see if there are any outliers. 1  1.5(5)  6.5 6  1.5(5)  13.5 There are no outliers.

1

0

2

3

4

5

6

7

8

9

11. x3  8x2  16x  x(x2  8x  16)  x(x  4) (x  4)  x(x  4) 2 2 12. 6x  x  2  0 (3x  2)(2x  1)  0 3x  2 or 2x  1

10

19. Order the data from least to greatest. 9 9 9 10 11 12 12 12 13 14 14 15 16 16 18 22 c c c 10  11 12  13 15  16 Q1   10.5 Q2   12.5 Q3   15.5 2 2 2

The interquartile range is 22  9  13. Check to see if there are any outliers. 10.5  1.5(13)  9 15.5  1.5(13)  35 There are no outliers.

2

x  3

1

x2

13. 2419  14  3  213 1

9

14. Each hour, Maren can complete 4 of the job. 1 Juliana can complete 6 of the job in the same amount of time.

10 11 12 13 14 15 16 17 18 19 20 21 22

20. B; The interquartile range for A is 57  49 or 8. The interquartile range for B is 45  28 or 17. The interquartile range for C is 30  20 or 10. B has the greatest interquartile range.

2

116 2  14 x  16 x  1 1 3

1

1

 4x  6x  1 1 x 4

1

2

 6x  3

3x  2x  8 5x  8

Chapter 13 Standardized Test Practice

8

x5 8

It will take 5 hours for the two to complete the job working together.

Pages 750–751 1. B; The equation y  4x  6 has a slope of m  4. 1

The equation y  4 x  2 has a slope of m  making it perpendicular. 2. B; 3x  5x  8  32 8x  40 x5 5x  5(5)  25

15. (AB) 2  (AC) 2  (BC) 2 (5) 2  (3) 2  (BC) 2 25  9  BC2 BC2  16 BC  4 miles The distance between points B and C is 4 miles. 16. B; Column A: The next three terms of the sequence are 749, 1082, and 1415. Their sum is 749  1082  1415  3246. Column B: The 67th term of the arithmetic sequence is

1 , 4

3. B; (x  8) 2  (x  8)(x  8)  x2  8x  8x  64  x2  16x  64 4. D; The least value of the graph occurs when x  0. As a result, y  x2  4  (0) 2  4  4

a67  a1  (67  1)  r, where r  2  (49)  51. Therefore, a  49  66  51  3317. 67 Column B is greater.

5. C; 3172  312  314  9  2  312  3  2  312  312  1812  312  1512 6. B;

12 9

x

 27

9x  324 x  36 m

Chapter 13

608

17. A; Column A: The root of y  0.25x2  x  1 is

19.

2

0.25x  x  1  0 x2  4x  4  0 (x  2) 2  0 x2 Column B: The roots of b 

3a2

 5a  2 are

2

a3 2

The sum of the roots is 1  3  13. Column A is greater. 3x  1 14

18. C; Column A:

x

4

3  4x  4  14x 12x  4  14x 4  2x 2x Column B:

45 4y  1

6 4 2 0

4050

50- 60- 70- 80- 9060 70 80 90 100

20. There are 1  0  4  5  7  3  20 pieces of data. The interval 80  90 has 7 pieces of data, 7 representing 20  0.35  35% of the data. 21. Order the data from least to greatest. Use this list to determine the quartiles. 22, 24, 25, 30, 32, 34, 36, 38, 38, 39, 40, 40, 40, 41, 42, 45, 45, 47, 47, 48, 49 Q1  33 Q2  40 Q3  45 Determine the interquartile range. IQR  45  33  12 Check to see if there are any outliers. Q1  1.5(12)  33  18  15; Q3  1.5(12)  45  18  63 There are no outliers. Complete the box-and-whisker plot.

3a2  5a  2  0 (3a  2) (a  1)  0 3a  2 or a  1 2

Frequency

PQ249-6481F-13[586-609] 26/9/02 6:38 PM Page 609 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

10

 y 45y  4y  10  10 45y  40y  10 5y  10 y2

So, x  y.

20

609

30

40

50

Chapter 13

Chapter 14 Probability Page 753

3. 5! is the product of 5 and all positive integers less than 5. 5!  5  4  3  2  1  120 4. Spin 1 Spin 2 Spin 3 Outcomes

Getting Started

1. There are 6 red cubes and 14 cubes total. 6

3

P(red)  14 or 7 2. There are 3 blue cubes and 14 cubes total. 3

P(blue)  14 3. There are 4 yellow cubes and 14 cubes total. 4

R

2

P(yellow)  14 or 7 4. There are 8 cubes that are not red and 14 cubes total. 8 14

P(not red) 

or

R

1

5.

4 5

3 4



4 5

1

3 4

6.

1

5 12

6  11

3



2

4

7

8.

5

4 32

7



13 52 13



G

8

7

 256

1

9.

7

4

 32  32  32

7

4  52

Y

1

4

 19  20  19  95

13 52

6  11

 22 1

7 20

5 12 5

5 7.

B

4 7

1

14

4  52

10.

4

56 100

24  100

1 52



56 100



84 625

25

R

6

24  100 25

B

11. 0.725  (0.725  100) %  72.5% 12. 0.148  (0.148  100) %  14.8% 13. 0.4  (0.4  100) %  40% 14. 0.0168  (0.0168  100) %  1.68% 15.

7 8

17.

107 125

 0.875

B Y

G

16.

33 80

18.

625 1024

 87.5%

 0.4125

R

 41.3%

 0.856  85.6%

 0.6103515625

B

 61% Y

Y

14-1 Counting Outcomes Page 756

G

Check for Understanding

1. Sample answer: choosing 2 books from 7 on a shelf 2. Toss 1 Toss 2 Toss 3 Outcomes

H H T

H T H T

HHH HHT HTH HTT

H T H T

THH THT TTH TTT

R

B G Y

H T T

Chapter 14

G

610

R B Y G R B Y G R B Y G R B Y G R B Y G R B Y G R B Y G R B Y G R B Y G R B Y G R B Y G R B Y G R B Y G R B Y G R B Y G R B Y G

RRR RRB RRY RRG RBR RBB RBY RBG RYR RYB RYY RYG RGR RGB RGY RGG BRR BRB BRY BRG BBR BBB BBY BBG BYR BYB BYY BYG BGR BGB BGY BGG YRR YRB YRY YRG YBR YBB YBY YBG YYR YYB YYY YYG YGR YGB YGY YGG GRR GRB GRY GRG GBR GBB GBY GBG GYR GYB GYY GYG GGR GGB GGY GGG

13. 11!  11  10  9  8  7  6  5  4  3  2  1  39,916,800 14. 13!  13  12  11  10  9  8  7  6  5  4  3  2  1  6,227,020,800 15. There are 6 possible outcomes for each die. Multiply. red white blue possible die die die outcomes 123 123 123 14 424 43

5. The tree diagram shows that there are 64 possible outcomes. 6. Of the 64 possible outcomes shown on the tree diagram, we observe that 18 involve both green and blue. 7. 8!  8  7  6  5  4  3  2  1  40,320 8. Multiply to find the number of ways students can choose to do their assignments. project paper presentation number choices choices choices of ways 14243 14243 144 424 443 14243

6  6  6  216 There are 216 possible outcomes. 16. Multiply to find the number of outfits. 5  3  3  4  180 There are 180 possible outfits. 17. Multiply to find the number of ways. Seattle Denver number to to of Denver St. Louis ways 14243 14243 14243

15  6  8  720 There are 720 different ways that students can choose to do their assignments.

Pages 757–758 9. English

Practice and Apply Math

Science A B C A B C A B C A B C A B C A B C A B C A B C A B C

A A

B C A

B

B C A

C

Outcomes

B C

4  (4  2)  24 A traveler can book a flight from Seattle to St. Louis in 24 ways. 18. There are ten digits. Five of these, 1, 3, 5, 7, and 9, are odd. Multiply to find the number of area codes. odd 0 or 1 any digit area codes 123 123 14243 1442443

AAA AAB AAC ABA ABB ABC ACA ACB ACC BAA BAB BAC BBA BBB BBC BCA BCB BCC CAA CAB CAC CBA CBB CBC CCA CCB CCC

5  2  10  100 There are 100 possible area codes.

19.

0 4 1

0 5 1

Monitor 1 CD-ROM Monitor 2 Monitor 1 CD recorder Monitor 2 Monitor 1 DVD Monitor 2

Printer Scanner Printer Scanner Printer Scanner Printer Scanner Printer Scanner Printer Scanner

407 408 409 417 418 419 507 508 509 517 518 519

20. Let C represent a win for Columbus Crew and D a win for D.C. United. If either team wins all of the first three games, the 4th and 5th games will not be played. Two possible outcomes are C-C-C and D-D-D. If one of the teams achieves its third win in the 4th game, the 5th game will not be played. More possible outcomes are C-C-D-C, C-D-C-C, D-C-C-C, D-D-C-D, D-C-D-D, and C-D-D-D. If all five games are played, the possible outcomes are C-C-D-D-C, C-D-C-D-C, C-D-D-C-C, D-C-D-C-C, D-D-C-C-C, D-C-C-D-C, D-D-C-C-D, D-C-D-C-D, D-C-C-D-D, C-D-C-D-D, C-C-D-D-D, and C-D-D-C-D. 21. From the list of possible outcomes, we see that 6 outcomes require that exactly four games be played to determine the champion.

The tree diagram shows that there are 27 possible outcomes. 10.

7 8 9 7 8 9 7 8 9 7 8 9

CD-M1-P CD-M1-S CD-M2-P CD-M2-S CDR-M1-P CDR-M1-S CDR-M2-P CDR-M2-S DVD-M1-P DVD-M1-S DVD-M2-P DVD-M2-S

The tree diagram shows that there are 12 possible outcomes. 11. 4!  4  3  2  1 12. 7!  7  6  5  4  3  2  1  24  5040

611

Chapter 14

33. There are 8 data values greater than the median. The 4th and 5th of these are 71 and 76, respectively. 71  76 The upper quartile is or 73.5. 2

22. From the list of possible outcomes, we see that there are 10 ways D.C. United can win the championship. 23. Since there are five days in a school week, Tucker walked 3 days (60% of 5 days), rode his bike 1 day (20% of 5 days), and rode with a friend 1 day. Tyler could have taken his bicycle any one of the five days of the week, leaving any one of the remaining four days as the day he could have gone with a friend. The remaining three days he walked; there are no choices here. choices for choices for number day he rode day he rode of bike with friend outcomes 144 424 443

144 424 443

There are 8 data values less than the median. The 4th and 5th of these are 35 and 44, respectively. 35  44 The lower quartile is or 39.5. 2 The interquartile range is 73.5  39.5 or 34.0. 34. An outlier must be 1.5(34.0) or 51 less than the lower quartile, 39.5, or 51 greater than the upper quartile 73.5. There are neither data values less than 39.5  51 or 11.5 nor greater than 73.5  51 or 124.5. There are no outliers. 35.

14 424 43

5  4  20 This situation is represented by 20 outcomes. 24. Sample answer: You can make a chart showing all possible outcomes to help determine a football team’s record. Answers should include the following. • a tree diagram or calculations to show 16 possible outcomes 25. A; 9!  9  8  7  6  5  4  3  2  1  362,880 26. C; Multiply the number of model choices, the number of package choices, and the number of color choices. 4  6  12  288 There are 288 possibilities for this car.

Page 758

36.

37.

Maintain Your Skills

Chapter 14

56  59 2

x  4  2

x



(2x  1) (x  2) (x  4) (3x  1)  (x  2) (3x  1) (3x  1) (x  2) (2x  1) (x  2)  (x  4) (3x  1) (3x  1) (x  2)



2x2  3x  2  3x2  11x  4 (3x  1) (x  2)



5x2  8x  6 (3x  1) (x  2) 4n 3 4n 3  n  3  2(n  3)  n  3 2n  6 2n 3 n3n3 2n  3  n3 3z  2 z  2 3z  2 z  2  z2  4  3(z  2)  (z  2) (z  2) 3z  6 3z  2 1  3(z  2)  z  2 3z  2 3  3(z  2)  3(z  2) 3z  2  3  3(z  2) 3z  1  3(z  2) 3z  1  3z  6 m  n 1 m  n 1  m2  n2  m  n  (m  n) (m  n) m  n (m  n) (m  n) 1  (m  n) (m  n)  (m  n) (m  n) (m  n) (m  n)  1  (m  n) (m  n)



38.

27. For plot A, the least value is 32, the greatest value is 88, the lower quartile is 44, the upper quartile is 85, and the median is 60. For plot B, the least value is 38, the greatest value is 86, the lower quartile is 48, the upper quartile is 74, and the median is 64. 28. The least data value in either set is 32, which is contained in set A. 29. For plot A, the interquartile range is 85  44 or 41. For plot B, the interquartile range is 74  48 or 26. Plot B has the smaller interquartile range. 30. For plot A, the range is 88  32 or 56. For plot B, the range is 86  38 or 48. Plot A has the greater range. 31. The greatest data value is 109. The least data value is 30. Thus, the range of the data is 109  30 or 79. 32. There are 16 data values. In ascending order, the 8th and 9th data values are 56 and 59, respectively. The median is

2x  1 3x  1

 39.

m2  2mn  n2  1 m2  n2

522n2  28  20 1 (522n2 5

1

 28)  5 (20)

22n2  28  4 ( 22n2  28) 2  42 2n2  28  16 2n2  44 n2  22 n  122 The solutions are 122 and 122.

or 57.5.

612

40.

44. x2  11x  17  0 x2  11x  17

25x2  7  2x ( 25x2  7) 2  (2x) 2 5x2  7  4x2 x2  7  0 x2  7 x  17

121 4 11 2 2

x2  11x 

1x  2 x

25x  7  2x

Check:

2

x

?

15(7)  7  217

11 2

11 2



3

189 4

189 4



45.

25x2  7  2x ?

189 4

11 2



3

189 4

 1.4

2p2  10p  3  0 1 (2p2 2

25( 17) 2  7  2( 17)

3

x

or

 12.4 {1.4, 12.4}

?

128  217 217  217 ✕

 10p  3)  0 3

p2  5p  2  0

?

25(7)  7  217 ?

3

p2  5p   2

128  217 217  217 ✓

25 4 5 2 2

p2  5p 

1p  2

Since 17 does not satisfy the original equation, 17 is the only solution. 1x  2  x  4 ( 1x  2) 2  (x  4) 2 x  2  x2  8x  16 0  x2  9x  14 0  (x  7)(x  2) x  7  0 or x  2  0 x7 x2 Check:

 3

x

?

25(17)  7  2(17)

41.

189 4



11 2

2

121 4

 17 

3

 2  

25 4

19 4

5

p  2  3

19 4

5

p  2  5

19

p  2 

34

or

19

34 5

p  2 

19

34

 0.3  4.7 {4.7, 0.3} 46. There are four 10s and a total of 52 cards.

1x  2  x  4 ?

4

1

P(10)  52 or 13

17  2  7  4 ?

19  3 33✓ 1x  2  x  4

47. There are four aces and a total of 52 cards. 4

1

P(ace)  52 or 13 48. There are two red 5s and a total of 52 cards.

?

12  2  2  4

2

1

P(red 5)  52 or 26

?

14  2 2  2 ✕ Since 2 does not satisfy the original equation, 7 is the only solution. 42. b2  6b  4  0 b2  6b  4 2  6b  9  4  9 b (b  3) 2  5 b  3  15 b  3  15 b  3  15 or b  3  15  5.2  0.8 {0.8, 5.2}

49. There is one queen of clubs and a total of 52 cards. 1

P(queen of clubs)  52 50. There are 20 even numbers, four each of 2, 4, 6, 8, and 10, and a total of 52 cards. 20

5

P(even number)  52 or 13 51. There are four 3s and four kings. Thus, there are eight ways to get a 3 or king and a total of 52 cards. 8

2

P(3 or a king)  52 or 13

Page 759

43. n2  8n  5  0 n2  8n  5 2  8n  16  5  16 n (n  4) 2  21 n  4  121 n  4  121 n  4  121 or n  4  121  0.6  8.6 {8.6, 0.6}

Algebra Activity (Follow-Up of Lesson 14-1)

1. Yes; Sample answer: Front St., Main St., Second Ave., State St., Elm St., First Ave., Town St. 2. Besides sample answer in Exercise 1: (2) Second Ave., Main St., Front St., State St., Elm St., First Ave., Town St. (3) Second Ave., Main St., Front St., State St., Town St., First Ave., Elm St. (4) Front St., Main St., Second Ave., State St., Town St., First Ave., Elm St. There are 4 traceable routes that begin at Alek’s house.

613

Chapter 14

3. Also: (5) Town St., First Ave., Elm St., State St., Front St., Main St., Second Ave. (6) Elm St., First Ave., Town St., State St., Front St., Main St., Second Ave. (7) Town St., First Ave., Elm St., State St., Second Ave., Main St., Front St. (8) Elm St., First Ave., Town St., State St., Second Ave., Main St., Front St. There are 8 traceable routes now. 4. Yes; all nodes can be connected without retracing an edge. 5. No; all nodes cannot be connected without retracing an edge. 6. Yes; all nodes can be connected without retracing an edge. 7a.



n! (n  r)! 8! (8  5)! 8! 3!



87654321 321

6. P  n r P 

8 5

1

1

 8  7  6  5  4 or 6720 7. C  n r C 

7 5



n! (n  r)! r! 7! (7  5)! 5! 7! 2! 5! 1

7654321

 2154321 1

76

 2  1 or 21 10!

3!

8. ( P )( P )  (10  5)!  (3  2)! 10 5 3 2 10! 3!  5!  1! 

10  9  8  7  6 1



321 1

 181,440 6!

4!

9. ( C ) ( C )  (6  2)! 2!  (4  3)! 3! 6 2 4 3 6! 4!  4! 2!  1! 3!

7b. Yes; start at any node and first follow the edges to adjacent nodes until you come back to where you started. Then follow the edges that form a star until you again come back to where you started. 7c. Yes; starting at any building you can follow the sidewalk to the building two buildings away in the clockwise direction. If you repeat this, you will eventually reach all five buildings without using any sidewalk more than once. 8. Sample answer: If you follow the edges of a graph, you should cover each edge only once.



65 2!

4

1

 60 10. Permutation; the order of the digits is important. 11. We find the number of permutations of 10 digits taken 3 at a time.



n! (n  r)! 10! (10  3)! 10! 7!



10  9  8  7  6  5  4  3  2  1 7654321

P 

n r

P 

10 3

1

1

 10  9  8 or 720 720 different codes are possible. 12. We begin by finding the number of codes whose digits are all odd. Since there are five odd digits, we find the number of permutations of 5 odd digits taken 3 at a time.

14-2 Permutations and Combinations Page 764

Check for Understanding

1. Sample answer: Order is important in a permutation but not in a combination. Permutation: the finishing order of a race Combination: toppings on a pizza 2. C  n n P  n n

n! (n  n)! n! n! (n  n)!

n! 0!

or n!, since 0!  1



54321 21

1

3. Alisa; both are correct in that the situation is a combination, but Alisa’s method correctly computes the combination. Eric’s calculations find the number of permutations. 4. Combination; order is not important. 5. Permutation; order is important.

Chapter 14



P 

5 3

n!

 0!n! or 1, since 0!  1.



n! (n  r)! 5! (5  3)! 5! 2!

P 

n r

1

 5  4  3 or 60 There are 60 codes whose digits are all odd and a total of 720 possible codes. 60

1

P(all odd)  720 or 12

614

13. B; Since order is not important, we find the number of combinations of 8 items taken 2 at a time. C 

n r

C 

8 2



27.

C 

n r

C 

n! (n  r)!r! 8! (8  2)!2! 8! 6!2!

20 8

  

1 87654321

 65432121 87

28.

1

 2  1 or 28

P 

n r

P 

15 3



Pages 764–766 14. 15. 16. 17. 18. 19. 20. 21. 22.



Practice and Apply

P 

P 

12 3

 

29.

P   

P 

4 1

7!

 



C 

7 3



20!

24. nCr  C 

6 6

 



n! (n  r)!r! 6! (6  6)!6! 6! 0!6! 6! 1  6!

3

 1

1

   

16  15  14  13 1 7!  4)!4!

765 3!

 105 8!

5!

33. ( C ) ( P )  (8  5)!5!  (5  5)! 8 5 5 5 8! 5!  3!5!  0! 

876 3!



5! 1

 6720 34. Permutation; order is important. 35. We find the number of permutations of 9 players taken 9 at a time.

1 765

C 



3!

 3  2  1 or 35

15 3

20  19 1

32. ( C ) ( C )  (3  2)!2!  (7 3 2 7 4 3! 7!  1!2!  3!4!

1 7654321

C 

16!

 16,598,400

n! (n  r)!r! 7! (7  3)!3! 7! 4!3!

n r

7

1

 35,280

 4321321

26.

7! 1

31. ( P )( P )  (20  2)!  (16  4)! 20 2 16 4 20! 16!  18!  12!

4 25. C  n r

7!

30. ( P )( P )  (7  7)!  (7  1)! 7 7 7 1 7! 7!  0!  6!

1 12  11  10  9  8  7  6  5  4  3  2  1 987654321 1

4! (4  1)! 4! 3! 1 4321 321 1

n! (n  r)! 16! (16  5)! 16! 11! 16  15  14  13  12  11! 11!

 524,160

n! (n  r)! 12! (12  3)! 12! 9!

n! (n  r)!

P 

n r

16 5

 12  11  10 or 1320 23. P  n r

n! (n  r)! 15! (15  3)! 15! 12! 15  14  13  12! 12!

 2730

Combination; order is not important. Permutation; order is important. Combination; order is not important. Permutation; order is important. Permutation; order is important. Combination; order is not important. Combination; order is not important. Combination; order is not important. n r

n! (n  r)!r! 20! (20  8)!8! 20! 12!8! 20  19  18  17  16  15  14  13  12! 12!8! 5,079,110,400 or 125,970 40,320

P 

n! (n  r)!r! 15! (15  3)!3! 15! 12!3! 15  14  13  12! 12!3! 15  14  13 3! 2730 or 455 6

n r

P 

9 9



n! (n  r)! 9! (9  9)! 9! 0!

 9! or 362,880 The manager can make 362,880 different lineups. 36. Combination; order does not matter.

615

Chapter 14

There are 3780 ways to choose 5 women and 177,100 possible outcomes.

37. We find the number of combinations of 12 people taken 4 at a time. C 

n r

C 

12 4

 

n! (n  r)!r! 12! (12  4)!4! 12! 8!4! 12  11  10  9 4!

3780

45. We find the number of permutations of 8 girls taken 3 at a time. P 

8 3



 495 Thus, 495 different groups of students could be selected. 38. We find the number of permutations of 12 people taken 4 at a time. P 

n r

P 

12 4

 

( 4P2 )( 4P1 )   

 11,880 Thus, 11,880 different groups of students could be selected. 39. There are 12 students, one of which will be the chairperson.

48

P(1st, 2nd: West; 3rd: Central)  336 1

 7 or about 14% 47. There are 4 choices of entree, 3 choices of side dish, and 3 choices of beverage. Use the Fundamental Counting Principle. 4  3  3  36 There are 36 possible meal combinations. 48. There are 3 possible side dishes, one of which is soup.

1

40. Permutation; order matters. 41. Use the Fundamental Counting Principle. On each die, you have 6 possible outcomes. 6  6  6  6  6  7776 There are 7776 possible outcomes. 42. There are 6 different ways that all five dice could show the same number and 7776 possible outcomes. 6 7776

1

P(soup)  3 or about 33%. 49. There are 4  3 or 12 ways that a student can choose an entree and a side dish, one of which is to choose a sandwich and soup.

1 1296

P(all same number)  or 43. We find the number of combinations of 15 men taken 4 at a time and 10 women taken 2 at a time and multiply.



P(sandwich and soup) 

C   

P(tap shoe) =

C 

6 4

 

 177,100 Next, we find the number of ways to choose 1 man and 5 women.  

1 30,240

6! (6  4)! 4! 6! 2! 4! 65 or 15 2!

The coach can form 15 different teams. 53. We find the number of permutations of 4 swimmers taken 4 at a time.

15! 10!  (15  1)!1! (10  5)!5! 15! 10!  14!1! 5!5! 15 10  9  8  7  6  1 5!

P 

4 4



4! (4  4)! 4! 0!

 4! or 24 The four swimmers had to swim 24 relays.

 3780 Chapter 14

or about 8%

52. We find the number of combinations of 6 swimmers taken 4 at a time.

25! (25  6)!6! 25! 19!6! 25  24  23  22  21  20 6!

( 15C1 ) ( 10C5 ) 

1 12

50. A two-word arrangement is composed of the seven letters and one space. There are 7! ways to arrange the letters and 6 possible positions for the space, so there are 7!  6 or 30,240 two-word arrangements. 51. Tap shoe is one possible arrangement and there are a total of 30,240 possible arrangements.

15! 10!  (15  4)!4! (10  2)!2! 15! 10!  11!4! 8!2! 15  14  13  12 10  9  2! 4!

 61,425 This can be done in 61,425 ways. 44. First, we find the number of ways 6 people can be chosen from the total of 25 men and women. 25 6

4! 4!  (4  2)! (4  1)! 4! 4!  2! 3! 43 4  1 or 48 1

There are 48 ways for the described event to happen and 336 possible outcomes.

P(being chairperson)  12



8! (8  3)! 8! 5!

 8  7  6 or 336 The runners can place first, second, and third in 336 ways. 46. First, we find the number of permutations of 4 girls taken 2 at a time and 4 girls taken 1 at a time and multiply.

n! (n  r)! 12! (12  4)! 12! 8! 12  11  10  9 1

( 15C4 ) ( 10C2 ) 

27

P(5 women)  177,100 or 1265

616

54. There are positions, one of which is the third leg. 1 4

P(third leg) 

66. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

or 25%

 2(16  12) 2  (34  20) 2

55. Sample answer: Combinations can be used to show how many different ways a committee can be formed by various members. Answers should include the following. • Order of selection is not important. • Order is important due to seniority, so you need to find the number of permutations. 56. B; We find the number of permutations of 12 songs taken 10 at a time. P

12 10

 

 242  142  1212  2153 or about 14.56 units 67. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(2  (18)) 2  (15  7) 2  2202  82  1464  4129 or about 21.54 units

12! (12  10)! 12! 2!

68. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

P 

4 4



 

s

4  242  4(1) (2) 2(1) 4  18 2 4  2 12 2

m  2  12  3.41

b  2b2  4ac 2a

1  212  4(2) (15) 2(2) 1  1121  4 1  11  4 1  11 1  11 s or s  4 4 10 12  4  4 5 2  3 5 3, 2



71.

5

6

2n2

2n2  n  4 n40

n

b  2b2  4ac 2a

1  3



1 (x  7) (x  7 ) (x  7 ) (x  5)

 n

1

x  7  5

(1)  2(1) 2  4(2) (4) 2(2) 1  133 4 1  133 1  133 or n  4 4

 1.69 {1.19, 1.69}

1



b  2b2  4ac 2a

70. 2s2  s  15  0

x 65.

1

 2  12 m  2  12 or  0.59 {3.41, 0.59}

1

n2  n  20 n2  9n  20

25

34



x  3 (x  3) (x  3 )



9

m

1

x2  49 x2  2x  35

 (2) 2

69. m2  4m  2  0

50 60 70 80 90 100 110 120130140150

64.

 (3  5) 2

34  4 5

60. The greatest data value is $148,000. The least data value is $55,800. Thus, the range of the data is $148,000  $55,800 or $92,200. 61. There are ten data values. Of the five smallest data values, $56,700 is in the middle. Thus, the lower quartile is $56,700. Of the five largest data values, $91,300 is in the middle. Thus, the upper quartile is $91,300. 62. The interquartile range is $91,300  $56,700 or $34,600. An outlier must be 1.5($34,600) or $51,900 less than the lower quartile or greater than the upper quartile. No data values are less than $56,700  $51,900 or $4800. One data value, $148,000, is greater than $91,300  $51,900 or $143,200. Thus, $148,000 is the only outlier.

x

2

 2 or 22 units

Maintain Your Skills



3 2



58. There are two possibilities, boy or girl, for each year and there are four years. Use the Fundamental Counting Principle. 2  2  2  2  16 There are 16 different ways. 59.

x  3 x2  6x  9

3 12 2



 4! or 24

63.

1



4! (4  4)! 4! 0!

Page 767

3 1 2  (2) 2



 12  11  10  9  8  7  6  5  4  3  239,500,800 57. C; We find the number of permutations of 4 numbers taken 4 at a time.

(n  5) (n  4) (n  5) (n  4)

72.

1

n  5  5

8 52

4

12

 1.19 3

 52  52 or 13

n

617

Chapter 14

73.

7 32

5

7

20

Page 768

 8  32  32 27

 32 74. 75.

76.

77.

5 6 2 9 3  15  15  15 or 5 15 15 11 3 15 11 18  24  4  24  24  24 24 8 1  24 or 3 2 15 1 24 15 9  36  4  36  36  36 3 30 5  36 or 6 16 3 1 64 30 25  10  4  100  100  100 25 69  100

Page 767

Practice Quiz 1

1. There are 6 possible outcomes for the die and 2 possible outcomes for each of the coins. Use the Fundamental Counting Principle. 6  2  2  24 There are 24 possible outcomes. 2. Use the Fundamental Counting Principle. 5  4  2  5  200 There are 200 possible mountain bikes. 3.

C 

n r

C 

13 8

  

n! (n  r)! r! 13! (13  8)!8! 13! 5! 8! 13  12  11  10  9  8! 5! 8! 13  12  11  10  9 5!

14-3 Probability of Compound Events Pages 772–773

 1287 4. P  n r P 

9 6

 

 987654  60,480 5. First, find the number of combinations of 14 flowers taken 4 at a time. C   

14! (14  4)!4! 14! 10!4! 14  13  12  11 4!

 1001 Now, find the number of ways to choose 2 roses out of 6 and 2 lilies out of 3. ( 6C2 ) ( 3C2 )   

6! (6  2)! 2! 6! 3!  4!2! 1!2! 65 3 1 2!

3!

5

 (3  2)! 2!

Chapter 14

6

 21  21 30

10

 441 or 147 6. Since the first chip is not replaced, the events are dependent. P(yellow, yellow)  P(yellow)  P(yellow)

 45 There are 45 four-flower bouquets with two roses and two lilies and 1001 possible four-flower bouquets. P(two roses and two lilies) 

Check for Understanding

1. A simple event is a single event, while a compound event involves two or more simple events. 2. Sample answer: The probability of rolling a number less than or equal to six on a number cube and tossing heads or tails on a coin. 3. Sample answer: With dependent events, a first object is selected and not replaced. With independent events, a first object is selected and replaced. 4. Chloe; sample answer: Since it is possible for the person chosen to be a girl and a senior, the events are inclusive. So, add the probability that a girl 14 is chosen, 34, and the probability that a senior is 15 chosen, 34, then subtract the probability that a 6 senior girl is chosen, 34. 5. Since the first chip is replaced, the events are independent. P(red, green)  P(red)  P(green)

n! (n  r)! 9! (9  6)! 9! 3! 9  8  7  6  5  4  3! 3!

14 4

Reading Mathematics

1. Sample answer: Yes; combine can mean placing many things together, as you do in a combination. A mutation is a change in genes and the order in which they appear as in a permutation. 2. Sample answer: Both permutations and combinations involve selecting items. However, a permutation considers the order of the selected items. 3. Sample answer: factorial—the product of all the positive integers from 1 to n—symbol n! factor—any of the numbers or symbols in mathematics that when multiplied together form a product factorization—the operation of resolving a quantity into factors The meanings all involve products. 4. probability—the quality or state of being probable; something (as an event or circumstance) that is probable probus—upright, liberal, generous probare—to test, approve, prove Sample answer: The words all involve something being true or approved.

2

1

 21  20 2

1

 420 or 210

45 1001

618

17. Since the first marble is not replaced, the events are dependent. P(blue, then red)  P(blue)  P(red)

7. Since the chips are replaced, the events are independent. P(green, blue, red)  P(green)  P(blue)  P(red) 6

8

5

 18  17

6

80

 306 or 51

 21  21  21 240

12

 9261 or 3087

8

5

 21  20  19 240

7

12

1

126

6

5 8

84

or

7

20. P(3 and D)  P(3)  P(D) 1

3

3

2

 65

 

12 24 24 24



22. P(a prime number and A)  P(a prime number)  P(A) 3

7 8

1

 65

12 24

3

1

 30 or 10

or 1

23. P(2 and A, B, or C)  P(2)  P(A, B, or C)

12. Since there are students who are both male and not 11th graders, these events are inclusive. P(male or not 11th grader)  P(male)  P(not 11th grader)  P(male and not 11th grader) 21 24

1

 30 or 5

11. Since a student cannot be both male and female, the events are mutually exclusive. P(male or female)  P(male)  P(female)

18

1

3

 65 3

1

 30 or 10 24. P(both even)  P(even)  P(even)

9

 24  24  24 

or

15

210

1

 840 or 4

1

1

25. P(both 7 20 and 6 30)  P( 7 20 and 6 30)  P( 7 20 and 6 30) 9

9

 30  28 81

27

 840 or 280

3

 5  5  5 or 5 15. The last one tested is tested third out of 3 possibilities, first, second, or third. P(last) 

14

 30  28

13. These are independent events, since each DVD 1 player has the same probability, 5, of being the defective player. 14. P(defective, defective, or defective)  P(defective)  P(defective)  P(defective) 1

1

21. P(an odd number and a vowel)  P(an odd number)  P(vowel) 6

12

1

 6  5 or 30

 24  24  24 

2

 4896 or 408

10. Since there are students who are both 10th graders and females, these events are inclusive. P(10th grader or female)  P(10th grader  P(female) P(10th grader and female) 15 24

7

 18  17  16

 24 or 2

12

7

19. Since the marbles are not replaced, the events are dependent. P(blue, then yellow, then red)  P(blue)  P(yellow)  P(red)

 24  24

6

3

 4896 or 272

9. Since a student cannot be both a 9th and 12th grader, the events are mutually exclusive. P(9th or 12th grader)  P(9th grader)  P(12th grader) 6

6

 18  17  16

4

 7980 or 133

6

2

18. Since the marbles are not replaced, the events are dependent. P(2 yellows in a row then orange)  P(yellow)  P(yellow)  P(orange)

8. Since the chips are not replaced, the events are dependent. P(green, blue, red)  P(green)  P(blue)  P(red) 6

2

26. P(first 7 10, second 6 40 or odd)  P( 7 10)  P( 6 40 or odd) 20

 30 

1 3

2

11928  1428  1028 2

23

 3  28 46

Pages 773–776

Practice and Apply

27. P(first 7 12 or prime, second multiple of 6 or 4)  P(7 12 or prime)  P(multiple of 6 or 4)

16. Since the first marble is not replaced, the events are dependent. P(2 orange)  P(orange)  P(orange) 3



11830  1030  305 2  1285  287  283 2

23

9

 30  28

2

 18  17 6

23

 84 or 42

207

69

 840 or 280

1

 306 or 51

619

Chapter 14

40. In the 16–24 age group, 1145  2080 or 3225 thousand earn no more than minimum wage. P(no more than minimum wage)

28. A

B 0.90

0.06

3225

 15,793 or about 0.20

0.02

In the 25 age group, 970  2043 or 3013 thousand earn no more than minimum wage. P(no more than minimum wage) 0.02

3013

 55,287 or about 0.05

29. P(A or B)  P(A)  P(B)  P(A and B)  0.96  0.92  0.90  0.98 or 98% 30. From the Venn diagram, we see that this probability is 0.02 or 2%. 31. No; P(A and B)  P(A)  P(B) P(A and B)  0.90 P(A)  P(B)  0.96  0.92 or 0.8832 32. There is one combination of genes, bb, for blue eyes and four combinations.

A worker in the 16–24 age group is more likely to earn at most minimum wage, since the probability is greater for this group. 41. There are 6C2 or 15 ways to choose 2 of the six angles. 150, 30 150, 130 150, 50 150, 20 150, 160 30, 130 30, 50 30, 20 30, 160 130, 50 130, 20 130, 160 50, 20 50, 160 20, 160 Every pair listed above contains either an angle inside ABC or an obtuse angle. P(inside or obtuse)  1 42. Among the six angles, there is no straight angle, and there is no right angle inside ABC. P(straight or right)  0 43. P(20 or 130)  P(20)  P(130)  P(20 and 130)

1

P(blue eyes)  4 33. The eye colors of the children are independent events. P(both brown eyes)  P(brown eyes)  P(brown eyes) 3

3

9

 4  4 or 16 34. P(both brown)  P(first or second blue)  1 9 16

 P(first or second blue)  1 P(first or second blue) 

9

3

1

44. Since there are regions that are both a triangle and red, these events are inclusive. The area of the dartboard is 12  10 or 120 in2. The area of 1 each trapezoid is 2 (6  12)(5) or 45 in2. The area 1

of each triangle is 2 (3)(5) or 7.5 in2. The total area of the four triangles is 4  7.5 or 30 in2. The total area of the red regions is 45  7.5  7.5 or 60 in2. The total area that is both a triangle and red is 7.5  7.5 or 15 in2. P(a triangle or a red region)  P(triangle)  P(red)  P(triangle and red) 30

60

75

5

15

 120  120  120  120 or 8

2115 71,080

45. Since there is a region that is both a trapezoid and blue, these events are inclusive. The total area of the two trapezoids is 45  45 or 90 in2. The total area of the blue regions is 45  7.5  7.5 or 60 in2. The total area that is both a trapezoid and blue is 45 in2. P(a trapezoid or a blue region)

 0.03 There are 2080  2043 or 4123 thousand who earn less than minimum wage. 4123

P(less than minimum wage)  71,080  0.06 39. These are mutually exclusive events. P(less than or equal to minimum wage)  P(less than)  P(equal to)  0.06  0.03 or 0.09

Chapter 14

5

 15 or 5

35. See students’ work. 36. These events are mutually exclusive. P(public transportation or walks)  P(public transportation)  P(walks)  0.049  0.015  0.064 or 6.4% 37. We estimate that 88.9% of the 400 employees will use a motor vehicle. 88.9% of 400  0.889(400)  355.6 There should be at least 356 parking spaces. 38. There are 1145  970 or 2115 thousand who earn minimum wage and a total of 15,793  55,287 or 71,080 thousand hourly workers. P(earn minimum wage) 

5

 15  15  15

7 16

 P(trapezoid)  P(blue)  P(trapezoid and blue) 90

60

105

7

45

 120  120  120  120 or 8

620

46. Since no region is both blue and red, these events are mutually exclusive. P(a blue triangle or a red triangle)  P(blue triangle)  P(red triangle)  

15 120 30 120

Page 776

C 

n r

15  120 1 or 4

C 

5 3



47. Since no region is both a square and a hexagon, these events are mutually exclusive. There are no regions that are squares. There is one hexagon that is formed by the two trapezoids; its area is 45  45 or 90 in2. P(a square or a hexagon)  P(square)  P(hexagon)

 

90

3

4

C 

48. Find the sum of the numbers in the regions of the Venn diagram. 36  38  8  25  2  5  3  3  120 Thus, 120 students were surveyed. 49. Find the sum of the numbers in the circle representing Event A. 36  38  25  2  101 Thus, 101 students said that they drive a car to school. 50. There are two students in the overlap of the three circles, indicating the students who do all three. There is a total of 120 students.

n r

C 

4 2

  

6

57. Use the Fundamental Counting Principle. There are 10 choices for the first store, 9 choices for the second store, and so on. 10  9  8  7  6  5  4  604,800 There are 604,800 different arrangements. 3 6 2 4 3  (2) (6)  4 58. c d  c d  c d 1 2 1 5 (1)  1 25

1

or

 c

39 40

59. c

52. Sample answer: Meteorologists use probabilities to forecast the weather. Answers should include the following. • You can use compound probabilities to forecast the weather over an extended period of time. • 80% 53. C; Since the marbles are not replaced, these events are not independent. P(first three red)  P(first red)  P(second red)  P(third red) 8

7

60.

2m2  7m  15 m  5



5 2 d 4 1

9m2  4 3m  2



2m2  7m  15 3m  2  9m2  4 m  5 (2m  3) (m  5) 3m  2  (3m  2) (3m  2) m  5



(2m  3) (m  5) m  5



1

1

6

1 2 d 0 7

4 5 9 7 (4)  (9) (5)  (7) d c d c d 8 8 4 9 84 89 c

 24  23  22 336

3

P(any one person)  10 or 5

51. Find the sum of the numbers within the three circles. 36  38  8  25  2  5  3  117 There are 117 students who drive or are involved in activities or have a job. There is a total of 120 students. P(A or B or C) 

n! (n  r)!r! 4! (4  2)!2! 4! 2!2! 4  3  2! 2!2! 43 or 6 2!

Thus, there are 6 different possible committees with our particular person and 10 possible committees.

P(all three)  120 or 60

117 120

n! (n  r)!r! 5! (5  3)!3! 5! 2!3! 5  4  3! 2!3! 54 or 10 2!

There are 10 possible committees. 56. Consider a particular person. Any committee for which this person is selected must also have two of the remaining four people. Find the number of combinations of the remaining four people taken two at a time.

 0  120

2

Maintain Your Skills

55. Find the number of combinations of five people taken three at a time.

1

3m  2

 (3m  2) (3m  2) 1

2m  3

 3m  2

7

 12,144 or 253

61. 145  13  3  5  232  15  315

54. A; We can eliminate answer choices B, C, and D with the following observation. The probabilty of Yolanda making a free throw in one attempt is 0.8. Three attempts would increase the probability that she makes a free throw. Thus, the probability of Yolanda making at least one free throw in three attempts is greater than 0.8 or 80%.

62. 1128  18  8  2  282  12  812

63. 240b4  223  5  b4  222  12  15  2b4  2  12  15  b2  2b2 110

621

Chapter 14

There are 5 ways to roll a sum of 6 and 36 distinct rolls.

64. 2120a3b  223  3  5  a3  b  222  12  13  15  2a2  1a  1b  2  12  13  15  0a 0  1a  1b

5

P(X  6)  36 6. First, find the probability that the sum is greater than 6 on a single roll. P(X 7 6)  P(X  7)  P(X  8)  P(X  9)  P(X  10)  P(X  11)  P(X  12)

 2 0a 0 130ab

65. 317  612  (3  6)( 17  12)  1817  2  18114

67.

9 24

69.

63 128

 9 24

68.

2 15

70.

5 52

 0.375  0.492 71.

8 36

73.

81 2470

74.

18 1235

 0.222

75.

 18 1235  0.015

14-4 Probability Distributions Check for Understanding

1. The probability of each event is between 0 and 1 inclusive. The probabilities for each value of the random variable add up to 1. 2. Sample answer: The probability of tossing a coin and getting a head versus getting a tail is the same. No matter what is tossed, the same probability is multiplied three times. 3. Sample answer: the number of possible correct answers on a 5-question multiple-choice quiz, and the probability of each 1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

Pages 779–780

Practice and Apply

10. There are two possible outcomes on each spin, red (R) and blue (B). Thus, for three spins, the possible outcomes are RRR, RRB, RBR, RBB, BRR, BRB, BBR, and BBB. 11. Each spin is an independent event. P(X  0)  P(RRR)  P(R)  P(R)  P(R) 1

1

1

 444

6 7 8 9 10 11 12

1

 64 P(x  1)  P(RRB)  P(RBR)  P(BRR) 1

1

3

1

3

3

3

1

1

3

9

 64 P(X  2)  P(RBB)  P(BRB)  P(BBR) 1

3

3

3

1

3

3

3

1

 444  444  444

1

9

9

9

 64  64  64

There are 4 ways to roll a sum of 5 and 36 distinct rolls.

 64

27

P(X  3)  P(BBB)

1

P(X  5)  36 or 9

3

3

3

 444 27

 64 Chapter 14

3

 64  64  64

P(X  4)  36 or 12

4

1

 444  444  444

5. There are 3 ways to roll a sum of 4 and 36 distinct rolls. 3

7

7. For each value of X, the probability is greater than or equal to 0 and less than or equal to 1. 0.05  0.10  0.40  0.40  0.05  1, so the probabilities add up to 1. 8. The probability of passing the course is the sum of the probabilities of earning an A, B, C, or D. P(X 1.0)  P(X  1.0)  P(X  2.0)  P(X  3.0)  P(X  4.0)  0.10  0.40  0.40  0.05  0.95 9. P(B or better )  P(X 3.0)  P(X  3.0)  P(X  4.0)  0.40  0.05  0.45

 0.036

4. Dice 1 2 3 4 5 6

7

343

 128 3570

Page 779

1

 1728

 11 38

 0.033 128 3570

7

 0.289

 81 2470

2

 12  12  12

 5 52

11 38

7

3

P(X 7 6 on three rolls)  P(X 7 6 on first roll)  P(X 7 6 on second roll)  P(X 7 6 on third roll)

 2 15

72.

21

4

Each roll is an independent event.

 0.096

 8 36

5

 36 or 12

 0.133

 63 128

6

 36  36  36  36  36  36

66. 13( 13  16)  13  13  13  16  232  13  13  12  3  232  12  3  312

622

12.

27 64

P(X )

32 64 24 64 16 64

23a. Let G represent a girl and B a boy. For a single 1 1 birth, P(G)  2 and P(B)  2. If X  1, then the first child is a girl. P(X  1)  P(G)

27 64

9 64

8 64

1

2

1 64

0

1 2 X = Blue

If X  2, then the first child is a boy and the second is a girl. P(X  2)  P(BG)  P(B)  P(G)

3

1

1

If X = 3, then the first two children are boys and the third is a girl. P(X  3)  P(BBG)  P(B)  P(B)  P(G) 1

1

1

1

 2  2  2 or 8 If X  4, then the first three children are boys and the fourth is a girl. P(X  4)  P(BBBG)  P(B)  P(B)  P(B)  P(G) 1

1

1

1

1

 2  2  2  2 or 16 23b. P(X  1)  P(X  2)  P(X  3)  P(X  4)  P(X 7 4)  1 1 2

1

1

1

 4  8  16  P(X 7 4)  1 15 16

 P(X 7 4)  1 1

P(X 7 4)  16 24. Sample answer: A pet store owner could use probability distributions to plan sales and special events. Answers should include the following. • Determine the probability of each outcome of an event and list them in a table. • The owner could look at the probability of a customer owning more than one pet and create special discounts for larger purchases. 25. A; P(X 2)  P(X  0)  P(X  1)  P(X  2)  0.0625  0.25  0.375  0.6875 26. B; P(X 6 5)  P(X  2)  P(X  3)  P(X  4)

0.3 0.25 P(X )

1

 2  2 or 4

13. No; the probability of X  1 is less than the probability of X  2, indicating that X  1 is less likely to occur. 14. Let X  the number of CDs. X can have values of 100, 200, 300, 400, and 500. 15. For each value of X, the probability is greater than or equal to 0 and less than or equal to 1. 0.10  0.15  0.40  0.25  0.10  1, so the probabilities add up to 1. 16. P(X 400)  P(X  100)  P(X  200)  P(X  300)  P(X  400)  0.10  0.15  0.40  0.25  0.90 17. P(X 7 200)  P(X  300)  P(X  400)  P(X  500)  0.40  0.25  0.10  0.75 18. P(at most some college)  P( some high school)  P(high school graduate)  P(some college)  0.167  0.333  0.173  0.673 19. 0.35

0.2 0.15 0.1 0.05

1

2

6

1

3

 36  36  36

0 me

ee gr De e re ed nc D e g va e A d l o r ' s egre e sD ch Ba t e' c i a e ge so ll As Co m e o o l ool So ch Sch S g h i gh H

Hi

So

 36 or 6 or about 0.17

Page 781

Maintain Your Skills

27. These are mutually exclusive events. P(ace or ten)  P(ace)  P(ten)

20. Sample answer: Add the values for the bars representing bachelor’s and advanced degrees. 21. No; 0.221  0.136  0.126  0.065  0.043  0.591. The sum of the probabilities does not equal 1. 22. We would expect about 6.5% of the 35 women to say they watch figure skating. 6.5% of 35  0.065(35)  2.275 We would expect about 2 women to say they watch figure skating.

4

4

8

2

 52  52  52 or 13 28. These are inclusive events. P(3 or diamond)  P(3)  P(diamond)  P(3 and diamond) 4

13

16

4

1

 52  52  52  52 or 13

623

Chapter 14

 612  712  1312

29. These are inclusive events. P(odd number or spade)  P(odd number)  P(spade)  P(odd number and spade)

30.



16 52



25 52



C 

n r

C 

10 7

  

13 52

38. 213  112  213  222  3  213  ( 222  13)  213  213  413 39. 317  2128  317  2222  7  317  2( 222  17)  317  2(217)  317  417  17

4 52



n! (n  r)!r! 10! (10  7)!7! 10! 3!7! 10  9  8  7! 3!7! 10  9  8 3!

C 

31.

n r

C 

12 5

  

 120 32. ( P ) ( P )  6 3 5 3  

n! (n  r)!r! 12! (12  5)!5! 12! 7!5! 12  11  10  9  8  7! 7!5! 12  11  10  9  8 5!

1

 792

 c

2 4 d 3 12

34. B  A  c  c

1

1.44 1.8

2

0.0825 12(4) 12

 900(1.006875) 48  1250.46 The balance is $1250.46. 42. Sample answer: Monthly; the balance after four years is greater with monthly compounding.

40 d 75

3 0 1 4 d d  c 5 7 2 5

43.

16 80

45.

30 114

 0.2

44.

20 52

46.

57 120

 20%

3  1 0  4 d 2  5 5  7

 0.3846  38%

 0.2632  26%

47.

72 340

 0.2118

Page 781

 0.475  48%

48.

54 162

 21%

 0.3333  33%

Practice Quiz 2

1. For each value of X, the probability is greater than or equal to 0 and less than or equal to 1. 0.25  0.32  0.18  0.15  0.07  0.02  0.01  1, so the probabilities add up to 1. 2. P(X 4)  P(X  4)  P(X  5)  P(X  6)  P(X  7)  0.15  0.07  0.02  0.01  0.25 or 25% 3. 0.35

or 0.8

0.3

P(X )

0.25

4

 3

0.2 0.15 0.1

4

x3

0.05

4

x  3 when y  3.

0

37. 318  712  3222  2  7 12  3( 222  12)  712  3(212)  712 Chapter 14

1

A  900 1 

y  0.8 when x  1.8. 36. First find k. xy  k (1) (4)  k 4  k The equation is xy  4. Now find x when y  3. xy  4 x(3)  4 3x 3

2

r 41. A  P 1  n nt

4 4  c d 7 2 35. First find k. xy  k (0.6) (2.4)  k 1.44  k The equation is xy  1.44. Now find y when x  1.8. xy  1.44 1.8y  1.44 y

2

0.0825 4(4) 4

 900(1.020625) 16  1247.68 The balance after 4 years is $1247.68.

 7200 3 0 1 4 33. A  B  c d  c d 2 5 5 7 1  (3) 5  (2)

1

A  900 1 

6! 5!  (6  3)! (5  3)! 6! 5!  3! 2! 654 543  1 1

 c

2

r 40. A  P 1  n nt

1 2 3 4 5 6 7 X  Number of People

624

5. Sample answer: 5 marbles of two colors where three of the marbles are one color to represent making a free throw, and the other two are a different color to represent missing a free throw. Randomly pick one marble to simulate a free throw 25 times. 6. See students’ work. 7. See students’ work. 8. See students’ work. 9. Yes; 70% of the marbles in the bag represent water and 30% represent land. 10. Since 30% of the Earth’s surface is land, the theoretical probability that a meteorite reaching the Earth’s surface hits land is 30%. 11. Brown represents hitting land. There are 19 and a total of 56  19 or 75 occurrences.

4. These events are inclusive. P(odd or greater than 4)  P(odd)  P(greater than 4)  P(odd and greater than 4) 5

6

8

4

3

 10  10  10  10 or 5

5. These events are mutually exclusive. P(less than 3 or greater than 7)  P(less than 3)  P(greater than 7) 2

3

5

1

 10  10  10 or 2

14-5 Probability Simulations Page 783

19

P(hits land)  75  0.25 or about 25% 12. We would expect 25% of the next 500 meteorites to hit land. 25% of 500  0.25(500)  125 We would expect 125 of the next 500 meteorites to hit land.

Algebra Activity

1. Rolling a 2 is one of six possible outcomes. P(rolling a 2) 

1 6

2. These events are mutually exclusive. P(rolling a 1 or a 6)  P(rolling a 1)  P(rolling a 6) 1

1

2

1

66  6 or 3

Pages 785–787

1

1

3

1

1

666  6 or 2 4. Sample answer: The class’s probability is closer since there are more trials. 5. Sample answer: The class’s probability should be closer to the theoretical probability. 6. Sample answer: The more times an experiment is performed, the closer the experimental probability is to the theoretical probability.

Page 785

Practice and Apply

13. Sample answer: a coin tossed 15 times 14. Sample answer: a spinner divided into 3 sections 1 1 where 2 represents cola, 3 represents diet cola, 1 and 6 represents root beer 15. Sample answer: a coin and a die since there are 12 possible outcomes 16. Sample answer: toss a coin and roll a die 100 times each 17. See students’ work. 18. See students’ work. 19. See students’ work. 20. See students’ work. 21. As you roll the dice more times, the experimental probabilities should get closer to the theoretical probabilities. Consider the following theoretical probabilities.

3. P(rolling a value less than 4)  P(rolling a 1)  P(rolling a 2)  P(rolling a 3)

Check for Understanding

4

P(sum of 5)  36 or about 0.11

1. An empirical study uses more data than a single study, and provides better calculations of probability. 2. As the number of trials increases, the experimental probabilities tend toward the theoretical probabilities. 3. Sample answer: a survey of 100 people voting in a two-person election where 50% of the people favor each candidate; 100 coin tosses 4. If you flip a coin, the theoretical probability of 1 getting heads is 2. If you flip a coin twice and get 1 head and 1 tail, the experimental probability of 1 getting heads is also 2. However, if you flip a coin twice and get two heads, then the experimental 2 probability of heads is 2 or 1. Thus, the theoretical and experimental probabilities of an event are sometimes the same.

4

P(sum of 9)  36 or about 0.11 Since these theoretical probabilities are close to 10%, we would expect to get a sum of 5 about 10% of the time and a sum of 9 about 10% of the time. 22. There are a total of 787 people enrolled. 47

P(Preschool)  787 or about 0.060 46

P(Kindergarten)  787 or about 0.058 378

P(Elementary School)  787 or about 0.480 201

P(High School)  787 or about 0.255 115

P(College)  787 or about 0.146

625

Chapter 14

45. These events are independent. P(3 dimes)  P(dime)  P(dime)  P(dime)

23. These events are mutually exclusive. P(Elementary or High School)  P(Elementary)  P(High School)  0.480  0.255  0.735 or about 0.74 or 74% 24. We would expect about be in kindergarten.

46 787

25

29. 31. 32. 33.

34.

15,625

18

1

125

46. These events are dependent. P(nickel, quarter, dime)  P(nickel)  P(quarter)  P(dime)

of the new students to

12

25

 55  54  53

We would expect about 105 of the new students to be in kindergarten. Sample answer: 3 coins 26. See students’ work. See students’ work. MMMMM, MMMMF, MMMFM, MMMFF, MMFMM, MMFMF, MMFFM, MMFFF, MFMMM, MFMMF, MFMFM, MFMFF, MFFMM, MFFMF, MFFFM, MFFFF, FMMMM, FMMMF, FMMFM, FMMFF, FMFMM, FMFMF, FMFFM, FMFFF, FFMMM, FFMMF, FFMFM, FFMFF, FFFMM, FFFMF, FFFFM, FFFFF See students’ work. 30. See students’ work. See students’ work. No; there were 181 heads out of the 300 tosses. The experimental probability of heads is about 60%. Sample answer: Probability can be used to determine the likelihood that a medication or treatment will be successful. Answers should include the following. • Experimental probability is determining probability based on trials or studies. • To have the experimental probability more closely resemble the theoretical probability the researchers should perform more trials. D; These are independent events. P(2 tails and a 3)  P(tail)  P(tail)  P(3) 1

25

 166,375 or 1331

46 1787 2 (1800)  105.2

25. 27. 28.

25

 55  55  55

5400

20

 157,410 or 583 47. Find the number of combinations of 55 coins taken 3 at a time. C 

55 3

 

55! (55  3)!3! 55! 52!3! 55  54  53 3!

 26,235 Now find the number of combinations of 2 dimes and 1 quarter. ( 25C2 )( 12C1 )   

25! 12!  (25  2)!2! (12  1)!1! 25! 12!  23!2! 11!1! 25  24 12  1 2!

 3600 There are 3600 ways to pick 2 dimes and 1 quarter and a total of 26,235 ways to choose 3 coins. 3600

48.

1

1

2a  3 2 a  3 2a  3 2 a  3

1



1

35. B; There are three ways that the one head can occur—first, second, or third—and there are eight possible outcomes for flipping a coin three times.

(a  3) (a  3) 1

(a  3) (2a  3)  2(a  3)(a  3)  12(a  3) (2a2  3a  9)  (2a2  18)  12a  36 2a2  3a  9  2a2  18  12a  36

3

P(exactly one head)  8

3a  9  12a  36

37. See students’ work. 39. See students’ work. 41. See students’ work.

9a  45 a5

The solution is 5. r2 r  7 r2 r  7

49.

Maintain Your Skills

42. For each value of X, the probability is greater than or equal to 0 and less than or equal to 1. 0.579  0.276  0.107  0.038  1, so the probabilities add up to 1. 43. P(X 2)  P(X  2)  P(X 3)  0.107  0.038  0.145 44. P(X 1)  P(X  0)  P(X  1)  0.579  0.276  0.855

Chapter 14

12  3

1

1

 24

Page 788

a

2  (a  3)(a  3) 1a 12 3 2 1 (a  3)1(a  3)  2aa  33 2  1 (a  3)1(a  3)  21 2 (a  3)(a  3)

 226

36. See students’ work. 38. See students’ work. 40. See students’ work.

80

P(2 dimes and 1 quarter)  26,235 or 583

1

(r  7) 1

r  7 1

r

r2  7 1

50  r 50 r  7

7

 14



 14

1r r 7  r 50 7 2  (r  7)14

2 1 

2

1

r  7 1

r

50  7 1

2  14(r  7)

r2  50  14r  98 r2  14r  48  0 (r  6)(r  8)  0 r  6  0 or r  8  0 r6 r8 The solutions are 6 and 8.

626

12

a  3 1

50.

1

1

x(x  6) 1

x (x  6) 1



x  2 x 1

2 1 

x  2 x  3 x6 x x  2 x  3 x6 x 1

x(x  6) 1

x  3  6

x

1



1 x



x (x  6) 1

54. Let c be the length of the ladder which will be the hypotenuse of the right triangle formed by the house, the ground, and the ladder. The angle formed by the ladder and the ground can be at most 75; this angle is opposite the side of the triangle formed by the house, which is 24 feet.

2  x(x  6) 11x 2 2

1

1

x 1

(x  6)(x  2)  x(x  3)  x  6 (x2  8x  12)  (x2  3x)  x  6 x2  8x  12  x2  3x  x  6 5x  12  x  6 6x  18 x3 The solution is 3. 51. 14

1

2x  3 x 2 7 2x  3 x 2 7



x  3 14 x  3 14 14

2 1

sin75  sin75 

c sin75  24 24

c  sin75 c  24.8466 The shortest ladder he should buy is 25 feet long. 55. Since the measure of the longest side is 9, let c  9, a  5, and b  7.

2

1141  2x 7 3 2  1141  2x 2  141  x 14 3 2

7

1

1

1

c2  a2  b2 ? 92  52  72 ? 81  25  49 81  74 Since c2  a2  b2, the triangle is not a right triangle.

1

2(2x  3)  7x  x  3 4x  6  7x  x  3 3x  6  x  3 4x  9 9

x  4

56. Since the measure of the longest side is 3134, let c  3134, a  9, and b  15. c2  a2  b2

9

The solution is 4. 52.

1

n(n  1) 1

n(n  1) 1

5n n  1 1

21

1

5n 1 n n  1 5n 1 n n  1

5

2  n(n  1)(5)

1

n (n  1) 1

1 n 1

? (3134) 2  92  152 ?

2  5n(n  1)

306  81  225 306  306 Since c2  a2  b2, the triangle is a right triangle. 57. Since the measure of the longest side is 93.6, let c  93.6, a  36, and b  86.4. c2  a2  b2 ? 93.62  362  86.42 ? 8760.96  1296  7464.96 8760.96  8760.96 Since c2  a2  b2, the triangle is a right triangle. Exercises 58–63 For checks, see students’ work. 58. (x  6) 2  4 x  6  14 x  6  2 x6  2 x62 x62 8 4 {4, 8}

n(5n)  (n  1)  5n2  5n 5n2  n  1  5n2  5n n  1  5n 1  4n

56

1 4

n



7 3

1 4

53.

a  2 a  2

3(a  2) (a  2)

a

2  2

1aa  22  a 2 2 2  3(a  2) (a  2) 173 2

13(a  2)1(a  2)  aa  22 2  13(a  2)1(a  2)  (a 2 2) 2 1

1

1

1

1



opposite leg hypotenuse 24 c

3 (a  2) (a  2) 1



7 3 1

3(a  2) (a  2)  3(a  2) (2)  (a  2)(a  2) (7) 3(a2  4a  4)  6(a  2)  7(a2  4) 3a2  12a  12  6a  12  7a2  28 3a2  6a  24  7a2  28 10a2  6a  4  0 2(5a2  3a  2)  0 2(5a  2) (a  1)  0 5a  2  0 or a  1  0 5a  2 a  1

59.

x2  121  22x x  22x  121  0 x2  2(x)(11)  112  0 (x  11) 2  0 x  11  0 x  11 {11} 2

2

a5 2

The solutions are 1 and 5.

627

Chapter 14

60.

4x2  12x  9  0 (2x)  2(2x) (3)  32  0 (2x  3) 2  0 2x  3  0 2x  3

11. List the possible outcomes. Let J represent a win for the Jackals and T a win for the Tigers. If either team wins the first three games, the last two games will not be played. Thus, two possible outcomes are JJJ and TTT. If either team achieves its third victory in the fourth game, the fifth game will not be played. Six more possible outcomes are JJTJ, JTJJ, TJJJ, TTJT, TJTT, and JTTT. If all five games are played, the possible outcomes are JJTTJ, JTJTJ, JTTJJ, TJJTJ, TJTJJ, TTJJJ, TTJJT, TJTJT, TJJTT, JTTJT, JTJTT, and JJTTT. There are 20 possible outcomes.

2

61.

3

x  2

532 6

25x2  20x  4 25x  20x  4  0 (5x) 2  2(5x) (2)  22  0 (5x  2) 2  0 5x  2  0 5x  2 2

62.

12. P  n r

2

P 

x  5

525 6

4 2

 

49x2  84x  36  0 (7x)  2(7x) (6)  62  0 (7x  6) 2  0 7x  6  0 7x  6 2

C 

4 4

6 7

 2

63. 180x  100  81x 0  81x2  180x  100 0  (9x) 2  2(9x)(10)  102 0  (9x  10) 2 0  9x  10 10  9x

5 6



7

654 3! 7!

7!

16. ( P )( P )  (7  3)!  (7  2)! 7 3 7 2 7! 7!  4!  5!  76576  8820 3!

2. 4. 6. 8.

4!

17. ( C ) ( P )  (3  2)!2!  (4  1)! 3 2 4 1 3! 4!  1!2!  3! 3

 14

factorial 1 mutually exclusive Theoretical

 12 18. P(blue, red, green)  P(blue)  P(red)  P(green) 22

30

22

 74  73  72 14,520

605

 388,944 or 16,206

Lesson-by-Lesson Review

19. P(red, red, blue)  P(red)  P(red)  P(blue)

9. There are a total of 10 videos, so she has 10 choices for the first video. Then after she has chosen the first video, she has only 9 videos from which to choose the second video. Then she has 8 choices for the third video. Use the Fundamental Counting Principle. 10  9  8  720 There are 720 possible outcomes. 10. Use the Fundamental Counting Principle. 12  8  10  5  4800 There are 4800 possible outcomes. Chapter 14

6!  3)!3!

 140

Vocabulary and Concept Check

Pages 789–792



n! (n  r)!r! 4! (4  4)!4! 4! 0!4! 4! or 1 4!

 1

x

permutation independent are not 1



7!

Chapter 14 Study Guide and Review 1. 3. 5. 7.



15. ( C ) ( C )  (7  1)!1!  (6 7 1 6 3 7! 6!  6!1!  3!3!

10 9

Page 789

C 

8 3

n! (n  r)!r! 8! (8  3)!3! 8! 5!3! 8  7  6  5! 5! 3! 876 3!

 56 14. C  n r

6

10 9

13. C  n r

 4  3 or 12

x7

56

n! (n  r)! 4! (4  2)! 4! 2! 4  3  2! 2!

30

29

22

 74  73  72 19,140

1595

 388,944 or 32,412 20. P(red, green, not blue)  P(red)  P(green)  P(not blue) 30

22

50

 74  73  72 33,000

1375

 388,944 or 16,206

628

21. P(diamond or club)  P(diamond)  P(club) 13

13

26 52

1 2

Chapter 14 Practice Test

 52  52 

or

Page 793 1. permutation 3. random variable 4.

22. P(heart or red)  P(heart)  P(red)  P(heart and red) 13

26

Clifton-2 Bakersville-1 Clifton-3

1

 52 or 2

Clifton-4 Ashville

23. P(10 or spade)  P(10)  P(spade)  P(10 and spade)  

4 52 16 52

13  52 4 or 13



Clifton-1 Clifton-2 Bakersville-2

1 52

Clifton-3 Clifton-4

Probability

Derry–1 Derry–2 Derry–1 Derry–2 Derry–1 Derry–2 Derry–1 Derry–2 Derry–1 Derry–2 Derry–1 Derry–2 Derry–1 Derry–2 Derry–1 Derry–2

A-B1-C1-D1 A-B1-C1-D2 A-B1-C2-D1 A-B1-C2-D2 A-B1-C3-D1 A-B1-C3-D2 A-B1-C4-D1 A-B1-C4-D2 A-B2-C1-D1 A-B2-C1-D2 A-B2-C2-D1 A-B2-C2-D2 A-B2-C3-D1 A-B2-C3-D2 A-B2-C4-D1 A-B2-C4-D2

5. We can either count the outcomes from the tree diagram or use the Fundamental Counting Principle. 2  4  2  16 There are 16 different routes from Ashville to Derry. 6. Combination; order is not important. We find the number of ways the students could choose 6 of 9 chairs.

24. For each value of X, the probability is greater than or equal to 0 and less than or equal to 1. 0.04  0.12  0.37  0.30  0.17  1, so the probabilities add up to 1. 25. P(1 X 3)  P(X  1)  P(X  2)  P(X  3)  0.12  0.37  0.30  0.79 or 79% 26. Extracurricular Activities 0.4 0.3 0.2 0.1 0

Clifton-1

13

 52  52  52 26

2. experimental

C 

n r

C 

9 6

 

0 1 2 3 4 X  Number of Activities



n! (n  r)!r! 9! (9  6)!6! 9! 3!6! 9  8  7  6! 3!6! 987 3!

 84 There are 84 ways to choose 6 of the 9 chairs. 7. Permutation; order is important. We find the number of permutations of 10 participants taken 4 at a time.

27. Let R represent red and P represent pink. For a single flower produced, P(R)  0.75 and P(P)  0.25. P(4 red, 1 pink)  P(RRRRP)  P(RRRPR)  P(RRPRR)  P(RPRRR)  P(PRRRR)  P(R)P(R)P(R)P(R)P(P)  P(R)P(R)P(R)P(P)P(R)  P(R)P(R)P(P)P(R)P(R)  P(R)P(P)P(R)P(R)P(R)  P(P)P(R)P(R)P(R)P(R)  (0.75)(0.75)(0.75)(0.75)(0.25)  (0.75)(0.75)(0.75)(0.25)(0.75)  (0.75)(0.75)(0.25)(0.75)(0.75)  (0.75)(0.25)(0.75)(0.75)(0.75)  (0.25)(0.75)(0.75)(0.75)(0.75)  0.07910  0.07910  0.07910  0.07910  0.07910  0.396 or 39.6% 28. Sample answer: There are six possible outcomes, so you could use a die. 29. Of the 80 total observations, 23 have 3 red and 2 pink.

P 

n r

P 

10 4

 

n! (n  r)! 10! (10  4)! 10! 6! 10  9  8  7  6! 6!

 10  9  8  7 or 5040 There are 5040 different ways that 10 participants can finish first, second, third, and fourth. 8. Permutation; since each has a separate responsibility, order is important. We find the number of permutations of 15 girls taken 2 at a time and 19 boys taken 2 at a time and use the Fundamental Counting Principle.

115P22119P22  (1515! 2)!  (1919! 2)! 15!

19!

 13!  17! 

23

P(3 red, 2 pink)  80

15  14 1



19  18 1

 71,820 There are 71,820 possible committees.

 0.2875 or about 28.8%

629

Chapter 14

9. These are dependent events. P(blue, green)  P(blue)  P(green) 6

19. First find P(queen or red). These events are inclusive. P(queen or red)  P(queen)  P(red)  P(queen and red)

2

 16  15 12

1

 240 or 20 10. These are dependent events. P(yellow, yellow)  P(yellow)  P(yellow) 4

1

 240 or 20

7

7

20. First find P(ace or heart) after a black 10 has been selected and not replaced. P(ace or heart)  P(ace)  P(heart)  P(ace and heart)

4

1

 3360 or 35 12. These are dependent events.

4

4

16

P(black 10, ace or heart)  P(black 10)  P(ace or heart)

3

 3360 or 35

13. These are independent events. P(yellow, 4)  P(yellow)  P(4) 1

2

32

21. To find the probability for each number of heads, divide the number of outcomes given by 16, the total number of outcomes.

1

 48 14. These are independent events. P(red, even)  P(red)  P(even)

Four Coins Tossed X  Number of Heads Probability 0 0.0625 1 0.25 2 0.375 3 0.25 4 0.0625

3

 86 1

 48 or 16 15. These are independent events. P(purple or white, not prime)  P(purple or white)  P(not prime) 2

3

 86 6

22. From the probability distribution, we see that P(X  0)  0.0625 or 6.25%. 23. P(X 2)  P(X  2)  P(X  3)  P(X  4)  0.375  0.25  0.0625  0.6875 or 68.75% 24. When there are two tails, there are also two heads. P(two tails)  P(two heads)  P(X  2)  0.375 or 37.5% 25. D; Three numbers a, b, and c can be arranged in 6 different ways. a, b, c a, c, b b, a, c c, a, b b, c, a c, b, a

1

 48 or 8 16. These are independent events. P(green, even or less than 5)  P(green)  P(even or less than 5) 1

5

 86 5

 48 17. These are dependent events. P(club, heart)  P(club)  P(heart) 13

13

 52  51 169

13

 2652 or 204 18. These are dependent events. P(black 7, diamond)  P(black 7)  P(diamond) 2

13

 52  51 26

1

 2652 or 102

Chapter 14

8

 2652 or 663

1

3

16

 52  51

 86

1

1

 51

12

 16  15  14 288

13

 51  51  51

P(blue, red, not green)  P(blue)  P(red)  P(not green) 6

1

 663

 16  15  14 96

7

 13  51

11. These are dependent events. P(red, blue, yellow)  P(red)  P(blue)  P(yellow) 6

28

2

P(queen or red, jack of spades)  P(queen or red)  P(jack of spades)

3

4

26

 52 or 13

 16  15 12

4

 52  52  52

630

9. D; There are 26 choices for each character. Use the Fundamental Counting Principle. 26  26  26  17,576 There are 17,576 possible passwords. 10. Solve the system of equations. x  4y  0 2x  3y  11 Solve the first equation for x. x  4y  0 x  4y Substitute 4y for x in the second equation. 2x  3y  11 2(4y)  3y  11 8y  3y  11 11y  11 y  1 Use x  4y to find x. x  4y  4(1) or 4 The intersection point is (4, 1).

Chapter 14 Standardized Test Practice Pages 794–795 1. D; Since the average of a and b is 20, we have a  b  20 or a  b  40. Since the average of a, 2 a  b  c b, and c is 25, we have  25 or 3 a  b  c  75. Substitute 40 for a  b and solve for c. a  b  c  75 40  c  75 c  35 2. D; Since the formula for the volume of a cube is V  s3 and the volume of the cube is 27 cubic inches, s3  27 where s is the length of an edge of the cube. Thus, s  3. surface area  6s2  6(3) 2  6(9) or 54 The surface area of the cube is 54 square inches. 3. B; Assuming that the truck is traveling at a constant speed, the truck is halfway between the 1 two towns after 2 hour. Thus, the car has traveled 1 for 2 hour when the truck reaches Newton. d  rt

1

1

2

2

 4x  4x  1  1  4x2  4x The equation is true for all values of x. 12. Determine whether c2  a2  b2. c2  a2  b2

1

 60  2  30 The car has traveled 30 miles. 4. A; One point on the graph appears to have coordinates (70, 50). Check (70, 50) in each equation. The only equation given that has 1 (70, 50) as a solution is y  2x  15. 5. C; There are eight possible outcomes. Three of the outcomes have exactly one boy, since the boy could be born first, second, or third.

? ( 174) 2  52  72 ?

74  25  49 74  74 The triangle is a right triangle with hypotenuse c. Thus, the angle opposite side c measures 90. 13. There are 9 possible digits for each of the first and last digits of the four digits. There are 10 possible digits for each of the remaining two digits. Use the Fundamental Counting Principle. 9  10  10  9  8100 There are 8100 telephone numbers available.

3

P(exactly one boy)  8 6. C; 52  52 or 25 1

2

1 1 11. 4 x  2 2  1  4 x2  x  4  1

1

x2  x  20 x  x  20  0 (x  5)(x  4)  0 or x  4  0 x50 x  5 x4 The solutions are 5 and 4. 8. B; The ordered pairs (2, 2) and (7, 8) describe the locations of the planes relative to the airport. Use the distance formula to find the distance between the planes.

7. D;

2

1

14. On a single roll, P(left)  6 or 3 and 4 2 P(right)  6 or 3. These are independent events. There are two ways to reach the goal in two turns: first go left then right or first go right then left. P(reach goal in two turns)  P(left the right)  P(right then left)  P(left) · P(right)  P(right) · P(left)

2

1

2

2

1

 33  33 2

2

4

 9  9 or 9

d  2(x2  x1 ) 2  (y2  y1 ) 2

15. A; 3x  15 7 3x 7 x 7 2y  3 7 2y 7 y 6 Thus, x 7 10

d  2(8  2) 2  (7  2) 2 d  262  52 d  161 or about 7.8 The planes are about 7.8 miles apart.

631

45 30 10 17 20 10 7 y.

Chapter 14

18a. 10 Random Specials are possible You can list all of the combinations, using letters for each topping. PS PO PM PG SO SM SG OM OG MG There are 10 possible combinations. 18b. Of the ten combinations, four include mushrooms. The probability of mushrooms on a Random 4 Special pizza is 10 or 40%.

12!  4)!

16. A; 12P4  (12 12!  8!

 12  11  10  9 or 11,880 C 

10 6

 

10! (10  6)!6! 10! 4!6! 10  9  8  7 4!

or 210

Thus, 12P4 7 10C6. 17. B; These are dependent events. P(blue, green)  P(blue)  P(green) 7

2

 14  13 14

 182 or about 0.08 P(red, red)  P(red)  P(red) 5

4

 14  13 20

 182 or about 0.11 Thus, P(blue, green) P(red, red).

Chapter 14

632

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Prerequisite Skills Page 799

Operations with Fractions: Adding and Subtracting

1.

2 5

5

1

3.

4 3

4 3

 

 

5.

5 16

2  1 5 3 5

 7.

6 9

9.

28 40

11.

27 99

13.

2 9

2 3



7 2

10.

27  9

 99  9

12.

24 180

3 11

1

3

1

14.

2 15

3

16.

17.



1 9

 

19.

1 2

1

2

4 3

5

9 

23.

1 4

1

1

20.

5

5 4

3 4

1

5

4

2 3

1

26.

8 9

27.

3 7

5

2

5

28.

8

3

15.

 

31.

3 4

2

9



2

30.



3 5

8

32.

4 15

3

1

33.



4 15

  

55 60 39 60 13 20





34.



1 8



7

 15  

3 2

1

15 2 7 2

9 4

1

12.

1

1

7

 15 1

1

or 32

2

3

3  23

1 3

6

1

6

5  35 1

2

5 1

9

1

 18  4  18

14.

2

11 3

9

 44 

3

11 3

9

 44

1

4

3

4

2 7

14 3



9 10 3 10

24 88 35 88

2

 7 1

4 3

10

14 3

or

1 3

16.

2 11



110 17

1 13

2

 11  1



12

1

20 17

18.

1

1 3



15 2

4 19

1

30

1

 3 1



10

30 11

110 17 3

or 117

5

12

 19  3  19

1

 3  11  3

20.

1

8

6 5

10

5 2

15 2 1

or 22

1

2

6

10

1

2

 12  5  12

10

2

2

 11

1 21. 

6 10

45

7 6

1

3 2

23.

11 88

1 22



22 1

7

1

3 2

is 3.

14 23

2

1 1 22

is 22.

14 23

is 14 or 114.

23

 14  1

The reciprocal of

633

is 6 or 16.

2

The reciprocal of 24.

6 7

3  1

The reciprocal of

1



6 7

The reciprocal of 22.

16



15 2

1 9

 19.

61

3 11

10.

1

6

 60 or 160 16 60

1

1

4

17.

 4  60  60

 12

21

1

 11  3  11 2

2

3

13



 3  12  12

11 12

1

5

9 10

2 3

1

or 54

 33

2  92



1

29. 1 

2 9

 20 1 19

8.

1

4 19 19 18 19

12

2

21 4



5  35

 5  20  20

11

73

3

2  22

2

3

8

 14 1 19

1 3

399

13 20

7 2

8

2

6

6.

1

13.

9

 14  14  14

51

1

1 6

1

2

6

1

 4  12  12 2

23

3

7  57

2

11

4

2 5

 35

4  24

0 24.

4.

1

 12

22

5 2

 11.

3

3

9

3 2

3 5

22. 12  2  2  2

 5  20  20

13

3

 20  5  20

1

9.

366

 20 25.



8

2

 21

2

1 2 2 4 1 2

or 19

21

1

3  73

 15

4

1

9

7.

9



12 9 17 9

2 7

5

3

3 4

2.

8

88

1 2

5.

 15  15



1 5

2 15



444 

21.

18.

3.

24  12



3 9 1 3

31

1

5  45 3

 180  12

1 4 9

3 4

4 25

7 8

125

Operations with Fractions: Multiplying and Dividing

 20

16  4



18

143

3

7

5

 6  150  150  150

 100

1 3

33

1.

 100  4



99

Page 801

1 2



3 25

36.

1

7  7

7 10

44

2

 14  7

16 100

94 50

7  4 2 3 1 or 12 2

4

7 14

11

 25  100  100  100

3  4 9 7 9



2



28  4

 15.

8.

 40  4



4 9

94 100

2  1 7 1 7

1



6  3  3 2 3



4.

3 9

6.

9 

7



5  4 16 1 16

4

2 7



4  4 3 8 2 or 23 3

 16 

2.

35.

23

9

Prerequisite Skills

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3

24 

25. 11 4

11 4

11. The part is 16, and the base is 40. Let p represent the percent.

4

 11  1 3

a b 16 40

4

The reciprocal of 24 is 11. 26.

1 53 16 3  3 16



16 3

1

2 3

1

1600 40

3

2

3

 3  31

28.

16 9

30.

3 7

6

3 2

1

3

2

 2  21

1

5

6

 2 or 3 31.

9 10

3

9

7

 7  10  3

32.

1 2

3

5

63

 30 21

16 9  9 4 144  36 or 4 3 5  71 15 1  7 or 27 1 5  23 5 6

a b 14 b

1

33.

35.

37.

1 2

9 1  42 9 2  41 18  4 9 1  2 or 42 11 2 11 5  13  12  3 12 11 3  12  5 33 11  60 or 20 1 1 1 6  15  3  5 3 1 5  36 5  18

Page 803 1. 3. 5.



2 3

  

36.

38.

3 8



3 25

1 4



1400 20

4 2 3 3 4 3  3 2 12 or 2 6

3 4  81 12  8 3  2 or 2 3  25 15 45  50

0.9

2. 4. 6.

8.

9  1000 1400 1400%  100

a b 80 b

8000 50

a b a 18

60 3 60%  100 or 5 120 6 120%  100 or 5 2.5 2.5%  100 25 1  1000 or 40 0.4 0.4%  100 4 1  1000 or 250

100a 100

p

 100 50

 100



50b 50

p

 100 25

 100

450

 100

a  4.5 4.5 is 25% of 18. 15. The percent is 10, and the base is 95. Let a represent the part. a b a 95

or 14

p

 100 10

 100

100a  95(10) 100a  950

p

100a 100

p

 100

950

 100

a  9.5 9.5 is 10% of 95.

125p 125

20  p 25 is 20% of 125.

Prerequisite Skills

20b 20

100a  18(25) 100a  450

 100





160  b 80 is 50% of 160. 14. The percent is 25, and the base is 18. Let a represent the part.

9

or 10

25(100)  125p 2500  125p 2500 125

20

 100

80(100)  50b 8000  50b

1 12 15  2

10. The part is 25, and the base is 125. Let p represent the percent. a b 25 125

p

 100

70  b 14 is 20% of 70. 13. The part is 80, and the percent is 50. Let b represent the base.

The Percent Proportion

5 1 5%  100 or 20 11 11%  100 78 39 78%  100 or 50

7. 0.9%  100

9.

34.

1 13

40p 40

14(100)  20b 1400  20b

 10 or 210 1 24



40  p 16 is 40% of 40. 12. The part is 14, and the percent is 20. Let b represent the base.

4

9

 3 or 2 29.

p

 100

16(100)  40p 1600  40p

1

The reciprocal of 53 is 16. 27.

p

 100

634

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21. The part is 49, and the percent is 200. Let b represent the base.

16. The part is 30, and the base is 48. Let p represent the percent. a b 30 48

p

a b 49 b

 100 p

 100

30(100)  48p 3000  48p 3000 48

4900 200

62.5  p 30 is 62.5% of 48. 17. The percent is 150, and the base is 32. Let a represent the part. a b a 32

200

 100

49(100)  200b 4900  200b

48p 48



p

 100



200b 200

24.5  b 49 is 200% of 24.5. 22. The part is 15, and the base is 12. Let p represent the percent.

p

 100

a b 15 12

100a  32(150) 100a  4800

15(100)  12p 1500  12p

100a 100

 100 150



1500 12

4800 100

a b 48 32

p

 100 5

 100



4800 32

5b 5

a b a 20

p

 100 p

 100



100a 100

400p 400

 

a b 36 40

p 100 0.5 100



32p 32

p

 100 85

 100 

1700 100

p

 100 p

 100

3600  40p

100a  250(0.5) 100a  125 100a 100

p

 100

a  17 Madeline will likely make 17 of the 20 shots. 25. The part is 36, and the base is 40. Let p represent the percent.

0.25  p 1 is 0.25% of 400. 20. The percent is 0.5, and the base is 250. Let a represent the part. a b a 250

p

 100

100a  1700

1(100)  400p 100  400p 100 400

12p 12

150  p 48 is 150% of 32. 24. The percent is 85, and the base is 20. Let a represent the part.

70  b 3.5 is 5% of 70. 19. The part is 1, and the base is 400. Let p represent the percent. a b 1 400



48(100)  32p 4800  32p

3.5(100)  5b 350  5b 350 5

p

 100

125  p 15 is 125% of 12. 23. The part is 48, and the base is 32. Let p represent the percent.

a  48 48 is 150% of 32. 18. The part is 3.5, and the percent is 5. Let b represent the base. a b 3.5 b

p

 100

3600 40



40p 40

90  p Brian answered 90% of the questions correctly.

125

 100

a  1.25 1.25 is 0.5% of 250.

635

Prerequisite Skills

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26. The part is 4, and the percent is 80. Let b represent the base. a b 4 b

7.

3 10

80

 100 

25

5

13. Let N  0.6 or 0.666 p . Then 10N  6.666 p 10N  6.666 p  1N  0.666 p 9N  6 6

2

So, 0.6  3.

15

 100

34

17

14. 0.0034  10,000 or 5000 8

2

4

1

15. 2.08  2100 or 225

15b 15

16. 0.004  1000 or 250

20  b The recommended number of grams of saturated fat is 20. 29. The part is 470, and the percent is 20. Let b represent the base. p

 100 20

 100

47,000  20b 20b 20

2350  b The recommended daily value of sodium is 2350 mg or 2.35 g. 30. The part is 110, and the base is 250. Let p represent the percent.

17. 19. 21. 23. 25. 27. 29. 31.

0.4  40% 2.5  250% 0.065  6.5% 0.005  0.5% 45%  0.45 68%  0.68 200%  2 5.2%  0.052

18. 20. 22. 24. 26. 28. 30. 32.

0.08  8% 0.33  33% 5  500% 0.3  33.3% 3%  0.03 115%  1.15 0.1%  0.001 10.5%  0.105

33.

3 4

34.

9 20

 0.75  75%

35.

1 2

37.

1 3

p

 0.5

p

39.

11,000  250p

 0.3333 p

6 5

 1.2

40.

70

7

43. 52%  0.52

Expressing Fractions as Decimals and Percents

38

2.

2 5

 0.375 3.

2 3

23  0.666 p or 0.6

Prerequisite Skills



25

4.

1

 1000

34  0.75

636

 0.76

3

 100 25

1

 100 or 4 46. 135%  1.35

3 50

47. 0.1%  0.001

 0.4 3 4

or

19 25

44. 25%  0.25 13

45. 6%  0.06 6 100

 0.875

 76% 42. 3%  0.03

 100 or 10 52

1.

38.

7 8

 0.1666 p

 87.5%

 100 or 25

3 8

1 6

 16.7%

 120% 41. 70%  0.70

250p 250

44  p 44% of the Calories come from fat.

Page 805

36.

 33.3%

 100

 0.45  45%

 50%

 100



2

N  9 or 3

p

11,000 250

5

So, 0.45  11.

 100

a b 110 250

6

45

120p 120



1

N  99 or 11

p

47,000 20

56

10. 0.25  100 or 4

 100

a b 470 b

5 6

12. Let N  0.45 or 0.4545 p . Then 100N  45.4545 p 100N  45.4545 p  1N  0.4545 p 99N  45

p



8.

 0.8333 p or 0.83

9 10 24

300  15b 300 15

59

11. 5.24  5100 or 525

5p The solution is 5% glucose. 28. The part is 3, and the percent is 15. Let b represent the base. a b 3 b

5 9

 0.555 p or 0.5

 3  10

9. 0.9 

80b 80

 100



6.

 0.3

600  120p 600 120

12  0.5

5b José played 5 games of solitaire. 27. The part is 6, and the base is 120. Let p represent the percent. a b 6 120

1 2

p

 100

400  80b 400 80

5.

35

7

 1100 or 120 48. 0.5%  0.005 

5 1000

1

or 200

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Page 807

Making Bar and Line Graphs

Page 809

1. A line graph is the better choice since we are looking at how one quantity, the size of the plant, changes over time. 2. A bar graph is the better choice since we are comparing three different quantities, the populations of Idaho, Montana, and Texas. 3. A bar graph is the better choice since we are making a comparison of the number of students in several classes. 4. A line graph is the better choice since we are looking at how a single quantity, your height, has changed over time. 5. A bar graph is the better choice since we are comparing the number of people in two categories, those who shower in the morning and those who shower at night. 6. Hours of Sleep 10 Number of Hours

Making Circle Graphs

1. Find the number of degrees for each continent or region. North America: 7.9% of 360  0.079  360  28 South America: 5.7% of 360  0.057  360  21 Europe: 12.0% of 360  0.12  360  43 Asia: 60.7% of 360  0.607  360  219 Africa: 13.2% of 360  0.132  360  48 Australia: 0.5% of 360  0.005  360  2 Antarctica: 0% of 360  0 Due to rounding, the angles do not total 360. Make the circle graph.

8 6 4 2

World Population, 2000

0 Alana Kwam Tomas Nick Student

Kate Sharla

Lawn Care Profits

7.

Asia 60%

North America 8% Europe South 12% America 6% 2. Find the number of degrees for each part of the body. There are a total of 206 bones in the body.

Profits ($)

200 150 100 50 0 1 2 3 4 5 6 7 8 9 10111213 1415 Week

8.

29  360  51 206 26 Spine: 206  360  45

Skull:

Play Attendance Number of People

Africa Australia 13% 1%

25 206

 360  44

50

Ribs and Breastbone:

40

Shoulders, Arms, and Hands: 64 206

30

 360  112

Pelvis, Legs, and Feet:

20

62 206

 360  108

Make the circle graph. 10

Bones in Parts of the Human Body

0 Under 20-39 40-59 60 and 20 over Age

Shoulders, Arms, Hands, 64

Ribs, Breast bones 25

637

Pelvis, Legs, Feet, 62 Skull, 29 Spine, 26

Prerequisite Skills

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Page 811

Identifying Two-Dimensional Figures

Page 814

1. The triangle has all acute angles and no congruent sides. It is an acute scalene triangle. 2. The triangle has all acute angles and all sides congruent. It is an acute equilateral triangle. 3. The triangle has one obtuse angle and two congruent sides. It is an obtuse isosceles triangle. 4. The triangle has one right angle and no congruent sides. It is a right scalene triangle. 5. The figure has six sides. It is a hexagon. 6. The figure has five sides. It is a pentagon. 7. The figure has four sides with opposite sides congruent and four right angles. It is a quadrilateral, a parallelogram, and a rectangle. 8. The figure has four sides with opposite sides parallel, four congruent sides, and four right angles. It is a quadrilateral, a parallelogram, a rectangle, a rhombus, and a square. 9. The figure has four sides with opposite sides parallel. It is a quadrilateral and a parallelogram. 10. The figure has four sides with one pair of opposite sides parallel. It is a quadrilateral and a trapezoid. 11. The figure has four sides with opposite sides congruent and four right angles. It is a quadrilateral, a parallelogram, and a rectangle. 12. The figure has four congruent sides. It is a quadrilateral, a parallelogram, and a rhombus. 13. The figure has eight sides. It is an octagon. 14. The figure has four sides, none of which are either parallel or congruent. It is a quadrilateral. 15. The figure has five sides. It is a pentagon.

Page 812

1. P  2(/  w)  2(3  2)  2(5)  10 A  /w  32 6 The perimeter is 10 centimeters, and the area is 6 square centimeters. 2. P  4s  4(1) 4 A  s2  12 1 The perimeter is 4 inches, and the area is 1 square inch. 3. P  2(/  w)  2(1  7)  2(8)  16 A  /w  17 7 The perimeter is 16 yards, and the area is 7 square yards. 4. P  4s  4(7)  28 A  s2  72  49 The perimeter is 28 kilometers, and the area is 49 square kilometers. 5. P  2(6  4)  2(10)  20 A  64  24 The perimeter is 20 feet, and the area is 24 square feet. 6. P  2(12  9)  2(21)  42 A  12  9  108 The perimeter is 42 centimeters, and the area is 108 square centimeters. 7. P  4(3)  12

Identifying Three-Dimensional Figures

1. The figure has two parallel, congruent faces that are rectangles. It is a rectangular prism. 2. The figure has one base that is a triangle and three faces that are triangles. It is a triangular pyramid. 3. The figure is a sphere. 4. The figure has two parallel, congruent faces that are triangles. It is a triangular prism. 5. The figure has one base that is a rectangle and four faces that are triangles. It is a rectangular pyramid. 6. The figure has a circular base and one vertex. It is a cone. 7. The figure is a rectangular prism in which all of the faces are squares. It is a cube. 8. The figure has two parallel, congruent faces that are triangles. It is a triangular prism. 9. The figure has one base that is a rectangle and four faces that are triangles. It is a rectangular pyramid.

Prerequisite Skills

Perimeter and Area of Squares and Rectangles

A  32 9 The perimeter is 12 meters, and the area is 12 square meters.

638

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8. P  4(15)  60

Page 816

A  152  225 The perimeter is 60 inches, and the area is 225 square inches.

1  112 1  2 1 192 2 39  21 2 2

9. P  2

1 82

 39 1

A  82  11  

17 11  1 2 187 or 2

1

932

The perimeter is 39 inches, and the area is 1 932 square inches. 1 1 1 2 1 2  2 1 124  144 2 3  2 1 264 2 107  21 4 2

10. P  2 124  142



107 2

1

or 532

1



49 29  2 4 1421 or 8

is about 18.8 meters.

is about 31.4 inches.

is about 75.4 centimeters.

is about 9.4 kilometers.

is about 3.1 yards.

1 12 1  1 54 2 

1

A  124  142 

Area and Circumference of Circles

1. C  2r  2(3)  6  18.8 The circumference 2. C  d  (10)  10  31.4 The circumference 3. C  2r  2(12)  24  75.4 The circumference 4. C  2r  2(1.5)  3  9.4 The circumference 5. C  d  (1)   3.1 The circumference 6. C  d   54

5

1778

 16.5 The circumference is about 16.5 feet. 7. C  2r

1

The perimeter is 532 feet, and the area is 5 1778 square feet.

1 12

11. P  4(2.4)  9.6 A  (2.4) 2  5.76 The perimeter is 9.6 centimeters, and the area is 5.76 square centimeters. 12. P  4(5.8)  23.2 A  (5.8) 2  33.64 The perimeter is 23.2 meters, and the area is 33.64 square meters. 13. P  2(18  90)  2(108)  216 A  18  90  1620 The perimeter for each plot of the garden is 216 feet, and the area is 1620 square feet.

 2 242

 49  153.9 The circumference is about 153.9 inches. 8. A  r2  (5) 2  25  78.5 The area is about 78.5 square inches. 9. A  r2  (2) 2  4  12.6 The area is about 12.6 square feet. 10. The radius is one-half times the diameter, or 1 km. A  r2  (1) 2   3.1 The area is about 3.1 square kilometers.

639

Prerequisite Skills

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11. The radius is one-half times the diameter, or 2 yd. A  r2  (2) 2  4  12.6 yd2 The area is about 12.6 square yards.

Page 817

1

 252

2.

12. A  r2  (1) 2   3.1 The area is about 3.1 square meters. 13. A  r2  (1.5) 2  2.25  7.1 The area is about 7.1 square feet. 14. The radius is one-half the diameter, or 7.5 cm. A  r2  (7.5) 2  56.25  176.7 The area is about 176.7 square centimeters. 15. C  2r 25,000  2r 25,000 2



Volume

1. V  /  w  h

3.

4.

5.

6.

2r 2

7.

3979  r The tunnel would be about 3979 miles. 16. C  d  (27)  27  84.82 The bicycle will travel about 84.82 inches for one rotation of the tire. Multiply by 10. 10  84.82  848.2 The bicycle will travel about 848.2 inches in 10 rotations of the tire. 17. A  r2  (2) 2  4  13 The area of the region that will benefit from the system is about 13 square miles. 18. The radius is one-half times the diameter, or 125 feet. A  r2  (125) 2  15,625  49,087.4 The grass and sidewalk cover about 49,087.4 square feet of area.

8.

5 The volume is 5 cubic inches. V  /wh  12  3  2  72 The volume is 72 cubic centimeters. V  /wh  621  12 The volume is 12 cubic yards. V  /wh  100  1  10  1000 The volume is 1000 cubic meters. V  /wh  225  20 The volume is 20 cubic meters. V  /wh  12  6  2  144 The volume is 144 cubic inches. V  /wh  8  5  5.5  220 The volume of the tank is 220 cubic feet. V  /wh 1

 18  10  112  2070 The volume of the microwave oven is 2070 cubic inches. 9. V  /  w  h  222 8 The volume of the cube is 8 cubic meters. 10. 8  4  4  128 The volume of a full cord of firewood is 128 cubic feet. 1

11. 8  4  22  80 1

The volume of a short cord of 22-foot logs is 80 cubic feet. 12. 12  2  h  128 24  h  128 h h

128 24 16 or 3

1

53 feet or 5 ft 4 in.

The stack will be 5 feet 4 inches high.

Prerequisite Skills

640

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Page 819

The numbers 201 and 199 each occur twice, and the number 200 occurs only once. The numbers that appear most often in the set are 201 and 199. The modes are 201 and 199. 7. {4, 5, 6, 7, 8}

Mean, Median, and Mode

1. {1, 2, 3, 5, 5, 6, 13} mean  

1  2  3  5  5  6  13 7 35 or 5 7

mean 

The set is arranged in order. The middle number in the set is 5. The median is 5. The number that appears most often in the set is 5. The mode is 5. 2. {3, 5, 8, 1, 4, 11, 3} mean  



The set is arranged in order. The middle number in the set is 6. The median is 6. All numbers in the set occur exactly once. There is no mode. 8. {3, 7, 21, 23, 63, 27, 29, 95, 23}

3  5  8  1  4  11  3 7 35 or 5 7

mean 

Order the numbers from least to greatest. 1, 3, 3, 4, 5, 8, 11 The middle number in the set is 4. The median is 4. The number that appears most often in the set is 3. The mode is 3. 3. {52, 53, 53, 53, 55, 55, 57} mean  





52  53  53  53  55  55  57 7 378 or 54 7

mean  



8  7  5  19 4 39 or 9.75 4

15 2

2  2.5 2



or 7.5

mean  



or 2.25

5  7  5  4  3  3  8 7 35 or 5 7

Order the numbers from least to greatest. 3, 3, 4, 5, 5, 7, 8 The middle number in the set is 5. The median is 5. The numbers 3 and 5 each occur twice, and all other numbers occur only once. The numbers that appear most often in the set are 3 and 5. The modes are 3 and 5. 11. 17.6, 16.0, 14.1, 13.7, 13.5, 12.9, 12.3, 11.6, 11.4, 11.4 mean

3  11  26  4  1 5 45 or 9 5

Order the numbers from least to greatest. 1, 3, 4, 11, 26 The middle number in the set is 4. The median is 4. All numbers in the set occur exactly once. There is no mode. 6. {201, 201, 200, 199, 199} mean 

4.5 2



The median is $2.25. The amount that appears most often is $2.00. The mode is $2.00. 10. 5, 7, 5, 4, 3, 3, 8

The median is 7.5. All numbers in the set occur exactly once. There is no mode. 5. {3, 11, 26, 4, 1} mean 

0.5  2  2  1.25  5.25  3  2.5  3.5 8 20 or $2.50 8

Order the numbers from least to greatest. 0.5, 1.25, 2, 2, 2.5, 3, 3.5, 5.25 There is an even number of items. Find the mean of the middle two.

Order the numbers from least to greatest. 5, 7, 8, 19 There is an even number of items. Find the mean of the middle two. 7  8 2

3  7  21  23  63  27  29  95  23 9 291 1 or 32 9 3

Order the numbers from least to greatest. 3, 7, 21, 23, 23, 27, 29, 63, 95 The middle number in the set is 23. The median is 23. The number that appears most often in the set is 23. The mode is 23. 9. $0.50, $2.00, $2.00, $1.25, $5.25, $3.00, $2.50, $3.50

The set is arranged in order. The middle number in the set is 53. The median is 53. The number that appears most often in the set is 53. The mode is 53. 4. {8, 7, 5, 19} mean 

4  5  6  7  8 5 30 or 6 5

201  201  200  199  199 5 1000 or 200 5



17.6  16.0  14.1  13.7  13.5  12.9  12.3  11.6  11.4  11.4 10



134.5 10

or 13.45

Order the numbers from least to greatest. 11.4, 11.4, 11.6, 12.3, 12.9, 13.5, 13.7, 14.1, 16.0, 17.6 There is an even number of items. Find the mean of the middle two.

Order the numbers from least to greatest. 199, 199, 200, 201, 201 The middle number in the set is 200. The median is 200.

12.9  13.5 2



26.4 2

or 13.2

The median is 13.2. The number that appears most often is 11.4. The mode is 11.4.

641

Prerequisite Skills

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12. Let x represent Bill’s score on the fifth test. mean  88  88 

14. mean 

sum of the first four scores  fifth score 5 86  90  84  91  x 5 351  x 5

  

440  351  x 89  x Bill must earn 89 on the fifth test.



Olivia’s new average score will be about 92.7.

13. Let x represent Sue’s score for the tenth game. mean  110  110 

9(average score of first 9 games)  tenth score 10 9(108)  x 10 972  x 10

1100  972  x 128  x Sue needs to score 128 in the tenth game.

Prerequisite Skills

5(average score of first 5 tests)  sixth score 6 5(92)  96 6 460  96 6 556 6 2 923 or about 92.7

642

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Extra Practice Page 820

5. (8  1)  3  (7)  3  21

Lesson 1-1

1. The word sum implies add, so the expression can be written as b  21. 2. The word product implies multiply, so the expression can be written as 7x. 3. The words increased by imply add, so the expression can be written as t  6. 4. Sum implies add, and times implies multiply. So the expression can be written as 4  6z. 5. Increased by implies add, and times implies multiply. So the expression can be written as 10  4a. 6. Sum implies add, and times implies multiply. So the expression can be written as 8  (2n). 7. The word cube means to raise to the third power, 1 so the expression can be written as 2x3. 8. The word square means to raise to the second 4 power, so the expression can be written as 5m2.

6. 4(5  3) 2  4(2) 2  4(4)  16 7. 3(12  3)  5  9  3(15)  5  9  45  45 0 8. 53  63  52  125  216  25  341  25  316 9. 16  2  5  3  6  8  5  3  6  40  3  6  120  6  20 10. 7(53  32 )  7(125  9)  7(134)  938

9. 24  2  2  2  2  16 2 10. 10  10  10  100 3 11. 7  7  7  7  343 12.

203

11.

  

36  12 64 48 64 48 24

2 12. 25 

 20  20  20  8000

13. 36  3  3  3  3  3  3  729

13.

14. 45  4  4  4  4  4  1024 15. Sample answer: two times n 16. Sample answer: ten to the seventh power 17. Sample answer: m to the fifth power 18. Sample answer: the product of x and y 19. Sample answer: five times n squared minus 6 20. Sample answer: nine times a cubed plus 1 21. Sample answer: x cubed times y squared 22. Sample answer: c to the fourth power times d to the sixth power 23. Sample answer: three times e plus 2 times e squared

Page 820

94  26 64

14.

15.

16.

1 (18 3

1

 9)  25  3 (27)

 25  9  16 8a  b  8  2  5  16  5  21 48  ab  48  2  5  48  10  58 a(6  3n)  2(6  3  10)  2(6  30)  2(24)  48 bx  an  5  4  2  10  20  20  40

17. x2  4n  42  4  10  16  4  10  16  40  24 18. 3b  16a  9n  3  5  16  2  9  10  15  32  90  47  90  43

Lesson 1-2

1. 3  8  2  5  3  4  5 75 2 2. 4  7  2  8  4  14  8  18  8  26 3. 5(9  3)  3  4  5(12)  3  4  60  12  48

19. n2  3(a  4)  102  3(2  4)  102  3(6)  100  3(6)  100  18  118

4. 9  32  9  9 0

643

Extra Practice

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4. Replace y in 5y  4  11 with each value in the replacement set for y.

20. (2x) 2  an  5b  (2  4) 2  2  10  5  5  (8) 2  2  10  5  5  64  2  10  5  5  64  20  25  84  25  59

5y  4  11

y 1 3

21. [a  8(b  2) ] 2  4  [2  8(5  2) ] 2  4  [2  8(3) ] 2  4  [2  24] 2  4  [26] 2  4  676  4  169

5 7 9

5  1  4  11 S 1  11 5  3  4  11 S 11  11 5  5  4  11 S 21  11

0 2 4 6 8

?

0  4  4 S 4  4 ?

2  4  4 S 2  4 ?

444 S 04 ?

644 S 24 ?

844 S 44

5  7  4  11 S 31  11

25  y  18 ?

1

25  1  18 S 24  18

3

25  3  18 S 22  18

5 7 9

? ?

25  5  18 S 20  18 ?

25  7  18 S 18  18 ?

25  9  18 S 16  18

0

True or False? false false false false

5  9  4  11 S 41  11

3x  1  25

0 2

?

3  0  1  25 S 1  25 ?

3  2  1  25 S 7  25 ?

4

3  4  1  25 S 13  25

6

3  6  1  25 S 19  25

8

3  8  1  25 S 25  25

? ?

14 

true ✓

96 x

96 2

4

96 4

6

96 6

96 x

 2 with each value in the

2

True or False?

 2  14 S undefined

false

?

 2  14 S 50  14

false

?

 2  14 S 26  14

false

?

 2  14 S 18  14

false

?

96 8

 2  14 S 14  14

The solution of 14 

96 x

true ✓

 2 is 8.

y 3

6. Replace y in 0   3 with each value in the replacement set for y. y

033

y

True or False? false false false true ✓ false

?

True or False?

1

1 3

30 S

3

3 3

 3  0 S 2  0

5

5 3

7

7 3

9

9 3

2 23

0

false

? ?

false

1

false

0

false

 3  0 S 13  0 ?

30 S

2 3

?

true ✓

30 S 00 y

The solution of 0  3  3 is 9. 7. x 

True or False?

x

27  9 2

8.

36 2

x  18 The solution is 18.

false false

6(5)  3  3 30  3 8  3 33 11

false

9. n  2(4)

false

n n

true ✓

11.

72  9(2  1) 2(10)  1 49  9(3) 20  1 49  27 19 76 19

t

y

11 11

y

5(4)  6 22  3 20  6 4  3 14 7

z z z

2z The solution is 2. 33  52

12. a  2(3  1)

t

a

t

a

t 4t The solution is 4.

644

18  7 13  2

1y The solution is 1. 10.

n3 The solution is 3.

The solution of 3x  1  25 is 8.

Extra Practice

false

?

96 0

2

8

The solution of 25  y  18 is 7. 3. Replace x in 3x  1  25 with each value in the replacement set for x. x

false

?

x

The solution of x  4  4 is 8. 2. Replace y in 25  y  18 with each value in the replacement set for y. y

false

?

replacement set for x.

1. Replace x in x  4  4 with each value in the replacement set for x. x44

true ✓

?

The solution of 5y  4  11 is 3.

Lesson 1-3

x

false

?

5. Replace x in 14 

Page 820

True or False?

?

27  25 2(2) 52 4

a  13 The solution is 13.

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x

13. Replace x in x  2  7 with each value in the replacement set for x. x

x27

4

42 7 7 S 67

?

?

5

52 7 7 S 77

6

62 7 7 S 8 7 7

7

72 7 7 S 9 7 7

8

8  2 7 7 S 10 7 7

17. Replace x in 3 6 2 with each value in the replacement set for x.

True or False? false

6

6 2 S 2 2

true ✓

7

7 3

6 2 S 23 2

8

8 3

6 2 S 23 2

true ✓

?

true ✓

?

true ✓

?

true ✓

71 6 8 S 6 6 8

8

81 6 8 S 7 6 8

2x  15

True or False?

?

4

2  4 15 S 8 15

true ✓

?

5

true ✓

6 7 8

2  5 15 S 10 15 ?

true ✓

?

true ✓

2  6 15 S 12 15 2  7 15 S 14 15 ?

2  8 15 S 16  15

3y  36

10

3  10 36 S 30 36

12 14 16

?

true ✓

?

true ✓

3  14 36 S 42 36 3  16 36 S 48 36

2

false

5y 4

x 3

6 2 is {4, 5}.

 20

True or False?

10

5  10 ?

4

20 S 12.5 20

false

12

5  12 ?

4

20 S 15 20

false

14

5  14 ?

4

20 S 17.5 20

false

16

5  16 ?

4

20 S 20 20

true ✓

Page 821

5y 4

20 is {16}.

Lesson 1-4

1. Reflexive Property of Equality n  3, since 4  3  4  3. 2. Additive Identity Property 5

n  4, since

5 4

5

 4  0.

3. Multiplicative Identity Property n  1, since 15  15  1. 4. Multiplicative Inverse Property 3

2

3

n  2, since 3  2  1. 5. Commutative Property of Addition n  1.3, since 2.7  1.3  1.3  2.7. 6. Substitution Property of Equality, Multiplicative Inverse Property, and Multiplicative Identity Property

false true ✓

?

The solution set for

True or False?

?

3  12 36 S 36 36

false

y

false

?

1

5y

The solution set for 2x 15 is {4, 5, 6, 7}. 16. Replace y in 3y 36 with each value in the replacement set for y. y

false

?

The solution set for

The solution set for x  1 8 is {4, 5, 6, 7, 8}. 15. Replace x in 2x 15 with each value in the replacement set for x. x

?

18. Replace y in 4 20 with each value in the replacement set for y.

True or False? true ✓

7

true ✓

6 3

?

61 6 8 S 5 6 8

2

6

?

51 6 8 S 4 6 8

true ✓

?

true ✓

The solution set for x  2  7 is {6, 7, 8}. 14. Replace x in x  1 8 with each value in the replacement set for x.

5

1

6 2 S 13 6 2 6 2 S 13 6 2

?

41 6 8 S 3 6 8

True or False?

5 3

?

4

2

?

5

true ✓

x18

4 3

4

false

?

x

x 3

x

1

2

1 n  4, since 4 62  36  4.

The solution set for 3y 36 is {12, 14, 16}.

7. Multiplicative Property of Zero n  0, since 8  0  0. 8. Multiplicative Inverse Property 1 n  1, since 1  9  9. 9. Reflexive Property of Equality n  7, since 5  7  5  7. 10. Substitution Property of Equality n  2, since (13  4)(2)  9(2).

645

Extra Practice

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1 12

2



2 [15 3

34

 10] Substitution; 12  2  10

2 3 2

3

3

Substitution; 15  10  2

1

12.

7 4

3 4  118  824 

2

1 12

3

Multiplicative Inverse; 3  2  1

7

Multiplicative Inverse; 8  8  1

7 [4 ] 4

Multiplicative Identity; 4  1  4

 4 [4  1 ]

1

16.

Substitution; 4  4  7

13. [ (18  3)  0]  10  [6  0 ]  10 Substitution; 18  3  6 Multiplicative Property of  [0 ]  10 Zero; 6  0  0 Multiplicative Property of 0 Zero; 0  10  0

Page 821

17. 18. 19. 20.

1

2

8. 32 x  8  32(x)  32

 36(5)  36  180  9  189

1

1

2

114 2

 (4)18 

1181 218

 72  1  73 13a  5a  (13  5)a  18a 21x  10x  (21  10)x  11x 8(3x  7)  8(3x)  8(7)  24x  56 There are no like terms. 4m  4n is simplified. 3(5am  4)  3(5am)  3(4)  15am  12

22. 9y2  13y2  3  (9  13)y2  3  22y2  3 23. 11a2  11a2  12a2  (11  11  12)a2  12a2 24. 6a  7a  12b  8b  (6  7)a  (12  8)b  13a  20b

Page 821

Lesson 1-6

1. 23  8  37  12  23  37  8  12  (23  37)  (8  12)  60  20  80 2. 19  46  81  54  19  81  46  54  (19  81)  (46  54)  100  100  200 3. 10.25  2.5  3.75  10.25  3.75  2.5  (10.25  3.75)  2.5  14  2.5  16.5 4. 22.5  17.6  44.5  22.5  44.5  17.6  (22.5  44.5)  17.6  67  17.6  84.6

118 2

 32x  4 9. c(7  d)  c(7)  c(d)  7c  cd 10. 6  55  6(50  5)  6(50)  6(5)  300  30  330 11. 15(108)  15(100  8)  15(100)  15(8)  1500  120  1620 12. 1689  5  (1700  11)5  (1700)5  (11)5  8500  55  8445 13. 7  314  7(300  14)  7(300)  7(14)  2100  98  2198

Extra Practice

2

21. 15x2  7x2  (15  7)x2  22x2

Lesson 1-5

1. 5(2  9)  5(2)  5(9)  10  45  55 2. 8(10  20)  8(10)  8(20)  80  160  240 3. 20(8  3)  20(8)  20(3)  160  60  100 4. 3(5  w)  3(5)  3(w)  15  3w 5. (h  8)7  (h)7  (8)7  7h  56 6. 6(y  4)  6(y)  6(4)  6y  24 7. 9(3n  5)  9(3n)  9(5)  27n  45

1

1

15. 418  18  4  18 18

7

7

1

14. 36 54  36 5  4

11. 3 [ 15  (12  2) ]

1

2

1

2

5. 23  6  33  4  23  33  6  4

1

1

2

2

 23  33  (6  4)  6  10  16 6

1

1

6

1

2

6. 57  15  47  25  57  47  (15  25)  10  40  50 7. 6  8  5  3  6  5  8  3  (6  5)  (8  3)  30  24  720

646

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8. 18  5  2  5  (18  5)  (2  5)  90  10  900 9. 0.25  7  8  0.25  8  7  (0.25  8)  7  27  14 10. 90  12  0.5  90  (12  0.5)  90  6  540 1

21. 12b3  12  12b3  12b3  12b3  12  (12b3  12b3 )  12  (12  12)b3  12  24b3  12 22. 7  3(uv  6)  u  7  3(uv)  3(6)  u  7  3uv  18  u  7  18  3uv  u  (7  18)  3uv  u  11  3uv  u 23. 3(x  2y)  4(3x  y)  3(x)  3(2y)  4(3x)  4(y)  3x  6y  12x  4y  3x  12x  6y  4y  (3x  12x)  (6y  4y)  (3  12)x  (6  4)y  15x  10y 24. 6.2(a  b)  2.6(a  b)  3a  6.2(a)  6.2(b)  2.6(a)  2.6(b)  3a  6.2a  6.2b  2.6a  2.6b  3a  6.2a  2.6a  3a  6.2b  2.6b  (6.2a  2.6a  3a)  (6.2b  2.6b)  (6.2  2.6  3)a  (6.2  2.6)b  11.8a  8.8b 25. 3  8(st  3w)  3st  3  8(st)  8(3w)  3st  3  8st  24w  3st  3  (8st  3st)  24w  3  (8  3)st  24w  3  11st  24w 26. 5.4(s  3t)  3.6(s  4)  5.4(s)  5.4(3t)  3.6(s)  3.6(4)  5.4s  16.2t  3.6s  14.4  5.4s  3.6s  16.2t  14.4  (5.4s  3.6s)  16.2t  14.4  (5.4  3.6)s  16.2t  14.4  9s  16.2t  14.4 27. 3[4  5(2x  3y) ]  3[ 4  5(2x)  5(3y) ]  3[ 4  10x  15y ]  3[4]  3[ 10x]  3[ 15y]  12  30x  45y

1

11. 53  4  6  53  6  4

1

1

2

 53  6  4  32  4  128 5

5

12. 46  10  12  46  12  10

1

2

5

 46  12  10  58  10  580 13. 5a  6b  7a  5a  7a  6b  (5a  7a)  6b  (5  7)a  6b  12a  6b 14. 8x  4y  9x  8x  9x  4y  (8x  9x)  4y  (8  9)x  4y  17x  4y 15. 3a  5b  2c  8b  3a  5b  8b  2c  3a  (5b  8b)  2c  3a  (5  8)b  2c  3a  13b  2c 2

2

16. 3x2  5x  x2  3x2  x2  5x

123x2  x22  5x 2  1 3  1 2 x2  5x 

5

 3x2  5x

17. (4p  7q)  (5q  8p)  4p  7q  5q  8p  4p  8p  5q  7q  (4p  8p)  (5q  7q)  (4  8)p  (5  7)q  4p  (2q)  4p  2q 18. 8q  5r  7q  6r  8q  7q  5r  6r  (8q  7q)  (5r  6r)  (8  7)q  (5  6)r  1q  (1r) qr 19. 4(2x  y)  5x  4(2x)  4(y)  5x  8x  4y  5x  8x  5x  4y  (8x  5x)  4y  (8  5)x  4y  13x  4y

Page 822

Lesson 1-7

1. Hypothesis: an animal is a dog Conclusion: it barks 2. Hypothesis: a figure is a pentagon Conclusion: it has five sides 3. Hypothesis: 3x  1  8 Conclusion: x  3 4. Hypothesis: 0.5 is the reciprocal of 2 Conclusion: 0.5  2  1 5. Hypothesis: a figure is a square Conclusion: it has four congruent sides. If a figure is a square, then it has four congruent sides. 6. Hypothesis: a  4 Conclusion: 6a  10  34 If a  4, then 6a  10  34.

20. 9r5  2r2  r5  9r5  r5  2r2  (9r5  r5 )  2r2  (9  1)r5  2r2  10r5  2r2

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7. Hypothesis: it is night Conclusion: the video store is open If it is night, then the video store is open. 8. Hypothesis: it is Thursday Conclusion: the band does not have practice If it is Thursday, then the band does not have practice. 9. It can snow in May in some locations. 10. You may live in Portland, Maine. 11. y  3; 2(3)  4  6  4  10 but 3 3 If y  3, then 2y  4  10 is true, but y 3 is false.

Page 823 1. The The 2. The The 3. The The 4. The The 5. The The 6. The The 7.

12. Sample answer: a  1; (1) 2  1 7 0 but 1  0

Page 822

6

Lesson 1-8

3

2

1

0

1

6 5 4 3 2 1

0

2 1

4

1

2

3

0

1

2

3

10. 10 9 8 7 6 5 4 3 2 1

0

1

11. 22 is twenty-two units from zero in the positive direction. 022 0  22 12. 2.5 is two and five-tenths units from zero in the negative direction. 02.5 0  2.5 13.

2 3

is two-thirds unit from zero in the positive direction.

0 23 0  23 7

14. 8 is seven-eighths unit from zero in the negative direction.

0 78 0  78

Lesson 1-9

Page 823

Lesson 2-2

1. 3  16  19 2. 27  19  ( 027 0  019 0 )  (27  19)  8 3. 8  (13)  ( 013 0  08 0 )  (13  8)  5 4. 14  (9)  ( 014 0  09 0 )  (14  9)  23 5. 25  47  ( 047 0  025 0 )  (47  25)  22 6. 97  (79)  ( 097 0  079 0 )  (97  79)  18

 0.16

About 16% of U.S. Presidents have been born in Ohio.

Extra Practice

4

9.

1. The bar for Virginia shows 8 presidents and the bar for Texas shows 2 presidents. Since 8 equals 4 times 2, there were 4 times more presidents born in Virginia than in Texas. 2. The bar for Massachusetts shows 4 presidents and the bar for New York shows 4 presidents. So, Massachusetts and New York have the same number of presidents. 3. No; you need to have the birthplaces of all presidents to compare parts to the whole in a circle graph. 4. The bar for Ohio shows 7 presidents. What percent of 43 is 7? 7 43

5

8.

1. Sample answer: The temperatures increase from January through the summer and then begin to decrease again. 2. Sample answer: The roller coaster goes down a small hill, coasts at about the same speed, increases in speed on the way down the hill, decreases again on the way up the hill, increases down another hill, and then slows down for the end of the ride. 3. Sample answer: The jogger increases in speed, runs about the same speed, increases again, runs at a faster pace for a while, decreases, maintains a speed, and finally slows down at the finish of the run. 4. Sample answer: The hiker walks away from the camp, stops for a rest, hikes a little further, and then returns to camp.

Page 822

Lesson 2-1

dots indicate each point on the graph. coordinates are {3, 2, 1, 0, 1, 2, 3, 4}. dots indicate each point on the graph. coordinates are {2, 0, 2, 3, 6}. dots indicate each point on the graph. coordinates are {2, 3, 4}. dots indicate each point on the graph. coordinates are {7, 8, 9, 10, 11, 12, 13}. dots indicate each point on the graph. coordinates are {6, 4, 2, 0}. dots indicate each point on the graph. coordinates are {2, 1, 0, 1, 2, 3, 4}.

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7. 4.8  3.2  ( 04.8 0  03.2 0 )  (4.8  3.2)  1.6 8. 1.7  (3.4)  ( 01.7 0  03.4 0 )  (1.7  3.4)  5.1 9. 0.009  0.06  ( 00.06 0  00.009 0 )  (0.06  0.009)  0.051

1 72

11

22.

5 11

6

1

 11 23.

2 7

3

4

3

 14  14  14 1

 14

1 0 11 0 0 7 0 2 11 7  1 9  9 2

1

7

5

2

28

1 0 28 0 0 2560 0 2 28 25   1 60  60 2   60 

25

11. 5  6  30  30

1 0 2530 0  0 1830 0 2 25 18   1 30  30 2 

12.

3 8

1

7

7

2

1

9

14

 12  24  24

2

3

 60 1

 20

10 0 0 02 14 9   1 24  24 2 14 24



25

 60  60

 2

 30

1 25 2 28 25  60  1 60 2 28

24. 15  12  60  60

18

18

2

1 0 0  0 115 0 2 6 5   1 11  11 2

 9 5

6

6 11



10.  9  9    9  9

3

1

5

 11  11  11



Page 823

Lesson 2-3

1. 5(12)  60 2. (6)(11)  66 3. (7)(5)  35 4. (6)(4)(3)  (24)(3)  72

9 24

5

 24

13. 27  14  13 14. 8  17  8  (17)  ( 017 0  08 0 )  (17  8)  9 15. 12  (15)  12  (15)  12  15  27 16. 35  (12)  35  (12)  35  12  ( 035 0  012 0 )  (35  12)  23 17. 2  (1.3)  2  (1.3)  2  1.3  ( 02 0  01.3 0 )  (2  1.3)  0.7 18. 1.9  (7)  1.9  (7)  1.9  7  8.9 19. 4.5  8.6  4.5  (8.6)  ( 04.5 0  08.6 0 )  (4.5  8.6)  13.1 20. 89.3  (14.2)  89.3  (14.2)  89.3  14.2  103.5 21. 18  (1.3)  18  (1.3)  18  1.3  ( 018 0  01.3 0 )  (18  1.3)  16.7

5. 7.

1

178 2113 2  247 1 42

21 2  1 21 2 1 23

  

9 7 2 3 63 6 21 2 1 102

1 22

10

6. (5) 5   5 8.

1

2 17

 2

21359 2  197 21329 2   

288 63 32 7 4 47

9. (5.34)(3.2)  17.088 10. (6.8)(5.415)  36.822 11. (4.2)(5.1)(3.6)  (21.42)(3.6)  77.112 12. (3.9)(1.6)(8.4)  (6.24)(8.4)  52.416 13. 5(3a)  6a  5(3)a  6a  15a  6a  (15  6)a  21a 14. 8(x)  3x  8(1)x  3x  8x  3x  (8  3)x  5x 15. 2(6y  2y)  2(6y)  2(2y)  2(6)y  2(2)y  12y  4y  (12  4)y  8y 16. (c  7c)(3)  c(3)  7c(3)  1(3)c  7(3)c  3c  (21c)  [3  (21)]c  24c

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17. 3n(4b)  2a(3b)  3(4)nb  2(3)ab  12nb  6ab  12bn  6ab 18. 7(2m  3n)  7(2m)  (7)(3n)  7(2)m  (7)(3)n  14m  (21n)  14m  (21n)  14m  21n

18.

2a  10b 2

 (2a  10b)  (2)

1 12 1 1  2a 1 2 2  10b 1 2 2

 (2a  10b) 2

 1a  (5b)  a  5b 19.

27c  (99b) 9

 [ 27c  (99b) ]  9

119 2 1 1  27c 1 9 2  (99b) 1 9 2

 [27c  (99b) ]

Page 824 1. 3. 5. 6.

Lesson 2-4

49  (7)  7 2. 52  (4)  13 66  (0.5)  132 4. 25.8  (2)  12.9 55.25  (0.25)  221 82.1  (16.42)  5 2

2

1

7. 5  5  5  5 2

 25

1

7

2

8.

1

10

9. 4  10  4   7

10.

3 2

 

1 12

3

40 7 5 57

1 22

 2  2  1

7 8

12.



   13.

32a 4

7

2

1 52

8

 32a   8a 15.

5n  15 5

8

64

114 2

1 82

Page 824

14

 225

 

14.

12x 2

   

 12x  (2)

1 12

 12x  2  6x

 4

1 12 1 1  5n 1 5 2  15 1 5 2

1 12 1 1  2b 1 2 2  10 1 2 2

8

60

148



152

156

12



 16

20

 

64

68

  

   

76

80

72

4. The lowest value is 111, and the highest value is 175, so use a scale that includes those values. Place an  above each value for each occurrence.

 1b  (5) b5

x

 (65x  15y)  (5)

12 1 1  65x 1 5 2  15y 1 5 2

 (65x  15y)

   

  

 (2b  10)  (2)

x

x

x x

x

x x xx

xx

x x x

110 115 120 125 130 135 140 145 150 155 160 165 170 175

1 5

5. The greatest common place value is tens, so the digits in the tens place are the stems. Stem Leaf 1 0 7 7 8 8 9 9 2 2 2 3 3 3 5 5 5 9 4 0 4 5 7 1|0  10

 13x  3y

Extra Practice

144



3. The lowest value is 62, and the highest value is 80, so use a scale that includes those values. Place an  above each value for each occurrence.

 (2b  10) 2

65x  15y 5

140



2. The lowest value is 4, and the highest value is 19, so use a scale that includes those values. Place an  above each value for each occurrence.

 1n  (3)  n  3

17.

 

 

136

 (5n  15)  (5)

2b  10 2

Lesson 2-5

1. The lowest value is 134, and the highest value is 156, so use a scale that includes those values. Place an  above each value for each occurrence.

 (5n  15) 5

16.

1 12 1 1  3n 1 3 2  (3m) 1 3 2

 1n  1m nm

11. 5  8  5  5  25

 [ 3n  (3m) ]  (3)  [3n  (3m) ] 3

13 25 15  3 325  45 65 9 2 79

 32a  4

3n  (3m) 3

 32

6



1 12

7

 2

3 25

20.

 (4)  8  4

 3 13 15

 3c  (11b)  3c  11b

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9. There are 3 letters that are s and 7 total letters.

6. The greatest common place value is ones, so the digits in the ones place are the stems. Stem Leaf 1 0 1 2 2 3 8 9 2 0 1 2 5 3 0 2 4 1 1 5 5 6 1|0  1.0 7. The greatest common place value is hundreds, but the hundreds digit in every number is a one. Thus, we use the digits in the tens place as the stems. Stem Leaf 10 1 1 4 5 8 9 11 1 12 1 3 9 13 0 3 4 5 7 9 10|1  101

3

P(the letter s)  7  0.43 3

The probability of selecting the letter s is 7 or about 43%. 10. There is no letter b in the word success and 7 total letters. 0

P(the letter b)  7  0 The probability of selecting the letter b is 0 or 0%. 11. There are 2 letters that are vowels and 7 total letters. 2

P(vowel)  7  0.29 2

The probability of selecting a vowel is 7 or about 29%. 12. There are 3 letters that are u or c and 7 total letters. 3

P(the letters u or c)  7  0.43

Page 824

Lesson 2-6

The probability of selecting the letters u or c is or about 43%. 13. There is 1 outcome that is a 4 and 6  1 or 5 outcomes that are not a 4.

1. There is one way for the coin to land tails up and two ways for the coin to land. 1

P(tails up)  2 or 0.50

1

1

odds of a 4  5

The probability a coin lands tails up is 2 or 50%. 2. At some point during this month you will eat something. Therefore, the probability you eat this month is 1 or 100%. 3. There is one outcome that is a girl and two possible outcomes. P(a girl) 

1 2

The odds of rolling a 4 are 1:5. 14. There are 3 outcomes that are numbers greater than 3 and 6  3 or 3 outcomes that are numbers not greater than 3. 3

odds of a number greater than 3  3

or 0.50

1

1

1 2

The probability a baby will be a girl is or 50%. 4. There are no blue elephants. Therefore, the probability you will see a blue elephant is 0 or 0%. 5. You are studying algebra from an algebra book, and the statement is written in that book. Therefore, the probability this is an algebra book is 1 or 100%. 6. There is 1 day in the week that is Wednesday and 7 days in a week.

The odds of rolling a number greater than 3 are 1:1. 15. There are 2 outcomes that are numbers that are a multiple of 3 and 6  2 or 4 outcomes that are numbers that are not a multiple of 3. 2

odds of a multiple of 3  4 1

2 The odds of rolling a number that is a multiple of 3 are 1:2. 16. There are 4 outcomes that are numbers less than 5 and 6  4 or 2 outcomes that are numbers that are not less than 5.

1

P(a day is Wednesday)  7  0.14 The probability a day is Wednesday is 14%. 7. There is 1 e and 7 total letters.

1 7

or about

4

odds of a number less than 5  2 2

1

1

P(the letter e)  7  0.14

The odds of rolling a number less than 5 are 2:1. 17. There are 3 outcomes that are odd numbers and 6  3 or 3 outcomes that are not odd numbers.

1

The probability of selecting the letter e is 7 or about 14%. 8. There are 5 letters that are not c and 7 total letters. P(not c) 

5 7

3 7

3

odds of an odd number  3

 0.71

1

1

The probability of selecting a letter that is not c is 5 or 71%. 7

The odds of rolling an odd number are 1:1. 18. There is 1 outcome that is a 6 and 6  1 or 5 outcomes that are not a 6. 5

odds against a 6  1 The odds against rolling a 6 are 5:1.

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Page 825

18. You can use a calculator to find an approximation for 115. 3.88  3.88 115  3.87298334… Therefore, 3.88 7 115. 19. You can use a calculator to find the value of 1529. 1529  23.0 20  20.0 Therefore, 1529 6 20. 20. You can use a calculator to find the value of 10.25. 10.25  0.5 0.5  0.55555555… Therefore, 10.25 7 0.5.

Lesson 2-7

1. 1121 represents the positive square root of 121. 121  112 S 1121  11 2. 136 represents the negative square root of 36. 36  62 S 136  6 3. 12.89 represents the positive square root of 2.89. 2.89  1.72 S 12.89  1.7 4. 1125 represents the negative square root of 125. 125  11.182 S 1125  11.18 5.

81

81

3 100 represents the positive square root of 100. 81 9 2 81 9  1 10 2 S 3 100  10 100 36

6.  3 196 represents the negative square root of 36 196



36 6  14 1146 22 S 3 196

36 . 196

21. You can use a calculator to find an approximation 13 for 3 .

3

 7

1 3 13 3

7. 19.61 represents the positive and negative square roots of 9.61. 9.61  3.12 and 9.61  (3.1)2 19.61  3.1 7

roots of 8.  0.94 and

7 8

 (0.94)

1 13 13 3

2

7 8

 3  0.94

13 . 3

 0.577350269… 1 13



23 . 3 1

23. You can use a calculator to find the value of  3 4. 1

 3 4  0.5 1

4  0.25 1

1

Therefore,  3 4 6 4.

8

3 2  2, this number is a natural

24. You can use a calculator to find an approximation 1 for  16.

number, a whole number, an integer, and a rational number. 12. Because 66 and 55 are integers and (66  55)  1.2 is a terminating decimal, this number is a rational number. 13. Because 1225  15, this number is a natural number, a whole number, an integer, and a rational number.

1

6  0.16666666… 1

 16  0.408248290… 1

1

Therefore, 6 7  16.

Page 825

Lesson 3-1

is the divided 1. A number z times 2 minus 6 same as m by 3. 1442443 123 { 1 424 3 { 14243 { 14243 { z  2  6  m  3 The equation is 2z  6  m  3. 2. The cube decreased the square c. by of a of b 3 is equal to 14243 1442443 1 44244 1 44244 3{

3

14. Because  3 4  0.866025403…, which is not a repeating or terminating decimal, this number is irrational. 15. Because 1 and 7 are integers and 1  7  0.142857142… is a repeating decimal, this number is a rational number.

a3  b2  c The equation is a3  b2  c. 3. the product is the Twenty-nine decreased by of x and y same as z . 144424443 144424443 1442443 14243 { 29  xy  z The equation is 29  xy  z.

16. Because 10.0016  0.04 is a terminating decimal, this number is a rational number. 17. Write the numbers as decimals. 6.16  6.16161616… 6  6.0 Therefore, 6.16 7 6.0.

Extra Practice

6

 0.577350269…

Therefore,

9. Because 1149  12.2065556…, which is not a repeating or terminating decimal, this number is irrational. 10. Because 5 and 6 are integers and 5  6  0.8333333… is a repeating decimal, this number is a rational number. 11. Because

1 3

22. You can use a calculator to find approximations 1 13 for 13 and 3 .

7

2

 0.577350269…

Therefore,

8.  3 8 represents the positive and negative square

7 8

 0.33333333…

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4. The perimeter P P

144424443

3.

the sum of twice the length of the leg a and the length of the base b. is { 14 4444444424444444 443 2a  b 

The equation is P  2a  b. the quotient Thirty increased by of s and t is equal to v . 1 424 3 1442443 1442443 1442443 { v 30  (s  t)  The equation is 30  (s  t)  v. 6. The area half the product of lengths is of the diagonals a and b. A 14243 { 144444424444443 1 A  ab

?

5.

4  (5)  9 9  9 ✓ The solution is 5. 4. m62 m6626 m  4 m62 Check: ? 4  6  2 22✓ The solution is 4. t  (4)  10 5. t  (4)  4  10  4 t  14 t  (4)  10 Check:

2

1

The equation is A  2ab or 0.5ab. 7. Sample answer: 0.5x  3



14243

10

123

{

The sum of five-tenths times x and three 8. Sample answer: n

is equal to

negative ten.



6 123

?

14  (4)  10 10  10 ✓ The solution is 14. 6. v  7  4 v  7  7  4  7 v3 v  7  4 Check: ? 3  7  4 4  4 ✓ The solution is 3. 7. a  (6)  5 a  (6)  (6)  5  (6) a  11 a  (6)  5 Check:

2n  1

14243

{

The quotient of n is the same as the sum of two and negative six times n and one. 9. Sample answer: 18  123



5h

123

{

13h

123

{

Eighteen decreased five times h is the thirteen by same as times h. 10. Sample answer: n2  16 { 123

123

The square of n is equal to sixteen. 11. Sample answer: 2x2  3  21 { 123 14243

The sum of three is equal to twenty-one. and twice x squared 12. Sample answer: m

4



n 14243

{

The sum of four and the quotient of m and n

Page 825 1.

4  y  9 4  y  4  9  4 y  5 Check: 4  y  9

?

11  (6)  5 5  5 ✓ The solution is 11. 8. 2  x  8 2  x  x  8  x 2  8  x 2  8  8  x  8 6x Check: 2  x  8 ? 2  6  8 8  8 ✓ The solution is 6. 9. d  (44)  61 d  (44)  44  61  44 d  17 Check: d  (44)  61 ? 17  (44)  61 61  61 ✓ The solution is 17. 10. e  (26)  41 e  (26)  (26)  41  (26) e  15 e  (26)  41 Check: ? 15  (26)  41 41  41 ✓ The solution is 15.

12

123

is equal to twelve.

Lesson 3-2

2  g  7 2  g  2  7  2 g9 2  g  7 Check: ?

2  9  7 77✓ The solution is 9. 2. 9  s  5 9  s  9  5  9 s  14 9  s  5 Check: ? 9  (14)  5 5  5 ✓ The solution is 14.

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Extra Practice

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11.

p  47  22 p  47  47  22  47 p  69 Check: p  47  22

18.

?

69  47  22 22  22 ✓ The solution is 69. 12. 63  f  82 63  f  f  82  f 63  82  f 63  82  f 19  f Check: 63  f  82

t  (46.1)  3.673 t  (46.1)  (46.1)  3.673  (46.1) t  49.773 t  (46.1)  3.673 Check: ? 49.773  (46.1)  3.673 3.673  3.673 ✓ The solution is 49.773. 7 10

19. 7 10

1

aa2a 7 10 7 1 2 10 2 10 1 5

?

63  19  82 82  82 ✓ The solution is 19. 13. c  5.4  11.33 c  5.4  5.4  11.33  5.4 c  16.73 c  5.4  11.33 Check: ? 16.73  5.4  11.33 11.33  11.33 ✓ The solution is 16.73. 14. 6.11  b  14.321 6.11  b  6.11  14.321  6.11 b  20.431 6.11  b  14.321 Check:

Check:

1

2a 1

1

2a2 a a 7 10

a2

1

7 10

5

1 ? 1 2 5 ? 1

20.



10 1 2 1 . 5

The solution is



2 1 2



1 12 3 1 1 3 1 f  1 8 2  1 8 2  10  1 8 2 f  8  10 7

f  40

?

6.11  20.431  14.321 14.321  14.321 ✓ The solution is 20.431. 5  y  22.7 15. 5  22.7  y  22.7  22.7 17.7  y 5  y  22.7 Check: 5  17.7  22.7 5  5 The solution is 17.7. 16. 5  q  1.19 5  q  q  1.19  q 5  1.19  q 5  1.19  1.19  q  1.19 6.19  q 5  q  1.19 Check: ? 5  (6.19)  1.19 1.19  1.19 ✓ The solution is 6.19. 17. n  (4.361)  59.78 n  (4.361)  4.361  59.78  4.361 n  64.141 Check: n  (4.361)  59.78

Check:

1 12 3 7 1 ? 3  1 8 2  10 40 f  8  10

12 ? 3  10 40 3 3  10 10

21.



7 . 40

The solution is

1 12 5 1 1 1 412  1 1036 2  t  1 1036 2  1 1036 2 5

412  t  1036 16

1436  t 4

149  t Check:

412  5

412  The solution is

?

654

1

5

1

412  t  10 36 5 ? 412  5 ?

64.141  (4.361)  59.78 59.78  59.78 ✓ The solution is 64.141.

Extra Practice

1

a2

4 149 15 436 5 412

4 149.

1

2

1

 10 36



2

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3

1

3

1

3. 2y  3

x84

22.

3

2y 2

3

x8848

y

1

x  8

y  12 or 1.5 2y  3 Check:

1

3 ? 1 4 2 ? 1

1

?

8  8  8 1 4

23.



4 1 4



2(1.5)  3 3  3 ✓ The solution is 1.5. 4. 62y  2356



62y 62

1 The solution is 8. 7 9 116  s  8 7 7 9 7 116  s  116  8  116 5 s  16

1

2

2

? 9 8 ? 9

9 8

8 9 8

5.

116  16 

The solution is 8

116  

6

8 17 9



5 26

13 20 18

d

8

1

8 ? 179  8 ?

13 2018

8 179

8 179



16

179  1718

5



1

6.

2

5 26

59

2

 7

1 2  59(7) c  413

2

c 59

 7

413 ?  59

7

7  7 ✓ The solution is 413.

✓ 7.

Lesson 3-3

f 14 f 14

 63

1 2  14(63) f  882

1. 7p  35

f 14

Check:

35 7

 63

882 ?  14

p5 Check:

63

63  63 ✓ The solution is 882.

7p  35 ? 7(5)  35 35  35 ✓ The solution is 5. 2. 3x  24 

c 59 c 59

Check:

14

3x 3

 2

2  2 ✓ The solution is 12.

13



a 6

12 ?  6

The solution is 2018.

7p 7

1 2  6(2)

Check:

1 2 5 5  d  1 26 2  26 179  d  26

Page 826

 2

5 16.

5

Check:

a 6 a 6

a  12



17 9  d  26

24.

62y  2356

62(38)  2356 2356  2356 ✓ The solution is 38.

9

5

2356 62

?

7

7



y  38 Check:

116  s  8

Check:

3 2 3 2 1

3

x84

Check:



8.

x

84  97 97(84)  97 8148  x

24 3

Check:

x8 Check:

197x 2

84 

197x 2

? 8148 97

84 

3x  24 ? 3(8)  24 24  24 ✓ The solution is 8.

84  84 ✓ The solution is 8148.

655

Extra Practice

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9. 5

w 5 w 5

3

1

1 2  5(3)

1

26 3

9

12

1

26 26 5 26 5 12 5

j j

3

33✓ The solution is 15. q 9 q 9

55



1

w 5

15 ?  5

10.

1

26 j

w  15

Check:

1

14. 26 j  55

j

13 6

 6

 13 2

or 25 1

1 22 1 13 12 ? 1  55 6152

q  27 q  9 27 ?  9

26 ?  5 1 55 

3 3

3  3 ✓ The solution is 27. 11.

2 x 5 5 2 x 2 5

7

3 7

11

3  18  q 11 6 5 16

2

33✓ 5

3

5

16. 14 p  8 5

3

14p 3

5

14



8 3

14

1 72 5 4 p  8  1 7 2 5

p  8  4

1

1 12 9 15 r  5 1 2 2

5

9

p  14

5 9r  5 72

3

27

1

r   2 or 132

1

5

5

1

1

2

?

135 ?  18 1 72  1 132.

1

The solution is

5

5 . 14

17. 57k  0.1824

1

72 1 72

?

8  8 ✓

9r  72

9 132  72

The solution is

1 2  58 ? 7 5 5 4 1 14 2  8 3 5 14

14

135

5

5

14p  8

Check:

r   10

Check:

7

The solution is 16.

9r  72 5

1 52 ? 18 11 3  11 1 6 2 ?

1

1

7

3  111q

3  111 16

The solution is 22. 9

q

2

5

5

q

Check:

12  12 ✓

13.

7

111

18

1 2

1

111q

3  11  q

3

or 17

2 4 x7 5 ? 2 3 4 1 7 5 7 4 4 7✓ 7 3 The solution is 17. z 5  12 6 z 5 6 6  6 12 5 1 z  2 or 22 z 5 Check:  12 6 1 22 ? 5  12 6

12



111

20

x  14

12.

7

3  111q

15.

1 2  52 147 2

Check:

1

55 ✓

2

4

10 7

1

55

The solution is 25.

7

x

?

1

26 25  55

 9(3)

Check:

1

26 j  55

Check:

 3

57k 57





0.1824 57

k  0.0032 Check:

57k  0.1824 ?

57(0.0032)  0.1824 0.1824  0.1824 ✓ The solution is 0.0032. Extra Practice

656

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18. 0.0022b  0.1958 0.0022b 0.0022



1

24. 184  2.5x

0.1958 0.0022

1

184

b  89 0.0022b  0.1958 Check: ? 0.0022(89)  0.1958 0.1958  0.1958 ✓ The solution is 89. 19. 5j  32.15 5j 5



2.5 18.25 2.5

7.3  x 1 ? 1 ?

184  18.25 1

w 2 w 2

The solution is 7.3.

Page 826

 2.48

1 2  2(2.48) w  2 4.96 ?  2

Check:

2.48

2x 2

2.48

1 2  2.8(6.2) z  17.36

z  2.8 17.36 ?  2.8

4.

6.2 6.2

0.063

1

x 0.063 x 0.063

Check:

2  0.063(0.015)

23.

32 4

t8

8g 8

3c 3

0.015 6.

0.015

 5p



15 3

c5 5k  7  52 5k  7  7  52  7 5k  45 5k 5



45 5

k  9 7. 5s  4s  72 9s  72

5p  5 5 123 1  5  p 8

1 2

9s 9

3

340  p 3

158  5p

1

3 ?

3

158  5 340 3 ?

15

3

3

158  1540



72 9

s  8 8. 3x  7  2 3x  7  7  2  7 3x  9

123

 40  p

Check:



x  0.000945

0.015  0.015 ✓ The solution is 0.000945. 3 158 3 158



5  g 5. 3c  9  24 3c  9  9  24  9 3c  15

 0.015

x  0.063 0.000945 ?  0.063

4t 4

a  6 47  8g  7 47  7  8g  7  7 40  8g 40 8

6.2  6.2 ✓ The solution is 17.36. 22.

8

2

x4 3. 7a  6  36 7a  6  6  36  6 7a  42 7a 42  7 7

 6.2

Check:

Lesson 3-4

Exercises 1–21 For checks, see students’ work. 1. 2. 2x  5  3 4t  5  37 2x  5  5  3  5 4t  5  5  37  5 2x  8 4t  32

2.48  2.48 ✓ The solution is 4.96. z 2.8 z 2.8

1

184  184 ✓

5j  32.15

w  4.96

2.8

1

184  2.5x 184  2.5(7.3)

?

21.

x

Check:

5(6.43)  32.15 32.15  32.15 ✓ The solution is 6.43. 2

2.5x 2.5

32.15 5

j  6.43 Check:

20.



2

3x 3

9

3

x3

9.

8  3x  5 8  3x  8  5  8 3x  3 3x 3



3 3

x  1

158  158 ✓ 3

The solution is 340.

657

Extra Practice

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10.

3y  7.569  24.069 3y  7.569  7.569  24.069  7.569 3y  16.5 3y 3



21. 4.8a  3  1.2a  9 6a  3  9 6a  3  3  9  3 6a  12

16.5 3

6a 6

y  5.5 11. 7  9.1f  137.585 7  9.1f  7  137.585  7 9.1f  130.585 9.1f 130.585  9.1 9.1 12.

Page 826

Lesson 3-5

Exercises 1–21 For checks, see students’ work. 1. 5x  1  3x  3 5x  1  3x  3x  3  3x 2x  1  3 2x  1  1  3  1 2x  4

2.4m 2.4



4.75  m e 5

13. e 5

 6  2

 6  6  2  6

5

e 5 e 5

d 4

14. d 4

 8  5

 8

4

e  40

15.

2x 2

d 4 d 4

3

1 2  4(3) d  12

13n 13

76

4

13y  7  7  6  7 13

1

4

3.

13y  13 4

2

13

 4 13y   4 (13)

16. 10

1

169 4

y

p  3 10 p  3 10

1

or 424

4

2  10(4)

17. 6

p  3  40 p  3  3  40  3 p  37 18. 8

1

5f  1 8 5f  1 8

 3

2  8(3)

19. 2



25 5

2

1

3t  4 2 3t  4 2

2 h 3

1 a 2 1 a 2

1

3 a 4

1

 4  3  4a 1

1

4434 7

1 2  43 (7) a

6.

20

 3 2

t   3 or 63

28 3

1

or 93

6(y  5)  18  2y 6y  30  18  2y 6y  30  2y  18  2y  2y 8y  30  18 8y  30  30  18  30 8y  48 8y 8



48 8

y6

658

1

43

3 a 4 4 3 a 3 4

8

3t  4  16 3t  4  4  16  4 3t  20

Extra Practice

1

 4  4a  3  4a  4a 3 a 4

16 4

n  4

20

1

 5  3h  4  3h  3h

5.

2  2(8)

3t 3

1

 5  4  3h

h  5  4 h  5  5  4  5 h  9

 12



2 h 3

4.

2  2(12)

4n 4

0

 5

z0

4n  8  24 4n  8  8  24  8 4n  16

f  5 20.

1

13

 13

5z 5

2  6(1)

4n  8 2 4n  8 2

4 2

n  1 3z  5  2z  5 3z  5  2z  2z  5  2z 5z  5  5 5z  5  5  5  5 5z  0

1

h76 h7767 h  13

5f  1  24 5f  1  1  24  1 5f  25 5f 5

1

h  7 6 h  7 6



x  2 2. 6  8n  5n  19 6  8n  5n  5n  19  5n 6  13n  19 6  13n  6  19  6 13n  13

 8  8  5  8

1 2  5(8)

4 13y

12 6

a2

f  14.35 6.5  2.4m  4.9 6.5  4.9  2.4m  4.9  4.9 11.4  2.4m 11.4 2.4



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7.

14. 3(x  1)  5  3x  2 3x  3  5  3x  2 3x  2  3x  2 Since the expressions on each side of the equation are the same, this equation is an identity. The statement 3(x  1)  5  3x  2 is true for all values of x.

28  p  7( p  10) 28  p  7p  70 28  p  7p  7p  70  7p 28  6p  70 28  6p  28  70  28 6p  42 6p 6



42 6

15.

p7 1 (b  9) 3 1 b3 3

8.

1 b 3

x 2

b9



b9 x 6

3bb9b



x 2 1 3 x 6 1 3

2

6

2

16.

3

2 3b  2 (12)

3

b  18 9. 4x  6  0.5(x  30) 4x  6  0.5x  15 4x  6  0.5x  0.5x  15  0.5x 4.5x  6  15 4.5x  6  6  15  6 4.5x  9 4.5x 4.5





1

x

1

1

1

1

x 6 x 6

 6

1

1

1 2  6116 2

1

6v  9 3 6v  9 3

2  3(v)

9 3

17. 4

1

3t  1 4 3t  1 4

3

 4t  5

3t  1  3t  20 3t  1  3t  3t  20  3t 1  20 Since 1  20 is a false statement, this equation has no solution. 18. 0.4(x  12)  1.2(x  4) 0.4x  4.8  1.2x  4.8 0.4x  4.8  1.2x  1.2x  4.8  1.2x 0.8x  4.8  4.8 0.8x  4.8  4.8  4.8  4.8 0.8x  0 0.8x 0.8

0

 0.8

x0

s  1 12. 2.85y  7  12.85y  2 2.85y  7  12.85y  12.85y  2  12.85y 10y  7  2 10y  7  7  2  7 10y  5

3y 

19. 3y 

4 5

4 5



 3y  4

5 

3

1 42

8 5 

5

3 10

 10

y  0.5 13. 2.9m  1.7  3.5  2.3m 2.9m  1.7  2.3m  3.5  2.3m  2.3m 0.6m  1.7  3.5 0.6m  1.7  1.7  3.5  1.7 0.6m  1.8 

3v 3

2  4134t  52

2.9 2.9

0.6m 0.6



3v

9 4.5

10y 10

v

6v  9  3v 6v  9  6v  3v  6v 9  3v

x  2 10. 4(2y  1)  8(0.5  y) 8y  4  4  8y 8y  4  8y  4 Since the expressions on each side of the equation are the same, this equation is an identity. The statement 4(2y  1)  8(0.5  y) is true for all values of y. 11. 1.9s  6  3.1  s 1.9s  6  s  3.1  s  s 2.9s  6  3.1 2.9s  6  6  3.1  6 2.9s  2.9 2.9s 2.9

x

x  1

3b  12 2

x

 3  2  3

3b  3  3  9  3

1

1

 3  2

2

3

x

3323

3b  3  9 2

1

332

3 x 4

20. 3 x 4

y

1

2

1

 4  7  2x 1

1

1

 4  2x  7  2x  2x 1 x 4

1 x 4

47

4474

1.8 0.6

m3

1 y 3 1 y  3y 3 8 3y 3 8 8 3y

4

1 x 4 1 x 4

 11

1 2  4(11) x  44

659

Extra Practice

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21.

0.2(1  x)  2(4  0.1x) 0.2  0.2x  8  0.2x 0.2  0.2x  0.2x  8  0.2x  0.2x 0.2  8 Since 0.2  8 is a false statement, this equation has no solution.

Page 827 4 5

1.

2y 2

x

 20

2.



5x 5

7b 7

y 5

3

4



y y 5.

10.

4.

7 4

7a 7



a

12 7 12 7



x 9





5

or 17

4a 4

12.

4n 4



22 2

13p 13

n  4 7

12 4

14.

q

210 168 5 4 or

65

 13



k  3 2



9k 9

k  k or 2.5  k 5m  3 4



5m  3 6

(5m  3)6  4(5m  3) 30m  18  20m  12 30m  18  20m  20m  12  20m 10m  18  12 10m  18  18  12  18 10m  30

30

 14



2 9

23 9 23 9 5 29

12q(14)  7(30) 168q  210 168q 168

5p  7 8

2(2)  9(k  3) 4  9k  27 4  27  9k  27  27 23  9k

n3 12q 7



p5 13.

 3(n  4)  3n  12  3n  12  3n  12 

36 4

(6p  2)8  7(5p  7) 48p  16  35p  49 48p  16  35p  35p  49  35p 13p  16  49 13p  16  16  49  16 13p  65

2.16 3

n(7) 7n 7n  3n 4n

8.

6p  2 7

0.24 3

n 3



a9

x  0.72 7.

3

4

(a  3)4  8(3) 4a  12  24 4a  12  12  24  12 4a  36

x(3)  9(0.24) 3x  2.16 3x 3

34.84

 17.42 a  3 8

11.

t  11 6.

4

 17.42

x2

3

a

(t  5)2  4(3) 2t  10  12 2t  10  10  12  10 2t  22 2t 2

x 8.71

17.42x 17.42

7(a)  4(3) 7a  12

15 4 15 4 3 34 or 3.75 t  5 3 2 4

4

 2

x(17.42)  8.71(4) 17.42x  34.84

189 7

b  27

y(4)  5(3) 4y  15 4y 4



3  5

y2

3

7

b(7)  63(3) 7b  189

16  x 3.

b 63

y

1(y  5)  (y  3)3 y  5  3y  9 y  5  3y  3y  9  3y 2y  5  9 2y  5  5  9  5 2y  4

Lesson 3-6

4(20)  5(x) 80  5x 80 5

1 y  3

9.

1

14

10m 10

30

 10

m3

Extra Practice

660

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w  5 4

15.



w  3 3

2. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 98  62  36 Find the percent using the original number, 62, as the base.

(w  5)3  4(w  3) 3w  15  4w  12 3w  15  4w  4w  12  4w w  15  12 w  15  15  12  15 w  27 w 1

36 62

27

 1

36(100)  62(r) 3600  62r

w  27 16.

96.8 t

3600 62

12.1 7



12.1t 12.1



56  t r  1 r  1

17.

3

5

(r  1)5  (r  1)3 5r  5  3r  3 5r  5  3r  3r  3  3r 2r  5  3 2r  5  5  3  5 2r  8 2r 2

33 322

3300 322

8



2n  7 7

(4n  5)7  5(2n  7) 28n  35  10n  35 28n  35  10n  10n  35  10n 18n  35  35 18n  35  35  35  35 18n  0 18n 18

42 78

0

 18

4200 78

Lesson 3-7

r

 100

33(100)  100(r) 3300  100r 3300 100



322r 322

r

 100



78r 78

54  r The percent of decrease is about 54%. 5. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 230  212  18 Find the percent using the original number, 212, as the base.

1. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 100  67  33 Find the percent using the original number, 100, as the base. 33 100



42(100)  78(r) 4200  78r

n0

Page 827

r

 100

10  r The percent of decrease is about 10%. 4. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 78  36  42 Find the percent using the original number, 78, as the base.

r4 18.

62r 62

33(100)  322(r) 3300  322r

2

4n  5 5



58  r The percent of increase is about 58%. 3. The percent of change is a percent of decrease because the new amount is less than the original. Find the change. 322  289  33 Find the percent using the original number, 322, as the base.

96.8(7)  t(12.1) 677.6  12.1t 677.6 12.1

r

 100

18 212

r

 100

18(100)  212(r) 1800  212r

100r 100

33  r The percent of decrease is 33%.

1800 212



212r 212

8r The percent of increase is about 8%.

661

Extra Practice

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The tax is 4% of the discounted price of the jacket. 4% of $45.50  0.04  45.50  1.82 Add this amount to the discounted price. $45.50  $1.82  $47.32 The total price of the jacket is $47.32. 12. The discount is 10% of the original price. 10% of $28.95  0.10  28.95  2.895 Subtract $2.90 from the original price. $28.95 $2.90  $26.05 The discounted price of the backpack is $26.05. The tax is 5% of the discounted price of the backpack. 5% of $26.05  0.05  26.05  1.3025 Round $1.3025 to $1.31 since tax is always rounded up to the nearest cent. Add this amount to the discounted price. $26.05  $1.31  $27.36 The total price of the backpack is $27.36.

6. Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase. 65  35  30 Find the percent using the original number, 35, as the base. 30 35

r

 100

30(100)  35r 3000  35r 3000 35

7.

8.

9.

10.

11.



35r 35

86  r The percent of increase is about 86%. The discount is 20% of the original price. 20% of $299  0.20  299  59.80 Subtract $59.80 from the original price. $299.00  $59.80  $239.20 The discounted price of the television is $239.20. The tax is 7% of the price of the book. 7% of $15.95  0.07  15.95  1.1165 Round $1.1165 to $1.12 since tax is always rounded up to the nearest cent. Add this amount to the original price. $15.95  $1.12  $17.07 The total price of the book is $17.07. The tax is 6.25% of the price of the software. 6.25% of $36.90  0.0625  36.90  2.30625 Round $2.30625 to $2.31 since tax is always rounded up to the nearest cent. Add this amount to the original price. $36.90  $2.31  $39.21 The total price of the software is $39.21. The discount is 15% of the original price. 15% of $49.99  0.15  49.99  7.4985 Subtract $7.50 from the original price. $49.99  $7.50  $42.49 The discounted price of the boots is $42.49. The tax is 3.5% of the discounted price of the boots. 3.5% of $42.49  0.035  42.49  1.48715 Round $1.48715 to $1.49 since tax is always rounded up to the nearest cent. Add this amount to the discounted price. $42.49  $1.49  $43.98 The total price of the boots is $43.98. The discount is 30% of the original price. 30% of $65  0.3  65  19.50 Subtract $19.50 from the original price. $65.00  $19.50  $45.50 The discounted price of the jacket is $45.50.

Extra Practice

Page 827

Lesson 3-8

xrq xrrqr xqr The value of x is q  r. 2. ax  4  7 ax  4  4  7  4 ax  3 1.

ax a

3

a 3

xa 3

The value of x is a. Since division by 0 is undefined, a  0. 3.

2bx  b  5 2bx  b  b  5  b 2bx  5  b 5  b 2b 5  b x  2b 5  b of x is 2b .

2bx 2b



The value Since division by 0 is undefined, 2b  0 or b  0. 4. (c  a)

1

x  c c  a x  c c  a

a

2  (c  a)a

x  c  ca  a2 x  c  c  ca  a2  c x  ca  a2  c or a2  ac  c The value of x is a2  ac  c.

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5. c

1

x  y c x  y c

d

11.

2  c(d)

2 (A) h 2A h 2A y h 2A y h

x  y  cd x  y  y  cd  y x  cd  y The value of x is cd  y. 6. 2

1

ax  1 2 ax  1 2

b

12.

2b  1 a 2b  1 x a 2b  1 value of x is a .

The undefined, a  0.

Page 828

x

d  n  d  n d  a . n  b

Since division by 0 is

a b a b

0.60x 0.60

85(4)  95(1) 4  1

9 (y) 5 9 y 5 9 y 5 9 y 5

3

9 5

 5 9 (x  32)  x  32

 

1.5

 0.60

340  95 5 435 5

 87 The student’s average is 87%.

The value of division by 0 is undefined, 3  r  0 or 3  r. 5

Amount of Grape Juice 0.10 (5) 1.00 x 0.40(5  x)

x  2.5 2.5 qt of pure grape juice should be added. 2. The student’s average is determined using a weighted average.

3

y  9 (x  32)

Lesson 3-9

0.10(5)  1.00x  0.40(5  x) 0.50  x  2  0.40x 0.50  x  0.40x  2  0.40x  0.40x 0.50  0.60x  2 0.50  0.60x  0.50  2  0.50 0.60x  1.5

2r  r 2r x3r 2r x is 3  r. Since

10.

x

Amount of amount of amount of grape juice grape juice grape juice in 10% sol plus in 100% sol equals in 40% sol. 1442443 123 1442443 1 424 3 1442443 0.10(5)  1.00x  0.40(5  x)

The value of x is Since division by 0 is undefined, n  b  0 or n  b. 3x  r  r(3  x) 9. 3x  r  3r  rx 3x  r  rx  3r  rx  rx 3x  r  rx  3r 3x  r  rx  r  3r  r 3x  rx  2r (3  r)x  2r (3  r)x 3  r

2rx 2r

Amount of Solution 10% grape juice 5 100% grape juice x 40% grape juice 5x

nx  a  bx  d nx  a  bx  bx  d  bx nx  a  bx  d nx  a  bx  a  d  a nx  bx  d  a (n  b)x  d  a 



1. Let x  the amount of 100% grape juice to be added.



The value undefined, d  0.

A  2r2  2rx A  2r2  2r2  2rx  2r2 A  2r2  2rx

A

d(x  3)  5 dx  3d  5 dx  3d  3d  5  3d dx  5  3d

(n  b)x n  b

x

The value of x is 2r  r. Since division by 0 is undefined, 2r  0 or r  0.

Since division by 0 is

5  3d d 5  3d x d 5  3d of x is d .

8.

xyy

A  2r2 2r A r 2r



dx d

xy

4

The value of x is h  y. Since division by 0 is undefined, h  0.

ax  1  2b ax  1  1  2b  1 ax  2b  1

7.

3

2 1

 h 2h(x  y)

2A

2  2(b)

ax a

1

A  2h(x  y)

4

 32  x  32  32  32  x 9

The value of x is 5y  32.

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Extra Practice

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3. Let x  the number of adult tickets sold. Number of Tickets

Cost of Tickets

Total Cost

x

$5.50

5.50x

21  x

$3.50

3.50(21  x)

Adult Tickets Child Tickets

Page 828

1. 2. 3. 4. 5. 6. 7. 8. 9.

Total cost total cost of adult of child equals tickets plus tickets43 14 $83.50 . 1442443 123 14 44244 424 3 14243 83.50 5.50x  3.50(21  x)  5.50x  3.50(21  x)  83.50 5.5x  73.5  3.5x  83.5 2x  73.5  83.5 2x  73.5  73.5  83.5  73.5 2x  10 2x 2



10 2

Peanuts Chocolate Mixture

Price per Unit $4.50 $6.50 $5.25

Total Price 4.50(5) 6.50x 5.25(5  x)

Price of price of price of peanuts plus chocolate equals mixture. 14243 123 14 4244 31 424 3 1442443 4.50(5)  6.50x  5.25(5  x) 4.50(5)  6.50x  5.25(5  x) 22.5  6.5x  26.25  5.25x 22.5  6.5x  5.25x  26.25  5.25x  5.25x 22.5  1.25x  26.25 22.5  1.25x  22.5  26.25  22.5 1.25x  3.75 1.25x 1.25

3.75

 1.25

x3 3 lb of chocolate should be mixed with 5 lb of peanuts. 5. Let t  the number of hours until they meet. Sheila Casey

r 55 65

t t t

d  rt 55t 65t

Distance distance traveled traveled by Sheila plus by Casey equals 210 miles. 14 4244 3 123 14 4244 31 4 424 4 3 14 424 43 55t  65t  210 55t  65t  210 120t  210 120t 120

210

 120

t  1.75 65t  65(1.75) or 113.75 They will meet in 1.75 hours, and they will be 113.75 miles from Bozeman. Extra Practice

xPoint Coordinate B 1 T 5 P 6 Q 0 A 2 K 4 J 2 L 4 S 3

yCoordinate 2 0 2 6 2 5 5 0 5

Ordered Pair (1, 2) (5, 0) (6, 2) (0, 6) (2, 2) (4, 5) (2, 5) (4, 0) (3, 5)

Quadrant I none IV none III I IV none II

10. A(2, 3) • Start at the origin. • Move right 2 units and down 3 units. • Draw a dot and label it A. (See coordinate plane after Exercise 18.) 11. B(3, 6) • Start at the origin. • Move right 3 units and up 6 units. • Draw a dot and label it B. (See coordinate plane after Exercise 18.) 12. C(4, 0) • Start at the origin. • Move left 4 units. • Since the y-coordinate is 0, the point is on the x-axis. • Draw a dot and label it C. (See coordinate plane after Exercise 18.) 13. D(4, 3) • Start at the origin. • Move left 4 units and up 3 units. • Draw a dot and label it D. (See coordinate plane after Exercise 18.) 14. E(5, 5) • Start at the origin. • Move left 5 units and down 5 units. • Draw a dot and label it E. (See coordinate plane after Exercise 18.) 15. F(1, 1) • Start at the origin. • Move left 1 unit and up 1 unit. • Draw a dot and label it F. (See coordinate plane after Exercise 18.) 16. G(0, 2) • Start at the origin. • Since the x-coordinate is 0, the point is on the y-axis. • Move down 2 units. • Draw a dot and label it G. (See coordinate plane after Exercise 18.)

x5 21  x  21  5 or 16 5 adult tickets and 16 child tickets were sold. 4. Let x  the number of pounds of chocolate in the mixture. Units 5 x 5x

Lesson 4-1

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5b.

17. H(2, 3) • Start at the origin. • Move right 2 units and up 3 units. • Draw a dot and label it H. (See coordinate plane after Exercise 18.) 18. J(0, 3) • Start at the origin. • Since the x-coordinate is 0, the point is on the y-axis. • Move up 3 units. • Draw a dot and label it J. 10–18. y

J

A' A

U

S' S

Q'

Q x

O

6a. To dilate the triangle by a scale factor of 2, multiply the coordinates of each vertex by 2. (x, y) S (2x, 2y) R(2, 1) S R¿(2  2, 2  1) S R¿(4, 2) E(3, 1) S E¿(2  (3), 2  (1) ) S E¿(6, 2) D(2, 4) S D¿(2  2, 2  (4) ) S D¿(4, 8) )

B D

y

U'

H

6b.

y

R'

F C

R x

O

G

E

A

E'

E

Page 828

D

D'

Lesson 4-2

7a. To reflect the pentagon over the x-axis, multiply the y-coordinate of each vertex by 1. (x, y) S (x, y) B(3, 5) S B¿(3, 5) L(4, 5) S L¿(4, 5) A(4, 1) S A¿(4, 1) C(0, 4) S C¿(0, 4) K(4, 1) S K¿(4, 1) 7b. y B' L'

1. The figure has been flipped over a line. This is a reflection. 2. The figure has been shifted horizontally to the left. This is a translation. 3. The figure has been increased in size. This is a dilation. 4a. To translate the quadrilateral 1 unit up, add 1 to the y-coordinate of each vertex. To translate the quadrilateral 2 units right, add 2 to the x-coordinate of each vertex. (x, y) S (x  2, y  1) A(2, 2) S A¿(2  2, 2  1) S A¿(4, 3) B(3, 5) S B¿(3  2, 5  1) S B¿(1, 6) C(4, 0) S C¿(4  2, 0  1) S C¿(2, 1) D(2, 2) S D¿(2  2, 2  1) S D¿(4, 1) 4b. y B' B

C

x

O

A'

K' C' B

A

L

8a. To rotate the triangle 90 counterclockwise about the origin, switch the coordinates of each vertex and then multiply the new first coordinate by 1. (x, y) S (y, x) A(2, 1) S A¿(1, 2) N(4, 1) S N¿(1, 4) G(3, 4) S G¿(4, 3)

C' x

O

A

K

A'

C

x

O

D' D 5a. To reflect the square over the y-axis, multiply the x-coordinate of each vertex by 1. (x, y) S (x, y) S(1, 1) S S¿(1, 1) Q(4, 1) S Q¿(4, 1) U(4, 4) S U¿(4, 4) A(1, 4) S A¿(1, 4)

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8b.

2. Table

y

G

N'

x 4 2 0 2

G' A' N

A

x

O

y 2 0 2 4

Graph

Mapping y

9a. To translate the parallelogram 2 units down, add 2 to the y-coordinate of each vertex. To translate the parallelogram 1 unit left, add 1 to the x-coordinate of each vertex. (x, y) S (x  1, y  2) G(3, 2) S G¿(3  1, 2  2) S G¿(4, 4) R(4, 2) S R¿(4  1, 2  2) S R¿(3, 4) A(6, 4) S A¿(6  1, 4  2) S A¿(5, 2) M(1, 4) S M¿(1  1, 4  2) S M¿(2, 2) 9b.

y

M

A'

x 7 2 4 5 9

x

O

R

G R'

G'

Y

4 2 0 2

0 2 4

The domain of this relation is {4, 2, 0, 2}. The range is {0, 2, 4}. 3. Table

A

M'

x

O

X

y 5 3 0 7 2

Graph y

Page 829

Lesson 4-3

1. Table x 5 0 9

y 2 0 1

O

x

Graph y

O

x

Mapping

Mapping

X

Y

5 0 9

2 0 1

Y

7 2 4 5 9

5 3 0 7 2

The domain of this relation is {9, 2, 4, 5, 7}. The range is {7, 3, 0, 2, 5}.

The domain of this relation is {9, 0, 5}. The range is {1, 0, 2}.

Extra Practice

X

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4. Table x 3.1 4.7 2.4 9

Page 829 1.

y 1 3.9 3.6 12.12

Graph y

Lesson 4-4

x 0

y 1

4

2

2

4

2

5

y  3x  1 1  3(0)  1 1  1 2  3(4)  1 2  11 4  3(2)  1 45 5  3(2)  1 55

True or False? true ✓ false false true ✓

The solution set is {(0, 1), (2, 5)}. 2.

O

x

x 1

y 8

0

7

2

3

5

4

3y  x  7 3(8)  1  7 24  8 3(7)  0  7 21  7 3(3)  2  7 99 3(4)  5  7 12  12

True or False? false false true ✓ true ✓

The solution set is {(2, 3), (5, 4)}. 3.

Mapping

X 3.1 4.7 2.4 9

5. 6. 7. 8. 9.

10.

Y 1 3.9 3.6 12.12

The domain of this relation is {9, 4.7, 2.4, 3.1}. The range is {3.6, 1, 3.9, 12.12}. relation: {(1, 3), (2, 4), (3, 5), (4, 6), (5, 7)} inverse: {(3, 1), (4, 2), (5, 3), (6, 4), (7, 5)} relation: {(4, 1), (2, 3), (0, 1), (2, 3), (4, 1)} inverse: {(1, 4), (3, 2), (1, 0), (3, 2), (1, 4)} relation: {(1, 5), (2, 5), (2, 4), (2, 1), (6, 1)} inverse: {(5, 1), (5, 2), (4, 2), (1, 2), (1, 6)} relation: {(3, 7), (5, 2), (9, 1), (3, 2)} inverse: {(7, 3), (2, 5), (1, 9), (2, 3)} relation: {(4, 3), (2, 2), (2, 1), (0, 0), (1, 1), (2, 3), (2, 1)} inverse: {(3, 4), (2, 2), (1, 2), (0, 0), (1, 1), (3, 2), (1, 2)} relation: {(3, 1), (3, 3), (2, 2), (2, 0), (1, 3), (1, 1), (0, 2), (1, 3), (1, 1), (2, 0), (2, 2), (3, 1), (3, 3)} inverse: {(1, 3), (3, 3), (2, 2), (0, 2), (3, 1), (1, 1), (2, 0), (1, 1), (3, 1), (0, 2), (2, 2), (1, 3), (3, 3)}

x 2

y 0

0

4

0

2

4

12

4x  8  2y 4(2)  8  2(0) 88 4(0)  8  2(4) 00 4(0)  8  2(2) 04 4(4)  8  2(12) 16  16

True or False? true ✓ true ✓ false true ✓

The solution set is {(2, 0), (0, 4), (4, 12)}. 4.

3x  10  4y True or False? 3(10)  10  4(5) false 30  10 1 3(2)  10  4(1) true ✓ 66 0.25 3(3)  10  4(0.25) true ✓ 99 5 3(5)  10  4(5) false 15  10

x y 10 5 2 3 5

The solution set is {(2, 1), (3, 0.25)}. 5. First solve the equation for y in terms of x. xy3 xyx3x y3x x 2 1 0 1 2

3 3 3 3 3

3x  (2)  (1) 0 1 2

y 5 4 3 2 1

(x, y) (2, 5) (1, 4) (0, 3) (1, 2) (2, 1)

The solution set is {(2, 5), (1, 4), (0, 3), (1, 2), (2, 1)}.

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6.

x 2 1 0 1 2

10. First solve the equation for y in terms of x. 2x  y  4 2x  y  2x  4  2x y  4  2x

(x, y) (2, 2) (1, 1) (0, 0) (1, 1) (2, 2)

y 2 1 0 1 2

x 2 1 0 1 2

The solution set is {(2, 2), (1, 1), (0, 0), (1, 1), (2, 2)}. 7.

x 2 1 0 1 2

5x  1 5(2)  1 5(1)  1 5(0)  1 5(1)  1 5(2)  1

y 9 4 1 6 11

(x, y) (2, 9) (1, 4) (0, 1) (1, 6) (2, 11)



y

11.

13  4x 3

2

13  4(2) 3

1

13  4(1) 3

17 3

0

13  4(0) 3

13 3

1

13  4(1) 3

2

13  4(2) 3

y 7

(x, y)

5 3

5 (2, 7), 11, 173 2, 10, 133 2, (1, 3), 12, 53 26.

11, 173 2 10, 133 2 12, 53 2

y

2

8  4(2) 5

16 5

1

8  4(1) 5

12 5

0

8  4(0) 5

8 5

1

8  4(1) 5

4 5

2

8  4(2) 5

0

y 2 3 4 5 6

(x, y) (2, 2) (1, 3) (0, 4) (1, 5) (2, 6)

10  2x 3 10  2x 3

x

10  2x 3

2

10  2(2) 3

1

10  2(1) 3

y 14 3

4

(x, y)

12, 165 2 11, 125 2 10, 85 2 11, 45 2 (2, 0)

512, 165 2, 11, 125 2, 10, 85 2, 11, 45 2, (2, 0) 6. 668

(x, y)

12, 143 2 (1, 4)

0

10  2(0) 3

10 3

1

10  2(1) 3

8 3

2

10  2(2) 3

2

512, 143 2, (1, 4), 10, 103 2, 11, 83 2 (2, 2) 6.

The solution set is

Extra Practice

4x  (2)  (1) 0 1 2

The solution set is

8  4x 5 8  4x 5



y

9. First solve the equation for y in terms of x. 5y  8  4x 5y 8  4x  5 5

x

4 4 4 4 4

3y 3

(1, 3)

The solution set is

y

x 2 1 0 1 2

(2, 7)

3

(x, y) (2, 8) (1, 6) (0, 4) (1, 2) (2, 0)

y 8 6 4 2 0

The solution set is {(2, 2), (1, 3), (0, 4), (1, 5), (2, 6)}. 12. First solve the equation for y in terms of x. 2x  3y  10 2x  3y  2x  10  2x 3y  10  2x

13  4x 3 13  4x 3

x

4  2x  2(2)  2(1)  2(0)  2(1)  2(2)

The solution set is {(2, 8), (1, 6), (0, 4), (1, 2), (2, 0)}.

The solution set is {(2, 9), (1, 4), (0, 1), (1, 6), (2, 11)}. 8. First solve the equation for y in terms of x. 4x  3y  13 4x  3y  4x  13  4x 3y  13  4x 3y 3

4 4 4 4 4

10, 103 2 11, 83 2 (2, 2)

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16. First solve the equation for y in terms of x. x  4y  2 x  4y  x  2  x 4y  2  x

13. First solve the equation for y in terms of x. 2y  3x  1 2y 2



y

3x  1 2 3x  1 2

4y 4

x

3x  1 2

y

2

3(2)  1 2

2

1

3(1)  1 2

0

3(0)  1 2

1

3(1)  1 2

1

12,  52 2

(1, 1)

10, 12 2

1 2

2

3(2)  1 2

2

y

(x, y)

5

(1, 2) 7 2

512, 52 2, (1, 1), 10, 12 2, (1, 2), 12, 72 26.

The solution set is

12, 72 2

x1 2  1 1  1 01 11 21

y 3 2 1 0 1

x

2  x 4

8

2  (8) 4

2.5

(8, 2.5)

4

2  (4) 4

1.5

(4, 1.5)

0

2  0 4

0.5

(0, 0.5)

4

2  4 4

0.5

(4, 0.5)

8

2  8 4

1.5

(8, 1.5)

(x, y)

y

y

x

O

(x, y) (2, 3) (1,2) (0, 1) (1, 0) (2, 1)

17. First solve the equation for y in terms of x. y3x y33x3 yx3

Graph the solution set {(2, 3), (1, 2), (0, 1), (1, 0), (2, 1)}.

x 5 1 3 7 9

y

x

O

2  x 4 2  x 4

Graph the solution set {(8, 2.5), (4, 1.5), (0, 0.5), (4, 0.5), (8, 1.5)}.

14. First solve the equation for y in terms of x. xy1 x1y11 x1y x 2 1 0 1 2



x3 5  3 1  3 33 73 93

(x, y) (5, 2) (1, 2) (3, 6) (7, 10) (9, 12)

y 2 2 6 10 12

Graph the solution set {(5, 2), (1, 2), (3, 6), (7, 10), (9, 12)}. y

15.

x1 3  1 1  1 01 11 31

x 3 1 0 1 3

y 2 0 1 2 4

(x, y) (3, 2) (1, 0) (0, 1) (1, 2) (3, 4)

Graph the solution set {(3, 2), (1, 0), (0, 1),(1, 2), (3, 4)}. O

y

O

x

x

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20. First solve the equation for y in terms of x.

18. First solve the equation for y in terms of x. x  y  2 x  y  x  2  x y  2  x 2  x 2  (4) 2  (3) 2  0 2  1 2  3

x 4 3 0 1 3

2

3y  3x  4 1 (3y) 3

y

(x, y) (4, 2) (3, 1) (0, 2) (1, 3) (3, 5)

y 2 1 2 3 5

2 x 9



4 3

2

2

4

y

4 3

8 3

9 x  3

x 6

2 (6) 9



3

2 (3) 9

3

Graph the solution set {(4, 2), (3, 1), (0, 2), (1, 3), (3, 5)}. y

4

0

2 (0) 9



1

2 (1) 9

3

2 (3) 9

3

(x, y)

2

16, 83 2

(3, 2)

4 3

4 3

3

4

9

4

3

10 2

10, 43 2 11, 109 2 13, 23 2

516, 83 2, (3, 2), 10, 43 2, 11, 109 2, 13, 23 26.

Graph the solution set

x

O

1

1 2

 3 3x  4

y

x

O

19. First solve the equation for y in terms of x. 2x  3y  5 2x  3y  2x  5  2x 3y  5  2x 3y 3



y

1

5

5  2(5) 3

3

3

5  2(3) 3

3

5  2(0) 3

5

5  2(5) 3 5  2(6) 3

1

y 5 1 5 3

5

y  4 

(x, y)

15, 53 2 13, 13 2 10, 53 2

17 3

515, 53 2, 13, 13 2, 10, 53 2, (5, 5), 16, 173 26.

Graph the solution set

(5, 5)

16, 173 2

3 x 4

2

3

4  4x

4

4  4 (4)

0

4  4 (0)

4

4  4 (4)

6

4 

3 (6) 4

0.5

4 

3 (8) 4

2

3

8

(x, y)

y 7

(4, 7)

3

4

(0, 4)

3

1

(4, 1) (6, 0.5) (8, 2)

Graph the solution set {(4, 7), (0, 4), (4, 1), (6, 0.5), (8, 2)}. y

O

Extra Practice

3

x

y

O

1

2 (2y)  2 8  2x

x

0

3

2y  8  2x

5  2x 3

6

21. First solve the equation for y in terms of x.

5  2x 3 5  2x 3

x

670

x

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Page 829

Lesson 4-5

y

1. First rewrite the equation so that the variables are on one side of the equation. 3x  2y 3x  2y  2y  2y 3x  2y  0 The equation is now in standard form where A  3, B  2, and C  0. This is a linear equation. 2. Since the term y2 has an exponent of 2, the equation cannot be written in the form Ax  By  C. Therefore, this is not a linear equation. 3. First rewrite the equation so that the variables are on one side of the equation and a constant is on the other side. 4x  2y  8 4x  2y  2y  8  2y 4x  2y  8 Since the GCF of 4, 2, and 8 is not 1, divide each side by 2. 2(2x  y)  8 2(2x  y) 2

8. Select five values for the domain and make a table. 3x  1 3(2)  1 3(1)  1 3(0)  1 3(1)  1 3(2)  1

x 2 1 0 1 2

(x, y) (2, 5) (1 2) (0, 1) (1, 4) (2, 7)

y

8

2

1 x

9. Solve the equation for y. 3x  2y  12 3x  2y  3x  12  3x 2y  12  3x 2y 2



y

12  3x 2 3x  12 2

Select five values for the domain and make a table.

5 y

y 7 4 1 2 5

x

O

6. Since each of the terms and has a variable in the denominator, the equation cannot be written in the form Ax  By  C. Therefore, this is not a linear equation. 7. Solve the equation for y. 3x  y  4 3x  y  3x  4  3x y  4  3x Select five values for the domain and make a table. 4  3x 4  3(1) 4  3(0) 4  3(1) 4  3(2) 4  3(3)

y 5 2 1 4 7

Graph the ordered pairs and draw a line through the points.

2x  y  4 The equation is now in standard form where A  2, B  1, and C  4. This is a linear equation. 4. First rewrite the equation so that the variables are on one side of the equation and a constant is on the other side. 5x  7y  2x  7 5x  7y  2x  2x  7  2x 3x  7y  7 The equation is now in standard form where A  3, B  7, and C  7. This is a linear equation. 5. First simplify. Then rewrite the equation so that the variables are on one side of the equation. 2x  5x  7y  2 7x  7y  2 7x  7y  7y  2  7y 7x  7y  2 The equation is now in standard form where A  7, B  7, and C  2. This is a linear equation.

x 1 0 1 2 3

x

O

x

3x  12 2

y

(x, y)

1

3(1)  12 2

7.5

(1, 7.5)

0

3(0)  12 2

6

1

3(1)  12 2

4.5

2

3(2)  12 2

3

4

3(4)  12 2

0

(0, 6) (1, 4.5) (2, 3) (4, 0)

Graph the ordered pairs and draw a line through the points. y

(x, y) (1, 7) (0, 4) (1, 1) (2, 2) (3, 5)

O

x

Graph the ordered pairs and draw a line through the points.

671

Extra Practice

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12. The only value in the range is 2. Since there is no x in the equation, the value of x does not depend on y. Therefore, x can be any real number. Select five values for the domain and make a table.

10. Solve the equation for y. 2x  y  6 2x  y  2x  6  2x y  6  2x 1(y)  1(6  2x) y  2x  6 Select five values for the domain and make a table. 2x  6 2(1)  6 2(0)  6 2(1)  6 2(2)  6 2(3)  6

x 1 0 1 2 3

y 8 6 4 2 0

x 3 1 0 1 4

(x, y) (1, 8) (0, 6) (1, 4) (2, 2) (3, 0)

y 2 2 2 2 2

(x, y) (3, 2) (1, 2) (0, 2) (1, 2) (4, 2)

Graph the ordered pairs and draw a line through the points.

Graph the ordered pairs and draw a line through the points.

y

y x

O

O

13. To find the x-intercept, let y  0. y  5x  7 0  5x  7 0  7  5x  7  7 7  5x

11. Solve the equation for y. 2x  3y  8 2x  3y  2x  8  2x 3y  8  2x 3y 3



y

7 5 7 5

8  2x 3 2x  8 3

2x  8 3

2

2(2)  8 3

(x, y)

y 4

(2, 4)

0

2(0)  8 3

1

2(1)  8 3

2

2(2)  8 3

3

4

2(4)  8 3

0

8

3 2

10, 83 2

(1, 2) 4

12, 43 2

x

O

(4, 0)

y

Extra Practice

5x 5

17 2

y

Graph the ordered pairs and draw a line through the points.

O



The graph intersects the x-axis at 5, 0 . To find the y-intercept, let x  0. y  5x  7 y  5(0)  7 y  7 The graph intersects the y-axis at (0, 7). Plot these points and draw the line that connects them.

Select five values for the domain and make a table. x

x

x

672

x

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16. To find the x-intercept, let y  0. 5x  2y  8 5x  2(0)  8 5x  8

14. The only value in the domain is 4. Since there is no y in the equation, the value of y does not depend on x. Therefore, y can be any real number. Select five values for the range and make a table. x 4 4 4 4 4

y 4 2 0 1 3

5x 5

(x, y) (4, 4) (4, 2) (4, 0) (4, 1) (4, 3)

8

x5 The graph intersects the x-axis at To find the y-intercept, let x  0. 5x  2y  8 5(0)  2y  8 2y  8

Graph the ordered pairs and draw a line through the points.

2y 2

y

185, 02.

8

 2

y  4 The graph intersects the y-axis at (0, 4). Plot these points and draw the line that passes through them. x

O

8

5

y

x

O

15. To find the x-intercept, let y  0. 1

x  3y  2 1

x  3 (0)  2 x2 The graph intercepts the x-axis at (2, 0). To find the y-intercept, let x  0.

17. To find the x-intercept, let y  0. 4.5x  2.5y  9 4.5x  2.5(0)  9 4.5x  9

1

x  3y  2 1

0  3y  2

3

1 y 3 1 y 3

1 2

4.5x 4.5

2

9

 4.5

x2 The graph intersects the x-axis at (2, 0). To find the y-intercept, let x  0. 4.5x  2.5y  9 4.5(0)  2.5y  9 2.5y  9

 3(2)

y6 The graph intersects the y-axis at (0, 6). Plot these points and draw the line that passes through them.

2.5y 2.5

y

9

 2.5

y  3.6 The graph intersects the y-axis at (0, 3.6). Plot these points and draw the line that passes through them. y

O

x

O

673

x

Extra Practice

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18. To find the x-intercept, let y  0. 1 x 2 1 x 2

8. Graph the equation. For each value of x, a vertical line passes through no more than one point on the graph. Thus, the line represents a function.

 3y  12

 3(0)  12

2

1 x 2 1 x 2

y

 12

1 2  2(12)

x  24 The graph intersects the x-axis at (24, 0). To find the y-intercept, let x  0. 1 x 2

 3y  12

1 (0) 2

 3y  12

9. Graph the equation. For each value of x, a vertical line passes through no more than one point on the graph. Thus, the graph represents a function.

3y  12 3y 3



12 3

y4 The graph intersects the y-axis at (0, 4). Plot these points and draw the line that passes through them.

12 9 6 3

y

–12 –9 –6 –3 –3 –6 –9 –12

10. x

O

Page 830

y

O 3 6 9 12x

f(x)  2x  5 f(4)  2(4)  5  8  5  3

11. g(x)  3x2  1 g(2)  3(2) 2  1  3(4)  1  12  1  11 12. f(x)  2x  5 f(3)  5  [ 2(3)  5]  5  (6  5)  5  11  5 6

Lesson 4-6

1. Since an element of the domain is paired with more than one element in the range, the relation is not a function. Both 3 and 7 in the range are paired with x  1. 2. The mapping does not represent a function since the element 2 in the domain is paired with both 4 and 2 in the range. Also, 3 in the domain is paired with both 2 and 0 in the range. 3. For each value of x, a vertical line passes through no more than one point on the graph. Thus, the relation graphed represents a function. 4. Since an element of the domain is paired with more than one element in the range, the relation is not a function. Both 3 and 4 in the range are paired with x  1. 5. Since each element of the domain is paired with exactly one element of the range, the relation is a function. 6. Since each element of the domain is paired with exactly one element of the range, the relation is a function. 7. Since an element of the domain is paired with more than one element in the range, the relation is not a function. Both 7 and 5 in the range are paired with x  4.

Extra Practice

x

O

g(x)  3x2  1 g(2)  4  [3(2) 2  1]  4  [3(4)  1]  4  (12  1)  4  11  4  15 14. f(x)  2x  5 f(b2 )  2(b2 )  5  2b2  5 13.

15.

g(x)  3x2  1 g(a  1)  3(a  1) 2  1  3(a2  2a  1)  1  (3a2  6a  3)  1  3a2  6a  2

16. f(x)  2x  5 and g(x)  3x2  1 f(0)  g(3)  [ 2(0)  5]  [ 3(3) 2  1]  (0  5)  [ 3(9)  1]  5  (27  1)  5  26  31

674

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17. f(x)  2x  5 and g(x)  3x2  1 f(n)  g(n)  [2(n)  5]  [3(n) 2  1]  (2n  5)  (3n2  1)  3n2  2n  4

Page 830 1. 2

0

1

2. 3

5

8

12

3. 2

4

8

16

 2 4 8

4. 21 16 11 6

 5

5

5

5. 0 0.25 0.5 0.75

1 3

1 9

1 81

1 27

9

27

13

81

23

33

?

?

10

10

10

10

11.

8

9 8

1

?

?

?

  1 1 1 1 1 1 8

5 4

1

8

11 8

8

8

8

8

1

3 2

8

5

12. 3

4

6

11

3

6

?

?

?

     1 1 1 1 1 1 2

2

2

2

2

2

1 2.

1

The common difference is Add 2 to the last 1 term of the sequence and continue adding 2 until the next three terms are found. 11

7

6

17

3

10

6

3

   1 1 1 2

2

2

7

17

10

The next three terms are 3,  6 ,  3 . 13. Use the formula for the nth term of an arithmetic sequence with a1  3, d  6, and n  12. an  a1  (n  1)d a12  3  (12  1)6 a12  3  66 a12  69

10

The 12th term of the arithmetic sequence is 69. 14. Use the formula for the nth term of an arithmetic sequence with a1  2, d  4, and n  8. an  a1  (n  1)d a8  2  (8  1)4 a8  2  28 a8  26

2

The 8th term of the arithmetic sequence is 26. 15. Use the formula for the nth term of an arithmetic sequence with a1  1, d  3, and n  10. an  a1  (n  1)d a10  1  (10  1) (3) a10  1  (27) a10  28

    0.6

8

5 11 3 , . 8 2

The next three terms are 12, 14, 16. 9. 2 1.4 0.8 0.2 ? ? ? 0.6

7 8

8

   2 2 2

0.6

8

The next three terms are 4,

The common difference is 2. Add 2 to the last term of the sequence and continue adding 2 until the next three terms are found. 10 12 14 16

0.6

8

 1 1 1

?

2

3 4

9 8

    2

?

1

The next three terms are 43, 53, 63. 8. 4 6 8 10 ? ? ? 2

?

The common difference is 8. Add 8 to the last term 1 of the sequence and continue adding 8 until the next three terms are found.

 10 10 10

2

8

8

The common difference is 10. Add 10 to the last term of the sequence and continue adding 10 until the next three terms are found. 33 43 53 63

2

?

The next three terms are 37, 45, 53.



10

29

  8 8 8

This is not an arithmetic sequence because the difference between terms is not constant.

  2 2 2

7. 3

8

This is an arithmetic sequence because the difference between terms is constant. The common difference is 0.25.

 0.25 0.25 0.25

21

The common difference is 8. Add 8 to the last term of the sequence and continue adding 8 until the next three terms are found. 29 37 45 53

This is an arithmetic sequence because the difference between terms is constant. The common difference is 1. This is not an arithmetic sequence because the difference between terms is not constant. This is not an arithmetic sequence because the difference between terms is not constant. This is an arithmetic sequence because the difference between terms is constant. The common difference is 5.

 2 3 4

13



Lesson 4-7 1

  1 1 1

6.

10. 5

0.6 0.6

The 10th term of the arithmetic sequence is 28.

The common difference is 0.6. Add 0.6 to the last term of the sequence and continue adding 0.6 until the next three terms are found. 0.2 0.4 1.0 1.6

  0.6 0.6 0.6

The next three terms are 0.4, 1.0, 1.6.

675

Extra Practice

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20. 25

16. Use the formula for the nth term of an arithmetic sequence with a1  2.2, d  1.4, and n  5. an  a1  (n  1)d a5  2.2  (5  1)(1.4) a5  2.2  5.6 a5  7.8

40

  5 5

The first term is 2. The common difference is 5. Use the formula for the nth term of an arithmetic sequence with a1  2, d  5, and n  12. an  a1  (n  1)d a12  2  (12  1)(5) a12  2  (55) a12  57

n

15n  10

an

(n, an)

1 2 3 4 5

15(1) 15(2) 15(3) 15(4) 15(5)

 10  10  10  10  10

25 40 55 70 85

(1, (2, (3, (4, (5,

The 12th term of the arithmetic sequence is 57. 1

1

3

28

90 80 70 60 50 40 30 20 10

3

14

18

 3 3 3 8

8

8

1

3

The first term is 22. The common difference is 8. Use the formula for the nth term of an arithmetic 1 3 sequence with a1  22, d  8, and n  10. an  a1  (n  1)d 1

1 32

O

a10  22  (10  1) 8 1

1

3

a10  22  38 7

a10  8

2

21. 9

The first term is 3. The common difference is 4. Use the formula for the nth term of an equation with a1  3 and d  4. an  a1  (n  1)d an  3  (n  1)4 an  3  4n  4 an  3  4n  4 an  4n  7

20 15 10 5 O 5

3

9

n

6n  15

an

(n, an)

6(1)  15 6(2)  15 6(3)  15 6(4)  15 6(5)  15

9 3 3 9 15

(1, 9) (2, 3) (3, 3) (4, 9) (5, 15)

4n  7

an

(n, an)

7 7 7 7 7

3 1 5 9 13

(1, 3) (2, 1) (3, 5) (4, 9) (5, 13)

15 12 9 6 3 O 3 6 9

an

1 2 3 4 5n

Extra Practice

1 2 3 4 5n

3

1 2 3 4 5

4(1) 4(2) 4(3) 4(4) 4(5)

an

The first term is 9. The common difference is 6. Use the formula for the nth term to write an equation with a1  9 and d  6. an  a1  (n  1)d an  9  (n  1)6 an  9  6n  6 an  6n  15

7

 4 4 4

1 2 3 4 5

25) 40) 55) 70) 85)

   6 6 6

The 10th term of the arithmetic sequence is 8. 19. 3 1 5 9

n

70

The first term is 25. The common difference is 15. Use the formula for the nth term to write an equation with a1  25, d  15. an  a1  (n  1)d an  25  (n  1)15 an  25  15n  15 an  15n  10

The 5th term of the arithmetic sequence is 7.8. 17. 2 7 12

18. 22

55

  15 15 15

676

an

1 2 3 4 5n

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22. 3.5

2 0.5

3. 12

1

The first term is 3.5. The common difference is 1.5. Use the formula for the nth term to write an equation with a1  3.5 and d  1.5. an  a1  (n  1)d an  3.5  (n  1)1.5 an  3.5  1.5n  1.5 an  1.5n  5 n

1.5n  5

an

(n, an)

1 2 3 4 5

1.5(1)  5 1.5(2)  5 1.5(3)  5 1.5(4)  5 1.5(5)  5

3.5 2 0.5 1 2.5

(1, 3.5) (2, 2) (3, 0.5) (4, 1) (5, 2.5)

3 2 1 O 1 2 3 4

23

34

45

 11 11 11

  1.5 1.5 1.5

The difference between each pair of terms is always 11. The sequence is arithmetic with a common difference of 11. Each term is 11 more than the term before it. Add 11, 11, and 11. 45 56 67 78

 11 11 11

The next three terms are 56, 67, and 78. 4. 39 33 27 21

  6 6 6

The difference between each pair of terms is always 6. The sequence is arithmetic with a common difference of 6. Each term is 6 less than the term before it. Add 6, 6, and 6. 21 15 9 3

 6 6 6

an

The next three terms are 15, 9, and 3. 5. 6.0 7.2 8.4 9.6

  1.2 1.2 1.2

1 2 3 4 5n

The difference between each pair of terms is always 1.2. The sequence is arithmetic with a common difference of 1.2. Each term is 1.2 more than the term before it. Add 1.2, 1.2, and 1.2. 9.6 10.8 12.0 13.2

  1.2 1.2 1.2

Page 830

The next three terms are 10.8, 12.0, and 13.2. 6. 86 81.5 77 72.5

Lesson 4-8

   4.5 4.5 4.5

1. The pattern consists of circles with one-tenth shaded. The section that is shaded is the fourth section in a clockwise direction from the previously-shaded section. The next two figures in the pattern are shown:

The difference between each pair of terms is always 4.5. The sequence is arithmetic with a common difference of 4.5. Each term is 4.5 less than the term before it. Add 4.5, 4.5, and 4.5. 72.5 68 63.5 59

  4.5 4.5 4.5

The next three terms are 68, 63.5, and 59. 7. 4 8 16 32

2. The pattern consists of a figure that is alternately flipped (or reflected) then enlarged and flipped. Continue the pattern by flipping the last figure without changing its size or shape, then enlarging and flipping the figure you just sketched. The next two figures in the pattern are shown.

  4 8 16

The difference between each pair of terms doubles for each successive pair. Continue doubling each successive difference. Add 32, 64, and 128. 32 64 128 256

   32 64 128

The next three terms are 64, 128, and 256. 8. 3125 625 125 25

 100

2500 500

The difference between each pair of successive 1 terms is 5 of the difference between the previous 1 pair of terms. Continue taking 5 of each successive difference. Add 20, 4, and 0.8. 25 5 1 0.2

  20 4 0.8

The next three terms are 5, 1, and 0.2.

677

Extra Practice

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9. 15

16

18

21

25

14. Make a table of ordered pairs for several points on the graph.

30

 1 2 3 4 5

1

The difference between each pair of terms increases by 1 for each successive pair. Continue increasing each successive difference by 1. Add 6, 7, and 8. 30 36 43 51

1

0

1

0

2

1

1

2

x

y

 1

2

1

0

1

2

1

1 2 1 2

0

2

1

1

1

3 2

2

0

y is always 1 1 less than 2x.

1  2x  1. 1

If x  2, then y  2 (2) 1 or 0. 1

3

If x  1, then y  2 (1)  1 or 2. 1

1

Thus, y  2x  1 describes this relation. Since this relation is also a function, we can write the 1 equation in function notation as f(x)  2x  1. 15. Make a table of ordered pairs for several points on the graph.

  1

1

1 1

1

  0 0

  1 1

1

1 1



2 2

3

 1

x y

The difference of the x values is 1, and the difference of the y values is 1. The difference in y values is the opposite of the difference in x values. This suggests y  x. Check: If x  1, then y  (1) or 1. If x  2, then y  (2) or 2. Thus, y  x describes the relation. Since this relation is also a function, we can write the equation in function notation as f(x)  x. 13. Make a table of ordered pairs for several points on the graph. 1

1 3

 0

  0

1

0 3

  0

3 0

0 2

3 4

 2 2

The difference of x values is 3, and the difference of 2 y values is 2. The difference of the y values is 3 the 2 difference of the x values. This suggests y  3x. 2

If x  3, then y  3 (3) or 2. But the y value for x  3 is 0. This is a difference of 2. Try other values in the domain to see if the same difference occurs.

Check:

2

1 3

3





3 3

x

3

0

3

2 x 3

2

0

2

2

4

y

 0

0

Check y 

2 x 3

y is always 2 more than 2 x. 3



2

 2. 2

If x  3, then y  3 (3)  2 or 0.

There is no difference in y values, no matter what the difference of x values. The value of y remains constant at 3. This suggests y  3. Check: If x  1, then y  3. If x  3, then y  3. Thus, y  3 describes the relation. Since this relation is also a function, we can write the equation in function notation as f(x)  3. Extra Practice

2

Check y

The next three terms are 7, 4, and 5. 12. Make a table of ordered pairs for several points on the graph.

3 3

1

2



1

 2x

  1 3 1

x y



2

2

1

 2

The difference between successive pairs of terms alternates between 3 and 1. Continue alternating the difference. Add 1, 3, and 1. 6 7 4 5



3

If x  2, then y  2 (2) or 1. But the y value for x  2 is 0. This is a difference of 1. Try other values in the domain to see if the same difference occurs.

Check:

 3 1 3 1 3

2 2

1

2



The difference of x values is 1, and the difference 1 of y values is 2. The difference of the y values is 1 the opposite of the difference of the x values. 2 1 This suggests y  2x.

The next three terms are w  10, w  12, and w  14. 11. 13 10 11 8 9 6

x y



1

 2

 2

1

2

2

The difference between each pair of terms is always 2. The sequence is arithmetic with a common difference of 2. Each term is 2 less than the term before it. Add 2, 2, and 2. w8 w  10 w  12 w  14

 2

1



y



The next three terms are 36, 43, and 51. 10. w  2 w4 w6 w8

 2



x

  6 7 8  2

1



2

If x  3, then y  3 (3)  2 or 4. 2

Thus, y  3x  2 describes this relation. Since this relation is also a function, we can write the 2 equation in function notation as f(x)  3x  2.

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Page 831

11. Let (1, r)  (x1, y1 ) and (1, 4)  (x2, y2 ).

Lesson 5-1

y2  y1

1. Let (4, 4)  (x1, y1 ) and (2, 0)  (x2, y2 ).

mx

y2  y1

mx

2

2

5 

 x1

0  (4)  (4) 4 or 2 2

m  2 m

5 

5(2)  4  r 10  4  r 10  4  4  r  4 6  r (1)(6)  (1) (r) 6r 12. Let (r, 2)  (x1, y1 ) and (7, 1)  (x2, y2 ).

2. Let (4, 2)  (x , y ) and (0, 1)  (x , y ). 1 1 2 2 y2  y1

mx

2

 x1

1  2 (4) 3 3 or 4 4

m0 m

y2  y1

3. Let (2, 2)  (x1, y1 ) and (3, 3)  (x2, y2 ).

mx

y2  y1

mx

2

2

1 4 1 4

 x1

3  2 (2) 5 or 1 5

m3 m

y2  y1 2

m m

 x1

4  (8) 1  (2) 12 or 4 3 y2  y1 2

 x1

2

m  1 or 2 6. Let (5, 4)  (x1, y1 ) and (1, 11)  (x2, y2 ).

26 3 26 3

y2  y1 2

m m

 x1

11  4 1  (5) 7 4

7. Let (18, 4)  (x1, y1 ) and (6, 10)  (x2, y2 ).

Page 831

2

m m

 x1

m

r

y2  y1 2

m

 x1

2  0 3  0

2

or 3 3

2. The constant of variation is 2. y2  y1

y2  y1

m

3r 3

Lesson 5-2

mx

8. Let (4, 6)  (x1, y1 ) and (4, 8)  (x2, y2 ). 2



1. The constant of variation is 3.

10  (4) 6  18 6 1 or 2 12

mx



2

y2  y1

mx

1

r  2 (3) r  2 10

7

2(10)  3(r  2) 20  3r  6 20  6  3r  6  6 26  3r

6  4

m43

mx



2

2 3 2 3

5. Let (3, 4)  (x , y ) and (4, 6)  (x , y ). 1 1 2 2 mx



 x1

1  (2) 7  r 1 7  r

1(7  r)  4(1) 7r4 7r747 r  3 13. Let (3, 2)  (x1, y1 ) and (7, r)  (x2, y2 ). y2  y 1 mx x

4. Let (2, 8)  (x , y ) and (1, 4)  (x , y ). 1 1 2 2 mx

 x1

4  r 1  (1) 4  r 2

mx

 x1

2

8  (6) 4  (4) 2 0

m

 x1

3  0 2  0

3

or 2 1

3. The constant of variation is 5. y2  y1

mx

Since division by zero is undefined, the slope is undefined.

2

m

9. Let (0, 0)  (x , y ) and (1, 3)  (x , y ). 1 1 2 2

 x1

1  0 5  0

1

or 5

y2  y1

mx

2

 x1

3  0  0

m  1

or 3

10. Let (8, 1)  (x1, y1 ) and (2, 1)  (x2, y2 ). y2  y1

mx

2

m

 x1

1  1 2  (8)

or 0

679

Extra Practice

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8. Find the value of k. y  kx 7  k(1) (1)(7)  (1)(k) 7k Therefore, y  7x. Find x when y  84. y  7x 84  7x

4. Write the slope as a ratio. 5

5 1

Graph (0, 0). From the point (0, 0) move up 5 units and right 1 unit. Draw a dot. Draw a line containing the points. y

y  5x

84 7

450 6

5. Write the slope as a ratio. 6



k(6) 6

75  k Therefore, y  75x. Find y when x  10. y  75x y  75(10) y  750 10. Find the value of k. y  kx 6  k(48)

6  1 Graph (0, 0). From the point (0, 0) move up 6 units and left 1 unit. Draw a dot. Draw a line containing the points. y

6 48 1 8

y  6x x

O

7x 7

12  x 9. Find the value of k. y  kx 450  k(6)

x

O





k(48) 48

k 1

Therefore, y  8x. Find y when x  20.

6. Write the slope as a ratio.

1

y  8x

4

4

3  3 Graph (0, 0). From the point (0, 0) move down 4 units and right 3 units. Draw a dot. Draw a line containing the points.

1

y  8 (20) y  2.5

y

Page 831 4 3

y x O

x

7. Find the value of k. y  kx 45  k(9) 45 9



k(9) 9

4

5

4. Replace m with 3 and b with 3.

5k Therefore, y  5x. Find y when x  7. y  5x y  5(7) y  35

Extra Practice

Lesson 5-3

1. Replace m with 5 and b with 15. y  mx  b y  5x  (15) y  5x  15 2. Replace m with 6 and b with 3. y  mx  b y  6x  3 3. Replace m with 0.3 and b with 2.6. y  mx  b y  0.3x  (2.6) y  0.3x  2.6 y  mx  b 4

5

y  3x  3 2

5. Replace m with 5 and b with 2. y  mx  b 2

y  5 x  2

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6. Replace m with

7 4

11. The y-intercept is 3. So, graph (0, 3). The slope is 2 2 or 1 . From (0, 3), move down 2 units and right 1 unit. Draw a dot. Draw a line connecting the points.

and b with 2.

y  mx  b 7

y  4x  (2) 7

y  4x  2

y

7. Find the slope. y2  y1

mx

2

m m

 x1

1  3 2  0 2 or 2

y  2x  3

1

The line crosses the y-axis at (0, 3). So, the y-intercept is 3. y  mx  b y  1x  3 y  x  3 8. Find the slope.

12. Solve for y to find the slope-intercept form. 3x  y  6 3x  y  3x  6  3x y  3x  6 (1)(y)  (1) (3x  6) y  3x  6 The y-intercept is 6. So, graph (0, 6). The slope 3 is 3 or 1. From (0, 6), move up 3 units and right 1 unit. Draw a dot. Draw a line connecting the points.

y2  y1

mx

2

m m

 x1

2  (3) 2  0 1 1 or 2 2

The line crosses the y-axis at (0, 3). So, the y-intercept is 3. y  mx  b

y

1

y  2x  (3) y

1 2x

x

O

O

x

3

9. Find the slope. y2  y1

mx

2

3x y  6

 x1

2  1 (3)

m0 1

m3 The line crosses the y-axis at (0, 2). So, the y-intercept is 2. y  mx  b

Page 832

1

y  3x  2 10. The y-intercept is 1. So, graph (0, 1). The slope 5 is 5 or 1. From (0, 1), move up 5 units and right 1 unit. Draw a dot. Draw a line connecting the points. y

y  5x  1 O

Lesson 5-4

1. The point (0, 0) lies on the y-axis. The y-intercept is 0. y  mx  b y  2x  0 y  2x 2. Find the y-intercept. y  mx  b 2  4(3)  b 2  12  b 2  12  12  b  12 14  b Write the slope-intercept form. y  mx  b y  4x  14 3. The point (0, 5) lies on the y-axis. The y-intercept is 5. y  mx  b y  1x  5 y  x  5

x

681

Extra Practice

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4. Find the y-intercept. y  mx  b

8. Find the slope. y2  y1

mx

1

m

1

32b 1

1

5 2

b

5

9. Find the slope. y2  y1

5 2



mx

2

5. Find the y-intercept. y  mx  b

m m

2

5  3 (1)  b 2 2

2

5  3  3  b  3 17

3  b Write the slope-intercept form. y  mx  b 2

1

17

y  3x   3 y

2 x 3



17 3

2

y2  y1

mx

6. Find the y-intercept. y  mx  b 1 4 1 4 1 4

2

m

1 2b 1 2

8

44b4 15

4  b Write the slope-intercept form. y  mx  b

1

15

y  8x 

15 4

2

2

m m

m m

 x1

2

7  3 (5)  b

2  7 8  (1) 9 or 1 9

7 7

Find the y-intercept. y  mx  b 7  1(1)  b 71b 711b1 6b Write the slope-intercept form. y  mx  b y  1x  6 y  x  6

Extra Practice

 x1

3  7 1  5 4 2 or 3 6

Find the y-intercept. y  mx  b

y2  y1 2

or 1

y2  y1

mx

7. Find the slope. mx

 x1

1  0 0  1

The point (0, 1) lies on the y-axis. The y-intercept is 1. y  mx  b y  1x  1 y  x  1 11. Find the slope.

4b

y  8x   4

 x1

1  (1) 7  8 0 or 0 1

Find the y-intercept. y  mx  b 1  0(8)  b 1  b Write the slope-intercept form. y  mx  b y  0x  (1) y  1 10. Find the slope.

5  3  b 2

5

or 4

y  4x  5

Write the slope-intercept form. y  mx  b y

5  0 0  4

The point (0, 5) lies on the y-axis. The y-intercept is 5. y  mx  b

1

322b2

1 4x

 x1

2

3  4 (2)  b

10 3 11 3



10 3 10 3

b b

10 3

b

Write the slope-intercept form. y  mx  b 2

y  3x 

682

11 3

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12. Find the slope.

Write the slope-intercept form. y  mx  b

y2  y1

mx

 x1

2

m m

1

16. Find the slope of the line containing the points (2, 0) and (0, 1). y2  y1

Find the y-intercept. y  mx  b

mx

2

5

1

y  2x  1 17. Find the slope of the line containing the points (1, 0) and (0, 4). y2  y1

5

y  3x  (10)

mx

2

5

y  3x  10

m

13. Find the slope.

m

 x1

3  3 1  (2) 0 or 0 3

Find the y-intercept. y  mx  b 3  0(2)  b 3b Write the slope-intercept form. y  mx  b y  0x  3 y3 14. Find the slope.

2

3  0  0

m  4

Write the slope-intercept form. y  mx  b y  1x  5 y  x  5 19. Find the slope of the line containing the points (1, 0) and (0, 3). y2  y1

mx

3

2

or 4

m

The point (0, 0) lies on the y-axis. The y-intercept is 0. y  mx  b 3 3

y  4x

2

y2  y1

m

m

m

 x1 3 4

1 12

 2

1 4 3 4

or 3

1 1

y  4x  1

1

1 2



1

21. Find the slope of the line containing the points (3, 0) and (0, 3). y2  y1

1 12

mx

2

 3 2  b

m

1

 6  b 1

1

or 4

y  4x  (1)

Find the y-intercept. y  mx  b 1 2 1 2 1 6 2 3

 x1

1  0 0  (4)

Write the slope-intercept form. y  mx  b

1

2

1 4

or 3

y2  y1

mx

15. Find the slope. 2

 x1

3  0 0  (1)

Write the slope-intercept form. y  mx  b y  3x  3 20. Find the slope of the line containing the points (4, 0) and (0, 1).

y  4x  0

mx

 x1

5  0

m  0  5 or 1

 x1

2

or 4

y2  y1

mx

y2  y1

mx

 x1

4  0 0  1

Write the slope-intercept form. y  mx  b y  4x  (4) y  4x  4 18. Find the slope of the line containing the points (5, 0) and (0, 5).

y2  y1

m

1

Write the slope-intercept form. y  mx  b

5  5  b 5  5  5  b  5 10  b Write the slope-intercept form. y  mx  b

2

 x1

1  0

m  0  2 or 2

5  3 (3)  b

mx

2

y  3x  3

15  (5) 3  (3) 10 5 or 3 6

 x1

3  0 0  3

or 1

Write the slope-intercept form. y  mx  b y  1x  (3) yx3

1

 6  b  6 b

683

Extra Practice

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Page 832 1.

2. 3.

4.

5.

11.

Lesson 5-5

y  y1  m(x  x1 ) y  (2)  3(x  5) y  2  3(x  5) y  y  m(x  x ) 1 1 y  4  5(x  5) y  y1  m(x  x1 ) y  6  2(x  0) y  6  2x y  y1  m(x  x1 ) y  1  0 [x  (3) ] y10 y  y  m(x  x ) 1 1

12.

2

y  0  3 [x  (1) ] 2

y  3 (x  1) 6.

y  y  m(x  x ) 1 1 3 y  (4)  4 [ x  (2) ]

13.

3

y  4  4 (x  2) 7.

8.

y  3  2(x  4) y  3  2x  8 y  3  3  2x  8  3 y  2x  11 y  2x  2x  11  2x 2x  y  11 (1)(2x  y)  (1) (11) 2x  y  11

14.

15.

1

y  3  2 (x  6)

1 12

y  1  1.5(x  3) 2(y  1)  2(1.5) (x  3) 2y  2  3(x  3) 2y  2  3x  9 2y  2  2  3x  9  2 2y  3x  11 2y  3x  3x  11  3x 3x  2y  11 (1)(3x  2y)  (1)11 3x  2y  11 y  6  3.8(x  2) 5(y  6)  5(3.8)(x  2) 5y  30  19(x  2) 5y  30  19x  38 5y  30  30  19x  38  30 5y  19x  8 5y  19x  19x  8  19x 19x  5y  8 y  1  2(x  5) y  1  2x  10 y  1  1  2x  10  1 y  2x  9 y  3  4(x  1) y  3  4x  4 y  3  3  4x  4  3 y  4x  7 y  6  4(x  2) y  6  4x  8 y  6  6  4x  8  6 y  4x  14 4

y  1  5 (x  5)

16.

2(y  3)  2 2 (x  6)

4

y  1  5x  4

2y  6  1(x  6) 2y  6  x  6 2y  6  6  x  6  6 2y  x  12 2y  x  x  12  x x  2y  12 9.

y4 3(y  4) 

2 3 (x 2 3 3

4

y  1  1  5x  4  1 4

y  5x  3

 5)

3

3

7

y y

2

1 4 1 4

2 x 3 2 x 3 2 x 3

1 4



 

y

143 2 (x  6)

1

1

1

y43 x2

18.

Page 832

3y  6  4(x  6) 3y  6  4x  24 3y  6  6  4x  24  6 3y  4x  30 3y  4x  4x  30  4x 4x  3y  30 (1)(4x  3y)  (1)(30) 4x  3y  30

Extra Practice

3

y  4x  2

y  2  3 (x  6) 3(y  2)  3

3

y  2  2  4x  2  2

3y  12  2(x  5) 3y  12  2x  10 3y  12  12  2x  10  12 3y  2x  22 3y  2x  2x  22  2x 2x  3y  22 10.

3

y  2  4x  2

1 2 (x  5)

4

3

y  2  4 (x  2)

17.

  

2

1 3 1 1 4 3 1 12

Lesson 5-6

1. The line parallel to y  4x  2 has the same slope, 4. y  y1  m(x  x1 ) y  6  4(x  1) y  6  4x  4 y  6  6  4x  4  6 y  4x  2

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2. The line parallel to y  2x  7 has the same slope, 2. y  y1  m(x  x1 ) y  6  2(x  4) y  6  2x  8 y  6  6  2x  8  6 y  2x  2

8. Find the slope of the given line. 6x  y  4 6x  y  6x  4  6x y  6x  4 The slope of the line perpendicular to this line is 1 the opposite reciprocal of 6 or 6. y  y1  m(x  x1 )

2

3. The line parallel to y  3x  1 has the same 2 slope, 3. y  y1  m(x  x1 )

1

y  3  6 [ x  (2) ] 1

y  3  6 (x  2) 1

1

1

10 3

2

y  3  3  6x  3  3 y  6x 

2

y  3x  2

3

9. The slope of the given line is 4. So, the slope of the line perpendicular to this line is the opposite 3 4 reciprocal of 4 or 3.

4. The line parallel to y  3x  7 has the same slope, 3. y  y1  m(x  x1 ) y  (2)  3(x  5) y  2  3x  15 y  2  2  3x  15  2 y  3x  13 5. Solve the given equation for y. 3x  8y  4 3x  8y  3x  4  3x 8y  3x  4 8y 8



y

y  y1  m(x  x1 ) 4 y  0  3 (x  0) 4

y  3x 10. Find the slope of the given line. 4x  3y  2 4x  3y  4x  2  4x 3y  4x  2

3x  4 8 3 1 8x  2

3y 3 1

3

y  0  4 (x  4)

4

3



y

y  4 (x  4) 3

y  4x  3 11. Find the slope of the given line. 3x  5y  1 3x  5y  3x  1  3x 5y  3x  1

x  7 5 1 7 x 5 5 1

5y 5

7

The line parallel to y  5x  5 has the same 1 slope, 5. y  y1  m(x  x1 )



y

3x  1 5 3 1 x 5 5

The slope of the line perpendicular to this line is 3 5 the opposite reciprocal of 5 or 3.

1

y  3  5 (x  2) 1

2

y  y  m(x  x ) 1 1

1

2

y  7  3 (x  6)

1 x 5

13 5

y  3  5x  5

5

y  3  3  5x  5  3 y

4x  2 3 4 2 x3 3

The slope of the line perpendicular to this line is 4 3 the opposite reciprocal of 3 or 4. y  y  m(x  x ) 1 1

6. Solve the given equation for y. x  5y  7 x  5y  x  7  x 5y  x  7 5y 5



y 3

The line parallel to y  8x  2 has the same 3 slope, 8. The point (0, 4) lies on the y-axis. The y-intercept is 4. y  mx  b y

1

y  3  6x  3

y  3 (x  3)

3 8x

1

2

y  0  3 [x  (3) ]



5

y  7  3  10 5

y  7  7  3x  10  7

3 5.

7. The slope of the given line is So, the slope of the line perpendicular to this line is the opposite 3 5 reciprocal of 5 or 3. The point (0, 1) lies on the y-axis, so the y-intercept is 1. y  mx  b

5

y  3x  17

5

y  3x  (1) 5

y  3x  1

685

Extra Practice

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2. d  (3) 6 d3 6 d33 6 d 6 The solution

12. Find the slope of the given line. 8x  4y  15 8x  4y  8x  15  8x 4y  8x  15 4y 4



y

8x  15 4 15 2x  4

0

The slope of the line perpendicular to this line is 1 the opposite reciprocal of 2 or 2.

3.

y  y1  m(x  x1 ) 1 y  (1)  2 (x  5) 1

5

1 x 2 1 x 2

5 2 7 2

y  1  2x  2 y11 y

Page 833

 

0

Lesson 5-7

1. The graph shows a negative correlation. As the age of the car increases, the value decreases. 2. The graph shows a positive correlation. As the decade increases, the number of students increases. 3. The graph shows no correlation. 4. The graph shows a positive correlation. As the year increases, the number of metric tons of fish caught in China increases. 5. Find the slope.

6.

38  24

m  1998  1994 m

14 4

7

or 2 1

7.

1

7

y  24  2 (x  1994) 7

y  24  2x  6979 7

y  24  24  2x  6979  24 7

y  2x  6955 7

6. Use the equation y  2x  6955. y y

7 x  6955 2 7 (2005)  2

6955

y  62.5 There should be about 62.5 millions of metric tons of fish caught in China in 2005.

Lesson 6-1

Exercises 1–16 For checks, see students’ work. 1. c93 c9939 c  6 The solution set is {c|c  6}. 6

Extra Practice

4

2

10

16

20

24

4

2

0

4

2

0

2x 7 x  3 2x  x 7 x  3  x x 7 3 The solution set is {x|x 7 3}. 2

0

2

4

2x  3  x 2x  3  x  x  x x30 x3303 x3 The solution set is {x|x  3}. 0

2

4

6

16  w 6 20 16  w  16 6 20  16 w 6 36 The solution set is {w|w 6 36}. 40 32 24 16 8

9.

Page 833

12

6

8

2

8.

8

2 2 2  7 9 set is {h|h 7 9}. 6

8

4

Use the point-slope form. y  y  m(x  x )

6

11 7 d  4 11  4 7 d  4  4 7 7 d 7 7 d is the same as d 6 7. The solution set is {d|d 6 7}. 10

 x1

2

8

4

10

5.

y2  y1

mx

4

2

z  4 7 20 z  4  4 7 20  4 z 7 24 The solution set is {z|z 7 24}.

4. h  (7) 7 h7 7 h77 7 h 7 The solution

1

13 13 13  3 10 set is {d|d 6 10}.

2

0

686

0

14p 7 5  13p 14p  13p 7 5  13p  13p p 7 5 The solution set is {p|p 7 5}. 0

2

4

6

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10.

7 6 16  z 7  z 6 16  z  z 7  z 6 16 7  z  7 6 16  7 z 6 23 The solution set is {z|z 6 23}. 0

11.

8

4

12

16

20

12. 1 t 2

11 12



2

0

1 1 t4 2 1 1  2t 4 1 4 1 2 3 4 11 12

 

3 t 2 3 t 2

4 2 3 2 3

 

6

5j 5

2

t33

0

1

11

2

3

2

1

0

11 12

6.

3.

1

0

1

3

4.

4

2

0

2

1

0

2

4

6

p 5 p (5) 5

8f 8

7 12 7 (3) (12)

6 8 6 (5)8

48 8

7

f 7 6 The solution set is { f|f 7 6}. 6. 0.25t  10 0.25t 0.25

2

10

 0.25

t  40 The solution set is {t|t  40}. 7.

g 8 g (8) 8

6 4 7 (8)4

g 7 32 The solution set is { g|g 7 32}. 8. 4.3x 6 2.58

4

5z  6 7 4z 5z  6  4z 7 4z  4z z6 7 0 z66 7 06 z 7 6 The solution set is {z|z 7 6}. 2

w 3 w (3) 3

p 6 40 The solution set is { p|p 6 40}. 5. 8f 6 48

a  2.3  7.8 a  2.3  2.3  7.8  2.3 a  5.5 The solution set is {a|a  5.5}. 6

60 5

7

w 7 36 The solution set is {w|w 7 36}.

2  9n  10n 2  9n  9n  10n  9n 2  n 2  n is the same as n  2. The solution set is {n|n  2}. 2

49 7

j 7 12 The solution set is { j|j 7 12}.

t

1



b  7 The solution set is {b|b  7}. 2. 5j 6 60

9x 6 8x  2 9x  8x 6 8x  2  8x x 6 2 The solution set is {x|x 6 2}.

3

16.

7b 7

2

5

4

15.

8

t3 2

Lesson 6-2

Exercises 1–16 For checks, see students’ work. 1. 7b  49

1

 t is the same as t  12.

2

Page 833

 2t

The solution set is t|t 

14.

24

1.1v  1 7 2.1v  3 1.1v  1  1.1v 7 2.1v  3  1.1v 1 7 v  3 1  3 7 v  3  3 2 7 v 2 7 v is the same as v 6 2. The solution set is {v|v 6 2}. 2

13.

17. Let n  the number. n  (6) 7 9 n6 7 9 n66 7 96 n 7 15 The solution set is {n|n 7 15}. 18. Let n  the number. 5n 6 6n  12 5n  6n 6 6n  12  6n n 6 12 The solution set is {n|n 6 12}.

4.3x 4.3

7

2.58 4.3

x 7 0.6 The solution set is {x|x 7 0.6}. 9. 4c  6 4c 4

8



6 4

c  1.5 The solution set is {c|c  1.5}.

687

Extra Practice

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10.

6  0.8n 6 0.8

Page 834

0.8n 0.8



7.5  n 7.5  n is the same as n  7.5. The solution set is {n|n  7.5}. 11.

2 m 3 3 2 m 2 3

12

3y 3

 22

132 2 (22)



7s 7

0.05a 0.05

6

28 15 28 or 15

7

a 7

5

13

3.

7

9x 6 42

197 2 179x2 7 197 242

6



y

32 0.375 256 or 3

6v 6

5

1

6

The solution set is y|y  853 . 16. 7y  91 7y 7

 10 

153 2 (10) 50

2

n   3 or 163

5

2

6

100

400 3

1

or 1333

5

1

6

The solution set is n|n  1333 .

Extra Practice

30 6

7

18 2

7

12 4

t 7 3 The solution set is {t|t 7 3}.

 0.75

n



4t 4

The solution set is n|n  163 . 19. Let n  the number. 0.75n  100 0.75n 0.75

15 5

k 7 9 The solution set is {k|k 7 9}. 6. 2x  1 6 16  x 2x  1  2x 6 16  x  2x 1 6 16  x 1  16 6 16  x  16 15 6 x 15 6 x is the same as x 7 15. The solution set is {x|x 7 15}. 7. 15t  4 7 11t  16 15t  4  11t 7 11t  16  11t 4t  4 7 16 4t  4  4 7 16  4 4t 7 12

91

y  13 The solution set is {y|y  13}. 17. Let n  the number. 1n 7 7 (1)(n) 6 (1)(7) n 6 7 The solution set is {n|n 6 7}. 18. Let n  the number.

12

6

2k 2

 7

3 n 5 5 3 n 3 5

6

v5 The solution set is {v|v  5}. 5. 2k  12 6 30 2k  12  12 6 30  12 2k 6 18

1

853

4

e 6 3 The solution set is {e|e 6 3}. 4. 6v  3  33 6v  3  3  33  3 6v  30

x 7 54 The solution set is {x|x 7 54}. 15. 0.375y  32 0.375y 0.375

5

5e  9 7 24 5e  9  9 7 24  9 5e 7 15 5e 5

The solution set is a|a 7 115 . 14.

4

or 37

The solution set is s|s 6 37 .

13

115

25 7 25 7

6

s 6

500 6 a 500 6 a is the same as a 7 500. The solution set is {a|a 7 500}. 13. 15a 6 28 15a 15

33 3

7

y 7 11 The solution set is {y|y 7 11}. 2. 7s  12 6 13 7s  12  12 6 13  12 7s 6 25

m  33 The solution set is {m|m  33}. 12. 25 7 0.05a 25 0.05

Lesson 6-3

Exercises 1–20 For checks, see students’ work. 1. 3y  4 7 37 3y  4  4 7 37  4 3y 7 33

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8.

13  y  29  2y 13  y  y  29  2y  y 13  29  3y 13  29  29  3y  29 16  3y 16 3 16 3 16 3

The 9.



14.

5

2q 2



15 5

6

18 6

4 2



7

5w 5

6t 6

6 (3)(9)

11 4 3 24 3 24

32 2

12 6



8k 8

6

4y 4

6 y 3

6 y is the same as y 7 24 or 2.75.

5

3

6

The solution set is y|y 7 24 .

t 6 16 The solution set is {t|t 6 16}.

19.

 7  5

 7  7  5  7 z 4 z (4) 4

7

1  k 1  k is the same as k  1. The solution set is {k|k  1}. 17. 9m  7 6 2(4m  1) 9m  7 6 8m  2 9m  7  8m 6 8m  2  8m m  7 6 2 m  7  7 6 2  7 m 6 9 The solution set is {m|m 6 9}. 18. 3(3y  1) 6 13y  8 9y  3 6 13y  8 9y  3  9y 6 13y  8  9y 3 6 4y  8 3  8 6 4y  8  8 11 6 4y

6 9

6

r 7 8 set is {r|r 7 8}. 7 c  17 7 c  17 7 c  17 7 c  17  c 7 17 7 17  5 7 12

8 8

2t  5 6 27 2t  5  5 6 27  5 2t 6 32

z 4

48 6

c 7 2 The solution set is {c|c 7 2}. 16. 5(k  4)  3(k  4) 5k  20  3k  12 5k  20  5k  3k  12  5k 20  8k  12 20  12  8k  12  12 8  8k

3 7 t 3 7 t is the same as t 6 3. The solution set is {t|t 6 3}.

z 4

7

6c 6

3w 3  w is the same as w  3. The solution set is {w|w  3}. 11. 4t  5 7 2t  13 4t  5  4t 7 2t  13  4t 5 7 6t  13 5  13 7 6t  13  13 18 7 6t

13.

6r 6

The solution 15. 8c  (c  5) 8c  c  5 7c  5 7c  5  c 6c  5 6c  5  5 6c

1 53.

q  2 The solution set is {q|q  2}. 10. 2(w  4)  7(w  1) 2w  8  7w  7 2w  8  2w  7w  7  2w 8  5w  7 8  7  5w  7  7 15  5w

2t 2

7r  37 7r  37  7r 37 37  11 48

y

16  y is the same as y   3 or 1 solution set is y|y  53 .

2t  5 3 2t  5 (3) 3

7 7 7 7 7

3y 3

5q  7  3(q  1) 5q  7  3q  3 5q  7  3q  3q  3  3q 2q  7  3 2q  7  7  3  7 2q  4

12.

13r  11 13r  11  7r 6r  11 6r  11  11 6r

5x  10(3x  4) 5x  30x  40 5x  30x  30x  40  30x 25x  40 25x 25

 12

40

 25 8

3

x  5 or 15 or 1.6

 (4)(12)

5

3

6

The solution set is x|x  15 .

z  48 The solution set is {z|z  48}.

689

Extra Practice

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1

2

2

4.

3a3 a1

20.

3a  2  a  1 3a  2  a  a  1  a 2a  2  1 2a  2  2  1  2 2a  3 2a 2



a

3 2 3 2

4v  1  3v 4v  1  4v  3v  4v 1  v (1)(1)  (1) (v) 1v

and

The solution set is the intersection of the two graphs. Graph 2  v or v  2.

1

or 12

5

2v  2  3v 2v  2  2v  3v  2v 2  v

1

6

–2

0

2

4

6

–2

0

2

4

6

–2

0

2

4

6

Graph 1  v or v  1.

The solution set is a|a  12 .

Find the intersection.

Page 834 1.

or 2x 7 5 2  x 6 5 2x2 7 52 2  x  2 6 5  2 x 6 7 x 7 3 The solution set is the union of the two graphs. –8

–6

–8

–4

–6

–8

2.

Lesson 6-4

–2

–4

–6

0

–2

–4

2

0

–2

0

16  4b 16 4

–2

0

–2

2

0

–2

4

2

0

6

4

2

8

6

4

10

8

6

10

8

10

12

12

Graph t  11.

12

Find the union.



2g 2

2g 2

0

2

4

6

–4

–2

0

2

4

6

–4

–2

0

2

4

6

–2

0

2

4

6

–6

–4

–2

0

2

4

6

–6

–4

–2

0

2

4

6

9 6 2z  7 9  7 6 2z  7  7 16 6 2z 16 2

4  b or b  4.

Find the intersection.

6

2z  7 6 10 2z  7  7 6 10  7 2z 6 3

and

2z 2

2z 2

8 6 z

6

3 2

z 6 1.5

The solution set is the intersection of the two graphs.

8

2

–8

–6

–4

–2

0

2

–8

–6

–4

–2

0

2

–8

–6

–4

–2

0

2

Graph 8 6 z or z 7 8. Graph z 6 1.5.

2  g or g  2. Graph g  4.

Find the intersection.

The solution set is {z|8 6 z 6 1.5}.

Find the intersection.

The solution set is {g|2  g  4}.

Extra Practice

–4

The solution set is {b|4  b  4}. 6. First express 9 6 2z  7 6 10 using and.

Graph

–2

–6

Graph b  4.

2  g g4 The solution set is the intersection of the two graphs. –4

b4

4b 4

Graph

The solution set is {t|t is a real number.}. and 2g  7  15 3  2g  7 3. 2g  7  7  15  7 3  7  2g  7  7 4  2g 2g  8 4 2

32 8

The solution set is the intersection of the two graphs.

The solution set is {x|x 6 7 or x 7 3}. or 4  t 6 7 4  t 7 5 4  t  4 6 7  4 4  t  4 7 5  4 t 7 1 t 6 11 The solution set is the union of the two graphs. Graph t  1.





4  b

Find the union.

4

8b 8

4  12  4b  12  12

Graph x  3.

4

2

3b  4  7b  12 and 8b  7  25 3b  4  3b  7b  12  3b 8b  7  7  25  7 4  4b  12 8b  32

Graph x  7.

4

2

The solution set is {v|v  1}. 5.

690

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5m  8  10  m

7.

5m  11 6 9

or

6m  8  10

5m 5

6m  18 

6

20 5

4p  6 6 8

4p  8  6 4p  8  8  6  8

4p 4

m3

Graph m  3. 2

4

–4

–2

0

2

4

6

2 4

p 6

1 2

4p 4

1

1

–2 –1 0 1

Graph p 6 2.

2

1

–2 –1 0 1

2

–2 –1 0 1

2

Graph p  2.

Find the union. –2

0

2

Find the intersection.

4

The solution set is {m|m 6 4 or m  3}. 8.



25 5

14 7

c2



11.

5c 5

7 7 7 7 7

16  2r and 16  2r  2r 16 16  8 8

4r 4

7

8 4

5  c

–6

–4

–2

–2

0

0

2

4

–4

–2

0

2

4

The solution set is {c|c is a real number.}. 9. First express 2h  2  3h  4h  1 using and. and 2h  2  3h 2h  2  2h  3h  2h 2  h

7r  21 7r  21  r 6r  21 6r  21  21 6r

6 6 6 6 6

r9 r9r 9 9  21 30

6r 6

6

30 6

r 6 5

Graph r 7 2.

Graph 5  c or c  5. Find the union.

–6

6.

r 7 2

4

2

1 2

The solution set is the intersection of the two graphs.

Graph c  2.

–4

1

2r  8 2r  8  2r 4r  8 4r  8  8 4r

The solution set is the union of the two graphs. –6

5

The solution set is p|2  p 6

12c  4  5c  10 4c  1  c  24 or 12c  4  5c  5c  10  5c 4c  1  4c  c  24  4c 7c  4  10 1  5c  24 7c  4  4  10  4 1  24  5c  24  24 7c  14 25  5c 7c 7

2 4

p  2

Graph m 6 4.

–4



The solution set is the intersection of the two graphs.

The solution set is the union of the two graphs. 0

4p  2

4p 6 2

18 6

–2

5p  8  p  6 5p  8  p  p  6  p

4p  6  6 6 8  6

m 6 4

–4

and

3p  6 6 8  p 3p  6  p 6 8  p  p

5m 6 20

6m  8  8  10  8

6m 6

10.

5m  11  11 6 9  11

5m  8  m  10  m  m

–6

–4

–2

0

2

–6

–4

–2

0

2

–6

–4

–2

0

2

Graph r 6 5. Find the intersection.

Since the graphs do not intersect, the solution set is the empty set .

3h  4h  1 3h  4h  4h  1  4h h  1 (1) (h)  (1) (1) h1

12.

The solution set is the intersection of the two graphs.

4j  3 4j  3  4j 3 3  22 19

6 6 6 6 6

j  22 and j  22  4j 5j  22 5j 5j

19 5

6

5j 5

Graph 2  h or h  2.

j3 j3j 3 3  15 12

6 6 6 6 6

2j  15 2j  15  j j  15 j  15  15 j

3.8 6 j

–2 –1 0 1 2 3

The solution set is the intersection of the two graphs.

Graph h  1. –2 –1 0 1 2 3

Graph

Find the intersection.

–4

0

4

8

12

–4

0

4

8

12

–4

0

4

8

12

–2 –1 0 1 2 3

The solution set is {h|h  1}.

3.8 6 j or j 7 3.8. Graph 12 6 j or j 7 12. Find the intersection.

The solution set is { j|j 7 12}.

691

Extra Practice

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13.

2(q  4)  3(q  2)

16. First express (2x  5)  x  5  2x  9 using and.

q84q

or

q8q4qq

2q  8  3q  6

2q  8  4

2q  8  2q  3q  6  2q

(2x  5)  x  5

2q  8  8  4  8

8  q  6 8  6  q  6  6

2q  12 2q 2

14  q



10  3x 10 3



3x 3

1

33  x

4 8

The solution set is the intersection of the two graphs.

Graph q  6. 0

14  x

5  5  3x  5  5

Graph 14  q or q  14.

–16 –12 –8 –4

5x9 59x99

5  3x  5

The solution set is the union of the two graphs. 0

x  5  x  2x  9  x

2x  5  2x  x  5  2x

12 2

q6

–16 –12 –8 –4

x  5  2x  9

and

2x  5  x  5

4 8

Graph

Find the union. –16 –12 –8 –4

0

1

1

–4

0

4

8

12

16

–4

0

4

8

12

16

14  x or x  14.

–4

0

4

8

12

16

Find the intersection.

33  x or x  33 .

4 8

Graph

The solution set is {q|q is a real number.}. 1 w 2

14. First express and. 1 w 2 1 w 2

5w2

5w2 1

1

1

1

The solution set is {x|x  14}.

1

w  2  2w  2w  9  2w

 5  2w  w  2  2w 1

1 w 2

5  2w  2 1

1 w 2

5  2  2w  2  2 1

29

1 (2) 2 w

1 (2) 2 w

6w

Page 834

2292 1 w 2

3  2w (2)3 

 9 using

w  2  2w  9

and 1

1 w 2

7  (2)7

w  14

The solution set is the intersection of the two graphs. 5

6 7 8

9 10 11 12 13 14 15

Graph 6  w or w  6.

5

6 7 8

9 10 11 12 13 14 15

Graph w  14.

5

6 7 8

9 10 11 12 13 14 15

Find the intersection.

12 8 4

or

n  6  n 7 10 2n  6 7 10

3n 3

7

8

12

16

20

24

28

30

0

2

4

The solution set is { g|g 6 14 or g 7 2}. 3. Write |t  5|  3 as t  5  3 and t  5  3. Case 1: Case 2: t53 t  5  3 t5535 t  5  5  3  5 t8 t2

21 3

n 6 7

2n 7 16 2n 2

6

4

16 14 12 10 8 6 4 2

3n  1 7 20 3n  1  1 7 20  1 3n 7 21

2n  6  6 7 10  6

0

The solution set is { y|10 6 y 6 28}. 2. Write |g  6| 7 8 as g  6 7 8 or g  6 6 8. Case 1: Case 2: g6 7 8 g  6 6 8 g66 7 86 g  6  6 6 8  6 g 7 2 g 6 14

Since the graphs do not intersect, the solution set is the empty set . 15. n  (6  n) 7 10

Lesson 6-5

1. Write |y  9| 6 19 as y  9 6 19 and y  9 7 19. Case 1: Case 2: y  9 6 19 y  9 7 19 y  9  9 6 19  9 y  9  9 7 19  9 y 6 28 y 7 10

16 2

n 7 8

The solution set is the union of the two graphs.

0

–4

0

4

8

–8

–4

0

4

8

–8

–4

0

4

8

4

6

8

10

The solution set is {t|2  t  8}. 4. Write |a  5|  0 as a  5  0 or a  5  0. Case 1: Case 2: a50 a50 a5505 a5505 a  5 a  5

Graph n 7 8. –8

2

Graph n 6 7. Find the union.

3

The solution set is {n|n 6 7 or n 7 8}.

2

1

0

1

2

3

The solution set is {a|a is a real number.}. Extra Practice

692

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5. Write |14  2z|  16 as 14  2z  16 or 14  2z  16. Case 1: Case 2: 14  2z  16 14  2z  14  16  14

14  2z  16 14  2z  14  16  14 2z  30

2z  2 2z 2

2z 2

2

 2

z  1 0

4

10. Write |3p  5|  23 as 3p  5  23 and 3p  5  23. Case 1: Case 2: 3p  5  23 3p  5  23 3p  5  5  23  5 3p  5  5  23  5 3p  18 3p  28



30 2

3p 3

12

16

20

10

The solution set is {1, 15}. 6. |a  5|  3 means that the distance between a and 5 is 3 units. Since distance cannot be negative, there is no real number that makes this statement true. The solution set is the empty set . 3

2

1

0

1

2

7

2m 2

6

8 6 4 2

0

2

4

6

8

8 4

0

4

8

20 24

5

12

y1 1

0

1

2



y

5

3

The solution set is 1,

1 45

6.

4

21 5 21 or 5

2

4

6

6

p6 .

6b  12  36 6b  12  12  36  12 6b  24

48 6

6b 6



24 6

b  4

0

3x 3

0

2

2

x 7

20 3 2 63

4

6

7

The solution set is

4

6

8

5

3x 3

6

30 3

x 6 10 8 2 x|63

10

12

6

6 x 6 10 .

13. Write |7  8x| 7 39 as 7  8x 7 39 or 7  8x 6 39. Case 1: Case 2: 7  8x 6 39 7  8x 7 39 7  8x  7 6 39  7 7  8x  7 7 39  7 8x 7 32 8x 6 46

28 32 36

5y 5

 5

0

1 p|93

28 3 1 93

The solution set is {b| 4  b  8}. 12. Write |25  3x| 6 5 as 25  3x 6 5 and 25  3x 7 5. Case 1: Case 2: 25  3x 6 5 25  3x 7 5 25  3x  25 6 5  25 25  3x  25 7 5  25 3x 6 20 3x 7 30

The solution set is {w|w  6 or w  34}. 9. Write |13  5y|  8 as 13  5y  8 or 13  5y  8. Case 1: Case 2: 13  5y  8 13  5y  8 13  5y  13  8  13 13  5y  13  8  13 5y  5 5y  21 5y 5



6 4 2

14  w  20 14  w  14  20  14 w  34 (1) (w)  (1) (34) w  34

12 16

5

2

b8

The solution set is {m|m 6 4 or m 7 9}. 8. Write |14  w|  20 as 14  w  20 or 14  w  20. Case 1: Case 2: 14  w  20 14  w  14  20  14 w  6 (1) (w)  (1) (6) w  6

4

6

6b 6

8 2

10

p

6b  12  36 6b  12  12  36  12 6b  48

m 6 4

m 7 9

8



11. Write |6b  12|  36 as 6b  12  36 and 6b  12  36. Case 1: Case 2:

3

18 2

3p 3

The solution set is

7. Write |2m  5| 7 13 as 2m  5 7 13 or 2m  5 6 13. Case 1: Case 2: 2m  5 7 13 2m  5 6 13 2m  5  5 7 13  5 2m  5  5 6 13  5 2m 7 18 2m 6 8 2m 2

18 3

p6

z  15

8



8x 8

7

32 8

8x 8

46 8

x 6 5.75

x 7 4 6 4 2

6

0

2

4

6

The solution set is {x|x 6 5.75 or x 7 4} .

1

45

5

693

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14. Write |4c  5|  25 as 4c  5  25 or 4c  5  25. Case 1: Case 2: 4c  5  25 4c  5  25 4c  5  5  25  5 4c  5  5  25  5 4c  20 4c  30 4c 4



20 4

4c 4

c5



18. Write

Case 1:

0

2

4

6

30 4

2b 2

6

5s 5

42 5

s 6 8.4 8 6 4 2

7

2

4

6

0

4

0

4

2

17. Write

0

Case 1: 2n  1 3 2n  1 (3) 3

0  10 as

2n  1 3 2n  1 (3) 3

 10  (3)10



2n 2

8

4

0

4

8

3

4

5

6

2

4

6

8

10

12

2b 2

7

 2

6 4 2

0



13 2

b  6.5 2

4

6

8

The solution set is {b|b  3.5 or b  6.5}.

 10  (3)(10)

Page 835



12

Lesson 6-6

1. Use a table to substitute the x and y values of each ordered pair into the inequality.

29 2

n  14.5

n  15.5 16 12

 10.

2n  1  30 2n  1  1  30  1 2n  29

31 2

13 2

b  6.5

b  3.5

2n  1 3

Case 2:

2n  1  30 2n  1  1  30  1 2n  31 2n 2

 10 or

0

2b 2

8

2n  1 3



The solution set is {x| 2  x  12}. 20. |3  (2b  6)|  10 |3  2b  6|  10 |3  2b|  10 Write |3  2b|  10 as 3  2b  10 or 3  2b  10. Case 1: Case 2: 3  2b  10 3  2b  10 3  2b  3  10  3 3  2b  3  10  3 2b  7 2b  13

The solution set is {x|x  13 or x  7}. 2n  1 3

2b 2

10

The solution set is {s|s 6 8.4 or s 7 10}. 16. |4  (1  x)|  10 |4  1  x|  10 |3  x|  10 Write |3  x|  10 as 3  x  10 or 3  x  10. Case 1: Case 2: 3  x  10 3  x  10 3  x  3  10  3 3  x  3  10  3 x7 x  13 16 12 8

1

2

 (2) (3)

7  2b  6 7  2b  7  6  7 2b  13

 2

1

 3

The solution set is {b|0.5  b  6.5}. 19. |2  (x  3)|  7 |x  5|  7 Write |x  5|  7 as x  5  7 and x  5  7. Case 1: Case 2: x57 x  5  7 x5575 x  5  5  7  5 x  12 x  2

50 5

8

 (2)3

b  0.5

s 7 10 0

7  2b 2 7  2b (2) 2

3

7  2b  6 7  2b  7  6  7 2b  1

The solution set is {c|c  7.5 or c  5}. 15. Write |4  5s| 7 46 as 4  5s 7 46 or 4  5s 6 46. Case 1: Case 2: 4  5s 7 46 4  5s 6 46 4  5s  4 7 46  4 4  5s  4 6 46  4 5s 7 42 5s 6 50 5s 5

Case 2:

7  2b 2 7  2b (2) 2

c  7.5

8 6 4 2

0 7 2 2b 0  3 as 7 2 2b  3 and 7 2 2b  3.

16

The solution set is {14.5, 15.5}.

x 0

y 0

1

3

2

2

3

3

xy0 000 00 1  (3)  0 2  0 220 40 3  (3)  0 00

True or False true false true true

The ordered pairs {(0, 0), (2, 2), (3, 3)} are part of the solution set.

Extra Practice

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2. Use a table to substitute the x and y values of each ordered pair into the inequality. x 0

y 0

1

1

3

2

8

0

2x  y  8 2(0)  0  8 08 2(1)  (1)  8 3  8 2(3)  (2)  8 48 2(8)  0  8 16  8

6. Since the boundary is not included in the solution, draw a dashed line. Test the point (0, 0). x 6 4 0 6 4 true The half plane that contains (0, 0) should be shaded.

True or False true true

y

true

x4

false

x 0 2 3 2

y 0 0 4 1

y 0 0 4 1

x 7 0 7 2 7 3 7 2

True or False false false true false

7. Since the boundary is not included in the solution, draw a dashed line. Test the point (0, 0). x  y 6 2 0  0 6 2 0 6 2 false The half plane that does not contain (0, 0) should be shaded.

The ordered pair {(3, 4)} is part of the solution set. 4. Use a table to substitute the x and y values of each ordered pair into the inequality. x 0

y 0

3

2

4

5

0

6

3x  2y  1 3(0)  2(0) 6 0 6 3(3)  2(2) 6 5 6 3(4)  2(5) 6 2 6 3(0)  2(6) 6 12 6

y

True or False 1 1 1 1 1 1 1 1

true x

O

false x y  2

true true

8. Since the boundary is included in the solution, draw a solid line. Test the point (0, 0). x  y  4 0  0  4 0  4 true The half plane that contains (0, 0) should be shaded.

The ordered pairs {(0, 0), (4, 5), (0, 6)} are part of the solution set. 5. Since the boundary line is included in the solution, draw a solid line. Test the point (0, 0). y  2 0  2 false The half plane that does not contain (0, 0) should be shaded.

y x y  4

y

O

x

O

The ordered pairs {(0, 0), (1, 1), (3, 2)} are part of the solution set. 3. Use a table to substitute the x and y values of each ordered pair into the inequality.

O

y  2

x

x

695

Extra Practice

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12. Since the boundary is not included in the solution, draw a dashed line. Test the point (1, 3). x 6 y 1 6 3 true The half plane that contains (1, 3) should be shaded.

9. Since the boundary is not included in the solution, draw a dashed line. Test the point (0, 0). y 7 4x  1 0 7 4(0)  1 0 7 1 true The half plane that contains (0, 0) should be shaded.

y

y y  4x  1

O

x

xy

x

O

13. Since the boundary is included in the solution, draw a solid line. Test the point (0, 0). 3x  y  4 3(0)  0  4 0  4 true The half plane that contains (0, 0) should be shaded.

10. Since the boundary is not included in the solution, draw a dashed line. Test the point (0, 0). 3x  y 7 1 3(0)  0 7 1 0 7 1 false The half plane that does not contain (0, 0) should be shaded.

y

y 3x  y  4

3x y  1

x

O

x

O

14. Since the boundary is not included in the solution, draw a dashed line. Test the point (0, 0). 5x  y 6 5 5(0)  0 6 5 0 6 5 true The half plane that contains (0, 0) should be shaded.

11. Since the boundary is included in the solution, draw a solid line. Test the point (0, 0). 3y  2x  2 3(0)  2(0)  2 0  2 true The half plane that contains (0, 0) should be shaded.

y

y O 5x  y  5 O

x

3y  2x  2

Extra Practice

696

x

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18. Since the boundary is included in the solution, draw a solid line. Test the point (0, 0). 3x  8y  4 3(0)  8(0)  4 0  4 true The half plane that contains (0, 0) should be shaded.

15. Since the boundary is included in the solution, draw a solid line. Test the point (0, 0). 2x  6y  12 2(0)  6(0)  12 0  12 false The half plane that does not contain (0, 0) should be shaded. y

y

2x  6y  12

x

O

x

O

3x  8y  4

19. Since the boundary is included in the solution, draw a solid line. Test the point (0, 0). 5x  2y  6 5(0)  2(0)  6 0  6 false The half plane that does not contain (0, 0) should be shaded.

16. Since the boundary is included in the solution, draw a solid line. Test the point (0, 0). x  3y  9 (0)  3(0)  9 0  9 true The half plane that contains (0, 0) should be shaded. y

y

x

O x  3y  9 O

5x  2y  6

x

17. Since the boundary is not included in the solution, draw a dashed line. Test the point (0, 0). y 7 3x  7 0 7 3(0)  7 0 7 7 false The half plane that does not contain (0, 0) should be shaded.

Page 835

Lesson 7-1

1. The graphs appear to intersect at (3, 9). Check in each equation. y  3x 4x  2y  30 Check: ? ? 4(3)  2(9)  30 9  3(3) ? 99✓ 12  18  30 30  30 ✓

y

y

y  3x  7

4 x  2 y  30

12

y  3x

(3, 9)

4 O

x O

4

12

x

There is one solution. It is (3, 9).

697

Extra Practice

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2. The graphs appear to intersect at (2, 1). Check in each equation. x  2y xy1 Check: ? ? 2(1) 2  (1)  1 2 22✓ 11✓ y

x  2 y

5. The graphs of the equations are parallel lines. Since they do not intersect, there are no solutions to this system of equations. y x y 6

x y 1 x

O (2, 1)

x

O 3x  3y  3

6. The graphs appear to intersect at (2, 6). Check in each equation. Check: y  3x 4x  y  2 ? ? 4(2)  (6)  2 6  3(2) ? 6  6 ✓ 862 22✓

There is one solution. It is (2, 1). 3. The graphs appear to intersect at (2, 6). Check in each equation. yx4 3x  2y  18 Check: ? ? 624 3(2)  2(6)  18 ? 66✓ 6  12  18 18  18 ✓

y 4x  y  2

x

O

y (2, 6)

y  3 x

3 x  2 y  18

y x4

There is one solution. It is (2, 6). 7. The graphs appear to intersect at (4, 0). Check in each equation. 2x  y  8 xy4 Check: ? ? 404 2(4)  0  8 88✓ 44✓

x

O

(2, 6)

There is one solution. It is (2, 6). 4. The graphs appear to intersect at (4, 2). Check in each equation. Check: xy6 xy2 ? ? 422 426 66✓ 22✓

y 2x  y  8

y (4, 0)

xy 6

O

x y 4

(4, 2)

x

O

There is one solution. It is (4, 0).

x y 2

There is one solution. It is (4, 2).

Extra Practice

x

698

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8. The graphs appear to intersect at (3, 3). Check in each equation. 1 x 5

Check: 1 (3) 5

12 5 ? 12  (3)  5 ? 12 3 5  3  5 12 12  5 ✓ 5

y

11. The graphs appear to intersect at (2, 2). Check in each equation.

3x  5y  6 ?

2

3(3)  5(3)  6

? 1 (2)  2 ?

?

9  15  6

?

264 22✓

(2, 2)

1 2

x y 3 x

x

O

(3, 3) 1 x  y  12 5

?

2  3(2)  4

3

y

y O

y  3x  4

213 33✓

?

9  (15)  6 66✓

3x  5y  6

1

x  2y  3

Check:

5

y  3x  4

There is one solution. It is (2, 2). 12. The graphs of the equations coincide. Since every point is a point of intersection, there are infinitely many solutions to this system of equations.

There is one solution. It is (3, 3). 9. The graphs appear to intersect at (6, 3). Check in each equation. x  2y  0 y  3  x Check: ? ? 3  3  (6) 6  2(3)  0 ? 66✓ 6  6  0 00✓

y 2 1 x y 2 3 2

4 x  3 y  12

y x

O

(6, 3)

x  2y  0 x

13. The graphs appear to intersect at (3, 1). Check in each equation.

O

y  3  x

1

yx4

Check:

?

1

1  3  4

? 5 2 ? 5 2 5 2✓

3  2 (1)  1

1  1 ✓

There is one solution. It is (6, 3). 10. The graphs appear to intersect at (1, 5). Check in each equation. x  2y  9 xy6 Check: ? ? 1(5)  6 1  2(5)  9 ? ? 156 1  (10)  9 9  9 ✓ 66✓

5

x  2y  2 32 5 2

y

O

(3, 1)

y  x4

x

y 1 2

x y

x

O

x  2 y  9

5 2

There is one solution. It is (3, 1). x y  6

(1, 5)

There is one solution. It is (1, 5).

699

Extra Practice

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2. Since y  7  x, substitute 7  x for y in the second equation. 2x  y  8 2x  (7  x)  8 2x  7  x  8 3x  7  8 3x  7  7  8  7 3x  15

14. The graphs of the equations coincide. Since every point is a point of intersection, there are infinitely many solutions to this system of equations. y 2x  y  3

4x  2y  6

3x 3

15. The graphs appear to intersect at (2, 3). Check in each equation. Check: 1 2 x  3 y  3 12x  y  21 2 ?

1 (2) 2

12(2)  (3)  21 ?

24  3  21 21  21 ✓

2

?

 3 (3)  3 ?

1  (2)  3 3  3 ✓

y 12 x  y  21

6y 6

x

O



y

1 2 x  y  3 2 3

(2, 3)

15 3

x5 Use y  7  x to find the value of y. y7x y75 y2 The solution is (5, 2). 3. Since x  5  y, substitute 5  y for x in the second equation. 3y  3x  1 3y  3(5  y)  1 3y  15  3y  1 3y  16  3y 3y  3y  16  3y  3y 6y  16

x

O



16 6 8 3

Use x  5  y to find the value of x. x5y 8

x53 7

There is one solution. It is (2, 3).

x3 The solution is

Page 835

4. Since y  2  x, substitute y  2 for x in the first equation. 3x  y  6 3(y  2)  y  6 3y  6  y  6 4y  6  6 4y  6  6  6  6 4y  0

Lesson 7-2

1. Since y  x, substitute x for y in the second equation. 5x  12y 5x  12(x) 5x  12x 5x  5x  12x  5x 0  7x 0 7



4y 4

7x 7

0

4

y0 Use y  2  x to find the value of x. y2x 02x 2x The solution is (2, 0).

0x Use y  x to find the value of y. yx y0 The solution is (0, 0).

Extra Practice

173, 83 2.

700

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7. Solve the first equation for x since the coefficient of x is 1. x  2y  10 x  2y  2y  10  2y x  10  2y Since x  10  2y, substitute 10  2y for x in the second equation. x  y  2 (10  2y)  y  2 10  2y  y  2 10  3y  2 10  3y  10  2  10 3y  12

5. Solve the first equation for x since the coefficient of x is 1. x  3y  3 x  3y  3y  3  3y x  3  3y Since x  3  3y, substitute 3  3y for x in the second equation. 2x  9y  11 2(3  3y)  9y  11 6  6y  9y  11 6  15y  11 6  15y  6  11  6 15y  5 15y 15

5

3y 3

 15

x3

1 3

for y in either equation to find

113 2  3

x13 x1131 x4

1 12

The solution is 4, 3 . 6. Solve the second equation for x since the coefficient of x is 1. x  3y  4 x  3y  3y  4  3y x  4  3y Substitute 4  3y for x in the first equation. 3x  18  2y 3(4  3y)  18  2y 12  9y  18  2y 12  9y  9y  18  2y  9y 12  18  11y 12  18  18  11y  18 30  11y 30 11 30 11

6 3

y

Now substitute the value of x. x  3y  4 x3

11y 11



30 11

13011 2  4 90 90

90

x  11  11  4  11 x

46 11

The solution is

1

46 30 11, 11



3x 3

2  x Now substitute 2 for x in either equation to find the value of y. 2y  12  x 2y  12  (2) 2y  12  2 2y  14

for y in either equation to find

x  11  4 90

12 3

y4 Now substitute 4 for y in either equation to find the value of x. x  2y  10 x  2(4)  10 x  8  10 x  8  8  10  8 x2 The solution is (2, 4). 8. Solve the first equation for y since the coefficient of y is 1. 2x  3  y 2x  y  3  y  y 2x  y  3 2x  y  2x  3  2x y  3  2x Since y  3  2x, substitute 3  2x for y in the second equation. 2y  12  x 2(3  2x)  12  x 6  4x  12  x 6  4x  4x  12  x  4x 6  12  3x 6  12  12  3x  12 6  3x

1

y3 Now substitute the value of x. x  3y  3



2y 2



14 2

y7 The solution is (2, 7).

2.

701

Extra Practice

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x  6y  1 x  6(1)  1 x  (6)  1 x61 x6616 x7 The solution is (7, 1). 12. Solve the second equation for y since the coefficient of y is 1.

9. Since y  3x, substitute 3x for y in the first equation. 6y  x  36 6(3x)  x  36 18x  x  36 19x  36 19x 19

36

 19 36

x  19 Use y  3x to find the value of y. y  3x y  3 y

13619 2

108  19

The solution is

3 x 2 3 x 2

3 (10 4 15 2

. 13619, 108 19 2

15 2

1

1 15 2

13

 2 

11213 21132 2

2x 2

y  6 Now substitute 6 for y in either equation to find the value of x. x  y  10 x  (6)  10 x  6  10 x  6  6  10  6 x4 The solution is (4, 6). 11. Solve the first equation for x since the coefficient of x is 1. x  6y  1 x  6y  6y  1  6y x  1  6y Since x  1  6y, substitute 1  6y for x in the second equation. 3x  10y  31 3(1  6y)  10y  31 3  18y  10y  31 3  28y  31 3  28y  3  31  3 28y  28 28y 28

5  3y 2 5 3  y 2 2



x 5

3

Since x  2  2y, substitute second equation. 4x  9y  9 4

5 2

3

 2y for x in the

152  32y2  9y  9

10  6y  9y  9 10  15y  9 10  15y  10  9  10 15y  1 15y 15

1

 15 1

y  15 1 15

Now substitute the value of x. 2x  3y  5 2x  3

for y in either equation to find

1151 2  5 1

2x  5  5 1

1

1

2x  5  5  5  5 24 5 1 1 24 2x  2 5 2 12 x 5 12 1 solution is 5 , 15

12

28

 28

y  1 Now substitute 1 for y in either equation to find the value of x.

Extra Practice

2

3x  3x  6  12 6  12 The statement 6  12 is false. This means that there is no solution to the system of equations. 13. Since no coefficient of x or y is 1 or 1, solve for either variable in either equation. For example, solve for x in the first equation. 2x  3y  5 2x  3y  3y  5  3y 2x  5  3y

 4y  3y  1

1 2

3

13

1

1

y

3x  2 2x  3  12

 y)  3y  1



3y3

3  y, substitute 2x  3 for y in the first equation. 3x  2y  12

1

15 13  12y 2 13 15 y 2 12 13 y 12 12 13 y 13 12

3y

3 x3 2 3 x3 2 3 Since 2x 

 3y  1

3

yy3y 3 x 2

10. Solve the second equation for x since the coefficient of x is 1. x  y  10 x  y  y  10  y x  10  y Since x  10  y, substitute 10  y for x in the first equation. 3 x 4

y3

The

702

2x 

12

1

2.

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2. Since the coefficients of the y terms, 1 and 1, are additive inverses, you can eliminate the y terms by adding the equations. 2x  y  32 () 2x  y  60 4x  92

14. Since x  4  8y, substitute 4  8y for x in the second equation. 3x  24y  12 3(4  8y)  24y  12 12  24y  24y  12 12  12 The statement 12  12 is true. This means that there are infinitely many solutions to the system of equations. 15. Since no coefficient of x or y is 1 or 1, solve for either variable in either equation. For example, solve for x in the first equation. 3x  2y  3 3x  2y  2y  3  2y 3x  2y  3 3x 3



x

4x 4

2

Since x   1, substitute 3y  1 for x in the second equation. 25x  10y  215

12

2

25 3 y  1  10y  215 50 y 3

2x 2

11 2 11 x 2 11 substitute 2

 25  10y  215 80 y 3

80 y 3

 25  215

 25  25  215  25 80 y 3 3 80 y 80 3

1 2

1803 2240

y

y9 Now substitute 9 for y in either equation to find the value of x. 3x  2y  3 3x  2(9)  3 3x  18  3 3x  18  18  3  18 3x  15 3x 3



y

5

The solution is

11 2

1112, 12 2.

4. Since the coefficients of the t terms, 2 and 2, are additive inverses, you can eliminate the t terms by adding the equations. s  2t  6 () 3s  2t  2 4s 8

15 3

4s 4

8

4

s2 Now substitute 2 for s in either equation to find the value of t. s  2t  6 2  2t  6 2  2t  2  6  2 2t  4

1. Since the coefficients of the y terms, 1 and 1, are additive inverses, you can eliminate the y terms by adding the equations. xy7 () x  y  9 2x  16 



5 1

Lesson 7-3

2x 2

11 2

11 2 11 2

for x in either equation to find

y  2

x5 The solution is (5, 9).

Page 836



Now the value of y. yx5

 240 

92 4

x  23 Now substitute 23 for x in either equation to find the value of y. 2x  y  60 2(23)  y  60 46  y  60 46  y  46  60  46 y  14 The solution is (23, 14). 3. Since the coefficients of the y terms, 1 and 1, are additive inverses, you can eliminate the y terms by adding the equations. y  x  6 () y  x  5 2x  11

2y  3 3 2 y1 3

2 y 3



2t 2

16 2

4

2

t2 The solution is (2, 2).

x8 Now substitute 8 for x in either equation to find the value of y. xy7 8y7 8y878 y  1 The solution is (8, 1).

703

Extra Practice

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8. Since the coefficients of the y terms, 1 and 1, are additive inverses, you can eliminate the y terms by adding the equations. xy 8 () 2x  y  6 3x  14

5. Since neither the coefficients of the x terms nor the coefficients of the y terms are the same or additive inverses, you cannot use elimination by addition or subtraction to solve this system. Use substitution. Since x  y  7, substitute y  7 for x in the second equation. 2x  5y  2 2(y  7)  5y  2 2y  14  5y  2 3y  14  2 3y  14  14  2  14 3y  12 3y 3

3x 3

14 3 14 x 3 14 substitute 3

Now the value of y. xy8 14 3

12

 3

14 3

y  4 Use x  y  7 to find the value of x. xy7 x  4  7 x  11 The solution is (11, 4). 6. Since the coefficients of the x terms, 3 and 3, are the same, you can eliminate the x terms by subtracting the equations. 3x  5y  16 () 3x  2y  2 7y  14 7y 7





14 3

8

14 3

10 3

1143, 103 2.

9. Rewrite the second equation so the system is in column form. s  7  3t s  3t  7  3t  3t s  3t  7 Since the coefficients of the t terms, 3 and 3, are additive inverses, you can eliminate the t terms by adding the equations. 2s  3t  4 () s  3t  7 3s  3

14 7

3s 3

3

3

s1 Now substitute 1 for s in either equation to find the value of t. s  7  3t 1  7  3t 1  7  7  3t  7 6  3t

6 3

6 3



3t 3

2t The solution is (1, 2). 10. Rewrite the second equation so the system is in column form. 6x  42  16y 6x  42  42  16y  42 6x  16y  42 6x  16y  16y  42  16y 6x  16y  42 Since the coefficients of the x terms, 6 and 6, are additive inverses, you can eliminate the x terms by adding the equations. 6x  16y  8 () 6x  16y  42 0  34 The statement 0  34 is false. This means that there is no solution to the system of equations.

6

2

x3 Now substitute 3 for x in either equation to find the value of y. xy3 3y3 3y333 y  0 (1)(y)  (1)0 y0 The solution is (3, 0).

Extra Practice

y

The solution is

x  2 The solution is (2, 2). 7. Since the coefficients of the y terms, 1 and 1, are additive inverses, you can eliminate the y terms by adding the equations. xy3 () x  y  3 2x 6 2x 2

for x in either equation to find

y8 y

y  2 Now substitute 2 for y in either equation to find the value of x. 3x  5y  16 3x  5(2)  16 3x  (10)  16 3x  10  16 3x  10  10  16  10 3x  6 3x 3



704

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Since the coefficients of the y terms, 1 and 1, are additive inverses, you can eliminate the y terms by adding the equations. xy0 () x  y  7 2x 7

11. Rewrite the second equation so the system is in column form. 3x  0.4y  4 3x  0.4y  0.4y  4  0.4y 3x  0.4y  4 Since the coefficients of the x terms, 3 and 3, are the same, you can eliminate the x terms by subtracting the equations. 3x  0.2y  7 () 3x  0.4y  4 0.6y  3 0.6y 0.6

2x 2

x  3.5 Now substitute 3.5 for x in either equation to find the value of y. xy 3.5  y The solution is (3.5, 3.5). 1 1 14. Since the coefficients of the y terms, 3 and 3, are additive inverses, you can eliminate the y terms by adding the equations.

3

 0.6

y5 Now substitute 5 for y in either equation to find the value of x. 3x  0.2y  7 3x  0.2(5)  7 3x  1  7 3x  1  1  7  1 3x  6 3x 3

1

4x  3y  8 ()

x

14 9 14 x 9 14 substitute 9 for

234 7

1

4

39 10.5 26 7

56 9

 26 

1

1

 3y  8

56 9 1 3y 1 3y



8 

56 9

16 9

2  (3)169 16

y  3

1149, 163 2.

15. Since the coefficients of the y terms, 1 and 1, are the same, you can eliminate the y terms by subtracting the equations. 2x  y  3

234 7

52

2y   7

112 22y  112 21527 2

()

2 x 3 4 x 3

 y  1

12

26

y  7

The solution is



56 9 1 y 3

The solution is

 2y  26 234 7

1149 2  13y  8

(3)

1267 2  2y  26

 2y 

x in either equation to find

4x  3y  8

Now substitute for x in either equation to find the value of y. 9x  2y  26 234 7



Now the value of y.

26 7

9

 14 9x 9

x2 The solution is (2, 5). 12. Since the coefficients of the y terms, 2 and 2, are additive inverses, you can eliminate the y terms by adding the equations. 9x  2y  26 () 1.5x  2y  13 10.5x  39 

1

5x  3y  6 9x

6

3

10.5x 10.5

7

2

1267, 267 2.

3 4 x 4 3

 

4

1 24 3 4

x3 Now substitute 3 for x in either equation to find the value of y. 2x  y  3 2(3)  y  3 6y3 6y636 y  3 (1)(y)  (1) (3) y3 The solution is (3, 3).

13. Rewrite the first equation so the system is in column form. xy xyyy xy0

705

Extra Practice

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Page 836

2

Lesson 7-4

4. Multiply the first equation by 5 so the coefficients of the y terms are additive inverses. Then add the equations.

1. Multiply the second equation by 2 so the coefficients of the y terms are additive inverses. Then add the equations. 3x  2y  10 2x  y  5

Multiply by 2.

3x  2y  10 () 4x  2y  10 7x

 7x 7

1 x 3

 y  1

2

1 x 5

 5 y  1

2

Multiply by 5 .

2

( )

0 0

 7

1 x 3

 y  1

1 (9) 3

 y  1

39

 13

3x  5y  8

Multiply by 4.

4x  7y  10

Multiply by 3.

12x  21y 

y2

16

37y  74 

x  0.5y 

74 37

y  2



x  0.5y  Multiply by 0.5.

()

1

0.2x  0.5y  1 

1.2x 1.2x 1.2



0 0 1.2

x0

Now substitute 0 for x in either equation to find the value of y. 0.4x  y  2 0.4(0)  y  2 y  2 The solution is (0, 2).

10 5

x2 The solution is (2, 2).

Extra Practice

18 3

1

0.4x  y  2

Now substitute 2 for y in either equation to find the value of x. 5x  3y  4 5x  3(2)  4 5x  6  4 5x  6  6  4  6 5x  10 5x 5



x6 The solution is (6, 2). 6. Multiply the second equation by 0.5 so the coefficients of the y terms are additive inverses. Then add the equations.

20x  25y  90 37y 37

30

Now substitute 2 for y in either equation to find the value of x. 3x  5y  8 3x  5(2)  8 3x  10  8 3x  10  10  8  10 3x  18 3x 3

20x  12y 

1 32

(1) (y)  (1) (2)

x  1 The solution is (1, 3). 3. Eliminate x. ()

3

y  2

2 2

Multiply by 5.

 5

12x  20y  32 ()

y3

Multiply by 4.

1 x 15

3  y  1 3  y  3  1  3 y  2 (1)(y)  (1)2 y  2 The solution is (9, 2). 5. Eliminate x.

Now substitute 3 for y in either equation to find the value of x. 2x  5y  13 2x  5(3)  13 2x  15  13 2x  15  15  13  15 2x  2

4x  5y  18

2

x  9

13y  39

5x  3y  4

 5 y  1

Now substitute 9 for x in either equation to find the value of y.

2x  5y  13 Multiply by 2. 4x  10y  26 4x  3y  13 () 4x  3y  13



1 x 5

1

Now substitute 0 for x in either equation to find the value of y. 2x  y  5 2(0)  y  5 y  5 (1)(y)  (1)(5) y5 The solution is (0, 5). 2. Multiply the first equation by 2 so the coefficients of the x terms are additive inverses. Then add the equations.

2x 2

2 5

(15) 15 x  (15) 5

x0

13y 13

2

15 x  5 y 

706

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7. Multiply the first equation by 4 so the coefficients of the x terms are additive inverses. Then add the equations. x  8y  3 Multiply by 4. 4x  2y  7

7

y  5  2

34y  5

y52



y 5 34

Now substitute the value of x. x  8y  3

1 32

4x  32y  12 ( ) 4x  2y  34y 34

x8

3

Now substitute 5 for x in either equation to find the value of y. yx2 3

5 34 5 34

3

3

7

y5

1

3 7

2

The solution is 5, 5 .

for y in either equation to find

10. Multiply the first equation by 2 so the coefficients of the x terms are additive inverses. Then add the equations.

1345 2  3

x  4y  30 Multiply by 2. 2x  8y  60 2x  y  6 () 2x  y  6

20

x  17  3 20

3

y5525

20

20

9y  66

x  17  17  3  17 x

9y 9

31 17

13117, 345 2.

The solution is

y

9x 9



x

4

x4

44 9

4 8

y  9

1223 2  30

x

11 9 11 9

x

88 3



88 3 88 3

 30  30  2

x3 The solution is

for x in either equation to find

44 9

8

The solution is

1 82

1

Now substitute the value of y. 4x  4y  5

1119, 89 2.

Multiply by 3.

5

 10

x2

9. Multiply the second equation by 3 so the coefficients of the y terms are additive inverses. Then add the equations. 3y  8x  9 yx2

123, 223 2.

10x 10

(1)(y)  (1) 9 y9

88 3

11. Multiply the first equation by 2 so the coefficients of the y terms are additive inverses. Then add the equations. 3x  2y  0 Multiply by 2. 6x  4y  0 4x  4y  5 () 4x  4y  5 10x 5

y4

y

22 3

22

1119 2  y  4 44 9

44 9

11 9

66 9

Now substitute 3 for y in either equation to find the value of x. x  4y  30

8. Multiply the first equation by 2 so the coefficients of the y terms are additive inverses. Then add the equations. 4x  y  4 Multiply by 2. 8x  2y  8 x  2y  3 () x  2y  3 9x  11

Now substitute the value of y. 4x  y  4



()

4

5x 5

for x in either equation to find

112 2  4y  5

2  4y  5 2  4y  2  5  2 4y  3

3y  8x  9 3y  3x  6 5x 

1 2

4y 4

3

3

4 3

3

 5

y4

3

x  5

The solution is

707

112, 34 2.

Extra Practice

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15. Eliminate y.

12. Multiply the second equation by 3 so the coefficients of the y terms are additive inverses. Then add the equations. 9x  3y  5 9x  3y  5 xy1 Multiply by 3. () 3x  3y  3 12x 8 12x 12

6x  3y  9 8x  2y  4

2 3 2 3

1

Now substitute 2 for x in either equation to find the value of y. 6x  3y  9

for x in either equation to find

2

The solution is 13. Eliminate x. 2x  7y  9 3x  4y  6

3y 3

123, 13 2.

Multiply by 3. Multiply by 2.

()

6x  21y  27 6x  8y  12

Page 836

39

 13

y  3

y 6

x 3

12 2

2x  6y  16

Multiply by 5.

5x  7y  18

Multiply by 2.

x

O

2. The solution includes the ordered pairs in the intersection of the graphs of y  2 and y  x  2. The region is shaded. The graphs of y  2 and y  x  2 are boundaries of this region. The graphs of both boundaries are dashed and are not included in the solution.

10x  30y  80 ( )

10x  14y 

36

44y  44 44y 44



44 44

y1

y

Now substitute 1 for y in either equation to find the value of x. 2x  6y  16 2x  6(1)  16 2x  6  16 2x  6  6  16  6 2x  10

y 2

O

y  x  2

10 2

x  5 The solution is (5, 1).

Extra Practice

Lesson 7-5

y

x  6 The solution is (6, 3). 14. Eliminate x.



112, 42.

1. The solution includes the ordered pairs in the intersection of the graphs of x  3 and y  6. The region is shaded. The graphs of x  3 and y  6 are boundaries of this region. The graphs of both boundaries are dashed and are not included in the solution.

Now substitute 3 for y in either equation to find the value of x. 2x  7y  9 2x  7(3)  9 2x  (21)  9 2x  21  9 2x  21  21  9  21 2x  12

2x 2

12 3

y4

13y 13





The solution is

13y  39

2x 2

112 2  3y  9

3  3y  9 3  3y  3  9  3 3y  12

y313 1

6

 12 1

y1 y3

 6

x2

8

 12

6

2

12x  6y  18 24x  6y  12 12x 12

2

2 3

()

12x

x3 Now substitute the value of y. xy1

Multiply by 2. Multiply by 3.

708

x

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6. The solution includes the ordered pairs in the intersection of the graphs of y  x  3 and y  x  2. The region is shaded. The graphs of y  x  3 and y  x  2 are boundaries of this region. The graphs of both boundaries are solid and are included in the solution.

3. The solution includes the ordered pairs in the intersection of the graphs of x  2 and y  3  5. The region is shaded. The graphs of x  2 and y  3  5, or y  2, are boundaries of this region. The graphs of both boundaries are solid and are included in the solution. y

y y 2

yx 3 yx 2

O

x 2

x O

7. The solution includes the ordered pairs in the intersection of the graphs of x  3y  4 and 2x  y  5. The region is shaded. The graphs of x  3y  4 and 2x  y  5 are boundaries of this region. The graph of x  3y  4 is solid and is included in the graph of x  3y  4. The graph of 2x  y  5 is dashed and is not included in the graph of 2x  y  5.

4. The solution includes the ordered pairs in the intersection of the graphs of x  y  1 and 2x  y  2. The region is shaded. The graphs of x  y  1 and 2x  y  2 are boundaries of this region. The graphs of both boundaries are solid and are included in the solution. y

2x  y  2

O

x

x

y

x  y  1

x

O

x  3y  4

2x  y  5

5. The solution includes the ordered pairs in the intersection of the graphs of y  2x  2 and y  x  1. The region is shaded. The graphs of y  2x  2 and y  x  1 are boundaries of this region. The graphs of both boundaries are solid and are included in the solution. y  x  1

8. The solution includes the ordered pairs in the intersection of the graphs of y  x  1 and y  2x  10. The region is shaded. The graphs of y  x  1 and y  2x  10 are boundaries of this region. The graph of y  x  1 is dashed and is not included in the graph of y  x  1. The graph of y  2x  10 is solid and is included in the graph of y  2x  10.

y

y  2x  2 O

y

y  2 x  10

x

y x1

O

709

x

Extra Practice

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12. The solution includes the ordered pairs in the intersection of the graphs of 4x  10y  5 and 2x  5y  1. The region is shaded. The graph of 2x  5y  1 is a boundary of this region. The graph of this boundary is dashed and is not included in the solution. (Note: The graph of 4x  10y  5 is not included in the solution since it is not a boundary of the shaded region.)

9. The solution includes the ordered pairs in the intersection of the graphs of 5x  2y  15 and 2x  3y  6. The region is shaded. The graphs of 5x  2y  15 and 2x  3y  6 are boundaries of this region. The graphs of both boundaries are dashed and are not included in the solution. y x

O 2x  3y  6

y 4x  10y  5

5 x  2 y  15

x

O 2x  5y  1

10. The solution includes the ordered pairs in the intersection of the graphs of 4x  3y  4 and 2x  y  0. The region is shaded. The graphs of 4x  3y  4 and 2x  y  0 are boundaries of this region. The graphs of both boundaries are dashed and are not included in the solution.

13. The solution includes the ordered pairs in the intersection of the graphs of y  x  0, y  3, and x  0. The region is shaded. The graphs of y  x  0, y  3 and x  0 are boundaries of this region. The graphs of these boundaries are solid and are included in the solution.

y

y y3

4x  3y  4 O

2x – y  0

x0

x

yx0

x

O

11. The solution includes the ordered pairs in the intersection of the graphs of 4x  5y  20 and y  x  1. The region is shaded. The graphs of 4x  5y  20 and y  x  1 are boundaries of this region. The graphs of both boundaries are solid and are included in the solution.

14. The solution includes the ordered pairs in the intersection of the graphs of y  2x, x  3, and y  4. The region is shaded. The graphs of y  2x, x  3, and y  4 are boundaries of this region. The graphs of these boundaries are dashed and are not included in the solution.

y

y

yx1

y 4

x  3 x

O

O

4x  5y  20

Extra Practice

710

y  2x

x

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15. The solution includes the ordered pairs in the intersection of the graphs of y  x, x  y  4 and y  3. The region is shaded. The graphs of y  x, x  y  4, and y  3 are boundaries of this region. The graph of x  y  4 is dashed and is not included in the graph of x  y  4. The graphs of y  x and y  3 are solid and are included in the graphs of y  x and y  3, respectively. y

15.

1

 4 w6 w8 1

 4w68 1

 4w14 16. [ (23 ) 3 ] 2  [ (8) 3 ] 2  (8) 6  262,144

yx

17. xy4 x

O

Lesson 8-1

1. No; it shows subtraction, not multiplication of variables and numbers. 2. Yes; it is a real number and therefore a monomial. 3. Yes; it is a product of a number and two variables. 4. No; it shows subtraction, not multiplication of variables and numbers. 5. a5 (a)(a7 )  a517  a13 3 4 4 4 6. (r t ) (r t )  (r3 r4 )(t4 t4 )  (r34 )(t44 )  r7t8 3 4 3 7. (x y )(xy )  (x3 x)(y4 y3 )  (x31 )(y43 )  x4y7 8. (bc3 )(b4c3 )  (b b4 )(c3 c3 )  (b14 )(c33 )  b5c6 2 9. (3mn )(5m3n2 )  (3 5)(m m3 )(n2 n2 )  15(m13 ) (n22 )  15m4n4 3 2 2 6 2 10. [ (3 ) ]  [3 ]  312 or 531,441 3 2 11. (3s t )(4s3t2 )  3(4) (s3 s3 )(t2 t2 )  12(s33 )(t22 )  12s6t4 12. x3 (x4y3 )  (x3 x4 )y3  (x34 )y3  x7y3 2 4 13. (1.1g h ) 3  (1.1) 3 (g2 ) 3 (h4 ) 3  1.331g6h12 14.

3 4a(a2b3c4 )

  

123y32 (3y2)3  123y32 (3)3(y2)3 2  1 3y3 2 (27) (y6 ) 2  1 3 27 2 (y3 y6 )

 18(y36 )  18y9 3 18. (10s t)(2s2t2 ) 3  (10s3t) (2) 3 (s2 ) 3 (t2 ) 3  (10s3t) (8) (s6 ) (t6 )  10(8) (s3 s6 ) (t t6 )  80(s36 ) (t16 )  80s9t7 3 4 3 19. (0.2u w )  (0.2) 3 (u3 ) 3 (w4 ) 3  0.008u9w12

y  3

Page 837

112w322(w4)2  112 22(w3)2(w4)2

Page 837 10

Lesson 8-2 107

1.

6 67

2.

b c b3c2

6

 63 or 216 6 5



1bb 21cc 2 6

5

3

2

63

 (b ) (c52 ) 3 3 b c 8

3.

(a) 4b a4b7

4 8

a b

 a4b7 

1aa 21bb 2 4

8

4

7

44

 (a ) (b87 ) 0 1 a b  1b or b 4.

(x) 3y x3y6

3



1xx 21yy 2 3

3

3

6

 (x33 ) (y36 )  (x0 ) (y3 )  (1)  5

5.

12ab 4a4b3



1 y3

1y1 2 3

1124 21aa 21bb 2 5

4

3

3(a14 ) (b53 )

  3a3b2

1 21 2

3 1 b2 a3 1

1  5

3 4 (a a2 )b3c4 3 4 (a12 )b3c4 3 4a3b3c4

6.

24x 8x2



3b2 a3

24 x 18 21x 2 5 2

3(x52)

  3x3

711

Extra Practice

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7.

9b2k4 18h5j3k4

h 1 k 19 18 21 h 21 j 21 k 2 1 1  2 (h25 ) 1 j 2 (k44 ) 1 1  2 (h3 ) 1 j 2 (k0 ) 1 1 1  2 1 h 21 j 2 (1) 2



4

5

3

15.

4

1ab 23  (a(b )) 3

3 3

2

2 3

1a1 21b1 2 1 b  1 a 21 1 2

3

8.

1

2



(2a b ) 2 (3a3b) 2

2 4 2a b 2 3a3b

9.

9a b c 2a5b4c5

b

 a9

16.

2

4



(2) 2 (a ) 2 (b ) 2 (3) 2 (a3 ) 2 (b) 2



4a4b8 9a6b2

 

4b 9a2



10.

5

17.

(r)s r3s4

3

18.

4 0

28a b 14a3b1



 

2

2 3

19.

x y z 115 10 21 x 21 y 21 z 2 5

7

6

4

4

4

(j k m) (jk4 ) 1

 2 (x1(4) )(y56 )(z7(4) ) 3

 2x5y11z11 3

1 21y1 21z1 2 5

11

3 x 1

2

4

11. 3 12.

156 2

2



1 34

or

11

3x z 2y11

21.

1 81

151 2161 2 1 6  1 5 21 1 2 2

2

22.

2



5b0a7

(u v ) (u3v) 3

j k m j1k4



8

12

1

4

8 12

4

1 jj 21kk 21m1 2 4

1

0

3

2

3

122aab b 21  (2(2aab b) ) 3 2

3 2

(u ) (v ) (u3 ) 3 (v) 3 v

1uu 21vv 2 9

3

1

2



2 (a ) 1 (b ) 1 (21 ) 1 (a5 ) 1 (b3 ) 1



21a3b2 2a5b3



6 6

6

1 5 3 1 3

122 21aa 21bb 2 1

3

2

5

3

 (211 ) (a35 ) (b2(3) )  22a8b5

3 2

6

3 2 1

1 5 3



121 21a1 21b1 2 5

2

b5

 4a8

23.

(u6(9) )(v6(3) )

  u3v9

Extra Practice

3

2 3

 u9v3 

2

18x a 18 x a  1 6 21 x 21 a 2 16x a 2

  (a2 ) (1)

u

7

2b a7

0 3

(a5(7) )(b0 )



121 21a1 21b1 2



0

1

14.

1

(j ) 4 (k ) 4 (m) 4 (j) 1 (k4 ) 1

2

 a2 3 3 2

0

3

 ( j8(1) )(k12(4) ) (m4 )  j9k16m4

36 25

or

4

  3x2a0  3x2

52

62 52

12814 21aa 21bb 2



 62



4

3(x0(2) ) (a3(3) )

2

13. a

y 12x 4y 2 4

20.

5 11



5

3

 2(a43 )(b0(1) )  2a7b1

1 21 21c1 2 

6

1rr 21ss 2



3

5 7

2

 (r1(3) )(s5(4) )  r4s9

1 21 21 21 2

15xy z 10x4y6z4

6

2

9 25 (a )(b74 )(c35 ) 2 9 3 3 2 a b c 2

9b3 2a3c2

2

1

9 a b c 2 a5 b4 c5

9 1 b a3 1

2

 4x2y6

6

7

x y6

121 21x1 21y1 2 1 1 1  1 2 21 x 21 y 2

1 21 2 2

2 2

2



6

2 

3 2

8

4 1 b 9 a2 1



2

3

1 21 2 4





1y2x 22  (y(2x)) 

4 a b a6 b2 4 46 (a )(b82 ) 9 4 2 6 a b 9



2 7 3

9

6

2 4

9

6

6

3



9



3

1 2h3j3

9

a b6



3

712

15n2nmm 20  1 1

2

2

8

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(3ab c) 3 (2a2bc2 ) 2 2

24.

 

25. (3.1 104 ) (4.2 105 )  (3.1 4.2)(104 105 )

3

3

(a) 3 (b2 ) 3 (c) 3 22 (a2 ) 2 (b) 2 (c2 ) 2

 13.02 101  1.302 101 101  1.302 100 or 1.302

33a3b6c3 22a4b2c4

131 2121 21aa 21bb 21cc 2 1 1  1 27 21 4 2 (a34 )(b62 )(c34 ) 3



3

6

3

4

2

4

2

26. (78 106 ) (0.01 105 )  (78 0.01)(106 105 )  0.78 1011  7.8 101 1011  7.8 1010 or 78,000,000,000

 108a7b8c7 1 1

 108

1a1 21b1 21c1 2 7

8

27.

7

1

 108a7b8c7

Page 837

2

2.31 10 3.3 103



10 12.31 3.3 21 10 2 2 3

 0.7 102(3)  0.7 101  0.7 101 101  7 100 or 7

Lesson 8-3

1. 2.6 105  260,000 2. 4 103  0.004

Page 838

3. 6.72 103  6720 4. 4.93 104  0.000493 5. 1.654 106  0.000001654 6. 7.348 107  73,480,000 7. 6500  6.5 103 8. 953.56  9.5356 102 9. 0.697  6.97 101

3.

12. 0.0000269  2.69 105 13. 0.121212  1.21212 101 14. 543 104  5.43 102 104  5.43 106 15. 739.9 105  7.399 102 105  7.399 103 2 16. 6480 10  6.48 103 102  6.48 101 7 17. 0.366 10  3.66 101 107  3.66 108 3 18. 167 10  1.67 102 103  1.67 105

7. The polynomial



10 14.8 1.6 21 10 2 3 1

10. The polynomial

 3 1031  3 102 or 300 21. (4 102 ) (1.5 106 )  (4 1.5)(102 106 )  6 108 or 600,000,000 2

22.

8.1 10 2.7 103



and

7.8 105 1.3 107



has only one term, whose a3

1 , 5

x2 3

10 18.1 2.7 21 10 2

x

1

 2  5 has three terms,

x2 , 3

x

2,

whose degrees are 2, 1, and 0, respectively.

Thus, the degree of

x2 3

x

1

 2  5 is 2, the greatest of

2, 1, and 0. 11. The polynomial 6 has only one term, whose degree is 0. Thus, the degree of 6 is 0. 12. The polynomial a2b3  a3b2 has two terms, a2b3 and a3b2, whose degrees are both 5. Thus, the degree of a2b3  a3b2 is 5. 13. 2x2  3x  4x3  x5  2x2  3x1  4x3  x5  3x  2x2  4x3  x5

2

3

 3 102(3)  3 105 or 300,000 23.

a3 4

degree is 3. Thus, the degree of 4 is 3. 8. The polynomial 10 has only one term, whose degree is 0. Thus, the degree of 10 is 0. 9. The polynomial 4h5 has only one term, whose degree is 5. Thus, the degree of 4h5 is 5.

19. (2 105 ) (3 108 )  (2 3) (105 108 )  6 103 or 0.006 3

 k2y

The term k is not a monomial. No, the expression is not a polynomial. 4. 3a2x  5a  3a2x  (5a) This expression is the sum of two monomials. Yes, it is a polynomial; it is a binomial. 5. The polynomial a  5c has two terms, a and 5c, whose degrees are both 1. Thus, the degree of a  5c is 1. 6. The polynomial 14abcd  6d3 has two terms, 14abcd and 6d3, whose degrees are 4 and 3, respectively. Thus, the degree of 14abcd  6d3 is 4, the greater of 4 and 3.

11. 568,000  5.68 105

4.8 10 1.6 101

5 k

5

10. 843.5  8.435 102

20.

Lesson 8-4

1. 5x2y  3xy  7  5x2y  3xy  (7) This expression is the sum of three monomials. Yes, it is a polynomial; it is a trinomial. 2. 0 This expression is a monomial. Yes, it is a polynomial.

10 17.8 1.3 21 10 2 5 7

 6 105(7)  6 102 or 600 24. (2.2 102 ) (3.2 105 )  (2.2 3.2) (102 105 )  7.04 103 or 7040

713

Extra Practice

PQ249J-6481F-20-23[686-725] 26/9/02 9:45 PM Page 714 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

14. x3  x2  x  1  x3  x2  x1  1x0  1  x  x2  x3 15. 2a  3ax2  4ax  2ax0  3ax2  4ax1  2a  4ax  3ax2 16. 5bx3  2bx  4x2  b3  5bx3  2bx1  4x2  b3x0  b3  2bx  4x2  5bx3 8 17. x  2x2  x6  1  x8  2x2  x6  1x0  1  2x2  x6  x8 18. cdx2  c2d2x  d3  cdx2  c2d2x1  d3x0  d3  c2d2x  cdx2 19. 5x2  3x3  7  2x  5x2  3x3  7x0  2x1  3x3  5x2  2x  7 20. 6x  x5  4x3  20  6x1  x5  4x3  20x0  x5  4x3  6x  20

7. (4a2  10b2  7c2 )  (5a2  2c2  2b)  (4a2  5a2 )  (10b2 )  (7c2  2c2 )  2b  a2  10b2  9c2  2b 2  6z  8)  (4z2  7z  5) 8. (z  (z2  6z  8)  (4z2  7z  5)  (z2  4z2 )  (6z  7z)  (8  5)  3z2  13z  3 9. (4d  3e  8f )  (3d  10e  5f  6)  (4d  3e  8f )  (3d  10e  5f  6)  (4d  3d)  (3e  10e)  (8f  5f )  6  7d  7e  3f  6 10. (7g  8h  9)  (g  3h  6k)  (7g  g)  (8h  3h)  9  6k  6g  5h  9  6k 11. (9x2  11xy  3y2 )  (x2  16xy  12y2 )  (9x2  11xy  3y2 )  (x2  16xy  12y2 )  (9x2  x2 )  (11xy  16xy)  (3y2  12y2 )  8x2  5xy  15y2

12. (3m  9mn  5n)  (14m  5mn  2n)  (3m  14m)  (9mn  5mn)  (5n  2n)  11m  4mn  7n 13. (4x2  8y2  3z2 )  (7x2  14z2  12)  (4x2  8y2  3z2 )  (7x2  14z2  12)  (4x2  7x2 )  8y2  (3z2  14z2 )  12  3x2  8y2  11z2  12 14. (17z4  5z2  3z)  (4z4  2z3  3z)  (17z4  5z2  3z)  (4z4  2z3  3z)  (17z4  4z4 )  2z3  5z2  (3z  3z)  13z4  2z3  5z2 15. (6  7y  3y2 )  (3  5y  2y2 )  (12  8y  y2 )

2

21. 5b  b3x2  3bx 2

 5bx0  b3x2  3bx1 2

 b3x2  3bx  5b 22. 21p2x  3px3  p4  21p2x1  3px3  p4x0  3px3  21p2x  p4 23. 3ax2  6a2x3  7a3  8x  3ax2  6a2x3  7a3x0  8x1  6a2x3  3ax2  8x  7a3 1

2

 (3y2  2y2  y2 )  (7y  5y  8y)  (6  3  12)  2y2  20y  3

16.

1

24. 3 s2x3  4x4  5 s4x2  4 x 1

2

1

2 4 2 s x 5

1 x 4

 (7c 2  16c 2  9c 2 )  (2c  9c  7c )  (5  6  3  7 )

 3 s2x3  4x4  5 s4x2  4 x1 4

 4x 

1 2 3 s x 3





(7c 2  2c  5 )  (9c  6 )  (16c 2  3 )  (9c 2  7c  7 )  1

Page 838 Page 838

Lesson 8-5

1. (3a2  5)  (4a2  1)  (3a2  4a2 )  (5  1)  7a2  4 2. (5x  3)  (2x  1)  (5x  2x)  (3  1)  3x  2 3. (6z  2)  (9z  3)  (6z  2)  (9z  3)  (6z  9z)(2  3)  3z  1 4. (4n  7)  (7n  8)  (4n  7)  (7n  8)  (4n  7n)  (7  8)  3n  15 5. (7t2  4ts  6s2 )  (5t2  12ts  3s2 )  (7t2  5t2 )  (4ts  12ts)  (6s2  3s2 )  12t2  8ts  3s2 6. (6a2  7ab  4b2 )  (2a2  5ab  6b2 )  (6a2  7ab  4b2 )  (2a2  5ab  6b2 )  (6a2  2a2 )  (7ab  5ab)  (4b2  6b2 )  4a2  12ab  10b2

Extra Practice

Lesson 8-6

1. 3(8x  5)  3(8x)  (3) (5)  24x  15 2. 3b(5b  8)  3b(5b)  3b(8)  15b2  24b 3. 1.1a(2a  7)  1.1a(2a)  1.1a(7)  2.2a2  7.7a 1

1

1

4. 2x(8x  6)  2x(8x)  2x(6)  4x2  3x 5. 7xy(5x  y2 )  7xy(5x2 )  7xy(y2 )  35x3y  7xy3 6. 5y(y2  3y  6)  5y(y2 )  5y(3y)  5y(6)  5y3  15y2  30y 2

7. ab(3b2  4ab  6a2 )  ab(3b2 )  ab(4ab)  ab(6a2 )  3ab3  4a2b2  6a3b 8. 4m2 (9m2n  mn  5n2 )  4m2 (9m2n)  4m2 (mn)  4m2 (5n2 )  36m4n  4m3n  20m2n2

714

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11(n  3)  5  2n  44 11(n)  11(3)  5  2n  44 11n  33  5  2n  44 11n  28  2n  44 9n  28  44 9n  72 n8 24. a(a  6)  2a  3  a(a  2) a(a)  a(6)  2a  3  a(a)  a(2) a2  6a  2a  3  a2  2a a2  4a  3  a2  2a 4a  3  2a 2a  3

9. 4st2 (4s2t3  7s5  3st3 )  4st2 (4s2t3 )  4st2 (7s5 )  4st2 (3st3 )  16s3t5  28s6t2  12s2t5 1

1

1

23.

1

10. 3x(9x2  x  5)  3x(9x2 )  3x(x)  3x(5) 1

5

 3x3  3x2  3x 11. 2mn(8m2  3mn  n2 )  2mn(8m2 )  2mn(3mn)  2mn(n2 )  16m3n  6m2n2  2mn3 3

11

4

12. 4ab2 3b2  9b  1  

1 2

2

1

3 1 3 4 4 ab2 3b2  4 ab2 9b 1 1 3 4ab4  3ab3  4ab2

2  34 ab2(1)

3

a  2 or 1.5

13. 3a(2a  12)  5a  3a(2a)  3a(12)  5a  6a2  36a  5a  6a2  41a 14. 6(12b2  2b)  7(2  3b)  6(12b2 )  6(2b)  7(2)  7(3b)  72b2  12b  14  21b  72b2  33b  14 15. x(x  6)  x(x  2)  2x  x(x)  x(6)  x(x)  x(2)  2x  x2  6x  x2  2x  2x  2x2  6x 16. 11(n  3)  2(n2  22n)  11(n)  11(3)  2(n2 )  2(22n)  11n  33  2n2  44n  2n2  55n  33 17. 2x(x  3)  3(x  3)  2x(x)  2x(3)  3(x)  3(3)  2x2  6x  3x  9  2x2  3x  9 18. 4m(n  1)  5n(n  1)  4m(n)  4m(1)  5n(n)  5n(1)  4mn  4m  5n2  5n 19. 7xy  x(7y  3)  7xy  x(7y)  x(3)  7xy  7xy  3x  3x 20. 5(c  3a)  c(2c  1)  5(c)  5(3a)  c(2c)  c(1)  5c  15a  2c2  c  2c2  6c  15a 21. 9n(1  n)  4(n2  n)  9n(1)  9n(n)  4(n2 )  4(n)  9n  9n2  4n2  4n  5n  13n2 22. 6(11  2x)  7(2  2x) 6(11)  6(2x)  7(2)  7(2x) 66  12x  14  14x 66  26x  14 26x  52 x2

25.

q(2q  3)  20  2q(q  3) q(2q)  q(3)  20  2q(q)  2q(3) 2q2  3q  20  2q2  6q 3q  20  6q 9q  20  0 9q  20 20

q  9

w(w  12)  w(w  14)  12 w(w)  w(12)  w(w)  w(14)  12 w2  12w  w2  14w  12 12w  14w  12 2w  12 w  6 x(x  3)  4x  3  8x  x(3  x) 27. x(x)  x(3)  4x  3  8x  x(3)  x(x) x2  3x  4x  3  8x  3x  x2 x2  x  3  11x  x2 x  3  11x 10x  3  0 10x  3 26.

3

x  10 28.

3(x  5)  x(x  1)  x(x  2)  3 3(x)  3(5)  x(x)  x(1)  x(x)  x(2)  3 3x  15  x2  x  x2  2x  3 4x  15  x2  x2  2x  3 4x  15  2x  3 6x  15  3 6x  12 x  2

29.

n(n  5)  n(n  2)  2n(n  1)  1.5 n(n)  n(5)  n(n)  n(2)  2n(n)  2n(1)  1.5 n2  5n  n2  2n  2n2  2n  1.5 2n2  3n  2n2  2n  1.5 3n  2n  1.5 n  1.5 n  1.5

Page 839

Lesson 8-7

1. (d  2)(d  5)  d(d)  d(5)  2(d)  2(5)  d2  5d  2d  10  d2  7d  10 2. (z  7)(z  4)  z(z)  z(4)  7(z)  7(4)  z2  4z  7z  28  z2  3z  28

715

Extra Practice

PQ249J-6481F-20-23[686-725] 26/9/02 9:45 PM Page 716 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

3. (m  8) (m  5)  m(m)  m(5)  8(m)  8(5)  m2  5m  8m  40  m2  13m  40 4. (a  2) (a  19)  a(a)  a(19)  2(a)  2(19)  a2  19a  2a  38  a2  17a  38 5. (c  15) (c  3)  c(c)  c(3)  15(c)  15(3)  c2  3c  15c  45  c2  12c  45 6. (x  y)(x  2y)  x(x)  x(2y)  y(x)  y(2y)  x2  2xy  xy  2y2  x2  xy  2y2 7. (2x  5)(x  6)  2x(x)  2x(6)  5(x)  5(6)  2x2  12x  5x  30  2x2  7x  30 8. (7a  4) (2a  5)  7a(2a)  7a(5)  4(2a)  4(5)

19. (x  2)(x2  2x  4)  x(x2  2x  4)  2(x2  2x  4)  x3  2x2  4x  2x2  4x  8  x3  8 20. (3x  5)(2x2  5x  11)  3x(2x2  5x  11)  5(2x2  5x  11)  6x3  15x2  33x  10x2  25x  55  6x3  5x2  8x  55 21. (4s  5)(3s2  8s  9)  4s(3s2  8s  9)  5(3s2  8s  9)  12s3  32s2  36s  15s2  40s  45  12s3  47s2  4s  45 22. (3a  5)(8a2  2a  3)  3a(8a2  2a  3)  5(8a2  2a  3)  24a3  6a2  9a  40a2  10a  15  24a3  34a2  19a  15 23. (a  b)(a2  ab  b2 )  a(a2  ab  b2 )  b(a2  ab  b2 )  a3  a2b  ab2  a2b  ab2  b3  a3  b3 24. (c  d)(c2  cd  d2 )  c(c2  cd  d2 )  d(c2  cd  d2 )  c3  c2d  cd2  c2d  cd2  d3  c3  d 3

 14a2  35a  8a  20  14a2  43a  20

9. (4x  y)(2x  3y)  4x(2x)  4x(3y)  y(2x)  y(3y)  8x2  12xy  2xy  3y2  8x2  10xy  3y2 10. (7v  3)(v  4)  7v(v)  7v(4)  3(v)  3(4)  7v2  28v  3v  12  7v2  31v  12 11. (7s  8) (3s  2)  7s(3s)  7s(2)  8(3s)  8(2)  21s2  14s  24s  16  21s2  38s  16 12. (4g  3h)(2g  5h)  4g(2g)  4g(5h)  3h(2g)  3h(5h)  8g2  20gh  6gh  15h2  8g2  14gh  15h2 13. (4a  3) (2a  1)  4a(2a)  4a(1)  3(2a)  3(1)  8a2  4a  6a  3  8a2  2a  3 14. (7y  1) (2y  3)  7y(2y)  7y(3)  1(2y)  1(3)  14y2  21y  2y  3  14y2  23y  3 15. (2x  3y)(4x  2y)  2x(4x)  2x(2y)  3y(4x)  3y(2y)  8x2  4xy  12xy  6y2  8x2  16xy  6y2 16. (12r  4s)(5r  8s)  12r(5r)  12r(8s)  4s(5r)  4s(8s)  60r2  96rs  20rs  32s2  60r2  76rs  32s2 17. (a  1)(3a  2)  a(3a)  a(2)  1(3a)  1(2)  3a2  2a  3a  2  3a2  a  2 18. (2n  4)(3n  2)  2n(3n)  2n(2)  4(3n)  4(2)  6n2  4n  12n  8  6n2  8n  8

Extra Practice

25. (5x  2)(5x2  2x  7)  5x(5x2  2x  7)  2(5x2  2x  7)  25x3  10x2  35x  10x2  4x  14  25x3  20x2  31x  14 26. (n  2)(2n2  n  1)  n(2n2  n  1)  2(2n2  n  1)  2n3  n2  n  4n2  2n  2  2n3  5n2  3n  2 27. (x2  7x  4)(2x2  3x  6)  x2 (2x2  3x  6)  7x(2x2  3x  6)  4(2x2  3x  6)  2x4  3x3  6x2  14x3  21x2  42x  8x2  12x  24  2x4  17x3  23x2  30x  24

28. (x2  x  1) (x2  x  1)  x2 (x2  x  1)  x(x2  x  1)  1(x2  x  1)  x4  x3  x2  x3  x2  x  x2  x  1  x4  x2  2x  1

29.

(a2  2a  5)(a2  3a  7)  a2 (a2  3a  7)  2a(a2  3a  7)  5(a2  3a  7)  a4  3a3  7a2  2a3  6a2  14a  5a2  15a  35  a4  a3  8a2  29a  35

30.

(5x4  2x2  1)(x2  5x  3)  5x4 (x2  5x  3)  2x2 (x2  5x  3)  1(x2  5x  3)  5x6  25x5  15x4  2x4  10x3  6x2  x2  5x  3  5x6  25x5  13x4  10x3  5x2  5x  3

Page 839

Lesson 8-8

1. (t  7) 2  t2  2(t) (7)  72  t2  14t  49 2. (w  12)(w  12)  w2  122  w2  144 2 2 3. (q  4h)  q  2(q) (4h)  (4h) 2  q2  8hq  16h2

716

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28. (x  2)(x  2) (2x  5)  (x2  22 ) (2x  5)  (x2  4) (2x  5)  x2 (2x)  x2 (5)  4(2x)  4(5)  2x3  5x2  8x  20 29. (4x  1)(4x  1) (x  4)  [ (4x) 2  12 ] (x  4)  (16x2  1) (x  4)  16x2 (x)  16x2 (4)  1(x)  1(4)  16x3  64x2  x  4 30. (x  5)(x  5) (x  4) (x  4)  (x2  52 ) (x2  42 )  (x2  25)(x2  16)  x2 (x2 )  x2 (16)  25(x2 )  25(16)  x4  16x2  25x2  400  x4  41x2  400 31. (a  1)(a  1) (a  1) (a  1)  (a  1) (a  1) (a  1) (a  1)  (a2  12 ) (a2  12 )  (a2  1) 2  (a2 ) 2  2(a2 ) (1)  12  a4  2a2  1 32. (n  1)(n  1) (n  1)  (n2  12 ) (n  1)  (n2  1) (n  1)  n2 (n)  n2 (1)  1(n)  1(1)  n3  n2  n  1 33. (2c  3)(2c  3) (2c  3) (2c  3)  (2c  3) (2c  3) (2c  3) (2c  3)  [ (2c) 2  32 ] [ (2c) 2  32 ]  (4c2  9) (4c2  9)  (4c2  9) 2  (4c2 ) 2  2(4c2 ) (9)  92  16c4  72c2  81 34. (4d  5e)(4d  5e) (4d  5e) (4d  5e)  (4d  5e) (4d  5e) (4d  5e) (4d  5e)  [ (4d) 2  (5e) 2 ] [ (4d) 2  (5e) 2 ]  (16d2  25e2 ) (16d2  25e2 )  (16d2  25e2 ) 2  (16d2 ) 2  2(16d2 ) (25e2 )  (25e2 ) 2  256d2  800e2d2  625e4

4. (10x  11y)(10x  11y)  (10x) 2  (11y) 2  100x2  121y2 2 2 5. (4e  3)  (4e)  2(4e)(3)  32  16e2  24e  9 6. (2b  4d)(2b  4d)  (2b) 2  (4d) 2  4b2  16d2 2 2 7. (a  2b)  a  2(a)(2b)  (2b) 2  a2  4ab  4b2 8. (3x  y) 2  (3x) 2  2(3x)(y)  y2  9x2  6xy  y2 9. (6m  2n) 2  (6m) 2  2(6m) (2n)  (2n) 2  36m2  24mn  4n2 2 10. (3m  7d)  (3m) 2  2(3m) (7d)  (7d) 2  9m2  42md  49d2 11. (5b  6) (5b  6)  (5b) 2  62  25b2  36 12. (1  x) 2  1  2(1)(x)  x2  1  2x  x2 13. (5x  9y) 2  (5x) 2  2(5x)(9y)  (9y) 2  25x2  90xy  81y2 14. (8a  2b)(8a  2b)  (8a) 2  (2b) 2  64a2  4b2 15.

114 x  422  114 x22  2114 x2 (4)  (4)2 1

 16 x2  2x  16

16. (c  3d) 2  c2  2(c)(3d)  (3d) 2  c2  6cd  9d2 17. (5a  12b) 2  (5a) 2  2(5a)(12b)  (12b) 2  25a2  120ab  144b2 18.

112 x  y22  112 x22  2112 x2 (y)  y2 1

 4x2  xy  y2

19. (n2  1) 2  (n2 ) 2  2(n2 ) (1)  (1) 2  n4  2n2  1 2 2 20. (k  3j)  (k2 ) 2  2(k2 )(3j)  (3j) 2  k4  6k2j  9j2 2  5) (a2  5)  (a2 ) 2  (5) 2 21. (a  a4  25 3  7)(2x3  7)  (2x3 ) 2  (7) 2 22. (2x  4x6  49 3 3 23. (3x  9y)(3x  9y)  (3x3 ) 2  (9y) 2  9x6  81y2 2 2 24. (7a  b) (7a  b)  (7a2 ) 2  (b) 2  49a4  b2 25.

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112x  102112x  102  112x22  (10)2 1

 4x2  100

26.

Lesson 9-1

1. List all pairs of numbers whose product is 23. 1 23 factors: 1, 23 prime 2. List all pairs of numbers whose product is 21. 1 21 3 7 factors: 1, 3, 7, 21 composite 3. List all pairs of numbers whose product is 81. 1 81 3 27 9 9 factors: 1, 3, 9, 27, 81 composite

113n  m2113n  m2  113n22  m2 1

 9n2  m2

27. (a  1)(a  1)(a  1)  (a  1) [ a2  2(a)(1)  12 ]  (a  1) [a2  2a  1]  a(a2  2a  1)  1(a2  2a  1)  a3  2a2  a  a2  2a  1  a3  3a2  3a  1

717

Extra Practice

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21. Factor each monomial and circle the common prime factors. 45  3 3 5 80  2 2 2 2 5 GCF: 5 22. Factor each monomial and circle the common prime factors. 29  1 29

4. List all pairs of numbers whose product is 24. 1 24 2 12 3 8 4 6 factors: 1, 2, 3, 4, 6, 8, 12, 24 composite 5. List all pairs of numbers whose product is 18. 1 18 2 9 3 6 factors: 1, 2, 3, 6, 9, 18 composite 6. List all pairs of numbers whose product is 22. 1 22 2 11 factors: 1, 2, 11, 22 composite 7. 42  2 21  2 3 7 8. 267  3 89 9. 72  1 72  1 2 36  1 2 3 12  1 2 3 3 4  1 23 32 10. 164  2 82  2 2 41  22 41 11. 57  1 57  1 3 19 12. 60  1 60  1 2 30  1 2 2 15  1 22 3 5 13. 240mn  2 120 m n  2 2 60 m n  2 2 2 30 m n  2 2 2 2 15 m n  2 2 2 2 3 5 m n 14. 64a3b  1 2 32 a a a b  1 2 2 16 a a a b  1 2 2 2 8 a a a b  1 2 2 2 2 2 2 a a a b 15. 26xy2  1 2 13 x y y 16. 231xy2z  1 3 77 x y y z  1 3 7 11 x y y z 17. 44rs2t3  2 22 r s s t t t  2 2 11 r s s t t t 18. 756m2n2  1 2 378 m m n n  1 2 2 189 m m n n  1 2 2 3 63 m m n n  1 2 2 3 3 21 m m n n  1 2 2 3 3 3 7 m m n n 19. Factor each monomial and circle the common prime factors. 16  2 2 2 2 60  2 2 3 5 GCF: 2 2 or 4 20. Factor each monomial and circle the common prime factors. 15  3 5 50  2 5 5 GCF: 5

Extra Practice

23.

24.

25.

26.

27.

28.

29.

30.

31.

718

58  2 29 GCF: 29 Factor each monomial and circle the prime factors. 55  5 11 305  5 61 GCF: 5 Factor each monomial and circle the prime factors. 126  2 3 3 7 252  2 2 3 3 7 GCF: 2 3 3 7 or 126 Factor each monomial and circle the prime factors. 128  2 2 2 2 2 2 2 245  5 7 7 GCF: 1 Factor each monomial and circle the prime factors. 7y2  7 y y 14y2  2 7 y y GCF: 7 y y or 7y2 Factor each monomial and circle the prime factors. 4xy  2 2 x y 6x  1 2 3 x GCF: 2x Factor each monomial and circle the prime factors. 35t2  5 7 t t 7t  7 t GCF: 7t Factor each monomial and circle the prime factors. 16pq2  2 2 2 2 p q q 12p2q  2 2 3 p p q 4pq  2 2 p q GCF : 2 2 p q or 4pq Factor each monomial and circle the prime factors. 5 5 1 15  3 5 10  2 5 GCF: 5 Factor each monomial and circle the prime factors. 12 mn  2 2 3 m n 10 mn  2 5 m n 15 mn  3 5 m n GCF: mn

common

common

common

common

common

common

common

common

common

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17. d(d  11)  0 d  0 or d  11  0 d  11 {11, 0} 18. z(z  2.5)  0 z  0 or z  2.5  0 z  2.5 {0, 2.5} 19. (2y  6)(y  1)  0 2y  6  0 or y  1  0 y  3 y1 {3, 1} 20. (4n  7)(3n  2)  0 4n  7  0 or 3n  2  0

32. Factor each monomial and circle the common prime factors. 14xy  2 7 x y 12y  2 2 3 y 20x  2 2 5 x GCF: 2 33. Factor each monomial and circle the common prime factors. 26jk4  2 13 j k k k k 16jk3  2 2 2 2 j k k k 8j2  2 2 2 j j GCF: 2j

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523, 74 6

10a  40a  10a(a  4) 15wx  35wx2  5wx(3  7x) 27a2b  9b3  9b(3a2  b2 ) 11x  44x2y  11x(1  4xy) 16y2  8y  8y(2y  1) 14mn2  2mn  2mn(7n  1) 25a2b2  30ab3  5ab2 (5a  6b) 2m3n2  16mn2  8mn  2mn(m2n  8n  4) 2ax  6xc  ba  3bc  (2ax  6xc)  (ba  3bc)  2x(a  3c)  b(a  3c)  (2x  b)(a  3c) 10. 6mx  4m  3rx  2r  (6mx  4m)  (3rx  2r)  2m(3x  2)  r(3x  2)  (2m  r)(3x  2) 11. 3ax  6bx  8b  4a  (3ax  6bx)  (8b  4a)  3x(a  2b)  4(2b  a)  3x(a  2b)  4(1) (2b  a)  3x(a  2b)  (4)(a  2b)  (3x  4)(a  2b) 12. a2  2ab  a  2b  (a2  2ab)  (a  2b)  a(a  2b)  1(a  2b)  (a  1)(a  2b) 13. 8ac  2ad  4bc  bd  (8ac  2ad)  (4bc  bd)  2a(4c  d)  b(4c  d)  (2a  b)(4c  d) 14. 2e2g  2fg  4e2h  4fh  (2e2g  2fg)  (4e2h  4fh)  2g(e2  f )  4h(e2  f )  (2g  4h)(e2  f )  2(g  2h)(e2  f ) 2  xy  xy  y2  (x2  xy)  (xy  y2 ) 15. x  x(x  y)  y(x  y)  x(x  y)  y(1)(x  y)  (x  y)(x  y)  (x  y) 2 Exercises 16–27 For checks, see students’ work. 16. a(a  9)  0 a  0 or a  9  0 a9 {0, 9} 1. 2. 3. 4. 5. 6. 7. 8. 9.

7

2

n4

Lesson 9-2

2

n  3

21. (a  1)(a  1)  0 a  1  0 or a  1  0 a  1 a1 {1, 1} 22. 10x2  20x  0 10x(x  2)  0 10x  0 or x  2  0 x0 x2 {0, 2} 23. 8b2  12b  0 4b(2b  3)  0 4b  0 or 2b  3  0 3 b0 b2

50, 32 6

or {0, 1.5}

2

24. 14d  49d  0 7d(2d  7)  0 7d  0 or 2d  7  0 d0

7

d  2

572, 06 or {3.5, 0}

15a2  60a 15a2  60a  0 15a(a  4)  0 15a  0 or a  4  0 a0 a4 {0, 4} 33x2  22x 26. 2  22x  0 33x 11x(3x  2)  0 11x  0 or 3x  2  0 25.

x0

27.

523, 06

2

x  3

32x2  16x 32x  16x  0 16x(2x  1)  0 16x  0 or 2x  1  0 2

x0

50, 12 6 719

1

x2

Extra Practice

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Page 840

18. k2  27kj  90j2  k2  (27j)k  90j2

Lesson 9-3

1. Among all pairs of factors of 14, choose 7 and 2, the pair of factors whose sum is 9. x2  9x  14  (x  7)(x  2) 2. Among all pairs of factors of 36, choose 12 and 3, the pair of factors whose sum is 9. a2  9a  36  (a  12)(a  3) 3. Among all pairs of factors of 15, choose 5 and 3, the pair of factors whose sum is 2. x2  2x  15  (x  5)(x  3) 4. Among all pairs of factors of 15, choose 5 and 3, the pair of factors whose sum is 8. n2  8n  15  (n  5)(n  3) 5. Among all pairs of factors of 21, choose 21 and 1, the pair of factors whose sum is 22. b2  22b  21  (b  21)(b  1) 6. Among all pairs of factors of 3, choose 3 and 1, the pair of factors whose sum is 2. c2  2c  3  (c  3)(c  1) 7. Among all pairs of factors of 24, choose 8 and 3, the pair of factors whose sum is 5. x2  5x  24  (x  8)(x  3) 8. Among all pairs of factors of 7, choose 7 and 1, the pair of factors whose sum is 8. n2  8n  7  (n  7)(n  1) 9. Among all pairs of factors of 39, choose 13 and 3, the pair of factors whose sum is 10. m2  10m  39  (m  13)(m  3) 10. Among all pairs of factors of 36, choose 12 and 3, the pair of factors whose sum is 15. z2  15z  36  (z  12)(z  3) 11. s2  13st  30t2  s2  (13t)s  30t2 Among all pairs of factors of 30t2, choose 15t and 2t, the pair of factors whose sum is 13t. s2  13st  30t2  (s  15t) (s  2t) 12. Among all pairs of factors of 35, choose 5 and 7, the pair of factors whose sum is 2. y2  2y  35  ( y  5)( y  7) 13. Among all pairs of factors of 40, choose 5 and 8, the pair of factors whose sum is 3. r2  3r  40  (r  5)(r  8) 14. Among all pairs of factors of 6, choose 1 and 6, the pair of factors whose sum is 5. x2  5x  6  (x  6)(x  1) 15. x2  4xy  5y2  x2  (4y)x  5y2 Among all pairs of factors of 5y2, choose y and 5y, the pair of factors whose sum is 4y. x2  4xy  5y2  (x  y)(x  5y) 16. Among all pairs of factors of 63, choose 9 and 7, the pair of factors whose sum is 16. r2  16r  63  (r  9) (r  7) 17. Among all pairs of factors of 52, choose 26 and 2, the pair of factors whose sum is 24. v2  24v  52  (v  26)(v  2)

Extra Practice

Among all pairs of factors of 90j2, choose 30j and 3j, the pair of factors whose sum is 27j. k2  27kj  90j2  (k  30j) (k  3j) Exercises 19–33 For checks, see students’ work. 19. a2  3a  4  0 (a  4)(a  1)  0 a  4  0 or a  1  0 a  4 a1 {4, 1} 20. x2  8x  20  0 (x  2)(x  10)  0 x  2  0 or x  10  0 x  2 x  10 {2, 10} 21. b2  11b  24  0 (b  3)(b  8)  0 b  3  0 or b  8  0 b  3 b  8 {8, 3} 22. y2  y  42  0 (y  7)(y  6)  0 y  7  0 or y  6  0 y  7 y6 {7, 6} 23. k2  2k  24  0 (k  4)(k  6)  0 k  4  0 or k  6  0 k4 k  6 {6, 4} 24. r2  13r  48  0 (r  16)(r  3)  0 r  16  0 or r  3  0 r  16 r  3 {3, 16} 25. n2  9n  18 2 n  9n  18  0 (n  6)(n  3)  0 n  6  0 or n  3  0 n6 n3 {3, 6} 26. 2z  z2  35 2  2z  35  0 z (z  7)(z  5)  0 z  7  0 or z  5  0 z  7 z5 {7, 5} 27. 20x  19  x2 2 x  20x  19  0 (x  1)(x  19)  0 x  1  0 or x  19  0 x1 x  19 {1, 19} 28. 10  a2  7a 2 a  7a  10  0 (a  5)(a  2)  0 a  5  0 or a  2  0 a  5 a  2 {5, 2}

720

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29.

z2  57  16z z  16z  57  0 (z  3)(z  19)  0 z  3  0 or z  19  0 z  3 z  19 {3, 19} x2  14x  3 2  14x  33  0 x (x  11) (x  3)  0 x  11  0 or x  3  0 x  11 x  3 {11, 3} 22x  x2  96 2  22x  96  0 x (x  6)(x  16)  0 x  6  0 or x  16  0 x6 x  16 {6, 16} 144  q2  26q 2  26q  144  0 q (q  8)(q  18)  0 q  8  0 or q  18  0 q8 q  18 {8, 18} x2  84  20x 2  20x  84  0 x (x  6)(x  14)  0 x  6  0 or x  14  0 x6 x  14 {6, 14}

6. Among all positive pairs of factors of 24, choose 3 and 8, the pair of factors whose sum is 11. 4y2  11y  6  4y2  3y  8y  6  y(4y  3)  2(4y  3)  (4y  3) (y  2) 7. Among all pairs of factors of 18, choose 2 and 9, the pair of factors whose sum is 7. 6n2  7n  3  6n2  2n  9n  3  2n(3n  1)  3(3n  1)  (2n  3) (3n  1) 8. Among all pairs of negative factors of 70, choose 7 and 10, the pair of factors whose sum is 17. 5x2  17x  14  5x2  7x  10x  14  x(5x  7)  (2) (5x  7)  (5x  7) (x  2) 9. Among all pairs of negative factors of 26, there are no pairs whose sum is 11. The trinomial is prime. 10. Among all pairs of factors of 24, choose 2 and 12, the pair of factors whose sum is 10. 8m2  10m  3  8m2  12m  2m  3  4m(2m  3)  1(2m  3)  (2m  3) (4m  1) 11. Among all pairs of factors of 12, there are no pairs whose sum is 2. 6y2  2y  2  2(3y2  y  1) 12. Among all pairs of factors of 28, choose 4 and 7, the pair of factors whose sum is 3. 2r2  3r  14  2r2  4r  7r  14  2r(r  2)  7(r  2)  (2r  7) (r  2) 13. Among all pairs of negative factors of 75, there are no pairs whose sum is 3. The trinomial is prime. 14. Factor out the GCF, 6. 18v2  24v  12  6(3v2  4v  2) Now factor 3v2  4v  2. Among all pairs of positive factors of 6, there are no pairs whose sum is 4. Thus, 18v2  24v  12  6(3v2  4v  2) . 15. Factor out the GCF, 2. 4k2  2k  12  2(2k2  k  6) Now factor 2k2  k  6. Among all pairs of factors of 12, choose 3, and 4, the pair of factors whose sum is 1. (2k2  k  6)  2k2  3k  4k  6  k(2k  3)  2(2k  3)  (2k  3) (k  2) Thus, 4k2  2k  12  2(2k  3) (k  2) .

2

30.

31.

32.

33.

Page 840

Lesson 9-4

1. Among all pairs of factors of 252, choose 14 and 18, the pair of factors whose sum is 4. 4a2  4a  63  4a2  14a  18a  63  2a(2a  7)  9(2a  7)  (2a  7) (2a  9) 2. Among all pairs of factors of 18, choose 9 and 2, the pair of factors whose sum is 7. 3x2  7x  6  3x2  9x  2x  6  3x(x  3)  2(x  3)  (3x  2)(x  3) 3. Among all pairs of negative factors of 24, choose 1 and 24, the pair of factors whose sum is 25. 4r2  25r  6  4r2  r  24r  6  r(4r  1)  (6)(4r  1)  (4r  1)(r  6) 4. Among all pairs of negative factors of 30, choose 5 and 6, the pair of factors whose sum is 11. 2z2  11z  15  2z2  5z  6z  15  z(2z  5)  (3) (2z  5)  (2z  5)(z  3) 5. Among all pairs of factors of 63, choose 7 and 9, the pair of factors whose sum is 2. 3a2  2a  21  3a2  9a  7a  21  3a(a  3)  7(a  3)  (3a  7) (a  3)

721

Extra Practice

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16. Factor out the GCF, 10. 10x2  20xy  10y2  10(x2  2xy  y2 ) Now factor x2  2xy  y2. Among all pairs of factors of y2, choose y and y, the pair of factors whose sum is 2y. 10(x2  2xy  y2 )  10[x2  yx  yx  y2 ]  10[x(x  y)  y(x  y) ]  10[ (x  y)(x  y) ]  10(x  y) 2 2  20xy  10y2  10(x  y)(x  y) Thus, 10x or 10(x  y)2. 17. Among all pairs of factors of 60, choose 4 and 15, the pair of factors whose sum is 11. 12c2  11cd  5d2  12c2  15cd  4cd  5d2  3c(4c  5d)  d(4c  5d)  (3c  d)(4c  5d) 18. Among all pairs of factors of 30, choose 5 and 6, the pair of factors whose sum is 1. 30n2  mn  m2  30n2  5mn  6mn  m2  5n(6n  m)  m(6n  m)  (5n  m)(6n  m) Exercises 19–33 For checks, see students’ work. 19. 8t2  32t  24  0 8(t2  4t  3)  0 t2  4t  3  0 (t  3)(t  1)  0 t  3  0 or t  1  0 t  3 t  1 {3, 1} 20. 6y2  72y  192  0 6(y2  12y  32)  0 y2  12y  32  0 (y  4)(y  8)  0 y  4  0 or y  8  0 y  4 y  8 {8, 4} 21. 5x2  3x  2  0 (5x  2)(x  1)  0 5x  2  0 or x  1  0 2

x5

51, 6

24.

3

n2

25.

516, 32 6

1 6

n

12x2  x  35  0  21x)  (20x  35)  0 3x(4x  7)  5(4x  7)  0 (4x  7)(3x  5)  0 4x  7  0 or 3x  5  0 (12x2

7

5

x4

26.

553, 74 6

x  3

18x2  36x  14  0 2(9x2  18x  7)  0 9x2  18x  7  0 9x2  21x  3x  7  0 (9x2  21x)  (3x  7)  0 3x(3x  7)  1(3x  7)  0 (3x  7)(3x  1)  0 3x  7  0 or 3x  1  0 7

x  3

27.

573, 13 6

1

x3

15a2  a  2  0  5a)  (6a  2)  0 5a(3a  1)  2(3a  1)  0 (3a  1)(5a  2)  0 3a  1  0 or 5a  2  0 (15a2

1

2

a3

28.

x  1

2 5

22. 9x2  18x  27  0 9(x2  2x  3)  0 x2  2x  3  0 (x  1)(x  3)  0 x  1  0 or x  3  0 x1 x  3 {3, 1} 23. 4x2  4x  4  4 4x2  4x  8  0 4(x2  x  2)  0 x2  x  2  0 (x  1)(x  2)  0 x  1  0 or x  2  0 x  1 x2 {1, 2}

Extra Practice

12n2  16n  3  0  18n)  (2n  3)  0 6n(2n  3)  1(2n  3)  0 (2n  3)(6n  1)  0 2n  3  0 or 6n  1  0 (12n2

525, 13 6

a  5

14b2  7b  42  0 7(2b2  b  6)  0 2b2  b  6  0 2b2  4b  3b  6  0 2b(b  2)  3(b  2)  0 (2b  3)(b  2)  0 2b  3  0 or b  2  0 3

b2

29.

52, 32 6

b  2

13r2  21r  10  0 (13r  5r)  (26r  10)  0 r(13r  5)  2(13r  5)  0 (13r  5)(r  2)  0 13r  5  0 or r  2  0 2

5

r  13

52, 6 5 13

722

r  2

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30.

31.

35y2  60y  20  0 5(7y2  12y  4)  0 7y2  12y  4  0 2 7y  14y  2y  4  0 7y(y  2)  2(y  2)  0 (7y  2)(y  2)  0 7y  2  0 or y  2  0

5

2

y  7 2 7,

2

6

y2

13. 9x2  100y2  (3x) 2  (10y) 2  (3x  10y) (3x  10y)

16x2  4x  6  0 2(8x2  2x  3)  0 8x2  2x  3  0 2 8x  6x  4x  3  0 2x(4x  3)  1(4x  3)  0 (2x  1) (4x  3)  0 2x  1  0 or 4x  3  0 1

x  2

32.

9. 75r2  48  3(25r2  16)  3[ (5r) 2  (4) 2 ]  3(5r  4) (5r  4) 10. x2  36y2  (x) 2  (6y) 2  (x  6y) (x  6y) 11. 3a2  16 prime 12. 12t2  75  3(4t2  25)  3[ (2t) 2  (5) 2 ]  3(2t  5) (2t  5)

512, 34 6

14. 49  a2b2  (7) 2  (ab) 2  (7  ab) (7  ab) 15. 5a2  48 prime 16. 169  16t2  (13) 2  (4t) 2  (13  4t) (13  4t) 17. 8r2  4  4(2r2  1)

3

x4

18. 45m2  5  5(9m2  1)  5[ (3m) 2  (1) 2 ]  5(3m  1) (3m  1) Exercises 19–33 For checks, see students’ work. 4x2  16 19. 2  16  0 4x 4(x2  4)  0 x2  4  0 2  (2) 2  0 x 4(x  2)(x  2)  0 or x  2  0 x20 x  2 x2 { 2} 2x2  50 20. 2  50  0 2x 2(x2  25)  0 x2  25  0 x2  52  0 (x  5)(x  5)  0 x  5  0 or x  5  0 x5 x  5 { 5} 9n2  4  0 21. (3n) 2  (2) 2  0 (3n  2)(3n  2)  0 3n  2  0 or 3n  2  0 3n  2 3n  2

28d2  5d  3  0 28d  12d  7d  3  0 4d(7d  3)  1(7d  3)  0 (4d  1)(7d  3)  0 4d  1  0 or 7d  3  0 2

1

3

d4

33.

537, 14 6

d  7

30x2  9x  3  0 3(10x2  3x  1)  0 10x2  3x  1  0 2  5x  2x  1  0 10x 5x(2x  1)  1(2x  1)  0 (5x  1) (2x  1)  0 5x  1  0 or 2x  1  0 1

x  5

515, 12 6 Page 841

1

x2

Lesson 9-5

1. x2  9  x2  32  (x  3)(x  3) 2. a2  64  a2  82  (a  8)(a  8) 3. 4x2  9y2  (2x) 2  (3y) 2  (2x  3y) (2x  3y) 4. 1  9z2  12  (3z) 2  (1  3z)(1  3z) 5. 16a2  9b2  (4a) 2  (3b) 2  (4a  3b) (4a  3b) 6. 8x2  12y2  4(2x2  3y2 ) 7. a2  4b2  (a) 2  (2b) 2  (a  2b)(a  2b) 8. x2  y2  (x  y)(x  y)

22.

2

2

n3

5 23 6

n  3 25

a2  36  0

156 22  0 1a  56 21a  56 2  0 a2 

5

a60 5

a6

5 56 6 723

or

5

a60 a

5 6

Extra Practice

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23.

16 9 4 2 3

1 2  b2  0 143  b2143  b2  0

4 3

b0 b

24.

5 43 6

72  2z2  0 2(36  z2 )  0 36  z2  0 (6) 2  z2  0 (6  z)(6  z)  0 6  z  0 or 6  z  0 z6 z  6 { 6} 30. 25a2  1 2 25a  1  0

 b2  0

4 3

or

29.

b0

4 3

4

b  3 1

1

18  2x2  0

1 (36 2

1  25 1 2 a2  5

36  x2  0 (6) 2  x2  0 (6  x)(6  x)  0 6  x  0 or 6  x  0 x6 x  6 { 6} 25. 20  5g2  0 5(4  g2 )  0 4  g2  0 (2) 2  g2  0 (2  g)(2  g)  0 2  g  0 or 2  g  0 g2 g  2 { 2}

1

a50

 p2 )  0

1 2 c 4

1 4

4

90

1c2  169 2  0 16 9 4 2 3

c2 

c2



12

0 0

1c  43 21c  43 2  0 4

c30

or

4

c  3

28.

5 43 6

or

5 15 6

1

a50

1

1

a  5

4

c30

Page 841

4

c3

Lesson 9-6

1. Yes; The first term is a perfect square: x2 The last term is a perfect square: 62 The middle term is 2(x)(6). x2  12x  36  x2  2(x) (6)  62  (x  6) 2 2. No; the middle term is not 2(n)(6). 3. Yes; The first term is a perfect square: a2 The last term is a perfect square: 22 The middle term is 2(a)(2). a2  4a  4  a2  2(a) (2)  22  (a  2) 2

3z2  48  0 3(z2  16)  0 z2  16  0 2 z  (4) 2  0 (z  4)(z  4)  0 z  4  0 or z  4  0 z4 z  4 { 4}

Extra Practice

0

2q3  2q  0 2q(q2  1)  0 2q(q  1)(q  1)  0 or q  1  0 2q  0 or q  1  0 q0 q  1 q1 {1, 0, 1} 32. 3r3  48r 3  48r  0 3r 3r(r2  16)  0 3r [ r2  (4) 2 ]  0 3r(r  4)(r  4)  0 3r  0 or r  4  0 or r  4  0 r0 r4 r  4 {4, 0, 4} 33. 100d  4d3  0 4d(25  d2 )  0 4d [ (5) 2  d2 ]  0 4d(5  d)(5  d)  0 4d  0 or 5  d  0 or 5  d  0 d0 d5 d  5 {5, 0, 5} 31.

64  p2  0 (8) 2  p2  0 (8  p)(8  p)  0 8  p  0 or 8  p  0 p8 p  8 { 8} 27.

12

0

a5

1

1 (64 4

a2

1a  15 21a  15 2  0

16  4p2  0

26.

2

1 25 a2  25  0

 x2 )  0

724

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4. No; the last term, 100, is not a perfect square. 5. No; the first term, 2n2, is not a perfect square. 6. Yes; The first term is a perfect square: (2a)2 The last term is a perfect square: 52 The middle term is 2(2a)(5). 4a2  20a  25  (2a) 2  2(2a)(5)  52  (2a  5) 2 2 2 7. 3x  75  3(x  25)  3(x  5)(x  5) 8. n2  8n  16  (n) 2  2(4)(n)  (4) 2  (n  4) 2 9. 4p2  12pr  9r2  (2p) 2  2(2p)(3r)  (3r) 2  (2p  3r) 2 10. 6a2  72  6(a2  12) 11. s2  30s  225  (s) 2  2(15)(s)  (15) 2  (s  15) 2 12. 24x2  24x  9  3(8x2  8x  3) 13. 1  10z  25z2  (1) 2  2(1)(5z)  (5z) 2  (1  5z) 2 14. 28  63b2  7(4  9b2 )  7(2  3b)(2  3b) 2 15. 4c  2c  7 is prime. Exercises 16–21 For checks, see students’ work. x2  22x  121  0 16. 2 x  2(x)(11)  (11) 2  0 (x  11) 2  0 x  11  0 x  11 {11} 17. 343d2  7

20.

21.

594 6

573 6

p3

9p2  42p  20  29 9p2  42p  49  0 2  2(3p) (7)  72  0 (3p) (3p  7) 2  0 3p  7  0 3p  7 7

Lesson 10-1

1. y  x2  6x  8 Sample answer: x 0 1 2 3 4 5 6

y 8 3 0 1 0 3 8

y

O

x

y  x 2  6x  8

2. y  x2  3x Sample answer:

7

1

d2  49

5 17 6

9

s4

Page 841

d2  343 d

16s2  81  72s 16s  72s  81  0 (4s) 2  2(4s) (9)  92  0 (4s  9) 2  0 4s  9  0 4s  9 2

x 1 0 1 2 3 4 5

1 7

2

18. (a  7)  5 a  7  15 a  7 15 {7 15} c2  10c  36  11 19. c2  10c  25  0 c2  2(c)(5)  52  0 (c  5) 2  0 c50 c  5 {5}

725

y 4 0 2 2 0 4 10

y y  x 2  3x O

x

Extra Practice

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3. y  x2 Sample answer: x 3 2 1 0 1 2 3

7. In y  x2  2x  3, a  1 and b  2. x

y 9 4 1 0 1 4 9

y

b 2a (2) 2(1)

or 1 is the equation of the axis of x symmetry. y  (1) 2  2(1)  3 y  1  2  3 y  2 The vertex is at (1, 2). Since the coefficient of the x2 term is negative, the vertex is a maximum.

O

x y  x 2

y O

y  x 2  2x  3

4. y  x2  x  3 Sample answer: x 3 2 1 0 1 2 3

y 9 5 3 3 5 9 15

y

8. In y  3x2  24x  80, a  3 and b  24. x y  x2  x  3

5. y  x2  1 Sample answer: y 10 5 2 1 2 5 10

y

Extra Practice

y 40 25 16 13 16 25 40

140 120 100 80 60 40 20

y  x2  1 x

O

y  3x 2  24x  80 8

6. y  3x2  6x  16 Sample answer: x 4 3 2 1 0 1 2

b 2a (24) 2(3)

or 4 is the equation of the axis of x symmetry. y  3(4) 2  24(4)  80 y  48  96  80 y  32 The vertex is at (4, 32). Since the coefficient of the x2 term is positive, the vertex is a minimum.

x

O

x 3 2 1 0 1 2 3

x

y 28 20 12 y  3x 2  6x  16 4 2

O

2

x

726

6

4

2

O

y

2x

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9. In y  x2  4x  4, a  1 and b  4. x

Since the coefficient of the x2 term is positive, the vertex is a minimum.

b 2a (4) 2(1)

y

x or 2 is the equation of the axis of symmetry. y  (2) 2  4(2)4 y484 y  8 The vertex is at (2, 8). Since the coefficient of the x2 term is positive, the vertex is a minimum.

y  3x 2  6x  3 x

O

y

12. In y  2x2  12x, a  2 and b  12. x x

O

b 2a (12) 2(2)

x or 3 is the equation of the axis of symmetry. y  2(3) 2  12(3) y  18  36 y  18 The vertex is at (3, 18). Since the coefficient of the x2 term is positive, the vertex is a minimum.

y  x 2  4x  4

10. In y  5x2  20x  37, a  5 and b  20. x

12 8 4

b 2a (20) 2(5)

x or 2 is the equation of the axis of symmetry. y  5(2) 2  20(2)  37 y  20  40  37 y  17 The vertex is at (2, 17). Since the coefficient of the x2 term is positive, the vertex is a minimum. 70 60 50 40 30 20 10 2

O

6

y  2x 2  12x

x

2x

b 2a (6) 2(1)

x or 3 is the equation of the axis of symmetry. y  (3) 2  6(3)  5 y  9  18  5 y  4 The vertex is at (3, 4). Since the coefficient of the x2 term is positive, the vertex is a minimum.

y  5x 2  20x  37 4x

y

11. In y  3x2  6x  3, a  3 and b  6. x

2 4 O 8 12 16 20

13. In y  x2  6x  5, a  1 and b  6.

y

2

4

y

b 2a (6) 2(3)

or 1 is the equation of the axis of x symmetry. y  3(1) 2  6(1)  3 y363 y0 The vertex is at (1, 0).

O

x

y  x 2  6x  5

727

Extra Practice

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14. In y  x2  6x  9, a  1 and b  6. x

Since the coefficient of the x2 term is positive, the vertex is a minimum.

b 2a (6) 2(1)

y

x or 3 is the equation of the axis of symmetry. y  (3) 2  6(3)  9 y  9  18  9 y0 The vertex is at (3, 0). Since the coefficient of the x2 term is positive, the vertex is a minimum.

x y  4x 2  1

O

y

17. In y  2x2  2x  4, a  2 and b  2. x x

b 2a (2) 2(2)

1

or 2 1

The equation of the axis of symmetry is x  2.

1 2

1 2

1 1 y  2 2 2  2 2  4

O

x

y

y  x 2  6x  9

y

15. In y  x2  16x  15, a  1 and b  16. x

1

1

2

y y  2x 2  2x  4

x

O

2 y y  x  16x  15

18. In y  6x2  12x  4, a  6 and b  12. x 4

12

b 2a (12) 2(6)

x or 1 is the equation of the axis of symmetry. y  6(1) 2  12(1)  4 y  6  12  4 y  10 The vertex is at (1, 10). Since the coefficient of the x2 term is positive, the vertex is a minimum.

x

16. In y  4x2  1, a  4 and b  0. x

1

Since the coefficient of the x2 term is negative, the vertex is a maximum.

x or 8 is the equation of the axis of symmetry. y  (8) 2  16(8)  15 y  64  128  15 y  49 The vertex is at (8, 49). Since the coefficient of the x2 term is negative, the vertex is a maximum.

O 10 20 30

14

The vertex is at 2, 42 .

b 2a (16) 2(1)

50 40 30 20 10

1 2 1 42

b 2a (0) 2(4)

x or 0 is the equation of the axis of symmetry. y  4(0) 2  1 y01 y  1 The vertex is at (0, 1).

y O

x

y  6x 2  12x  4

Extra Practice

728

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19. In y  x2  1, a  1 and b  0. x

Since the coefficient of the x2 term is negative, the vertex is a maximum.

b 2a (0) 2(1)

x or 0 is the equation of the axis of symmetry. y  (0) 2  1 y01 y  1 The vertex is at (0, 1). Since the coefficient of the x2 term is negative, the vertex is a maximum.

y  5x 2  3x  2 2 1 1

1x

O 1

y

22. In y  x2  x  20, a  1 and b  1.

x y  x 2  1

O

y

3

x

b 2a (1) 2(1)

1

or 2 is the equation of the axis of x symmetry. y

112 22  112 2  20

1

1

y  4  2  20 1

20. In y  x

x2

b 2a (1) 2(1)

y  204

 x  1, a  1 and b  1.

The vertex is at

Since the coefficient of the x2 term is negative, the vertex is a maximum.

1

or 2 is the equation of the axis of x symmetry. y

112 22  12  1

1

20 15 10 5

1

y  4  2  1 1

y  14 The vertex is at

112, 114 2.

4

Since the coefficient of the x2 term is negative, the vertex is a maximum. y

O

2

y  x 2  x  20

y

O

2

4

x

23. In y  2x2  5x  2, a  2 and b  5. x

y  x 2  x  1

b 2a (5) 2(2)

or 1.25 is the equation of the axis of x symmetry. y  2(1.25) 2  5(1.25)  2 y  3.125  6.25  2 y  5.125 The vertex is at (1.25, 5.125). Since the coefficient of the x2 term is positive, the vertex is a minimum.

x

21. In y  5x2  3x  2, a  5 and b  3. x

112, 2014 2.

3 2 1

b 2a (3) 2(5)

x or 0.3 is the equation of the axis of symmetry. y  5(0.3) 2  3(0.3)  2 y  .45  0.9  2 y  2.45 The vertex is at (0.3, 2.45).

3

729

2

1

y

O

1x 1 2 3 y  2x 2  5x  2 4 5

Extra Practice

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24. In y  3x2  18x 15, a  3 and b  18. x

3. Graph d2  36  0. Sample answer:

b 2a (18) 2(3)

d 2 1 0 1 2

x or 3 is the equation of the axis of symmetry. y  3(3) 2  18(3)  15 y  27  54  15 y  12 The vertex is at (3, 12). Since the coefficient of the x2 term is negative, the vertex is a maximum. y  3x 2  18x  15

6

4

2

15 10 5

5 10 20 25 30

f (d ) 60 50 40 30 20 f (d )  d 2  36 10 O 4321 1 2 3 4d

The graph has no d-intercept. Thus there are no real number solutions for this equation. 4. Graph b2  18b  81  0. Sample answer:

y

b 7 8 9 10 11

2x

O

f(d) 40 37 36 37 40

f(b) 4 1 0 1 4

f (b )  b 2  18b  81 5 4 3 2 1 O

Page 842

Lesson 10-2

f(a) 0 16 24 25 24 16 0

15 10 5 5

5 10 15 20 25

f (a )

x 3 2 1 0

5a

O

f(n) 0 12 15 16 15 12 0

16 12 8 4 2 4 8 12 16

6

4

2

f (x ) 40 35 30 25 20 f (x )  x 2  3x  27 15 10 5 x O 2

The graph has no x-intercept. Thus there are no real number solutions for this equation. 6. Graph y2  3y  10  0. Sample answer:

f (n)

y 7 5 3 2 1 0 2

O 2 4 6 8 10 n

f (n )  n 2  8n

The n-intercepts of the graph are 0 and 8. Thus, the solutions of the equation are 0 and 8.

Extra Practice

f(x) 27 25 25 27

f (a )  a 2  25

The a-intercepts of the graph are 5 and 5. Thus, the solutions of the equation are 5 and 5. 2. Graph n2  8n  0. Sample answer: n 0 2 3 4 5 6 8

b 2 4 6 8 10 12

The graph has one d-intercept, 9. Thus the solution of the equation is 9. 5. Graph x2  3x  27  0. Sample answer:

1. Graph a2  25  0. Sample answer: a 5 3 1 0 1 3 5

f (b )

f(y) 18 0 10 12 12 10 0

f (y )  y 2  3y  10 f (y ) 12 10 8 6 4 2 54321O

1 2y

The y-intercepts of the graph are 5 and 2. Thus, the solutions of the equation are 5 and 2.

730

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7. Graph x2  2x  3  0. Sample answer: x 5 3 2 1 0 1 3

11. Graph 3x2  6x  9  0. Sample answer:

f(x) 12 0 3 4 3 0 12

x 1 0 1 2 3

f (x )

x

O

f (x )  x 2  2x  3

f(x) 21 5 3 4 3 5 21

c 3 2 1 0 1 2 4

2 ( ) f (x ) f x  x  6x  5

x

O

f(a) 21 5 3 4 3 5 21

t 2 1 0 1 2

f (a ) f (a )  a 2  2a  3

f(r) 15 5 1 3 1 5 15

f (c )

2 1 2

1

O

1c

f (c )  c 2  c 1

f(t) 14 5 2 5 14

f (t )

f (t )  3t 2  2 O

t

The graph has no t-intercept. Thus, there are no real number solutions for this equation. 14. Graph b2  5b  2  0. Sample answer: b 1 0 1 2 3 4 5 6

f (r )

O

3

a

O

The a-intercepts of the graph are 3 and 1. So, the solutions of the equation are 3 and 1. 10. Graph 2r2  8r  5  0. Sample answer: r 1 0 1 2 3 4 5

f(c) 6 2 0 0 2 6 20

The c-intercepts of the graph are 0 and 1. So, the solutions of the equation are 0 and 1. 13. Graph 3t2  2  0. Sample answer:

The x-intercepts of the graph are 1 and 5. So, the solutions of the equation are 1 and 5. 9. Graph a2  2a  3  0. Sample answer: a 6 4 2 1 0 2 4

O f (x ) 21 1 2 3x 2 4 f (x )  3x 2  6x  9 6 8 10 12 14 16

The graph has no x-intercept. Thus, there are no real number solutions for this equation. 12. Graph c2  c  0. Sample answer:

The x-intercepts of the graph are 3 and 1. So, the solutions of the equation are 3 and 1. 8. Graph x2  6x  5  0. Sample answer: x 2 0 2 3 4 6 8

f(x) 18 9 6 9 18

r

f(b) 4 2 6 8 8 6 2 4

8 6 4 2 1 2 4 6 8

f (b )

f (b )  b 2  5b  2

O 1 2 3 4 5 6b

The b-intercepts of the graph are between 1 and 0 and between 5 and 6. So, one root is between 1 and 0, and the other is between 5 and 6.

f (r )  2r 2  8r  5

The r-intercepts of the graph are between 0 and 1 and between 3 and 4. So, one root is between 0 and 1 and the other is between 3 and 4.

731

Extra Practice

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15. Graph 3x2  7x  1. x 4 3 2 1 0 1 2

f(x) 19 5 3 5 1 9 25

19. Graph a2  12a  36  0. Sample answer:

f (x )

a 8 7 6 5 4

x

O

f (x )  3x 2  7x  1

f(x) 0 18 28 30 30 28 18 0

f(n) 8 1 4 7 8 7 4 1

O f (x ) 8

4

x 3 2 1 0 1 2 3

4x

5 10 15 20 25 30 f (x )  x 2  5x  24

f (n )  8  n 2

f(x) 0 24 30 30.25 30 24 0

10 5 2 5 10 15 20 25 30

x 3 2 1 0 1 2 3 n

z 4 3 2 1 0 1 2

x

O a

f(x) 55 60 63 64 63 60 55

f (x )  64  x 2

60 50 40 30 20 10

f (x )

8642O

2 4 6 8x

20

f(x) 41 19 5 1 1 11 29

f (x )  4x 2  2x  1 1 1

f (x )

1

O

x

1 2

f(z) 47 20 3 4 1 12 35

f (z ) 4 3 2 1 2

1

1 2 3 4 f (z )  5z 2  8z  1

O

1z

The z-intercepts of the graph are between 2 and 1 and between 0 and 1. So, one root is between 2 and 1, and the other is between 0 and 1.

f (x )  x 2  7x  18

The x-intercepts of the graph are 2 and 9. So, the solutions of the equation are 2 and 9. Extra Practice

2

The x-intercepts of the graph are between 1 and 0 and between 0 and 1. So, one root is between 1 and 0, and the other is between 0 and 1. 22. Graph 5z2  8z  1. Sample answer:

f (x ) O 2 4 6 8

4

The x-intercepts of the graph are 8 and 8. So, the solutions of the equation are 8 and 8. 21. Graph 4x2  2x  1. Sample answer:

f (n )

O

6

The graph has only one a-intercept, 6, So, the solution of the equation is 6. 20. Graph 64  x2  0. Sample answer:

The n-intercepts of the graph are between 3 and 2 and between 2 and 3. So, one root is between 3 and 2, and the other is between 2 and 3. 18. Graph x2  7x  18. Sample answer: x 2 1 3 3.5 4 6 9

6 5 4 3 2 1

f (a )  a 2  12a  36

The x-intercepts of the graph are 8 and 3. So, the solutions of the equation are 8 and 3. 17. Graph 8  n2  0. Sample answer: n 4 3 2 1 0 1 2 3

f (a )

8

The x-intercepts of the graph are between 3 and 2 and between 0 and 1. So, one root is between 3 and 4 and the other is between 0 and 1. 16. Graph x2  5x  24  0. Sample answer: x 8 6 4 3 2 1 1 3

f(a) 4 1 0 1 4

732

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3. b2  10b  25  11 (b  5) 2  11 2(b  5) 2  111 0b  5 0  111 b  5  111 b  5  5  111  5 b  5  111 b  5  111 or b  5  111  1.7  8.3 The solution set is {8.3, 1.7}. 4. a2  22a  121  3 (a  11) 2  3 2(a  11) 2  13 0a  11 0  13 a  11  13 a  11  11  13  11 a  11  13 a  11  13 or a  11  13  9.3  12.7 The solution set is {9.3, 12.7}. 5. x2  2x  1  81 (x  1) 2  81 2(x  1) 2  181 0x  1 0  181 x  1  181 x  1  1  181  1 x  1  181 x  1  181 or x  1  181 8  10 The solution set is {10, 8}. 6. t2  36t  324  85 (t  18) 2  85 2(t  18) 2  185 0t  18 0  185 t  18  185 t  18  18  185  18 t  18  185 t  18  185 or t  18  185  27.2  8.8 The solution set is {8.8, 27.2}. 7. a2  20a  c

23. Graph p  27  p2. Sample answer: p 7 6 5 4 3 1 0 1 2 3 4 5 6

f(p) 15 3 7 15 21 27 27 25 21 15 7 3 15

15 10 5

f (p )

O

8 642 5 10 15 20 25

2 4 6p

f (p )  p 2  p  27

The p-intercepts of the graph are between 6 and 5 and between 4 and 5. So, one root is between 6 and 5, and the other is between 4 and 5. 24. Graph 6w  15  3w2. Sample answer: x 3 2 1 0 1

f(x) 24 15 12 15 24

f (w ) 21 18 15 12 9 f (w )  3w 2  6w  15 6 3 321O

1w

The graph has no w-intercept. Thus, there are no real number solutions for this equation.

Page 842

Lesson 10-3

1. x2  4x  4  9 (x  2) 2  9 2(x  2) 2  19 0x  2 0  19 x  2  19 x  2  2  19  2 x  2  19 x  2  19 or x  2  19 5  1 The solution set is {1, 5}. 2. t2  6t  9  16 (t  3) 2  16 2(t  3) 2  116 0t  3 0  116 t  3  116 t  3  3  116  3 t  3  116 t  3  116 or t  3  116  1 7 The solution set is {1, 7}.

Find c

1 of 2 20 2 2

1 2

20 and square the result.

 102 or 100 8. x  10x  c 2

Find c

1 of 2 10 2 2

1 2

10 and square the result.

 52 or 25 9. t  12t  c 2

Find c

1 of 2 12 2 2

1 2

12 and square the result.

 62 or 36

733

Extra Practice

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10. y2  9y  c Find

1 2

16. 2y2

of 9 and square the result.

1 9 22

2y2  7y  4  0  7y  4  4  0  4 2y2  7y 2 7y y2  2 7y 49 y2  2  16 7 y42 7 y4 7 7 y44

c  2 

81 4

or 20.25

11. p2  14p  c 1 of 14 2 14 2 2 196 or 49 4

Find c 

1 2

1

and square the result.

1 of 13 and 2 13 2 2 169 or 42.25 4

c  13.

1 2

2 49

 2  16

2

81

 16 9

 4

7 9  4 4 7  9 y 4 7  9 7  9 y 4 or y  4 2 1 16  4 or 2   4 or 1 The solution set is 4, 2 .

12. b2  13b  c Find

4

2

square the result.

a2  8a  84  0 2  8a  84  84  0  84 a a2  8a  84 2  8a  16  84  16 a (a  4) 2  100 a  4  10 a  4  4  4  10 a  4  10 a  4  10 or a  4  10  14  6 The solution set is {6, 14}.

17. t2  3t 

1t

t

2

9 4

3 2 2 3 t2 3 3 2 2



 40 

5

9 4

6

169 4 13  2 3 13 t  2  2 3  13 t 2 3  13 3  13  or t 2 2



t5 t  8 The solution set is {8, 5}. x2  8x  9  0 18. 2 x  8x  9  9  0  9 x2  8x  9 2 x  8x  16  9  16 (x  4) 2  25 x  4  5 x  4  4  4  5 x  4  5 x  4  5 or x  4  5 1  9 The solution set is {9, 1}.

c2  6  5c c2  6  6  5c  5c  5c  6 c2  5c  6 c2  5c  6.25  6  6.25 (c  2.5) 2  0.25 c  2.5  0.5 c  2.5  2.5  2.5  0.5 c  2.5  0.5 c  2.5  0.5 or c  2.5  0.5  2  3 The solution set is {3, 2}. 15. p2  8p  5  0 2  8p  5  5  0  5 p p2  8p  5 2  8p  16  5  16 p ( p  4) 2  11 p  4  111 p  4  4  4  111 p  4  111 p  4  111 or p  4  111  0.7  7.3 The approximate solution set is {0.7, 7.3}.

14.

19.

y2  5y  84  0 y  5y  84  84  0  84 y2  5y  84 2

25 4 5 2 2 5 y2 5 5 2 2

y2  5y 

1y

y y

5  19 2

or

2

 84 

25 4

361 4 19  2 5 19  2  2 5  19 y 2 5  19 y 2



y  12 y7 The solution set is {12, 7}.

Extra Practice

4

734

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t2  12t  32  0 t  12t  32  32  0  32 t2  12t  32 2  12t  36  32  36 t (t  6) 2  4 t  6  2 t  6  6  6  2 t  6  2 t  6  2 or t  6  2  4  8 The solution set is {8, 4}. 2x  3x2  8 21. 3x2  2x  8

20.

23.

2

3x2  2x 3 2x x2  3 2x 4 x2  3  36 2 2 x6 2 x6 2 2 x66

1

2

z

4

 3  36 

2

5

6

1

4

3

2 25

 2  16 57

 16  

 157 4 5  157 4 5  157 4

z

or

5  257 4

2

0

8 1

08 1

8 1

t

9

02

3  211 4

11

 16  

or

 1.6

9

2

9

 8  16

t

0

 111 4 3  111 4 3  111 4

t

3  211 4

 0.08

The approximate solution set is {1.6, 0.08}.

1

 2  16 73

 16

Page 842

173 4 1 173  4 4 1  173 4

 

y y

5  257 4

8t2  12t  1 8 12t 1 1 2 t  8 88 3t t2  2 3t 9 t2  2  16 3 t42 3 t4 3 3 t44

2

9

02

 3.1  0.6 The approximate solution set is {0.6, 3.1}. 24. 8t2  12t  1  0

8 8

2

0

2

z

3

2y2  y  9 2 y 9 9 y2  2  2  2 y y2  2 y 1 y2  2  16 1 y42 1 y4 1 1 y44

1  273 4

1

8

2y2  y  9  0

1

2z2  5z  4 2 5z z2  2  2  2 5z z2  2 5z 25 z2  2  16 5 z42 5 z4 5 5 z44

 3

100 36 10  6 2 10 6  6 2  10 x 6 2  10 2  10 x 6 or x  6 8 2  6 or 4 The solution set is 3, 2 .

22.

2z2  5z  4  0

or

y

Lesson 10-4

1. x2  8x  4  0 x  

1  273 4



 2.4  1.9 The approximate solution set is {1.9, 2.4}.

b  2b2  4ac 2a (8)  2(8) 2  4(1) (4) 2(1) 8  164  16 2 8  180 2 8  280 8  280 or x  2 2

x  8.5  0.5 The approximate solution set is {0.5, 8.5}. 2. x2  7x  8  0 x     x

b  2b2  4ac 2a (7)  2(7) 2  4(1) (8) 2(1) 7  149  32 2 7  181 2 7  9 2 7  9 7  9 or x  2 2

1  8 The solution set is {8, 1}.

735

Extra Practice

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3. x2  5x  6  0 x     x

7. m2  4m  2  0

b  2b2  4ac 2a (5)  2(5) 2  4(1) (6) 2(1) 5  125  24 2 5  11 2 5  1 2 5  1 5  1 or x  2 2

m    m

 0.6  3.4 The approximate solution set is {3.4, 0.6}. 8. 2t2  t  15  0

3 2 The solution set is {2, 3}. 4. y2  7y  8  0 y     y

t

b  2b2  4ac 2a (7)  2(7) 2  4(1) (8) 2(1) 7  149  32 2 7  181 2 7  9 2 7  9 7  9 or y  2 2

    t

m2  2m  35  35  35 m2  2m  35  0

    x

x

b  2b2  4ac 2a



(2)  2(2) 2  4(1) (35) 2(1) 2  14  140 2 2  1144 2 2  12 2 2  12 2  12 or x  2 2

   x

    y

b  2b2  4ac 2a (0)  2(0) 2  4(5) (125) 2(5) 0  10  2500 10  12500 10 50 10 50 50 or x  10 10

5  5 The solution set is {5, 5}. 10. t2  16  0

7  5 The solution set is {5, 7}. 6. 4n2  20n  0 y

b  2b2  4ac 2a (1)  2(1) 2  4(2) (15) 2(2) 1  11  120 4 1  1121 4 1  11 4 1  11 1  11 or t  4 4

 2.5 3 The solution set is {2.5, 3}. 9. 5t2  125 2  125  125  125 5t 5t2  125  0

 1 8 The solution set is {1, 8}. 5. m2  2m  35

x

b  2b2  4ac 2a (4)  2(4) 2  4(1) (2) 2(1) 4  116  8 2 4  18 2 4  18 4  18 or m  2 2

t 

b  2b2  4ac 2a (20)  2(20) 2  4(1) (0) 2(4) 20  1400  0 8 20  1400 8 20  20 8 20  20 20  20 or y  8 8

 

b  2b2  4ac 2a (0)  2(0) 2  4(1) (16) 2(1)  10  64 2  164 2

The solution set is . 11. 4x2  8x  3 4x2  8x  3  3  3 4x2  8x  3  0 x

5 0 The solution set is {0, 5}.

   x

b  2b2  4ac 2a (8)  2(8) 2  4(4) (3) 2(4) 8  164  48 8 8  1112 8 8  1112 8  1112 or x  8 8

 0.3  2.3 The approximate solution set is {0.3, 2.3}.

Extra Practice

736

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12.

16. s2  8s  7  0

3k2  2  8k 3k  2  2  8k  8k  8k  2 3k2  8k  2  0 2

k    k

s 

b  2b2  4ac 2a (8)  2(8) 2  4(3) (2) 2(3) 8  164  24 6 8  140 6 8  140 8  140 or k  6 6

   s

 7  1 The solution set is {7, 1}. 17. d2  14d  24  0

 0.3  2.4 The approximate solution set is {2.4, 0.3}. 13. 8t2  10t  3  0

14.

5x 1  4 20 b  2b2  4ac  2a

x

    x

1 52

5 2



3

 12 2 The solution set is {2, 12}. 18. 3k2  11k  4 2 3k  11k  4  4  4 3k2  11k  4  0 k

1 12

 4(3) 2

2(3)

25 16

s

6

3 1 4 2

 4  5 4

5



5 4



(11)  2(11) 2  4(3) (4) 2(3) 11  1121  48  6 11  1169  6 11  13 11  13 k or k  6 6 1 3  4 1 The solution set is 4, 3 .

6

121

3 16 6 11 4

6 5 4

 6

11 4

or

x

2

3 The solution set is

5 4



19.

11 4

 

n2

 3n  1  0

d

6 1  4



523, 14 6.

 

15. 5b  3b2  1  0 b

b  2b2  4ac 2a



6 5 4

b  2b2  4ac 2a (14)  2(14) 2  4(1) (24)  2(1) 14  1196  96  2 14  1100  2 14  10  2 14  10 14  10  or s  2 2

d

b  2b2  4ac 2a (10)  2(10) 2  4(8) (3)  2(8) 10  1100  96  16 10  14  16 10  2  16 10  2 10  2 t  16 or t  16 1 3  2  4 3 1 The solution set is 4, 2 .

t

3x2

b  2b2  4ac 2a (8)  2(8) 2  4(1) (7) 2(1) 8  164  28 2 8  136 2 8  6 2 8  6 8  6 or s  2 2

b  2b2  4ac 2a (3)  2(3) 2  4(5) (1) 2(5) 3  19  20 10 2  111 1

d

5

6

b  2b2  4ac 2a (3)  2(3) 2  4(1) (1) 2(1) 3  19  4 2 3  15 2 3  15 3  25 or d  2 2

 2.6  0.4 The approximate solution set is {0.4, 2.6}.

 The solution set is .

737

Extra Practice

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20. 2z2  5z  1  0 z    z

27.

 0.2  2.7 The approximate solution set is {2.7, 0.2}. 21.

25t2  30t  9 25t  30t  9  9  9 25t2  30t  9  0 b2  4ac  (30) 2  4(25) (9) 0 Since the discriminant is 0, the equation has one real root. 2

b  2b2  4ac 2a (5)  2(5) 2  4(2) (1) 2(2) 5  125  8 4 5  132 4 5  132 5  232 or z  4 4

Page 842

3h2  27 3h  27  27  27 3h2  27  0 2

h    h

b  2b2  4ac 2a (0)  2(0) 2  4(3) (27) 2(3) 0  2324 6 18 6 18 18 or h  6 6

2

72

y

70 71 72 73

3f 2  2f  6 3f  2f  6  6  6 3f 2  2f  6  0 b2  4ac  (2) 2  4(3) (6)  76 Since the discriminant is positive, the equation has two real roots.

2. y 

1 7

1 7 49 343

1

y y  7x

O

2x

1

113 2x, 113 25.6

Sample answer: x 3 2

2x2  0.7x  0.3 2 2x  0.7x  0.3  0.7x  0.7x  0.3  0.3 2x2  0.7x  0.3  0 b2  4ac  (0.7) 2  4(2)(0.3)  2.89 Since the discriminant is positive, the equation has two real roots.

1 0 1 2

113 2x 113 23 113 22 113 21 113 20 113 21 113 22

y 14 12 10 8 6 1 4 y 3 2

y 27 9 3 1

x

( )

2 1 O

1 2x

1 3 1 9

113 25.6  0.002 3 3 3. y  1 5 2 x, 1 5 2 4.2

The y-intercept is 1.

Sample answer:

25. 4r2  12r  9  0 b2  4ac  (12) 2  4(4) (9) 0 Since the discriminant is 0, the equation has one real root.

x 3 2

x2  5x  9 2  5x  9  9  9 x x2  5x  9  0 b2  4ac  (5) 2  4(1)(9)  11 Since the discriminant is negative, the equation has no real roots.

1 0 1 2

135 2x 135 23 135 22 135 21 135 20 135 21 135 22

135 24.2  8.5

y

738

14 12 10 8 6 4 2

125 27 25 9 5 3

1 3 5 9 25

The y-intercept is 1.

Extra Practice

70 60 50 40 30 20 10

1 49

The y-intercept is 1. 71.5  18.5

24. 3w2  2w  8  0 b2  4ac  (2) 2  4(3)(8)  92 Since the discriminant is negative, the equation has no real roots.

26.

7x

0 1 2 3

2

23.

x

1 71

3  3 The solution set is {3, 3}. 22.

Lesson 10-5

1. y  7x, 71.5 Sample answer:

x

( 3)

y 5

4

2

O

y

1x

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4. y  3x  1 Sample answer: x

3x  1

8. y 

Sample answer: y

2 32  1

10 9

1 31  1

4 3

0 1 2 3

30  1 31  1 32  1 33  1

14 12 10 8 6 4 2

2 4 10 28

x 2 1 y

1O

x

2x  5

2

22  5

19 4

1

21  5

9 2

3x 

0

1

1 2 3x

1 2 3

y

20  5 21  5 22  5 23  5

10 8 6 4 2

4 3 1 3

8

4

y

x

O 4x 2 4 6 y  2x  5

2 1

1 y 28 24 20 16 12 8 4 y  2x  3 4

2

2 3

2x

O

x

7. y  3x + 1 3x  1

2

31

1 0 1 2 3

30 31 32 33 34

1 3

1 3 9 27 81

4

2

O

2

3 2

1

x

( 2)

y 3

1 2

2 3

1

O

2x

1

4 9 8 27

125 2x

125 2x 2 5 1 5 2 2 2 5 1 5 2 1 2 515 20 2 515 21 2 515 22 2 515 23

5

5(3x)

y

5

y

14 12 10 8 6 4 2

125 4 25 2

2

2

y

x

( 2)

y 5 5

O

x

2

4 5 8 25

2 5(32 )

5 9

1 5(31)

5 3

0 5(30 ) 1 5(31) 2 5(32)

y 14 12 10 8 6 4 2 y  3x  1

y

9 4

y

The y-intercept is 5. 10. y  5(3x ) Sample answer:

The y-intercept is 8.

x

3

Sample answer:

0

6

y

The y-intercept is 1. 9. y  5

The y-intercept is 4. 6. y  2x  3 Sample answer: x 2x  3 y 2 21 2 1 22 4 3 0 2 8 1 24 16 2 25 32 3 26 64

123 2x 123 22 123 21 123 20 123 21 123 22 123 23

y

The y-intercept is 2. 5. y  2x  5 Sample answer:

0 1 2 3

123 2x

5 15 45

16 14 12 10 8 6 4 2

y  5(3)x

The y-intercept is 5.

3 21O

y

1 2x

11. y  4(5)x Sample answer:

2x

x The y-intercept is 3.

4(5)x

y

2 4(5)2

4 25

1 4(5)1

4 5

0 4(5)0 1 4(5)1 2 4(5)2

4 20 100

35 30 25 20 15 10 5 2

O

y

y  4(5)x 2x

The y-intercept is 4.

739

Extra Practice

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12. y  2 (5)x  1 x 2

Page 843

x

2(5)  1 2(5)2

y

1

1 2(5)1  1 0 2(5)0  1 1 2(5)1  1 2 2(5)2  1

7 5

3 11 51

21O

x 2 1 0 1 2

112 2x1

112 2x1 y 112 221 2 112 211 1 112 201 12 112 211 14 112 221 18

 8500(1.00604) 48  11,349.73 He will have about $11,349.73.

1 2x

The y-intercept is 14. y 

118 2x

y 7 6 5 4 3 1 2 y 2 1

2a. V  C(1  r) t V  21,500 (1  0.08) 5 2b. V  21,500(0.92) 5  14,170.25 The value of the car will be $14,170.25. 3a. P  C(1  r) t P  3422(1  0.049) 8

x1

3b. P  3422(1.049) 8  5017 The population was 5017.

( )

2

O

x

2

Page 843 1 . 2

1. 12

2 1 0 1 2

118 2x 118 22 118 21 118 20 118 21 118 22

y 14 12 10 8 6 1 4 y 8 2

y 64 8

  1.2 1.2

2

O

No; the difference between consecutive terms is constant. This sequence is arithmetic, not geometric. 4. 86 68.8 55.04 44.032

2x

1 64

 0.8

2 1 0 1 2

1 2 2 134 22  2 134 21  2 134 20  2 134 21  2 134 22  2

 0.8

   2 2 2

y

y 2 9

2

2 3

1

1 5 4

4 3 x

( )

2

y  4 2

O

Yes; the common ratio is 2. 6. 13 10 11 8 9

6

     3 1 3 1 3

No; the difference between consecutive terms is a constant pattern. This sequence is arithmetic, not geometric. 7. 3125, 625, 125, 25, ... Divide the second term by the first.

2x

1

23 16

625 3125

The y-intercept is 1. 16. No; the domain values are at regular intervals and the range values have a common difference of 4. 17. Yes; the domain values are at regular intervals and the range values have a common ratio of 5.

Extra Practice

 0.8

Yes; the common ratio is 0.8. 5. 4 8 16 32

Sample answer: x

 1.2

   3 3 3

134 2x  2 3 x 4

45

Yes; the common ratio is 1.2. 3. 39 33 27 21

x

The y-intercept is 1. 15. y 

34

No; the difference between consecutive terms is constant. This sequence is arithmetic, not geometric. 2. 6 7.2 8.64 10.368

( )

1 1 8

Lesson 10-7 23

   11 11 11

Sample answer: x

2

M  8500 1 

The y-intercept is 3. 13. y 

1

12(4) 1 0.0725 12 2 0.0725 1b. M  8500 1 1  12 2 12(4)

14 12 10 8 6 4 2 y  2(5)x  1

27 25

Lesson 10-6

r 1a. M  P 1  n nt

y

 0.2

Multiply by 0.2 three more times. The next three terms are 5, 1, and 0.2. 8. 15, 45, 135, 405, ... Divide the second term by the first. 45 15

 3

Multiply by 3 three more times. The next three terms are 1215, 3645, and 10,935.

740

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22. an  a1  rn1 a3  a1  r31

9. 243, 81, 27, 9, ... Divide the second term by the first. 81 243

1

3

81  1  r2 81  r2 9  r The geometric mean is 1(9)  9 or 1(9)  9.

1

Multiply by 3 three more times. The next three 1 terms are 3, 1, and 3. 10. 15, 7.5, 3.75, 1.875, ... Divide the second term by the first. 7.5 15

23. an  a1  rn1 a3  a1  r31

 0.5

9  81  r2

Multiply by 0.5 three more times. The next three terms are 0.9375, 0.46875, and 0.234375. 11. 25, 15, 9, 5.4, ... Divide the second term by the first. 15 25

9 81 1 9 1 3

 0.6

12.

14 504 1 36 1 6

2

terms are

three more times. The next three

4 8 , , 625 3125

16

and 15,625.

13. an  a1  rn1 a10  1  6101

15. an  a1  rn1 a4  6  0.441 a4  6  0.43 a4  0.384

25.

a  1  (4) 6 7 a7  4096

504r2 504

 r2 r

1 2

116 2  84

an  a1  rn1 a3  a1  r31

162  0.5  r2 162 0.5

16. an  a1  rn1 a10  100  0.1101

17. an  a1  rn1 a5  750  (1.5) 51



The geometric mean is 504 1 or 504 6  84.

14. an  a1  rn1 a7  1  (4) 71

a10  1  69 a10  10,077,696

113 2  27

14  504  r2

3

Multiply by

r

24. a  a  rn1 n 1 a3  a1  r31

p

2 3

 r2

1 2

Divide the second term by the first. 1 10 1 4

81r2 81

The geometric mean is 81 1 or 81 3  27.

Multiply by 0.6 three more times. The next three terms are 3.24, 1.944, and 1.1664. 1 1 1 2 , , , , 4 10 25 125





0.5r2 0.5

324  r2 18  r The geometric mean is 0.5(18)  9 or 0.5(18)  9.

a10  100  0.19 a10  0.0000001

26. an  a1  rn1 a3  a1  r31

a5  750  (1.5) 4 a5  3796.875

4  1  r2 4 1

 rn1

18. an  a1 a5  64  851



1r2 1

4  r2 2  r The geometric mean is 1(2)  2 or 1(2)  2.

a5  64  84 a5  262,144

27.

 rn1

19. an  a1 a9  0.5  (10) 91 8

a9  0.5  (10) a9  50,000,000

 rn1

20. an  a1 a5  20  (2.5) 61

an  a1  rn1 a3  a1  r31

0.36  0.25  r2

5

a5  20  (2.5) a5  1953.125

0.36 0.25

0.25

 0.25r2

1.44  r2 1.2  r The geometric mean is 0.25(1.2)  0.3 or 0.25(1.2)  0.3.

n1

21. an  a1  r a4  350  (0.9) 41 a  350  (0.9) 3 4 a4  255.15

741

Extra Practice

PQ249J-6481F-24-26[726-749] 26/9/02 9:03 PM Page 742 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

28. a  a  rn1 n 1 a3  a1  r31 1 8 1 8 1 2 1 4 1 2

11.

1

 2  r2 1 2 r 2 1 2



 r2 r  rn1

29. an  a1 a3  a1  r31 32 27 32 27 2 3 16 9 4 3

1 2

1 1 2 2

1 4

or

1 2

1 2 1 2



1 2  89 or 23 143 2  89.

an  a1  rn1 a3  a1  r31 6.25

 6.25r2

21.



13 15



115 5



3



2. 1200  1100  2  1012 4. 1700  1100  7  1017

23.

24.

72 6

9.

25.

17

8.



14  2  7 7



2 114 7

3 3

2 6

5 24



3



15 124



15 2 16



130 12

Extra Practice



232c5



132  2c5 19  2d2



4 12  c2 1c 30d0



3 7  17  17

5 8

32c5

3 9d2

15 15

18

8

3 16 0g0

22. 299x3y7  199  2x3  2y7  3111  x1x  y3  1y  3 0 xy3 0 111xy

 112  14  3  213 7.

154

54

3 g2  2g2 

Lesson 11-1

1. 150  125  2  512 3. 1162  181  2  912

172 16

150 2z2 125  2 0z0

5 12 0z0

15. 612  13  616

20. 2175a4b6  1175  2a4  2b6  517  a2  0 b3 0  5a2 0 b3 0 17

0.36  r2 0.6  r The geometric mean is 6.25(0.6)  3.75 or 6.25(0.6)  3.75.

6.



19. 212ts3  112  1t  2s3  213  1t  s1s  2 0 s 0 13st

2.25  6.25  r2

5.



18. 2200m2y3  1200  2m2  2y3  1100  2  0 m 0  2y2  1y  10  12  0 m 0  y  1y  10 0 my 0 12y

2 4 3

The geometric mean is 3

13 15



115 115

17. 24x4y3  14  2x4  2y2  1y  2x2 0 y 0 1y

r

Page 844



115x 15

50

3 z2 

16. 516  213  10118  1019  2  10(312)  3012

 r2

2.25 6.25



1x 115

12.

14. 17  13  121 1 4.

2  r2 3 2  r2 3 2 3



x

13. 110  120  1200  1100  2  1012

The geometric mean is

30.

2x

3 30  3 15



10.

17

7

27p4

3 3p2

1 3  15

3 32  132

2 3

 19  2p2  30p0 1

3 

 

17 4 12



114 8



3 2

 11



1



3 3



 26.

16 16

742

2 13  5

4c2 12c 30d0

 29p2

17 116  2 12 12

29d2



15 3  15 9  5 3  15 4

3  15 15

3 

2 13  5  13  5 13  5 2( 13  5) 3  25 2( 13  5) 22 13  5 11

PQ249J-6481F-24-26[726-749] 26/9/02 9:03 PM Page 743 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

27.

13 13  5

  

28.

16 7  2 13

1

5

 

15  3 15  12 1 19 2 1 11 15 2 1 15 15  10 1 15  15 2  3 15 12 1 3 2 15  10 1 5 2  3 15  4 15

19. 10 3 5  145  12 3 9  10

16 7  2 13   2 13 7  2 13 16(7  2 13) 49  12 7 16  2 118 37 7 16  6 12 37

7 

Page 844

18. 4175  6127  4(513)  6(313)  2013  1813  3813

13 13  5  13  5 13  5 13( 13  5) 3  25 3  5 13 22

 2 15  3 15  4 15  5 15

20. 115 

13

3

3 5  115  15  115 

Lesson 11-2



1. 3111  6111  2111  (3  6  2) 111  7111 2. 6113  7113  (6  7) 113  13113 3. 2112  513  2(213)  513  413  513  913 4. 917  412  312  517  (9  5) 17  (4  3) 12  1417  12 5. 315  513 in simplest form 6. 418  315  4(212)  315  812  315 7. 2127  4112  2(313)  4(213)  613  813  213 8. 8132  4150  8(412)  4(512)  3212  2012  5212 9. 145  6120  315  6(215)  315  1215  1515 10. 2163  6128  8145  2(317)  6(217)  8(315)  617  1217  2415  6 17  2415 11. 1413t  813t  2213t 12. 716x  1216x  516x 13. 517  3128  517  3(217)  517  617  17 14. 718  118  7(212)  312  1412  312  1112 15. 7198  5132  2175  7(712)  5(412)  2(513)  4912  2012  1013  6912  1013 16. 416  312  215 in simplest form 17. 3120  2145  17  3(215)  2(315)  17  615  615  17  17

 1 3

21. 3 3  9 3

1 12

5 115 5 4 115 5



13 15  15 15 115 5

11  9 1 112 2  913 1 11 13 2 1 13 1 13  3 1 13  13 2  9 1 2 13  13 2  913 13 13  3 1 3 2  9 1 6 2  913

 1243  3

 13 

3 13 2

 913



2 13 2

3 13 2





17 13 2



18 13 2

22. 13( 15  2)  115  213 23. 12( 12  315)  2  3110 24. ( 12  5) 2  ( 12) 2  2( 12) (5)  52  2  1012  25  27  1012 25. (3  17)(3  17)  32  ( 17) 2 97 2 26. ( 12  13) ( 13  12)  16  14  19  16  216  2  3  216  5 27. (417  12) ( 13  315)  4121  12135  16  3110

Page 844 1.

Lesson 11-3

15x  5 ( 15x) 2  52 5x  25 x5 ?

15(5)  5 ? 125  5 55✓ 2. 417  1m (417) 2  ( 1m) 2 112  m 112  m Check:

Check:

743

?

417  1(112) ? 417  1112 417  417 ✓

Extra Practice

PQ249J-6481F-24-26[726-749] 26/9/02 9:03 PM Page 744 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

3. 1t  5  0 1t  5 ( 1t) 2  52 t  25

9. 1a  2  0 1a  2 ( 1a) 2  22 a4

?

125  5  0 ? 550 00✓

Check:

Check:

10. 12j  4  8 12j  12 ( 12j) 2  122 2j  144 j  72

4. 13b  2  0 13b  2 ( 13b) 2  (2) 2 3b  4 4

b3

3 31

Check:

4 3

220

Check:

? ?

14  2  0 ? 220 40✕

Since does not satisfy the original equation, there is no solution. 1x  3  6 ( 1x  3) 2  62 x  3  36 x  39

Check:

?

12.

139  3  6 ? 136  6 66✓

Check:

6. 5  13x  1 13x  4 (13x) 2  (4) 2 3x  16 16 3

x

5

Check:

3 31 3 2  1 ?

5  116  1 ? 541 11✓ 7. 2  31y  13 31y  11

4

c5 Check:

11 3 11 2 3

121 ?  9 11 ?

14. 13

1 3 2  13 ?

2  11  13 13  13 ✓ 13g  6

?

13(12)  6 ?

136  6 66✓

Extra Practice

?

215t  10 15t  5 ( 15t) 2  52 5t  25 t5 Check:

( 13g) 2  62 3g  36 g  12 Check:

4

?

2  33 23

3 515 2  9

7  14  9 ? 729 99✓

121 9

y

8.

7

1 2

( 1y) 2 

Check:

?

15(9)  4  7 ? 145  4  7 ? 149  7 77✓

13. 7  15c  9 15c  2 ( 15c) 2  22 5c  4

?

1y 

?

5  116  9 ? 549 99✓

15y  4  7 ( 15y  4) 2  72 5y  4  49 5y  45 y9 Check:

16

?

12(72)  4  8 ? 1144  4  8 ? 12  4  8 88✓

11. 5  1x  9 1x  4 ( 1x) 2  42 x  16

4 3

5.

?

14  2  0 ? 220 00✓

744

?

215(5)  10 ? 2125  10 ? 2(5)  10 10  10 ✓

PQ249J-6481F-24-26[726-749] 26/9/02 9:03 PM Page 745 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

15.

144  21p 2111  21p 111  1p ( 111) 2  ( 1p) 2 11  p

?

144  2111 2111  2111 ✓ 41x  5  15

Check: 16.

20.

1x  5  ( 1x  5) 2  x5 x Check:

225 ?  16 15 ?  4

1 2

15 15

15  15 ✓

17. 4  1x  3  9 1x  3  5 (1x  3) 2  52 x  3  25 x  28

?

22(2 13) 2  12  213

12(12)  12  213

12(12)  12  213

124  12  213

124  12  213

? ?

Check:

?

?

?

?

110(5)  5  315

110(5)  5  315

145  315 315  315 ✓

?

145  315 315  3 15 ✕

?

22a2  144  a ( 22a2  144) 2  a2 2a2  144  a2 a2  144 2a2  1144 a  12 ?

22(12) 2  144  12

?

12(144)  144  12

?

1288  144  12

?

1144  12

12  12 ✓

?

2( 12) 2  16  2( 12)  512 ? 12  16  212  512 ? 118  212  512 ? 312  212  512 512  512 ✓

2(12) 2  16  2(12)  5(12) ? 12  16  212  512 ? 118  212  512 ? 312  212  512 12  512 ✕ Since 12 does not satisfy the original equation, 12 is the only solution.

Since 15 does not satisfy the original equation, 15 is the only solution.

1144  12

?

112  213 213  213 ✕

22. 2b2  16  2b  5b 2b2  16  3b ( 2b2  16) 2  (3b) 2 b2  16  9b2 8b2  16 b2  2 2b2  12 b  12

210( 15) 2  5  315 210(15) 2  5  3(15)

1288  144  12

?

Since 213 does not satisfy the original equation, 213 is the only solution.

210x2  5  3x ( 210x2  5) 2  (3x) 2 10x2  5  9x2 x2  5 2x2  15 x  15

12(144)  144  12

?

112  213 213  213 ✓

?

22(12) 2  144  12

?

22(213) 2 12  213 ?

4  128  3  9 ? 4  125  9 ? 459 1  9 ✕ Since 28 does not satisfy the original equation, there is no solution.

Check:

124  1  5 ? 125  5 55✓

?

Check:

19.

?

21. 22x2  12  x 22x2  12  x2 2x2  12  x2 x2  12 2x2  112 x  213 Check:

?

 5  15

4

Check:

?

Since 1 does not satisfy the original equation, 8 is the only solution.

43

18.

13(8)  1  8  3

13  1  2 ? 14  2 2  2 ✕

1 2

305 16

?

13(1)  1  1  3

Check:

15 4 15 2 4 225 16 305 16

43

13y  1  y  3 ( 13y  1) 2  (y  3) 2 3y  1  y2  6y  9 0  y2  9y  8 0  ( y  1) (y  8) y  1  0 or y  8  0 y1 y8

? ? ? ?

12  12 ✕

Since 12 does not satisfy the original equation, 12 is the only solution.

745

Extra Practice

PQ249J-6481F-24-26[726-749] 26/9/02 9:04 PM Page 746 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

23. 1m  2  m  4 1m  2  4  m ( 1m  2) 2  (4  m) 2 m  2  16  8m  m2 0  m2  9m  14 0  (m  2)(m  7) m  2  0 or m  7  0 m2 m7 ?

c2  a2  b2 ( 1200) 2  a2  102 200  a2  100 100  a2 1100  2a2 10  a The length of the leg is 5. c2  a2  b2 (312) 2  32  b2 18  9  b2 9  b2 19  2b2 3  b The length of the leg is 6. c2  a2  b2 142  62  b2 196  36  b2 160  b2 1160  2b2 12.65  b The length of the leg is 7. c2  a2  b2 ( 147) 2  ( 111) 2  b2 47  11  b2 36  b2 136  2b2 6  b The length of the leg is 4.

?

12  2  2  4 17  2  7  4 ? ? 14  2  4 19  7  4 ? ? 224 374 44✓ 10  4 ✕ Since 7 does not satisfy the original equation, 2 is the only solution. 24. 13  2c  3  2c 13  2c  2c  3 ( 13  2c) 2  (2c  3) 2 3  2c  4c2  12c  9 0  4c2  10c  6 0  2(2c2  5c  3) 0  2(2c  3)(c  1) 2c  3  0 or c  1  0 2c  3 c1 Check:

3

c2 Check:

3 3  212 2  3  212 2 3

3

13  3  3  3

10  3  3 33✓

?

13  2(1)  3  2(1) ?

13  2  3  2 ?

11  3  2 42✕

Since 1 does not satisfy the original equation, the only solution.

Page 845

3 2

is

about 12.65 units.

6 units.

b2

  c2  ( 113) 2  b2 c2  13  36 c2  49 2c2  149 c  7 The length of the hypotenuse is 7 units.

9.

c2  a2  b2 c2  ( 16) 2  32 c2  6  9 c2  15 2c2  115 c  3.87 The length of the hypotenuse is about 3.87 units.

1.

Extra Practice

a2

3 units.

8.

Lesson 11-4

c2  a2  b2 292  a2  202 841  a2  400 441  a2 1441  2a2 21  a The length of the leg is 21 units. c2  a2  b2 2. c2  72  242 c2  49  576 c2  625 2c2  1625 c  25 The length of the hypotenuse is 25 units. c2  a2  b2 3. c2  22  62 c2  4  36 c2  40 2c2  140 c  6.32 The length of the hypotenuse is about 6.32 units.

c2

10 units.

10.

c2  a2  b2 102  a2  ( 175) 2 100  a2  75 25  a2 125  2a2 5  a The length of the leg is 5 units.

11.

746

c2  a2  b2 ( 1130) 2  a2  92 130  a2  81 49  a2 149  2a2 7  a2 The length of the leg is 7 units.

PQ249J-6481F-24-26[726-749] 26/9/02 9:04 PM Page 747 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

12.

13.

14.

15.

c2  a2  b2 152  92  b2 225  81  b2 144  b2 1144  2b2 12  b The length of the leg is 12 units.

3. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(1  4) 2  (2  (2) ) 2  2(3) 2  42  19  16  125 5 4. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

c2  a2  b2 112  a2  52 121  a2  25 96  a2 196  2a2 9.80  a The length of the leg is about 9.80 units. c2

a2

 2[4  (2) ] 2  (2  4) 2  262  (6) 2  136  36  172  612 or about 8.49 5. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

b2

  c2  ( 133) 2  42 c2  33  16 c2  49 2c2  149 c  7 The length of the hypotenuse is 7 units. c2

a2

 2(2  3) 2  (1  1) 2  2(5) 2  (2) 2  125  4  129 or about 5.39 6. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2[7  (2) ] 2  (8  4) 2

b2

   52  b2 34  25  b2 9  b2 19  2b2 3  b The length of the leg is 3 units. Yes; 142  482  502 No; 202  302  402 Yes; 212  722  752 Yes; 52  ( 1119) 2  122 Yes; 152  362  392 2 2 2 No; ( 15)  12  13 2 2 2 No; 10  ( 122)  12

 292  (12) 2  181  144  1225  15

( 134) 2

7. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2[ 9  (5) ] 2  (6  0) 2  2(4) 2  62  116  36  152  2113 or about 7.21

16. 17. 18. 19. 20. 21. 22. 23. No; 22  32  42 2 2 2 24. Yes; ( 17)  8  ( 171)

Page 845

8. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(5  5) 2  (13  (1) ) 2  202  142  1196  14 9. d  2(x2  x1 ) 2  (y2  y1 ) 2  2(10  2) 2  [ 8  (3) ] 2

Lesson 11-5

 282  112  164  121  1185 or about 13.60

1. d  2(x2  x1 ) 2  (y2  y1 ) 2  2(2  4) 2  (10  2) 2  2(6) 2  82  136  64  1100  10

10. d  2(x2  x1 ) 2  (y2  y1 ) 2  2[2  (7) ] 2  (7  5) 2  292  (12) 2  181  144  1225  15

2. d  2(x2  x1 ) 2  (y2  y1 ) 2  2(7  (5)) 2  (6  1) 2  2122  52  1144  25  1169  13

11. d  2(x2  x1 ) 2  (y2  y1 ) 2  2[ 5  (6) ] 2  [ 4  (2) ] 2  212  62  11  36  137 or about 6.08

747

Extra Practice

PQ249J-6481F-24-26[726-749] 26/9/02 9:04 PM Page 748 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

12. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

20. d  2(x2  x1 ) 2  (y2  y1 ) 2

 2(3  8) 2  [ 2  (10) ] 2

 2[3  (9) ] 2  (3  2) 2

 2(5) 2  122  125  144  1169  13

 2122  (5) 2  1144  25  1169  13

13. d  2(x  x ) 2  (y  y ) 2 2 1 2 1 2

 2(7  4)  [ 9  (3) ] 2

21. d  2(x2  x1 ) 2  (y2  y1 ) 2 2

 2(512  312) 2  (9  7) 2

2

 23  (6)  19  36  145  315 or about 6.71

 2(212) 2  22  18  4  112  213 or about 3.46

14. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

22. d  2(x2  x1 ) 2  (y2  y1 ) 2

 2(9  6) 2  (7  3) 2 232

 2(10  6) 2  (0  3) 2

42

   19  16  125 5

 242  (3) 2  116  9  125 5 23. d  2(x2  x1 ) 2  (y2  y1 ) 2

15. d  2(x  x ) 2  (y  y ) 2 2 1 2 1

 2(5  3) 2  (5  6) 2

 2(9  10) 2  (7  0) 2 2

 222  (11) 2  14  121  1125  515 or about 11.18

2

 2(1)  7  11  49  150  512 or about 7.07

24. d  2(x2  x1 ) 2  (y2  y1 ) 2

16. d  2(x  x ) 2  (y  y ) 2 2 1 2 1  2(3  2) 2  (3  (1)) 2

 2[5  (4) ] 2  (4  2) 2

 2(5) 2  42  125  16  141 or about 6.40

 292  22  181  4  185 or about 9.22

17. d  2(x2  x1 ) 2  (y2  y1 ) 2

25.

d  2(x2  x1 ) 2  (y2  y1 ) 2

 2 [3  (5) ] 2  (2  4) 2

5  2(a  0) 2  (3  0) 2

 282  (6) 2  164  36  1100  10

5  2a2  32 (5) 2  ( 2a2  9) 2 25  a2  9 16  a2 116  2a2 4  a

18. d  2(x2  x1 ) 2  (y2  y1 ) 2  2(0  0) 2  [ 7  (9) ] 2 26.

 202  162  1256  16

10  2(6  2) 2  (a  (1) ) 2 10  2(8) 2  (a  1) 2

19. d  2(x2  x1 ) 2  (y2  y1 ) 2  2 [8  292

(1) ] 2

 (4 

10  264  a2  2a  1

7) 2

102  ( 2a2  2a  65) 2 100  a2  2a  65 0  a2  2a  35 0  (a  7)(a  5) a70 or a  5  0 a  7 a5

(3) 2

   181  9  190  3110 or about 9.49

Extra Practice

d  2(x2  x1 ) 2  (y2  y1 ) 2

748

PQ249J-6481F-24-26[726-749] 26/9/02 9:04 PM Page 749 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

d  2(x2  x1 ) 2  (y2  y1 ) 2

27.

6.

161  2(a  1) 2  (6  0) 2 161  2a2  2a  1  62 ( 161) 2  ( 2a2  2a  37) 2 61  a2  2a  37 0  a2  2a  24 0  (a  6)(a  4) a  6  0 or a  4  0 a6 a  4 185  2(5 

(2)) 2

6

 (10 

1274  2(0  15)  (4  a)



AC DF 18 e

8.

AB DE c 17.5

AC DF b 11

BC

 EF

9.

1.7

 8.5

10.

2

1274  2(15) 2  16  8a  a2

d  2(x2  x1 )  (y2  y1 )

15 d

AB

 DE 20

 10

AC DF 5.6 7

1.7

BC

 EF 4

d

5.6d  28 d5 AB DE 7.2 f

 8.5

BC EF 80 100



AC

 DF 

5.6 7

5.6f  50.4 f9

AC

 DF b

 125

100b  10,000 b  100

( 1274) 2  ( 2225  16  8a  a2 ) 2 274  a2  8a  241 0  a2  8a  33 0  (a  11)(a  3) or a  3  0 a  11  0 a  11 a  3 2

BC

 EF

8.5b  18.7 b  2.2

2

BC

 EF

20e  180 e9

8.5c  29.75 c  3.5

d  2(x2  x1 )  (y2  y1 ) 2



AB DE 20 10

20d  150 d  7.5

1 93 BC EF 6 4.5

4.5c  45 c  10

a) 2

2

30.

7.

 4.5

b AB DE c 7.5

185  272  100  20a  a2 ( 185) 2  ( 2149  20a  a2 ) 2 85  a2  20a  149 0  a2  20a  64 0  (a  4)(a  16) a  4  0 or a  16  0 a4 a  16 29.

BC

 EF

4.5b  42

d  2(x2  x1 ) 2  (y2  y1 ) 2

28.

AC DF b 7

BC EF 80 100

AB

 DE c

 218.75

100c  17,500 c  175

2

1136  2(a  3) 2  (9  3) 2

Page 846

1136  2a2  6a  9  62 2

2

( 1136)  ( 2a  6a  9  36) 136  a2  6a  45 0  a2  6a  91 0  (a  7)(a  13) a70 or a  13  0 a  7 a  13

Lesson 11-7

1. sin N 

2



opposite leg hypotenuse 36 39

 0.9231 cos N  

adjacent leg hypotenuse 15 39

 0.3846

Page 845

tan N 

Lesson 11-6

1. No; corresponding angles do not have equal measures. 2. Yes; corresponding angles have equal measures. 3. Yes; corresponding angles have equal measures. 4.

AB DE 7 f

 

BC EF 5 10

5f  70 f  14 AC DE 8 e

 

BC EF 5 10

5e  80 e  16

5.

AB DE 4 f

 



 2.4 3. sin N 

BC EF 2 3



 



opposite leg hypotenuse 14 50

 0.28 cos N  

adjacent leg hypotenuse 48 50

 0.96 opposite leg

tan N  adjacent leg 14

 48  0.2917

opposite leg hypotenuse 40 50

 0.8 adjacent leg cos N  hypotenuse

2f  12 f6 AC DF 3 e

opposite leg adjacent leg 36 15

2. sin N 

30

 50

BC EF 2 3

 0.6 opposite leg

tan N  adjacent leg

2e  9 e  4.5



40 30

 1.3333 4. cos 25  0.9063 6. sin 71  0.9455

749

5. tan 31  0.6009 7. cos 64  0.4384 Extra Practice

PQ249-6481F-P27-28[750-765] 26/9/02 8:58 PM Page 750 Sahuja 79:PQ249:PQ249J:PQ249J-ch15-28-Repro:

8. tan 9  0.1584 10. tan B  0.5427 B  tan1 (0.5427)  28 11. cos A  0.8480 A  cos1 (0.8480)  32 12. sin J  0.9654 J  sin1 (0.9654)  75 13. cos Q  0.3645 Q  cos1 (0.3645)  69 14. sin R  0.2104 R  sin1 (0.2104)  12 15. tan V  11.4301 V  tan1 (11.4301)  85 16. 180  90  60  30 The measure of ∠ A is 30.

9. sin 2  0.0349

Page 846 1.

Lesson 12-1

xy  k (7.5)(10)  k 75  k y

75 x

5 15

x y

3 25

1 75

1 75

1 15

3 5

1 12

1 12

3 25

5 15

2 6

3 4

y 50 ⫺2

O

2

x

⫺50

2.

xy  k (3)(5)  k 15  k y 3 5

x y

15

tan 60  BC BC(1.7321)  15 BC  8.7 cm 15 sin 60  AB

15

AB(0.8660)  15 AB  17.3 cm 17. 180  90  55  35 The measure of ∠ Z is 35.

15 x

1 15 y

5 ⫺2

⫺5

x O

2

⫺15

21

sin 55  XZ XZ(0.8192)  21 XZ  25.6 ft

3.

21

tan 55  XY

y

XY(1.4281)  21 XY  14.7 ft 18.

12

2 6

y

4 ⫺2

16 8

M  tan1 (2)  63 180  90  63  27 The measure of  L is 27.

Extra Practice

12 x

3 4

x y

c2  a2  b2 LM 2  162  82 LM 2  256  64 LM 2  320 2LM 2  1320 LM  17.0 m tan M 

xy  k (2)(6)  k 12  k

⫺4 ⫺12

750

O

2

x

PQ249-6481F-P27-28[750-765] 26/9/02 8:58 PM Page 751 Sahuja 79:PQ249:PQ249J:PQ249J-ch15-28-Repro:

4.

xy  k (0.5) (1)  k 0.5  k y x y

9.

y

0.5 x

1 0.5

9

0.5 1

1.0

0.5 1

1 0.5

x

y

O

1

2

12.

x1y1  x2 y2

113 2 (27)  x2 y2 y

y

y  6 13.

xy  k (3)(2.5)  k 7.5  k y x y

7.5 x

3 2.5 15

⫺5

y

1.5 5

1 7.5

1 7.5

1.5 3 5 2.5

15.

y

y

O

⫺15

6.

Page 846

x y

2 x

2 1

1 2

1 2

2 1

2 ⫺1

O

1

2x

⫺2 ⫺4

7.

x1y1  x2 y2 (4) (54)  x2 y2 y 27  x

216 x 216 x 216 27

4 16.

18.75 x 18.75 2.5

x  2.25 x1y1  x2 y2 (3.2)(0.4)  x2 y2 y 0.2 

8.

x1y1  x2 y2 (6)(18)  x2 y2 y 12 

1.28 x 1.28 x

x  6.4

Lesson 12-2

3. Exclude the values for which c2  4  0. (c  2)(c  2)  0 or c  2  0 c20 c  2 c2 The excluded values are 2 and 2. 4. Exclude the values for which b2  8b  15  0. (b  5)(b  3)  0 b  5  0 or b  3  0 b5 b3 The excluded values are 5 and 3.

y 4

⫺2

9 x 9 x

1. Exclude the values for which x  1  0. x10 x  1 The excluded value is 1. 2. Exclude the values for which n  0. n0 The excluded value is 0.

xy  k (1)(2)  k 2k y

3 4

x1y1  x2 y2 (3) (3)  x2 y2

y  7.5

2

9

y

y  12 x1y1  x2 y2 (2.5)(7.5)  x2 y2

x

14.

24 x 24 2

y

5 ⫺2

x1y1  x2 y2 (8)(3)  x2 y2 y

9 x

y  12

⫺1.0

5.

64 x 64 16

y y4

24 4

y

⫺0.5

x1y1  x2 y2 (8) (8)  x2 y2 y

(8)(3)  x2 y2 24 y x

x ⫺1

10.

288 x 288 x 288 9

x  32 x1y1  x2 y2

11.

0.5 ⫺2

x1y1  x2 y2 (24) (12)  x2 y2

5.

13a 39a2

13a (13a) (3a)



13a (13a ) (3a)

 2x (21y)

1

 21y

1

108 x 108 x

1

 3a

x9

a0

x8

751

38x2 42xy



6.

2x(19x)

 2x(21y) 1

2x (19x) 1

19x

x  0 and y  0

Extra Practice

PQ249-6481F-P27-28[750-765] 26/9/02 8:58 PM Page 752 Sahuja 79:PQ249:PQ249J:PQ249J-ch15-28-Repro:

1

p  5 2( p  5)

7.

13.

p  5  5)

 2(p 

r1

8.

14.



a  b (a  b) (a  b)



a  b (a  b ) (a  b) 1

1

ab 2

15.

2

Exclude the values for which a  b  0. (a  b)(a  b)  0 or a  b  0 ab0 a  b ab a  b y  4 y2  16



y  4 (y  4) (y  4)



y  4 (y  4 ) (y  4)

4t2  8 4t  4



4(t2  2) 4(t  1)



t2  2 t  1

Exclude the values for which 4t  4  0. 4t  4  0 4t  4 t1 t1

1

9.

6y3  12y2 12y2  18

12y2  18  0 6(2y2  3)  0 2y2  3  0 2y2  3 3

y2  2 3

1  4

y  3 2

Exclude the values for which y2  16  0. ( y  4)(y  4)  0 or y  4  0 y40 y  4 y4 y  4 c2  4 c2  4c  4



(c  2) (c  2) (c  2) (c  2)



(c  2 ) (c  2) (c  2 ) (c  2)



c  2 c  2

y

16.



Exclude the values for which c2  4c  4  0. (c  2)(c  2)  0 c20 c  2 c  2 a  a a  1



a(a  1) a  1



a(a  1 ) a  1

Page 847 1.

a2b b2c

a2bc

a2  b

 bbd a2

 bd

1

a Exclude the values for which a  1  0. a1 12.



Lesson 12-3

c

 d  b2cd

1

x2  4 x4  16

2.

6a2n 8n2



12n 9a



3234aann 2433nna

a

x2  4 (x2  4) (x2  4)

3.

2

2a d 3bc

1

2

9b c

 16ad2 

1



x2  4



(x2  4 ) (x2  4) 1

1

 x2  4

4.

10n3 6x3

5.

120n5x4 5

6m3n 10a2



4a2m 9n3

 

752

8

3ab 8d 4

12n2x4



(x2  4)(x2  4)  0 x2  4  0 x2  4  0 2 x  4 x2  4 x  2 not possible x  2

3

2aad9bbc 3  b  c  16  a  d  d

 25n2x2  150n2x5

Exclude the values for which x4  16  0.

Extra Practice

5(x2  2x  1)  1) 5 3

 3(x2  2x

Exclude the values for which 3x2  6x  3  0. 3x2  6x  3  0 x2  2x  1  0 (x  1) 2  0 x  1 x  1

1

11.

16 2 5x2  10x  5 3x2  6x  3

16 2

y

1

2

y2 (y  2) 2y2  3

Exclude the values for which 12y2  18  0.

1

10.

6y2 (y  2)

 6(2y2  3) 

1

y

r2 (r  1) r  1



 r2

1

1 2

2( p  5)  0 p50 p  5 p  5. a  b a2  b2

r3  r2 r  1

4n3 5x 24a2m4n 90a2n3 4m4 15n2

PQ249-6481F-P27-28[750-765] 26/9/02 8:58 PM Page 753 Sahuja 79:PQ249:PQ249J:PQ249J-ch15-28-Repro:

6. 7. 8.

(a  5) (a  1) (a  7) (a  6) a  a (a  1) (a  7) (a  8) (a  5) x  1 x  2   (x (x  2) (x  3) (x  3) (x  1) 5n  5 3

n

9  1

5(n  1) 3



1

n

 6  8 1  3) 2

4.

9.

a a  b

3a  3b  a



x2y 6

9  1

 12z2

x4

p2 14qr3



2r2p 7q

2(a  2b) 5

25

 6a  8b 

1

p

 4r5

11. 12.

3 x  y x  5 3x

(x  y) 2 6





2

4

12x



 10

x  5 3x 1

 (x

2

5e  f 5e  f

7.

t2  2t  15 t  5

12x  5) (x  2)

8.

4x  2

x

4

13.

a2  b2 4

a  b 

14.

4a  8 a2  25

a  5  5a  10

15.

r2 r  s

16

(a  b) (a  b) 4

a

16  b

 4a  4b



6.

x  y 2

2

 x2  7x

25

 2(3a  4b)

5a  10b 3a  4b

 1

r2  s2 s2



10.

4(a  2) a  5  (a  5) (a  5) 5(a  2) 4  5a  25 r2 (r  s) (r  s)  r  s s2



r3  r2s s2 a2  b2 7 (a  b) (a  b) a  b  a  b a  b

a

17.

x2  3x  2  x2  7x  6

12.

7  b

(x  9) (x  1) (x  9) (x  2)



(x  2) (x  1)  6) (x  1)

 (x

x  1  6

x

18.

x2  6x  5 x2  7x  12

Page 847 1.

5m2n 12a2



x2  14x  40  x2  5x  50

13.



(x  5) (x  1) (x  4) (x  3)



x  1 x  3



(x  4) (x  10) (x  10) (x  5)

Lesson 12-4 30m4 18an



5m2n 12a2 2



14.

31

18an

 30m4 6

2

ammnn 4aammmm

15.

n2

 4am2 2.

25g7h 28t3



5g5h2 42s2t3

63

5



25g7h 28t3 4

16.

2

15g7hs2t3 2g5h2t3



15g2s2 2h 5

3.

6a  4b 36



3a  2b 45



2(3a  2b) 36



5 2

18

t5

5x  10

(x  2) x  2 5(x  2) 1  x  2 x  2 5 x2 3v2  27 v  3

v2 15v 3(v2  9) v2  15v  v  3

9.

11.

(v  3) (v  3)  v2 5v (v  3) v(v  3)  5 b2  9

(b  3) 4b (b  3) (b  3) 1  b  3 4b b  3  4b p p2

2y y2  4 p2 2  y  (y  2) (y  2)  p p2 1(y  2)  (y  2) (y  2)  p p y2 k2  81 k  9

k6 k2  36 (k  9) (k  9) k  6  (k  6) (k  6)  k  9 k  9 k6 2a3 a2

a1 a  1 2a3 a  1  a  1  a2

9  3 3d 2d  3  d(2d  3)  9 3 1 3 g  5 3g2  15g

g2 4 3g(g  5) g2  g  5 4

3d 2d2  3d



2d

3g3 4

 2a

42  5g5h2





5

7 x2  10x  9 x2  11x  18

5e  f 1   f 25e2  f 2 5e  f 1  5e  f  (5e  f ) (5e  f ) 1  (5e  f ) 2 t  3 (t  5) (t  3) t  5  t  3 t  5 t  5

(25e2  f 2 )  5e





16.

7q

2

p2

5

2a  4b 5

p2

 14qr3  2r2p  4r5p

 3a 10.

3x2

 18z  2yz x4y

3(a  b)  a

a a  b

2yz 3x2

 12yz2

5. 2



3

 15 2

x2y 18z

45

 3a  2b

2

753

7 x2  16

x 16  x2 (x  4) (x  4) x  (x2  16)  7 (x  4) (x  4) x  (x  4) (x  4)  7 x  1  7 x  7

Extra Practice

PQ249-6481F-P27-28[750-765] 26/9/02 8:58 PM Page 754 Sahuja 79:PQ249:PQ249J:PQ249J-ch15-28-Repro:

17.

18.

y 5

y2  25 5  y y 5  y  5  y2  25 y 1(y  5)  5  (y  5) (y  5) y  5(y  5) y  5y  25

3m m  1



t6 8.

3m

(2r2

 3r  35) (2r  7)  

9.

1  2

12n2  36n  15 6n  3

20.

21.

a2  3a  10 a2  3a  10

a2  2a  3 a2  3a  2 (a  5) (a  2) (a  3) (a  1)  (a  2) (a  1)  (a  5) (a  2) a  3 a2 x2  x  2 x2  6x  8

x2  x  12 x2  4x  3 (x  2) (x  1) (x  4) (x  3)  (x  3) (x  1)  (x  4) (x  2)

1

Lesson 12-5

x  7  2x 1. 2x  32x2  11x  20 () 2x2  3x 14x  20 () 14x  21 1

1  3



x2  2x  35 x  7 (x  7) (x  5) x  7

4



c2  6c  27 c  9 (c  9) (c  3) c  9

c3 y  13  2  6y  25 y  7y 6. () y2  7y 13y  25 () 13y  91 66

Extra Practice

4  2

4b2  b  b  2 16. b  24b3  7b2  2b  4 () 4b3  8b2 b2  2b () b2  2b 04

x5 5. (c  6c  27) (c  9) 

(2m  3) (2m  5)

 2m  3  2m  5

3c  2  9c 2 15. 9c  227c  24c  8 () 27c2  6c 18c  8 () 18c  4 4

m2  4m  5 m  5 (m  5) (m  1) m  5

m1

2

4m2  4m  15 2m  3

3

3. (m2  4m  5) (m  5) 



13.

(5x  7) (2x  3)

3t2  2t  3  2t  3 3 14. 2t  36t  5t2  0t  12 () 6t3  9t2 4t2  0t () 4t2  6t 6t  12 () 6t  9 3

a 7 2. a  3a2  10a  21 () a2  3a 7a  21 () 7a  21 0

4. (x2  2x  35) (x  7) 

10.

10x2  29x  21 5x  7

 5x  7  2x  3 t2  4t  1 11. 4t  14t3  17t2  0t  1 () 4t3  t2 16t2  0t () 16t2  4t 4t  1 () 4t  1 0 2a2  3a  4 12. a  32a3  9a2  5a  12 () 2a3  6a2 3a2  5a () 3a2  9a 4a  12 () 4a  12 0

m2  6m  16 2m  16

m2  m  6 m  2 2(m  8) (m  3) (m  2)  m  2  (m  8) (m  2) 2m  6  m2

Page 847



(6n  3) (2n  5) 6n  3

 2n  5

3m

 m2  m  2 19.

2r2  3r  35 2r  7 (2r  7) (r  5) 2r  7

r5

(m  2)

 m  1m

3t2  14t  24 3t  4 (3t  4) (t  6) 3t  4

7. (3t2  14t  24) (3t  4) 



66 y  7

754

PQ249-6481F-P27-28[750-765] 26/9/02 8:58 PM Page 755 Sahuja 79:PQ249:PQ249J:PQ249J-ch15-28-Repro:

17. t

t2  4t  3  t  19t  9

 4t3  0t2 () t3  4t2

3  4

12.

18. 3x

 

6y  y



6x  6y x  y



6(x  y) x  y



6(x  y) x  y 1

6 13.

3x2  2x  2  3x 0x2 6x2

x

1

4t2  19t () 4t2  16t 3t  9 () 3t  12 3

 29x3 () 9x3

6x x  y

5x 24

3x

 24  

6  2

5x  3x 24 2x 24 1

 2x  10

2x

 24 12

6x2  2x () 6x2  4x 6x  10 () 6x  4 6

x

 12 14.

7p 3



8p 3

 

7p  8p 3 p 3

15.

8k 5m

3k

 5m  

8k  3k 5m 5k 5m k

m

Page 848 1.

4 z



16.

Lesson 12-6

3 z

4  3 z 7 z

 

2.

a 12



2a 12

 

a  2a 12 3a 12

2

17.

7 2t



  

5  7 2t 2 2t

4.

y 2

18.

a 4 y  y 2 2y 2

y

2 

6.

y 2

2 x

 



y  6 2

20.

 

2a 2a  5 1 4z  1

 2a

7.

x x  1



1 x  1

 

x x x x

   

1 1 1 1

22.

8.



  

2n 2n  2n 2n  2n  2n 

3a a  2 n n  1

3a  2

a 1

1  n

5 5 5 5

 

23.

5 (2n  5) 5 2n  5

x  y 2  y

x  y  2

y

10.

r2 r  s

11.

12n 3n  2

r

s2  s

 3n

 

r2  s2 r  s

8  2

a

7

a

7  7

a  7

a7 1 24.

2a 6a  3

3

(1)  6a

2a

1

2a

1

 6a  3  3  6a  6a  3  6a  3 2a  1

 6a  3

x  y  2

y

2a  1

 3(2a  1) 1

2a  1

 3(2a  1 )



12n  8 3n  2 4(3n  2) 3n  2



4(3n  2) 3n  2



(7)

7aa77a a7a

1

y  x  2 2y y  2

y

a a  7

n 1 n1  1 n  (1) n  1 n  1 n  1

n 

1

1

0



2n  5  5

 2n

9.

1  4z

(4z)

1

5 5  2n

 2a

 4z  1  4z  1

1

y3 2n 2n  5

2a  5  5

5  5

4z  1

21.

y  (y  6) 2 2y  6 2 2(y  3) 2

a  2  a  3 6 1 6



 4z  1 1

b  2 x



a  3 6

y

1

b x



1

 t 5.

19.

2y 2



1

a  2 6



1

21 2t

y  2y b  6 y b  6

2y

b6 

 5 2t

y b  6

3a 12 4

3.

8  6

6

m2m2 m2

1



8 m  2

1

1

3

1

4

1

755

Extra Practice

PQ249-6481F-P27-28[750-765] 26/9/02 8:58 PM Page 756 Sahuja 79:PQ249:PQ249J:PQ249J-ch15-28-Repro:

Page 848

11. 3t  2  3t  2 t2  4  (t  2)(t  2) LCD  (3t  2)(t  2) (t  2)

Lesson 12-7

27a2bc

 333aabc 36ab2c2  2  2  3  3  a  b  b  c  c LCM  2  2  3  3  3  a  a  b  b  c  c  108a2b2c2 2. 3m  1  3m  1 6m  2  2(3m  1) LCM  2(3m  1)  6m  2 1.

t  2 3t  2  t2  4 3t  2 (3t  2) (t  2) (t  2) (t  2) (3t  2)  (3t  2) (t  2) (t  2)  (3t  2) (t  2) (t  2) (t  2) [ (3t  2) (t  2)  (3t  2) ]  (3t  2) (t  2) (t  2) (3t  2) (t  2)  3t  2  (3t  2) (t  2) 3t2  6t  2t  4  3t  2  3t2  6t  2t  4 3t2  t  6  3t2  8t  4

3. x2  2x  1  (x  1) (x  1) x2  2x  3  (x  3) (x  1) LCM  (x  1) (x  1) (x  3)  (x  1) 2 (x  3)

4. LCD  21 s 3



2s 7

  

7(s) 3(2s)  21 21 7s  6s 21 13s 21

12. a  5  a  5 a2  5a  a(a  5) LCD  a(a  5) or a2  5a 3 a  5

5. 2a  2a 6a  3(2a) LCD  3(2a) or 6a 5 2a



3 6a

3(5)

 3(2a) 

6

 2n 5

3 6a 15  3 6a 12 2 or a 6a



3m 4

15





7 10x2

 

6

14. 7w  7  w  w ww LCD  7  w  w or 7w2 3z 7w2

 

t  3 s

 

 2b

s t2



a

3  3

a a2  4

 

Extra Practice

4

4(a  2) a2  4 a  (4a  8) a2  4 3a  8 a2  4 a

 a  2  a2  4   

17. m  n  m  n mm LCD  m(m  n)

a(2b  3a) b(a  b)  3a2  ab  2b2  2b2 2ab  3a2  ab  b2 3a2  ab  2b2 3a2  3ab  b2 3a2  ab  2b2

m m  n

5

m(m)

5(m  n)

 m  m(m  n)  m(m  n)  

m2  (5m  5n) m(m  n) m2  5m  5n m(m  n)

18. y  5  y  5 y2  25  (y  5)(y  5) LCD  (y  5)(y  5) or y2  25

4a 3(2)  2a  6  6 4a  6 2a  6 2(2a  3) 2(a  3) 2a  3 a  3

 2a 

3(s) t(r)  3t2 3t2 3s  rt 3t2

16. a  4  (a  2)(a  2) a2a2 LCD  (a  2)(a  2) or a2  4

10. 2a  6  2(a  3) a3a3 LCD  2(a  3) or 2a  6 4a 2a  6

r

 3t  

 3a2  ab 

7w(2z) 7w2 3z  14wz 7w2

2

2(s) t(t  3)  st st 2 2s  t  3t st

b  3a

3z

 7w2 

15. t  t  t 3t  3  t LCD  3  t  t  3t2

9. a  b  a  b 2b  3a  2b  3a LCD  (a  b)(2b  3a)  2ab  3a2  2b2  3ab  3a2  ab  2b2 a a  b

2z w

2

8. LCD  st 2 t





2x(6) 7  10x2 10x2 12x  7 10x2

5(z) 6(x)  xyz xyz 5z  6x xyz

 yz 

4(2n) 5(3m)  20 20 8n  15m 20

 

7. LCD  xyz 5 xy

3a  6 a2  5a

2

6. 5x  5x 10x2  2  5  x  x LCD  2  5  x  x or 10x2 6 5x

6

13. LCD  20

3 6a

 6a  

3(a)

 a2  5a  a2  5a  a2  5a

y  5 y  5

2y

 y2  25   

756

(y  5) (y  5) 2y  y2  25 y2  25 y2  10y  25  2y y2  25 y2  8y  25 y2  25

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19. t2  100  (t  10)(t  10) 10  t  1(t  10) LCD  1(t  10)(t  10) t  10 t2  100

 10

1  t

10.

1(t  10)  10) (t  10)

 1(t

x  3 x  1 x2 y2

t  10  10) (t  10)

 1(t

t  10  t  10  10) (t  10)

x  3 y2   1 x2 y2 (x  3) x2 (x  1)



1(2t  20) 1(t  10) (t  10)

t

y2

x

 1(t 

x2

x  3  1

x

11.

2  10

y 3



2y  5 6 2y  5 y 2y  5 6



2y  5 6

5

6

2

20. a2  3a  10  (a  5)(a  2) a2  4a  12  (a  6)(a  2) LCD  (a  2)(a  5)(a  6)

5 y





2a2  6a  12a  36  (3a2  5a  15a  25) (a  2) (a  5) (a  6)



2a2  18a  36  3a2  10a  25 (a  2) (a  5) (a  6)



a2  8a  61 (a  2) (a  5) (a  6)

1. 4 

 5

2. 8  3t   3.

b  1 2b

12.

1 x 1 y

 

1 y 1 x



y  x xy x  y xy



y  x xy



x  y xy



y  x xy

x

xy  y



y  x x  y

1

4x 2 x x 4x  2 x

 4. 3z 

z  2 z

 

5.

2 a  2

1

7.

3

44

 

t2



3z2 z  2  z z 3z2  z  2 z

2

14.

aa1 3

aa2



7 2 19 4 7 19

4 2

Page 849 1.

4

6

14

 19 x2 y

x3



x2 y x5 y2

x3 y





a2 (a  2) a  2 2  a3  2a2  a  2 a3  2a  2  a  2 4 3r2 (2r  1) 4   2r  1 2r  1 2r  1 6r3  3r  4  2r  1



y

9.

t  2 t  3

t t3

 2  19

8.

t2 t  2  3 t  3 2t  2

2

7

t2

(t  2) (t  2) (t  2) (t  3)

t2 t

b  1 3b  2b  2b 2b 6b2  b  1 2b

2

x2 y y x3



t2  4 t2  5t  6

 a2  a  2 

6. 3r2 

32

13.

8  3t 5  3t 3t 24t  5 3t

 3b 

1

y

Lesson 12-8 

y

 2y  5

6

1

2 x

2y  5 y

1

3a  5 2a  6  a2  4a  12 a2  3a  10 (a  6) (2a  6) (a  5) (3a  5)  (a  2) (a  5) (a  6)  (a  2) (a  5) (a  6)

Page 848



t4 u t3 u2



t4 u

t3

u2 3. 1



4

t u 1



 tu

2

u t3

b

1

1

a(a  1)  2 a  1 a(a  2)  3 a  2 a2  a  2 a  1



a2  2a  3 a  2 a2  a  2 a2  2a  3

a  1 a  2



a2  a  2 a  1



(a2  a  2) (a  2) (a  1) (a2  2a  3)

Lesson 12-9

k 2k  3 6 k 2k  3 6



5 2 5 2

2  1 26

2.

3

4.

70

k  4k  15 5k  15 k  3 18 b 18 b

b3

1 21

3 b

2

3b

18  3  3b 15  3b 5b

757

a  2  3

 a2  2a

1

2x 27  10 7 2x 27  10 7



4x 5 4x 5

2  1 270

20x  189  56x 189  36x 5.25  x

10x

1

3 7  2x 5x 3 7  2x 5x

1

2  10x(1)

6  35  10x 41  10x 41 x 10

Extra Practice

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5. 6

1

2a  3 6 2a  3 6



2a 3

2  61

1

2 2a 3

1 2



2a  3  4a  3 6  2a 3  a

7. 14

1

2b  3 b 2 7 2b  3 b 2 7

2

b  3 14 b  3 14 14



2(2b  3)  7b  b  3 4b  6  7b  b  3 3b  6  b  3 9  4b

8. 5(y  4)

1

x  3 x 4x  5 x



5

13.

5

4x  5  5x 5x

2 1

9 4

3x  2 x

6.

(x  3)

2

t(t  3)

1

3

15.

2

2  t(t  3) (2)

x(x  1)

1

5x 1 x x  1 5x 1 x x  1

(r  2)(r  9)

(3x  10) (x  5)

1

(m  1) (m  1)

1

2x x  5

2

2  (3x  10)(x  5) (2)

 5x  100  0 x2  x  20  0

(x  5)(x  4)  0 x  5 or x  4 2a  3 a  3

17. (a  2) (a  3)

12

2a2

12aa 33 22  (a  2)(a  3) 1a 12 2 2

2a2  4a  3a  6  2a2  6a  4a  12  12a  36

6

3a  6  12a  36 42  9a

5  1

5 m  1

18. (z  1) (z  3)

1

2  (m  1) (m  1) (1)

m

1

z  3 z  1 z  3 z  1





z  1 z  3

z  1 z  3

42 9

a

14 3

a

2

2  2(z  1) (z  3)

(z  3) (z  3)  (z  1)(z  1)  (2z  2) (z  3) z2  9  z2  1  2z2  6z  2z  6 2z2  10  2z2  8z  6 8z  16 z2

m(m  1)  5(m  1)  m2  1 m2  m  5m  5  m2  1 4m  5  1 4m  6

Extra Practice



2x  5

11x2  45x  6x2  50x  100

2  (r  2) (r  9)6



5x 3x  10

x

(a  2) (2a  3)  2(a  2) (a  3)  (a  3)12

r  2 2r r9 r  2 r  2 2r r9 r  2

m m  1

1

5x 3x  10

5x2

x

m

2

a

5x2  25x  6x2  20x  6x2  30x  20x  100

5

m m  1

 3  2

5x(x  5)  2x(3x  10)  (6x  20)(x  5)

(r  9) (r  2)  2r(r  2)  6(r2  11r  18) r2  7r  18  2r2  4r  6r2  66r  108 r2  3r  18  6r2  66r  108 0  7r2  63r  126 0  r2  9r  18 0  (r  6) (r  3) r  6 or r  3 12.

1

16.

x(5x)  x  1  5x(x  1) 5x2  x  1  5x2  5x x  1  5x 1  4x

11.

6

1b 14 6 2  2(b  6)(b  8) 112  b 6 8 2

a 4a  15

2  x(x  1)(5)

1 4

1

2b8

(4a  15) 4a  15  3  (4a  15)(2) a  3(4a  15)  8a  30 a  12a  45  8a  30 11a  45  8a  30 15  3a 5  a

t(2t)  3(t  3)  2t(t  3) 2t2  3t  9  2t2  6t 3t  9  6t 9  3t 3t 10.

14 b  6

2(b  8) (14)  (b  6) (b  8)  2(b  6)(6) 28(b  8)  b2  14b  48  12(b  6) 28b  224  b2  14b  48  12b  72 28b  224  b2  2b  24 0  b2  30b  200 0  (b  10)(b  20) b  10 or b  20

2  5(y  4)(3)

2t 3 t t  3 2t 3 t t  3

 12

2x  4x  12x  36 6x  12x  36 6x  36 x6

2(b  6) (b  8)

10y  3(y  4)  15( y  4) 10y  3y  12  15y  60 7y  12  15y  60 72  8y 9y 9.

 12

2  12(x  3)

14.

b

2y 3 5 y  4 2y 3 5 y  4

1

2x 4x 3x x  3 2x 4x x3 x  3 2x 4x x3 x  3

3 2

758

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Page 849

4. The matrix has 1 row and 1 column. Therefore, it is a 1 by 1 matrix. 5. Matrices A and B have different dimensions; therefore, it is impossible to add them. 2 4 1 0 6. A  C  c d  c d 3 5 0 1

Lesson 13-1

1. The sample is 10 dogs from a county. The population is all dogs from a county. The sample is biased because not every dog in the county has an equal chance of being selected. Since the sheriff checked the first ten dogs near his office, the sample is a convenience sample. 2. The sample is 25 students. The population is all students at the school. The sample is biased, since the students are self-selecting. The sample is a voluntary response sample, since the students respond to the school bulletin. 3. The sample is ice cream cones made during three shifts. The population is all ice cream cones made during three shifts. The sample is unbiased since every ice cream cone has an equal chance of being selected. The sample is a systematic random sample since every tenth cone is selected for weighing. 4. The sample is cars from an assembly line. The population is all cars from an assembly line. The sample is unbiased since every car is equally likely to be selected. Since the cars are identified by the day of the week, the sample is a stratified random sample. 5. The sample is people who return the survey. The population is all people entering the department store. The sample is biased since the respondents are self-selecting. Since people can decide whether to return the survey, the sample is a voluntary response sample. 6. The sample is some residents in a community. The population is all residents of the community. The sample is unbiased since every resident is equally likely to be selected. The sample is a systematic random sample since every twentieth person is surveyed. 7. The sample is malformed frogs identified by residents and reported. The population is all malformed frogs in the state’s lakes. The sample is biased, since not every frog has an equal chance of being selected. The sample is a voluntary response sample, since residents choose to report sightings of malformed frogs. 8. The sample is 10 cartons of strawberries. The population is all cartons of strawberries in the store. The sample is biased since not every carton of strawberries has an equal chance of being selected. Since the top ten cartons of strawberries were taken from the shelf, the sample is a convenience sample.

Page 849

 c

2  1 4  0 d 3  0 51

 c

3 4 d 3 6 1 1 4 5 1 4 d  c d 0 3 2 3 0 2

7. B  D  c  c

1  (5) 0  (3)

 c

4 0 0 d 3 3 0 5 1 4 1 1 4 d  c d 3 0 2 0 3 2

8. D  B  c  c

5  1 1  (1) 4  4 d 3  0 0  3 2  (2)

 c

6 2 8 d 3 3 4

9. 2B  2 c

1 3

1 0

2(1) 2(0)

 c

2 2 8 d 0 6 4

3(0) d 3(1)

 c

3(1) 3(0)

 c

3 0 d 0 3 2 4 1 0 d  c d 3 5 0 1

 c

2  1 4  0 d 3  0 51

 c

1 4 d 3 4

12. 5c  5 c

1 0 d 0 1

 c

5(1) 5(0)

 c

5 0 d 0 5

13. 2A  C  2 c

759

2(4) d 2(2)

2(1) 2(3)

1 0 d 0 1

11. A  C  c

1. The matrix has 1 row and 4 columns. Therefore, it is a 1 by 4 matrix. 2. The matrix has 2 rows and 2 columns. Therefore, it is a 2 by 2 matrix. 3. The matrix has 3 rows and 3 columns. Therefore, it is a 3 by 3 matrix.

4 d 2

 c

10. 3c  3 c

Lesson 13-2

1  1 4  (4) d 30 2  2

5(0) d 5(1)

2 4 1 0 d  c d 3 5 0 1 2(4) 1 0 d  c d 2(5) 0 1

 c

2(2) 2(3)

 c

4  1 8  0 d 6  0 10  1

 c

5 8 d 6 11

Extra Practice

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5 3

1 4 1 1 4 d  c d 0 2 0 3 2

5. Sample answer:

3(5) 3(3)

 c

15  1 3  (1) 12  4 d 9  0 0  3 6  (2)

 c

16 9

3(1) 3(0)

Number of Defective Light Bulbs Per Shift

3(4) 1 1 4 d  c d 3(2) 0 3 2

 c

Frequency

14. 3D  B  3 c

4 16 d 3 8

15. Matrices B and C have different dimensions, therefore, it is impossible to add them. 1 0 2 4 16. 2C  3A  2 c d  3c d 0 1 3 5 2(1)  c 2(0)

2(0) 3(2) d  c 2(1) 3(3)

26  c 0  (9)

10 8 6 4 2 0 02

35

68

911

1214

Number of Bulbs

3(4) d 3(5)

Page 850

0  (12) d 2  15

Lesson 13-4

1. Order the data from least to greatest. 10 33 34 37 43 45 45 50 The range is 56–10 or 46. The median is the middle value, or 43.

8 12  c d 9 17

56

33  34 or 33.5. 2 45  50 is or 47.5. 2

The lower quartile is

Page 850

Lesson 13-3

The upper quartile

1. The data set has a total of 2  6  1  2  0.5  11.5 values, so the median is the average of the sixth and seventh values. Both values occur in the range 1000–1500, so the median occurs in that range as well. The data appears to be skewed to the left. 2. The data set has 2  4  7  5  14  11  4  4  51 values, so the median is the 26th value. The median occurs in the range 70–75. About half of the data lie in the 70–80 percent range. 3. Sample answer:

The interquartile range is 47.5  33.5 or 14. The outliers would be less than 33.5  1.5(14) or 12.5 or greater than 47.5  1.5(14) or 68.5. Since 10  12.5, 10 is the only outlier. 2. Order the data from least to greatest. 65 65 68 77 78 84 84 95 96 99 The range is 99  65 or 34. The median is

Frequency

Prices of Notebooks

2 1 0 2741

4256

5771

7286

87101

The median is

Price in Cents

The

4. Sample answer:

The

Frequency

Number of Fish in Each Tank

4 2 0 916

1724

2532

3340

4148

4956

Number of Fish

Extra Practice

60  70 2

or 65.

40  50 lower quartile is or 45. 2 80  90 or 85. upper quartile is 2

The interquartile range is 85  45 or 40. The outliers would be less than 45  1.5(40) or 15 or greater than 85  1.5(40) or 145. There are no outliers. 4. Order the data from least to greatest. 1 1.3 2.4 3 3.7 4 5.2 6 7.1 8 9 The range is 9  1 or 8. The median is the middle value or 4. The lower quartile is 2.4. The upper quartile is 7.1. The interquartile range is 7.1  2.4 or 4.7. The outliers would be less than 2.4  1.5(4.7) or 4.65 or greater than 7.1  1.5(4.7) or 14.15. There are no outliers.

14 12 10 8 6

18

or 81.

The lower quartile is 68. The upper quartile is 95. The interquartile range is 95  68 or 27. The outliers would be less than 68  1.5(27) or 27.5 or greater than 95  1.5(27) or 135.5. There are no outliers. 3. Order the data from least to greatest. 30 40 50 60 70 80 90 100 The range is 100  30 or 70.

5 4 3

1226

78  84 2

760

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3. Order the set from least to greatest.

5. Order the data from least to greatest. 0 4 13 25 29 31 37 44 56 66 73 The range is 73  0 or 73. The median is the middle value or 31. The lower quartile is 13. The upper quartile is 56. The interquartile range is 56  13 or 43. The outliers would be less than 13  1.5(43) or 51.5 or greater than 56  1.5(43) or 120.5. There are no outliers. 6. Order the data from least to greatest. 112 234 268 369 400 406 527 648 775 The range is 775  112 or 663. The median is the middle value or 400.

1.2

Q1 

234  268 or 251. 2 527  648 or 587.5. is 2

55

1

49  55 2

3.2

1.7  1.8 2

 1.75 Q2 

3.5

4.2

5.5

c

2.2  2.6 2

 2.4 Q3 

3.2  3.5 2

 3.35

3

4

5

6

0 10 20 30 40 50 60 70 80 90 100

55

c Q 

3.0

35  35

5. Order the set from least to greatest. A: 21 24 27 34 40 46 50 58 61 67 70 72 c c c Q1 

59 c

 52 Q  2

55  59 2

63

27  34 2

Q1 

69

69

69

69  69 2

46  50 2

 48 Q3 

61  67 2

 76.5 Q3 

81  82 2

 64

69  72 2

 70.5 Q2 

75  78 2

 81.5

The interquartile range for set B is 81.5  70.5  11. Check to see if there are any outliers. 70.5  1.5(11)  54 81.5  1.5(11)  98 There are no outliers.

89

c  57 Q3 

 30.5 Q2 

The interquartile range for set A is 64  30.5  33.5. Check to see if there are any outliers. 30.5  1.5(33.5)  19.75 64  1.5(33.5)  114.25 There are no outliers. Order the set from least to greatest. B: 67 69 69 72 74 75 78 79 81 82 83 83 c c c

2. Order the data from least to greatest. 55

2.6

Q1 Q2   35 Q3 2 The interquartile range is 52  22 or 30. Check to see if there are any outliers. 22  1.5(30)  23 52  1.5(30)  97 There is one outlier, 99.

0 2 4 6 8 10 12

49

2.2

4. Order the set from least to greatest. 15 15 18 22 25 25 35 35 37 50 52 65 69 99 c c c

Lesson 13-5

45

1.8

c

2

1

1. Order the data from least to greatest. 1 2 2 3 3 4 4 5 7 8 8 9 10 11 12 c c c Q2 Q3 Q1 The interquartile range is 9  3 or 6. Check to see if there are any outliers. 3  1.5(6)  6 9  1.5(6)  18 There are no outliers.

40

1.8

The interquartile range is 3.35  1.75 or 1.6. Check to see if there are any outliers. 1.75  1.5(1.6)  0.65 3.35  1.5(1.6)  5.75 There are no outliers.

The interquartile range is 587.5  251 or 336.5. The outliers would be less than 251  1.5(336.5) or 253.75 or greater than 587.5  1.5(336.5) or 1092.25. There are no outliers.

Page 850

1.7 c

The lower quartile is The upper quartile

1.2

 69

The interquartile range is 69  52 or 17. Check to see if there are any outliers. 52  1.5(17)  26.5 69  1.5(17)  94.5 There are no outliers.

A B 20

30

40

50

60

70

80

90

The A data are much more diverse than the B data. In general, the B data are greater than the A data.

40 50 60 70 80 90

761

Extra Practice

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6. Order the set from least to greatest.

8. Order the set from least to greatest.

A: 60 65 70 72 75 76 83 85 88 91 92 92 92 99 100 c

c

c

Q1

Q2

Q3

A:

Q1 

5.25

6.25

6.50

c

3.25  4.25 2

 3.75

Q

7.95

8.25

c

2

Q3 

6.50  7.95 2

 7.225

7.225  1.5(3.475)  12.4375

There are no outliers. Order the set from least to greatest. B:

3.25

4.25

4.50

5.25

5.50

c Q1 

4.25  4.50 2

5.75

6.95

c  4.375

8.65

9.50

c

Q

2

Q3 

6.95  8.65 2

 7.8

The interquartile range is 7.8  4.375  3.425. Check to see if there are any outliers. 4.375  1.5(3.425)  0.7625

7.8 + 1.5(3.425)  12.9375

There are no outliers.

A B 70

60

80

90

100

3.00 4.00 5.00 6.00 7.00 8.00 9.00

The B data are more diverse than the A data. The data sets have approximately the same range. 7. Order the set from least to greatest. 0.5

0.8

0.8

1.1

1.5

c Q1 

0.8  0.8 2

2.2

2.2

2.3

c  0.8 Q2 

3.0

3.6

The distribution of both sets are similar. The values for B are somewhat greater than the values for A.

3.8

c

1.5  2.2 2

 1.85 Q  2.3 3

 3.0 2

 2.65

Page 851

The interquartile range is 2.65  0.8  1.85. Check to see if there are any outliers. 0.8  1.5(1.85)  1.975 2.65  1.5(1.85)  5.425 There are no outliers. Order the set from least to greatest 0.9

4.75

3.75  1.5(3.475)  1.4625

B

B:

4.25

The interquartile range is 7.225  3.75  3.475. Check to see if there are any outliers.

A

0.4

3.25 c

The interquartile range is 92  72  20. Check to see if there are any outliers. 72  1.5(20)  42 92  1.5(20)  122 There are no outliers. Order the set from least to greatest. B: 62 65 73 77 77 81 82 85 85 88 91 92 95 98 99 c c c Q1 Q2 Q3 The interquartile range is 92  77  15. Check to see if there are any outliers. 77  1.5(15)  54.5 92  1.5(15)  114.5 There are no outliers.

A:

2.65

1.2

1.6

1.8

1.9

2.5

c Q1 

3.3

3.8

4.0

c

1.6  1.8 2

 1.7 Q2 

5.4

5.7

Lesson 14-1

1.

ice cream chicken

cookies ice cream 6.0

lettuce

beef

c

2.5  3.3 2

 2.9 Q3 

4.0  5.4 2

pudding

pudding cookies

 4.7

The interquartile range is 4.7  1.7  3.0. Check to see if there are any outliers. 1.7  1.5(3)  2.8 4.7  1.5(3)  9.2 There are no outliers.

ice cream fish

pudding cookies ice cream

A

chicken

pudding cookies

B

ice cream 0

1

2

3

4

5

6

coleslaw

beef

The B data are more diverse than the A data. In general, the B data are greater than the A data.

pudding cookies ice cream

fish

pudding cookies

The tree diagram shows that there are 18 possible outcomes.

Extra Practice

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2.

H H T H H T T H H T T H T T

4.

H T H T H T H T H T H T H T H T

C

M C B

W

Y

B

R

G

S M

The tree diagram shows that there are 12 possible outcomes. 5. There are ten digits and 26 letters. Multiply to find the number of possible outcomes. 10  10  26  26  67,600 There are 67,600 possible license plates. 6. Multiply to find the number of ways.

The tree diagram shows that there are 16 possible outcomes. 3.

S

V

W P W P W P W P W P W P

W Y B R G W Y B R G W Y B R G W Y B R G W Y B R G

rocking or non-rocking 1442443

swivel or non-swivel 1442443

cotton, leather or plush cover 144424443

green, blue, maroon, or black 1442443

2 2 3    There are 48 possible lounge chairs. 7. Multiply to find the number of ways. boots skis poles 123 123 123

4

 48

3  4  5  60 There are 60 possible selections. 8. There are 4 possible outcomes for each die. Multiply. red die blue die white die 1 424 3 14243 14 424 43

9. 10. 11. 12. 13. 14.

The tree diagram shows that there are 25 possible outcomes.

15. 16.

4 4 4    64 There are 64 possible outcomes. 8!  8  7  6  5  4  3  2  1  40,320 1!  1 0!  1 5!  5  4  3  2  1  120 2!  2  1 2 9!  9  8  7  6  5  4  3  2  1  362,880 3!  3  2  1 6 14!  14  13  12  11  10  9  8  7  6  5  4  3  2  1  87,178,291,200

Page 851

Lesson 14-2

1. This is a combination because order is not important. 2. This is a permutation because order of runners can make a difference.

763

Extra Practice

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3. This is a combination because order is not important. 4. This is a permutation because order of winning is important. 5. This is a permutation because order is important with letters. 6. This is a permutation because order is important with a lock. 7. This is a combination because order is not important. 8. 5P2   

8!



3!

3!

6

7! (7  7)! 7! 0!

Page 851

1

1

C

 

8

40

39

1560

dependent. P(2 black):  55  54  2970 or

8

52 99

8. Since the first duck is not replaced, the events are 2

1

2

dependent. P(2 yellow)  55  54  2970 or

1 1485

9. Since the first duck is not replaced, the events are 40 40 1 dependent. P(black, then gold)  55  54  2970 4 or 297

13! (13  13)!13! 13! 13!

40

39

38

8

474,240

10. P(3 blacks, then red)  55  54  53  52  8,185,320 304 or 5247

1 9!

11. P(yellow, then blue, then gold) 2 4 1 8 4  55  54  53  157,410 or 78,705 1

0

4

3

12. P(2 gold)  55  54  0

987654 1

2

1

24

13. P(4 blue)  55  54  53  52  8,185,320 or

 60,480 7!

4!

4

16. ( P )( P )  (7  3)!  (4  2)! 7 3 4 2 7! 4!  4!  2! 765 1



3

2

1

1

1 341,055 24

14. P(4 blue, then gold)  55  54  53  52  51  417,451,320 1 or 17,393,805

43 1

 2520

Extra Practice

1

7. Since the first duck is not replaced, the events are

18! (18  10)!10! 18! 8!10! 18  17  16  15  14  13  12  11 8!



1 18

40  55 ducks. P(red, then gold)  55  54  2970 4 or 1485

15. P  (9  6)! 9 6 9!  3! 

2

6. Since the first duck is not replaced, the events are dependent. There are a total of 8  2  1  4 

 43,758 13 13

2

1 4

5. P(red greater than 2, blue greater than 3) 4 3 1 12  6  6  36 or 3

 56

14.

1 4

9

4. P(red 6, blue greater than 4)  6  6  36 or

8! (8  2)! 8! 6! 87 1



3

3. There are 3 prime numbers (2, 3, 5) on a die. 3 3 9 P(red prime number, blue even)  6  6  36 or

6! (6  5)!5!



1

3

10  9 2!

C  18 10

1

2. P(red even, blue even)  6  6  36 or

6

13.

Lesson 14-3

1. P(red 1, blue 1)  6  6  36

 45



10!

19. ( P )( C )  (3  2)!  (10  10)!(10!) 3 2 10 10 3! 10!  1!  10!

10!



10!

 3  10!  10,886,400

10. 10C2  (10  2)!2! 10!  8!2!

12. 8P2 

76 2!

18. ( C ) ( P )  (3  2)!2!  (10  10)! 3 2 10 10 3! 10!  2!  0!

 7!  5040

11. 6C5 



 28  21  588

5! (5  2)! 5! 3! 54 1



87 2!



 20 9. P  7 7

7!

17. ( C ) ( C )  (8  6)!6!  (7  5)!5! 8 6 7 5 8! 7!  2!6!  2!5!

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Page 852 1.

1 2 3 4 5 6

4. See students’ work.

Lesson 14-4 1 1 2 3 4 5 6

2 2 4 6 8 10 12

3 3 6 9 12 15 18

4 4 8 12 16 20 24

5 5 10 15 20 25 30

5.

6 6 12 18 24 30 36

1 36

There are 4 ways to roll a product of 12. P(X  12) 

4 36

or

2 36

or

Ways

Product

Ways

8

2

18

2

2

2

9

1

20

2

3

2

10

2

24

2

4

3

12

4

25

1

5

2

15

2

30

2

6

4

16

1

36

1

P(product of 12) 

1 18

P(product greater than 15) 

11 36

11

4 36 4 36

or about 0.11 or about 0.11

Since these theoretical probabilities are close to 10%, we would expect to get a product of 6 about 10% of the time and a product of 12 about 10% of the time.

3. There are 1  2  2  2  1  2  1  11 different ways to roll a product greater than 15. 121

 36  1296

6. P(product is 2) 

4. Let X  the number of customers. X can have the values of 500, 1000, 1500, 2000, and 2500. 5. For each value of X, the probability is greater than or equal to 0 and less than or equal to 1. 0.05  0.25  0.35  0.30  0.05  1, So the probabilities add up to 1. 6. P(X 1000)  P(0 X 500)  P(501 X 1000)

2 36

or

1 18

7. There are a total of 1351 households. P(1)  P(2)  P(3)  P(4) 

172 1351 293 1351 482 1351 256 1351

or about 12.7% or about 21.7% or about 35.7% or about 18.9%

P(5 or more) 

 0.05  0.25  0.30

7.

Product

1

P(product of 6) 

1 9

There are 2 ways to roll a product of 24. P(X  24) 

Ways

1

As you roll the dice more times, the experimental probabilities should get closer to the theoretical probabilities. Consider the following theoretical probabilities.

2. There is 1 way to roll a product of 9. P(X  9) 

Product

8. P(5 or more) 

P(X 7 500)  P(501 X 1000)  P(1001 X 1500) 

148 1351 148 1351

or about 11.0% or about 0.11 or 11%

9. These events are mutually exclusive. P(1 or 2)  P(1)  P(2)

P(1501 X 2000)  P(2001 X 2500)  0.25  0.35  0.30  0.05

172

293

 1351  1351

 0.95

465

 1351 or about 0.34 or 34%

Page 852

Lesson 14-5

1. See students’ work. 2. See students’ work. 1

1

1

1

1

3. P(4 heads)  2  2  2  2  16

765

Extra Practice

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Mixed Problem Solving Page 853

8. Sample answer: 4 tank tops 8 shorts 4 t-shirts at $6.99 at $7.99 plus at $4.99 plus

Chapter 1 The Language of Algebra

1. Sum implies add, and product implies multiply. So the expression can be written as r2  r/.

14243 123 14 42443 123 1442443

2. Replace / in r2  r/ with r.

3.



8(7.99)

4(4.99)



4(6.99)

r2  r/  r2  r  r  r2  r2  (1  1)r2  2r2

8 shorts 4 t-shirts 4 tank tops at $5.99 plus at $8.99 plus at 2.99 14243 123 14243 123 1442443 8(5.99)  4(8.99)  4(2.99)

Money for money for money for small tables large tables small tables for first market plus for first market plus for second market 144424443 123 144424443 123 14444244443

14243 123 14243 123 14243 123 1442443



25(5)

10(8.5)



4 tank tops 4 shorts 4 shorts 4 t-shirts at $7.99 plus at $5.99 plus at $4.99 plus at $2.99 4(7.99)

35(5)

money for large tables plus for second market 123 14444244443 

12(8.5)

The expression is 25(5)  10(8.5)  35(5)  12(8.5). 4. 25(5)  10(8.5)  35(5)  12(8.5)  125  85  175  102  487 $487 was collected at the two markets. 5. Let a represent adult price, c represent child price, and g represent observer price. The cost of the cost of the cost of is less than 2 adults plus 2 children plus 1 observer or equal to $55. 

2(c)



1(g)



2a  2c  g  55

16.95, 12.95, 4.95

2(16.95)  2(12.95) ?  4.95  55 S 64.75  55

4(4.99)



4(2.99)

8(5.99)  8(2.99) 8(5.99)  8(2.99)  47.92  23.92  71.84 The discount is 15% of the purchase. 15% of $71.84  0.15  71.84  10.776 Subtract $10.78 from the purchase. $71.84  $10.78  $61.06 The least amount you can spend is $61.06. amount of sales 11. amount of sales of 25 blankets plus of 25 rabbits 144424443 123 144424443

True or False? false

No, the family cannot go for a full day.

Mixed Problem Solving



8(7.99)  8(8.99) 8(7.99)  8(8.99)  63.92  71.92  135.84 The discount is 15% of the purchase. 15% of $135.84  0.15  135.84  20.376 Subtract $20.38 from the purchase. $135.84  $20.38  $115.46 The greatest amount you can spend is $115.46. 8144 shorts at4443 $5.99 1 plus 8 tops at $2.99 4424 23 144424443

55

The inequality is 2a  2c  g  55. 6. To find the cost for a full day, evaluate 2a  2c  g if a  16.95, b  12.95, and g  4.95. 2a  2c  g  2(16.95)  2(12.95)  4.95  33.90  25.90  4.95  64.75 A full day would cost $64.75. To find the cost for a half day, evaluate 2a  2c  g if a  10.95, b  8.95, and g  3.95. 2a  2c  g  2(10.95)  2(8.95)  3.95  21.9  17.9  3.95  43.75 A half day would cost $43.75. 7. Replace a, c, and g in 2a  2c  g  55 with the prices for a full day. a, c, g

4(5.99)

144 4424 4443 123 144424443

1442443 123 1442443 123 1442443 1442443 123

2(a)



9. Sample answer: 8(7.99)  4(4.99)  4(6.99)  63.92  19.96  27.96  111.84 8(5.99)  4(8.99)  4(2.99)  47.92  35.96  11.96  95.84 4(7.99)  4(5.99)  4(4.99)  4(2.99)  31.96  23.96  19.96  11.96  87.84 The cost of the 16 items could be $111.84, $95.84, or $87.84. All totals end in $0.84. 10. 8 shorts at $7.99 plus 8 tops at $8.99

25(28)  25(18) 25(28)  25(18)  25(28  18)  25(46)  1150 The total amount of sales was $1150.

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12. 10 tickets

10 tickets 10 tickets at $18.95 plus at $12.95 plus at $9.95 plus 30 combos

Page 854

10(18.95)  10(12.95) 

10(9.95)



30(5.50)

10(18.95)  10(12.95)  10(9.95)  30(5.50)  189.5  129.5  99.5  165  583.5 The total cost is $583.50. 13. If two lines are perpendicular, then they meet to form four right angles. If two lines meet to form four right angles, then they are perpendicular. 14. The domain contains the number of lawns Laurie mows each week. She mows at most 30 lawns each week. Therefore, a reasonable domain would be values from 0 to 30 lawns. The range contains the weekly profit from $0 to 30  $15 or $450. Thus, a reasonable range is $0 to $450. Graph the ordered pairs (0, 0) and (30, 450). Since she can mow any amount of lawns up to 30 lawns, connect the two points with a line to include those points. Profit ($)

y

(30, $450)

400 300 200 100

x 5 10 15 20 25 30 Number of Lawns

0

15. Sample answer: The second graph is misleading because the intervals for the years are not equal and the intervals for the y-axis are not equal.

4

x 1980 2000

1960

0 1920 1940

Population Density

y 6 5 3 2 1

144424443

y 6 5 4

1980 2000

1960

x 1920

Population Density

14243

144424443

29,035 (1312)  29,035  (1312)  29,035  (1312)  29,035  1312  30,347 The difference is 30,347 feet. 5. To find the change in temperature, multiply the number of 1000-foot increments in 10,000 feet by the temperature drop in 1000 feet. 10(3.6)  36 The change in temperature is 36 F. 6. To find the altitude, find the difference of 70 and 38, then divide by 3.6. [70  (38) ] 3.6  [ 70  (38) ] 3.6  [ 70  38] 3.6  108 3.6  30 The altitude is 30  1000 or 30,000 feet.

Year

3

Chapter 2 Real Numbers

1. Since 18 15 13 11 2 14 19 31 34 38 39, the monthly normal temperatures ordered from least to greatest are 18, 15, 13, 11, 2, 2, 14, 19, 31, 34, 38, 39. 2. 18 is eighteen units from 0 in the negative direction. |18|  18 15 is fifteen units from 0 in the negative direction. |15|  15 13 is thirteen units from 0 in the negative direction. |13|  13 11 is eleven units from 0 in the negative direction. |11|  11 2 is two units from 0 in the negative direction. |2|  2 2 is two units from 0 in the negative direction. |2|  2 14 is fourteen units from 0 in the positive direction. |14|  14 19 is nineteen units from zero in the positive direction. |19|  19 31 is thirty-one units from zero in the positive direction. |31|  31 34 is thirty-four units from zero in the positive direction. |34|  34 38 is thirty-eight units from zero in the positive direction. |38|  38 39 is thirty-nine units from zero in the positive direction. |39|  39 3. No; the lowest temperatures should be at the beginning and end of the table for January and December. 4. Write 29,035 as a positive number to represent above sea level, and 1312 as a negative number to represent below sea level. Subtract to find the difference. elevation of elevation of Mount Everest minus the Dead Sea

14 424 43 123 14 424 43 123 14 424 43 123 14 424 43

Year

767

Mixed Problem Solving

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14. There are 20  39  39  39  44  35  40 or 256 million people predicted to not be 15–24 years old and 299 million people predicted as a total.

7. By finding the values obtained by using all twodigit whole numbers, we conclude that the least result is obtained from the number 19. 19 1  9

19

 10

256,000,000

P(not be 15–24)  299,000,000

 1.9 8. The greatest common place value is ones, so the digits in the ones place are the stems. Stem Leaf 7 1 7 8 1 6 9 9 0 1 1 6 10 2 4 5 11 0 9 12 5 8 7|1  7.1 9. The least value is 7.1 and the greatest value is 12.8. 12.8  7.1  5.7 10. mean 

256

 299  0.86 The probability that a person selected will not be 256 15–24 years old is 299 or about 86%. 15. 2202  452  1400  2025  12425  49.24 The hypotenuse of the garden is about 49 ft. 16. Let x  the length of the other leg. The situation can be represented by the equation 55  2x2  452. Estimate to find reasonable values for the replacement set. Start by letting x  25 and then adjust values up or down as needed.

sum of wind speeds number of speeds 8.9  7.1  9.1  9.0  10.2  12.5  11.9  11.0



 12.8  10.4  10.5  8.6  7.7  9.6  9.1  8.1 16

 9.78125 The mean is about 9.8. To find the median, order the numbers from least to greatest. The median is in the middle. 7.1, 7.7, 8.1, 8.6, 8.9, 9.0, 9.1, 9.1, 9.6, 10.2, 10.4, 10.5, 11.0, 14243 11.9, 12.5, 12.8 9.1  9.6 2

55  2x2  452

x



55 S 51.5 55

too low



? 452 

55 S 54.1 55

too low

35 235 

? 452 

55 S 57.0 55

too high

25 30

2302 2

2 ?

31 2312  45  55 S 54.6 55

 9.35

2 ?

32 2322  45  55 S 55.2  55

The median is 9.35. The mode is the number that occurs most frequently. 9.1 occurs twice and all the other speeds occur once. The mode is 9.1. 11. Sample answer: Yes, many of the data values are around 9.1. 12. There are 20 million people predicted to be under age 5 and 299 million people predicted as a total.

Reasonable?

? 452 

2252

almost reasonable

The length of the other leg should be about 32 ft. 17. To find the speed, divide the number of meters by the number of seconds. 400 44  9.1 The speed of the 400 meter run was about 9 meters per second. 18. To find the speed, divide the number of meters by the number of seconds. 400 3 min 41 sec  400 221 sec  400 221  1.8 The speed of the 400-meter freestyle was about 2 meters per second. 19. Since 2  4.5  9, the running speed is 4.5 times faster than the swimming speed.

20,000,000

P(under age 5)  299,000,000 20

 299  0.07 The probability that a person selected will be 20 under 5 is 299 or about 7%. 13. There are 40 million people predicted to be 65 or over and there are 299  40 or 259 million people predicted to be under 65. 40,000,000

odds of 65 or over  259,000,000

Page 855

40

Chapter 3 Solving Linear Equations

 259 1.

The odds of selecting a person that will be 65 or over are 40:259.

The lateral surface area is

the product of the two times  times radius and the height.

144424443 123 123 123 123 123 144444 42444 4443

L



2







The formula is L  2rh. 2. L  2rh  2(3.14)(4.5)(7)  197.82 The lateral area is about 197.8 in2.

Mixed Problem Solving

768

rh

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3.

1

the area the lateral The total surface area equals two times of the base plus surface area.

The Wilson family bought 2 (30)  1 or 16 bags. 1 The Martinez family bought 2 (30  16)  1 or 8 bags. The Brightfeather family bought 1 (30  24)  1 or 4 bags. The Wimberly family 2 bought 2 bags. 11. Let n  the least even integer. Then n  2  the next greater even integer, and n  4  the greatest of the three even integers.

144424443 14243 123 123 1442443 123 144 424443 2



T

2



r



2rh

The formula is T  2r2  2rh. 4. T  2r2  2rh  2(3.14)(4.5) 2  2(3.14)(4.5)(7)  127.17  197.82  324.99 The total surface area is about 325.0 in2. 5. Let n  the length of the Niger River.

Five times the twice the sum of greatest even integer is equal to the other two integers plus 42.

44244 3 1444442444443 123 123 1444442444443 1





310

2900

The equation is n  310  2900. 6. n  310  2900 n  310  310  2900  310 n  2590 The length of the Niger River is 2590 miles. 7. Let g  the length of an average common goldfish. the length of an the average length of a 4.8 times average common goldfish equals yellow-banded angelfish.

123 123 144444424444443 1 4244 444443 424 3 1444444

4.8





g

12

The equation is 4.8g  12. 8. 4.8g  12 4.8g 4.8

12

 4.8

x  (x  10)  2[ x  (x  10) ]

g  2.5 The length of an average common goldfish is 2.5 in. 9. Let x  the number of registered Labrador Retrievers. The number of the number of Labrador Retrievers was fifteen times Great Danes plus 6997. 

15



9860



6x 6

6997

Undo the Statement 2

The Wimberlys bought half the remaining bags plus one. The Brightfeathers bought half the remaining bags plus one. The Martinezes bought half the remaining bags plus one. The Wilsons bought half the remaining bags plus one. The starting number of bags

2  2x  1 S x  2

42



180



150 6

x  25 x  10  25  10 or 35 2[x  (x  10)  2[ 25  (25  10) ] or 120 The measures of the three angles are 25 , 35 , and 120 . 13. Let c represent the amount of chemicals needed.

x  15(9860)  6997 x  147,900  6997 x  154,897 There were 154,897 registered Labrador Retrievers. 10. Start at the end of the problem and undo each step. Statement The Wimberlys bought 2 bags.



x  (x  10)  2[ x  (x  10) ]  180 2x  10  2[ 2x  10]  180 2x  10  4x  20  180 6x  30  180 6x  30  30  180  30 6x  150

14 444424444 43 123 1 424 3 123 144424443 123 123

x

2[ n  (n  2) ]

5(n  4)  2[n  (n  2) ]  42 5n  20  2[ 2n  2]  42 5n  20  4n  4  42 5n  20  4n  46 5n  20  4n  4n  46  4n n  20  46 n  20  20  46  20 n  26 n  2  26  2 or 28 n  4  26  4 or 30 The consecutive even integers are 26, 28, and 30. 12. Let x  the measure of the second angle. Then x  10  the measure of the first angle, and 2[x  (x  10)]  the measure of the third angle. The sum of the measures of the three angles of a triangle equals 180.

1444444442444444443 14243 123

1444 4244443 123 1442443 14243 1444 4244443

n



5(n  4)

the length of The length of the Niger River plus 310 miles equals the Congo river.

1.5 5000

c

 12,500

1.5(12,500)  5000(c) 18,750  5000c

1

18750 5000



5000c 5000

3.75  c The amount of chemicals needed is 3.75 pounds. 14. Find the change. 2999  1999  1000 Find the percent using the original number, 2999, as the base.

2(2  1)  6

2(6  1)  14

1000 2999

r

 100

1000(100)  2999(r) 100,000  2999r

2(14  1)  30

100,000 2999



2999r 2999

33  r The percent of decrease was about 33%.

30

769

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3.

15. The tax is 7% of the cost of the camp. 7% of $1254  0.07  1254  87.78 Add this amount to the original cost. $1254.00  $87.78  $1341.78 The total cost of the camp is $1341.78. 16. The refund is 1 2

of $87.78 

O



Since it has five sides, the design is in the shape of a pentagon. 4. To dilate the pentagon by a scale factor of 0.75, multiply the coordinates of the vertices by 0.75.

Prt rt

(x, y) S (0.75x, 0.75y) (1, 1) S (0.75  1, 0.75  (1)) S (0.75, 0.75) (2, 2) S (0.75  2, 0.75  2) S (1.5, 1.5) (0, 3) S (0.75  0, 0.75  3) S (0, 2.25) (2, 2) S (0.75  (2), 0.75  2) S (1.5, 1.5) (1, 1) S (0.75  (1), 0.75  (1) ) S (0.75, 0.75)

P I

18. P  rt 1848.75

P  0.0725(6) P  4250 The amount of money invested is $4250. 19. Let x  the amount of 40% iodine solution to be added. Amount of Solution 40 x 40  x

15% Solution 40% Solution 20% Solution

5. The area of the original design can be found by dividing the pentagon into five polygons as follows: • The area of the rectangle with vertices (1, 1), (1, 2), (1, 2), and (1, 1) is bh  2(3) or 6 units2. • The area of the triangle with vertices (1, 1), 1 1 (2, 2), and (1, 2) is 2 bh  2(1)(3) or 2 1.5 units . The area of the triangle with vertices (1, 1), (1, 2), and (2, 2) is also 1.5 units2. • The area of the triangle with vertices (2, 2), 1 1 (0, 3), and (0, 2) is 2bh  2(2)(1) or 1 unit2. The area of the triangle with vertices (2, 2), (0, 3), and (0, 2) is also 1 unit2. The total area of the original design is the sum of the areas of the rectangle and four triangles described above. This is 6  2(1.5)  2(1) or 11 units2. Similarly, the area of the dilated design can be found by dividing the dilated pentagon into five polygons as follows: • The area of the rectangle with vertices (0.75, 0.75), (0.75, 1.5), (0.75, 1.5), and (0.75, 0.75) is bh 1.5(2.25) or 3.375 units2. • The area of the triangle with vertices (0.75, 0.75), (1.5, 1.5), and (0.75, 1.5) is 1 1 bh  2(0.75)(2.25) or 0.84375 units2. The area 2 of the triangle with vertices (0.75, 0.75), (1.5, 1.5), and (0.75, 1.5) is also 0.84375 units2. • The area of the triangle with vertices (1.5, 1.5), (0, 2.25), and (0, 1.5) is 1 1 bh  2 (1.5) (0.75) or 0.5625 units2. The area 2 of the triangle with vertices (1.5, 1.5), (0, 2.25), and (0, 1.5) is also 0.5625 units2. The total area of the dilated design is the sum of the areas of the rectangle and four triangles described above. This is 3.375  2(0.84375)  2(0.5625) or 6.2 units2.

Amount of Iodine 0.15(40) 0.40x 0.20(40  x)

amount of Amount of amount of iodine in iodine in iodine in 15% solution plus 40% solution equals 20% solution. 44244 43 14 4424443 123 144424443 1 4243 14 0.15(40)



0.40x



0.20(40  x)

0.15(40)  0.40x  0.20(40  x) 6  0.4x  8  0.2x 6  0.4x  0.2x  8  0.2x  0.2x 6  0.2x  8 6  0.2x  6  8  6 0.2x  2 0.2x 0.2

2

 0.2

x  10 Isaac should add 10 gallons of the 40% solution to the 40 gallons of the 15% solution.

Page 856

x

1 of the tax. 2 1  87.78 2

 43.89 The amount of the refund is $43.89. 17. I  Prt I rt I rt

y

Chapter 4 Graphing Relations and Functions

1. • Start at the origin (the building). • Move right (east) 2 units and down (south) 1 unit. • The x-coordinate is 2, and the y-coordinate is 1. The coordinates of the pool are (2, 1). 2. • Start at the origin (the building). • Move left (west) 3 units and up (north) 5 units. • The x-coordinate is 3, and the y-coordinate is 5. The entrance to the community has coordinates (3, 5).

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6. Graph the ordered pairs (60, 102), (62, 109), (64, 116), (66, 124), (68, 131), (70, 139), (72, 147), and (74, 155).

Graph the ordered pairs and draw a line through the points. c 100

y

Weight

160

120

0

Cost

140

75 50 25

100 0

64

68 72 Height

76

7. No; for each 2 unit increase in height, the increase in weight is not constant. 8. Sample answer: 171 pounds; from 72 to 74 inches, there is an increase of 8 pounds, so 8  2  16 and added to 155 is 171.

y 230 228

9.

Distance from Sun, d

d 93,000,000

AU

36,000,000

36,000,000 93,000,000

0.387

Mars

141,650,000

141,650,000 93,000,000

1.523

Jupiter

483,750,000

483,750,000 93,000,000

5.202

222

39.223

220

Mercury

Pluto

3,647,720,000 93,000,000

3,647,720,000

226 Time

Planet

To the nearest thousandth, Mercury is 0.387 AU from the Sun, Mars is 1.523 AU from the Sun, Jupiter is 5.202 AU from the Sun, and Pluto is 39.223 AU from the Sun. 10. If the number of AU is less than 1, the planet is closer than Earth. If the number of AU is greater than 1, the planet is farther than Earth. 11.

x 2 4 6 8 Number of Blinds

13. If Pam has 8 blinds installed, then x  8. Substituting into the given equation, we have c  25  6.5(8) or 77. Thus, the cost of 8 blinds is $77. 14. Graph the ordered pairs (0, 229), (4, 227), (8, 218), (12, 230), (16, 224), and (20, 222).

x 60

c  25  6.5x

d 93,000,000 d (93,000,000) 93,000,000

218 0

 270,000

C 25 38 51 64 77

8 16 24 Years Since 1980

      5 5 5 5 5 5

 (93,000,000)270,000

25  6.5x 25  6.5(0) 25  6.5(2) 25  6.5(4) 25  6.5(6) 25  6.5(8)

x

15. Yes; each value of x is paired with only one value of y. 16. Sample answer: 218 min 17. 1 6 11 16 21 26 31

d  25,110,000,000,000 So Alpha Centauri is about 25,110,000,000,000 mi from the Sun. 12. Select five values for the domain and make a table. x 0 2 4 6 8

224

The first term is 1. The common difference is 5. Use the formula for the nth term to write an equation with a1  1 and d  5. an  a1  (n  1)d an  1  (n  1)5 an  1  5n  5 an  5n  4

x, C (0, 25) (2, 38) (4, 51) (6, 64) (8, 77)

18. Use the formula from Exercise 17 with n  20. an  5n  4 a20  5(20)4 a20  100  4 a  96 20 Row 20 would contain 96 beads.

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19. Side 1 2 3 4 5 6 7 8 9 10

7. The graph passes through (0, 453,589) with slope 2890.

Area 1 4 9 16 25 36 49 64 81 100

Population of Wyoming Population (thousands)

500

Chapter 5 Analyzing Linear Equations

change in price change in time

2.00  0.67

 1950  1940 

2

m m

4820x 4820

rate of number of years population Population equals change times after 1990 plus in 1990. 1442443 14243 14243 123 14444244443 123 1442443 y



2890

(0, 453,589)



1

2

3 4 5 6 7 8 Years Since 1990

9 10

x



 x1

136  112 66  60 24 or 4 6

Find the y-intercept. y  mx  b 112  4(60)  b 112  240  b 112  240  240  b  240 128  b Write the slope-intercept form. y  mx  b y  4x  (128) y  4x  128 10. y  4x  128 y  4(72)  128 y  160 A person who is 72 inches tall should be 160 pounds. 11. Write the point-slope form of the equation for a line that passes through (1990, 3,600,000) with slope 300,000. y  y1  m(x  x1 ) y  3,600,000  300,000(x  1990) 12. y  3,600,000  300,000(x  1990)

15  x It will take 15 seconds for sound to travel 72,300 feet. 6.

450

y2  y1

r 4820   1 Solve for the rate. 4820  r(1) 4820  r Therefore, the equation is y  4820x. 5. y  4820x 72,300  4820x 

460

mx

1.33 10

 0.133 Over this 10-year period, the price increased by $1.33, for a rate of change of $0.133 per year. 3. The slope of the line segment that represents the 10-year period from 1980 to 1990 is negative. This represents a drop in price. rate times time. 4. Distance equals 1 14243 14243 23 123 123

72,300 4820

470

8. The year 2005 is 15 years after 1990. y  2890x  453,589 y  2890(15)  453,589 y  496,939 The population of Wyoming should be about 496,939 in 2005. 9. Let x represent the height of a person. Let y represent the weight of the person. Write an equation of the line that passes through (60, 112) and (66, 136). Find the slope.

1. The greatest rate of change occurred in the 10-year period from 1970–1980, which is represented by the steepest segment. The least rate of change occurred in the 9-year period from 1990–1999, which is represented by the least steep segment. 2.

(10, 482,489)

480

0

The values for the area are the squares of consecutive integers.

Page 857

490

453,589

Therefore, the equation is y  2890x  453,589.

y  3,600,000  300,000x  597,000,000 y  3,600,000  3,600,000  300,000x  597,000,000  3,600,000 y  300,000x  593,400,000

13. y  300,000x  593,400,000 y  300,000(2010)  593,400,000 y  9,600,000 The number of people who will take a cruise in 2010 should be about 9,600,000.

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14.

17. We use the points (0, 324) and (2, 1530). Find the slope.

y y  12 x  6

y2  y1

mx

2

m

y  6  2x y  2x

m

x

O

The y-intercept is 324. y  mx  b y  603x  324 18. The year 2005 is 13 years after 1992. y  603x  324 y  603(13)  324 y  8163 There should be about 8163 children adopted from Russia in 2005.

y  12 x  92

Solve each equation for y to find the slopeintercept form. Eq. 1: y  2x 2x  y  6 Eq. 2: 2x  y  2x  6  2x y  2x  6

Page 858

1

Eq. 3: y  2x  6 Eq. 4:



y

x  9 2 1 9 x 2 2

The slope of y  2x is 2. The slope of y  2x  6 is 2. Therefore, these two lines are parallel. 1

1

The slope of y  2x  6 is 2. 1

9

1

The slope of y  2x  2 is 2. Therefore, these two lines are parallel. 1 Since 2 is the opposite reciprocal of 2, y  2x and y  2x  6 are both perpendicular to 1 1 9 y  2x  6 and y  2x  2. Therefore, the figure is a rectangle. 15. Opposite sides have the same slopes, so they are parallel. Consecutive sides have slopes that are opposite reciprocals, so they are perpendicular. 16. Draw a line that passes close to the points. Sample answer:

0 4

Number of Children

6

4w 4



40 4

0 6 w  10 A female bustard can weigh up to 10 lb. 6. 1.50a  0.30a 75 7. 1.50a  0.30a 75 1.20a 75

Adopted Russian Children 4000

1.20a 1.20

3000

75

1.20

a 62.50 8. Jennie must make and sell 63 or more apples to make at least $75.

2000 1000

0

Chapter 6 Solving Linear Inequalities

1. Let a  the amount Scott can spend after buying a concert ticket. a  26  50 a  26  26  50  26 a  24 Scott can spend no more than $24. 2. Let a  the amount Scott can spend after buying a concert ticket, lunch, and a CD. a  26  2.99  12.49  50 a  41.48  50 a  41.48  41.48  50  41.48 a  8.52 Scott can spend no more than $8.52. 3. Let   the length of a male bustard. 0 6 /4 4. Let w  the weight of a male bustard. 0 6 w  40 5. Let w  the weight of a female bustard. Then 4w represents the weight of a male bustard. 0 6 4w  40

x  2y  9 x  2y  x  9  x 2y  x  9 2y 2

 x1

1530  324 2  0 1206 or 603 2

1

2 3 4 5 6 7 Years Since 1992

773

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9. Let p  the price of a pair of shoes for a customer with a $15 coupon. Then p  15 represents the original price of a pair of shoes. 24.95  p  15  149.95 First express 24.95  p  15  149.95 using and. 24.95  p  15 24.95  15  p  15  15 9.95  p

and

The half plane that contains (72, 72) should be shaded. The dimensions cannot be negative. Therefore, the domain and range contain only nonnegative numbers.



p  15  149.95 p  15  15  149.95 p  134.95

The solution set is {p|9.95  p  134.95}. The price for a pair of shoes for a customer who has a coupon is between $9.95 and $134.95, inclusive. 10. Let p  the price of the pair of shoes using a $15 coupon. p  109.95  15 p  94.95 Using a coupon, the price of the shoes is $94.95. Let d  the price of the pair of shoes after a 15% discount on the price. d  109.95  0.15(109.95) (to the nearest cent) d  109.95  16.49 d  93.46 With a 15% discount, the price of the shoes is $93.46. You should choose the 15% discount. 11. Let p  the regular price of a pair of shoes. p  0.15p  p  15 0.85p  p  15 0.85p  p  p  15  p 0.15p  15 0.15p 0.15

100 50

w O

4/ 4

p  100 A 15% discount is the same as $15 off if the regular price is $100. 12. Let m  the mean of the temperatures. m

927 12

m  77.25

13. 14.

15.

16.

17.

To the nearest degree, the mean of the temperatures is 77 . mean  lowest temperature  77  73 or 4 The lowest temperature varies by 4 from the mean. highest temperature  mean  81  77  4 The highest temperature varies by 4 from the mean. Let t  the temperature. The difference between the actual temperature and the mean is within 4 degrees, so |t  77|  4. Let w  the width of the quilt. Let /  the length of the quilt. Since the perimeter of a rectangle is given by P  2/  2w, the inequality 2/  2w  318 represents this situation. Since the boundary is included in the solution, draw a solid line. Test the point (72, 72). 2w  2/  318 2(72)  2(72)  318 144  144  318 288  318 true

Mixed Problem Solving

100

150



318 4

/  79.5 So, the area is greatest when the length / and width w are each 79.5 in. Since A  /w, the greatest area would be found by substituting 79.5 for both / and w. A  /w A  (79.5)(79.5) A  6320.25 The greatest area, 6320.25 in2, occurs when the dimensions of the quilt are 79.5 in. by 79.5 in. 19. Let m  the area of a state in square miles, 1045  m  570,473 20. Let m  the area of a state in acres. 677,120  m  365,481,600 21. There are 640 acres in one square mile. So, the number of square miles multiplied by 640 is the number of acres. Alaska: 570,473 (640)  365,102,720 Rhode Island: 1045 (640)  668,800 Alaska is listed in the table as having 365,481,600 acres, but 570,473 mi2 is 365,102,720 acres. While this difference of 378,880 acres is large, it actually represents only about a 0.1% difference. Rhode Island is listed in the table as having 677,120 acres, but 1045 mi2 is 668,800 acres. While this difference of 8320 acres is significantly less than Alaska’s difference of 378,800 acres, it actually represents about a 1.2% difference. Thus, Alaska is very close, but Rhode Island is a little off.

15

73  73  74  76  78  79  81  81  81  80  77  74 12

50

Sample answers: Since (70, 75) is in the shaded region, the quilt could be 70 in. by 75 in. Similarly, since (72, 72) is in the shaded region, the quilt could be 72 in. by 72 in. 18. The area of the quilt is greatest when the length / and width w are equal. 2/  2w  318 2/  2/  318 4/  318

 0.15

m

2w  2  318

150

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4. Since there are 60 min in each hour and 60 sec in each min, 1:48:29 is about 1 h 48 min  60 min  48 min, or 108 min. 1:51:41 is about 1 h 52 min  60 min  52 min, or 112 min. 2:04:58 is about 2 h 5 min  120 min  5 min, or 125 min. 1:54:43 is about 1 h 55 min  60 min  55 min, or 115 min.

Page 859 Chapter 7 Solving Systems of Linear Equations and Inequalities 1. Since the year 1980 is 0, the year 1990 is 10. Use the ordered pairs (0, 19.3) and (10, 17.6) to write the equation. First find the slope. y2  y1

mx

2

m m

 x1

17.6  19.3 10  0 1.7 or 0.17 10

Since the first coordinate of the ordered pair (0, 19.3) is 0, the y-intercept is 19.3. Use the slope-intercept form to write the equation. y  mx  b y  0.17x  19.3 2. Use the ordered pairs (0, 8.2) and (10, 8.4) to write the equation. First find the slope.

Year 1995 2000

2

m m

Women’s 125 115

5. Since the year 1995 is 0, the year 2000 is 5. Use the ordered pairs (0, 108) and (5, 112) to write the equation. First find the slope.

y2  y1

mx

Men’s 108 112

y2  y1

mx

 x1

2

8.4  8.2 10  0 0.2 or 0.02 10

m m

Since the first coordinate of the ordered pair (0, 8.2) is 0, the y-intercept is 8.2. Use the slopeintercept form to write the equation. y  mx  b y  0.02x  8.2 3. Solve the system of equations. y  0.17x  19.3 y  0.02x  8.2 Substitute 0.17x  19.3 for y in the second equation. y  0.02x  8.2 0.17x  19.3  0.02x  8.2 0.17x  19.3  0.17x  0.02x  8.2  0.17x 19.3  0.19x  8.2 19.3  8.2  0.19x  8.2  8.2 11.1  0.19x 11.1 0.19



 x1

112  108 5  0 4 or 0.8 5

Since the first coordinate of the ordered pair (0, 108) is 0, the y-intercept is 108. Use the slope-intercept form to write the equation. y  mx  b y  0.8x  108 6. Use the ordered pairs (0, 125) and (5, 115) to write the equation. First find the slope. y2  y1

mx

2

m m

 x1

115  125 5  0 10 or 2 5

Since the first coordinate of the ordered pair (0, 125) is 0, the y-intercept is 125. Use the slope-intercept form to write the equation. y  mx  b y  2x  125 7. Solve the system of equations. y  0.8x  108 y  2x  125 Substitute 0.8x  108 for y in the second equation. y  2x  125 0.8x  108  2x  125 0.8x  108  2x  2x  125  2x 2.8x  108  125 2.8x  108  108  125  108 2.8x  17

0.19x 0.19

58.4  x Therefore, the percent of working men and women will be the same about 58 yr after 1980, or during the year 2038.

2.8x 2.8

17

 2.8

x  6.1 Therefore, you might expect the men’s and women’s winning times to be the same about 6 yr after 1995, or in 2001.

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8. Let t  the time it took Mrs. Sumner to travel to Fullerton in hours. Let d  the distance between Mrs. Sumner’s home and Fullerton. Rate 40 56

To Fullerton Return trip

Time t t2

14. Solve the system of equations. g  s  64 3g  2s  168 Multiply the first equation by 3 so the coefficients of the g terms are additive inverses. Then add the equations. 3g  3s  192 () 3g  2s  168 s  24 (1)(s)  (1) (24) s  24 Now substitute s  24 in either equation to find the value of g. g  s  64 g  24  64 g  24  24  64  24 g  40 So, the U.S. won 40 gold and 24 silver medals. 15. Since the total points scored for gold and silver medals was 168, and the total number of points scored was 201, then the number of points earned for bronze medals was 201  168 or 33 points. Since each bronze medal is worth 1 point, the U.S. had 1 bronze medal for each point, or 33 bronze medals. 16. The cost the number the number the cost

Distance d d

Use the formula rate  time  distance, or rt  d. 40t  d 56(t  2)  d Substitute 40t for d in the second equation. 56(t  2)  d 56(t  2)  40t 56t  112  40t 56t  112  56t  40t  56t 112  16t 112 16



16t 16

7t Use d  40t to find the value of d. d  40t d  40(7) d  280 Therefore, Mrs. Sumner lives 280 mi from Fullerton. 9. Let t  the number of $2 bills and let f  the number of $50 bills. t  f  1,500,888,647 f  t  366,593,903 10. Substitute t  366,593,903 for f in the first equation.

of a child ticket times 14243 123 15

t  f  1,500,888,647

c 7 So, the inequality is c  2a. 18. a 0; c 0 19. c

2t  366,593,903  366,593,903  1,500,888,647  366,593,903 2t  1,134,294,744 1,134,294,744 2

t  567,147,372

Use f  t  366,593,903 to find the value of f. f  t  366,593,903 f  567,147,372  366,593,903 f  933,741,275 Therefore, there were 567,147,372 $2 bills and 933,741,275 $50 bills in circulation. 11. 567,147,372 $2 bills are worth 567,147,372  2 or $1,134,294,744. 933,741,275 $50 bills are worth 933,741,275  50 or $46,687,063,750. So the total in circulation in $2 bills and $50 bills was $1,134,294,744  $46,687,063,750 or $47,821,358,494. 12. Let g represent the number of gold medals, and let s represent the number of silver medals. g  s  64 13. 3g  2s  168

Mixed Problem Solving

20



1442443

a

is no more than $800. 123 123 

800

1442443 144424443 14444244443

2t  366,593,903  1,500,888,647





c

of adult tickets sold

So, the inequality is 15c  20a  800. 17. The number of child twice the number tickets is greater than of adult tickets.

t  (t  366,593,903)  1,500,888,647

2t 2



of child of an tickets adult sold plus ticket times 1442443 123 14243 123

2a

60

c  2a 40

15c  20a  800

20 O

20

40

60

a

20. Sample answer: The station can buy 5 adult and 40 child tickets, or 10 adult and 30 child tickets, or 12 adult and 32 child tickets since the graphs of the ordered pairs (5, 40), (10, 30), and (12, 32) lie in the region that represents the intersection of the inequalities.

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Page 860

11. R  N  (133.5x  10,278.5)  (37.9x  1315.9)

Chapter 8 Polynomials

 (133.5x  10,278.5)  (37.9x  1315.9)  (133.5x  (37.9x))  (10,278.5  1315.9)   95.6x  8962.6

1. Yes; each is the product of variables and/or a real number. 2. Volume  s3  33  27 The volume is 27 ft3.

12. Since 2015 is 18 years after 1997, substitute 18 for x.   95.6x  8962.6  95.6(18)  8962.6  10,683.4 There will be about 10,683 such stations in 2015. 13. 0.5n(n  3)  0.5n(n)  0.5n(3)  0.5n2  1.5n 2 14. 3: 0.5(3)  1.5(3)  0 4: 0.5(4)2  1.5(4)  2 5: 0.5(5)2  1.5(5)  5 6: 0.5(6)2  1.5(6)  9 7: 0.5(7)2  1.5(7)  14 8: 0.5(8)2  1.5(8)  20 9: 0.5(9)2  1.5(9)  27 10: 0.5(10)2  1.5(10)  35 15. You add one more each time to the previous number. For example, 0  2  2, 2  3  5, 5  4  9, and so on. 16. V  /wh  x(x  3) (2x  5)  (x2  3x) (2x  5)  x2 (2x)  x2 (5)  3x(2x)  3x(5)  2x3  5x2  6x2  15x  2x3  11x2  15x 17. See students’ work. 18. 4000 is the amount of the investment, 1 will add the amount of the investment to the interest, 0.05 is the interest rate as a decimal, and 2 is the number of years of the investment. 19. 4000(1  0.05)2  $4410 20. 10,000(1  0.0625)4  $12,744

Surface Area  6s2  6(3) 2  6(9)  54 The surface area is 54 ft2. 3.

s3  6s2 s  6s2  0 s2 (s  6)  0 3

s2  0 or s  6  0 s  0 or s6 When the side length is 6 units, the volume and surface area have the same numerical value. 4.

r2h (2r) 2 (2h)

r2h

   22  r2  2h 

1 212 1 2 21rr 21hh 2 2 2

2

1 8

 or 1:8 5. vacuum: 3.00  108  300,000,000 air: 3.00  108  300,000,000 ice: 2.29  108  229,000,000 glycerine: 2.04  108  204,000,000 crown glass: 1.97  108  197,000,000 rock salt: 1.95  108  195,000,000 6.

3.00  108 1.95  108



10 13.00 1.95 21 10 2 8 8

 (1.54)(100 )  1.54 Light travels about 1.54 times faster through a vacuum. 2.29  108  108

7. ice; 1.95



10 12.29 1.95 21 10 2 8 8

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 (1.17)(100 )  1.17 8. 0.003y2  0.086y  0.708  0.003y2  (0.086y)  0.708 This polynomial is the sum of three monomials; it is a trinomial. 9. The degrees of the terms, 0.003y2, 0.086y, and 0.708, are 2, 1, and 0, respectively. Thus, the degree of the polynomial is 2, the greatest of 2, 1, and 0. 10. Since 2010 is 90 years after 1920, substitute 90 for y and simplify. 0.003y2  0.086y  0.708  0.003(90)2  0.086(90)  0.708  17.3 The population should be about 17.3 people/square mile.

Chapter 9 Factoring

1. 1-foot by 1-foot squares can be used since both 10 and 12 are divisible by 1. 2-foot by 2-foot squares can be used since both 10 and 12 are divisible by 2. 2-foot by 3-foot squares can be used since 10 is divisible by 2 and 12 is divisible by 3. 1 by 1, 2 by 2, and 2 by 3; The 3-foot squares will not cover the 10-foot dimension without cutting.

777

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10  12

11. Let x  the number of feet being added and subtracted to 24. (24  x)(24  x)  512 576  x2  512 x2  64 x2  64 x8 24  8  16 24  8  32 The dimensions of the deck are 16 ft by 32 ft. 12. The area in the plans is 24 ft  24 ft  576 sq ft. 1 The new area is 2 (576 sq. ft)  288 sq ft.

2. For 1-foot by 1-foot squares, there are 1  1 or 120 squares. The cost is 120  $3.75  $450. 10  12 For 2-foot by 2-foot squares, there are 2  2 or 30 squares. The cost is 30  $15  $450. For 2-foot by 3-foot rectangular pieces, there are 10  12 or 20 squares. 2  3 The cost is 20  $21  $420. Therefore, the 2  3 tiles are the least expensive at a total cost of $420. 3. The height of the rocket is 0 when it returns to the ground. 4. 125t  16t2  h 125t  16t2  0 t(125  16t)  0 t  0 or 125  16t  0 125  16t t 5. 6.

7.

8.

125 16

y2  288 y  1288 or about 16.97 Since the new area is not a perfect square, the equation is not factorable, so there are no whole number solutions. 13. When the object hits the ground, it will have dropped 1483 ft. Let h  1483. 1483  16t2 92.6875  t2 9.6  t It will take about 9.6 seconds for the object to hit the ground. 14. Let h  386. 386  16t2 24.125  t2 4.9  t It will take the object dropped from the Petronas Tower I building about 9.6  4.9  4.7 seconds longer to hit the ground. 15. Let t  12.

or 7.8125

{0, 7.8125} Since h  0 when the rocket returns to the ground, it will take about 7.8 seconds. Let w  the width of the field; therefore, 2w  60  the length of the field. The area is length times width, or (2w  60)w  w(2w  60). w(2w  60)  36,000 2w2  60w  36,000 2  60w  36,000  0 2w 2(w2  30w  18,000)  0 w2  30w  18,000  0 (w  120) (w  150)  0 or w  150  0 w  120  0 w  120 w  150 The width must be 120 feet since it is a measurement. The length is 2w  60  2(120  60), or 300 feet. The dimensions of the field are 120 feet by 300 feet. There are 3 feet in a yard. Divide the length and width by 3. 120 3

 40 yd

300 3

h  16(12) 2 h  16(144) h  2304 ft. The building is 2304 feet tall. 16. The height of the pool in feet is V  /w  h 42

1750  /w  12 500  /w The area of the water that is exposed to the air is 500 ft2. 17. If w is the width of the pool, the length is equal to w  5. w(w  5)  500 w2  5w  500  0 (w  25)(w  20)  0 w  25  0 or w  20  0 w  25 w  20 The width must be 20 ft since it is a measurement. The length is 20  5  25 ft. The dimensions of the pool are 20 ft by 25 ft. 18. Model B holds 1750  2  3500 cubic feet of water.   w  h  3500 Sample answer: 20 ft by 50 ft by 42 in.

 100 yd

The dimensions of the field are 40 yd by 100 yd. 9. 48 is the initial velocity of the ball and 506 is the height from which the ball is thrown, which is 500 feet plus 6 feet, Teril’s height. 10. 16t2  48t  506  218 16t2  48t  288  0 16(t2  3t  18)  0 t2  3t  18  0 (t  3)(t  6)  0 or t  6  0 t30 t6 t  3 The ball was in the air for 6 seconds.

Mixed Problem Solving

42 . 12

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7. Solve 0  16t2  210.

19. The measurements for Model C are width  2(20)  40 ft length  2(25)  50 ft height  42 in. 42 Volume of Model C  40  50  12  7000 ft3. Model A Model C

1750

0  16t2  210 16t2  210 t2 

t  3

1

 7000  4

Chapter 10 Quadratic and Exponential Functions

1. In h  16t2  112t  6, a  16 and b  112. b

t  2a 112

t  2(16) or 3.5

2.

3.

4.

5.

The equation of the axis of symmetry is t  3.5. h  16t2  112t  6 h  16(3.5) 2  112(3.5)  6 h  196  392  6 h  202 The vertex is at (3.5, 202). Since the coefficient of the x2 is negative, the vertex is a maximum point. The h-coordinate of the vertex represents the maximum height. The maximum height the ball reaches is 202 feet. The t-coordinate of the vertex represents the time when the ball reaches its maximum height. The ball reaches its maximum height after 3.5 seconds. The ball hits the ground when h  0. When t  7, h  16(7)2  112(7)  6 or 6. When t  8, h  16(8)2  112(8)  6 or 122. Thus, the ball hits the ground between 7 and 8 seconds after it is thrown. The ball is in the air about 7 seconds. Replace h with 40 in the equation.

1 (10x2 10

x2  16.4x  19.8 x  16.4x  67.24  19.8  67.24 (x  8.2) 2  47.44 x  8.2  147.44 x  8.2  147.44 x  8.2  147.44 or x  8.2  147.44  15.1  1.3 {1.3, 15.1} 10. The poster board is not wide enough to have a side margin of 15.1 inches. Thus, the side margins are each about 1.3 inches wide. The top margin is about 3  1.3 or 3.9 inches, and the bottom margin is about 2  1.3 or 2.6 inches. 11.



4.95  116.8861 0.22

4.95  116.8861

f (t )  16t 2  210 f (t ) 200 100 2

4.95  2(4.95) 2  4(0.11) (17.31) 2(0.11)

O

4.95  116.8861

x or x  0.22 0.22  41.2  3.8 The values mean that 3.8 years and 41.2 years after 1977, 30% of households had cable. 12. No; the parabola only reaches a maximum height of about 68, meaning that no more than 68% of homes will ever have cable, which is not realistic.

Graph f(t)  16t2  210.

4

30  0.11x2  4.95x  12.69 30  30  0.11x2  4.95x  12.69  30 0  0.11x2  4.95x  17.31 x

40  16t2  250 40  40  16t2  250  40 0  16t2  210

f(t) 46 66 194 210 194 66 46

1

 164x)  10 (198)

2

h  16t2  250 40  16t2  250 6. Rewrite the equation.

t 4 3 1 0 1 3 4

210 16

t  3.6 Ignore the negative solution. It takes about 3.6 seconds to complete the ride. 8. The width of the part covered is 22  2x inches and the height is 27  5x. Thus, the area of the part covered can be represented by 2 (22  2x)(27  5x). This area is 3 the area of the 2 poster board, 3 (22)(27) or 396 square inches. Write the equation. (22  2x)(27  5x)  396 9. (22  2x) (27  5x)  396 594  110x  54x  10x2  396 10x2  164x  594  396 2 10x  164x  594  594  396  594 10x2  164x  198

The ratio of the volume of Model A to Model C is 1:4.

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210 16

2

4t

The equation has two roots. One is between 4 and 3, and the other is between 3 and 4.

779

Mixed Problem Solving

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13. P(x)  335(1.007)x x 0 100 200 300 400

Page 863

P(x) 335 673 1352 2716 5455

P (x )

6000 5000 4000 3000 2000 1000

1. 200 km  200,000 m Thus, the radius of the satellite’s orbit is 6,380,000  200,000 or 6,580,000 meters. P (x )  355(1.007)x

O

200

400

x

The y-intercept is 335. 14. The y-intercept represents the population of Asia in year 0 or 1650. Thus, the population of Asia in 1650 was 335,000,000. 15. The year 2050 is 2050  1650 or 400 years after 1650. Thus, x  400.

1

16. A  P 1  n

2

1

A  10,000 1 

9 32

27 64

2

0.0745 1(4) 1

81 128

243 256

729 512

2187 1024

       3

3

2 2

3

3

2 2  rn1

3

2

3

2

3

2

6561 2048

 3

2

(6.67  10  11 ) (5.97  1024 ) 6.58  106



3

39.8199  1013 6.58  106



3

39.8199 6.58

 107

2r v 2(6,580,000) 7779

4. d 

3

2hv2 g



3

2(10) (10) 2 9.8



3

2000 9.8

5. d 

3

2hv2 g



3

2(10) (20) 2 9.8



3

8000 9.8

 28.57 d is about 28.57 meters. This value is twice as great as the value in Exercise 4. 6. Yes; because 12  1.4

19,683 4096

7. r 

 3

2



V

3 h 162

3 (8.25)

 2.5 The radius of the container is about 2.5 inches. 8. First convert 98 acres to square miles. 98 98 acres  640 or 0.153125 mi2

2187 or about 1024 6561 sequence is 2048 or

21. The eighth term of the sequence is

2.1. The ninth term of the about 3.2. Laurie will first exceed 3 miles during her ninth session. 22. No; since Laurie is running every other day, her ninth session will be on the 17th day.

Mixed Problem Solving

3

 14.29 d is about 14.29 meters.

20. an  a 1 1 3 an  8 2 n1

12



 5315 The orbital period of the satellite is about 5315 seconds, which is about 5315 3600 or 1.5 hour.

In 2003, T(4)  2575(1.06) 4  3250.88. In 2004, T(5)  2575(1.06) 5  3445.93. In 2005, T(6)  2575(1.06) 6  3652.69. In 2006, T(7)  2575(1.06) 7  3871.85. The total cost of his tuition will be $3250.88  $3445.93  $3652.69  $3871.85 or $14,221.35. 18. Sample answer: Aaron could reinvest the money he does not need each year and earn additional interest during the time he is attending college. 3 16

GmE r



17. y  C(1  r) t y  2575(1  0.06) t y  2575(1.06) t

1 8

3

3. T 

A  10,000(1.0745) 4 A  13,329.86 Aaron will have $13,329.86 at the end of 4 years.

19.

2. v 

 7779 The orbital velocity of the satellite is about 7779 meters per second.

P(400)  335(1.007) 400  5455.48076 There will be about 5,455,480,760 people in Asia in 2050. r nt

Chapter 11 Radical Expressions and Triangles

The area of the square is 0.153125 square miles. A  s2 0.153125  s2 10.153125  s 0.3913  s The length of a side of the square is about 0.3913 mile, which is 0.3913  5280 or about 2066 feet.

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9. c2  a2  b2 c2  (2066) 2  (2066) 2 c2  8,536,712 c  18,536,712 c  2922 The diagonal distance is about 2922 feet. 10. Tyrone’s distance from the store is the length of the hypotenuse of a right triangle with legs 32 and 45.

16. Let A represent the angle of elevation.

c2  a2  b2 c2  (32) 2  (45) 2 c2  3049 c  13049 c  55 No; the distance is about 55 blocks which is about 55 12 (there are 12 blocks per mile) or 4.6 mi. 11. Sample answer: 40 blocks south and 45 blocks west; 38 blocks north and 47 blocks west

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377

sin A  745 A  sin1

 30

The angle of elevation is about 30 .

1. p 

Chapter 12 Rational Expressions and Equations

1 f

f

2

1

1

2

p

1 2

1

1

1 2

P 2 1

12. d  2(x  x ) 2  (y  y ) 2 2 1 2 1 side 1: d  2(3  1) 2  (2  1) 2

4

 2(4) 2  12  117

2

1m

 (5 

2

 2(8)  (6)  164  36  1100  10

2. f  20 cm  100 cm  0.2 m

1) 2

1

p  0.2  5 diopters

2

1m

3. f  40 cm  100 cm  0.4 m 1

p  0.4  2.5 diopters 4. Sample answer: One value is negative and the other is positive.

The perimeter is 10  117  165 or about 22.19 units. 13. The new vertices are (2, 2), (6, 4), and (14, 10). d  2(x2  x1 ) 2  (y2  y1 ) 2 side 1: d  2(6 

2) 2

 (4 

1

5. y  x2

2) 2

x

2

1

1

2

y

1 4

1

1

1 4

 2(8) 2  22  168 or 2117

4 3 2 1

side 2: d  2(14  (6)) 2  (10  4) 2  2(8) 2  (14) 2  1260 or 2165

2

side 3: d  2(14  2) 2  (10  2) 2

O

y

2x

The graph looks like an inverse variation graph except it lies in Quadrants I and II, consecutive quadrants, not opposite quadrants. 6. The values of x must be positive since x represents distance.

 2(16) 2  (12) 2  1400 or 20 The perimeter is 20  2117  2165 or about 44.37 units. 14. Yes; the ratio of each pair of corresponding sides is 1:2. 15. 745 ft

4f

2

 2(4) 2  (7) 2  165 side 3: d  2(7 

2

O 1

side 2: d  2(7  (3) ) 2  (5  2) 2

1) 2

1377 745 2

8.

d t d t

9.

157 ft 60 min 1 mi  1 h  5280 ft 1 min

7.



250 5

 50  157 ft/min  1.8 mi/h

377 ft

781

Mixed Problem Solving

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10. 0.232x  13,063,000 x  56,306,034 vehicles 11.

5.79  1010 mi

5.79

Mercury: 1.86  105 mi/s  1.86  105 s  Venus: Earth: Mars:

1.08  1011 mi 1.86  105 mi/s 11

1.496  10 mi 1.86  105 mi/s

2.28  1011 mi 1.86  105 mi/s



1.08 1.86



1.496 1.86



11

7.78  10 mi  105 mi/s

Jupiter: 1.86

2.28 1.86



 13,405 min  20,430 min

7.78

 1.86  106 sec 

1.43  1012 mi

1.43

2.87  10

mi

2.87

Pluto:

12.



4.50 1.86



1 min 60 s

1 min 60 s

Uranus: 1.86  105 mi/s  1.86  107 s  4.50  1012 mi 1.86  105 mi/s

 9677 min

1 min 106 sec  60 s

1 min 60 s

1 min 107 s  60 s

3.

226,338 257,808 191,310 215,180 95,195 91,738 181,596 198,484 77,691 78,282 Y Y; B  I AI 96,946 100,377 111,486 106,621 291,079 321,586 370,522 436,790 85,487 80,785 4. Each matrix has ten rows and one column. Both matrices are 10  1. 5. P  A  B

 69,713 min

 128,136 min  257,168 min

 403,226 min

9400  6900 9400  6900  3800  17,400  11,700  3300  5400 16,300 163  57,900  579

13. 35  5  7 50  2  5  5 75  3  5  5 LCM  2  3  5  5  7 or 1050 Celeste should order 1050 bricks. 14.

1 2

5

257,808 226,338 215,180 191,310 91,738 95,195 198,484 181,596 78,282 77,691 I YI Y 100,377 96,946 106,621 111,486 321,586 291,079 436,790 370,522 80,785 85,487

1

yd  8 yd  14 yd 4

5

5

8  8  4  

4  5  10 8 19 3  28 yd 8 3

15. No; they need 28 yd for one of each type. Then 3

30 28 yd is not a whole number, so they cannot

257,808  226,338 215,180  191,310 91,738  95,195 198,484  181,596 78,282  77,691 I Y 100,377  96,946 106,621  111,486 321,586  291,079 436,790  370,522 80,785  85,487

use the entire bolt by making an equal number of each type. 16. Rick rate: Phil rate: 1 x 6 2 x 12

1 house 6 days 1 house 4 days

1

 4x  1 3

 12x  1 5 x 12

1

x

31,470 23,870 3457 16,888 591 Y I 3431 4865 30,507 66,268 4702 6. Matrix P represents the difference in the populations of the ten cities between 1999 and 1990.

12 5

x  2.4 It will take 2.4 days.

Mixed Problem Solving

Chapter 13 Statistics

1. Sample answer: The sample contained students ages 14–18. 2. Sample answers: How were the students chosen? Was it random or were they volunteers? Were the students given only certain careers as choices? How many students responded?

 5188 min

1 min 60 s

Saturn: 1.86  105 mi/s  1.86  107 s  12

1 min 60 s

1 min 106 s  60 s

 106 s 



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9a. Order the low temperatures from least to greatest. 80, 70, 69, 66, 61, 60, 60, 60, 58, 54, 54, 52, 51, 50, 50, 50, 48, 48, 47, 47, 46, 45, 42, 40, 40, 40, 39, 37, 37, 36, 36, 35, 34, 34, 32, 32, 30, 29, 29, 27, 27, 25, 23, 19, 19, 17, 17, 16, 2, 12 The greatest temperature is 12, and the least temperature is 80. Thus, the range is 12  (80) or 92 F. 9b. There are 50 temperatures total. The lower quartile is the median of the 25 smallest temperatures. The lower quartile is 51 F. The upper quartile is the median of the 25 greatest temperatures. The upper quartile is 29 F. 9c. The interquartile range is 29  (51) or 22 F. 9d. An outlier would be 1.5(22) or 33 F less than the lower quartile or greater than the upper quartile. There are no temperatures in the list less than 51  33 or 84 F. There is one temperature, 12 F, greater than 29  33 or 4 F. There is one outlier, 12 F. 10. Low Temperatures: Since there are an even number of temperatures, the median is the mean of the 25th and 26th temperatures.

7. The matrix P shows that three cities had a decrease in population. For each of these cities, find the percent decrease using the absolute value of the number from matrix P as the change and the 1990 population as the base. Elmira, NY: 3457 95,195

r

 100

345,700  95,195r 345,700 95,195

r

3.6  r Lawton, OK: 4865 111,486

r

 100

486,500  111,486r 486,500 111,486

r

4.4  r Pine Bluff, AR: 4702 85,487

r

 100

470,200  85,487r 470,200 85,487

r

5.5  r Pine Bluff, AR has the greatest percent decrease. 8a. Order the high temperatures from least to greatest. 100, 100, 104, 105, 105, 106, 106, 107, 108, 108, 109, 110, 110, 110, 110, 111, 111, 112, 112, 112, 112, 113, 113, 114, 114, 114, 114, 114, 115, 116, 117, 117, 117, 118, 118, 118, 118, 118, 118, 119, 120, 120, 120, 120, 121, 121, 122, 125, 128, 134 There are an even number of temperatures, so find the mean of the 25th and 26th temperatures, which are both 114. 114  114 2

40  (40) 2

The median is 40. We found the other information necessary to construct the plot for the low temperatures in Exercise 9. High Temperatures: We found the median in Exercise 8. The greatest temperature is 134, and the least temperature is 100. Thus, the range of the high temperatures is 134  100 or 34 F. The lower quartile is the median of the 25 smallest temperatures. The lower quartile is 110 F. The upper quartile is the median of the 25 greatest temperatures. The upper quartile is 118 F. The interquartile range is 118  110 or 8 F. An outlier would be 1.5(8) or 12 F less than the lower quartile or greater than the upper quartile. There are no high temperatures less than 110  12 or 98 F. There is one high temperature, 134 F, greater than 118  12 or 130 F. Complete the parallel box-and-whisker plot.

 114

The median high temperature is 114 F. 8b. Sample answer:

Frequency

Record High Temperatures in United States 25 20 15 10 5 0 100109

109118

118127

 40

127136

Temperature (˚F)

8c. Sample answer: More than half of the states had high temperatures of less than 118 F. Only two states had temperatures over 126 F. 8d. Yes; to find the median, you must order the data from least to greatest. Then it is easier to place the values into measurement classes.

80 60 40

783

20

0

20

100

120

140

Mixed Problem Solving

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3. The four most expensive arrangements are listed. deluxe, velvet, orchids, large deluxe, velvet, orchids, small deluxe, satin, orchids, large deluxe, satin, orchids, small Find the cost of each of these. $12.00  $3.00  $35.00  $2.50 or $52.50 $12.00  $3.00  $35.00  $1.75 or $51.75 $12.00  $2.00  $35.00  $2.50 or $51.50 $12.00  $2.00  $35.00  $1.75 or $50.75 4. This problem involves combinations since the order of the letters is not important. Find the number of combinations of 12 letters taken 4 at a time.

10a. Sample answer: The range of the low temperatures is much greater than the range of the high temperatures. Both data sets have one outlier at the high end of the data set. 10b. The range of the low temperatures is almost 3 times greater than the range of the high temperatures. 11. AL 139 LA 130 OH 152 AK 180 ME 153 OK 147 AZ 157 MD 149 OR 173 AR 149 MA 142 PA 153 CA 179 MI 163 RI 129 CO 179 MN 174 SC 130 CT 138 MS 134 SD 178 DE 127 MO 158 TN 145 FL 110 MT 187 TX 143 GA 129 NE 165 UT 186 HI 88 NV 175 VT 155 ID 178 NH 152 VA 140 IL 153 NJ 144 WA 166 IN 152 NM 172 WV 149 IA 165 NY 160 WI 168 KS 161 NC 144 WY 180 KY 151 ND 181 11a. Sample answer:

C 

n r

C 

12 4

  

P 

n r

Frequency

P 

10 8 6

4 3



130139

140- 150149 159

160169

170179

180189

Temperature (˚F)

11b. Sample answer: More than half of the states had temperature differences between 129 F and 170 F.

Page 866

Chapter 14 Probability

1. There are 3 choices for the vase, 2 choices for the ribbon, 3 choices for the flowers, and 2 choices for the card. Use the Fundamental Counting Principle. 3  2  3  2  36 There are 36 possible floral arrangements. 2. The most expensive vase is $12.00, the most expensive ribbon is $3.00, the most expensive flowers are $35.00, and the most expensive card is $2.50. So, the most expensive arrangement costs $12.00  $3.00  $35.00  $2.50 or $52.50. The least expensive vase is $5.00, the least expensive ribbon is $2.00, the least expensive flowers are $12.00, and the least expensive card is $1.75. So, the least expensive arrangement costs $5.00  $2.00  $12.00  $1.75 or $20.75.

Mixed Problem Solving

n! (n  r)! 4! (4  3)! 4! 1!

 4! or 24 There are 24 different three-letter arrangements of the four letters. 6. The 24 different arrangements of the four letters are listed. ATR ART ATE AET ARE AER TAR TRA TAE TEA TRE TER RAT RTA RAE REA RTE RET EAT ETA EAR ERA ETR ERT Of these, 9 are words: ART, ATE, ARE, TAR, TEA, RAT, EAT, EAR, and ERA. 7. Hidalgo had 175 hits and a total of 558 times at bat.

4 2 0 120129

495

Melissa can choose the four letters in 495 different ways. 5. This problem involves permutations since the order of the letters is important. Find the number of permutations of 4 letters taken 3 at a time.

Difference in High and Low Temperatures

80119

n! (n  r)!r! 12! (12  4)!4! 12! 8!4! 12  11  10  9  8! 8!4! 12  11  10  9 or 4!

P(Hidalgo gets hit) 

175 558

or about 0.31

8. Compute the probabilities for the other three players. 161

P(Alou gets hit)  454 or about 0.36 68

P(Ward gets hit)  264 or about 0.26 73

P(Cedeno gets hit)  259 or about 0.28 Alou has the greatest probablity of getting a hit his next time at bat. 9. These events are independent. P(both get hit)  P(Ward gets hit)  P(Cedeno gets hit)  0.26  0.28 or about 0.07

784

PQ249J-6481F-P01-14[766-785] 26/9/02 9:07 PM Page 785 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:

14. There are six possible outcomes for each of the five digits. Use the Fundamental Counting Principle. 6  6  6  6  6  7776 There are 7776 possible numbers. 15. Sample answer: Roll five dice and record the results each time. Each roll represents one person picking a winning number. 16. See students’ work. The theoretical probability is 1 or about 0.00013. 7776

10. He should choose Alou and Hidalgo, the two players with the highest individual probabilities of getting the next hit. P(both get hit)  P(Alou gets hit)  P(Hidalgo gets hit)  0.36  0.31 or about 0.11 11. Yes; the probability for each value of X is greater than or equal to 0 and less than or equal to 1, and 0.053  0.284  0.323  0.198  0.142  1, so the probabilities add up to 1. 12. P(under 20)  0.053 13. P(50 or older)  P(50  64)  P(65 and over)  0.198  0.142 or 0.34

785

Mixed Problem Solving

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