Algebra I_Solutions Manual
January 26, 2017 | Author: Hector I. Areizaga Martinez | Category: N/A
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Description
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
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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
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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
Chapter 3
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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
Chapter 3
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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
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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
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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
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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
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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
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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)
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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).
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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.
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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).
Page 196
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.
129
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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
130
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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
Chapter 4
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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
Chapter 4
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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
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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
Chapter 4
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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
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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
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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)}.
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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
Chapter 4
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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
160
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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
Chapter 4
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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
162
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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
Chapter 4
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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)
164
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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)
165
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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
166
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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.
167
Chapter 4
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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
1 3 1 2
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
PQ249-6481F-05[217-243] 7/31/02 10:04 PM Page 218
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
PQ249-6481F-05[217-243] 7/31/02 10:04 PM Page 220
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
Chapter 5
PQ249-6481F-05[217-243] 7/31/02 10:04 PM Page 222
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
PQ249-6481F-05[217-243] 7/31/02 10:04 PM Page 223
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
PQ249-6481F-05[217-243] 7/31/02 10:04 PM Page 224
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
PQ249-6481F-05[217-243] 7/31/02 10:04 PM Page 225
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
PQ249-6481F-05[217-243] 7/31/02 10:04 PM Page 226
.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
PQ249-6481F-05[217-243] 7/31/02 10:04 PM Page 227
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
Chapter 5
PQ249-6481F-05[217-243] 7/31/02 10:05 PM Page 228
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
PQ249-6481F-05[217-243] 7/31/02 10:05 PM Page 230
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
Chapter 5
PQ249-6481F-05[217-243] 7/31/02 10:05 PM Page 232
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
PQ249-6481F-05[217-243] 7/31/02 10:05 PM Page 233
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
Chapter 5
PQ249-6481F-05[217-243] 7/31/02 10:05 PM Page 234
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
PQ249-6481F-05[217-243] 7/31/02 10:05 PM Page 235
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
PQ249-6481F-05[217-243] 7/31/02 10:05 PM Page 236
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
PQ249-6481F-05[217-243] 7/31/02 10:05 PM Page 237
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
PQ249-6481F-05[217-243] 7/31/02 10:06 PM Page 238
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
PQ249-6481F-05[217-243] 7/31/02 10:06 PM Page 239
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
PQ249-6481F-05[217-243] 7/31/02 10:06 PM Page 240
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
PQ249-6481F-05[217-243] 7/31/02 10:06 PM Page 241
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
PQ249-6481F-05[217-243] 7/31/02 10:06 PM Page 242
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.
PQ249-6481F-05[217-243] 7/31/02 10:06 PM Page 243
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.
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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
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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
Chapter 6
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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
Chapter 6
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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
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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
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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
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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
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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
4
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
2
3
4
5
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
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
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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
2
3
4
5
6
7
8
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
4
5
6
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
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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
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
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
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
–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
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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.
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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
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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
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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
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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
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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
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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
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0
1
2
3
4
5
–5
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0
1
2
3
4
5
–5
–4
–3
–2
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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
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–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
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–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.
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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
x
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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).
307
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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.
308
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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).
309
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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.
311
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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
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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
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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
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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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Chapter 7
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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
330
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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
331
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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.
333
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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.
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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
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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
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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).
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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
344
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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
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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
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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
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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).
351
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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
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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.
357
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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
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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).
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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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Chapter 7
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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)
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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
Chapter 8
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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
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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
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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
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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
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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
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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.
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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
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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
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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.
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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)
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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}
426
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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
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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
2
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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
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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
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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
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O
440
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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
443
x
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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
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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
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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.
Chapter 10
452
2s
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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.
Chapter 10
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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
458
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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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Chapter 10
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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
˛˛˛
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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
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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.
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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
Chapter 10
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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
Chapter 11
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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
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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
Chapter 11
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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
Chapter 11
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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
Chapter 11
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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
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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
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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
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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
✓
PQ249-6481F-12[541-568] 26/9/02 6:33 PM Page 559 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:
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
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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
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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
PQ249-6481F-12[541-568] 26/9/02 6:33 PM Page 563 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:
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.
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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
598
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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
Chapter 13
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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
Chapter 13
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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
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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
Chapter 13
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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
PQ249-6481F-PR[633-642] 26/9/02 8:20 PM Page 633 Sahuja Ahuja_QXP_06:Desktop Folder:Chandra:Algebra_FNL_Delivery:
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
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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
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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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Extra Practice
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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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Extra Practice
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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.
651
Extra Practice
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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…
652
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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.
653
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
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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
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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%.
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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
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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.
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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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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
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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
669
Extra Practice
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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
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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
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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.
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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
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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
Extra Practice
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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
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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
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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.
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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.
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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).
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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
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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
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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
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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
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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
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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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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
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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
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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
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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
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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
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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 ✓
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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
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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.
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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
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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
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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
762
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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
Mixed Problem Solving
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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).
Mixed Problem Solving
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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
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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
Page 861
(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.
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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
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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.
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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
Page 864
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
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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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