HonAlg2AnsKey
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Answer Key Transparencies Provides transparencies with answers for each lesson in the Student Edition
ISBN 0-07-828001-X
90000
9 780078 280016
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Chapter 1 Solving Equations and Inequalities Lesson 1-1 Expressions and Formulas Pages 8–10 1. First, find the sum of c and d. Divide this sum by e. Multiply the quotient by b. Finally, add a.
2. Sample answer:
3. b; The sum of the cost of adult and children tickets should be subtracted from 50. Therefore, parentheses need to be inserted around this sum to insure that this addition is done before subtraction.
4. 72
5. 6
6. 23
7. 1
8. 2
9. 119
10. 0
11. 23
12. 18
13. $432
14. $1875
15. $1162.50
16. 20
17. 3
18. 29
19. 25
20. 54
21. 34
22. 19
23. 5
24. 11
25. 31
26. 7
27. 14
28. 15
29. 3
30. 52
31. 162
32. 15.3
33. 2.56
34. 7
35. 25
1 3
36. about 1.8 lb
37. 31.25 drops per min
38. 3.4
39. 2
40. 45
©Glencoe/McGraw-Hill
14 4 5
1
Algebra 2
Chapter 1
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41. 4.2
42. 5.3
43. 4
44. 75
45. 1.4
46. 4
47. 8
48. 36.01
49. 2
y5 2 b 2
50. a
1 6
51. 16
52. 30
53. $8266.03
54. 400 ft
55. Sample answer: 44441 44442 14 4 42 4 3 4 14 42 4 4 14 4 42 4 5 14 42 4 4 6 44 4 4 7 14 42 14 42 8 44449 144 42 4 10
56. Nurses use formulas to calculate a drug dosage given a supply dosage and a doctor’s drug order. They also use formulas to calculate IV flow rates. Answers should include the following. • A table of IV flow rates is limited to those situations listed, while a formula can be used to find any IV flow rate. • If a formula used in a nursing setting is applied incorrectly, a patient could die.
57. C
58. D
59. 3
60. 4
61. 10
62. 13
63. 2
64. 5
65.
2 3
©Glencoe/McGraw-Hill
66.
2
6 7
Algebra 2
Chapter 1
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Lesson 1-2 1a. 1b. 1c. 1d. 1e. 1f.
Sample Sample Sample Sample Sample Sample
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Properties of Real Numbers Pages 14–18
answer: 2 answer: 5 answer: 11 answer: 1.3 answer: 12 answer: 1.3
2. A rational number is the ratio of two integers. Since 13 is 13 not an integer, is not a 2 rational number.
3. 0; Zero does not have a multiplicative inverse since
4. Z, Q, R 1 0
is undefined. 5. N, W, Z, Q, R
6. Q, R
7. Multiplicative Identity
8. Associative Property () 1 8
10. 8,
9. Additive Identity 1 3
2 3
11. , 3
12. 1.5,
13. 2x 4y
14. 13p
15. 3c 18d
16. 17a 1
17. 1.5(10 15 12 8 19 22 31) or 1.5(10) 1.5(15) 1.5(12) 1.5(8) 1.5(19) 1.5(22) 1.5(31)
18. $175.50
19. W, Z, Q, R
20. Q, R
21. N, W, Z, Q, R
22. Q, R
23. I, R
24. Z, Q, R
25. N, W, Z, Q, R
26. I, R
27. Q, R; 2.4, 2.49, 2.49, 2.49, 2.9
28. Additive Inverse
29. Associative Property ()
30. Additive Identity
31. Associative Property ()
32. Commutative Property ()
33. Multiplicative Inverse
34. Distributive
35. Multiplicative Identity
36. 0
37. m; Additive Inverse
38.
©Glencoe/McGraw-Hill
3
1 ; m
Multiplicative Inverse Algebra 2
Chapter 1
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39. 1
40. natural numbers
41. 12 units
42. The square root of 2 is irrational and therefore cannot be described by a natural number.
1 10
43. 10;
44. 2.5; 0.4
45. 0.125; 8
46.
5 ; 8
3 5
4 3 3 4
8 5 5 23
47. ,
48. 4 ,
49. 3a 2b
50. 10x 2y
51. 40x 7y
52. 11m 10a
53. 12r 4t
54. 32c 46d
55. 3.4m 1.8n
56. 4.4p 2.9q
57. 8 9y
58.
59. true
60. false; 3
61. false; 6
62. true
63. 6.5(4.5 4.25 5.25 6.5 5) or 6.5(4.5) 6.5(4.25) (6.5)5.25 6.5(6.5) 6.5(5)
64. 3.6; $327.60
©Glencoe/McGraw-Hill
4
9 x 10
19 y 6
Algebra 2
Chapter 1
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65. 3 a2 b 2 a1 b 1 4
1 8
66. 50(47 47); 50(47) 50(47)
3 a2 b 2 a1 b 1 4
1 8 Def. of a mixed number 1 1 3 122 3 a b 2 112 2 a b 4 8 Distributive 3 1 6 2 Multiply. 4 4
3 1 4 4 3 1 8 4 4 1 3 8a b 4 4
Comm. ()
8 1 or 9
Add.
62
Add. Assoc. ()
67. 4700 ft2
68. $113(0.36 0.19); $113(0.36) $113(0.19)
69. $62.15
70. Yes; 7;dividing 2 2 2 by a number is the same as multiplying by its reciprocal.
71. Answers should include the following. • Instead of doubling each coupon value and then adding these values together, the Distributive Property could be applied allowing you to add the coupon values first and then double the sum.
72. B
©Glencoe/McGraw-Hill
68
5
6
8
Algebra 2
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• If a store had a 25% off sale on all merchandise, the Distributive Property could be used to calculate these savings. For example, the savings on a $15 shirt, $40 pair of jeans, and $25 pair of slacks could be calculated as 0.25(15) 0.25(40) 0.25(25) or as 0.25(15 40 25) using the Distributive Property. 73. C
74. true
75. False; 0 1 1, which is not a whole number.
76. true
2 3
77. False; 2 3 , which is not
78. 9
a whole number. 79. 6
80. 5
81. 2.75
82. 358 in2
83. 11
84.
85. 4.3
86. 36
7 10
Chapter 1 Practice Quiz 1 Page 18 1. 14
2. 9
3. 6
4. 1
5. 2 amperes
6. Q, R
7. N, W, Z, Q, R
8. Additive Inverse
6 7 7 6
9. ,
©Glencoe/McGraw-Hill
10. 50x 64y
6
Algebra 2
Chapter 1
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Lesson 1-3 Solving Equations Pages 24–27 1. Sample answer: 2x 14
2. Sometimes true; only when the expression you are dividing by does not equal zero.
3. Jamal; his method can be confirmed by solving the equation using an alternative method.
4. 5 4n
C C C 9 cC 5
5 1322 9 5 1322 d 9
9 C 5
5 1F 322 9 5 5 F 1322 9 9 5 F 9
F
32 F
5. 2n n3
6. Sample answer: 9 times a number decreased by 3 is 6.
7. Sample answer: 5 plus 3 times the square of a number is twice that number.
8. Reflexive Property of Equality 10. 21
9. Addition Property of Equality 11. 14
12. 4
13. 4.8
14. 1.5
15. 16
16. y
17. p
I rt
9 2n 4
18. D
19. 5 3n
20. 10n 7
21. n 2 4
22. 6n3
23. 519 n2
24. 21n 82
25.
a
26. 1n 72 3
n 2 b 4
©Glencoe/McGraw-Hill
7
Algebra 2
Chapter 1
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27. 2rh 2r 2
28. 2r 1h r 2
29. Sample answer: 5 less than a number is 12.
30. Sample answer: Twice a number plus 3 is 1.
31. Sample answer: A number squared is equal to 4 times the number.
32. Sample answer: Three times the cube of a number is equal to the number plus 4.
33. Sample answer: A number divided by 4 is equal to twice the sum of that number and 1.
34. Sample answer: 7 minus half a number is equal to 3 divided by the square of x.
35. Substitution Property ()
36. Subtraction Property ()
37. Transitive Property ()
38. Addition Property ()
39. Symmetric Property ()
40. Multiplication Property ()
41. 7
42. 8
43. 3.2
44. 2.5
45.
3 4
1 12
46.
47. 8
48. 11
49. 7
50.
51. 1
52. 12
53.
1 4
55.
2 3
54. 19 10 17
55 2
56.
r
58. a
57.
d t
59.
3V r 2
h
61. b
x 1c 32 a
2
b 2x
60.
2A h
62.
4x 1x
ab y
63. n number of games; 2(1.50) n(2.50) 16.75; 5
64. s length of a side; 8s 124, 15.5 in.
65. x cost of gasoline per mile; 972 114 105 7600x 1837; 8.5¢/mi
66. n number of students that can attend each meeting; 2n 3 83; 40 students
©Glencoe/McGraw-Hill
8
Algebra 2
Chapter 1
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67. a Chun-Wei’s age; a (2a 8) (2a 8 3) 94; Chun-Wei: 15 yrs old, mother: 38 yrs old, father: 41 yrs old
68. c cost per student; 50 50130 c2 1452 5 1800; $3
69. n number of lamps broken; 12(125) 45n 1365; 3 lamps
70. h height of can A;
71. 15.1 mi/month
72. Central: 690 mi.; Union: 1085 mi
73. The Central Pacific had to lay their track through the Rocky Mountains, while the Union Pacific mainly built track over flat prairie.
74. $295
75. the product of 3 and the difference of a number and 5 added to the product of four times the number and the sum of the number and 1
76. To find the most effective level of intensity for your workout, you need to use your age and 10-second pulse count. You must also be able to solve the formula given for A. Answers should include the following. • Substitute 0.80 for I and 27 for P in the formula I 6 P 1220 A2 and solve for A. To solve this equation, divide the product of 6 and 28 by 0.8. Then subtract 220 and divide by 1. The result is 17.5. This means that this 1 person is 17 years old.
11.22 2h 122 23; 8 units 1 3
2
©Glencoe/McGraw-Hill
9
Algebra 2
Chapter 1
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• To find the intensity level for different values of A and P would require solving a new equation but using the same steps as described above. Solving for A would mean that for future calculations of A you would only need to simplify an expression, 220
6P , I
rather
than solve an equation. 77. B
78. D
79. 6x 8y 4z
80. 11a 8b
81. 6.6
82. 7.44
83. 105 cm2
84. 5
85. 3
86. 2.5 1 4
87.
88. 3x
89. 5 6y
Lesson 1-4
Solving Absolute Value Equations Pages 30–32
1. 0 a 0 a when a is a negative number and the negative of a negative number is positive.
2a. 0 x 0 4 2b. 0 x 6 0 2 4. Sample answer: 0 4 6 0 ; 2
3. Always; since the opposite of 0 is still 0, this equation has only one case, ax b 0. The solution is
b . a
6. 9
5. 8
8. 521, 136
7. 17
10. 511, 296
9. 518, 126
11. 532, 366 ©Glencoe/McGraw-Hill
12. 10
Algebra 2
Chapter 1
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13. 586
14. 0 x 160 0 2
15. least: 158 F; greatest: 162 F
16. 162 F; This would ensure a minimum internal temperature of 160 F.
17. 15
18. 24
19. 0
20. 4
21. 3
22. 13
23. 4
24. 7.8
25. 9.4
26. 5
27. 55
28. 22
29. {8, 42}
30. 512, 306
33. 52, 166
34.
32. 528, 206
31. 545, 216
35.
3 e f 2
37.
e 2,
e 2,
16 f 3
36. 38. 54, 16
9 f 2
39.
40.
41. {5, 11}
42. {3, 15}
43.
11 e , 3
3 f
44.
5 f 3
46. 546
45. {8}
48. 0 x 16 0 0.3; heaviest: 16.3 oz, lightest: 15.7 oz
47. 0 x 200 0 5; maximum: 205 F; minimum: 195 F 49. 0 x 13 0 5; maximum: 18 km, minimum: 8 km
50. sometimes; true only if a 0 and b 0 or if a 0 and b 0
51. sometimes; true only if c 0
©Glencoe/McGraw-Hill
e 3,
52. Answers should include the following. • This equation needs to show that the difference of the estimate E from the originally stated magnitude of 6.1 could be plus 0.3 or 11
Algebra 2
Chapter 1
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minus 0.3, as shown in the graph below. Instead of writing two equations, E 6.1 0.3 and E 6.1 0.3, absolute value symbols can be used to account for both possibilities, 0 E 6.1 0 0.3. 0.3 units 5.6
5.7
5.8
5.9
6.0
0.3 units 6.1
6.2
6.3
6.4
6.5
6.6
6.7
• Using an original magnitude of 5.9, the equation to represent the estimated extremes would be 0 E 5.9 0 0.3. 54. A
53. B
55. 0 x 1 0 2 x 4; 0 x 1 0 2 1x 42
56. x 1 2 x 4; x 1 2 x 4; x 1 2 x 4; x 1 2 x 4
57. 51.56
58. 8
59. 21n 112 61.
60. 5n 2
16 3
62. 2
63. 14
64. Commutative Property ()
65. Distributive Property
66. Multiplicative Inverse
67. Additive Identity
68. false; 23
69. true
70. true
71. false; 1.2
72.
73. 364 ft2
74. 2
75. 8
76. 2
77.
2 3
1 1x 2
321x 52
78. 6 3 4
79.
©Glencoe/McGraw-Hill
12
Algebra 2
Chapter 1
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Lesson 1-5 Solving Inequalities Pages 37–39 2. Sample answer: 2n 6
1. Dividing by a number is the same as multiplying by its inverse.
4. 5a 0 a 1.56 or 1 , 1.52
3. Sample answer: x2 x1
⫺2
5 3
0
1
2
⫺2
3
7. 5y 0 y 66 or 16, 2
2
⫺6
17. 5x 0 x 76 or 1 , 72
24
26
⫺4
⫺6
©Glencoe/McGraw-Hill
⫺5
⫺4
⫺3
⫺8
⫺6
⫺4
⫺2
0
10 11 12 13 14 15 16 17 18 19 20 21
18. 5d 0 d 86 or 18, 2 ⫺10
28
⫺8
⫺6
⫺4
⫺2
0
20. 5p 0 p 36 or 1 , 34 ⫺6
30
21. 5k 0 k 3.56 or 33.5, 2 ⫺7
8
16. 5b 0 b 186 or 1 , 184
19. 5g 0 g 276 or 1 , 274 22
6
14. at least 92
⫺1 0 1 2 3 4 5 6 7 8 9 10
20
4
4
15. 5n 0 n 116 or 311, 2 ⫺8
2
12. 12n 36; n 3
13. 2n 3 5; n 4
⫺14 ⫺12 ⫺10
3
⫺30 ⫺28 ⫺26 ⫺24 ⫺22 ⫺20
11. all real numbers or 1 , 2 0
2
10. 5n 0 n 246 or 1 , 244
8 9 10 11 12 13 14 15 16 17 18 19
⫺2
0
⫺10
9. 5p 0 p 156 or 115, 2
⫺4
1
8. 5w 0 w 76 or 1 , 72
⫺1 0 1 2 3 4 5 6 7 8 9 10
⫺6
0
6. 5c 0 c 36 or 3 3, 2
5. e x ` x f or a , d f 5 3
⫺1
⫺4
⫺2
0
2
4
22. 5y 0 y 56 or 1 , 52
⫺2
⫺4
13
⫺2
0
2
4
Algebra 2
6
Chapter 1
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23. 5m 0 m 46 or 14, 2 ⫺6
⫺4
⫺2
0
2
24. e b ` b f or c b 2 3
4
⫺1
⫺2
0
2
4
⫺2
6
27. 5n 0 n 1.756 or 3 1.75, 2 0
0.5
1
1.5
2
2.5
⫺7
20
⫺2
⫺4
⫺2
0
0
2
4
6
or a , b 1 20
⫺1 ⫺ 3 ⫺ 1 4
20
1 20
20
2.0
2.2
2.4
2.6
2.8
5 7
4
5 7
0 1 2 3 4 5 6 1 8 9 10 11
1 5
1 5
3 5
⫺6
1
⫺4
⫺2
0
2
6
⫺4
⫺2
0
2
4
⫺4
⫺3
⫺2
3 2
⫺1
0
40. no more than 14 rides
41. n 8 2; n 6
42. 4n 35; n 8.75
1 n 2
4
3 2
39. at least 25 h
43.
7 7
38. e n ` n f or a , d
37. ⫺6
7 7
36. 5p 0 p 06 or 10, 2
35. e y ` y f or a , b 1 5
3.0
34. e a ` a f or c , b 7 7 7 7 7 7
3 5
3 20
32. 5z 0 z 2.66 or 12.6, 2
2
2
1 5
8
⫺20 ⫺18 ⫺16 ⫺14 ⫺12 ⫺10
33. 5g 0 g 26 or 1 , 22 ⫺6
1
30. 5c 0 c 186 or 118, 2
31. 5d 0 d 56 or 35, 2 ⫺4
2
1 f 20
⫺286 ⫺284 ⫺282 ⫺280 ⫺278 ⫺276
⫺6
0
28. ew ` w
29. 5x 0 x 2796 or 1 , 2792
⫺8
0
26. 5r 0 r 66 or 1 , 64
25. 5t 0 t 06 or 1 , 04 ⫺4
2 3
7 5; n 24
44. 3n 1 16; n 5 n 2
46. n 9 ; n 18
45. 21n 52 3n 11; n 1 17
47. 217m2 17; m , at least 14 2 child-care staff members ©Glencoe/McGraw-Hill
1
48. $24,000 0.015130,500n2
40,000 14
Algebra 2
Chapter 1
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85 91 89 94 s 5
90
49. n 34.97; She must sell at least 35 cars.
50.
51. s 91; Ahmik must score at least 91 on her next test to have an A test average.
52a. It holds only for or ; 2 2. 52b. 1 2 but 2 1 52c. For all real numbers a, b, and c, if a b and b c then a c.
53. Answers should include the following. • 150 400 • Let n equal the number of minutes used. Write an expression representing the cost of Plan 1 and for Plan 2 for n minutes. The cost for Plan 1 would include a $35 monthly access fee plus 40¢ for each minute over 150 minutes or 35 0.41n 1502. The cost for Plan 2 for 400 minutes or less would be $55. To find where Plan 2 would cost less than Plan 1 solve 55 35 0.41n 1502 for n. The solution set is 5n 0 n 2006, which means that for more than 200 minutes of calls, Plan 2 is cheaper.
54. D
55. D
56. x 3
57. x 2
58. x 1
59. 514, 206
60.
5 11 e , f 4 4
61.
62. b online browsers each year; 6b 19.2 106.6; about 14.6 million online browsers each year
63. N, W, Z, Q, R
64. Q, R
©Glencoe/McGraw-Hill
15
Algebra 2
Chapter 1
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66. 4.25(5.5 8); 4.25(5.5) 4.25(8)
65. I, R 67. 57, 76 69.
e 4,
68. 513, 236 70. 511, 256
4 f 5
71. 511, 16
72. 518, 106
Chapter 1 Practice Quiz 2 Page 39 2s t2
g
1. 0.5
2.
3. 14
4. e , 5 f 19 3
5. e m ` m f or a , b 4 9
⫺2 9
0
4 9
2 9
4 9
2 3
8 1 9
Lesson 1-6 Solving Compound and Absolute Value Inequalities Pages 43–46 1. 5 c 15
2. Sample answer: x 3 and x2 4. 0 n 0 8
3. Sabrina; an absolute value inequality of the form 0 a 0 b should be rewritten as an or compound inequality, a b or a b.
⫺12
5. 0 n 0 3 ⫺6
⫺4
7. 0 n 0 2
⫺2
0
2
⫺4
0
4
8
6. 0 n 0 4
4
8. 5y 0 y 4 or y 16 ⫺4
©Glencoe/McGraw-Hill
⫺8
16
⫺2
0
2
4 Algebra 2
6 Chapter 1
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9. 5d 0 2 d 36 ⫺4
⫺2
0
10. 5a 0 a 5 or a 56 2
4
11. 5g 0 13 g 56 ⫺16 ⫺12
⫺8
⫺4
0
⫺8
6
⫺2
⫺2
4
0
2
4
8
6
12
8
c
0
2
4
6
16. 0 n 0 7
⫺8
⫺4
0
4
8
12
⫺4
⫺2
0
2
4
6
⫺8
⫺4
0
4
8
12
17. 0 n 0 4 19. 0 n 0 8 21. 0 n 0 1
4
8
12
⫺8
⫺4
0
4
8
12
⫺1.4 ⫺1.2
0
1.2
1.4
1.6
4
6
8
2
4
6
2
4
6
24. 0 n 0 6
0
4
29. 5x 0 2 x 46 0
2
31. 5f 0 7 f 56
26. 0 n 1 0 3
28. 5t 0 1 t 36 8
4
12
0
⫺4
⫺2
2
0
32. all real numbers
⫺8
⫺6
⫺4
⫺2
0
⫺8
⫺4
0
4
8
12
⫺4
⫺2
0
2
4
6
33. 5g 0 9 g 96
⫺2
30. 5c 0 c 2 or c 16
6
⫺10
⫺4
⫺2
0
34. 5m 0 m 4 or m 46 ⫺4
⫺2
0
2
4
6
⫺8
⫺4
0
4
8
12
36. 5y 0 7 y 76
35.
©Glencoe/McGraw-Hill
0
22. 0 n 0 5
27. 5p 0 p 2 or p 86
⫺2
⫺4
20. 0 n 0 1.2
25. 0 n 1 0 1 ⫺4
⫺8
18. 0 n 0 6
23. 0 n 0 1.5
⫺4
4
14. 55 60; 343.75 c 6.25 375; between $343.75 and $375
15. 0 n 0 5
⫺8
0
12. 5k 0 3 k 76
13. all real numbers ⫺4
⫺4
17
Algebra 2
Chapter 1
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38. 5r 0 3 r 46
37. 5b 0 b 10 or b 26 ⫺4
0
4
8
12
16
39. e w ` w 1 f 7 3
⫺2
⫺1
0
43.
⫺2
en ` n
0
1
0
⫺2
0
2
4
6
⫺4
⫺2
0
2
4
6
0
2
4
6
40. 1
42. 5n 0 n 06
41. all real numbers ⫺4
⫺4
2
4
6
⫺4
⫺2
44. 5n 0 n 1.56
7 f 2 2
3
4
⫺2
5
45. 6.8 x 7.4
⫺1
0
1
2
3
46. 45 s 65
47. 45 s 55
48. 0 t 98.6 0 8; 5b 0 b 106.6 or b 90.66
49. 108 in. L D 130 in.
50. 84 in. L 106 in.
51. a b c, a c b, bca
52. a b c a b 54. Compound inequalities can be used to describe the acceptable time frame for the fasting state before a glucose tolerance test is administered to a patient suspected of having diabetes. Answers should include the following. • Use the word and when both inequalities must be satisfied. Use the word or when only one or the other of the inequalities must be satisfied. • 10 h 16
53a. ⫺4
⫺2
0
2
4
6
⫺4
⫺2
0
2
4
6
⫺4
⫺2
0
2
4
6
53b. 53c. 53d. 3 0 x 2 0 8 can be rewritten as 0 x 2 0 3 and 0 x 2 0 8. The solution of 0 x 2 0 3 is x 1 or x 5. The solution of 0 x 2 0 8 is 10 x 6. Therefore, the union of these two sets is 1x 1 or x 52
©Glencoe/McGraw-Hill
18
Algebra 2
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Page 19 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-01:
and 110 x 6). The union of the graph of x 1 or x 5 and the graph of 10 x 6 is shown below. From this we can see that solution can be rewritten as 110 x 52 or 11 x 62. ⫺12
⫺8
⫺4
0
4
• 12 hours would be an acceptable fasting state for this test since it is part of the solution set of 10 h 16, as indicated on the graph below. 8 9 10 11 12 13 14 15 16 17 18 19
8
55. x 5 or x 6
56. D
57.
58. 2 x 3
59. 15x 2 32 or 15x 2 32; 5x 0 x 0.2 or x 16
60. abs12x 62 10; 5x x 2 or x 86
61. d 6 or 36, 2 ⫺8
⫺6
⫺4
⫺2
0
⫺2
0
2
⫺4
2
63. n 1 or 1 , 12 ⫺4
62. x 4 or 1 , 42
4
⫺2
0
2
4
6
64. 0 x 587 0 5; highest: 592 keys, lowest: 582 keys 6
66. 511, 46
65. {10, 16} 67.
68. Addition Property of Equality
69. Symmetric Property ()
70. Transitive Property of Equality
71. 3a 7b
72. 2m 7n 18
73. 2
74. 92
75. 7
©Glencoe/McGraw-Hill
19
Algebra 2
Chapter 1
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Page 20 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
Chapter 2 Linear Relations and Functions Lesson 2-1 Relations and Functions Pages 60–62 2. Sample answer:
1. Sample answer: {(4, 3), (2, 3), (1, 5), (2, 1)}
y
x
O
3. Molly; to find g(2a), replace x with 2a. Teisha found 2g(a), not g(2a).
4. yes
5. yes
6. no
7. D {7}, R {1, 2, 5, 8}, no
8. D {3, 4, 6}, R {2.5}, yes y
(7, 8)
y
(7, 5)
(4, 2.5) (3, 2.5)
(6, 2.5)
(7, 2)
x
O O
(7, 1)
x
10. D 5x 0 x 06, R all reals, no
9. D all reals, R all reals, yes
y
y
x y2 O
x
O
x
y 2x 1
12. 7
11. 10
©Glencoe/McGraw-Hill
20
Algebra 2
Chapter 2
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13. D {70, 72, 88}, R {95, 97, 105, 114}
14. {(88, 97), (70, 114), (88, 95), (72, 105)}
15. Record High Temperatures
16. No; the domain value 88 is paired with two range values.
115
July
110 105 100 95 0
70 80 January
90
17. yes
18. no
19. no
20. yes
21. yes
22. no
23. D {3, 1, 2}, R {0, 1, 5}; yes
24. D {3, 4, 6}, R {5}; yes y
y
(3, 5) (4, 5)
(1, 5)
(6, 5)
(2, 1) (3, 0)
x
O
x
O
26. D {3, 4, 5, 6}, R {3, 4, 5, 6}; yes
25. D {2, 3}, R {5, 7, 8}; no (2, 8)
y
(3, 7)
y (5, 6) (3, 4)
(2, 5)
(6, 5)
(4, 3) O
©Glencoe/McGraw-Hill
x
O
21
x
Algebra 2
Chapter 2
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Page 22 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
27. D {3.6, 0, 1.4, 2}, R {3, 1.1, 2, 8}; yes
28. D {2.5, 1, 0}, R {1, 1}; no y
y (3.6, 8)
(2.5, 1) (1, 1) (0, 1)
x
O (1, 1)
(1.4, 2)
x O (0, 1.1) (2, 3)
30. D all reals, R all reals; yes
29. D all reals, R all reals; yes
y
y
y 3x x
O
x
O
y 5x
31. D all reals, R all reals; yes
32. D all reals, R all reals; yes y
y
O O
x
x
y 7x 6
y 3x 4
©Glencoe/McGraw-Hill
22
Algebra 2
Chapter 2
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34. D 5x 0 x 36, R all reals; no
33. D all reals, R 5y 0 y 06, yes y
y
O
yx
x
2
x 2y 2 3
x
O
36. D {47, 48, 52, 56}, R {145, 147, 148, 157, 165}
35. American League Leaders 170 165 RBI
160 155 150 145 140 0
48
50
52
54
56
HR
37. No; the domain value 56 is paired with two different range values.
38. {(1997, 39), (1998, 43), (1999, 48), (2000, 55), (2001, 61), (2002, 52)}
39.
40. D {1997, 1998, 1999, 2000, 2001, 2002}, R {39, 43, 48, 52, 55, 61}
Stock Price 70 60 Price ($)
50 40 30 20 10 0 1996
1998
2000 2002 Year
2004
42. {(1987, 12), (1989, 13), (1991, 11), (1993, 12), (1995, 9), (1997, 6), (1999, 3)}
41. Yes; each domain value is paired with only one range value.
©Glencoe/McGraw-Hill
23
Algebra 2
Chapter 2
PQ245-6457F-P02[020-055].qxd
43.
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Page 24 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
44. D {1987, 1989, 1991, 1993, 1995, 1997, 1999}, R {3, 6, 9, 11, 12, 13}
30+ Years of Service Representatives
14 12 10 8 6 4 2 0
’87
’91 ’95 Year
’99
45. Yes; no; each domain value is paired with only one range value so the relation is a function, but the range value 12 is paired with two domain values so the function is not one-to-one.
46. 14
47. 6
48.
49. 3
50. 3a 5
51. 25n 2 5n
52. 4
53. 11
54. 39
55. f(x) 4x 3
56. Relations and functions can be used to represent biological data. Answers should include the following. • If the data are written as ordered pairs, then those ordered pairs are a relation. • The maximum lifetime of an animal is not a function of its average lifetime.
57. B
58. C
59. discrete
60. continuous
61. discrete
62. continuous
65. 5x 0 x 5.16
66. $2.85
2 9
63. 5y 0 8 y 66
©Glencoe/McGraw-Hill
64. 5m 0 4 m 66
24
Algebra 2
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67. $29.82
68. 43
69. 31a 10b
70. 1
71. 2
72. 6
73. 15
Lesson 2-2 Linear Equations Pages 65–67 2. 5, 2
1. The function can be written as 1 2
f(x) x 1, so it is of the form f(x) mx b, where m
1 2
and b 1.
3. Sample answer: x y 2
4. No, the variables have an exponent other than 1.
5. yes
6. 3x y 5; 3, 1, 5
7. 2x 5y 3; 2, 5, 3
8. 2x 3y 3; 2, 3, 3
5 3
9. , 5
10. 2, 2 y
y
x
O
x
O
y 3x 5
xy20
11. 2, 3
12. 3, y
3 2 y
3x 2y 6 O
x
x
O 4x 8y 12
©Glencoe/McGraw-Hill
25
Algebra 2
Chapter 2
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13. $177.62
14. 563.00 euros
15. yes
16. No; x appears in a denominator.
17. No; y is inside a square root.
18. No; x has exponents other than 1.
19. No; x appears in a denominator.
20. yes
21. No; x has an exponent other than 1.
22. No; x is inside a square root.
23. x 2 5y 0
24. h(x ) x 3 x 2 3x
25. 7200 m
26. Sound travels only 1715 m in 5 seconds in air, so it travels faster underwater.
27. 3x y 4; 3, 1, 4
28. 12x y 0; 12, 1, 0
29. x 4y 5; 1, 4, 5
30. x 7y 2; 1, 7, 2
31. 2x y 5; 2, 1, 5
32. x 2y 3; 1, 2, 3
33. x y 12; 1, 1, 12
34. x y 6; 1, 1, 6
35. x 6; 1, 0, 6
36. y 40; 0, 1, 40
37. 25x 2y 9; 25, 2, 9
38. 5x 4y 2; 5, 4, 2
39. 3, 5
40. 6, 2 y
y
2x 6y 12
5x 3y 15
x O
O
©Glencoe/McGraw-Hill
26
x
Algebra 2
Chapter 2
PQ245-6457F-P02[020-055].qxd
41.
10 , 3
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Page 27 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
5 2
42. 5, 2 y
y
3x 4y 10 0
x
O
x
O
2x 5y 10 0
43. 0, 0
44. y
1 , 2
2 y
yx O
x
x
O
y 4x 2
45. none, 2
46. none, 4
y
y y4 x
O
x
O
y 2
48. 1, none
47. 8, none 8 6 4 2 8 64 2 2 4 6 8
©Glencoe/McGraw-Hill
y
y
x1
x8 O 2 4 6
O
x
x
27
Algebra 2
Chapter 2
PQ245-6457F-P02[020-055].qxd
49.
1 , 4
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Page 28 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
1
50. 6, 3 g (x )
f (x ) f (x ) 4x 1
g (x ) 0.5x 3 O
x
O
51.
y
x
52. Sample answer: x y 2
x y 5
x O
xy0 x y 5
The lines are parallel but have different y-intercepts. 53. 90C
54. 4 km
55.
160 120 80 40
56. 1.75b 1.5c 525
T (d )
4 32
O1 2 3 4 d 40 80 120 T (d ) 35d 20 160
57.
c 350 300 250 200 150 100 50 0
58. Yes; the graph passes the vertical line test.
1.75b 1.5c 525
100
©Glencoe/McGraw-Hill
200
400b
28
Algebra 2
Chapter 2
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21 2
59. no
60.
61. A linear equation can be used to relate the amounts of time that a student spends on each of two subjects if the total amount of time is fixed. Answers should include the following. • x and y must be nonnegative because Lolita cannot spend a negative amount of time studying a subject. • The intercepts represent Lolita spending all of her time on one subject. The x-intercept represents her spending all of her time on math, and the y-intercept represents her spending all of her time on chemistry. 63. B
62. B
units2
64. D {1, 1, 2, 4}, R {4, 3, 5}; yes y (1, 5) ( 1, 3)
(4, 3)
x
O (2, 4)
66. 5x 0 1 x 26
65. D {0, 1, 2}, R {1, 0, 2, 3}; no y (1, 3) (0, 2) (1, 0) O
©Glencoe/McGraw-Hill
x (2, 1)
29
Algebra 2
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67. 5x 0 x 6 or x 26
68. $7.95
69. 3s 14
70. 4 1 4
1 3
72.
73. 2
74.
75. 5
76.
77. 0.4
78. 0.8
71.
3 2
4 15
Lesson 2-3 Slope Pages 71–74 1. Sample answer: y 1
2. Sometimes; the slope of a vertical line is undefined.
3. Luisa; Mark did not subtract in a consistent manner when using the slope formula. If y2 5 and y1 4, then x2 must be 1 and x1 must be 2, not vice versa.
4. 0
1 2
5. 7.
6. 1
O
©Glencoe/McGraw-Hill
8.
y
x
y
O
30
x
Algebra 2
Chapter 2
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9.
12:22 PM
Page 31 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
10.
y
y
x
O
O
11.
x
12. 5.5/hr
y
O
x
13. 1.25/hr
14. 2:00 P.M.–4:00 P.M.
5 2
15. 17.
16. 13
3 5
18. 4
19. 0
20. 1
21. 8
22. undefined
23. 4
24.
25. undefined
26. 0
27. 1
28. 9
29. about 0.6
30. about 1.3
31.
32.
5 4
y
y
O
O
©Glencoe/McGraw-Hill
x
x
31
Algebra 2
Chapter 2
PQ245-6457F-P02[020-055].qxd
33.
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Page 32 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
34.
y
y
x
O
O
35.
36.
y
x
O
x
y
O
x
37. about 68 million per year
38. about 32 million per year
39. The number of cassette tapes shipped has been decreasing.
40. 55 mph
41. 45 mph
42. speed or velocity
43.
O
45.
x
y
O
46.
y
O
©Glencoe/McGraw-Hill
44.
y
x
y
O
32
x
x
Algebra 2
Chapter 2
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12:22 PM
Page 33 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
y
47.
O
49.
48.
x
y
O
50.
y
x
y
x O O
51. Yes; slopes show that adjacent sides are perpendicular.
52. 1
53. The grade or steepness of a road can be interpreted mathematically as a slope. Answers should include the following. • Think of the diagram at the beginning of the lesson as being in a coordinate plane. Then the rise is a change in y-coordinates and the horizontal distance is a change in x-coordinates. Thus, the grade is a slope expressed as a percent.
54. D
©Glencoe/McGraw-Hill
33
x
Algebra 2
Chapter 2
PQ245-6457F-P02[020-055].qxd
7/24/02
•
12:22 PM
Page 34 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
y
x y 0.08x O
55. D
56. The graphs have the same y-intercept. As the slopes increase, the lines get steeper.
57. The graphs have the same y-intercept. As the slopes become more negative, the lines get steeper.
58. 10, 4 8 2x 5y 20 6 4 2 108 64 2 2 4 6 8
59. 2,
8 3
y
O 2 4x
60. 0, 0 y
y
O
y 7x
x O
4x 3y 8 0
x
61. 7
62. 5
5 2
63.
64. 3a 4
65. 5x 0 1 x 36
66. 5z 0 z 7356
©Glencoe/McGraw-Hill
34
Algebra 2
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67. at least 8
68. 17a b
69. 9
70. y 9 x
71. y 4x 2
72. y 3x 7
5 2
73. y x
1 2
2 3
75. y x
3 5
74. y x
4 5
11 3
Chapter 2 Practice Quiz 1 Page 74 1. D {7, 3, 0, 2}, R {2, 1, 2, 4, 5}
2. 375
3. 6x y 4
4. 10, 6 y
3x 5y 30
x O
5.
y O
©Glencoe/McGraw-Hill
x
35
Algebra 2
Chapter 2
PQ245-6457F-P02[020-055].qxd
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12:22 PM
Lesson 2-4
Page 36 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
Writing Linear Equations Pages 78–80
1. Sample answer: y 3x 2
2. 6, 0
3. Solve the equation for y to get 2 3 y x . The slope of this
4. 2, 5
5
5
3 . 5
line is The slope of a parallel line is the same. 3 2
6. y 0.5x 1
5. , 5
5 2
3 4
8. y x 16
7. y x 2 3 5
9. y x
16 5
10. y x 2
5 4
11. y x 7
12. B
2 3
13. , 4 15.
1 , 2
14.
3 , 4
0 3 5
5 2
16. , 6
17. undefined, none
18. c, d
19. y 0.8x
20. y x
21. y 4
22. y 2
23. y 3x 6
24. y 0.25x 4
1 2
25. y x
5 3
3 2
7 2
26. y x
4 5
17 5
30. no slope-intercept form for x7 3 2
31. y 0
32. y x
33. y x 4
34. y x
©Glencoe/McGraw-Hill
17 2
28. y 4x
27. y 0.5x 2 29. y x
29 3
3 4
36
1 4
Algebra 2
Chapter 2
PQ245-6457F-P02[020-055].qxd
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2 3
35. y x
12:22 PM
Page 37 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
10 3
1 15
37. y x
36. y 4x 3 23 5
38. y x 4
39. y 3x 2
40. y 2x 6
41. d 180c 360
42. 180, 360
43. 540
44. y 75x 6000
45. 10 mi
46. y x 32
9 5
80 60 40 30
10 20 40
y y 95 x 32 x O 10 20 30
47. 68F
48. 40
49. y 0.35x 1.25
50. $11.75
51. y 2x 4
52. A linear equation can sometimes be used to relate a company’s cost to the number they produce of a product. Answers should include the following. • The y-intercept, 5400, is the cost the company must pay if they produce 0 units, so it is the fixed cost. The slope, 1.37, means that it costs $1.37 to produce each unit. The variable cost is 1.37x. • $6770
53. C x y 55. 5 1
54. A
2
56.
5
57. 2
©Glencoe/McGraw-Hill
5 , 2
5
58. 3
37
Algebra 2
Chapter 2
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59. 0
60. 0.55 s
62. 5x 0 x 66
61.
63. 5r 0 r 66
64. 3
65. 6.5
66. 323.5
67. 5.85
Lesson 2-5 Modeling Real-World Data: Using Scatter Plots Pages 83–86 2. D {1, 1, 2, 4}, R {0, 2, 3}; Sample answer using (1, 0) and (2, 2): 4
1. d
3. Sample answer using (4, 130.0) and (6, 140.0): y 5x 110
4a.
Atmospheric Temperature Temperature (˚C)
16 14 12 10 8 6 4 2 0
1000
2000 3000 Altitude (ft)
4000
5000
4b. Sample answer using (2000, 11.0) and (3000, 9.1): y 0.0019x 14.8 4c. Sample answer: 5.3C 5a.
6a.
60 50 40 30 20 10 0 ’88 ’90 ’92 ’94 ’96 ’98 ’00 Year
©Glencoe/McGraw-Hill
Lives Saved by Minimum Drinking Age Lives (thousands)
Households (millions)
Cable Television 80 70
38
25 20 15 10 5 0 ’94 ’95 ’96 ’97 ’98 ’99 ’00 Year
Algebra 2
Chapter 2
PQ245-6457F-P02[020-055].qxd
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Page 39 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
5b. Sample answer using (1992, 57) and (1998, 67): y 1.67x 3269.64 5c. Sample answer: about 87 million
6b. Sample answer using (1996, 16.5) and (1998, 18.2): y 0.85x 1680.1 6c. Sample answer: 28,400
7a.
8a.
2000–2001 Detroit Red Wings
Bottled Water Consumption 14 12 10 Gallons
Assists
60 50 40 30 20
8 6 4
10 0
2 0
10 20 30 40 Goals
’91
’93
’95 Year
’97
’99
7b. Sample answer using (4, 5) and (32, 37): y 1.14x 0.44 7c. Sample answer: about 13
8b. Sample answer using (1993, 9.4) and (1996, 12.5): y 1.03x 2043.39 8c. Sample answer: about 26.9 gal
9a.
10. Sample answer using (1990, 563) and (1995, 739): y 35.2x 69,485
Revenue ($ millions)
Broadway Play Revenue 700 600 500 400 300 200 100 0
1 2 3 4 Seasons Since ’95–’96
9b. Sample answer using (1, 499) and (3, 588): y 44.5x 454.5, where x is the number of seasons since 1995–1996 9c. Sample answer: about $1078 million or $1.1 billion 11. Sample answer: $1091
©Glencoe/McGraw-Hill
12. The value predicted by the equation is somewhat lower than the one given in the graph. 39
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13. Sample answer: Using the data for August and November, a prediction equation for Company 1 is y 0.86x 25.13, where x is the number of months since August. The negative slope suggests that the value of Company 1’s stock is going down. Using the data for October and November, a prediction equation for Company 2 is y 0.38x 31.3, where x is the number of months since August. The positive slope suggests that the value of Company 2’s stock is going up. Since the value of Company 1’s stock appears to be going down, and the value of Company 2’s stock appears to be going up, Della should buy Company 2.
14. No. Past performance is no guarantee of the future performance of a stock. Other factors that should be considered include the companies’ earnings data and how much debt they have.
15.
16. Sample answer using (213, 26) and (298, 23): y 0.04x 34.52
World Cities Precipitation (in.)
40 35 30 25 20 15 10 5 0
200 400 600 Elevation (ft)
18. Sample answer: The predicted value differs from the actual value by more than 20%, possibly because no line fits the data very well.
17. Sample answer: about 23 in.
©Glencoe/McGraw-Hill
40
Algebra 2
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19. Sample answer using (1975, 62.5) and (1995, 81.7): 96.1%
20. Sample answer: The predicted percent is almost certainly too high. Since the percent cannot exceed 100%, it cannot continue to increase indefinitely at a linear rate.
21. See students’ work.
22. Data can be used to write a linear equation that approximates the number of Calories burned per hour in terms of the speed that a person runs. Answers should include the following. • Calories Burned While Running Calories
1000 800 600 400 200 0
5 6 7 8 Speed (mph)
9
• Sample answer using (5, 508) and (8, 858): y 116.67x 75.35 • about 975 calories; Sample answer: The predicted value differs from the actual value by only about 2%. 23. D
24. A
25. 1988, 1993, 1998; 247, 360.5, 461
26. y 21.4x 42,296.2
27. 354
28. about (1993, 356.17)
29. y 21.4x 42,294.03
30. about 613, about 720
31. y 4x 6
32. y x
33. 3
34. 7
©Glencoe/McGraw-Hill
3 7
41
6 7
Algebra 2
Chapter 2
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29 3
36.
37 3
37. 5x 0 x 7 or x 16
38. 3
39. 11
40. 0
41.
2 3
42. 1.5
Lesson 2-6 Special Functions Pages 92–95 2. 1
1. Sample answer: [[1.9]] 1
3. Sample answer: f(x) 0 x 10
4. A
5. S
6. D all reals, R all integers
7. D all reals, R all integers
8. D all reals, R all nonnegative reals
g (x )
h (x )
g (x ) 冀2x 冁
h (x ) |x 4|
O
x O
©Glencoe/McGraw-Hill
42
x
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10. D all reals, R 5y 0 y 26
9. D all reals, R all nonnegative reals
g (x )
x
O
11. D all reals, R all reals
12. step function
13.
14. $6
0
Time (hr)
15. C
16. A
17. S
18. S
19. A
20. P
21.
5 4 3 2 1 O
22.
y
60
x 180 300
1.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0
1 2 3 4 5 6 7 8 9 Minutes
C
©Glencoe/McGraw-Hill
43
C
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24. D all reals, R all integers
23. $1.00
f (x )
x
O
f (x ) 冀x 3冁
25. D all reals, R all integers
26. D all reals, R all even integers
g (x )
f (x )
g (x ) 冀x 2冁 x
O
x
O
f (x ) 2冀x 冁
27. D all reals, R {3a 0 a is an integer.} 12 9 6 O 4 3 2 1 3 6 9 12
28. D all reals, R all integers
h (x )
g (x )
h (x ) 3冀x 冁 x 1 2 3 4
x
O
g (x ) 冀x 冁 3
29. D all reals, R all integers
30. D all reals, R all nonnegative reals
f (x ) f (x ) 冀x 冁 1 O
©Glencoe/McGraw-Hill
x
44
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32. D all reals, R {y 0 y 3}
31. D all reals, R all nonnegative reals
g (x )
h (x )
h (x ) |x |
g (x ) |x | 3
x
O
x
O
33. D all reals, R {y 0 y 4}
34. D all reals, R all nonnegative reals
g (x )
h (x ) g (x ) |x | 4 x
O
O
h (x ) |x 3|
x
36. D all reals, R all nonnegative reals
35. D all reals, R all nonnegative reals
f (x )
f (x )
f (x ) |x 2| x
O
|
f (x ) x 1 4
|
O
x
38. D all reals, R {y 0 y 3}
37. D all reals, R all nonnegative reals
f (x )
f (x )
O
|
f (x ) x 1 2
©Glencoe/McGraw-Hill
|
O
x
x
45
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39. D {x 0 x 2 or x 2}, R {1, 1}
40. D all reals, R {y 0 y 0 or y 2} f (x )
h (x )
O
x
x
O
41. D all reals, R {y 0 y 2}
42. D all reals, R all nonnegative whole numbers
g (x )
f (x ) O
x O
x
f (x ) 冀|x |冁
43. D all reals, R all nonnegative whole numbers
44.
g (x )
O
2 if x 1 f(x) • 2x if 1 x 1 x if x 1
x
g (x ) |冀x 冁|
45. f (x) 0x 2 0
46. {x 0 x 0}
47.
48.
©Glencoe/McGraw-Hill
46
f (x) e
0 if 0 x 300 0.8 (x 300) if x 300
Algebra 2
Chapter 2
PQ245-6457F-P02[020-055].qxd
49.
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50. A step function can be used to model the cost of a letter in terms of its weight. Answers should include the following. • Since the cost of a letter must be one of the values $0.34, $0.55, $0.76, $0.97, and so on, a step function is the best model for the cost of mailing a letter. The gas mileage of a car can be any real number in an interval of real numbers, so it cannot be modeled by a step function. In other words, gas mileage is a continuous function of time. •
y |x | |y | 3
x
Cost ($)
O
2.10 1.80 1.50 1.20 0.90 0.60 0.30
0
1 2 3 4 5 6 7 Weight (oz)
51. B
52. D
53. Life Expectancy
54. Sample answer using (10, 69.7) and (47, 76.5): y 0.18x 67.9
78 76 74 72 70
Years Since 1950
56. y 3x 10
55. Sample answer: 78.7 yr
©Glencoe/McGraw-Hill
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58. {x 0 x 3}
57. y x 2
1 0 1 2 3 4 5 6
59. e y ` y f 5 6
60. yes
3 2 1 0 1 2 3
61. no
62. no
63. yes
64. no
65. yes
Chapter 2 Practice Quiz 2 Page 95 2 3
1. y x
11 3
2. Houston Comets 250 200 150 100 50 0
65 70 75 80 Height (in.)
4. Sample answer: 168 Ib
3. Sample answer using (66, 138) and (74, 178): y 5x 192 5. D all reals, R nonnegative reals f (x )
f (x ) |x 1| O
©Glencoe/McGraw-Hill
x
48
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Lesson 2-7
Graphing Inequalities Pages 98–99
1. y 3x 4
2. Substitute the coordinates of a point not on the boundary into the inequality. If the inequality is satisfied, shade the region containing the point. If the inequality is not satisfied, shade the region that does not contain the point.
3. Sample answer: y 0 x 0
4.
y
y 2 x
O
6.
5.
y
O
x
x y 0
7.
8.
y
y
y |2x |
x
O
O
x
x 2y 5
©Glencoe/McGraw-Hill
49
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10. 10c 13d 40
y
x O
11.
y 3|x | 1
12. No; (3, 2) is not in the shaded region.
d 10c 13d 40
c O
13.
14.
y
y 3 x 3y
x O
x
O
x y 5
y
15.
16.
y 6x 2 x
O
17.
18.
y
y
y 4x 3
O
x
O
x
y 2 3x
©Glencoe/McGraw-Hill
50
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20.
y
y
y 1 y 14 x
O
21.
x
O
22.
y
y
4x 5y 10 0
x
O
x
O
x 6y 3 0
23.
24.
y
y
y 1x 5 3
O
y 1x 5
x
2
O
25.
x
26.
y
y
y |4x | y |x | O
©Glencoe/McGraw-Hill
x
O
51
x
Algebra 2
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12:22 PM
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28.
y
y y |x 1| 2
x
O
x
O
y |x | 3
29.
30.
y
y y |x |
x y 1 O
x
x
O
x y 1 y |x |
31. x 2
32. y 3x 5 y
y x 2
O O
x
x y 3x 5
33.
34. yes
y 350 250 0.4x 0.6y 90
150 50 O
50
150 250 350 x
©Glencoe/McGraw-Hill
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35. 4a 3s 2000
36.
800
s
600
4a 3s 2000
400 200
a 200 400 600 800
O
38. 1.2a 1.8b 9000
37. yes
b 6000
1.2a 1.8b 9000
4000 2000 O
a 2000 4000 6000 8000
40.
39. yes
y
|y | x O
42. A
41. Linear inequalities can be used to track the performance of players in fantasy football leagues. Answers should include the following.
©Glencoe/McGraw-Hill
x
53
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• Let x be the number of receiving yards and let y be the number of touchdowns. The number of points Dana gets from receiving yards is 5x and the number of points he gets from touchdowns is 100y. His total number of points is 5x 100y. He wants at least 1000 points, so the inequality 5x 100y 1000 represents the situation. • y 12
5x 100y 1000
10 8 6 4 2 O 50
100 200
300 x
• the first one 43. B
44.
[10, 10] scl: 1 by [10, 10] scl: 1
45.
46.
[10, 10] scl: 1 by [10, 10] scl: 1
[10, 10] scl: 1 by [10, 10] scl: 1
©Glencoe/McGraw-Hill
54
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48. D all reals, R all integers
47.
f (x )
x
O [10, 10] scl: 1 by [10, 10] scl: 1
f (x ) 冀x 冁 4
49. D all reals, R {y 0 y 1}
50. D all reals, R all nonnegative reals
g (x )
g (x ) |x | 1
h (x )
O
x O
51.
x
52. Sample answer using (4, 6000) and (6, 8000): y 1000x 2000
Sales vs. Experience 10,000 Sales ($)
h (x ) |x 3|
8000 6000 4000 2000 0
1
2
3 4 Years
5
6
7
53. Sample answer: $10,000
54. 8
55. 3
56.
©Glencoe/McGraw-Hill
55
1 2
Algebra 2
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Chapter 3 Systems of Equations and Inequalities Lesson 3-1 Solving Systems of Equations by Graphing Pages 112–115 1. Two lines cannot intersect in exactly two points.
2. Sample answer: x y 4, x y 2
3. A graph is used to estimate the solution. To determine that the point lies on both lines, you must check that it satisfies both equations.
4.
y (2, 5)
y x 3
y 2x 9
x
O
5.
6.
y 3x 2y 10
y 4x 2y 22 O
(2, 2) 2x 3y 10
x x
8. inconsistent
7. consistent and independent y
y 2x 4y 8
yx4
x
y6x O
©Glencoe/McGraw-Hill
(4, 3)
6x 9y 3
O
O
x 2y 2
x
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10. y 0.08x 3.2, y 0.10x 2.6
9. consistent and dependent y
x
O
x 2y 8 1 xy4 2
11. The cost is $5.60 for both stores to develop 30 prints.
12. You should use Specialty Photos if you are developing less than 30 prints, and you should use The Photo Lab if you are developing more than 30 prints.
13.
14.
y
y x
O
y 3x 1
y 2x 4
15.
y 3x 8
x
O (1, 2)
(0, 8)
16.
y
yx8 y
2x 3y 12 (3, 2)
x 2y 6 (4, 1)
x
O 2x y 9
17.
y
x
O 2x y 4
18.
x 2y 11
y
(7, 6)
7x 1 8y (5, 3) 3x 7y 6
O
x
O
x
5x 11 4y
©Glencoe/McGraw-Hill
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y
19.
20.
x
(1.5, 5)
4x 2y 4
(3.5, 0)
2x 3y 7
y
O
x
O
8x 3y 3
2x 3y 7
21.
22.
y
y y 13 x 6
1 x 2y 5 4
(4, 2)
x
O
(9, 3)
2x y 6
O x 2 x y 3 3
23.
24.
y 1 xy0 2
y 4 x 15 y 3 3
x
O
x
O (4, 2)
25. inconsistent yx4
(3, 5)
2 x 35 y 5 3
1 x 12 y 2 4
26. consistent and independent
y
y yx3 x
O
x
O
yx4
©Glencoe/McGraw-Hill
y 2x 6
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27. consistent and independent
28. consistent and dependent
y
y 3x y 3
xy4
x
O
x O 4x y 9
6x 2y 6
29. inconsistent
30. consistent and dependent y
y yx5
4x 2y 6
x
O
x O
6x 3y 9
2y 2x 8
31. consistent and independent
32. inconsistent
y
y
2y x 2y 5 x
x
O
x
8y 2x 1
O 6y 7 3x
33. consistent and independent
34. consistent and dependent
y
y 1.6y 0.4x 1
0.8x 1.5y 10
O
1.2x 2.5y 4 O
©Glencoe/McGraw-Hill
x
0.4y 0.1x 0.25
x
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35. inconsistent y 13 x 2
36. consistent and independent
y
y 4 x y 2 3
O
x
x
O
3y x 2 2y 4x 3
38. (1, 3), (2, 1), (2, 3)
37. (3, 1)
y y 2x 1
x
O
2y x 4
39. y 52 0.23x, y 80
4x y 7
40. (120, 80) 120 Cost ($)
y 80 80
y 52 0.23x
40
0
40
80 120 Miles
160
41. Deluxe Plan
42. Supply, 200,000; demand, 300,000; prices will tend to rise.
43. Supply, 300,000; demand, 200,000; prices will tend to fall.
44. 250,000; $10.00
©Glencoe/McGraw-Hill
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45. y 304x 15,982, y 98.6x 18,976
Population (Thousands)
46. 2015 24,000
y 98.6x 18,976
20,000 16,000
y 304x 15,982
12,000 0
a b a 48b. b a 48c. b
4
d e d e d , e
8 12 16 Years After 1999
c b
f e
c b
f e
47. FL will probably be ranked third by 2020. The graphs intersect in the year 2015, so NY will still have a higher population in 2010, but FL will have a higher population in 2020.
48a. ,
49. You can use a system of equations to track sales and make predictions about future growth based on past performance and trends in the graphs. Answers should include the following. • The coordinates (6, 54) represent that 6 years after 1999 both the instore sales and online sales will be $54,000. • The in-store sales and the online sales will never be equal and in-store sales will continue to be higher than online sales.
50. A
51. C
52. (3.40, 2.58)
53. (5.56, 12)
54. (4, 3.42)
55. no solution
56. (9, 3.75)
©Glencoe/McGraw-Hill
61
Algebra 2
20
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58.
57. (2.64, 42.43)
y y 5 3x
O
59.
60.
y
x
y
2x y 4
x
O
O
x
2y 1 x
61. A
62. C
63. S
64. {13, 13}
65. {15, 9}
66.
67. {2, 3}
68. e5, f
69. {9}
70. 8 2n
71. x 2 6
72. 41a 52
73.
z 3
7 2
74. x 2
1
75. 9y 1
76. 3x 6y
77. 12x 18y 6
78. 15x 10y 10
79. x 4y
Lesson 3-2 Solving Systems of Equations Algebraically Pages 119–122 1. See students’ work; one equation should have a variable with a coefficient of 1. ©Glencoe/McGraw-Hill
2. There are infinitely many solutions.
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3. Vincent; Juanita subtracted the two equations incorrectly; y y 2y, not 0.
4. (4, 8)
5. (1, 3)
6. (4, 1)
7. (5, 2)
8. (9, 7)
9. (6, 20)
10. no solution
11. a3 , 2 b
12. C
13. (9, 5)
14. (2, 7)
15. (3, 2)
16. (6, 8)
17. no solution
18. (1, 1)
19. (4, 3)
20. (1, 8)
21. (2, 0)
22. (3, 1)
23. (10, 1)
24. (7, 9)
25. (4, 3)
26. (6, 5)
27. (8, 3)
28. (7, 1)
29. no solution
30. (5, 8)
31. a ,
32. a , 2b
1 3
1 2
2 3
3 b 2
1 3
33. (6, 11)
34. infinitely many
35. (1.5, 0.5)
36. (2, 4)
37. 8, 6
38. 2, 12
39. x y 28, 16x 19y 478
40. 18 members rented skis and 10 members rented snowboards.
41. 4 2-bedroom, 2 3-bedroom
42. (5, 2), (4, 4), (2, 8), (1, 10)
43. x y 30, 700x 200y 15,000
44. 18 printers, 12 monitors
45. 2x 4y 100, y 2x
46. 10 true/false, 20 multiplechoice
47. Yes; they should finish the test within 40 minutes.
48. a s 40, 11a 4s 335
©Glencoe/McGraw-Hill
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49. 25 min of step aerobics, 15 min of stretching
50. (4, 6)
51. You can use a system of equations to find the monthly fee and rate per minute charged during the months of January and February. Answers should include the following. • The coordinates of the point of intersection are (0.08, 3.5). • Currently, Yolanda is paying a monthly fee of $3.50 and an additional 8¢ per minute. If she graphs y 0.08x 3.5 (to represent what she is paying currently) and y 0.10x 3 (to represent the other long-distance plan) and finds the intersection, she can identify which plan would be better for a person with her level of usage.
52. C
53. A
54. inconsistent y yx2
x
O
yx1
©Glencoe/McGraw-Hill
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56. consistent and independent
55. consistent and dependent
y
y 4y 2x 4 3x y 1 O
x
O
y 2x 4
y 12 x 1
57.
x
58.
y
y
x
O
x
O
xy3 5y 4x 20
59.
60. 7x y 4; 7, 1, 4
y
3x 9y 15 O
x
61. x y 0; 1, 1, 0
62. 3x 5y 2; 3, 5, 2
63. 2x y 3; 2, 1, 3
64. x 2y 6; 1, 2, 6
65. 3x 2y 21; 3, 2, 21
66. 0.6 ampere
67. yes
68. no
69. no
70. yes
©Glencoe/McGraw-Hill
65
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Chapter 3 Practice Quiz 1 Page 122 1.
2.
y
y
y 3x 10
(1, 7)
(3, 2) 2x 3y 12
x
O
y x 6
2x y 4
x
O
4. (4, 1)
3. (2, 7) 5. Hartsfield, 78 million; O’Hare, 72.5 million
Lesson 3-3
Solving Systems of Inequalities by Graphing Pages 125–127
1. Sample answer: y x 3, y x 2
2. true
3a. 4 3b. 2 3c. 1 3d. 3
4.
y
y2 x
O
x4
5.
6.
y
xy2
y 2x 4 O
y
x 1 yx2
©Glencoe/McGraw-Hill
x
O
x
66
x3
Algebra 2
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8. (3, 3), (2, 2), (5, 3)
y y 2x 1
x1 O
x
x 2y 3
9. (4, 3), (1, 2), (2, 9), (7, 4)
10.
10 m
b2
Muffins
8
2.5b 3.5m 28
6 4 2
m3 b
0
11. Sample answer: 3 packages of bagels, 4 packages of muffins; 4 packages of bagels, 4 packages of muffins; 3 packages of bagels, 5 packages of muffins
12.
13.
14.
2
4
6 8 Bagels
12
y
y3 x
O
x2
y
y
x
O
10
y2x x
x 1
©Glencoe/McGraw-Hill
O
y 4
yx4
67
Algebra 2
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16.
y
y
y2 3x 2y 6
y 2 x
O
x
O 4x y 2
yx3
17.
18.
y 4x 3y 7
y 12 x 1
x
O
y
2y x 6
x
O
y 2x 3
19. no solution
20.
y y1
x3 x
O
x 3
21.
y 1
22. no solution
y x2 x 3y 6 O
x
x 4
23.
24. (0, 0), (0, 4), (8, 0)
y 2x y 4 2x 4y 7 O
x x 3y 2
©Glencoe/McGraw-Hill
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25. (3, 4), (5, 4), (1, 4)
26. (0, 4), (3, 0), (3, 5)
27. (6, 9), (2, 7), (10, 1)
28. (11, 3), (1, 3), (6, 4), a6, 5 b 1 2
30. 16 units2
29. (4, 3), (2, 7), (4, 1), a7 , 2 b 1 3
1 3
31. 64 units2
32. Hours Raking Leaves
16 14 12
8 6 4
0
16
x y 15
10
2
33. s 111, s 130, h 9, h 12
y
10x 12y 120 2
x
4 6 8 10 12 14 16 Hours Cutting Grass
34. category 4; 13-18 ft
h
Storm Surge (ft)
14
h 12 12
10
s 130
h9
8
s 111 s 0
80
©Glencoe/McGraw-Hill
100 120 140 Wind Speed (mph)
160
69
Algebra 2
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35.
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y 14
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36. Sample answer: 2 pumpkin, 8 soda; 4 pumpkin, 6 soda; 8 pumpkin, 4 soda
2x 1.5y 24
12 10
x 2.5y 26
8 6 4 2 0
2
4
6 8 10 12 14 Pumpkin
x
37. 6 pumpkin, 8 soda
38. 42 units2
39. The range for normal blood pressure satisfies four inequalities that can be graphed to find their intersection. Answers should include the following. • Graph the blood pressure as an ordered pair; if the point lies in the shaded region, it is in the normal range. • High systolic pressure is represented by the region to the right of x 140 and high diastolic pressure is represented by the region above y 90.
40. B
41. Sample answer: y 6, y 2, x 5, x 1
42. (3, 8)
43. (6, 5)
44. (8, 5)
45.
46. infinitely many
y
y 2x y 3
y 2x 1 (2, 3)
O
x O
y 12 x 4
©Glencoe/McGraw-Hill
x
6x 3y 9
70
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1 2
48. y x 6
y x 8y 12
(4, 2)
x
O
2x y 6
49. 5
50. 12
51. 8
52. 27
53. 5
54. 8.25
Lesson 3-4 Linear Programming Pages 132–135 1. sometimes
2. Sample answer: y x, y x 5, y 0
3.
4.
y
y
(1, 4) (5, 2)
(3, 1)
( 53 , 1)
(1, 2) O O
vertices: (3, 1), Q , 1R;
vertices: (1, 2), (1, 4), (5, 2); max: f (5, 2) 4, min: f (1, 4) 10
©Glencoe/McGraw-Hill
x
x
5 3
min: f (3, 1) 17; no maximum
71
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6.
y
y (7, 8.5)
(1, 3)
(2, 6)
(6, 3)
(0, 1) O
(10, 1)
x
x
O (2, 0)
vertices: (0, 1), (1, 3), (6, 3), (10, 1); max: f (10, 1) 31, min: f (0, 1) 1
(7, 5)
vertices: (2, 0), (2, 6), (7, 8.5), (7, 5); max: f(7, 8.5) 81.5, min: f(2, 0) 16 7.
8.
y
y
(2, 4)
(1, 2)
(2, 3)
(4, 1)
x
O
x
O
(3, 1)
(2, 3) (2, 3)
(3, 2)
vertices: (3, 1), (1, 2), (2, 3), (3,2); max: f(3, 2) 5, min: f(1, 2) 3
vertices: (2, 4), (2, 3), (2, 3), (4, 1); max: f(2, 3) 5; min: f (2, 4) 6 9. c 0, l 0, c 3l 56, 4c 2l 104
ᐉ 28
Leather Tote Bags
10.
24 20
(0, 18 23 )
16 (20, 12)
12 8 4 (0, 0) 0
4
(26, 0) 8
12 16 20 24 Canvas Tote Bags
11. (0, 0), (26, 0), (20, 12), a0, 18 b
12. f(c, l) 20c 35l
13. 20 canvas tote bags and 12 leather tote bags
14. $820
2 3
©Glencoe/McGraw-Hill
72
Algebra 2
28 c
Chapter 3
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16.
y
y
(3, 5)
(6, 13)
x
O
(0, 4) (3, 4) (0, 1)
vertices: (0, 4), (3, 5), (3, 4); max: f (3, 4) 7, min: f (3, 5) 2
(6, 1)
x
O
vertices: (0, 1), (6, 1), (6, 13); max: f (6, 13) 19; min: f (0, 1) 1 17.
18.
y
y
(5, 8) (2, 3)
(4, 4)
(2, 1)
(4, 1)
(1, 4)
x
O (1, 2)
(5, 2)
x
O
vertices: (2, 1), (2, 3), (4, 4), (4, 1); max: f (4, 4) 16; min: f (2, 1) 5
vertices: (1, 4), (5, 8), (5, 2), (1, 2); max: f (5, 2) 11, min: f (1, 4) 5 19.
y
y
20.
(3, 5)
12 8
(3, 1)
O
4
x 4
O
(6, 12) (2, 8) (2, 2) 4
8
x
4
vertices: (3, 1), (3, 5); min: f (3, 1) 9; no maximum
©Glencoe/McGraw-Hill
8
(6, 6)
vertices: (2, 2), (2, 8), (6, 12), (6, 6); max: f(6, 12) 30, min: f(6, 6) 24 73
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22.
y
(0, 2) O (0, 0)
y (0, 7) (4, 3)
(2, 1)
x (3, 0) (0, 0) O
vertices: (0, 0), (0, 2), (2, 1), (3, 0); max: f (0, 2) 6; min: f (3, 0) 12 23.
(2, 0) x
vertices: (0, 0), (0, 7), (4, 3), (2, 0); max: f (4, 3) 14; min: f (0, 7) 14 24.
y
y (8, 6) (0, 4)
(3, 0)
x
O
(4, 0)
(0, 3)
vertices: (3, 0), (0, 3); min: f (0, 3) 12; no maximum 25.
vertices: (0, 4), (4, 0), (8, 6); max: f (4, 0) 4; min: f (0, 4) 8 26.
y
(0, 2)
x
O
y
(0, 2)
(4, 3)
(4, 3) (2, 0)
x
O
O
x
( 73 , 13 ) vertices: (0, 2), (4, 3), (2, 0); max: f (4, 3) 13, min: f (2, 0) 2
vertices: (0, 2), (4, 3), 7 1 a , b; max: f (4, 3) 25, 3
3
min: f (0, 2) 6
©Glencoe/McGraw-Hill
74
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28.
y
y
(2, 5) (2, 2) (3, 0)
(4, 1)
(0, 1)
x
x
O
(0, 0) O
vertices: (0, 0), (0, 1), (2, 2), (4, 1), (5, 0); max: f (5, 0) 15, min: f (0, 1) 5
vertices: (2, 5), (3, 0); min: f (3, 0) 3, no maximum 29.
y
30a. Sample answer: f (x, y) 2x y 30b. Sample answer: f (x, y) 3y 2x 30c. Sample answer: f (x, y) x y 30d. Sample answer: f (x, y) x 3y 30e. Sample answer: f (x, y) x 2y
(4, 4) (2, 3)
(2, 1) O
(5, 0)
(5, 3) (4, 1)
x
vertices: (2, 1), (2, 3), (4, 1), (4, 4), (5, 3); max: f(4, 1) 0, min: f (4, 4) 12 31. g 0, c 0, 1.5g 2c
85, g 0.5c 40
Graphing Calculators
32.
50
g
40 30 (0, 42.5) 20 10 0
(0, 0)
(30, 20) (40, 0)
10 20 30 40 CAS Calculators
c 50
33. (0, 0), (0, 42.5), (30, 20), (40, 0)
34. f (g, c) 50g 65c
35. 30 graphing calculators, 20 CAS calculators
36. $2800
37. See students’ work.
38. c 0, s 0, c s 4500, 4c 5s 20,000
©Glencoe/McGraw-Hill
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39. (0, 0), (0, 4000), (2500, 2000), (4000, 0)
40. 2500 acres of corn, 2000 acres soybeans; $125,000
S 4000 3000 2000
(0, 4000)
(2500, 2000)
1000 (0, 0) 0
(4500, 0) 2000
4000
c
41. 4000 acres corn, 0 acres soybeans; $130,500
42. 3 chocolate chip, 9 peanut butter
43. There are many variables in scheduling tasks. Linear programming can help make sure that all the requirements are met. Answers should include the following. • Let x the number of buoy replacements and let y the number of buoy repairs. Then, x 0, y 0, x 8 and x 2.5y 24. The captain would want to maximize the number of buoys that a crew could repair and replace, so f (x, y) x y. • Graph the inequalities and find the vertices of the intersection of the graphs. The coordinate (0, 24) maximizes the function. So the crew can service the maximum number of buoys if they replace 0 and repair 24 buoys.
44. A
©Glencoe/McGraw-Hill
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46.
45. C
y
2y x 4
x
O
yx4
47.
48. (5, 8) y
3x 2y 6
x
O
y 3x 1 2
49. (2, 3) 51. c average cost each year; 15c 3479 7489 53. Additive Inverse 55. Multiplicative Inverse 57. 9 59. 16 61. 8
50. (5, 1) 52. about $267 per year 54. 56. 58. 60. 62.
Associative Property ( ) Distributive Property 5 3 4
Chapter 3 Practice Quiz 2 Page 135 1.
y
2.
yx0
y yx3
x
O
y 3x 4
yx4
©Glencoe/McGraw-Hill
O
77
x
Algebra 2
Chapter 3
PQ245-6457F-P03[056-080].qxd
3.
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y
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4.
4x y 16
y (1, 6) (0, 4)
x 3y 15
(0, 0)
x
O
O
(3, 0)
x
vertices: (0, 0), (0, 4), (1, 6), (3, 0); max: f (1, 6) 8, min: f (0, 0) 0 5.
y (5, 6)
(1, 3) (5, 1)
x
O (1, 3)
vertices: (1, 3), (1, 3), (5, 6), (5, 1); max: f (5, 1) 17, min: f (1, 3) 13
Lesson 3-5 Solving Systems of Equations in Three Variables Pages 142–144 1. You can use elimination or substitution to eliminate one of the variables. Then you can solve two equations in two variables.
2. No; the first two equations do represent the same plane, however they do not intersect the third plane, so there is no solution of this system.
3. Sample answer: x y z 4, 2x y z 9, x 2y z 5; 3 5 2 4, 2(3) 5 2 9, 3 2(5) 2 5
4. (6, 3, 4)
©Glencoe/McGraw-Hill
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5. (1, 3, 7)
6. infinitely many
7. (5, 2, 1)
8. no solution 10. 6c 3s r 42,
9. (4, 0, 8)
1 2
c s r 13 , r 2s 1 2
12. (3, 4, 7)
11. 4 lb chicken, 3 lb sausage, 6 lb rice 13. (2, 1, 5)
14. (2, 3, 6)
15. (4, 0, 1)
16. no solution
17. (1, 5, 7)
18. (1, 2, 1)
19. infinitely many
20. a , , b
21. a , , b
22. (8, 3, 6)
23. (5, 9, 4)
24. 3, 12, 5
25. 8, 1, 3
26. 1-$100, 3-$50, and 6-$20 checks
27. enchilada, $2.50; taco, $1.95; burrito, $2.65
28. $7.80
29. x y z 355, x 2y 3z 646, y z 27
30. 88 3-point goals, 115 2-point goals, 152 1-point free throws
1 3
1 3 9 2 2 2
1 1 2 4
4 3 4 2 x 3
1 3 1 x 3
31. a , b , c 3; y
©Glencoe/McGraw-Hill
32. You can write a system of three equations in three variables to find the number of each type of medal. Answers should include the following. • You can substitute b 6 for g and b 8 for s in the equation g s b 97. This equation is now in terms of b. Once you find b, you can substitute again to find g and s. The U.S. Olympians won 39 gold medals, 25 silver medals, and 33 bronze medals.
3
79
Algebra 2
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• Another situation involving three variables is winning times of the first, second, and third place finishers of a race. 33. D
34. A
35. 120 units of notebook paper and 80 units of newsprint
36.
y yx 2
x
O
y 7 2x
37.
38.
y
y
3x y 3
3x y 1
O O
x
x
2y x 4
4y 2x 4
39. Sample answer using (7, 15) and (14, 22): y x 8
40. Sample answer: about 47¢
41. x 3y
42. 2z 8
43. 9s 4t
44. 18a 16b
©Glencoe/McGraw-Hill
80
Algebra 2
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Chapter 4 Matrices Lesson 4-1 Introduction to Matrices Pages 156–158 2. Sample answers: row matrix, [1 2 3], 1 3;
1. The matrices must have the same dimensions and each element of one matrix must be equal to the corresponding element of the other matrix.
1 2
column matrix, B R, 2 1; 1 2 R, 3 4
square matrix, B
0 0 R, 0 0
zero matrix, B 3. Corresponding elements are elements in the same row and column positions.
4. 1 5
5. 3 4
6. (5, 6)
7. (3, 3)
8.
Fri High Low
B
10. 2 3
11. 3 1
12. 4 3
13. 3 3
14. 2 5
15. 3 2
16. (2.5, 1, 3)
17. a3, b
18. (5, 3)
19. (3, 5, 6)
20. (2, 5)
21. (4, 3)
22. (1.5, 3)
23. (14, 15)
24. (2, 7)
25. (5, 3, 2)
26. Child Senior
81
Sat Sun Mon Tue
Evening Matinee Twilight Adult
©Glencoe/McGraw-Hill
22
88 88 90 86 85 R 54 54 56 53 52
9. 2 5
1 3
2 2;
7.50 C 4.50 5.50
5.50 4.50 5.50
3.75 3.75 S 3.75
Algebra 2
Chapter 4
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27. 3 3
28.
Cost
Catalina Grill Oyster Club Casa di Pasta Mason’s Steakhouse
29. Sample answer: Mason’s Steakhouse; it was given the highest rating possible for service and atmosphere, location was given one of the highest ratings, and it is moderately priced.
30.
31.
32.
Double Suite
Weekend
60 70 75 B R 79 89 95
Atmosphere Location
* ** *** ****
* * *** ****
* ** ¥ *** ***
Weekday Weekend Single
Single Double Suite
Weekday
** ≥ *** **** **
Service
1 2 4 G 7 11 16 22
60 79 C 70 89 S 75 95
3 5 8 12 17 23 30
6 9 13 18 24 31 39
10 14 19 25 32 40 49
15 20 26 33 41 50 60
21 27 34 42W 51 61 72
33. row 6, column 9
34. Matrices are used to organize information so it can be read and compared more easily. Answers should include the following. • If you want the least expensive vehicle, the compact SUV has the best price; the large SUV has the most horsepower, towing capacity and cargo space, and the standard SUV has the best fuel economy. • Sample answer: Matrices are used to report stock prices in the newspaper.
35. B
36. C
37. (7, 5, 4)
38. (7, 3, 9)
©Glencoe/McGraw-Hill
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39. a3, 5, 11b 4 3
40.
y y 2x 15
yx 2
y3
x
O
vertices: (1, 3), (6, 3), a ,
13 19 b; 3 3
max: f a ,
13 19 b 3 3
83 , 3
min: f (1, 3)11 42.
41.
3
12
y
vertices: (3, 1), a , 15 2
5 b, 2
vertices: (2, 1), (6, 3); min: f(2, 1) 1, no maximum
3 17 a , b; 2 2 . 15 5 max: f a , b 35, 2 2 3 17 min: f a , b 1 2 2
43.
44. step function
Cost ($)
6 5 4 3 2 1 0
1
©Glencoe/McGraw-Hill
2 3 4 Hours
5
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Algebra 2
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45. $4.50
46. 2
47. 2
48. 0
49. 20
50. 3
51. 10
52. 6.2
53. 18
54. 17
55. 3
56. 75
57.
3 2
Lesson 4-2
Operations with Matrices Pages 163–166
1. They must have the same dimensions.
2. Sample answer: [3 1], [3 1]
4 4 3. C 4 4 S 4 4
4. impossible 18 3 15 6 R 21 9 6 24
1 10 R 7 5
6. B
22 8 R 3 24
8. B
5. B
10 6 R 1 7
7. B
21 29 R 12 22
3 30 R 26 11
9. B
16,763 14,620 11. Males E14,486 9041 5234 16,439 14,545 Females E12,679 7931 5450
©Glencoe/McGraw-Hill
10. B 549,499 477,960 455,305U, 321,416 83,411
1,006,372 883,123 12. E 795,785U 579,002 216,646
456,873 405,163 340,480U 257,586 133,235 84
Algebra 2
Chapter 4
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10 14. C 4 S 5
13. No; many schools offer the same sport for males and females, so those schools would be counted twice.
15 0 4 R 0 13 5
15. impossible
16. B
4 8 2 17. C 6 10 16 S 14 12 4
18. [15 29 65 2]
13 19. C 3 S 23
1.8 9.08 20. C 3.18 31.04 S 10.41 56.56 4 15
1.5 3 R 21. B 4.5 9 1
23. C
22. C
52
3
9
2 10 3
2 1 3
1 2 2
3 2
2
S
13 10 7S 24. C 4 7 5
S
2 1 25. C 4 1 S 7 4
0 16 26. C 8 20 S 28 4
38 4 27. C 32 6 S 18 42
12 13 28. C 3 8 S 13 37
2 29. D 1
4
2 3
120 97 64 75 30. Friday: C 80 59 36 60 S, 72 84 29 48
5T
6 1
112 87 56 74 Saturday: C 84 65 39 70 S 88 98 43 60
©Glencoe/McGraw-Hill
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Algebra 2
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232 184 120 149 31. C 164 124 75 130 S 160 182 72 108
8 10 8 1 6 3 10 S 32. C 4 16 14 14 12
245 228 33. E319U 227 117
15 41 34. E35U 27 51
35. 1996, floods; 1997, floods; 1998, floods; 1999, tornadoes; 2000, lightning
36. Residents: Child Adult Before 6 3.00 4.50 B R After 6 2.00 3.50 Nonresidents: Child Adult Before 6 4.50 6.75 B R After 6 3.00 5.25
1.50 2.25 37. B R 1.00 1.75
38. Before 6:00: Child Adult Residents 3.00 4.50 B R Nonresidents 4.50 6.75 After 6:00: Child Adult Residents 2.00 3.50 B R Nonresidents 3.00 5.25
1.00 1.00 39. B R 1.50 1.50
40. 0.5 0.75 3 1 1.5 6 R B R 2B 2 8 0.2 1 4 0.1 42. D
41. You can use matrices to track dietary requirements and add them to find the total each day or each week. Answers should include the following. 566 18 7 • Breakfast C 482 12 17 S , 530 10 11
©Glencoe/McGraw-Hill
86
Algebra 2
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785 22 19 Lunch C 622 23 20 S, 710 26 12 1257 40 26 Dinner C 987 32 45 S 1380 29 38 • Add the three matrices: 2608 80 52 £ 2091 67 82 § . 2620 65 61 43. A
44. 2 2
45. 1 4
46. 2 4
47. 3 3
48. 3 2
49. 4 3
50. (3, 4, 0)
51. (5, 3, 7)
52. a , 6, b
53. (2, 5)
54. (3, 1)
55. (6, 1)
56. 0.30p 0.15s 6
57.
58. No, it would cost $6.30.
59. Multiplicative Inverse
60. Associative Prop. ()
61. Distributive Property
62. Commutative Prop. ()
©Glencoe/McGraw-Hill
1 4
87
1 6
Algebra 2
Chapter 4
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Lesson 4-3 Multiplying Matrices Pages 171–174 1. Sample answer: 1 2 7 8 C3 4S B R 9 10 5 6
2. Never; the inner dimensions will never be equal.
3. The Right Distributive Property says that 1A B2C AC BC, but AC BC CA CB since the Commutative Property does not hold for matrix multiplication in most cases.
4. 3 2
5. undefined 15 5 20 7. B R 24 8 32
6. [19 15] 8. not possible
24 R 41
9. B
10. yes A(BC) 2 1 4 1 3 2 B R ¢B RB R≤ 3 5 8 0 1 2
13 6 2 1 RB R 3 5 24 16
B
50 28 R 81 62 (AB)C
B
2 1 4 1 3 2 ¢B R B R≤ B R 3 5 8 0 1 2
16 2 3 2 RB R 1 2 28 3
B
50 28 R 81 62
B
©Glencoe/McGraw-Hill
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Algebra 2
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11. [45
55
350 280 65], C 320 165 S 180 120
12. $74,525
13. 4 2
14. 2 2
15. undefined
16. 1 5
17. undefined
18. 3 5 8 11 20. B R 22 12
19. [6]
39 R 18
21. not possible
22. B
1 25 2 R 23. B 29 1 30
24. not possible
24 16 25. C 32 5 S 48 11
0 64 40 26. C 9 11 11 S 3 39 23
27. yes AC BC 1 2 5 1 B R B R 4 3 2 4
28. yes c (AB) 1 2 5 2 RB R≤ 4 3 4 3
3¢B
5 2 5 1 R B R 4 3 2 4
3B
1 9 21 13 RB R 26 8 26 8
B
13 4 R 8 17
B
39 12 R 24 51 A(cB)
B
20 4 R 52 16 (A B)C
B
¢B
1 2 5 R B 4 3 4
1 2 5 2 R ¢3 B R≤ 4 3 4 3
B
1 2 15 6 R RB 12 9 4 3
2 5 1 R≤ B R 3 2 4
B
5 1 4 0 R RB 2 4 8 6
B
39 12 R B 24 51
20 4 R 52 16
B
©Glencoe/McGraw-Hill
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Algebra 2
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29. no C (A B)
30. no ABC 1 2 5 2 5 1 RB RB R 4 3 4 3 2 4
5 1 1 2 5 2 B R ¢B RB R≤ 2 4 4 3 4 3
B
5 1 4 0 B RB R 2 4 8 6
B
12 6 R 40 24 AC BC 1 2 5 1 B RB R 4 3 2 4
B
5 2 5 1 B RB R 4 3 2 4
B
21 13 1 2 RB R 26 8 4 3
B
31 81 R 58 28
13 4 5 1 RB R 8 17 2 4
73 3 R 6 76 CBA
B
5 1 5 2 1 2 RB RB R 2 4 4 3 4 3
B
1 9 21 13 B RB R 26 8 26 8 20 4 R 52 16
B
290 165 210 31. C 175 240 190 S 110 75 0
22 32. C 25 S 18
14,285 33. C 13,270 S 4295
34. $31,850
35. any two matrices B
72 68 36. D 90 86
a b R c d
e f R g h
B
and
where bg cf, a d,
49 1.00 63 R T, B 56 0.50 62
and e h 96.50 99.50 37. D T 118 117 39. $431
38. Juniors
41. $26,360
42. $1460
©Glencoe/McGraw-Hill
40. $24,900
90
Algebra 2
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43. a 1, b 0, c 0, d 1; the original matrix
44. Sports statistics are often listed in columns and matrices. In this case, you can find the total number of points scored by multiplying the point matrix, which doesn’t change, by the record matrix, which changes for each season. Answers should include the following. • P R [479] • Basketball and wrestling use different point values in scoring.
45. B 12 6 47. B R 3 21 20 2 49. B R 28 12
46. A
51. (5, 9)
52. (2, 5, 7)
53. $2.50; $1.50
54.
55. 8; 16
56. 2; 5
©Glencoe/McGraw-Hill
48. impossible 50. (7, 4)
91
3 ; 2
3
Algebra 2
Chapter 4
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57.
58.
59.
60.
Chapter 4 Practice Quiz 1 Page 174 (5, 1) 120 80 64 75 4. B R, 65 105 77 53
1. (6, 3)
2.
3. (1, 3, 5)
112 79 56 74 R 69 95 82 50
B
3 5 R 3 13
232 159 120 149 R 134 200 159 103
6. B
4 3 R 1 3
8. B
5.
B
7.
B
9.
not possible
©Glencoe/McGraw-Hill
10 20 25 R 0 20 35 15 8 10 R 7 23 16
10. B
92
Algebra 2
Chapter 4
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Lesson 4-4
1.
Transformation reflection rotation translation dilation
p
Shape same same same changes
p
Size same same same same
Transformations with Matrices Pages 178–181
p
3 3 3 R 2 2 2
2. B
Isometry yes yes yes no
3 3 3 R 1 1 1
4. B
3. Sample answer: 4 4 4 B R 1 1 1 5. A¿(4, 3), B¿(5, 6),
6.
C¿(3, 7)
0 5 5 0 R 4 4 0 0
7. B
8. A¿(0, 12), B¿(15, 12), C¿(15, 0), D¿(0, 0) 10. A¿(0, 4), B¿(5, 4), C¿(5, 0), D¿(0, 0)
9. A¿(0, 4), B¿(5, 4), C¿(5, 0), D¿(0, 0)
4 4 4 R 2 2 2
11. B
12. B
13. D¿(3, 6), E¿(2, 3), F¿(10, 4)
14.
0 1.5 2.5 R 2 1.5 0
16. A¿(0, 6), B¿(4.5, 4.5), C¿(7.5, 0)
15. B
©Glencoe/McGraw-Hill
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Algebra 2
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17.
18. B
19. X¿(1, 1), Y¿(4, 2), Z¿(1, 7)
20.
2 5 4 1 R 4 4 1 1
21. B 23.
22. D¿(4, 2), E¿(4, 5), F¿(1, 4), G¿(1, 1)
y D
G O G'
1 2 7 R 1 4 1
24. E¿(6, 2), F¿(8, 9)
E
F x
D' F' 25.
E' 2 4 2 3 R (1) 3 3 5 2
26. B
J(5, 3), K(7, 2), L(4, 1)
2 4 2 3 R 3 3 5 2
B
©Glencoe/McGraw-Hill
94
Algebra 2
Chapter 4
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27.
28. 180 rotation
4 4 4 4 R 4 4 4 4
4 4 4 4 R 4 4 4 4
30. B
29. B
4 4 4 4 R 4 4 4 4
31. B
32. The figures in Exercise 29 and Exercise 30 have the same coordinates, but the figure in Exercise 31 has different coordinates.
33. (1.5, 1.5), (4.5, 1.5), (6, 3.75), (3, 3.75)
34. (3.75, 2.625)
3 35. B R 4
36. (6.5, 6.25)
37. (8, 7), (7, 8), and (8, 7)
38. The object is reflected over the x-axis, then translated 6 units to the right.
39. Multiply the coordinates
40. No; since the translation does not change the y-coordinate, it does not matter whether you do the translation or the reflection over the x-axis first. However, if the translation did change the y-coordinate, then order would be important.
1 0 R, 0 1
by B
then add the
6 0
result to B R.
41. (17, 2), (23, 2)
©Glencoe/McGraw-Hill
42. There is no single matrix to achieve this. However, you could reflect the object over the y-axis and then translate it 2(3) or 6 units to the right.
95
Algebra 2
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43. Transformations are used in computer graphics to create special effects. You can simulate the movement of an object, like in space, which you wouldn’t be able to recreate otherwise. Answers should include the following. • A figure with points (a, b), (c, d), (e, f ), (g, h), and (i, j) could be written in a 2 5 a c e g i R b d f h j
matrix B
44. B
and
multiplied on the left by the 2 2 rotation matrix. • The object would get smaller and appear to be moving away from you. 45. A
46. 2 2
47. undefined
48. 2 5
11 24 7 8S 49. C 18 13 33 8 21 51.
20 10 24 50. C 31 46 9 S 10 3 7 52.
D {all real numbers}, R {all real numbers}; yes
D 53, 4, 56, R 54, 5, 66; yes
©Glencoe/McGraw-Hill
96
Algebra 2
Chapter 4
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54. 0 x 0 4
53.
D 5x 0 x 06, R 5all real numbers6; no
55. 0 x 0 2.8
56. 0 x 1 0 2
57. 0 x 1 0 1
2 3
58. 513 mi
59. 6
60. 5
61. 28
62.
10 3
64.
5 3
63.
9 4
Lesson 4-5 Determinants Pages 185–188 2 1 R 8 4
1. Sample answer: B
2. Khalid; the value of the determinant is the difference of the products of the diagonals.
3. It is not a square matrix.
4. Sample answer: 3 1 4 3 B R, B R 6 5 1 3 2 3 5 6. † 0 1 4 † 9 7 2
5. Cross out the column and row that contains 6. The minor is the remaining 2 2 matrix.
2 `
1 4 0 4 0 1 ` 3` ` 5` ` 7 2 9 2 9 7
2(2 28) 3(0 36) 5(0 (9)) 2(30) 3(36) 5(9) 60 108 45 213 ©Glencoe/McGraw-Hill
97
Algebra 2
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2 3 5 2 3 † 0 1 4 † 0 1 9 7 2 9 7 4 108
0
45 56 0 2 3 5 2 3 † 0 1 4 † 0 1 9 7 2 9 7 4 108 0 (45) (56) 0 213 7. 38
8. 0 10. 28
9. 40 11. 43
12. 0
13. 45
14. 26 units2
15. 20
16. 22
17. 22
18. 0
19. 29
20. 14
21. 63
22. 6
23. 32
24. 37
25. 32
26. 11.3
27. 58
28. 0
29. 62
30. 60
31. 172
32. 265
33. 22
34. 21
35. 5
36. 49
37. 141
38. 123
39. 6
40.
41. 14.5 units2
42. 12
43. about 26 ft2
44. 2875 mi2
©Glencoe/McGraw-Hill
98
5 , 3
1
Algebra 2
Chapter 4
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1 1 1 45. Sample answer: † 1 1 1 † 1 1 1
46. Multiply each member in the top row by its minor and position sign. In this case the minor is a 3 3 matrix. Evaluate the 3 3 matrix using expansion by minors again.
47. If you know the coordinates of the vertices of a triangle, you can use a determinant to find the area. This is convenient since you don’t need to know any additional information such as the measure of the angles. Answers should include the following. • You could place a coordinate grid over a map of the Bermuda Triangle with one vertex at the origin. By using the scale of the map, you could determine coordinates to represent the other two vertices and use a determinant to estimate the area. • The determinant method is advantageous since you don’t need to physically measure the lengths of each side or the measure of the angles between the vertices.
48. C
49. C
50. 63.25
51. 36.9
52. 25.21
53. 493
54. 0 2 1 2 56. B R 1 2 3
55. 3252 ©Glencoe/McGraw-Hill
99
Algebra 2
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57. A¿(5, 2.5), B¿(2.5, 5), C¿(5, 7.5)
58.
2 26 R 9 12 62. undefined 7 69 64. B R 5 16 66. y x 2 60. B
59. [4] 61. undefined 8]
63. [14
65. 138,435 ft 4 3
67. y x
68. y 2x 1
1 2
69. y x 5
70. (0, 3)
71. (1, 9)
72. (2, 1)
73. (1, 1)
74. (2, 5)
75. (4, 7)
Lesson 4-6 Cramer’s Rule Pages 192–194 1. The determinant of the coefficient matrix cannot be zero.
2. Sample answer: 2x y 5 and 6x 3y 8
3. 3x 5y 6, 4x 2y 30
4. (5, 1)
5. (0.75, 0.5)
6. (6, 8)
7. no solution
8. a5, , b 2 3
9. a6, , 2b 1 2
©Glencoe/McGraw-Hill
1 2
10. s d 4000, 0.065s 0.08d 297.50
100
Algebra 2
Chapter 4
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11. savings account, $1500; certificate of deposit, $2500
12. (2, 1)
13. (12, 4)
14. (3, 5)
15. (6, 3)
16. (2.3, 1.4)
17. (0.75, 3)
18. (0.75, 0.625)
19. (8.5625, 19.0625)
20. a , 1b
21. (4, 8)
22. (3, 10)
23. a , b
24. (1.5, 2)
25. (3, 4)
26. (1, 3, 4)
27. (2, 1, 3)
28. a ,
2 3
2 5 3 6
29. a
141 , 29
11 39 , 19 19
102 244 b , 29 29
b 14 19
30. (11, 17, 14)
31. a
32. r s 8, 7r 5s 50
33. race car, 5 plays; snowboard, 3 plays
34. 8s 13c 604.79,
35. silk, $34.99; cotton, $24.99
36. p r c 5, 2r p 0, 3.2p 2.4r 4c 16.8
37. peanuts, 2 lb; raisins, 1 lb; pretzels, 2 lb
38. If the determinant is zero, there is no unique solution to the system. There is either no solution or there are infinitely many solutions. Sample answer: 2x y 4 and 4x 2y 8 has a det 0; there are infinitely many solutions of this system. 2x y 4 and 4x 2y 10 has a det 0; there are no solutions of this system.
155 143 673 b , , 28 70 140
©Glencoe/McGraw-Hill
1 2
5 s 14c 542.30
101
Algebra 2
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39. Cramer’s Rule is a formula for the variables x and y where (x, y) is a solution for a system of equations. Answers should include the following. • Cramer’s Rule uses determinants composed of the coefficients and constants in a system of linear equations to solve the system. • Cramer’s Rule is convenient when coefficients are large or involve fractions or decimals. Finding the value of the determinant is sometimes easier than trying to find a greatest common factor if you are solving by using elimination or substituting complicated numbers.
40. B
41. 111, 69
42. 16
43. 40
44. 53
1 1 1 R 3 3 3
45. B
46. A¿(1, 5), B¿(2, 2), C¿(1, 1)
47.
48.
y y 3x 5 (2, 1)
O
x
y 2x 5
(2, 1)
©Glencoe/McGraw-Hill
102
Algebra 2
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49.
50.
y
y 2x 4y 12
xy 7
x
O (4, 3)
x 2y 10
x O
1 x y 1 2
(4, 3)
no solution
51. c 10h 35 72 9 53. B R 66 23
52. [4 32] 21 54. B R 43
Chapter 4 Practice Quiz 2 Page 194 1 4 1 2 R 2 1 4 1
1. B
2. A¿(1, 2), B¿(4, 1), C¿(1, 4), D¿(2, 1)
3.
4. 22
5. 58
6. 105
7. 26
8. (1, 2)
9. (4, 5)
©Glencoe/McGraw-Hill
10. (1, 2, 1)
103
Algebra 2
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Lesson 4-7 1 0 1. D 0 0
0 1 0 0
0 0 1 0
Identity and Inverse Matrices Pages 198–201
0 0 T 0 1
2. Exchange the values for a and d in the first diagonal in the matrix. Multiply the values for b and c by 1 in the second diagonal in the matrix. Find the determinant of the original matrix. Multiply the negative reciprocal of the determinant by the matrix with the above mentioned changes. 3 3 R 3 3
3. Sample answer: B
4. no
5. yes
6. B
7. no inverse exists
8.
2 5 R 3 8 4 1 B 27 7
10. yes
9. See students’ work. 11. yes
12. no
13. no
14. yes
15. yes
16. true
17. true
18. true 1 1 0 R 20. B 5 0 5
19. false
1 3
21. no inverse exists 23.
1 1 B 7 4
25.
1 6 B 4 2
22. B
1 R 3
1 2 R 2 1
24. no inverse exists
7 R 3
©Glencoe/McGraw-Hill
1 R 5
26.
104
7 1 B 34 2
3 R 4
Algebra 2
Chapter 4
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6 1 B 12 5
27.
29.
0 R 2
1 1 B 32 6
5 R 2
3 4
31. 10 C
1 5
5 8
3 10
28. no inverse exists
30. 4 C
S
1 4
3 4
1 6
1 2
S
32a. no 32b. Sample answer: y
C A B x
O A'
A''
B'' C''
34. B
33a. yes 33b. Sample answer:
0 4 4 8 R 0 4 12 8
35. B
37. dilation by a scale factor of
B' C'
0 2 2 4 R 0 2 6 4
36. dilation by a scale factor of 2
1 2
1
38. B D
1 2
0
0
1 2
T;
the graph of the
inverse transformation is the original figure.
©Glencoe/McGraw-Hill
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39. MEET_IN_THE_LIBRARY
40. AT_SIX_THIRTY
41. BRING_YOUR_BOOK
42. See students’ work.
43. a 1, d 1, b c 0
44. A matrix can be used to code a message. The key to the message is the inverse of the matrix. Answers should include the following. • The inverse matrix undoes the work of the matrix. So if you multiply a numeric message by a matrix it changes the message. When you multiply the changed message by the inverse matrix, the result is the original numeric message. • You must consider the dimensions of the coding matrix so that you can write the numeric message in a matrix with dimensions that can be multiplied by the coding matrix. 46. D
45. A 5 9 R 47. B 6 11 3 5
1 5
1 5
2 5
49. C
48. no inverse exists
S
1
1
1 3
1 3
2 3
0V
51. F
7 3
8 3
1 3 2 5 16
1 2 2 1 8
1 4
0
5 32
3 16
52. F
1 3
53. (2, 4)
©Glencoe/McGraw-Hill
2 5
50. C
3 5
S 1 16
1 V 4 1 32
54. (0, 7)
106
Algebra 2
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55. (5, 4, 1)
56. 52
57. 14
58. 0
59. 1
60. 3
61. 5
62.
63.
1 3 3 8
5 2
64.
65. 7.82 tons/in2
66. 27 1 2
1 2
67. 5
68.
69. 3
70. 296
71. 300
72. 1
73. 2
74. 6
75. 4
76. 27
77. 34
Lesson 4-8
Using Matrices to Solve Systems of Equations Pages 205–207
1. 2r 3s 4, r 4s 2
2. Sample answer: x 3y 8 and 2x 6y 16
3. Tommy; a 2 1 matrix cannot be multiplied by a 2 2 matrix.
4. B
1 1 x 3 RB RB R 1 3 y 5
3 5 2 a 6. C 4 7 1S CbS 2 0 1 c
2 3 g 8 RB RB R 5. B 4 7 h 5
7. (5, 2)
9 C 3S 12 8. (1.5, 4)
9. (3, 5)
10. (1, 1.75)
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3 1 x 0 RB RB R 1 2 y 21
11. h 1, c 12
12. B
5 6 a 47 RB RB R 3 2 b 17
4 7 x 2 RB RB R 3 5 y 9
13. B
14. B
3 7 m 43 RB RB R 15. B 6 5 n 10
2 3 5 a 16. C 7 0 3S CbS 3 6 1 c 1 C 7S 5
3 5 2 x 17. C 1 7 3 S C y S 4 0 3 z
1 1 0 x 18. C 2 5 6 S C y S 9 10 1 z
9 C 11 S 1
8 C 27 S 54
3 5 6 r 19. C 11 12 16 S C s S 5 8 3 t
20. (5, 2)
21 C 15 S 7 21. (3, 4)
22. (2, 3)
23. (6, 1)
24. a , 3b
25. a , 4b
26. (2, 3)
27. (2, 2)
28. (7, 3)
29. (0, 9)
30. a1, b
1 2
1 3
9 2
31. a , b 3 1 2 3
32. 27 h of flight instruction and 23 h in the simulator 34. 80 mL of the 60% solution, and 120 mL of the 40% solution
33. 2010
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35. The solution set is the empty set or infinite solutions.
36. The food and territory that two species of birds require form a system of equations. Any independent system of equations can be solved using a matrix equation. Answers should include the following. • Let a represent the number of nesting pairs of Species A and let b represent the number of nesting pairs of Species B. Then, 140a 120b 20,000 and 500a 400b 69,000. a • B R b 400 1 B 4000 500
120 20,000 RB R; 140 69,000
a 70 and b 85, so the area can support 70 pairs of Species A and 85 pairs of Species B. 37. D
38. 17 small, 24 medium, 11 large
39. (6, 2, 5)
40. (1, 3, 2) 1
1 2
1
42. C
3 4
41. (0, 1, 3) 4 5 R 7 9 45. (4, 2) 43. B
S
44. no inverse exists 46. (4.27, 5.11)
47. (6, 8)
48. about 114.3 ft
49. {4, 10}
50. {5, 1}
51. {2, 7}
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Chapter 5 Polynomials Lesson 5-1 Monomials Pages 226–228 1. Sample answer: (2x 2)3 8x 6 since (2x 2)3 2x 2 2x 2 2x 2 2x x 2x x 2x x 8x 6
2. Sometimes; in general, x y x z x yz, so x y x z x yz when y z yz, such as when y 2 and z 2.
3. Alejandra; when Kyle used the Power of a Product Property in his first step, he forgot to put an exponent of 2 on a. Also, in his second step,
4. x 10
1 4
(2)2 should be , not 4. 5. 16b 4
6. 1
7. 6y 2
8.
ab 4 9
9. 9p 2q 3
10.
1 w z
9 c d
12.
1 4x 6
11.
2 2
12 6
13. 4.21 105
14. 8.62 104
15. 3.762 103
16. 5 100
17. about 1.28 s
18. a 8
19. b 4
20. n16
21. z10
22. 16x 4
23. 8c 3
24. an
25. y 3z 2
26.
27. 21b5c 3
28. ab
29. 24r 7s 5
30. 24x 4y 4
31. 90a4b4
32.
©Glencoe/McGraw-Hill
28x 4 y2
1 4y 4
110
Algebra 2
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a 2c 2 3b 4 m 4n 9 3
35.
34.
cd 4 5
36.
a4 16b 4
37.
8y 3 x6
38.
1 x y
39.
1 3 6 v w
40.
a 4b 2 2
41.
2x 3y 2 5z 7
42. 6
2 2
43. 7
44. 4.623 102
45. 4.32 104
46. 1.843 104
47. 6.81 103
48. 5.0202 108
49. 6.754 108
50. 1.245 1010
51. 6.02 105
52. 4.5 102
53. 6.2 1010
54. 4.225 109
55. 1.681 107
56. 6.08 109
57. 2 107 m
58. 1.67 1025
59. about 330,000 times
60. 10010 (102)10 or 1020, and 10100 1020, so 10100 10010.
61. Definition of an exponent
62. (ab)m m factors 6447448 ab ab p ab m factors m factors 64748 64748 aapabbpb a mb m
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63. Economics often involves large amounts of money. Answers should include the following. • The national debt in 2000 was five trillion, six hundred seventy-four billion, two hundred million or 5.6742 1012 dollars. The population was two hundred eighty-one million or 2.81 108. • Divide the national debt by the population. 5.6742 1012 2.81 108
64. D
$2.0193 104 or about $20,193 per person. 66. (1, 2) 2 5 68. c d 1 2
65. B 67. (3, 3) 69. C
1 2
3 2 S
70. 6
1 2
71. 7
72. (2, 3, 1)
73. (2, 0, 4)
74.
Median Age (yr)
Median Age of Vehicles y 8 7 6 5 4 0 0
10 20 30 Years Since 1970
75. Sample answer using (0, 4.9) and (28, 8.3): y 0.12x 4.9
76. Sample answer: 9.7 yr
77. 7
78. 3
79. 2x 2y
80. 3x 3z
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Algebra 2
x
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81. 4x 8
82. 6x 10
83. 5x 10y
84. 3y 15
Lesson 5-2 Polynomials Page 231–232 1. Sample answer: x 5 x 4 x 3 x
3. x
x
x 2
x
2. 4
x 2
x
4. yes, 1 2
x
x
x
x
x
x
2
5. yes, 3
6. no
7. 10a 2b
8. 3x 2 7x 8 10. 10p 3q 2 6p 5q 3 8p3q 5
9. 6xy 18x 11. y 2 3y 70
12. x 2 9x 18
13. 4z 2 1
14. 4m 2 12mn 9n 2
15. 7.5x 2 12.5x ft 2
16. yes, 2
17. yes, 3
18. no
19. no
20. yes, 6
21. yes, 7
22. 4x 2 3x 7
23. 3y 3y 2
24. r 2 r 6
25. 10m 2 5m 15
26. 4x 2 3xy 6y 2
27. 7x 2 8xy 4y 2
28. 4b 2c 4bdz
29. 12a 3 4ab
30. 15a3b3 30a4b3 15a5b6
31. 6x 2y 4 8x 2y 2 4xy 5
32. 6x 3 9x 2y 12x 3y 2
33. 2a4 3a3b 4a4b4
34. 46.75 0.018x
35. 0.001x 2 5x 500
36. $5327.50
37. p 2 2p 24
38. a 2 9a 18
39. b 2 25
40. 36 z 2
41. 6x 2 34x 48
42. 8y 2 16y 42
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43. a 6 b 2
44. 2m 4 7m 2 15
45. x 2 6xy 9y 2
46. 1 8c 16c 2
47. d 2 2
1 d4
48. xy 3 y
1 x
49. 27b 3 27b 2c 9bc 2 c 3
50. x 3 y 3
51. 9c 2 12cd 7d 2
52. 18x 2 27x 10
53. R 2 2RW W 2
54. 14; Sample answer: (x 8 1)(x 6 1) x 14 x 8 x 6 1
55. The expression for how much an amount of money will grow to is a polynomial in terms of the interest rate. Answers should include the following. • If an amount A grows by r percent for n years, the amount will be A(1 r )n after n years. When this expression is expanded, a polynomial results. • 13,872(1 r )3, 13,872r 3 41,616r 2 41,616r 13,872 • Evaluate one of the expressions when r 0.04. For example, 13,872(1 r)3 13,872(1.04)3 or $15,604.11 to the nearest cent. The value given in the table is $15,604 rounded to the nearest dollar.
56. D
57. B
58. 64d 6
59. 20r 3t 4
60.
61.
b2 4a 2
©Glencoe/McGraw-Hill
xz 2 y2
62. (1, 4)
114
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63.
Page 115 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-05:
64.
y
y
x y 2 x
O
x
O
y 13 x 2
65.
66. x 2
y
2x y 1
x
O
68. xy 2
67. 2y 3 69. 3a 2
Lesson 5-3 Dividing Polynomials Pages 236–238 1. Sample answer: (x 2 x 5) (x 1)
2. The divisor contains an x 2 term.
3. Jorge; Shelly is subtracting in the columns instead of adding.
4. 6y 3 2x
5. 5b 4 7a
6. x 12
7. 3a3 9a2 7a 6
8. z 4 2z 3 4z 2 5z 10
9. x 2 xy y 2
10. x 2 11x 34
11. b3 b 1
12. 2y 5
13. 3b 5
14. B
15. 3ab 6b 2
16. 5y
17. 2c 2 3d 4d 2
18. 4n 2 3mn 5m
©Glencoe/McGraw-Hill
115
6y 2 x
60 x2
3xy 2
Algebra 2
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2 b
19. 2y 2 4yz 8y 3z 4
20. a2b a
21. b 2 10b
22. x 15
23. n 2 2n 3
24. 2c 2 c 5
25. x 3 5x 2 11x 22
39 x2
26. 6w 4 12w 3 24w 2 30w 60
27. x 2
28. x 2 3x 9
29. y 2 y 1
30. m 2 7
31. a3 6a 2 7a 7
3 a1
32. 2m 3 m 2 3m 1
5 m3
34. 3c 4 c 3 2c 2 4c
33. x 4 3x 3 2x 2 6x 19
6 c2
56 x3
9
13 c2 4 b1
35. g 5
36. 2b 2 b 1
37. t 4 2t 3 4t 2 5t 10
38. y 4 2y 3 4y 2 8y 16
39. 3t 2 2t 3
40. h 2 4h 17
41. 3d 2 2d 3 43. x 3 x
2 3d 2
42. x 2 x 1
6 2x 3
44. 2x 3 x 2 1 46. x 2 1
47. x 2
48. x 3
49. x 2 x 3
50. 2y 2 3y 1
116
2 3x 1
3x 7 x2 2
45. x 3
©Glencoe/McGraw-Hill
51 2h 3
Algebra 2
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51. $0.03x 4
Page 117 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-05:
1000 x
52. Let x be the number. Multiplying by 3 results in 3x. The sum of the number, 8, and the result of the multiplication is x 8 3x or 4x 8. Dividing by the sum of the number and 2 gives
4x 8 x2
or 4. The end
result is always 4. 53. 170
170 t 1
54. 85 people
2
55. x 3 x 2 6x 24 ft
56. x 2 s
57. x 2 3x 12 ft /s
58. Sample answer: r 3 9r 2 27r 28 and r 3
59. Division of polynomials can be used to solve for unknown quantities in geometric formulas that apply to manufacturing situations. Answers should include the following. • 8x in. by 4x s in. • The area of a rectangle is equal to the length times the width. That is, A /w . • Substitute 32x 2 x for A, 8x for /, and 4x s for w. Solving for s involves dividing 32x 2 x by 8x. A /w 2 32x x 8x (4x s)
60. A
32x 2 x 8x 1 4x 8 1 8
4x s 4x s s
The seam is
1 8
inch. 62. x 2 4x 14
61. D
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63. y 4z 4 y 3z 3 3y 2z
64. y 2 2y 15
65. a 2 2ab b 2
66. 5 102 s or 8 min 20 s
67. y x 2
68. y x
69. 9
70. 12
71. 4
72. 3
73. 6
74. 5
2 3
4 3
Chapter 5 Practice Quiz 1 Page 238 1. 6.53 108
2. 7.2 103
3. 108x 8y 3
4.
5.
x2 z6
a3 b4c 3
6. 2x 5y
7. 3t 2 2t 8 9. m 2 3
8. n3 n 2 5n 2
19 m4
10. d 2 d 3
Lesson 5-4 Factoring Polynomials Pages 242–244 1. Sample answer: x 2 2x 1
2. Sample answer: If a 1 and b 1, then a 2 b 2 2 but 1a b2 2 4.
3. sometimes
4. 6x(2x 1)
5. a(a 5 b)
6. (x 7)(3 y)
7. (y 2)(y 4)
8. (z 6)(z 2)
9. 3(b 4)(b 4)
10. (4w 13)(4w 13)
11. (h 20)(h 2 20h 400)
©Glencoe/McGraw-Hill
12.
118
x4 x7
Algebra 2
Chapter 5
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13.
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2y y4
14. x y cm
15. 2x(y 3 5)
16. 6ab 2(a 3b)
17. 2cd 2(6d 4c 5c 4d )
18. prime
19. (2z 3)(4y 3)
20. (3a 1)(x 5)
21. (x 1)(x 6)
22. (y 1)(y 4)
23. (2a 1)(a 1)
24. (2b 1)(b 7)
25. (2c 3)(3c 2)
26. (3m 2)(4m 3)
27. 3(n 8)(n 1)
28. 3(z 3)(z 5)
29. (x 6)2
30. (x 3)2
31. prime
32. 3(m n)(m n)
33. (y 2 z)(y 2 z)
34. 3(x 3y)(x 3y)
35. (z 5)(z 2 5z 25)
36. (t 2)(t 2 2t 4)
37. (p 2 1)(p 1)(p 1)
38. (x 2 9)(x 3)(x 3)
39. (7a 2b)(c d )(c d )
40. (8x 3)(x y z)
41. (a b)(5ax 4by 3cz)
42. (a 3b)(3a 5)(a 1)
43. (3x 2)(x 1)
44. (2y 1)(y 4)
45. 30 ft by 40 ft
46.
x1 x4 x5 x2
47.
x5 x6
48.
49.
x4 x 2x 4
50. x
2
51. x 2
52. x 1 s
53. 16x 16 ft /s
54. x 8 cm
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55. (8pn 1)2
56. Factoring can be used to find possible dimensions of a geometric figure, given the area. Answers should include the following. • Since the area of the rectangle is the product of its length and its width, the length and width are factors of the area. One set of possible dimensions is 4x 2 by x 3. • The complete factorization of the area is 2(2x 1) (x 3), so the factor of 2 could be placed with either 2x 1 or x 3 when assigning the dimensions.
57. B
58. C
59. yes
60. no; (x 2)(x 2 2x 4)
61. no; (2x 1)(x 3)
62. yes
63. t 2 2t 1
64. y 3
65. x 2 2
66. x 3 x 2 2x 2
67. 4x 2 3xy 3y 2
68. 14x 2 26x 4
69. [2]
70. c
71. 15 in. by 28 in.
72. yes
73. no
74. Distributive Property
75. Associative Property ()
76. rational
77. irrational
78. rational
79. rational
80. irrational
1 3x 2
36 7 d 18 4
81. rational
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Lesson 5-5 Roots of Real Numbers Pages 247–249 1. Sample answer: 64
2. If all of the powers in the result of an even root have even exponents, the result is nonnegative without taking absolute value.
3. Sometimes; it is true when x 0.
4. 8.775
5. 2.668
6. 2.632
7. 4
8. 2
9. 3
10. not a real number
11. x
12. 0 y 0
15. about 3.01 mi
16. 11.358
17. 12.124
18. 0.933
19. 2.066
20. 3.893
21. 7.830
22. 4.953
23. 3.890
24. 4.004
25. 4.647
26. 26.889
27. 59.161
28. 15
29. 13
30. not a real number
31. 18
32. 3
33. 2
34.
14. 0 4x 3y 0
13. 6 0 a 0 b 2
1 4
1 5
36. 0.5
37. 0.4
38. z 2
35.
39. 0 x 0
40. 7 0 m3 0
41. 8a 4
42. 3r
43. c 2
44. 25g 2
46. 5x 2 0 y 3 0
45. 4z 2
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48. 13x 4y 2
47. 6x 2z 2
49. 3p 6 0 q 3 0
50. 2ab
52. 0 4x y 0
51. 3c 3d 4
54. 0 x 2 0
53. p q
56. 0 2a 1 0
55. 0 z 4 0 57. not a real number
58. 2
59. 5
60. about 127.28 ft
61. about 1.35 m
62. about 11,200 m s
63. x 0 and y 0, or y 0 and x 0
64. The speed and length of a wave are related by an expression containing a square root. Answers should include the following. • about 1.90 knots, about 3.00 knots, and 4.24 knots • As the value of / increases, the value of s increases.
65. B
66. D
67. 7xy 2(y 2xy 3 4x 2)
68. (a 3)(b 5)
69. (2x 5)(x 5)
70. (c 6)(c 2 6c 36)
71. 4x 2 x 5 73. c
8 x2
72. x 3 x 2 x
810 2320 d 1418 2504
74. (2, 2)
75. (1, 3)
76. (9, 4)
77. x 2 11x 24
78. y 2 3y 10
79. a 2 7a 18
80. a 2 3ab 2b 2
81. x 2 9y 2
82. 6w 2 7wz 5z 2
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Lesson 5-6 Radical Expressions Pages 254–256 1. Sometimes;
1 n
1a
n
1a only
2. Sample answer: 22 23 22
when a 1. 3. The product of two conjugates yields a difference of two squares. Each square produces a rational number and the difference of two rational numbers is a rational number.
4. 1527
4 5. 2x 0 y 0 2x
6.
7. 24235
8. 225
214y 4y 3
4 10. 523 323
9. 2a 2b 2 23 3 11. 22 22
12. 3 323 25 215
13. 2 25
14. about 49 mph
15. 923
16. 622
3 17. 322
4 18. 226
19. 5x 2 22
3 20. 2y22
21. 3 0 x 0 y 22y
22. 2ab 2 210a
3
3 24. 4mn23mn 2
4 02 c
26.
1 wz 2
23. 6y 2z 27 25.
1 c 0d 3
27.
26 2
28.
2 54 3
29.
a 2 2b b2
30.
2r 4 2t t5
4
3
32. 60230
31. 36 27 33.
5 2 wz 2
26 2
34.
210 5
35. 323
36. 522
37. 723 222
38. 425 23 26
©Glencoe/McGraw-Hill
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39. 25 522 526 223
40. 6 326 227 242
41. 13 2222
42. 8 2215
43.
28 7 23 13
44.
5 26 3 22 22
45.
1 23 2
46.
12 7 22 23
47.
2x 2 1 x1
48. 2x 1
49. 6 1622 yd, 24 622 yd2
50. The square root of a difference is not the difference of the square roots.
51. 0 ft /s
52. d v
53. about 18.18 m
54. 80 ft /s or about 55 mph
55. x and y are nonnegative.
56. The formula for the time it takes an object to fall a certain distance can be written in various forms involving radicals. Answers should include the following. • By the Quotient Property
24.9h 4.9
of Radicals, t Multiply by
2g 2g
22d 2g
.
to
rationalize the denominator. The result is h
22dg . g
• about 1.12 s 57. B
58. D
59. 12z 4
60. 6ab 3
61. 0 y 2 0
62.
63.
2 4 64. £ 9 15 § 3 5
x1 x4
©Glencoe/McGraw-Hill
x7 x4
124
Algebra 2
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1 4 d 5 4
66. 16, 15
67. consistent and independent
68. $4.20
69. 5
70. 2
71. 2, 4
72. , 1
73. 5x 0 x 66
74. 5x 0 x 76
7 3
75.
1 4
76.
1 2
77.
5 6
78.
13 12
79.
13 24
80.
19 30
81.
3 8
82.
5 12
Chapter 5 Practice Quiz 2 Page 256 1. x 2y (3x y 1)
2. prime
3. a(x 3)2
4. 8(r 2s 2)(r 2 2rs 2 4s 4) 6. 4a 2b 3
5. 6 0 x 0 0 y 3 0
7. 0 2n 3 0
8.
9. 1 17
10.
©Glencoe/McGraw-Hill
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x 2 2y y2 8 3 22 2
Algebra 2
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Lesson 5-7 Rational Exponents Pages 260–262 2. In radical form, the expression would be 216, which is not a real number because the index is even and the radicand is negative.
1. Sample answer: 64
n
3. In exponential1 form 2bm is equal to (b m)n . By the Power of a 1Power Property, m m m n n (b ) b . But, b n is also 1 equal to (b n )mby the Power of a Power Property. This last n expression is equal to ( 2b)m . n n Thus, 2bm (2b)m.
3 4. 27
3 3 5. 2x 2 or (2x)2
6. 264
1
5
1
7
8. 5
7. 63x 3y 3 9.
1 3
10. 9 11
12. a12
11. 2
2
13. x
2 3
14.
z3 2z
16.
m 3 n3 mn
1
2
3 2
15. a b
2 3
1
17.
z (x 2y)2 x 2y
18. 23x
19. 23
20. $5.11
5 21. 26
3 22. 24
5 5 23. 2c 2 or (2c)2
3 24. x 2 2x 2 1
1
26. 623
25. 232
1
1
2
1
27. 2 z 2
28. 53 x 3 y 3
29. 2
30. 6
31.
1 5
©Glencoe/McGraw-Hill
32.
126
1 27
Algebra 2
Chapter 5
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33.
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1 8
1 9
34.
35. 81
36. 4096
37.
2 3
38. 27
39.
4 3
40.
42. x 3
41. y 4 1
1
43. b 5
44. a9 5 x6 46. x
1
45.
1 2
w5 w
1
1
48. r 2
47. t 4
15
5
49.
a12 6a
51.
y 2 2y 2 y4
50.
2c 16 c
52.
x 3x 2 2 x1
1
3
53. 25
54. 23
6 55. 17 217
6 56. 5255 3
4
57. 25x 2y 2 59.
58. b29a 2b 3
xy 1z z
60.
ab 2 c 2 c 3
61. 12
62. 26
63. 216 5
64. x x 3z 3
3
1
2
1
1
65. 22 32
66. 2 3 3
67. 880 vibrations per second
68. about 262 vibrations per second
©Glencoe/McGraw-Hill
127
Algebra 2
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69. about 336
70. Rewrite the equation so that the bases are the same on each side. 1 9x 3x 2 1 (32 )x 3x 2 1 32x 3x 2 Since the bases are the same and this is an equation, the exponents must be equal. 1 2
Solve 2x x . The result 1 2
is x . 71. The equation that determines the size of the region around a planet where the planet’s gravity is stronger than the Sun’s can be written in terms of a fractional exponent. Answers should include the following. • The radical form of the equation is r D 5 a
Mp
B Ms
Mp2 . B Ms2
r D5
2
b
72. C
or
Multiply the
fraction under the radical by
Ms3 Ms3
.
Mp2 Ms3 B Ms2 Ms3
r D5
Mp2 Ms3 B Ms5
D5 D
5 2 Mp2Ms3 5 2 Ms5
5 D2 Mp2Ms3
Ms
The simplified radical form is r
5 D2 Mp2 Ms3
Ms
.
• If Mp and Ms are constant, then r increases as D increases because r is a linear function of D with positive slope. ©Glencoe/McGraw-Hill
128
Algebra 2
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74. 2x 0 y 0 2x
73. C 75. 36 22
76. 222
78. 4 0 x 5 0
77. 8 79.
1 2 x 2
80. 1440
81. x 2
82. 2x 3
83. x 22x 1
84. 4x 122x 9
Lesson 5-8
Radical Equations and Inequalities Pages 265–267
1. Since x is not under the radical, the equation is a linear equation, not a radical equation. The solution is x
2. The trinomial is a perfect square in terms of 1x . x 61x 9 (1x 3)2, so the equation can be written as (1x 3)2 0. Take the square root of each side to get 1x 3 0. Use the Addition Property of Equality to add 3 to each side, then square each side to get x 9.
23 1 . 2
3. Sample answer: 2x 2x 3 3
4. 2
5. 9
6. no solution
7. 15
8. 18 3 2
10. x 39
9. 31 11. 0 b 4
12. about 13.42 cm
13. 16
14. 49
15. no solution
16. no solution
17. 9
18. 5
19. 1
20.
21. 20
22. 5
©Glencoe/McGraw-Hill
129
27 2
Algebra 2
Chapter 5
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23. no solution
24. 9
25. x 1
26. 2 x 1
27. x 11
28. y 4
29. no solution
30. 4
31. 3
32. no solution
33. 0 x 2
34. 0 a 3
35. b 5
36. c
37. 3
38. 16
39. 1152 lb
40. t
41. 34 ft
42. 21.125 kg
43. Since 1x 2 0 and 12x 3 0, the left side of the equation is nonnegative. Therefore, the left side of the equation cannot equal 1. Thus, the equation has no solution.
44. If a company’s cost and number of units manufactured are related by an equation involving radicals or rational exponents, then the production level associated with a given cost can be found by solving a radical equation. Answers should include the following. 3 • C 102n 2 1500
79 16
4 2r 3 B GM
2
• 10,000 10n 3 1500
C 10,000
8500 10n 2
850 n 3 3
8502 n
2 3
Subtract 1500 from each side. Divide each side by 10. Raise each side to the 3 power. 2
24,781.55 n
Use a calculator.
Round down so that the cost does not exceed $10,000. The company can make at most 24,781 chips.
©Glencoe/McGraw-Hill
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Algebra 2
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45. D
46. C
3 7
1
48. (x 7)2
47. 5
50. 6 0 x 3 0 y 22y
2
49. (x 2 1)3 3
51.
2 100 10
52. 28 10 23 54. 4 x
53. x y 7, 30x 20y 160; (2, 5) y 30x 20y 160 (2, 5)
x y 7
x
O
55. 1 y
56. 2 4x
57. 11
58. 4 6z 2z 2
59. 3 10x 8x 2
Lesson 5-9 Complex Numbers Pages 273–275 1a. true 1b. true
2. all of them
3. Sample answer: 1 3i and 1 3i
4. 6i
5. 5i 0 xy 0 22
6. 12
7. 18023
8. i 10. 42 2i
9. 6 3i 11.
7 17
11 i 17
12. 3i
13. 2i22
14. i25
15. 3, 3
16. 5, 4
17. 10 3j amps
18. 12i
©Glencoe/McGraw-Hill
131
Algebra 2
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20. 8x 2i
19. 9i
21. 10a 2 0 b 0 i
22. 1322
23. 12
24. 48i
25. 75i
26. i
27. 1
28. 1
29. i
30. 9 2i
31. 6
32. 2
33. 4 5i
34. 25
35. 6 7i
36. 8 4i
37. 8 4i
38.
2 5
40.
39 17
39.
10 17
41.
2 5
6 i 17
1 5
43. 20 15i 1 3
14 i 17
42. 163 16i
i
45.
6 5
i
44.
2 22 i 3
11 14
5 23 i 14
46. (j 4)x 2 (3 i )x 2 4i
47. (5 2i )x 2 (1 i )x 7 i
48. i
49. 4i
50. i 26
51. 2i 23
52. i 23
53. 2i 210
54. 3i 25
55.
25 i 2
56. 4, 5 7 2
58. , 3
57. 4, 3 59.
5 , 3
61.
67 19 , 11 11
60. 3, 1
4
62. 5 2j ohms 64. 4 2j amps
63. 13 18j volts
©Glencoe/McGraw-Hill
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Algebra 2
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65. Case 1: i 0 Multiply each side by i to get i 2 0 i or 1 0. This is a contradiction. Case 2: i 0 Since you are assuming i is negative in this case, you must change the inequality symbol when you multiply each side by i. The result is again i 2 0 i or 1 0, a contradiction. Since both possible cases result in contradictions, the order relation “ ” cannot be applied to the complex numbers.
66. Some polynomial equations have complex solutions. Answers should include the following. • a and c must have the same sign. • i
67. C
68. C
69. 1, i, 1, i, 1, i, 1, i, 1
70. Examine the remainder when the exponent is divided by 4. If the remainder is 0, the result is 1. If the remainder is 1, the result is i. If the remainder is 2, the result is 1. And if the remainder is 3, the result is i.
71. 12
72. 11
73. 4
74. x 15
7
1
75. y 77. c
1 3
76.
2 1 2 d 3 2 1
©Glencoe/McGraw-Hill
a4 a
78. c
133
1 0 d 0 1
Algebra 2
Chapter 5
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2 1 2 d 3 2 1
79. c
80.
y
B' x
O
C' A'
81. sofa: $1200, love seat: $600, coffee table: $250
82.
y
y 2x 2 O
x
yx1
83.
84.
y
1 10
xy1 x 2y 4
O
x
85. 0
©Glencoe/McGraw-Hill
134
Algebra 2
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Chapter 6 Quadratic Functions and Inequalities Lesson 6-1 Graphing Quadratic Functions Pages 290–293 2a. (2, 1); x 2 2b. (3, 2); x 3
1. Sample answer: f (x ) 3x 2 5x 6; 3x 2, 5x, 6 3a. up; min. 3b. down; max. 3c. down; max. 3d. up; min.
4a. 0; x 0; 0 4b.
x f(x) 1 4 0 0 1 4
p
4c.
f (x) O (0, 0)
x f (x) 4x 2
5a. 0; x 1; 1
6a. 1; x 2; 2
5b.
6b.
x f(x) 3 3 2 0 1 1 0 0 1 3
p
5c.
6c.
f(x)
x 0 1 2 3 4
p
f(x) 1 2 3 2 1
f(x) (2, 3)
O f (x) x 2 2x
©Glencoe/McGraw-Hill
O
f (x) x 2 4x 1
x
x
(1, 1)
135
Algebra 2
Chapter 6
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7a. 3; x 4; 4
8a. 1; x 1; 1
7b.
8b.
x f(x) 1 7 1 0 1 1 2 1 3 7
8c.
f(x)
x f(x) 6 9 5 12 4 13 3 12 2 9
p
7c.
f(x) 10
8
4
p
x
O 4
f (x) 2x 2 4x 1
8 2
O
f (x) x 8x 3
12 (4, 13)
5 3
5 3
9a. 0; x ; 9b.
x 3 2
(1, 1)
x
10. max.; 7
f(x) 3 8
5 3
25 3
1 0
7 0
9c.
f(x) 4 2
4
O
2
x
4
(
)
5 , 25 3
3
8 f (x) 3x 2 10x 12
25 4
11. min.;
©Glencoe/McGraw-Hill
12. min.; 0
136
Algebra 2
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14a. 0; x 0; 0
13. $8.75
14b.
x f(x) 2 8 1 2 0 0 1 2 2 8
p
14c.
f (x)
f (x ) 2x 2 (0, 0)
x
O
15a. 0; x 0; 0
16a. 4; x 0; 0
15b.
16b.
x f(x) 2 20 1 5 0 0 1 5 2 20
p
15c.
x f(x) 2 8 1 5 0 4 1 5 2 8
p
16c.
f(x)
f(x)
O f (x ) 5x
2
(0, 0)
12
x
8 (0, 4) f (x ) x 2 4 4
2
O
17a. 9; x 0; 0
18a. 4; x 0; 0
17b.
18b.
x 2 1 0 1 2
p
f(x) 5 8 9 8 5
©Glencoe/McGraw-Hill
137
x 2 1 0 1 2
p
2
4x
f(x) 4 2 4 2 4 Algebra 2
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17c.
18c.
f(x)
f(x)
4 4
2
4x
2
O
x
O
4 (0, 9)
f (x ) 2x 2 4
f (x ) x 2 9
19a. 191; x 0; 0
20a. 4; x 2; 2
19b.
20b.
x f(x) 2 13 4 1 0 1 1 4 2 13
p
19c.
20c.
f (x)
f (x ) 3x 2 1
x 0 1 2 3 4
p
f(x) 4 1 0 1 4
f(x)
f (x ) x 2 4x 4
(0, 1) x
O
(0, 4)
O
x
(2, 0)
21a. 9; x 4.5; 4.5
22a. 5; x 2; 2
21b.
22b.
21c.
x 3 4 4.5 5 6
p
f(x) 9 11 11.25 11 9
p
f(x) 5 8 9 8 5
22c.
f(x)
f(x) x
O
2 O
x 0 1 2 3 4
4
8
12
x
4 8 12
©Glencoe/McGraw-Hill
f (x ) x 2 9x 9 f (x ) x 2 4x 5
(4 12 , 1114 )
(2, 9)
138
Algebra 2
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23a. 36; x 6; 6
24a. 1; x 1; 1
23b.
24b.
x f(x) 8 4 7 1 6 0 5 1 4 4
p
23c.
x 3 2 1 0 1
p
f(x) 8 1 4 1 8
24c.
f(x) 6
f(x) f (x ) 3x 2 6x 1
4 2
f (x ) x 12x 36
16 12
8 4 (6, 0)
O x
(1, 4)
2 3
2 3
25a. 3; x 2, 2
26a. 0; x ,
25b.
26b.
x 0 1 2 3 4
25c.
p
f(x) 3 3 5 3 3
f(x)
x 2 1
26c.
(2, 5)
f(x) 4 1
2 3
4 3
0 1
0 7
O
5 4
28a. 1; x 0; 0
27a. 0; x ;
©Glencoe/McGraw-Hill
x
f (x ) 3x 2 4x
x
5 4
f(x)
( 23 , 43 )
f (x ) 2x 2 8x 3
O
x
O
2
139
Algebra 2
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27b.
x 3 2
28b.
f(x) 3 2
5 4
x 2
25 8
1 0
0
3 0
27c.
p
1
p
1 2
f(x) 1 1 2
1 1 2
1
28c.
f(x)
f(x) f (x ) 0.5x 2 1
f (x ) 2x 2 5x O x
O x (0, 1)
( 54 , 258) 29a. 0; x 6; 6
30a.
29b.
30b.
29c.
x 8 7 6 5 4
f(x) 8 8.75 9 8.75 8
p
(6, 9)
8
9 ; 2
x 3, 3
x 5 4 3 2 1
p
f(x) 2 0.5 0 0.5 2
30c.
f(x)
f(x)
4 8
4
O
x
(3, 0)
4
f (x ) 1 x 2 3x 9
f (x ) 0.25x 2 3x
©Glencoe/McGraw-Hill
2
140
O
x
2
Algebra 2
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8 9
1 1 3 3
31a. ; x ; 31b.
x
f(x)
1
7 9 8 9
0 1 3
1
1
32. min.; 0
5 9 7 1 9
2 31c.
f(x)
2
2
8
f (x ) x 3 x 9 O x
(
1 , 1 3
)
33. max.; 9
34. min.; 14
35. min.; 11
36. max.; 5
37. max.; 12
38. min.;
7 8
9 2
39. max.;
40. max.; 5
41. min.; 11
42. max.; 5
43. min.; 10
1 3
44. x 40; (40, 40)
45. 40 m
46. 300 ft, 2.5 s
47. The y-intercept is the initial height of the object.
48. 120 2x
49. 60 ft by 30 ft
50. 1800 ft2
51. $11.50
52. $2645
©Glencoe/McGraw-Hill
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Algebra 2
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53. 5 in. by 4 in.
54. c; The x-coordinate of the 0 vertex of y ax 2 c is 2a or 0, so the y-coordinate of the vertex, the minimum of the function, is a(0)2 c or c ; 12.5
55. If a quadratic function can be used to model ticket price versus profit, then by finding the x-coordinate of the vertex of the parabola you can determine the price per ticket that should be charged to achieve maximum profit. Answers should include the following. • If the price of a ticket is too low, then you won’t make enough money to cover your costs, but if the ticket price is too high fewer people will buy them. • You can locate the vertex of the parabola on the graph of the function. It occurs when x 40. Algebraically, this is found
56. C
by calculating x
b 2a
which, for this case, is 4000 x or 40. Thus the 2(50)
ticket price should be set at $40 each to achieve maximum profit. 57. C
58. 2.08
59. 3.20
60. 0.88
61. 3.38
62. 0.43
63. 1.56
64. 1
65. 1 3i
66. 9 5i
©Glencoe/McGraw-Hill
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67. 23
68. 13
69. 4
70. [10 4 5] 28 20 44 d 72. c 8 16 36
71. [5 13 8] 6 73. C 14
0
24
2 3
8
S
74.
y y 3x yx4
(1, 3)
O
x
(1, 3); consistent and independent 75. 5
76. 8
77. 2
78. 1
Lesson 6-2
Solving Quadratic Equations by Graphing Pages 297–299 2. Sample answer: f (x) 3x 2 2x 1; 3x 2 2x 1 0
1a. The solution is the value that satisfies an equation. 1b. A root is a solution of an equation. 1c. A zero is the x value of a function that makes the function equal to 0. 1d. An x-intercept is the point at which a graph crosses the x-axis. The solutions, or roots, of a quadratic equation are the zeros of the related quadratic function. You can find the zeros of a quadratic function by finding the x-intercepts of its graph.
©Glencoe/McGraw-Hill
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3. The x-intercepts of the related function are the solutions to the equation. You can estimate the solutions by stating the consecutive integers between which the x-intercepts are located.
4. 4, 1
5. 2, 1
6. 4
7. 7, 0
8. 4, 6
9. 7, 4
10. 5
11. between 2 and 1, 3
12. between 1 and 0; between 1 and 2
13. 2, 7
14. 0, 6
15. 3
16. 2, 1
17. 0
18. , 3
19. no real solutions
20. 0, 3
21. 0, 4
22. between 5 and 4; between 0 and 1
23. between 1 and 0; between 2 and 3
24. 4, 5
25. 3, 6
26. 7
27. 6
28. 1 , 3
1 2
1 2
1 2
1 2
29. , 2
1 2
1 2
30. 4, 1
˛
1 2
31. 2 , 3
32. between 4 and 3; between 0 and 1
33. between 0 and 1; between 3 and 4
34. between 1 and 0, between 2 and 3
35. between 3 and 2; between 2 and 3
36. no real solutions
37. no real solutions
38. 8, 9
˛
©Glencoe/McGraw-Hill
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40. Let x be the first number. Then, 9 x is the other number. x(9 x) 24 2 x 9x 24 0
39. Let x be the first number. Then, 7 x is the other number. x(7 x) 14 2 x 7x 14 0
y
y 2
y x 7x 14 O
O
x
y x 9x 24
Since the graph of the related function does not intersect the x-axis, this equation has no real solutions. Therefore, no such numbers exist.
Since the graph of the related function does not intersect the x-axis, this equation has no real solutions. Therefore, no such numbers exist. 41. 2, 14
42. 4 s
43. 3 s
44. about 12 s
45. about 35 mph
46. about 8 s
©Glencoe/McGraw-Hill
x
2
145
Algebra 2
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47. 4 and 2; The value of the function changes from negative to positive, therefore the value of the function is zero between these two numbers.
48. Answers should include the following. • h (t ) 2 h (t ) 16t 185
180
160 140 120 100 80 60 40 20 0
1
2 3 4 5 t
• Locate the positive x-intercept at about 3.4. This represents the time when the height of the ride is 0. Thus, if the ride were allowed to fall to the ground, it would take about 3.4 seconds. 49. A
50. B
51. 1
52. 3
53. 3, 5
54. 9, 1
55. 1.33
56. no real solutions
57. 4, x 3; 3
58. 1; x 1; 1 f (x)
f(x)
(1, 3)
f (x) 4x 2 8x 1 x
O
O
x
f (x) x 2 6x 4 (3, 5)
©Glencoe/McGraw-Hill
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Algebra 2
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59. 4; x 6; 6
60.
1 5
62.
1 13
3 5
i
f(x) 8 f (x) 1 x 2 3x 4 4
4 O 12 8
4
x 4
(6, 5)
61.
10 13
2 i 13
5 i 13
63. 24
64. 8
65. 60
66. $500
67. x(x 5)
68. (x 10)(x 10)
69. (x 7)(x 4)
70. (x 9)2
71. (3x 2)(x 2)
72. 2(3x 2)(x 3)
Lesson 6-3
Solving Quadratic Equations by Factoring Pages 303–305
1. Sample answer: If the product of two factors is zero, then at least one of the factors must be zero.
2. Sample answer: roots 6 and 5; x 2 x 30 0
3. Kristin; the Zero Product Property applies only when one side of the equation is 0.
4. {0, 11}
5. {8, 2}
6. {7, 7}
7. {3}
8. e , 4 f 3 4
10. x 2 3x 28 0
9. {3, 4} 11. 6x 2 11x 4 0
12. 15x 2 14x 3 0
13. D
14. {8, 3}
15. {4, 7}
16. {5, 5}
17. {9, 9}
18. {6, 3}
©Glencoe/McGraw-Hill
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19. {3, 7}
20. e 0, f
21. e 0, f
22. {6}
23. {8}
24.
5 3
3 4
25.
e
1 , 4
e2,
1 f 4
26. e , f
4f
1 2
3 2
27. e , f
28. e , f
29. e ,
30. {2, 4}
31. {3, 1}
32. 0, 6, 5
33. 0, 3, 3
34. x 2 9x 20 0
35. x 2 5x 14 0
36. x 2 x 20 0
37. x 2 14x 48 0
38. 2x 2 7x 3 0
39. 3x 2 16x 5 0
40. 12x 2 x 6 0
41. 10x 2 23x 12 0
42.
43. 14, 16
44. 12 cm by 16 cm
45. B D 2 8D 16
46. 4; The logs must have a diameter greater than 4 in. for the rule to produce positive board feet values.
47. y (x p)(x q) y x 2 px qx pq y x 2 (p q)x pq a 1, b (p q), c pq axis of symmetry:
48. 1
2 3
8 3
3 2
3 9 f 4 4
1 4
2 3
s
b 2a (p q) 2(1) pq 2
x x x
©Glencoe/McGraw-Hill
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The axis of symmetry is the average of the x-intercepts. Therefore the axis of symmetry is located halfway between the x-intercepts. 49. 6
50. Answers should include the following. • Subtract 24 from each side of x 2 5x 24 so that the equation becomes x 2 5x 24 0. Factor the left side as (x 3) (x 8). Set each factor equal to zero. Solving each equation for x. The solutions to the equation are 3 and 8. Since length cannot be negative, the width of the rectangle is 3 inches, and the length is 3 5 or 8 inches. • To use the Zero Product Property, one side of the equation must equal zero.
51. D
52. B
53. 5, 1
54.
55. between 1 and 0; between 3 and 4
56. min.; 19
57. 322 223
58. 523
59. 33 20 22
60. (4, 4)
61. (3, 5)
62. a , 2b
63. 222
64. 225
65. 323
66. 5i 22
67. 2i 23
68. 4i23
1 2
1 3
˛
˛
©Glencoe/McGraw-Hill
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Chapter 6 Practice Quiz 1 Page 305 1. 4; x 2; 2
2. max.;
f(x) 4 O
37 4
or 9
1 4
f (x) 3x 2 12x 4 4
8
12
x
4 8
(2, 8)
4. e 5, f 1 2
1 2
3. 1 , 4 5. 3x 2 11x 4 0
Lesson 6-4 Completing the Square Pages 310–312 1. Completing the square allows you to rewrite one side of a quadratic equation in the form of a perfect square. Once in this form, the equation is solved by using the Square Root Property.
2. Never; the value of c that makes ax 2 bx c a perfect square trinomial is
3. Tia; before completing the square, you must first check to see that the coefficient of the quadratic term is 1. If it is not, you must first divide the equation by that coefficient.
4. {10, 4}
5. e 7.
b
the square of and the 2 square of a number can never be negative.
4 22 f 3
9 , ax 4
6. 36; (x 6)2
3 2 2
b
8. {6, 3}
9. 54 256 ©Glencoe/McGraw-Hill
10. 51 i 256 150
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11. e
3 233 f 4
12. Jupiter 14. {3, 7}
13. Earth: 4.5 s, Jupiter: 2.9 s 15. {2, 12}
16. 54 27, 4 276
17. 53 2226
18. e
19. e
5 211 f 3
7 25 f 2
20. e , 5 4
1 f 4
21. {1.6, 0.2}
22. 25 ft
23. about 8.56 s
24. 64; (x 8)2
25. 81; (x 9)2
26.
27.
49 ; ax 4
7 2 2
b
25 , ax 16
15 2 b 2
28. 0.09; (x 0.3)2
29. 1.44; (x 1.2)2 31.
225 ; ax 4
30.
5 2 4
b
16 ; ax 9
4 2 3
b
32. {3, 5}
33. {12, 10}
34. 51 276
37. {3 2i}
38. e , 1 f
39. e , 1 f 1 2
40. e
5 213 f 6
41. e
2 210 f 3
42. e
7 i 247 f 4
43. e
5 i 123 f 6
44. {2, 0.6}
35. 52 236
36. 52 i 6 5 2
45. {0.7, 4}
46. e 23 f
47. e 22 f
48. 5
1 3
3 4
49.
x 1 , 1 x1
©Glencoe/McGraw-Hill
50.
151
1 2
in. by 5
1 2
in.
1 25 2
Algebra 2
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51. Sample answers: The golden rectangle is found in much of ancient Greek architecture, such as the Parthenon, as well as in modern architecture, such as in the windows of the United Nations building. Many songs have their climax at a point occurring 61.8% of the way through the piece, with 0.618 being about the reciprocal of the golden ratio. The reciprocal of the golden ratio is also used in the design of some violins.
52a. n 0 52b. n 0 52c. n 0
53. 18 ft by 32 ft or 64 ft by 9 ft
54. To find the distance traveled by the accelerating race car in the given situation, you must solve the equation t 2 22t 121 246 or t 2 22t 125 0. Answers should include the following. • Since the expression t 2 22t 125 is prime, the solutions of t 2 22t 121 246 cannot be obtained by factoring. • Rewrite t 2 22t 121 as (t 11)2. Solve (t 11)2 246 by applying the Square Root Property. Then, subtract 11 from each side. Using a calculator, the two solutions are about 4.7 and 26.7. Since time cannot be negative, the driver takes about 4.7 seconds to reach the finish line.
55. D
56. D
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57. x 2 3x 2 0
58. x 2 6x 27 0
59. 3x 2 19x 6 0
60. 12x 2 13x 3 0
61. between 4 and 3; between 0 and 1
62. 6, 8
1 2
3
63. 4, 1
64. 57
65. (2, 5)
66. a , b 43 6 21 7
67. 0 x (257)0 2
68. greatest: 255C; least: 259C
69. 37
70. 16
71. 121
72. 0
Lesson 6-5
The Quadratic Formula and the Discriminant Pages 317–319
1a. Sample answer:
2. The square root of a negative number is a complex number.
y
x
O
1b. Sample answer: y
O
©Glencoe/McGraw-Hill
x
153
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1c. Sample answer: y
O
x
3. b 2 4ac must equal 0.
4a. 484 4b. 2 rational 4c.
2 22 2
1 2
6c.
8. 0, 8
7a. 3 7b. two complex 7c.
3 i 23 2
9. 3, 2 11.
5 2
6a. 0 6b. one rational
5a. 8 5b. 2 irrational 5c.
1 , 4
10. 1 23
5 i 22 2
12. at about 0.7 s and again at about 4.6 s
13. No; the discriminant of 16t 2 85t 120 is 455, indicating that the equation has no real solutions.
14a. 21 14b. 2 irrational
15a. 240 15b. 2 irrational 15c. 8 2215
16a. 16 16b. 2 complex 16c. 1 2i
17a. 23 17b. 2 complex
18a. 121 18b. 2 rational
17c.
14c.
1 i 223 2
1 2 4 3
18c. ,
19a. 49 19b. 2 rational ©Glencoe/McGraw-Hill
3 221 2
20a. 20 20b. 2 irrational 154
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19c. 2,
1 3
20c. 2 25
21a. 24 21b. 2 irrational
22a. 0 22b. one rational
21c. 1 26
22c.
23a. 0 23b. one rational
24a. 31 24b. 2 complex
5 2
1 3
9 i 231 8 28 9
23c.
24c.
25a. 135
26a.
25b. 2 complex
26b. 2 irrational
1 i 215 4
25c.
26c.
2 4 27 9
28. 2, 32
27a. 1.48 27b. 2 irrational 1 2 20.37 0.8
27c.
29. i
221 7
30. 2 i 23
31.
3 215 2
32. 22
33.
9 2
34. 3 i 27
35.
5 246 3
36. 4 27
3 10
37. 0,
38. 3 222
39. 2, 6
40.
41. This means that the cables do not touch the floor of the bridge, since the graph does not intersect the x-axis and the roots are imaginary.
42. domain: 0 t 25, range: 73.7 A(t ) 1201.2
43. 1998
44. about 40.2 mph
©Glencoe/McGraw-Hill
155
0.00288
Algebra 2
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45a. k 6 45b. k 6 or k 6 45c. 6 k 6
46. The person’s age can be substituted for A in the appropriate formula, depending upon their gender, and their average blood pressure calculated. See student’s work. • If a woman’s blood pressure is given to be 118, then solve the equation 118 0.01A2 0.05A 107 to find the value of A. Use the Quadratic Formula, substituting 0.01 for a, 0.05 for b, and 11 for c. This gives solutions of about 35.8 or 30.8. Since age cannot be negative, the only valid solution for A is 30.8.
47. D
48. C
49. 14, 4
50. 4 27
51.
1 2 22 2
52. 2, 0 2 , 3
53. 2, 7
54.
55. a 4b10
56. 10p6 0 q 0
5
58. 7.98 106
57. 4b 2c 2 59.
60.
y
y x1
xy9 8 yx4 6 4 2 6 4
O 4 6
yx
2 4 6 8 xy 3
x
x
O
y 1
61. no
62. yes; (x 7)2
63. yes; (2x 3)2
64. yes; (5x 2)2
65. no
66. yes; (6x 5)2
©Glencoe/McGraw-Hill
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Algebra 2
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Lesson 6-6
Analyzing Graphs of Quadratic Functions Pages 325–328
1a. 1b. 1c. 1d. 1e.
y 2(x 1)2 5 y 2(x 1)2 y 2(x 3)2 3 y 2(x 2)2 3 Sample answer: y 4(x 1)2 3 1f. Sample answer: y (x 1)2 3 1g. y 2(x 1)2 3
2. Substitute the x-coordinate of the vertex for h and the y -coordinate of the vertex for k in the equation y a(x h)2 k. Then substitute the x-coordinate of the other point for x and the y-coordinate for y into this equation and solve for a. Replace a with this value in the equation you wrote with h and k.
3. Sample answer: y 2(x 2)2 1
4. Jenny; when completing the square is used to write a quadratic function in vertex form, the quantity added is then subtracted from the same side of the equation to maintain equality.
5. (3, 1); x 3; up
6. y (x 4)2 19, (4, 19); x 4; up
7. y 3(x 3)2 38; (3, 38); x 3; down
8.
y
y 3(x 3)2 O x
9.
10.
y
y
O
y 2x 2 16x 31 x
1 y 3 (x 1)2 3
O
x
12. y (x 3)2 6
11. y 4(x 2)2 ©Glencoe/McGraw-Hill
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1 2
13. y (x 2)2 3
14. h(d ) 2d 2 4d 6
15. (3, 0); x 3 down
16. (1, 2); x 1; up
17. (0,6); x 0 up
18. (0, 3); x 0; down
19. y (x 2)2 12; (2, 12); x 2; down
20. y (x 3)2 8; (3, 8); x 3; up
21. y 3(x 2)2 12; (2, 12); x 2; down
22. y 4(x 3)2 36; (3, 36); x 3; up
23. y 4(x 1)2 7; (1, 7); x 1; up
24. y 2(x 5)2 15; (5, 15); x 5; down
1 2 2
3 2 2
25. y 3 ax b ; 1 a , 2
26. y 4 ax b 20;
7 4
b; x ; up 7 4
3 a , 2
1 2
27.
20b; x , up 3 2
28.
y
y O
y 4(x 3)2 1
29.
y (x 5)2 3
x
x
O
30.
y
y
x
O
1 y 4 (x 2)2 4
31.
1 y 2 (x 3)2 5
x
O
32.
y
y
O x
y x 2 8x 18 O
x
y x 2 6x 2
©Glencoe/McGraw-Hill
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33.
y
34.
y 4x 2 16x 11
y y 5x 2 40x 80 x
O
x
O
35.
36.
y
y
y 1 x 2 5x 27 2 2 x
O
O
x
y 1 x 2 4x 15 3
37. Sample answer: the graph of y 0.4(x 3)2 1 is narrower than the graph of y 0.2(x 3)2 1.
38. Sample answer: the graphs have the same shape, but the graph of y 2(x 4)2 1 is 1 unit to the left and 5 units below the graph of y 2(x 5)2 4
39. y 9(x 6)2 1
40. y 3(x 4)2 3
2 3
42. y 3(x 5)2 4
41. y (x 3)2
5 2
1 3
43. y x 2 5
44. y (x 3)2 2
45. y 2x 2
46. y (x 3)2 4
47. 34,000 feet; 32.5 s after the aircraft begins its parabolic flight
48. about 1.6 s
49. d (t ) 16t 2 8t 50
50. about 2.0 s
51. Angle A; the graph of the equation for angle A is higher than the other two since 3.27 is greater than 2.39 or 1.53.
52. Angle B; the vertex of the equation for angle B is farther to the right than the other two since 3.57 is greater than 3.09 or 3.22.
©Glencoe/McGraw-Hill
4 3
159
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53.
y ax 2 bx c
54. All quadratic equations are a transformation of the parent graph y x 2. By identifying these transformations when a quadratic function is written in vertex form, you can redraw the graph of y x 2. Answers should include the following. • In the equation y a(x h)2 k, h translated the graph of y x 2 h units to the right when h is positive and h units to the left when h is negative. The graph of y x 2 is translated k units up when k is positive and k units down when k is negative. When a is positive, the graph opens upward and when a is negative, the graph opens downward. If the absolute value of a is less than 1, the graph will be narrower than the graph of y x 2, and if the absolute value of a is greater than 1, the graph will be wider than the graph of y x 2. • Sample answer: y 2(x 2)2 3 is the graph of y x 2 translated 2 units left and 3 units down. The graph opens upward, but is narrower than the graph of y x 2.
y a ax 2 xb c b a
b 2 2a
y a cx 2 x a b d b a
b 2 2a
c aa b y a ax
b 2 b 2a
c
b2 4a
The axis of symmetry is x h or
b . 2a
55. D
©Glencoe/McGraw-Hill
56. B
160
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58. 225; 2 rational
57. 12; 2 irrational
60. 55 2226
59. 23; 2 complex 61. 53 3i 6 63. 2t 2 2t
62. e 3 t1
2 213 f 2
64. t 2 2t 1
65. n 3 3n 2 15n 21
66. y 3 1
67a. Sample answer using (1994, 76,302) and (1997, 99,448): y 7715x 15,307,408 67b. 161,167
68. yes
69. no
70. yes
4 y3
71. no
Chapter 6 Practice Quiz 2 Page 328 1. 57 2236
2. e
3. 11; 2 complex
4. 100; 2 rational
5. e
9 5 25 f 2
6. e
7. y 1x 22 2 5 2 3
2 2i 22 f 3
8. y (x 4)2 2; (4, 2), x 4; up
9. y 1x 62 2; 16, 02, x 6; down
©Glencoe/McGraw-Hill
1 3i f 2
10. y 2(x 3)2 5; (3, 5), x 3; up
161
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Lesson 6-7
Graphing and Solving Quadratic Inequalities Pages 332–335
1. y 1x 32 2 1
2. Sample answer: one number less than 3, one number between 3 and 5, and one number greater than 5
3a. x 1, 5 3b. x 1 or x 5 3c. 1 x 5
4.
y
O
5.
12 8 4 4
2
4 8 12
y
y 2x 2 4x 3
x
y
O
20
7.
6.
y
y x 2 10x 25
2
4x
O
x
y x 2 16
8. x 1 or x 5
y x 2 5x 6
12 8 4 2
O
2
4
6x
9. 5x 0 1 x 76
10. 5x 0 x 3 or x 46
12. 5x 0 23 x 236
11.
©Glencoe/McGraw-Hill
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13. about 6.1 s
14.
y 15 5 8
4
5
O
8x
4
15
y x 2 3x 18
25
15.
y
16.
y x 2 7x 8
y
12 8 4 4
O
17.
4
x
8
y x 2 4x 4
18.
y
5 8
O
x
O
y
4 O
8x
4
10
x
20 30
y x 2 4x
y x 2 36
19.
20.
y
y x 2 3x 10
14
y
10 O
x
6 2 6
y x 2 6x 5
©Glencoe/McGraw-Hill
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4
2 O
2x
Algebra 2
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21.
y x 2 7x 10
22.
y
y y x 2 10x 23
20 12
x
O 4 12 8
23.
y
4
O 4
4x
24.
y x 2 13x 36
y 4
6 2 O
2 2
6
10
x
2
O
4
x
4
4
8
y 2x 2 3x 5
8
25.
26. 5
y
x
O
y 2x 2 x 3
27. 2 x 6
28. x 3 or x 3
29. x 7 or x 3
30. 5x 0 x 3 or x 66
33. 5x 0 x 6 or x 46
34. 5x 0 4 x 36
31. 5x 0 7 x 46
32. 5x 0 1 x 56
35. 5x 0 x 7 or x 16
36. e x ` x 38.
37. all reals
39. 5x 0 x 76
40. all reals
42. 5x 0 4 x 1 or x 36
41.
43. 0 to 10 ft or 24 to 34 ft
©Glencoe/McGraw-Hill
1 f 3
44a. 0.98, 4.81; The owner will break even if he charges $0.98 or $4.81 per square foot. 164
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44b. 0.98 r 4.81; The owner will make a profit if the rent is between $0.98 and $4.81. 44c. 1.34 r 4.45; If rent is set between $1.34 and $4.45 per sq ft, the profit will be greater than $10,000 44d. r 1.34 or r 4.45; If rent is set between $0 and $1.34 or above $4.45 per sq ft, the profit will be less than $10,000. 45. The width should be greater than 12 cm and the length should be greater than 18 cm
46. P(n) n[15 1.5(60 n)] 525 or 1.5n 2 105n 525
47. 6
48. $1312.50; 35 passengers
49.
50. Answers should include the following. • 16t 2 42t 3.75 10 • One method of solving this inequality is to graph the related quadratic function h(t ) 16t 2 42t 3.75 10. The interval(s) at which the graph is above the x-axis represents the times when the trampolinist is above 10 feet. A second method of solving this inequality would be to find the roots of the related quadratic equation 16t 2 42t 3.75 10 0 and then test points in the three intervals determined by these roots to see if they satisfy the inequality. The interval(s) at which the inequality is satisfied represent the times when the trampolinist is above 10 feet.
y y x 2 4
O
x y x2 4
©Glencoe/McGraw-Hill
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51. C
52. A
55. 5x 0 x 9 or x 36
56. 5x 0 x 3.5 or x 2.56
53. 5x 0 all reals, x 26
54. 5x 0 7 x 76
57. 5x 0 1.2 x 0.46
58. no real solutions
60. y 21x 42 2; 14, 02, x 4; down
59. y (x 1)2 8; (1, 8), x 1; up 1 2
61. y (x 6)2; (6, 0),
62. 4, 8
x 6; up 63.
5 i 23 2
64.
65. 4a 2b 2 2a 2b 4ab 2 12a 7b 67. xy 3 y
66. 6x 3 4x 2y 13xy 2
1 x
68. 15a 2 14a 3
21 48 R 13 22
70. 354 64
69. B
71. 0x 0.008 0 0.002; 0.078 x 0.082
©Glencoe/McGraw-Hill
3 2 26 3
166
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7/24/02
12:02 PM
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Chapter 7 Polynomial Functions Lesson 7-1 Polynomial Functions Pages 350–352 1. 4 4x 0; x x 1
2. Sample answer: Evendegree polynomial functions with positive leading coefficients have graphs in which f (x ) S as x S and as x S. Odd-degree polynomial functions with positive leading coefficients have graphs in which f (x ) S as x S and f (x ) S as x S .
3. Sample answer given.
4. Sometimes; a polynomial function with 4 real roots may be a sixth-degree polynomial function with 2 imaginary roots. A polynomial function that has 4 real roots is at least a fourth-degree polynomial.
f (x)
O
x
5. 6; 5
6. 5; 3
7. 21; 3
8. 4; 12
9. 2a 9 6a 312
10. 100a 2 20
11. 6a 3 5a 2 8a 45
12a. f (x ) S as x S , f (x ) S as x S 12b. odd 12c. 3
13a. f(x ) S as x S , f(x ) S as x S 13b. even 13c. 0
14a. f (x ) S as x S , f (x ) S as x S 14b. odd 14c. 1
15. 109 lumens
16. 1; 1
17. 3; 1
18. No, the polynomial contains two variables, a and b.
19. 4; 6
20. 3; 5
©Glencoe/McGraw-Hill
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22. 2; 4
21. No, this is not a polynomial 1
because the term cannot c be written in the form x n, where n is a nonnegative integer. 23. 12; 18
24. 125; 37
25. 1008; 36
26. 166; 50
27. 86; 56
28. 100; 4
29. 7; 4
30. 27a3 3a 1
31. 12a 2 8a 20
32. 3a4 2a 2 5
33. 12a6 4a3 5
34. x 3 3x 2 4x 3
35. 3x 4 16x 2 26
36. 6x 2 44x 90
37. x 6 x 3 2x 2 4x 2
38. 9x 4 12x 2 8x 50
39a. f (x ) S as x S , f (x ) S as x S 39b. odd 39c. 3
40a. f (x) S as x S , f (x) S as x S 40b. even 40c. 4
41a. f (x) S as x S , f (x) S as x S 41b. even 41c. 0
42a. f (x ) S as x S , f (x ) S as x S 42b. odd 42c. 5
43a. f (x) S as x S , f (x) S as x S 43b. odd 43c. 1
44a. f (x ) S as x S , f (x ) S as x S 44b. even 44c. 2
45. 5.832 units
46. even
47. f (x ) S as x S ; f(x ) S as x S
48. Sample answer: Decrease; the graph appears to be turning at x 30 indicating a relative maximum at that point. So attendance will decrease after 2000.
49.
1 2
©Glencoe/McGraw-Hill
50. 1, 0, 4
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3 2
51. f(x) x 3 x 2 2x
52.
8
f (x )
3 2 f (x) 12 x 3 4 2 x 2x
4 2
O
2
x
4 8
53. 4
54. 16 regions
55. 8 points
56. Many relationships in nature can be modeled by polynomial functions; for example, the pattern in a honeycomb or the rings in a tree trunk. Answers should include the following. • You can use the equation to find the number of hexagons in a honeycomb with 10 rings and the number of hexagons in a honeycomb with 9 rings. The difference is the number of hexagons in the tenth ring. • Other examples of patterns found in nature include pinecones, pineapples, and flower petals.
57. C
58. C
60. 5x 0 x 9 or x 76
59. 5x 0 2 x 66
©Glencoe/McGraw-Hill
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61. e x ` 1 x f 4 5
62.
y y 2(x 2)2 3
x
O
63.
64.
y
y
2 12 8
O
x
2
y 1 (x 5)2 1 3
x
O
y 1 x2 x 3 2
2
4
65. 54 3226
66. e ,
67. 23,450(1 p); 23,450(1 p)3
68.
7 6
5 f 6 y
y x2 4
x
O
69.
70.
y
8
y
4 O
8
x
4 O
4
8x
4
1
y 2 x 2 2x 6
y x 2 6x 5
©Glencoe/McGraw-Hill
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Lesson 7-2
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Graphing Polynomial Functions Pages 356–358
1. There must be at least one real zero between two points on a graph when one of the points lies below the x-axis and the other point lies above the x-axis.
2. 4
3.
4.
f (x)
x O
x
p
3 20
8
2
4
1 0
f (x ) x 3 x 2 4x 4
f(x )
1 2 3
0
f (x )
6 4
4
2 O
0
2
4x
4
0
f(x) x 3 x 2 4x 4
10
6. between 1 and 0
5. x
p
f(x )
f (x)
8
3 20 2 9 1 2 0 5 1 0 2 5 3 26
f (x)
4 4
2 O
2
4x
x
O
4
f (x ) x 3 x 2 1
8
f (x ) x 4 7x 2 x 5
7. between 2 and 1, between 1 and 0, between 0 and 1, and between 1 and 2
8.
8
f (x)
4
f (x) 4
2
O
2
4x
4 O
8
x
f (x ) x 3 2x 2 3x 5
Sample answer: rel. max. at x 2, rel. min.: at x 0.5
f (x ) x 4 4x 2 2
©Glencoe/McGraw-Hill
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10.
f (x)
C (t ) 12000
8 4 4
2
O
2
Cable TV Systems
10000
4x
4
f (x ) x 4 8x 2 10
Sample answer: rel. max. at x 0, rel. min. at x 2 and at x 2
8000 6000
C(t) 43.2t 2 1343t 790
4000 2000 O
4
t
8 12 16 Years Since 1985
11. rel. max. between x 15 and x 16, and no rel. min.; f(x ) S as x S , f(x ) S as x S .
12. The number of cable TV systems rose steadily from 1985 to 2000. Then the number began to decline.
13a.
14a. x
f (x )
p
5 25 4 0 3 9 2 8 1 3 0 0 1 5 2 24
x 4 O 2
2
4x
4 8
8 4 4
2 O
2
4x
4
f (x) x 3 2x 2 6
14b. between 2 and 1 14c. Sample answer: rel. max. at x 0, rel. max. at x 1
f (x) x 3 4x 2
13b. at x 4 and x 0 13c. Sample answer: rel. max. at x 0, rel. min. at x 3
©Glencoe/McGraw-Hill
p
2 10 1 3 0 6 1 5 2 6 3 15 4 38
f (x )
f (x )
f (x )
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15a.
16a. x
p
f (x )
f (x )
2 18 1 2 0 2 1 0 2 2 3 2 4 18
x
p
5 9 4 7 3 9
x
O
2 3 1 5 0 9 1 3 2 19
f (x) x 3 3x 2 2
15b. at x 1, between 1 and 0, and between 2 and 3 15c. Sample answer: rel. max. at x 0, rel. min. at x 2
4 8
4
4
O
8x
4 8
f (x ) x 3 5x 2 9
16b. between 5 and 4, between 2 and 1, and between 1 and 2 16c. Sample answer: rel. max. at x 3, rel. min. at x 0 18a.
17a. x
f (x )
f (x )
p
f (x )
f (x )
1 75 0 16 1 3 2 0 3 7 4 0 5 39
x 2 1 0 1 2 3 4 5
4 4
2
O
2
4x
4 8
f (x) 3x 3 20x 2 36x 16
17b. between 0 and 1, at x 2, and at x 4 17c. Sample answer: rel. max. at x 3, rel. min. at x 1
©Glencoe/McGraw-Hill
f (x )
f (x )
p
29 8 1 2 5 4 7 34
x
O
f (x ) x 3 4x 2 2x 1
18b. between 3 and 4 18c. Sample answer: rel. max. at x 0.5, rel. min. at x 2.5
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19a.
20a. x
p
x
f (x )
3 73 2 8 1 7 0 8 1 7 2 8 3 73
3 0 2 15 1 0 0 9 1 0 2 15 3 0 4 105
4 4
2
2
O
f (x )
p
f (x )
x
4
f (x) x 4 8 8
19b. between 2 and 1 and between 1 and 2 19c. Sample answer: no rel. max., rel. min. at x 0
16
f (x )
8 4
2
x
2
O 8 16
f (x ) x 4 10x 2 9
20b. at x 3, x 1, x 1, and x 3 20c. Sample answer: rel. max. at x 0, rel. min. at x 2 and x 2 22a.
21a. x
x
f (x )
p
4 169 3 31 2 7 1 5 0 1 1 1 2 1 3 43
8
3 39 2 5 1 3 0 3 1 5 2 21 3 15 4 67
4 4
2
O
2
4x
4 8
f (x ) x 4 5x 2 2x 1
21b. between 3 and 2, between 1 and 0, between 0 and 1, and between 1 and 2 21c. Sample answer: rel. max. at x 2 and at x 1.5, rel. min. at x 0
©Glencoe/McGraw-Hill
f (x )
p
f (x )
24
f (x )
16 8 4
2
O
2
4x
8
f (x) x 4 x 3 8x 2 3
22b. between 3 and 2, between 1 and 0, between 0 and 1, and between 3 and 4 22c. Sample answer: rel. max. at x 1.5 and at x 2.5, rel. min. at x 0.
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23a.
24a.
x f (x )
p
1 65 0 6 1 1 2 2 3 3 4 10 5 11
x
f (x )
2 1 0 1 2 3
4 2
2
O
4
x
4 8
f (x) x 4 9x 3 25x 2 24x 6
p
f (x ) 45 4 5 6 7 40
4 2
f (x )
2
O
4
6x
4 8
f (x) 2x 4 4x 3 2x 2 3x 5
24b. between 2 and 1, and between 2 and 3 24c. Sample answer: rel. max. at x 0.5; rel. min. at x 0.5 and at x 1.5
23b. between 0 and 1, between 1 and 2, between 2 and 3, and between 4 and 5 23c. Sample answer: rel. max. at x 2, rel. min. at x 0.5 and at x 4
26a.
25a. f (x )
x f (x )
p
4 77 3 30 2 7 1 2 0 3 1 2 2 55
x f (x )
24
2 1
16
0
8 4
2
O
1
2
2
4x
3
f (x) x 5 4x 4 x 3 9x 2 3
4
25b. between 4 and 3, between 2 and 1, between 1 and 0, between 0 and 1, and between 1 and 2 25c. Sample answer: rel. max. at x 3 and at x 0, rel. min. at x 1 and at x 1
©Glencoe/McGraw-Hill
5
p
40
88 5
f (x )
20
6 5 20 3 10 269
4
2
O
2
4x
20 40
f (x) x 5 6x 4 4x 3 17x 2 5x 6
26b. between 2 and 1, between 1 and 0, between 0 and 1, between 2 and 3, and between 4 and 5 26c. Sample answer: rel. max. at x 1 and at x 2, rel. min. at x 0 and at x 3.5
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27. highest: 1982; lowest: 2000
28. Rel. max. between 1980 and 1985 and between 1990 and 1995, rel. min. between 1975 and 1980 and between 1985 and 1990; as the number of years increases, the percent of the labor force that is unemployed decreases.
29. 5
30. Sample answer: increase, based on the past fluctuations of the graph
31.
x 0 2 4 6 8 10 B(x) 25 34 40 45 50 54 G(x) 26 33 39 44 49 53
32. The growth rate for both boys and girls increases steadily until age 18 and then begins to level off, with boys averaging a height of 71 in. and girls a height of 60 in.
x 12 14 16 18 20 B(x) 59 64 68 71 71 G(x) 56 59 61 61 60
Average Height (in.)
y
B (x )
70 65 60 55
G (x )
50 45 40 35 30 25 20 0
2
4
6
8 10 12 14 16 18 x Age (yrs)
33. 0 and between 5 and 6
34. 5 s
35. 3 s
36.
y
O
©Glencoe/McGraw-Hill
176
x
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y
37.
x
O
O
x
O
y
39.
y
38.
40. The turning points of a polynomial function that models a set of data can indicate fluctuations that may repeat. Answers should include the following. • To determine when the percentage of foreign-born citizens was at its highest, look for the rel. max. of the graph, which is at t 5. The lowest percentage is found at t 75, the rel. min. of the graph. • Polynomial equations best model data that contain turning points, rather than a constant increase or decrease like linear equations.
x
41. D
42. B
43. 1.90; 1.23
44. 3.41; 0.59
45. 0; 1.22, 1.22
46. 0.52; 0.39, 1.62
47. 24a3 4a 2 2
48. 10c 2 25c 20
49. 8a4 10a 2 4
50. 3x 3 10x 2 11x 6
51. 2x 4 11x 2 16
52. 4x 4 9x 3 28x 2 33x 20
©Glencoe/McGraw-Hill
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53.
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54.
y
y x 2 4x 6
y
x
O
x
O
y x 2 6x 3
55.
56. (7, 4)
y
x
O 2
y x 2x
57. (3, 2)
58. (4, 2)
59. (1, 3)
60. 1 ft
61. (x 5)(x 6)
62. (2b 1)(b 4)
63. (3a 1)(2a 5)
64. (2m 3)(2m 3)
65. (t 3)(t 2 3t 9)
66. (r 2 1)(r 1)(r 1)
Lesson 7-3 Solving Equations Using Quadratic Techniques Pages 362–364 1. Sample answer: 16x 4 12x 2 0; 4[4(x 2)2 3x 2] 0
2. The solutions of a polynomial equation are the points at which the graph intersects the x-axis.
3. Factor out an x and write the equation in quadratic form so you have x[(x 2)2 2(x 2) 1] 0. Factor the trinomial and solve for x using the Zero Product Property. The solutions are 1, 0, and 1.
4. not possible
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5. 84(n 2 )2 62(n 2)
6. 0, 5, 4
7. 4, 1, 4, 1
8. 6, 3 3i13, 3 3i13
9. 64
10. 8 feet
11. 2(x 2)2 6(x 2) 10
12. not possible
13. 11(n 3 )2 44(n 3)
14. b[7(b 2)2 4(b 2 ) 2)]
15. not possible
16. 6 (x 5 )2 4 (x 5 ) 16 0
17. 0, 4, 3
18. 0, 1, 5
19. 13, 13, i 13, i 13
20. 0, 4, 4, 4i, 4i
21. 2, 2, 212, 212
22. 12, 12, 3, 3
23. 9,
1
9 9i 13 9 9i 13 , 2 2
1
24. 8, 4 4i13, 4 4i13
25. 81, 625
26. 343, 64
27. 225, 16
28. 400
29. 1, 1, 4
30. 8, i 23, i 23
31. w 4 cm, / 8 cm, h 2 cm
32. x 4 7x 2 9 27
33. 3 3 in.
34. 6 6 in.
35. h 2 4, 3h 2, h 3
36. The height increased by 3, the width increased by 2, and the length increased by 4.
37. Write the equation in quadratic form, u 2 9x 8 0, where u |a 3|. Then factor and use the Zero Product Property to solve for a; 11, 4, 2, and 5.
38. Answers should include the following. • Solve the cubic equation 4x 3 (164x 2) 1600x 3600 in order to determine the dimensions of the cut square if the desired volume is 3600 in3. Solutions are 10 in. and 31 2601 2
in. • There can be more than one square cut to produce the same volume because the height of the box is not specified and 3600 has a variety of different factors. ©Glencoe/McGraw-Hill
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39. D
40.
41.
42. x
f (x )
f (x )
p
2 21 1 1 0 5 1 3 2 1 3 1 4 9 5 35
O
x
1715 ; 3
p
f (x )
f (x )
1 15 0 3 1 1 2 3 3 3 4 25
x
x
O
f (x) x 4 6x 3 10x 2 x 3
f (x) x 3 4x 2 x 5
44. 262; 2 2 3 3 46. B R 1 3 1 y 48.
43. 17; 27 45.
1 18
135
47. A¿(1, 2), B¿(3, 3), C¿(1, 3)
C' A O
C x
A' B
B' 64 x2
49. x 2 5x 4
50. 4x 2 16x 27
51. x 3 3x 2 2
52. x 3 2x 2 10x 15
©Glencoe/McGraw-Hill
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Algebra 2
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Chapter 7
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Chapter 7 Practice Quiz 1 Page 364 1. 2a3 6a 2 5a 1
2. f(x ) S as x S , f(x ) S as x S ; odd; 3
3. Sample answer: maximum at x 2, minimum at x 0.5
4. (6x 3)2 3(6x 3) 5 or 3 3 36(2x)2 18(2x) 5
8
1
1
f (x )
4 4
2
O
4x
2
4 8
f (x) x 3 2x 2 4x 6
5. 3, 3, i13, i13
Lesson 7-4
The Remainder and Factor Theorems Pages 368–370
1. Sample answer: f(x ) x 2 2x 3
2. 4
3. dividend: x 3 6x 32; divisor: x 2; quotient: x 2 2x 10; remainder: 12
4. 7, 91
5. 353, 1186
6. x 1, x 3
7. x 1, x 2
8. 2x 1, x 4
9. x 2, x 2 2x 4
10. $2.894 billion
11. $2.894 billion
12. Sample answer: Direct substitution, because it can be done quickly with a calculator.
13. 9, 54
14. 37, 19
15. 14, 42
16. 55, 272
©Glencoe/McGraw-Hill
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17. 19, 243
18. 267, 680
19. 450, 1559
20. 422, 3110
21. x 1, x 2
22. x 4, x 2
23. x 4, x 1
24. x 3, x 1
25. x 3, x
1 2
26. x 1, x
or 2x 1
4 3
or 3x 4
27. x 7, x 4
28. x 1, x 6
29. x 1, x 2 2x 3
30. 2x 3, 2x 3, 4x 2 9
31. x 2, x 2, x 2 1
32.
33. 3
34. 8
35. 1, 4
36. 3
37.
5 1 14 69 140 100 5 45 120 100 1 9 24 20 0
1 4 29 24 8 32 24; 1 4 3 0 (x 3)(x 1) 8
38. Yes; 2 ft lengths. The binomial x 2 is a factor of the polynomial since f (2) 0.
39. 7.5 ft/s, 8 ft/s, 7.5 ft/s
40. 0; The elevator is stopped.
41. By the Remainder Theorem, the remainder when f(x ) is divided by x 1 is equivalent to f(1), or a b c d e. Since a b c d e 0, the remainder when f (x ) is divided by x 1 is 0. Therefore, x 1 is a factor of f(x ).
42. $31.36
43. $16.70
44. B(x) 2000x 5 340(x 4 x 3 x 2 x 1)
©Glencoe/McGraw-Hill
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45. No, he will still owe $4.40.
46. Using the Remainder Theorem you can evaluate a polynomial for a value a by dividing the polynomial by x a using synthetic division. Answers should include the following. • It is easier to use the Remainder Theorem when you have polynomials of degree 2 and lower or when you have access to a calculator. • The estimated number of international travelers to the U.S. in 2006 is 65.9 million.
47. D
48. x 2, x 2, x 1, x 2 1
49. (x 2)2 8(x 2) 4
50. 9(d 3)2 5(d 3) 2
51. not possible
52. Sample answer: rel. max, at x 0.5, rel. min. at x 3.5 f (x ) 16 f (x) x 3 6x 2 4x 3 8 2
O
2
4
x
8 16
54. T
53. Sample answer: maximum at x 1, rel. max. at x 1.5, rel. min. at x 1
2 2mrFc Fc
f (x ) 8 4 4
2
O
2
4x
4
f (x) x 4 2x 3 3x 2 7x 4
©Glencoe/McGraw-Hill
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55. (4, 2)
56. (3, 1)
57. A
58. C
59. S
60.
7 117 2
62.
3 i 17 4
61.
9 157 6
Lesson 7-5 Roots and Zeros Pages 375–377 1. Sample answer: p(x ) x 3 6x 2 x 1; p(x ) has either 2 or 0 positive real zeros, 1 negative real zero, and 2 or 0 imaginary zeros.
2. An odd-degree function approaches positive infinity in one direction and negative infinity in the other direction, so the graph must cross the x-axis at least once, giving it at least one real root.
3. 6
4. 2i, 2i; 2 imaginary
5. 7, 0, and 3; 3 real
6. 2 or 0; 1; 2 or 0
7. 2 or 0; 1; 2 or 4
8. 4, 1 2i, 1 2i 10. 2i, 2i, 3
9. 2, 1 i, 1 i
12. f (x ) x 3 2x 2 16x 32
11. 2 3i, 2 3i, 1 8 3
5 i 271 ; 4
13. ; 1 real
14.
15. 0, 3i, 3i; 1 real, 2 imaginary
16. 3i, 3i, 3i, and 3i; 4 imaginary
17. 2, 2, 2i, and 2i; 2 real, 2 imaginary
18. 2, 2, 0, 2, and 2, 5 real
19. 2 or 0; 1; 2 or 0
20. 2 or 0; 1; 2 or 0
21. 3 or 1; 0; 2 or 0
22. 1; 3 or 1; 2 or 0
23. 4, 2, or 0; 1; 4, 2, or 0
24. 5, 3, or 1; 5, 3, or 1; 0, 2, 4, 6, or 8
25. 2, 2 3i, 2 3i
26. 4, 1 i, 1 i
©Glencoe/McGraw-Hill
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2 imaginary
Algebra 2
Chapter 7
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i , 2
27. 2i, 2i,
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i 2
28. 5i, 5i, 7
3 2
1 , 2
29. , 1 4i, 1 4i
30.
31. 4 i, 4 i, 3
32. 3 i, 3 i, 4, 1
33. 3 2i, 3 2i, 1, 1
34. 5 i, 5 i, 1, 6
35. f(x ) x 3 2x 2 19x 20
36. f(x ) x 4 10x 3 20x 2 40x 96
37. f(x ) x 4 7x 2 144
38. f(x ) x 5 x 4 13x 3 13x 2 36x 36
39. f(x ) x 3 11x 2 23x 45
40. f(x) x 3 10x 2 32x 48
41a.
42. (3 x)(4 x)(5 x) 24
f (x )
x
O
41b.
4 5i, 4 5i
f (x )
x
O
41c.
f (x )
O
©Glencoe/McGraw-Hill
x
185
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43. 1 ft
44. V(r ) r 3 17 r 2
45. radius 4 m, height 21 m
46. 1; 2 or 0; 2 or 0
47. 24.1, 4.0, 0, and 3.1
48. Nonnegative roots represent times when there is no concentration of dye registering on the monitor.
[30, 10] scl: 5 by [20, 20] scl: 5
49. Sample answer: f(x) x 3 6x 2 5x 12 and g(x) 2x 3 12x 2 10x 24 each have zeros at x 4, x 2, and x 3.
50. One root is a double root. Sample graph: f (x )
O
©Glencoe/McGraw-Hill
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x
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51. If the equation models the level of a medication in a patient’s bloodstream, a doctor can use the roots of the equation to determine how often the patient should take the medication to maintain the necessary concentration in the body. Answers should include the following. • A graph of this equation reveals that only the first positive real root of the equation, 5, has meaning for this situation, since the next positive real root occurs after the medication level in the bloodstream has dropped below 0 mg. Thus according to this model, after 5 hours there is no significant amount of medicine left in the bloodstream. • The patient should not go more than 5 hours before taking their next dose of medication.
52. A
53. C
54. 127, 41
55. 254, 915
56. 36 in.
57. min.; 13
58. max.; 32
59. min.; 7
60. 5ab 2(3a c 2)
61. (6p 5)(2p 9) 3 2 63. C 3 4 S 2 9
62. 4y (y 3)2 11 5 64. C 7 0S 4 5
©Glencoe/McGraw-Hill
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29 8 65. £ 8 9§ 16 16 1 2
2 3
66. y x 1
5 2
68.
67. , 1, , 5 1 9
1 , 14
1 16
1 3
1 7
2 7
1 2
1 8
1 4
1 2
, , , 1, 2
70. , , , , 1, 2, 4
69. , , 1, 3
Lesson 7-6 Rational Zero Theorem Pages 380–382 1. Sample answer: You limit the number of possible solutions.
2. Sample answer: 2x 2 x 3
3. Luis; Lauren found numbers q p in the form , not as p q Luis did according to the Rational Zero Theorem.
4. 1, 2, 5, 10
1 2
1 3
1 6
2 3
5. 1, 2, , , ,
6. 4, 2, 7
7. 2, 4, 7
8. 2, 2, 3, 3
9. 2, 2,
7 2
10.
2 3 217 , 3 4
11. 10 cm 11 cm 13 cm
12. 1, 2
13. 1, 2, 3, 6
14. 1, 3, 5, 15, ,
15. 1, 2, 3, 6, 9, 18
16. 1, , 3
1 3
1 3
1 3
1 9
17. 1, , , 3, 9, 27
18. 6, 5, 10
19. 1, 1, 2
20. 1, 1
21. 0, 9
22.
23. 0, 2, 2
24. 0, 3
25. 2, 4
26. 7, 1, 3
©Glencoe/McGraw-Hill
5 3
188
1 , 2
1, 1
Algebra 2
Chapter 7
PQ245-6457F-P07[167-202].qxd
27.
1 , 2
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1 3
1 1 2 5
, 2
28. , , 2
1 1 1 3 2 3 2 4
30. 2, ,
4 3 i 3 2
29. , , , 31.
4 , 5
0,
5 i 13 2
2 3 213 3 2
2 3
32. 3, , , 1 3
4 3
33. 1, 2, 5, i, i
34. V r 3 r 2
35. 2, 2 i 13; 2
36. r 2 in., h 6 in.
37. V 2h3 8h 2 64h
38. / 36 in., w 48 in., h 32 in.
˛
˛
1 3
1 3
39. V / 3 3/ 2
40. 6300 /3 3/2
41. / 30 in., w 30 in., h 21 in.
42. k 3; 3, 6, 5
43. The Rational Zero Theorem helps factor large numbers by eliminating some possible zeros because it is not practical to test all of them using synthetic substitution. Answers should include the following. • The polynomial equation that represents the volume of the compartment is V w 3 3w 2 40w. • Reasonable measures of the width of the compartment are, in inches, 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 22, 28, 33, 36, 42, 44, 63, 66, 77, and 84. The solution shows that w 14 in., / 22 in., and d 9 in.
44. D
45. Sample answer: x 5 x 4 27x 3 41x 2 106x 120
46. 6, 3, 5
©Glencoe/McGraw-Hill
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47. 4, 2 i, 2 i
48. 5, 3i, 3i
49. 7, 5 2i, 5 2i
50. 4x 3, 5x 1
51. x 4, 3x 2 2
52. 725
54. 0 4x 5 0
53. 3xy 22x 55. 6 cm, 8 cm, 10 cm
56. x 3 4x 2 6
57. 4x 2 8x 3
58. x 3 5x 2 x 10
59. x 5 7x 4 8x 3 106x 2 85x 25
60. x 9
61. x 2 x 4
33 x7
5 x1
Chapter 7 Practice Quiz 2 Page 382 1. 930, 145
2. 0, 180
3. x 4 4x 3 7x 2 22x 24 0
4.
4 5
3 2
5.
Lesson 7-7
Operations on Functions Pages 386–389
1. Sometimes; sample answer: If f(x ) x 2, g(x ) x 8, then f g x 6 and g f x 6.
2. Sample answer: g(x ) {(2, 1), (1, 2), (4, 3)}, f (x ) {(1, 7),(2, 9), (3, 3)}
3. Danette; [g f ](x ) g [f (x )] means to evaluate the f function first and then the g function. Marquan evaluated the functions in the wrong order.
4. 4x 9; 2x 1; 3x 2 19x 20; 3x 4,
©Glencoe/McGraw-Hill
x 5
190
x5
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5. x 2 x 1; x 2 x 7; x 3 4x 2 3x 12; x2 3 , x4
6. {(5, 7), (4, 9)}; {(4, 12)}
x 4
7. {(2, 7)}; {(1, 0), (2, 10)}
8. 6x 8; 6x 4
9. x 2 11; x 2 10x 31
10. 30
11. 11
12. 1 3 4
13. p(x ) x; c (x ) x 5
14. $32.49; price of CD when 25% discount is taken and then the coupon is subtracted
15. $33.74; price of CD when coupon is subtracted and then 25% discount is taken
16. Discount first, then coupon; sample answer: 25% of 49.99 is greater than 25% of 44.99. 18. 6x 6; 2x 12; 8x 2 6x 27;
17. 2x ; 18; x 2 81;
x9 , x9
x 9
2x 3 , 4x 9
19. 2x 2 x 8; 2x 2 x 8; 2x 3 16x 2;
21.
2x 2 , 8x
20. x 2 8x 15; x 2 4x 3; 2x 3 18x 2 54x 54;
x 8
x3 , 2
x3 x2 1 , x 1 x1 x 3 x 2 2x 1 , x1
22.
x 3
x 3 x 2 7x 15 , x 2; x2 x 3 x 2 9x 9 , x 2; x2
x 2 6x 9, x 2; x 2 4x 4, x 2, 3
x 1; x 2 x ; x 1 x3 x2 x 1 , x
9 4
x
x 0
23. {(1, 3), (3, 1), (2, 1)}; {(1, 0), (0, 1)}
24. {(2, 4), (4, 4)}; {(1, 5), (3, 3), (5, 3)}
25. {(0, 0), (8, 3), (3, 3)}; {(3, 6), (4, 4), (6, 6), (7, 8)}
26. {(4, 5), (2, 5), (6, 12), (8, 12)}; does not exist
27. {(5, 1), (8, 9)}; {(2, 4)}
28. {(2, 3), (2, 2)}; {(5, 6), (8, 6), (9, 5)}
©Glencoe/McGraw-Hill
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Algebra 2
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29. 8x 4; 8x 1
30. 15x 5; 15x 1
31. x 2 2; x 2 4x 4
32. 3x 2 4; 3x 2 24x 48
33. 2x 3 2x 2 2x 2; 8x 3 4x 2 2x 1
34. 2x 2 5x 9; 2x 2 x 5
35. 12
36. 50
37. 39
38. 68
39. 25
40. 48
41. 2
42. 1
43. 79
44. 104
45. 226
46. 36
47. P(x ) 50x 1939
48. 939,000
49. p(x ) 0.70x; s(x ) 1.0575x
50. s[p(x )]; The 30% would be taken off first, and then the sales tax would be calculated on this price.
51. $110.30
52. [K C](F )
53. 373 K; 273 K
54. 309.67 K
55. $700, $661.20, $621.78, $581.73, $541.04
56. 244
57. Answers should include the following. • Using the revenue and cost functions, a new function that represents the profit is p(x ) r(c(x )). • The benefit of combining two functions into one function is that there are fewer steps to compute and it is less confusing to the general population of people reading the formulas.
58. A
59. C
60. 1, 2, 4, 8
©Glencoe/McGraw-Hill
1 2
192
5 9
(F 32) 273
Algebra 2
Chapter 7
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3 2
1 3
61. 1, , , 2, 3, ,
1 9
62. 1, ,
3 , 6 4
63. x 3 4x 2 17x 60
64. x 3 3x 2 34x 48
65. 6x 3 13x 2 9x 2
66. x 3 6x 2 4x 24
67. x 3 9x 2 31x 39
68. x 4 x 3 14x 2 26x 20 5 6 1 70. B R 2 7 8 72. does not exist
69. 10 2j 3 2 71. c d 1 1 1 2
1 2 R 3 4
75.
1 16
B
76. x
6 3y 2
77. y
1 4x 2 5x
78. x
2 3 7y
79. t
I pr
80. F C 32
73. B
74. does not exist
5 2 R 3 2
81. m
9 5
Fr 2 GM
Lesson 7-8
Inverse Functions and Relations Pages 393–394
1. no
2. Switch x and y in the equation and solve for y.
3. Sample answer: f (x) 2x, f 1(x) 0.5x; f [f 1(x )] f 1[f (x )] x
4. n is an odd whole number.
5. {(4, 2), (1, 3), (8, 2)}
6. {(3, 1), (1, 1), (3, 1), (1, 1)}
©Glencoe/McGraw-Hill
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Algebra 2
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1 3
8. g 1(x ) x
7. f 1(x ) x 4
f (x )
4
2
g (x )
f (x) x f 1(x ) x
2 O
g (x ) 3x 1 2
4x
2
4
2
2
O
2
4x
2
g1(x) 13 x 13
4
9. y 2x 10 12
1 3
10. yes
y y 1x 5 2
8 4 O
4
4
8
y
1
12
x
2x 10
11. no
12. 32.2 ft/s2
13. 15.24 m/s2
14. {(6, 2), (5, 4), (1, 3)}
15. {(8, 3), (2, 4), (3, 5)}
16. {(4, 7), (5, 3), (4, 1), (5, 7)}
17. {(2, 1), (2, 3), (4, 1), (6, 0)}
18. {(11, 6), (7, 2), (3, 0), (3, 5)}
19. {(8, 2), (5, 6), (2, 8), (6, 5)}
20. x 3 y 4 x 3 2 4
2 O
2
4x
2 4
©Glencoe/McGraw-Hill
194
y 3
Algebra 2
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1 2
22. f 1(x ) x 5
21. g1(x ) x g (x )
4
f 1(x) x 5
g (x ) 12 x 2 1
2
x
x 4
2
O
2
4
4
2 O
g (x ) 2x
24. f 1(x ) x 1
g (x ) 4
f (x ) 4
2 g (x ) x 4 O
2
2 f (x) 3x 3 4x
4
g1(x) x2 4
2 O
4
1 2
1 2
25. y x 1
2
y 1 3x
1x 1 2 2
O
f 1(x) 13 x 1
2
4
y
2 4
4x
2
O
2
4x
2
2 y 2x 1 4
©Glencoe/McGraw-Hill
4x
26. y 3x
y
4
4
2
2
4
y
4
1 3
23. g1(x ) x 4
2
2
2 f (x) x 5 4
2
4
f (x )
4
y 13 x
195
4
Algebra 2
Chapter 7
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8 5
27. f 1(x ) x
28. f 1(x ) 3x 12 f (x ) 8 7 f (x) 1 x 4 3 6 5
f (x )
4 2 4
2
O
4x
2
3 2 1
f (x) 58 x
2 4
O
1 2 3 4 5 6 7 8x
f 1(x ) 3x 12
f 1(x) 85 x
5 4
29. f 1(x ) x
35 4
30. g1(x ) 3x
f (x)1 54 x 35 4 40
4
f (x ) x O 30 20 10
4
4
2
O
20
2
30
4
f (x) 45 x 7 40
8 7
g (x )
g (x) 2x 6 32
10
31. f 1(x ) x
3 2
2
4x
g1(x) 3x 32
4 7
32. yes
f (x )
f 1(x) 87 x 2 47 4
2
O 2
2
4x
f (x) 7x 8 4
4
33. no
34. no
35. yes
36. yes
37. yes
38. y 1 2
39. y x
©Glencoe/McGraw-Hill
11 2
4(x 7) 6 2
40. 12
196
Algebra 2
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9 5
42. C 1(x ) x 32;
41. I(m) 320 0.04m; $4500
C [C1(x )] C 1[C (x )] x 43. It can be used to convert Celsius to Fahrenheit.
44. Sample answer: f(x ) x and f 1(x ) x or f(x ) x and f 1(x ) x
45. Inverses are used to convert between two units of measurement. Answers should include the following. • Even if it is not necessary, it is helpful to know the imperial units when given the metric units because most measurements in the U.S. are given in imperial units so it is easier to understand the quantities using our system. • To convert the speed of light from meters per second to miles per hour, 3.0 108 meters 1 second 3600 seconds 1mile 1 hour 1600 meters
46. A
f (x)
675,000,000 mi/hr 47. B
48. g [h(x)] 4x 20; h[g(x)] 4x 5
49. g[h(x )] 6x 10; h[g(x )] 6x
50. g [h(x)] x 2 3x 24; h[g(x)] x 2 5x 24
51. 7, 2, 3
52. , ,
53. 64
54. 32
55. 3
56. 4
57. 117
58. 196
©Glencoe/McGraw-Hill
1 4 5 4 3 2
197
Algebra 2
Chapter 7
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60.
59. 7 61.
25 4
Lesson 7-9
Square Root Functions and Inequalities Pages 397–399
1. In order for it to be a square root function, only the nonnegative range can be considered.
2. Both have the shape of the graph of y 2x, but y 2x 4 is shifted down 4 units, and y 2x 4 is shifted to the right 4 units.
3. Sample answer: y 22x 4
4.
8 7 6 5 4 3 2 1
y
y x 2 1 2 3 4 5 6 7 8x
O
D: x 0, R: y 2 5.
8 7 6 5 4 3 2 1
6.
y
4
y 3 x 2
x y 4x
O
O
8
12
1 2 3 4 5 6 7 8x
D: x 0; R: y 0 8 7 6 5 4 3 2 1
4
2
O
7.
y
D: x 0; R: y 3 8.
y
y x 1 3
1 2 3 4 5 6 7 8x
8 7 6 5 4 3 2 1 O
y
y x 4 1
1 2 3 4 5 6 7 8x
D: x 1; R: y 3 ©Glencoe/McGraw-Hill
198
Algebra 2
Chapter 7
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9.
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8
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10.
y
6
y 2x 4
4
1 O
O
11.
4 3 2 1 2
O
2
y
4
y y 3 5x 1 1 2 3 4 5 6 7x
2 3 4
2 2
4 3 2 1
6x
12. v 22gh
y x 2 1
1 2 3 4 5 6x
2 3 4
14.
13. Yes; sample answer: The advertised pump will reach a maximum height of 87.9 ft.
8 7 6 5 4 3 2 1
y
y 3x 1 2 3 4 5 6 7 8x
O
D: x 0, R: y 0 15.
16.
y O 1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8x
y 5x
1 2 3 4 5 6 7 8x
y 4x
D: x 0, R: y 0
D: x 0, R: y 0
©Glencoe/McGraw-Hill
y O 1 2 3 4 5 6 7 8
199
Algebra 2
Chapter 7
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18.
y
8 7 6 5 4 3 2 1
2
2 2
1 2 3 4 5 6 7 8x
20.
1 2 3 4 5 6 7 8x
D: x 7, R: y 0
1 2 3 4 5 6 7 8x
y 2x 1
22.
4
y
2
x
O 4
6
y 5x 3
8
D: x 6, R: y 3 24.
y
6 4
y 5 x 4
2 O 2
2
4x
8 7 6 5 4 3 2 1 O
D: x 4, R: y 5
©Glencoe/McGraw-Hill
2
y x 6 3 4
1 2 3 4 5 6 7 8x
23.
2 2
D: x 0.6, R: y 0
4
6x
4
D: x 0.5, R: y 0
y
O
2
y O 1 2 3 4 5 6 7 8
y x 7
O
O
D: x 2, R: y 0
y
8 7 6 5 4 3 2 1
8 7 6 5 4 3 2 1
y x 2
4
y 1 x
D: x 0, R: y 0
21.
y
6
O
19.
8
y
y 3x 6 4
1 2 3 4 5 6 7 8x
D: x 2, R: y 4
200
Algebra 2
Chapter 7
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25.
8
Page 201 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
26.
y
y O
y 23 4x 3 6 4 2 3
2
1
x
O
1 2 3 4 5 6 7 8x 2 4 6 8 10 12 y 6x 14 16
D: x 0.75, R: y 3 27.
8
y x 5
28.
y
8
6
6
4
4
2
2 O
4
29.
8 7 6 5 4 3 2 1 O
31.
8 7 6 5 4 3 2 1 O
2
y
y 2x 8
O 2
x
4
30.
y
y 5x 8 1 2 3 4 5 6 7 8x
8 7 6 5 4 3 2 1 O
2
2
4x
y
y x 3 4
1 2 3 4 5 6 7 8x
32. 125 ft
y
y 6x 2 1
1 2 3 4 5 6 7 8x
33. 317.29 mi
34. 119 lb
35. See students’ work.
36. If a is negative, the graph is reflected over the x-axis. The larger the value of a, the less steep the graph. If h is positive, the origin is
©Glencoe/McGraw-Hill
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Algebra 2
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translated to the right, and if h is negative, the origin is translated to the left. When k is positive, the origin is translated up, and when k is negative, the origin is translated down. 37. Square root functions are used in bridge design because the engineers must determine what diameter of steel cable needs to be used to support a bridge based on its weight. Answers should include the following. • Sample answer: When the weight to be supported is less than 8 tons. • 13,608 tons
38. C
39. D
40. yes
41. no
42. yes
43. 2x 2; 8; x 2 2x 15;
44. 11x 22; 9x18; 10x 2 40x 40; 10; x 2
45.
x5 ,x 3 x3 8x 3 12x 2 18x 26 , 2x 3 3 x ; 2 3 8x 12x 2 18x 28 , 2x 3 3 3 x ; 2x 3, x ; 2 2
46. 4; If x is your number, you can write the expression 3x x 8 , x2
which equals 4
after dividing the numerator and denominator by the GCF, x 2.
8x 3 12x 3 18x 27, 3 2
x
47. 2x 2 4x 16
48. 6p 2 2p 20
49. a3 1
©Glencoe/McGraw-Hill
202
Algebra 2
Chapter 7
Chapter 8 Conic Sections Lesson 8-1 Midpoint and Distance Formulas Pages 414–416 1. Since the sum of the x-coordinates of the given points is negative, the x-coordinate of the midpoint is negative. Since the sum of the y-coordinates of the given points is positive, the y-coordinate of the midpoint is positive. Therefore, the midpoint is in Quadrant II.
2. all of the points on the perpendicular bisector of the segment
4. a2,
3. Sample answer: (0, 0) and (5, 2)
13 b 2
5. (2.5, 2.25)
6. 10 units
7. 1122 units
8. 12.61 units 10. (12, 5)
9. D 11. (4, 2)
12. (2, 6)
13. a ,
14. (0.075, 3.2)
15. (3.1, 2.7)
16. a , b
17. a ,
18. a ,
17 27 b 2 2
1 24
5 12
5 24
1 13 5 b, a , 2 2 2
5 b 8
2b, a5, b 1 2
19. (7, 11)
20. around 8th St. and 10th Ave.
21. Sample answer: Draw several line segments across the U.S. One should go from the NE corner to the SW corner; another should go from the SW corner to the NW corner; another should go across the middle (east to west); and so on. Find the midpoints of these segments. Locate a point to represent all of these midpoints.
22. near Lebanon, Kansas
©Glencoe/McGraw-Hill
203
Algebra 2
Chapter 8
23. See students’ work.
24. 13 units
25. 25 units
26. 12 units
27. 3117 units
28. 0.75 unit
29. 170.25 units
30. 165 units
31. 1 unit
32. 1271 units
33.
1813 12
34. 6110 units, 90 units2
units
35. 712 158 units, 10 units2
36. 165 2 12 1122 1277 units
37. 1130 units
38. about 85 mi
39. about 0.9 h
40. 14 in.
41. The slope of the line through
42. The formulas can be used to decide from which location an emergency squad should be dispatched. Answers should include the following. • Most maps have a superimposed grid. Think of the grid as a coordinate system and assign approximate coordinates to the two cities. Then use the Distance Formula to find the distance between the points with those coordinates. • Suppose the bottom left of the grid is the origin. Then the coordinates of Lincoln are about (0.7, 0.3); the coordinates of Omaha are about (4.6, 3.3); and the coordinates of Fremont are about (1.5, 4.6). The distance from Omaha to Fremont is about
y y
1 (x1, y1) and (x2, y2) is 2 x2 x1 and the point-slope form of the equation of the line is
y2 y1 (x x1). x2 x1 x x2 y1 y2 Substitute ¢ 1 , ≤ 2 2
y y1
into this equation. The left
y2 y1 . 2 y y1 The right side is 2 x2 x1 y2 y1 x2 x1 x1 x2 ¢ x1≤ ≤ ¢ x2 x1 2 2 y y1 or 2 . Therefore, the 2
side is
y1 y2 2
y1 or
point with coordinates ¢
x1 x2 y1 y2 , ≤ 2 2
lies on the
line through (x1, y1) and (x2, y2). The distance from ¢
x1 x2 y1 y2 , ≤ 2 2
B
¢x1
or
to (x1, y1) is
y1 y2 2 x1 x2 2 ≤ ¢y1 ≤ 2 2
102(1.5 4.6)2 (4.6 3.3)2
or 34 miles. The distance from Lincoln to Fremont is
y1 y2 2 x1 x2 2 ≤ ¢ ≤ . The B 2 2 ¢
©Glencoe/McGraw-Hill
204
Algebra 2
Chapter 8
distance from ¢
x1 x2 y1 y2 , ≤ 2 2
about 102(1.5 0.7)2 (4.6 0.3)2
to (x2, y2) is B
¢x1
or 44 miles. Since Omaha is closer than Lincoln, the helicopter should be dispatched from Omaha.
y1 y2 x1 x2 b ¢y2 ≤ 2 2 2
2
y2 y1 2 x2 x1 2 ≤ ¢ ≤ B 2 2 ¢
or
y1 y2 2 x1 x2 2 ≤ ¢ ≤. B 2 2 ¢
Therefore, the point with coordinates ¢
x1 x2 y1 y2 , ≤ 2 2
is equidistant from (x1, y1) and (x2, y2). 43. C
44. B
45. on the line with equation y x
¿ is perpendicular to 46. 1; AA the line with equation y x, which has slope 1. 48. D 5x 0 x 06, R 5y 0 16
47. D 5x 0 x 26, R 5y 0 y 06 y
y
y 兹x 1
y 兹x 2
O
x
x
O
49. D 5x 0 x 06, R 5y 0 y 16
50. no
y
y 2兹x 1
O
x
51. 1 13i
52. 6 2i
53. 4 3i
54. y (x 3)2
©Glencoe/McGraw-Hill
205
Algebra 2
Chapter 8
55. y (x 2)2 3
56. y 2(x 5)2
57. y 3(x 1)2 2
58. y (x 2)2 10
59. y 3(x 3)2 17
Lesson 8-2 Parabolas Pages 423–425 1. (3, 7), a3, 6 b, x 3, 15 16
y 7
2. Sample answer: x y 2
1 16
3. When she added 9 to complete the square, she forgot to also subtract 9. The standard form is y (x 3)2 9 4 or y (x 3)2 5.
4. y 2(x 3)2 12
5. (3, 4), a3, 3 b, x 3,
6. (7, 3), a7, 3 b, x 7, 1 8
3 4
1 4
7 8
y 4 , upward, 1 unit
y 2 , upward,
y
1 2
unit
y 16 14 12 10 8 6 4 y 2(x 7)2 3 2
x
O
1412108 6 4 2
y (x 3) 2 4
©Glencoe/McGraw-Hill
206
2x
Algebra 2
Chapter 8
7. a , b, a , b, x , 4 3
y
2 3 7 , 12
4 3
3 4
downward,
y 3x 2 8x 6
8. a , b, a , b, y ,
4 3
1 3
3 9 2 2
9 9 8 2
15 8
x , right,
unit
9 2
3 2
units
y
y x
O
x 2 y 2 6y 12 3
x
O
1 8
1 8
9. y (x 3)2 6
10. x (y 1)2 5 y
y
8
8
x
O
11. x
x
O x 1 (y 1) 2 5
y 1 (x 3) 2 6
1 2 y 24
12. y (x 3)2 2
6
1 2
13. x (y 7)2 29 5 2
14. y (x 12)2 80
15. x 3 ay b 11 6
16. (0, 0), a0, b, x 0, y , 3 2
1 12
3 2
downward, 6 units y 6y x 2 O
©Glencoe/McGraw-Hill
207
x
Algebra 2
Chapter 8
17. (0, 0), a , 0b, y 0, x ,
18. (6, 3), a6, 3 b, x 6,
1 2
1 2
3 4
1 4
right, 2 units
y 2 , upward, 3 units
y
y y 2 2x x
O
3(y 3) (x 6)2 O x
19. (1, 4), a1, 3 b, x 1, y 4 , 1 2
1 2
20. (2, 3), (3, 3), y 3, x 1, right, 4 units
downward, 2 units
y
y
x
O
4(x 2) (y 3)2 O
2(y 4) (x 1)2
x
22. (6, 16), a6, 15 b, x 6, 3 4
21. (4, 8), (3, 8), y 8, x 5, left, 4 units 16 14 12 10 8 6 4 2 4 321
©Glencoe/McGraw-Hill
1 4
y 16 , upward, 1 unit
y
2 2 4 6 8 10 12 14 16
(y 8)2 4(x 4) O 1 2 3 4x
208
y O2 4 6 8 10 12 14 x
y x 2 12x 20
Algebra 2
Chapter 8
24. a
5 115 144 , b, a , 4 2 5 287 1 x , right, 10 5
23. (24, 7), a23 , 7b, y 7, 3 4
1 4
x 24 , right, 1 unit 24
y
y
x y 2 14y 25 16
8
8x
O
1 b, 12
5 2
unit
x 5y 2 25y 60
2 1
2 3 4 5 6
8
25. (4, 2), a4, 2
5 2
O 10 20 30 40 50 60 70 80 x
8 24 16
b, y ,
26. a , b, a , 7b, x , 5 4
x 4,
11 12
y 1 , upward,
1 3
y
unit
55 5 5 8 4 4 27 1 , downward, unit 4 2 y O
y 2
2
x
4
4 x 2x 2 5x 10 8 12
y 3x 2 24x 50
16
x
O
27. a , b, a , b, y , 17 3 67 3 4 4 16 4 69 1 x , left, 4 16
28. (3, 5), a3, 5 b, x 3, y 4 ,
3 4
1 2
1 2
upward, 2 units
unit
y
y x 4y 2 6y 2
y 1 x 2 3x 19
O
2
x
2
O
x
©Glencoe/McGraw-Hill
209
Algebra 2
Chapter 8
29. (123, 18), a122 , 18b, y 18,
1 4 3 x 123 , left, 4
20 120 60
O
30.
y x 3y 2 4y 1
3 units
y
O
x 120x
60
20 40 60
x 1 y 2 12y 15 3
1 3
32. 1 and
31. 1 33. y
34. a , b
35. 0.75 cm
36. y
1 3
2 3
2 3
1 2 x 16
1 y
y 1 x2 1 16
x
O
1 (y 24
37. x y
O
1 8
38. x (y 2)2 6
6)2 8
y
14 12 10 8 6 x 1 (y 6)2 8 24 4 2
x
O
x 1 (y 2)2 6 8
1 2 3 4 5 6 7 8x 2
©Glencoe/McGraw-Hill
210
Algebra 2
Chapter 8
39. y
1 (x 16
1 6
40. y (x 7)2 4
1)2 7 8 6 4 2
4 321 2 4 6 8
y
8
y 1 (x 7)2 4 6
y
4
y 1 (x 1)2 7 16
O 12
O1 2 3 4 5 6 x
8
4
x 4
XBox.
8
2 9
1 4
42. y x 2 2
41. x (y 3)2 4 y x 1 (y 3)2 4 4
x
O
1 (x 100
44. y
43. about y 0.00046x 2 325 1 x2 26,200
45. y
46. x (y 3)2 4
6550
47. A parabolic reflector can be used to make a car headlight more effective. Answers should include the following. • Reflected rays are focused at that point. • The light from an unreflected bulb would shine in all directions. With a parabolic reflector, most of the light can be directed forward toward the road.
48. B
49. A
50. 13 units
51. 10 units
52. 234 units
©Glencoe/McGraw-Hill
50)2 25
211
Algebra 2
Chapter 8
53.
54. 2.016 105
y
O
y 兹x 1
x
55. 4
56. 5
57. 9
58. 12
59. 223
60. 322
61. 423
62. 622
Lesson 8-3 Circles Pages 428–431 1. Sample answer: (x 6)2 (y 2)2 16
2. (x 3)2 (y 1)2 64; left 3 units, up 1 unit
3. Lucy; 36 is the square of the radius, so the radius is 6 units.
4. (x 3)2 (y 1)2 9
5. (x 1)2 (y 5)2 4
6. x 2 (y 2)2 25
7. (x 3)2 (y 7)2 9
8. (4, 1), 3 units y
O
x (x 4)2 (y 1)2 9
©Glencoe/McGraw-Hill
212
Algebra 2
Chapter 8
9. (0, 14), 234 units 24
10. (4, 0),
y
4 5
unit
y
x 2 (y 14)2 34
(x 4)2 y 2 16
16
25
8
x
O 16
8
16x
8
O 8
11. a , b
212 3
2 1 3 2
12. (4, 3), 5 units
unit
y
y
(x 4)2 (y 3)2 25
x
O 2
2
(x 23 ) (y 12 )
8
O
9
x
14. x 2 y 2 42,2002
13. (2, 0), 223 units y
x
O
(x 2)2 y 2 12
16. (x 1)2 (y 1)2 16
15. y Earth Satellite 35,800 km x 6400 km
©Glencoe/McGraw-Hill
42,200 km
213
Algebra 2
Chapter 8
17. (x 2)2 (y 1)2 4 19. (x 8)2 (y 7)2 1 2 2
21. (x 1)2 ay b
18. x 2 (y 3)2 49
1 4
20. (x 1)2 (y 4)2 20
1945 4
22. (x 8)2 (y 9)2 1130
23. (x 213)2 (y 42)2 1777
24. (x 8)2 (y 7)2 64
25. (x 4)2 (y 2)2 4
26. (x 1)2 (y 4)2 16
27. (x 5)2 (y 4)2 25
28. x 2 y 2 18
29. (x 2.5)2 (y 2.8)2 1600
30. (0, 2), 2 units y x 2 (y 2)2 4 O
32. (3, 1), 5 units
31. (0, 0), 12 units 16
y
x
y
x 2 y 2 144
8 16 8
O
8
16x
(x 3)2 (y 1)2 25
8
x
O
16
©Glencoe/McGraw-Hill
214
Algebra 2
Chapter 8
33. (3, 7), 9 units (x 3)2 (y 7)2 81 4 2
34. (3, 0), 4 units y
y
12108642 O2 4 6 8 x 2 4 6 8 10 12 14 16
(x 3)2 y 2 16
35. (3, 7), 522 units 2 64 2 2 4 6 8 10 12 14
x
O
36. (25, 4), 5 units y
y (x 3)2 (y 7)2 50 O 2 4 6 8 10 x
O
x
38. (7, 3), 222 units
37. (2, 23), 229 units y
y O x
O
©Glencoe/McGraw-Hill
x
215
Algebra 2
Chapter 8
40. (1, 0), 211 units
39. (0, 3), 5 units y
y
O
x
O
42. a , 4b, 9 2
41. (9, 9), 2109 units 18 16 14 12 10 8 6 4 2 2O 2
x
y
2129 2
units y
2 4 6 8 10 12 14 16 18 x
43. a , 4b, 3 2
4 2 6 42 2 4 6 8 10 12
©Glencoe/McGraw-Hill
3 217 2
O x
units
44. (6, 8), 4 units
y
16 14 12 10 8 6 4 2
O 2 4 6 8 10 x
2 2
216
y
O 2 4 6 8 10 12 14 x
Algebra 2
Chapter 8
45. (1, 2), 214 units
46. (2, 1), 22 units
y
O
y
x O
47. a0, b, 219 units 9 2
x
48. about 109 mi
y O
x
49. (x 1)2 (y 2)2 5
50. A circle can be used to represent the limit at which planes can be detected by radar. Answers should include the following. • x 2 y 2 2500 • The region whose boundary is modeled by x 2 y 2 4900 is larger, so there would be more planes to track.
51. A
52. D
53. y 216 (x 3)2
54. y 216 (x 3)2, y 216 (x 3)2
©Glencoe/McGraw-Hill
217
Algebra 2
Chapter 8
56. x 3 216 y 2; The equations with the symbol and symbol represent the right and left halves of the circle, respectively.
55.
[10, 10] scl:1 by [10, 10] scl:1
58. (3, 2), a3, 2 b, x 3,
57. (1, 0), a , 0b, y 0, x 1 , left,
1 3
1 4
1 12
11 12
3 4
y 1 , downward, 1 unit
unit y
y O
x 3y 2 1
y 2 (x 3)2
x
x
O
59. (2, 4), a2, 3 b, x 2, 3 4
60. (4, 4)
1 4
y 4 , upward, 1 unit y 2
y x 4x
O
x
61. (1, 2)
62. a , 6b
63. 4, 2, 1
64. , 2, 3
65. 28 in. by 15 in.
66. 12
67. 6
68. 4
69. 25
70. 225
3 2
1 2
71. 222 ©Glencoe/McGraw-Hill
218
Algebra 2
Chapter 8
Chapter 8 Practice Quiz 1 Page 431 1. 13 units
2. 2226 units
3. (0, 0), a1 , 0b, y 0,
4. (4, 4), a4, 4 b, x 4,
x
1 2 1 1 , 2
1 4
3 4
y 3 , upward, 1 unit
right, 6 units
y
y y 2 6x x
O
y x 2 8x 20 O x
5. (0, 4), 7 units 12 10 8 6 4 2 8642O 2 4
©Glencoe/McGraw-Hill
y
2 4 6 8x
x 2 (y 4)2 49
219
Algebra 2
Chapter 8
Lesson 8-4 Ellipses Pages 437–440 1. x 1, y 2
2. Let the equation of a circle be (x h)2 (y k)2 r 2. Divide each side by r 2 to get (y k)2 (x h)2 r2 r2
1. This
is the equation of an ellipse with a and b both equal to r. In other words, a circle is an ellipse whose major and minor axes are both diameters. 3. Sample answer: (x 4
5.
2)2
(y 4)2 36
(y 1
(x 2)2 4
5)2
4.
x2 36
6.
y2 100
1 1
7. (0, 0): (0, 3); 612; 6
y2 20
1
x2 36
1
8. (1, 2); (5, 2), (3, 2); 415; 4
y
y
O
x
O
y2 x2 1 18 9
x
( y 2)2 (x 1)2 1 4 20
10. (4, 2); (4 216, 2); 10; 2
9. (0, 0); (2, 0); 412; 4
y
y
x
O
x
O 2
2
4x 8y 32
©Glencoe/McGraw-Hill
220
Algebra 2
Chapter 8
11. about
x2 1.32 1015
12.
y2 64
14.
(x 5)2 64
(y 4)2 9
1
16.
(x 2)2 81
(y 5)2 16
1
1
18.
(y 2)2 100
(x 42 2 9
1
1
20.
(x 1)2 81
(y 2)2 56
1
22.
x2 324
24.
x2 193,600
y2 279,312.25
26.
(x 1)2 30
(y 1)2 5
y2 1.27 1015
x2 39
1
1 13.
x2 16
y2 7
15.
y2 16
(x 2)2 4
17.
(y 4)2 64
19.
(x 5)2 64
21.
x2 169
23. about
1
y2 25
1
(x 2)2 4 (y 4)2 81 4
1
x2 2.02 1016
y2 2.00 1016
y2 196
1 1
1 25.
y2 20
x2 4
1
27. (0, 0); (0, 15); 2110; 215
28. (0, 0); (4, 0); 10; 6
y
O
y
x
x
O
y2 x2 1 10 5
©Glencoe/McGraw-Hill
1
x2 y2 1 25 9
221
Algebra 2
Chapter 8
29. (8, 2); (8 3 17, 2); 24; 18 16
30. (5, 11); (5, 11 123); 24; 22
y
4 8
1284 O4 8 12 16 20 x 4 8 12 16 20 24 28 (y 11)2 (x 5)2 1
O 24 16
8
y
8x 8
(x 8) 2 (y 2)2 1 16 144 81
144
121
32. (0, 0); (0, 16); 6; 213
31. (0, 0); (16, 0); 6; 213
y
y
x
O
3x 2 9y 2 27
27x 2 9y 2 81
33. (0, 0); (0, 17); 8; 6
34. (0, 0); (315, 0); 18; 12
y
8 6 4 2 8642 2 4 6 8
x
O
2
x
O
2
16x 9y 144
y
O 2 4 6 8x
36x 2 81y 2 2916
36. (2, 7); (2 412, 7); 4110 ; 412
35. (3, 1); (3, 5), (3, 3); 416; 412 y
12
y
8 4 O O
8
x
4
4
x
4
©Glencoe/McGraw-Hill
222
Algebra 2
Chapter 8
37. (2, 2); (2, 4), (2, 0); 217; 213
38. (1, 3); (2, 3), (4, 3); 10; 8
y
y
x
O
x
O
39.
x2 12
y2 9
1
40. Knowledge of the orbit of Earth can be used in predicting the seasons and in space exploration. Answers should include the following. • Knowledge of the path of another planet would be needed if we wanted to send a spacecraft to that planet. • 1.55 million miles 42. B
41. C 43. about
x2 1.351019
44. (x 3)2 (y 2)2 25
y2 1.26 1019
1 45. (x 4)2 (y 1)2 101
46. (x 1)2 y 2 45
47. (x 4)2 (y 1)2 16
48. y (x 3)2 1
1 2
y
O
©Glencoe/McGraw-Hill
223
1 2 y 2 (x 3) 1 x
Algebra 2
Chapter 8
49.
50. Sample answer using (0, 104.6) and (10, 112.6): y 0.8x 104.6
People (millions)
Married Americans 120 118 116 114 112 110 108 106 104 0 0 2 4 6 8 10 12 14 16 18 20
51. Sample answer: 128,600,000
52.
y
O
x
y 2x
53.
54.
y
y y 12 x
55.
x
O
x
O
y 2x
56.
y
y
y 12 x
O
x
O
x
y 2 2(x 1)
©Glencoe/McGraw-Hill
224
Algebra 2
Chapter 8
57.
y
O
x y 2 2(x 1)
Lesson 8-5 Hyperbolas Pages 445–448 1. sometimes
3. Sample answer: 5.
x2 1
y2 15
2. As k increases, the branches of the hyperbola become wider. x2 4
y2 9
1
4.
y2 4
x2 21
1
6. (0, 3 22); (0, 238);
1
y
3110 x 10 8 6 4 2
8642 2 4 6 8
©Glencoe/McGraw-Hill
225
y
y2
x2 18 20 1 O2 4 6 8 x
Algebra 2
Chapter 8
7. (1, 6 2 25); (1, 6 3 25); y 6
8. (6, 0); (237, 0);
2 25 (x 5
1 6
y x 1)
16
y
2 2 8 x 36y 36
x
O
y
16
8
O
8
16x
8 2
(y 6)
2
20
(x 1) 25
16
1
3 4
9. (4 2 25, 2); (4 3 25, 2); y2
25 (x 2
10. (0, 15); (0, 25); y x 20 15 10 5
4)
y
16 12 8 4
13.
x2 4
y2 12
ay
11 2b
1
25 4
15.
x2 25
17.
(x 2)2 49
19.
x2 16
1
12.
(y 3)2 1
(x 2)2 4
1
14.
(x 3)2 4
(y 5)2 9
1
16.
y2 16
1
18.
(y 5)2 16
20.
y2 36
1
2
y2 36
x2 6
y2 9
1
1
©Glencoe/McGraw-Hill
(y 3)2 4
x2 400 1 225 y2
2015105 O5 10 15 20x 5 10 15 20
1284 O4 8 12 16 20 x 4 8 12 16
11.
y
1
226
x2 49
x2 4
(x 4)2 81
1
Algebra 2
Chapter 8
21. (9, 0); ( 2130, 0);
22. (0, 6); (0, 2210); y 3x
7 9
y x
8 6 4 2
2 2 y x y 1 81 49 16 12 8 4
4 5
24. (3, 0); ( 234, 0); y x
y
y x2
y2 9 25 1
8 6 42 O2 4 6 8 x 2 4 6 8 y2 x2 16 25 1
26. (2, 0); (222, 0); y x y
22 x 2
x 2 y 2 4
y x 2 2y 2 2
©Glencoe/McGraw-Hill
x
O
25. ( 22, 0); (23, 0);
O
x2 4 1
5 3
23. (0, 4); (0, 241); y x
y
y2 36
4 3 21 O1 2 3 4 x 2 4 6 8
161284 O4 8 12 16x 4 8 12 16
8 6 4 2
y
O
x
x
227
Algebra 2
Chapter 8
27. (0, 6); (0, 3 25); y 2x 16
28. (0, 22); (0, 222);
y
y
23 x 3
8
y y 2 36 4x 2
8
16
4
16x
8
O
6y 2 2x 2 12
12
30. (2, 2), (2, 8); (2, 3 241);
29. (2, 0), (2, 8); (2, 1),
5 4
4 3
y 3 (x 2)
(2, 9); y 4 (x 2) 12
y
y 10 8 (x 2)2 (y 36)2 1 4 16 25 2 O 2 4 6 8 10 x 6 42 2 4 6
8 )2
(y 4 16
8
(x 2)2
49 O 4
1 4
x
4
31. (3, 3), (1, 3);
32. (12, 3), (0, 3);
(1 213, 3); y3
x
O
3 (x 2
(6 3 25, 3); 1 2
1)
y 3 (x 6)
y O
6 4 2
x
y
O 2x 1412108 6 42 2 4 6 2 (y 3)8 (x 6)2 9 10 1 36 (y 3)2 (x 1)2 1 9 4
©Glencoe/McGraw-Hill
228
Algebra 2
Chapter 8
34. (4, 0), (6, 0); (1 229, 0);
33. (1, 3 2 26);
2 5
(1, 3 4 22);
y (x 1)
y 3 23(x 1)
8 6 4 2
y 6 4 2 8 6 42 O2 4 6 8 x 2 4 6 8 10
y
8 6 42 O2 4 6 8 x 2 4 4x 2 6 25y 2 8x 96 0 8
y 2 3x 2 6y 6x 18 0
35.
x2 1.1025
y2 7.8975
1
36.
y
Station
38.
37. 120 cm, 100 cm
©Glencoe/McGraw-Hill
229
(x 2)2 4
O
Station x
(y 3)2 4
1
Algebra 2
Chapter 8
39. about 47.32 ft
40. Hyperbolas and parabolas have different graphs and different reflective properties. Answers should include the following. • Hyperbolas have two branches, two foci, and two vertices. Parabolas have only one branch, one focus, and one vertex. Hyperbolas have asymptotes, but parabolas do not. • Hyperbolas reflect rays directed at one focus toward the other focus. Parabolas reflect parallel incoming rays toward the only focus.
41. C
42. B
43.
44. ( 22, 22), (22, 22)
y xy 2
x
O
45.
46. The graph of xy 2 can be obtained by reflecting the graph of xy 2 over the x-axis or over the y-axis. The graph of xy 2 can also be obtained by rotating the graph of xy 2 by 90 .
y xy 2 x O
47.
(x 5)2 16
©Glencoe/McGraw-Hill
(y 2)2 1
1
48.
230
(y 1)2 16
(x 3)2 9
1
Algebra 2
Chapter 8
49.
(x 1)2 25
(y 4)2 9
1
50. (5, 1), 2 units y
x
O
3 2
51. 4, 2 7 0 S 53. C 5 20 55. about 5,330,000 subscribers per year
52. 7,
57. 2x 17y
58. 2, 3, 5
59. 1, 2, 9
60. 3, 1, 2
61. 5, 0, 2
62. 1, 0, 0
54. [13 8 1] 56. 5, 4
63. 0, 1, 0
Chapter 8 Practice Quiz 2 Page 448 1.
(y 1)2 81
(x 3)2 32
1
2. (4, 2); (4 2 22, 2); 6; 2 y
O
x
(y 2)2 (x 4)2 1 1 9
©Glencoe/McGraw-Hill
231
Algebra 2
Chapter 8
3. (1, 1); (1, 1 211); 8;
4.
x2 9
y2 16
1
225 y
x
O
5.
(x 2)2 16
(y 2)2 5
1
Lesson 8-6 Conic Sections Pages 450–452 2. 2x 2 4x 7y 1 0
1. Sample answer: 2x 2 2y 2 1 0
3 2 2
4. y ax b , parabola
3. The standard form of the equation is (x 2)2 (y 1)2 0. This is an equation of a circle centered at (2, 1) with radius 0. In other words, (2, 1) is the only point that satisfies the equation.
5.
y2 16
x2 8
y
8642 2 4 6 8
©Glencoe/McGraw-Hill
x
O
1 2 2
6. ax b y 2 , circle
1, hyperbola 8 6 4 2
5 4
9 4
y
y
O 2 4 6 8x
O
232
x
Algebra 2
Chapter 8
7.
(x 1)2 4
(y 3)2 1
1,
8. parabola
ellipse y
x
O
9. ellipse 11.
10. hyperbola 12. x 2 y 2 27, circle
y 10 8 6 4 2 2 4 6
8
y
4 O2 4 6 8 10 12 14
x
8
4
O
8x
4
4 8
13.
y2 4
x2 2
1 8
14. y x 2, parabola
1, ellipse
y
y
O
©Glencoe/McGraw-Hill
O
x
233
x
Algebra 2
Chapter 8
15.
x2 4
y2 1
1, hyperbola
16.
(x 1)2 36
(y 4)2 4
1,
hyperbola
y
12
y
8
x
O
4 O 8
4
4
8
12x
4
1 9
18. x (y 4)2 4, parabola
17. y (x 2)2 4, parabola y
y
x
O
x
O
20. x 2 (y 3)2 36, circle
19. (x 2)2 (y 3)2 9, circle
y
y
4 O 8
4
4
8x
4 8 O
©Glencoe/McGraw-Hill
x
234
Algebra 2
Chapter 8
21.
(x 4)2 32
y2 32
1, hyperbola 8 6 4 2
22.
(x 1)2 9
y
24.
(y 1)2 25
y 8 6 4 2
x2 4
8642 2 4 6 8
x
(y 1)2 3
x
O
23. x 2 (y 4)2 5, circle
25.
1, ellipse
26.
(x 1)2 16
x2 9
O 2 4 6 8x
(y 1)2 4
1, ellipse
y
x O
©Glencoe/McGraw-Hill
1, hyperbola
y
y
O
1, ellipse
9 2
y
121086 42 O 2 4 x 2 4 6 8
O
y2
235
x
Algebra 2
Chapter 8
27. y (x 4)2 7, parabola 12
8
y O
4
28.
(x 2)2 5
(y 1)2 6
1,
hyperbola
4x
y
4 8
x
O
12 16
29.
(x 3)2 25
(y 1)2 9
1, ellipse
30. parabolas and hyperbolas
y
O
x
32.
31. hyperbola
y
x
O
33. circle
34. hyperbola
35. parabola
36. ellipse
37. ellipse
38. circle
39. parabola
40. hyperbola
41. b
42. a
43. c
44. 2 intersecting lines
©Glencoe/McGraw-Hill
236
Algebra 2
Chapter 8
45. The plane should be vertical and contain the axis of the double cone.
46. If you point a flashlight at a flat surface, you can make different conic sections by varying the angle at which you point the flashlight. Answers should include the following. • Point the flashlight directly at a ceiling or wall. The light from the flashlight is in the shape of a cone and the ceiling or wall acts as a plane perpendicular to the axis of the cone. • Hold the flashlight close to a wall and point it directly vertically toward the ceiling. A branch of a hyperbola will appear on the wall. In this case, the wall acts as a plane parallel to the axis of the cone.
47. D
48. C
49. 0 e 1, e 1
50.
51.
(x 3)2 9
(y 6)2 4
(y 4)2 36
(x 5)2 16
1
52. (3, 4); (3 25, 4); 6; 4
1
y x
O
54. m12n
53. x12 55.
x7 y4
56. 196 beats per min
57. (2, 6)
58. (3, 2)
59. (0, 2) ©Glencoe/McGraw-Hill
237
Algebra 2
Chapter 8
Lesson 8-7
Solving Quadratic Systems Pages 458–460
1a. (3, 4), (3, 4)
2. The vertex of the parabola is on the ellipse. The parabola opens toward the interior of the ellipse and is narrow enough to intersect the ellipse in two other points. Thus, there are exactly three points of intersection.
y
4x 3y 0 O
x
x 2 y 2 25
1b. (1, 4) y y 2x 2 2
y 5 x2 O
x
3. Sample answer: x 2 y 2 40, y x 2 x
4. (4, 5)
5. (4, 3), (3, 4)
6. no solution
7. (1, 5), (1, 5)
8.
14 12 10 8 6 4 2 108642 2 4 6
©Glencoe/McGraw-Hill
238
y
O 2 4 6 8 10 x
Algebra 2
Chapter 8
10. (40, 30)
y
9.
O
x
11. (2, 4), (1, 1)
12. a ,
13. (1 217, 1 217),
14. no solution
3 2
9 b, 2
(1, 2)
(1 217, 1 217) 15. ( 25, 25), (25, 25)
16. no solution
17. (5, 0), (4, 6)
18. (0, 3), a
19. (8, 0)
20. (0, 5)
21. no solution
22. (4, 3), (4, 3)
23. (5, 5), (5, 1), (3, 3)
24. (6, 3), (6, 1), (4, 4), (4, 0)
25. a , b, (1, 3)
26. (3, 4), (3, 4)
27. 0.5 s
28. Sample answer:
5 3
7 3
11 123 , b 2 4
x2 y2 1, 36 16 x2 (y 2)2 16 4 x2 y2 1 2 16 40 24 25 45 12 25 b , 5 5
29. a
©Glencoe/McGraw-Hill
1,
30. (39.2, 4.4)
239
Algebra 2
Chapter 8
32.
31. No; the comet and Pluto may not be at either point of intersection at the same time.
y
x
O
33.
34.
y
y
x
O
x
O
35.
36.
y
37.
O 8642 2 4 6 8
x
O
y
2 4 6 8x
38. k 3, 2 k 2, or k 3
y
O
8 6 4 2
x
39. none
40. k 2 or k 3
41. none
42. 3 k 2 or 2 k 3
©Glencoe/McGraw-Hill
240
Algebra 2
Chapter 8
43. Systems of equations can be used to represent the locations and/or paths of objects on the screen. Answers should include the following. • y 3x, x 2 y 2 2500 • The y-intercept of the graph of the equation y 3x is 0, so the path of the spaceship contains the origin. • (5 210, 15 210) or about (15.81, 47.43) 45. B
44. A
47. Sample answer:
48. Sample answer:
46. Sample answer: y x 2, x (y 2)2
x 2 y 2 36, (x 2)2 16
y2 4
1
49. Sample answer: x 2 y 2 81, x2 4
y2
100
x2 16
x 2 y 2 100,
50. Sample answer:
1
y2 x2 64 16
x2 64
y2 4
1
y2 16
1,
1
52. (x 2)2 (y 1)2 11, circle
51. impossible
y
O
©Glencoe/McGraw-Hill
241
x
Algebra 2
Chapter 8
53.
(y 3)2 9
x2 4
1, ellipse
54. (0, 2); (0, 4); y
y
23 x 3
y
6y 2 2x 2 24
x
O
O
x
55. 7, 0
56. 0, 3
57. 7, 3
58. 7, 5
4 3
3 4
1 2
59.
60. ,
61a. 40 61b. two real, irrational
62a. 48 62b. two imaginary
61c.
210 5
62c. 1
63. 2 9i 65.
8 5
2i 23 3
64. 29 28i
1 5
i
66. about 1830 times
67. 6
68. 2
69. 51
70. (5, 3, 7)
71. y 3x 2
72. y x
©Glencoe/McGraw-Hill
5 3
242
4 3
Algebra 2
Chapter 8
PQ245-6457F-P09[243-266].qxd 7/31/02 9:54 AM Page 243 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-09:
Chapter 9 Rational Expressions and Equations Lesson 9-1 Multiplying and Dividing Rational Expressions Pages 476–478 4 6
1. Sample answer: ,
4(x 2) 6(x 2)
2. To multiply rational numbers or rational expressions, you multiply the numerators and multiply the denominators. To divide rational numbers or rational expressions, you multiply by the reciprocal of the divisor. In either case, you can reduce your answer by dividing the numerator and the denominator of the results by any common factors.
3. Never; solving the equation using cross products leads to 15 10, which is never true.
4.
9m 4n4
5.
1 ab
6.
3y 2 y4
7.
3c 20b
8.
5 12x
9.
6 5
10.
p5 p1
11. cd 2x
12.
2y(y 2) 3(y 2)
13. D
14.
5c 2b
n2 7m
16. 3x 4y
15. 17.
s 3
18.
5 t1
19.
1 2
20.
y2 3y 1
21.
a1 2a 1
22.
3x 2 2y
Glencoe/McGraw-Hill
243
Algebra 2
Chapter 9
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4bc 27a
23.
24. f
25. 2p 2
26.
xz 8y
27.
b3 x 2y 2
28. 3
29.
4 3
30.
2 3
31. 1
32.
5(x 3) 2(x 1) 3(r 4) r3
33.
w3 w4
34.
35.
2(a 5) (a 2)(a 2)
36.
3n m
37. 2p
38.
mn m2 n2
39.
2x y 2x y
40. y 1
41.
4 3
42. d 2, 1 or 2
43. a b or b 45.
44.
6827 m 13,129 a
6827 13,129
46. 2x 1 units 1 a2
47. (2x 2 x 15)m 2
48.
49. A rational expression can be used to express the fraction of a nut mixture that is peanuts. Answers should include the following. • The rational expression 8x is in simplest form
50. C
13 x
because the numerator and the denominator have no common factors.
©Glencoe/McGraw-Hill
244
Algebra 2
Chapter 9
PQ245-6457F-P09[243-266].qxd 7/31/02 9:54 AM Page 245 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-09:
8x
• Sample answer: 13 x y could be used to represent the fraction that is peanuts if x pounds of peanuts and y pounds of cashews were added to the original mixture. 51. A
52. (1, 4), (5, 2)
53. (217, 222)
54. x
1 3
(y 3)2 1; parabola
y O
x x 13 (y 3)2 1
55.
(x 7)2 (y 2)2 9 1
1;
56. even; 2
hyperbola 8
y
4
x
O 4 4
8
(x 7) 9
2
12 2
(y 2) 1 1
8
AA
C A BLACK
57. odd; 3
58. even; 0
59. 1, 4
60. ,
61. 0, 5
62. 4.99 102 s or about 8 min 19 s
63.
64.
1 1 6 3
1 9
11 24
65. 1
©Glencoe/McGraw-Hill
3 19 , 2 16
66. 1 245
Algebra 2
Chapter 9
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4 15
11 16
68.
67. 1
11 18
69.
70.
Lesson 9-2
1 6
Adding and Subtracting Rational Epressions Pages 481–484
1. Catalina; you need a common denominator, not a common numerator, to subtract two rational expressions.
2. Sample answer: d 2 d, d 1
3a. Always; since a, b, and c are factors of abc, abc is always a common denominator of
4. 12x 2y 2
1 1 1 . a c b
3b. Sometimes; if a, b, and c have no common factors, then abc is the LCD of 1 1 1 . a c b
3c. Sometimes; if a and b have no common factors and c is a factor of ab, then ab is the 1 1 1 LCD of . a
b
c
3d. Sometimes; if a and c are factors of b, then b is the 1 1 1 LCD of . a
b
3e. Always; since
c
1 1 1 a c b
ac ab bc , the sum abc abc abc bc ac ab . always abc
is
6. x(x 2)(x 2)
5. 80ab3c 7.
2 x3 x 2y
©Glencoe/McGraw-Hill
8.
246
42a 2 5b 2 90ab 2
Algebra 2
Chapter 9
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37 42m
10.
5d 16 (d 2) 2
11.
3a 10 (a 5)(a 4)
12.
8 5
13.
13x 2 4x 9 2x (x 1)(x 1)
9.
14. 70s 2t 2
units
15. 180x 2yz
16. 420a3b3c 3
17. 36p 3q 4
18. 4(w 3)
19. x 2(x y)(x y)
20. (2t 3)(t 1)(t 1)
21. (n 4)(n 3)(n 2)
22.
6 8b ab
23.
31 12v
24.
5 7r r
25.
2x 15y 3y
26.
9x 2 2y 3 12x 2y
27.
25b 7a3 5a 2b 2
28.
29.
110w 423 90w
30.
13 y8
31.
a3 a4
32.
5m 4 3(m 2)(m 2)
33.
y (y 9) (y 3)(y 3)
34.
7x 38 2(x 7)(x 4)
35.
8d 20 (d 4)(d 4)(d 2)
36.
4h 15 (h 4)(h 5)2
37.
x2 6 (x 2)2(x 3)
38. 0
39.
2y 2 y 4 (y 1)(y 2)
40.
1 b1
42.
2s 1 2s 1
a7 a2
44.
3x 4 2x (x 2)
45. 12 ohms
46.
24 h x
3 20q
41. 1 43.
©Glencoe/McGraw-Hill
247
Algebra 2
Chapter 9
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47. 49.
24 h x4 2md (d L)2(d L) 2 2md 2 (d L2 ) 2
48. or
50. Sample answer: 1 1 , x1 x2
51. Subtraction of rational expressions can be used to determine the distance between the lens and the film if the focal length of the lens and the distance between the lens and the object are known. Answers should include the following. • To subtract rational expressions, first find a common denominator. Then, write each fraction as an equivalent fraction with the common denominator. Subtract the numerators and place the difference over the common denominator. If possible, reduce the answer. •
1 1 1 q 10 60
48(x 2) h x(x 4)
52. B
could be used
to determine the distance between the lens and the film if the focal length of the lens is 10 cm and the distance between the lens and the object is 60 cm. 54.
53. C
©Glencoe/McGraw-Hill
248
4 15xyz 2
Algebra 2
Chapter 9
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55.
a(a 2) a1
56.
y 8 x 2 y 2 16 8
8x
O
8 9x 2 y 2 81
57.
58. 2.5 ft
y 2
(y 3) x 2
x
O
x2 y 4
59.
60.
y
y 15
6
10 2 8
O
5
8x
2
10 5 O
x 6 y 1 20 2
2
5
10x
5
16
10
y2 x2 15 1
49
61.
10 8 6 4 2
25
y
O 2 2 4 6x
6
4 (x 2) (y 5) 1 8 16 25 2
©Glencoe/McGraw-Hill
2
249
Algebra 2
Chapter 9
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Chapter 9 Practice Quiz 1 Page 484 t2 t3
2.
c 6b 2
3.
y2 32
4.
7 2
5. (w 4)(3w 4)
6. x 1
1.
7.
4a 1 ab
9.
n 29 (n 6)(n 1)
8.
Lesson 9-3
10.
1 4
Graphing Rational Functions Pages 488–490
1. Sample answer: f(x)
6ax 20by 15a 2b3
2. Each of the graphs is a straight line passing through (5, 0) and (0, 5). However,
1 (x 5)(x 2)
the graph of f (x)
(x 1)(x 5) x1
has a hole at (1, 6), and the graph of g(x ) x 5 does not have a hole. 3. x 2 and y 0 are asymptotes of the graph. The y-intercept is 0.5 and there is no x-intercept because y 0 is an asymptote.
4. asymptote: x 2
5. asymptote: x 5; hole: x1
6.
f (x )
O
f (x )
©Glencoe/McGraw-Hill
250
x x x 1
Algebra 2
Chapter 9
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7.
8.
f (x ) 4
f (x ) 10
2 6 8
4
2
f (x )
4
9.
x 2 25
8x
4
O
2
6 (x 2)(x 3)
4 O
10.
2 8
O
2
10
x
4 ( x 1)2
8x
4
f (x )
x 5 x 1
4
O
C
11.
6
f (x ) f (x )
4
2
4
f (x )
4
f (x ) x 5
x
C
12. 100 mg
f (x)
O x f(x)
13.
x 2 x2 x 6
14. y 12, C 1; 0; 0
C 10 6
C
y y 12
8
16 y
2 16 8
O 4
15. y 0 and 0 C 1
16. asymptotes: x 2, hole: x3
17. asymptotes: x 4, x 2
18. asymptotes: x 4, hole: x 3
19. asymptotes: x 1, hole: x5
20. hole: x 4
©Glencoe/McGraw-Hill
251
Algebra 2
Chapter 9
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21. hole: x 1
22.
f (x ) 1
f (x ) x
x
O
23.
24.
f (x )
f (x ) f (x )
1 x2
x
O
x
O
3
f (x ) x
25.
26.
f (x ) 6
f (x )
5 x 1
2 8
4
f (x )
x O
4
x
O
8
4
f (x )
x x 3
8
27. 8
C
28.
f (x ) f (x )
C
f (x )
f (x )
5x x 1
3 ( x 2)2
x O
4 8
4
O
4
8x
4
29.
30.
f (x ) f (x )
1 ( x 3)2
f (x ) f (x )
x 4 x 1
6 2
8
4 O
4
8x
4 O
©Glencoe/McGraw-Hill
x
8
252
Algebra 2
Chapter 9
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31.
32.
f (x )
4 8 f(x)
x
O
f (x )
f(x)
4 O
x
4
x 2 36 4 x 6
8
x 1 x 3
12
33.
34.
f(x) f(x)
f (x )
x2 1 x 1
O x
O
x
f (x )
35.
36.
f (x )
f (x ) f (x )
O
x
1 ( x 2)( x 3)
37.
38.
f (x ) f (x )
x x2 1
x
O
f (x )
3 ( x 1)( x 5)
x 1 x2 4
f (x )
f (x ) O
x
O
©Glencoe/McGraw-Hill
6 (x 6)2
253
x
Algebra 2
Chapter 9
PQ245-6457F-P09[243-266].qxd 7/31/02 9:54 AM Page 254 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-09:
39.
40.
f (x ) f (x )
f (x )
1 (x 2)2
f (x )
64 x 2 16
x
O
x
O
41. The graph is bell-shaped with a horizontal asymptote at f(x) 0.
64 64 a 2 b, x 16 x 16 64 graph of f (x) 2 x 16
42. Since the
2
would be a reflection of the graph of f (x)
64 x 16 2
over
the x-axis. 43.
44. m1 7; 7; 5
Vf 20 12 4
Vf
m1 7 m1 7
5
O
16 8 4
8 m1
45. about 0.83 m/s
46. Sample answers: f(x) f (x) f (x)
47.
8
P (x ) 12
8
x2 , (x 2)(x 3) 2(x 2) , (x 2)(x 3) 5(x 2) (x 2)(x 3)
48. the part in the first quadrant
P (x )
6x 10 4 x
4 O
4x
4 8
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49. It represents her original freethrow percentage of 60%.
50. y 1; This represents 100%, which she cannot achieve because she has already missed 4 free throws.
51. A rational function can be used to determine how much each person owes if the cost of the gift is known and the number of people sharing the cost is s. Answers should include the following. • c
52. A
100 50
s0
c
150 s
O
100 50
50 100 s
50 c 0 100
• Only the portion in the first quadrant is significant in the real world because there cannot be a negative number of people nor a negative amount of money owed for the gift. 53. B 55.
3x 16 (x 3)(x 2)
54.
3m 4 mn
56.
5(w 2) (w 3) 2
58. (2, 0); 113
57. (6, 2); 5 y
y
(x 6)2 ( y 2)2 25 O
O
x
x x 2 y 2 4x 9
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59. $65,892
60. 4 2i
61. 12, 10
62.
63. 4.5
64. 1.4
65. 20
66. 12
Lesson 9-4
7 3 213 2
Direct, Joint, and Inverse Variation Pages 495–498
1a. inverse 1b. direct
2. Both are examples of direct variation. For y 5x, y increases as x increases. For y 5x, y decreases as x increases.
3. Sample answers: wages and hours worked, total cost and number of pounds of apples; distances traveled and amount of gas remaining in the tank, distance of an object and the size it appears
4. inverse; 20
5. direct; 0.5
6. joint;
7. 24
8. 45
9. 8
1 2
10. P 0.43d
11. 25.8 psi
12. about 150 ft
13.
14. direct; 1.5
Depth(ft) Pressure(psi) 0 0 1 0.43 2 0.86 3 1.29 4 1.72
Glencoe/McGraw-Hill
p
256
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P P 0.43d O
d
15. joint; 5
16. inverse; 18
17. direct; 3
18. inverse; 12
19. direct; 7
20. joint;
21. inverse; 2.5
22. V
23. V kt
24. directly; 2
25. 118.5 km
26. 60
27. 20
28. 216
29. 64
30. 25
31. 4
32. 1.25
33. 9.6
34. 12.6
35. 0.83
36. 2
37.
1 3
k p
1 4
1 6
38. 30 mph
39. 100.8 cm3
40. See students’ work.
41. m 20sd
42. joint
43. 1860 lb
44. / 15md
45. joint
46. See students’ work.
47. I
k d2
48.
I
I 162 d
O
Glencoe/McGraw-Hill
257
d
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1
0.02P1P2
49. The sound will be heard as 4 intensely.
50. 0.02; C
51. about 127,572 calls
52. about 601 mi
53. no; d 0
54. Sample answer: If the average student spends $2.50 for lunch in the school cafeteria, write an equation to represent the amount s students will spend for lunch in d days. How much will 30 students spend in a week? a 2.50sd; $375
55. A direct variation can be used to determine the total cost when the cost per unit is known. Answers should include the following. • Since the total cost T is the cost per unit u times the number of units n or C un, the relationship is a direct variation. In this equation u is the constant of variation. • Sample answer: The school store sells pencils for 20¢ each. John wants to buy 5 pencils. What is the total cost of the pencils? ($1.00)
56. D
57. C
58. asymptote: x 1; hole x 1
59. asymptotes: x 4, x 3
60. hole: x 3 t 2 2t 2 (t 2)(t 2)
61.
x yx
62.
63.
m (m 1) m5
64. 9.3 107
65. 0.4; 1.2
66. 3; 7
3 5
67. ; 3
68. C
69. A
70. S
©Glencoe/McGraw-Hill
d2
258
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71. P
72. A
73. C
Chapter 9 Practice Quiz 2 Page 498 1.
2.
f (x )
f (x ) O
1 f (x ) xx 4
x
x
O
f (x )
3. 49
2 x 6x 9 2
4. 4.4
5. 112
Lesson 9-5 Classes of Functions Pages 501–504 2. constant (y 1), direct variation (y 2x), identity (y x)
1. Sample answer: P
O
d
This graph is a rational function. It has an asymptote at x 1. 3. The equation is a greatest integer function. The graph looks like a series of steps.
4. greatest integer
5. inverse variation or rational
6. constant
©Glencoe/McGraw-Hill
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7. c
8. b
9. identity or direct variation
10. quadratic
y
y
y x 2 2
x
O
yx x
O
12. A r 2; quadratic; the graph is a parabola
11. absolute value y
y x2
O
x
13. absolute value
14. square root
15. rational
16. direct variation
17. quadratic
18. constant
19. b
20. e
21. g
22. a
23. constant
24. direct variation y
y
y 2.5x x
O
O
x
y 1.5
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25. square root
26. inverse variation or rational
y
y y 4x
y 兹9x
x
O
x
O
AA
C A BLACK
27. rational
28. greatest integer y
y
2 yx 1
x1
y 3[x ]
x
O
x
O
29. absolute value
30. quadratic y
y
y 2x O
y 2x 2 x
O
x
31. C 4.5 m
32. direct variation
33. a line slanting to the right and passing through the origin
34. similar to a parabola
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35.
36. The graph is similar to the graph of the greatest integer function because both graphs look like a series of steps. In the graph of the postage rates, the solid dots are on the right and the circles are on the left. However, in the greatest integer function, the circles are on the right and the solid dots are on the left.
y
Cost (cents)
160 120 80 40 x
0 2
4 6 Ounces
8
10
37a. absolute value 37b. quadratic 37c. greatest integer 37d. square root
38. A graph of the function that relates a person’s weight on Earth with his or her weight on a different planet can be used to determine a person’s weight on the other planet by finding the point on the graph that corresponds with the weight on Earth and determining the value on the other planet’s axis. Answers should include the following. • The graph comparing weight on Earth and Mars represents a direct variation function because it is a straight line passing through the origin and is neither horizontal nor vertical. • The equation V 0.9E compares a person’s weight on Earth with his or her weight on Venus. V
Venus
80 60 40 20
E
0 20
©Glencoe/McGraw-Hill
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40 60 Earth
80
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39. C
40. D
41. 22
42.
f (x ) 3 f (x ) x 2
x
O
43.
44.
f (x )
f (x ) 2 f (x ) x 5x 4
x4
x
O
f (x ) (
8 x 1)(x 3)
46. a3 , 1b ; a2 , 1b;
45. (8, 1); a8, b ; x 8; y 14 12 10 8 6 4 2 2 2
7 8 1 1 1 ; up; 8 2
x
O
1 4
y 1; x
unit
1 4 1 4 ; 4
right; 4 units
y
y
1
1
x 4 y2 2 y 3 1( y 1) (x 8)2 2
O 2 4 6
©Glencoe/McGraw-Hill
10 12
O
x
x
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47.
48. c
(5, 4); a5 , 4b ; y 4; 3 4
25 23 54 d 66 26 57
1 4
x 4 ; right; 3 units y x
O
3x y 2 8y 31
49. impossible
50. (7, 5)
51. a , 2b
52. (2, 2)
53. 1
54. 12
1 3
17 6
55.
56. 60a3b 2c 2
57. 45x 3y 3
58. 15(d 2)
59. 3(x y)(x y)
60. (a 3)(a 1)(a 2)
61. (t 5)(t 6)(2t 1)
Lesson 9-6
Solving Rational Equations and Inequalities Pages 509–511 2. 2(x 4); 4
1. Sample answer: 1 5
2 a2
1
3. Jeff; when Dustin multiplied by 3a, he forgot to multiply the 2 by 3a.
4. 3
5. 2, 6
6.
7. 6, 2
8. 2 c 2
©Glencoe/McGraw-Hill
264
2 3
Algebra 2
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2 9
1 6
9. v 0 or v 1
10. 2 h 4 3
11. 2
12.
13. 6, 1
14. 3, 2
15. 1 a 0
16. 1 m 1
17. 11
18. 3
19. t 0 or t 3
20. 0 b 1
21. 0 y 2
22. p 0 or p 2
23. 14
24.
25.
26.
27. 7
28.
7 3
30.
1 2145 4
29.
3 3 22 2
3 2
31. 32
32. 2 or 4
33. band, 80 members; chorale, 50 members
34. 4.8 cm/g
35. 24 cm
36. 15 km/h
37. 5 mL
38. 5
39. 6.15
40.
41. If something has a general fee and cost per unit, rational equations can be used to determine how many units a person must buy in order for the actual unit price to be a given number. Answers should include the following.
42. B
• To solve
500 5x x
1 2
b bc 1
6,
multiply each side of the equation by x to eliminate the rational expression. ©Glencoe/McGraw-Hill
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Then subtract 5x from each side. Therefore, 500 x. A person would need to make 500 minutes of long distance calls to make the actual unit price 6¢. • Since the cost is 5¢ per minute plus $5.00 per month, the actual cost per minute could never be 5¢ or less. 44. quadratic
43. C
y
y 2x 2 1 x
O
46. direct variation
45. square root
y
y
O
y 0.8x
y 2兹x O
x
47. 36
48. 33.75
49. 22130
50. 225
52. 5x 0 x 11 or x 36
51. 2137 53. 5x 0 0 x 46
©Glencoe/McGraw-Hill
x
54. e b ` 1 b 2 f 1 2
266
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Chapter 10 Esponential and Logarithmic Relations Lesson 10-1 Exponential Functions Pages 527–530 2a. 2b. 2c. 2d.
1. Sample answer: 0.8
quadratic exponential linear exponential
4. a
3. c
6. D {x 0 x is all real numbers.}, R {y 0 y 0}
5. b
y
y 3(4)x
O
7. D {x 0 x is all real numbers.}, R {y 0 y 0}
x
8. growth
y
( 13 )x
y2
O
x
10. growth
9. decay 1 x 2
11. y 3a b
12. y 18132 x
13. 2227 or 427
14. a4
15. 3322 or 2722
16. 9
17. x 0
18. 2
19. y 65,000(6.20)x
20. 22,890,495,000
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21. D {x 0 x is all real numbers.}, R {y 0 y 0}
22. D {x 0 x is all real numbers.}, R {y 0 y 0}
y
y
y 5(2)x y 2(3)x x
O
x
O
23. D {x 0 x is all real numbers.}, R {y 0 y 0}
24. D {x 0 x is all real numbers.}, R {y 0 y 0}
y
y
y4
)x
y 0.5(4
x
O
x
O
25. D {x 0 x is all real numbers.}, R {y 0 y 0}
x
( 13 )
26. D {x 0 x is all real numbers.}, R {y 0 y 0}
y
y
x O
O
x
y 2.5(5)x y
( 15 )x
27. growth
28. growth
29. decay
30. growth
31. decay
32. decay 1 x 4
33. y 2 a b
34. y 3(5)x
35. y 7(3)x
36. y 5 a b
37. y 0.2(4)x
38. y 0.3(2)x
©Glencoe/McGraw-Hill
1 x 3
268
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39. 54 or 625
40. x115
41. 7412
42. y 2 13
43. n 2
44. 25
45. n 5
46.
47. 1
48. n 2 8 3
2 3
49.
50. 0
51. n 3
52.
53. 3
54. p 2
55. 10
56. 3, 5
57. y 100(6.32)x
58. about 1,008,290
59. y 3.93(1.35)x
60. 9.67 million; 17.62 million; 32.12 million; These answers are in close agreement with the actual populations in those years.
61. 2144.97 million; 281.42 million; No, the growth rate has slowed considerably. The population in 2000 was much smaller than the equation predicts it would be.
62. Exponential; the base, 1 , n is fixed, but the exponent, nt, is variable since the time t can vary.
63. A(t ) 1000(1.01)4t 65. s . 4x
64. $2216.72
67. Sometimes; true when b 1, but false when b 1.
68. The number of teams y that could compete in a tournament with x rounds can be expressed as y 2x. The 2 teams that make it to the final round got there as a result of winning games played with 2 other teams, for a total of 2 2 22 or 4 games played in the previous rounds. Answers should include the following.
©Glencoe/McGraw-Hill
5 3
r
66. 1.5 three-year periods or 4.5 yr
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• Rewrite 128 as a power of 2, 27. Substitute 27 for y in the equation y 2x. Then, using the Property of Equality for Exponents, x must be 7. Therefore, 128 teams would need to play 7 rounds of tournament play. • Sample answer: 52 would be an inappropriate number of teams to play in this type of tournament because 52 is not a power of 2. 69. A
70. 780.25
71.
72.
[5, 5] scl: 1 by [1, 9] scl: 1
[5, 5] scl: 1 by [1, 9] scl: 1
The graphs have the same shape. The graph of y 3x1 is the graph of y 3x translated one unit to the left. The asymptote for the graph of y 3x and for y 3x1 is the line y 0. The graphs have the same domain, all real numbers, and range, y 0. The y-intercept of the graph of y 3x is 1 and for the graph of y 3x1 is 3.
The graphs have the same shape. The graph of y 2x 3 is the graph of y 2x translated three units up. The asymptote for the graph of y 2x is the line y 0 and for y 2x 3 is the line y 3. The graphs have the same domain, all real numbers, but the range of y 2x is y 0 and the range of y 2x 3 is y 3. The y-intercept of the graph of y 2x is 1 and for the graph of y 2x 3 is 4.
©Glencoe/McGraw-Hill
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73.
74.
[5, 5] scl: 1 by [1, 9] scl: 1
[5, 5] scl: 1 by [3, 7] scl: 1
The graphs have the same shape. The graph of
The graphs have the same shape. The graph of y
1 x2 a b 5
x
y a b 1 is the graph of 1 4
is the graph of
x
y a b translated two units
y a b translated one
to the right. The asymptote
unit down. The asymptote
1 5
x
1 4
1 5
1 5
1 4
the line y 0 and
x2
for y a b
is the line
for the graph of y
y 0. The graphs have the same domain, all real numbers, and range, y 0. The y-intercept of the graph of y
1 x a b 5
graph of y
1 x a b 4
1
is the line y 1. The graphs have the same domain, all real numbers, but the range of y
is 1 and for the 1 x2 a b 5
x
for the graph of y a b is
x
for the graph of y a b and
1 x a b 4 x
is y 0 and of y a b 1 1 4
is 25.
is y 1. The y-intercept x
of the graph of y a b is 1 1 4
and for the graph of y 75. For h 0, the graph of y 2x is translated 0 h 0 units to the right. For h 0, the graph of y 2x is translated |h| units to the left. For k 0, the graph of y 2x is translated 0k 0 units up. For k 0, the graph of y 2x is translated 0k 0 units down. ©Glencoe/McGraw-Hill
1 x a b 4
1 is 0.
76. 1, 15
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13 3
77. 1, 6
78. , 3
79. 0 x 3 or x 6
80. square root
81. greatest integer
82. constant y y8
O
1 0 R 0 1
83. B 85.
3 1 B 51 11
x
84. does not exist 6 R 5
86. about 23.94 cm 88. g [h(x)] x 2 6x 9; h [g(x)] x 2 3
87. g [h(x)] 2x 6; h [g(x)] 2x 11 89. g [h(x)] 2x 2; h [g(x)] 2x 11
©Glencoe/McGraw-Hill
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Logarithms and Logarithmic Functions Pages 535–538
1. Sample answer: x 5y and y log5 x
2. They are inverses.
3. Scott; the value of a logarithmic equation, 9, is the exponent of the equivalent exponential equation, and the base of the logarithmic expression, 3, is the base of the exponential equation. Thus x 39 or 19,683.
4. log5 625 4
5. log7
1 49
6. 34 81
2
1
7. 362 6
8. 4
9. 3
10. 21
11. 1
12. 27
13. 1000
14.
15.
1 , 2
1 2
x5
16. x 6
1
17. 3
18. 1013
19. 107.5
20. 105.5 or about 316,228 times
21. log8 512 3
22. log3 27 3
23. log5
1 125
24. log 31 9 2
3
25. log100 10
1 2
26. log2401 7 28. 132 169
27. 53 125 29. 41
1 4
30. 1002 1
1 2 5
31. 83 4
32. a b
33. 4
34. 2
2
©Glencoe/McGraw-Hill
1 4
273
1 10
25
Algebra 2
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1 2
36.
5 2
37. 5
38. 4
39. 7
40. 45
41. n 5
42. 3x 2
43. 3
44. 2x
45. 1018.8
46. 1010.67
47. 81
48. c 256
49. 0 y 8
50. 125
51. 7
52. 0 p 1
53. x 24
54. 3
55. 4
56. 11
57. 2
58. 25
59. 5
60. y 3
61. a 3
62. 8 ?
63. log5 25 2 log5 5 2 ?
log5 5 2 log5 5 ?
2 2(1)
2 2✓
1
64.
Original equation
?
log16 2 log2 16 1
25 5 and 5 51 2
1
?
log16 164 log2 24 1
Inverse Prop. of Exp. and Logarithms Simplify.
1 (4) 4
Original equation 1
2 164 and 16 24
?
1 Inverse Prop. of Exp. and Logarithms
1 1✓
©Glencoe/McGraw-Hill
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66a.
65. ?
log7 [log3 (log2 8)] 0
Original equation
?
log7 [log3 (log2 23)] 0
y
8 23
log7 (log3 3) 0
Inverse Prop. of Exp. and Logarithms
?
log7 (log3 31) 0
y log 1 x 2
66b. The graphs are reflections of each other over the line y x.
3 31
?
log7 1 0
x
( 12 )
x
O
?
?
y
Inverse Prop. of Exp. and Logarithms
log7 70 0 0 0✓
1 70 Inverse Prop. of Exp. and Logarithms
68. 103 or 1000 times as great
67a. y
y log2(x 2)
y log2x 3 y log2(x 1) O
x
y log2x 4
67b. The graph of y log2 x 3 is the graph of y log2 x translated 3 units up. The graph of y log2 x 4 is the graph of y log2 x translated 4 units down. The graph of log2 (x 1) is the graph of y log2 x translated 1 unit to the right. The graph of log2 (x 2) is the graph of y log2 x translated 2 units to the left. 69. 101.4 or about 25 times as great
©Glencoe/McGraw-Hill
70. 101.7 or about 50 times 275
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71. 2 and 3; Sample answer: 5 is between 22 and 23.
72. All powers of 1 are 1, so the inverse of y 1x is not a function.
73. A logarithmic scale illustrates that values next to each other vary by a factor of 10. Answers should include the following. • Pin drop: 1 100; Whisper: 1 102; Normal conversation: 1 106; Kitchen noise: 1 1010; Jet engine: 1 1012 •
74. B
Pin drop
Whisper (4 feet)
Normal conversation
2 10 11
0
4 10 11
Kitchen noise
6 10 11
Jet engine
8 10 11 1 10 12
• On the scale shown above, the sound of a pin drop and the sound of normal conversation appear not to differ by much at all, when in fact they do differ in terms of the loudness we perceive. The first scale shows this difference more clearly. 75. D
76. x 216
77. b12
78.
79. 3,
7 3
14 5
80.
81.
5 273 4
82.
83.
6x 58 (x 3)(x 3)(x 7)
84. $2400, CD; $1600, savings
43 30y
85. x10
86. y 24
87. 8a6b 3
88. an6
89.
x3 y 2z 3
©Glencoe/McGraw-Hill
90. 1
276
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Chapter 10 Practice Quiz 1 Page–538 1. growth
2. y 2(4)x
3. log4 4096 6
4. 92 27
3
5.
4 3
6. 15
7.
3 5
8. n 1
9. x 26
10. 3
Lesson 10-3
Properties of Logarithms Pages 544–546
1. properties of exponents
2. Sample answer: 2log3 x log3 5; log3 5x 2
3. Umeko; Clemente incorrectly applied the product and quotient properties of logarithms. log7 6 log7 3 log7 (6 3) or log7 18 Product Property of
4. 1.1402
Logarithms
log7 18 log7 2 log7 (18 2) or log7 9 Quotient Prop. of Logarithms
5. 2.6310
6. 0.3690
7. 6
8. 2
9. 3
10. 4
11. pH 6.1 log10
B C
12. 20:1
13. 1.3652
14. 1.2921
15. 0.2519
16. 0.2519
17. 2.4307
18. 2.1133
19. 0.4307
20. 0.0655
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21. 2
22. 3
23. 4
24. 2
25. 14
26. 12
27. 2
28. 4
29.
30. 6
31. 10
32. 12
33.
x3 4
34.
35. False; log2 (22 23) log2 12, log2 22 log2 23 2 3 or 5, and log2 12 Z 5, since 25 Z 12.
1 1x 2
12
36. ? n logb x m logb x (n m)logb x ? logb x n logb x m (n m)logb x
Power Prop. of Logarithms ?
logb (x n x m) (n m)logb x Product Prop. of Logarithms ?
logb (x nm) (n m)logb x Product of Powers Prop.
(n m)logb x (n m)logb ✓ Power Prop. of Logarithms
C2 C1
37. 2
38. E 1.4 log
39. about 0.4214 kilocalories per gram
40. about 0.8429 kilocalories per gram
41. 3
42. 3
43. About 95 decibels; L 10 log10 R, where L is the loudness of the sound in decibels and R is the relative intensity of the sound. Since the crowd increased by a factor of 3, we assume that the intensity also increases by a factor of 3. Thus, we need to find the loudness of 3R.
44. 5
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L 10 log10 3R L 10 (log10 3 log10 R ) L 10 log103 10 log10 R L 10(0.4771) 90 L 4.771 90 or about 95 45. 7.5
46. about 22
47. Let b x m and b y n. Then logb m x and logb n y.
48. Since logarithms are exponents, the properties of logarithms are similar to the properties of exponents. The Product Property states that to multiply two powers that have the same base, add the exponents. Similarly, the logarithm of a product is the sum of the logarithms of its factors. The Quotient Property states that to divide two powers that have the same base, subtract their exponents. Similarly, the logarithm of a quotient is the difference of the logarithms of the numerator and the denominator. The Power Property states that to find the power of a power, multiply the exponents. Similarly, the logarithm of a power is the product of the logarithm and the exponent. Answers should include the following. • Quotient Property:
bx by
b xy
m n m n
Quotient Prop.
m Prop. of n Equality for Logarithmic Equations m logb Inverse n Prop. of
logb b x y logb
xy
Exp. and Logarithms
logb m logb n logb
m Replace x n with log m b and y with logbn.
log2
32 a b 8
log2
25 a 3b 2
log2 2(53) 5 3 or 2
©Glencoe/McGraw-Hill
279
Replace 32 with 25 and 8 with 23. Quotient of Powers Inverse Prop. of Exp. and Logarithms
Algebra 2
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log2 32 log2 8 log2 25 log2 23 Replace 32 with 25 and 8 with 23.
5 3 or 2 Inverse Prop. of Exp. and Logarithms
So, log2 a
32 b 8
log2 32 log2 8
Power Property: log3 94 log3 (32)4
Replace 9 with 32.
log3 3(24) 2 4 or 8
Power of a Power Inverse Prop. of Exp. and Logarithms
4 log3 9 (log3 9) 4
Comm ( )
(log3 32) 4 2 4 or 8
Replace 9 with 32. Inverse Prop. of Exp. and Logarithms
So, log3 94 4 log3 9. • The Product of Powers Property and Product Property of Logarithms both involve the addition of exponents, since logarithms are exponents. 50. Let b x m, then logb m x. (b x)p m p b xp m p Product of Powers logb b xp logb m p Prop. of Equality
49. A
for Logarithmic Equations
xp logb m p plogb m logb m p 51. 4
52. 3
53. 2x
54. 6
55. 8
56. d 4
57. odd; 3
58. even; 4
©Glencoe/McGraw-Hill
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Inverse Prop. of Exp. and Logarithms Replace x with logb m.
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59.
3b a
60. 2
61.
5 3x
62. 3.06 s 64. 5
63. 1 65. x
3 4
5 3
66. x 2
Lesson 10-4 Common Logarithms Pages 549–551 1. 10; common logarithms
2. Sample answer: 5x 2; x 0.4307
3. A calculator is not programmed to find base 2 logarithms.
4. 0.6021
5. 1.3617
6. 0.3010
8. {n 0 n 0.4907}
7. 1.7325
10. 1.1615
9. 4.9824
12. {p 0 p 4.8188}
11. 11.5665 13.
log 5 ; log 7
0.8271
14.
15.
log 9 ; log 2
3.1699
16. at most 0.00003 mole per liter
log 42 ; log 3
3.4022
17. 0.6990
18. 1.0792
19. 0.8573
20. 0.3617
21. 0.0969
22. 1.5229
23. 11
24. 2.2
25. 2.1
26. 3.5
27. {x 0 x 2.0860}
28. 2.4550
29. {a 0 a 1.1590}
30. 0.5537
31. 0.4341
32. 4.8362
33. 4.7820
34. 8.0086
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36. 2.6281
35. 1.1909
37. {n 0 n 1.0178}
38. 1.0890
39. 3.7162
40. {p 0 p 1.9803}
41. 0.5873
42. 4.7095
43. 7.6377
44. 2.7674
45.
log 13 log 2
47.
log 3 log 7
49.
2log 1.6 log 4
3.7004
0.5646 0.6781
46.
log 20 log 5
48.
log 8 log 3
50.
0.5 log 5 log 6
1.8614
1.8928 0.4491
51. between 0.000000001 and 0.000001 mole per liter
52. 8
53. Sirius
54. Sirius: 1.45, Vega: 0.58
55. Vega
56a. 3;
1 3 3 2 56b. ; 2 3
56c. conjecture: loga b
1 ; logba
proof: ?
1 log b a
?
1 logb a 1 logb a
loga b logb b logb a 1 logb a
Change of Base Formula
✓
Inverse Prop. of Exponents and Logarithms
58. about 11.64 yr or 11 yr, 8 mo
57. about 3.75 yr or 3 yr 9 mo
©Glencoe/McGraw-Hill
Original statement
282
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59. Comparisons between substances of different acidities are more easily distinguished on a logarithmic scale. Answers should include the following. • Sample answer: Tomatoes: 6.3 105 mole per liter Milk: 3.98 107 mole per liter Eggs: 1.58 108 mole per liter • Those measurements correspond to pH measurements of 5 and 4, indicating a weak acid and a stronger acid. On the logarithmic scale we can see the difference in these acids, whereas on a normal scale, these hydrogen ion concentrations would appear nearly the same. For someone who has to watch the acidity of the foods they eat, this could be the difference between an enjoyable meal and heartburn.
60. A
61. C
62. 1.4248
63. 1.6938
64. 1.8416
65. 64
66. z
67. 62
68. 22
69. (d 2)(3d 4)
70. (7p 3)(6q 5)
71. prime
72. 2x 3
73. 32 x
74. 53 125
75. log5 45 x
76. log7 x 3
1 64
77. logb x y ©Glencoe/McGraw-Hill
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Algebra 2
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Lesson 10-5
Page 284 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-10:
Base e and Natural Logarithms Pages 557–559
1. the number e
2. e x 8
3. Elsu; Colby tried to write each side as a power of 10. Since the base of the natural logarithmic function is e, he should have written each side as a power of e; 10ln 4x Z 4x.
4. 403.4288
5. 0.0334
6. 0.1823
7. 2.3026
8. x ln 4
9. e0 1
10. 3
11. 5x
12. x 3.4012
13. 1.0986
14. 0.8047
15. 0 x 403.4288
16. 2.4630
17. 90.0171
18. h 26200 ln
19. about 15,066 ft
20. 54.5982
21. 148.4132
22. 0.3012
23. 1.6487
24. 1.0986
25. 2.3026
26. 1.6901
27. 3.5066
28. $183.21
29. about 49.5 cm
30. x ln 5
31. 2 ln 6x
32. e1 e
33. e x 5.2
34. 0.2
35. y
36. 4x
37. 45
38. 0.2877
39. 0.6931
40. x 1.5041
41. x 0.4700
42. 0.2747
43. 0.5973
44. x 0.6438
45. x 0.9730
46. 27.2991
©Glencoe/McGraw-Hill
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P 101.3
Algebra 2
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47. 49.4711
48. 1.7183
49. 14.3891
50. 232.9197
51. 45.0086
52. 2, 6
53. 1
54. about 19.8 yr
55. t
100 ln 2 r
56. 100 ln 2 70
57. t
110 r
58. about 7.33 billion
59. about 55 yr
60. about 32 students
61. about 21 min
62. always; log x log y
?
In x In y
Original statement
log x logx ? log e log y logy log e
63. The number e is used in the formula for continuously compounded interest, A Pe rt. Although no banks actually pay interest compounded continually, the equation is so accurate in computing the amount of money for quarterly compounding or daily compounding, that it is often used for this purpose. Answers should include the following.
©Glencoe/McGraw-Hill
log x log y
?
log x log e
log x log y
?
log x log y
Change of Base Formula
log e log y
Multiply log x by the log e reciprocal of log y . log e Simplify.
64. B
285
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• If you know the annual interest rate r and the principal P, the value of the account after t years is calculated by multiplying P times e raised to the r times t power. Use a calculator to find the value of e rt. • If you know the value A you wish the account to achieve, the principal P, and the annual interest rate r, the time t needed to achieve this value is found by first taking the natural logarithm of A minus the natural logarithm of P. Then, divide this quantity by r. 65. 1946, 1981, 2015; It takes between 34 and 35 years for the population to double. 67.
log 0.047 log 6
1.7065
66.
log 68 log 4
3.0437
68.
log 23 log 50
0.8015
69. 5
70. 4
71. inverse; 4
72. joint; 1
73. direct; 7
74. x
75. 3.32
76. 1.54
77. 1.43
78. 323.49
79. 13.43
80. 9.32
©Glencoe/McGraw-Hill
286
1 2 y 20
5
Algebra 2
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Chapter 10 Practice Quiz 2 Page 559 1.
log 5 ; log 4
2. e 2 3x
1.1610
4. x 5.3219
3. 3 5. 1.3863
Lesson 10-6
Exponential Growth and Decay Pages 563–565
1. y a(1 r)t, where r 0 represents exponential growth, and r 0 represents exponential decay.
2. Take the common logarithm of each side, use the Power Property to write log (1 r)t as t log(1 r), and then divide each side by the quantity log(1 r).
3. Sample answer: money in a bank
4. Decay; the exponent is negative.
5. about 33.5 watts
6. about 402 days
7. y 212,000e0.025t
8. about 349,529 people
9. C
10. $1600
11. at most $108,484.93
12. about 8.1 days
13. No; the bone is only about 21,000 years old, and dinosaurs died out 63,000,000 years ago.
14. more than 44,000 years ago
15. about 0.0347
16. y ae0.0347t
17. $12,565 billion
18. about 2025
19. after the year 2182
20. 4.7%
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21. Never; theoretically, the amount left will always be half of the previous amount.
22. Answers should include the following. • Find the absolute value of the difference between the price of the car for two consecutive years. Then divide this difference by the price of the car for the earlier year. • Find 1 minus the rate of decrease in the value of the car as a decimal. Raise this value to the number of years it has been since the car was purchased, and then multiply by the original value of the car.
23. about 19.5 yr
24. D
25. ln y 3
26. ln 29 4n 2
27. 4x 2 e8
28. 1.5323
29. p 3.3219
30. 9
31.
0.5 (0.08 p) 6
33.
p 150
0.5 (0.08 p) 4
32.
p 60
34. hyperbola
35. ellipse
36. parabola
37. circle
38. 2.06 108
39. 8 107
40. about 38.8%
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Chapter 11 Sequences and Series Lesson 11-1 Arithmetic Sequences Pages 580–582 1. The differences between the terms are not constant.
2. 95
3. Sample answer: 1, 4, 9, 14, . . .
4. 24, 28, 32, 36
5. 3, 5, 7, 9
6. 5, 8, 11, 14, 17
7. 14, 12, 10, 8, 6
8. 43
9. 112
10. 79
11. 15
12. an 11n 37
13. 56, 68, 80
14. $12,000
15. 30, 37, 44, 51
16. 10, 3, 4, 11
17. 6, 10, 14, 18
18. 1, 4, 7, 10
19.
7 , 3
3,
11 13 , 3 3
20.
12 , 5
8 6 5 5
2, ,
21. 5.5, 5.1, 4.7, 4.3
22. 8.8, 11.3, 13.8, 16.3
23. 2, 15, 28, 41, 54
24. 41, 46, 51, 56, 61
25. 6, 2, 2, 6, 10
26. 12, 9, 6, 3, 0
27.
4 , 3
2 1 3 3
1, , , 0
28.
5 , 8
1,
11 7 17 , , 8 4 8
29. 28
30. 49
31. 94
32. 175
33. 335
34. 340
35.
26 3
25 2
36.
37. 27
38. 47
39. 61
40. 173
41. 37.5 in.
42. 304 ft
43. 30th
44. 19th
45. 82nd
46. an 9n 2
47. an 7n 25
48. an 2n 1
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49. 13, 17, 21
50. pn 4n 3
51. Yes; it corresponds to n 100.
52. 70, 85, 100
53. 4, 2
54. 5, 2, 1, 4
55. 7, 11, 15, 19, 23
56. z 2y x
57. Arithmetic sequences can be used to model the numbers of shingles in the rows on a section of roof. Answers should include the following. • One additional shingle is needed in each successive row. • One method is to successively add 1 to the terms of the sequence: a8 9 1 or 10, a9 10 1 or 11, a10 11 1 or 12, a11 12 1 or 13, a12 13 1 or 14, a13 14 1 or 15, a14 15 1 or 16, a15 16 1 or 17. Another method is to use the formula for the nth term: a15 3 (15 1)1 or 17.
58. B
59. B
60. about 26.7%
61. 0.4055
62. 0.4621
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63. 146.4132
64. 15
65. 2, 5, 8, 11
66. 5, 4, 3, 2
67. 11, 15, 19, 23, 27
Lesson 11-2 Arithmetic Series Pages 586–587 2. Sample answer: 0 1 2 34
1. In a series, the terms are added. In a sequence, they are not. 4
3. Sample answer: a (3n 4)
4. 1300
5. 230
6. 1932
7. 552
8. 800
n1
9. 260
10. 63
11. 95
12. 11, 20, 29
13. 6, 0, 6
14. 28
15. 344
16. 663
17. 1501
18. 2646
19. 9
20. 88
21. 104
22. 182
23. 714
24. 225
25. 14
26.
27. 10 rows
28. 8 days
29. 721
30. 735
31. 162
32. 204
33. 108
34. 35
35. 195
36. 510
37. 315,150
38. 24,300
39. 1,001,000
40. 166,833
41. 17, 26, 35
42. 13, 8, 3
©Glencoe/McGraw-Hill
245 6
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43. 12, 9, 6
44. 13, 18, 23
45. 265 ft
46. True; for any series, 2a1 2a2 2a3 p 2an 2(a1 a2 a3 p an).
47. False; for example, 7 10 13 16 46, but 7 10 13 16 19 22 25 28 140.
48. Arithmetic series can be used to find the seating capacity of an amphitheater. Answers should include the following. • The sequence represents the numbers of seats in the rows. The sum of the first n terms of the series is the seating capacity of the first n rows. • One method is to write out the terms and add them: 18 22 26 30 34 38 42 46 50 54 360. Another method is to use the formula n Sn [2a1 (n 1)d ]: S10
2 10 2
[2(18)
(10 1)4] or 360. 49. C
50. C
51. 5555
52. 3649
53. 6683
54. 111
55. 135
56. about 3.82 days 16 3
9 2
58.
57. 59.
3 289 2
60. 23
61. 26 221
62. 2 22
63. 16
64. 54
65.
2 27
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Lesson 11-3 Geometric Sequences Pages 590–592 1a. Geometric; the terms have a common ratio of 2. 1b. Arithmetic; the terms have a common difference of 3.
2. Sample answer:
3. Marika; Lori divided in the wrong order when finding r.
4. 67.5, 101.25
5. 2, 4
6. 2, 6, 18, 54, 162
7.
2 4 8 , 3 9 27
1, , ,
15 64
8. 56
9. 4
10. an 4 2n1
11. 3, 9
12. A
13. 15, 5
14. 192, 256
15. 54, 81
16. 48, 32
17.
p
20 40 , 27 81
18.
125 625 , 24 48
19. 2.16, 2.592
20. 21.875, 54.6875
21. 2, 6, 18, 54, 162
22. 1, 4, 16, 64, 256
23. 243, 81, 27, 9, 3
24. 576, 288, 144, 72, 36
25.
3 16
26. 2592
27. 729
28. 1024
29. 243
30.
31. 1
32. 192
33. 78,125
34. 2
35. 8748
36.
37. 655.36 lb
38. $46,794.34 n1
39. an
5 72 n1
40. an 64 a b
1 36 a b 3
1 4
42. an 4(3)n 1
41. an 2(5)n 1
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1 4
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43. 18, 36, 72
44. 12, 36, 108
45. 16, 8, 4, 2
46. 6, 12, 24, 48
47. 8 days
48. 5 mg
49. False; the sequence 1, 4, 9, 16, p, for example, is neither arithmetic nor geometric.
50. False, the sequence 1, 1, 1, 1, p, for example, is arithmetic (d 0) and geometric (r 1).
51. The heights of the bounces of a ball and the heights from which a bouncing ball falls each form geometric sequences. Answers should include the following. • 3, 1.8, 1.08, 0.648, 0.3888 • The common ratios are the same, but the first terms are different. The sequence of heights from which the ball falls is the sequence of heights of the bounces with the term 3 inserted at the beginning.
52. A
53. C
54. 632.5
55. 203
56. 19, 23
57. 12, 16, 20
58. 5 22 3 210 units
59. 127
60.
61.
63 32
61 81
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Chapter 11 Practice Quiz 1 Page 592 11 2
1. 46
2.
3. 187
4. 816
5. 1
Lesson 11-4 Geometric Series Pages 596–598 2. The polynomial is a geometric series with first term 1, common ratio x, and 4 terms. The sum is
1. Sample answer: 421
1 2
1(1 x 4) x4 1 . 1x x1
4. 732
3. Sample answer: The first term is a1 2. Divide the second term by the first to find that the common ratio is r 6. Therefore, the nth term of the series is given by 2 6n1. There are five terms, so the series can be 5
written as a 2 6n1. n1
5. 39,063
6. 81,915
7. 165
8.
9. 129
10.
11.
1093 9
1330 9 31 4
12. 3
13. 3
14. 93 in. or 7 ft 9 in.
15. 728
16. 765
17. 1111
18. 300
©Glencoe/McGraw-Hill
295
Algebra 2
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19. 244
20. 1,328,600
21. 2101
22. 1441
23.
728 3
24.
215 4
25. 1040.984
26. 7.96875
27. 6564
28. 118,096
29. 1,747,625
30. $10,737,418.23
31. 3641
32. 206,668
33.
182 9
5461 16
34.
36. 364
35. 2555 37.
387 4
38.
58,975 256
39. 3,145,725
40. 86,093,440
41. 243
42. 1024
43. 2
44. 6
45. 80
46. 8
47. about 7.13 in.
48. If the first term and common ratio of a geometric series are integers, then all the terms of the series are integers. Therefore, the sum of the series is an integer.
49. If the number of people that each person sends the joke to is constant, then the total number of people who have seen the joke is the sum of a geometric series. Answers should include the following. • The common ratio would change from 3 to 4. • Increase the number of days the joke circulates so that it is inconvenient to find and add all the terms of the series.
50. A
©Glencoe/McGraw-Hill
296
Algebra 2
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51. C
52. 1,048,575
53. 3.99987793
54. 6.24999936 9
1 3 4 2
27
81
55. , , 9
56. 3, , , 2 4 8
57. 232
58. 192
59. Drive-In Movie Screens
60. Sample answer using (1, 826) and (3, 750): y 38x 864
Screens
1000 900 800 700 600 0 0
1 2 3 4 5 Years Since 1995
6
61. Sample answer: 294
62. 2
63. 2
64.
65.
2 3
1 4
66. 2
67. 0.6
Lesson 11-5
1. Sample answer:
Infinite Geometric Series Pages 602–604 n
1 a a 2b n1
2. 0.999999 . . . can be written as the infinite geometric 9 9 9 p. series 10
100
1000
The first term of this series is 9 and the common ratio is 10 9 1 10 , so the sum is 1 or 1. 10 1 10
©Glencoe/McGraw-Hill
297
Algebra 2
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3. Beth; the common ratio for the infinite geometric series
4. 108
is . Since ` ` 1, the 4 3
4 3
series does not have a sum a and the formula S 1 1r does not apply. 5. does not exist 7.
6. does not exist
3 4
8.
9. 100 11.
73 99
30 7
10.
5 9
12.
175 999
13. 96 cm
14. 14
15. does not exist
16. 7.5
17. 45
18. 64
19. 16
20. does not exist
21.
54 5
22. 3
23. does not exist
24. 1
25. 1
26. 7.5
27.
2 3
28. 144
29.
3 2
30. 6
31. 2
32. 30 ft
33. 40 20 22 20 p
34. 80 40 22 or about 136.6 cm
35. 900 ft
36. 27, 18, 12
37. 75, 30, 12
38. 24, 16 , 11 , 7
1 5
1 2
7 25
64 125
39. 8, 3 , 1 ,
©Glencoe/McGraw-Hill
40.
298
11 32
409 512
7 9
Algebra 2
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41.
1 9
42.
4 11
43.
82 99
44.
82 333
45.
427 999
46.
5 11
47.
229 990
48.
S a1 a1r a1r 2 a1r 3 p
()rS a1r a1r 2 a1r 3 a1r 4 p S rS a1 0 0 0 0 p S(1 r ) a1 a1 S 1r
49. The total distance that a ball bounces, both up and down, can be found by adding the sums of two infinite geometric series. Answers should include the following. • an a1 r n1, Sn a1(1 r n) , 1r
or S
50. D
a1 1r
• The total distance the ball falls is given by the infinite geometric series 3 3(0.6) 3(0.6)2 p . The sum of this series is 3 or 7.5. The total 1 0.6
distance the ball bounces up is given by the infinite geometric series 3(0.6) 3(0.6)2 3(0.6)3 p . The sum of this series is 3(0.6) 1 0.6
or 4.5. Thus, the total distance the ball travels is 7.5 4.5 or 12 feet. 52. 182
51. C 53.
8744 81
54. 32.768% 56.
55. 3 ©Glencoe/McGraw-Hill
299
3 2 Algebra 2
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57. x 5 59.
x 7 (x 3)(x 1)
61. (x 2)2 (y 4)2 36 1 2
63. ,
58.
2a 5b 2 a b
60.
3x 7 (x 4)(x 2)
62. (x 3)2 (y 1)2 32
3 7 , 2 2
1 2
1 3
64. , , 0,
1 2
65. x 2 36 0
66. x 2 9x 14 0
67. x 2 10x 24 0
68. about 180,724 visitors per year
69. The number of visitors was decreasing.
70. 2
71. 3
72. 2
1 2
74. 4
73.
75. 4
Lesson 11-6
Recursion and Special Sequences Pages 608–610
1. an an1 d; an r an1
2. Sample answer: an 2an1 an2
3. Sometimes; if f(x) x 2 and x1 2, then x2 22 or 4, so x2 x1. But, if x1 1, then x2 1, so x2 x1.
4. 12, 9, 6, 3, 0
5. 3, 2, 0, 3, 7
6. 0, 4, 4, 12, 20
7. 1, 2, 5, 14, 41
8. 5, 11, 29
9. 1, 3, 1
10. 3, 11, 123
11. bn 1.05bn1 10
12. $1172.41
13. 6, 3, 0, 3, 6
14. 13, 18, 23, 28, 33
15. 2, 1, 1, 4, 8
16. 6, 10, 15, 21, 28
17. 9, 14, 24, 44, 84
18. 4, 6, 12, 30, 84
19. 1, 5, 4, 9, 13
20. 4, 3, 5, 1, 9
©Glencoe/McGraw-Hill
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Algebra 2
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21.
7 7 7 7 7 , , , , 2 4 6 8 10
22.
3 3 15 25 425 , , , , 4 2 4 2 8
23. 67
24. 2.1
25. 1, 1, 2, 3, 5, . . .
26. the Fibonacci sequence
27. $99,921.21, $99,762.21, $99,601.29, $99,438.44,
28. 1, 3, 6, 10, 15
$99,841.95, $99,681.99, $99,520.11, $99,356.28
29. tn tn1 n
30. 20,100
31. 16, 142, 1276
32. 5, 17, 65
33. 7, 16, 43
34. 4, 19, 94
35. 3, 13, 333
36. 1, 1, 1
37.
5 37 1445 , , 2 2 2
38.
4 10 76 , , 3 3 3
39. $75.77
40. No; according to the first two iterates, f(4) 4. According to the second and third iterates, f(4) 7. Since f(x) is a function, it cannot have two values when x 4.
41. Under certain conditions, the Fibonacci sequence can be used to model the number of shoots on a plant. Answers should include the following. • The 13th term of the sequence is 233, so there are 233 shoots on the plant during the 13th month. • The Fibonacci sequence is not arithmetic because the differences (0, 1, 1, 2, . . .) of the terms are not constant. The Fibonacci sequence is not geometric because the ratios 3 Q1, 2, , . . .R of the terms
42. D
2
are not constant. ©Glencoe/McGraw-Hill
301
Algebra 2
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44. 27
43. C 1 6
46.
12 5
47. 5208
48.
1093 243
49. 3x 7 units
50. 120
51. 5040
52. 6
53. 20
54. 126
45.
55. 210
Lesson 11-7 The Binomial Theorem Pages 615–617 1. 1, 8, 28, 56, 70, 56, 28, 8, 1
2. n
3. Sample answer: (5x y)4
4. 40,320
5. 17,160
6. 66
7. p5 5p4q 10p3q 2 10p 2q 3 5pq 4 q 5
8. t 6 12t 5 60t 4
160t 3 240t 2 192t 64
9. x 4 12x 3y 54x 2y 2 108xy 3 81y 4
10. 56a5b3
11. 1,088,640a6b4
12. 10
13. 362,880
14. 6,227,020,800
15. 72
16. 210
17. 495
18. 2002
19. a 3 3a 2b 3ab 2 b 3
20. m4 4m 3n 6m 2n 2 4mn3 n4
21. r 8 8r 7s 28r 6s 2 56r 5s 3 70r 4s 4 56r 3s 5 28r 2s 6 8rs 7 s 8
22. m 5 5m 4a 10m 3a 2 10m 2a 3 5ma 4 a 5
23. x 5 15x 4 90x 3 270x 2 405x 243
24. a4 8a3 24a2 32a 16
©Glencoe/McGraw-Hill
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Algebra 2
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25. 16b4 32b3x 24b 2x 2 8bx 3 x 4
26. 64a6 192a5b 240a4b 2 160a3b3 60a2b4 12ab5 b6 28. 81x 4 216x 3y 216x 2y 2 96xy 3 16y 4
27. 243x 5 810x 4y 1080x 3y 2 720x 2y 3 240xy 4 32y 5 29.
a5 32
5a4 8
30. 243 135m 30m 2
5a3
10m 3 3
20a2 40a 32
5m 4 27
31. 27x 3 54x 2 36x 8 cm3
32. 1, 4, 6, 4, 1
33. 45
34. 126x 4y 5
35. 924x 6y 6
36. 280x 4
37. 5670a4
38. 1,088,640a6b4
39. 145,152x 6y 3
40.
35 4 x 27
42.
12! 7!5!
63 8
41. x 5
and
12! 6!6!
m5 243
represent the 6th
and 7th entries in the row for n 12 in Pascal’s triangle. 13! represents the seventh 7!6!
entry in the row for n 13. 13! 12! Since is below and 7!6! 7!5! 12! in Pascal’s triangle, 6!6! 12! 12! 13! . 7!5! 6!6! 7!6!
43. The coefficients in a binomial expansion give the numbers of sequences of births resulting in given numbers of boys and girls. Answers should include the following. • (b g)5 b5 5b4g 10b 3g 2 10b 2g 3 5bg 4 g 5;
©Glencoe/McGraw-Hill
44. D
303
Algebra 2
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There is one sequence of births with all 5 boys, five sequences with 4 boys and 1 girl, ten sequences with 3 boys and 2 girls, ten sequences with 2 boys and 3 girls, five sequences with 1 boy and 4 girls, and one sequence with all 5 girls. • The number of sequences of births that have exactly k girls in a family of n children is the coefficient of bnkgk in the expansion of (b g)n. According to the Binomial Theorem, this n! . coefficient is (n k)!k!
45. C
46. 7, 5, 3, 1, 1
47. 3, 5, 9, 17, 33
48. 125 cm
49.
log 5 ; log 2
2.3219
50.
51.
log 8 ; log 5
1.2920
52. asymptotes: x 2, x 3
1 ; log 3
2.0959
53. asymptotes: x 4, x 1
54. hole: x 3
55. hyperbola
56. parabola
57. yes
58. no
59. True;
1(1 1) 2
1(2) 2
60. False;
or 1.
2(3) 2
61. True;
12(1 1)2 4
©Glencoe/McGraw-Hill
1(4) 4
(1 1)(2 1 1) 2
or 3.
62. True; 31 1 2, which is even.
or 1.
304
Algebra 2
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Chapter 11 Practice Quiz 2 Page 617 1. 1,328,600
2. 364
3. 24
4.
5. 1, 5, 13, 29, 61
6. 2, 4, 8, 14, 22
7. 5, 13, 41
8. 243x 5 405x 4y 270x 3y 2 90x 2y 3 15xy 4 y 5
9. a6 12a5 60a4 160a3 240a 2 192a 64
Lesson 11–8
25 4
10. 4032a5b4
Proof and Mathematical Induction Pages 619–621
1. Sample answers: formulas for the sums of powers of the first n positive integers and statements that expressions involving exponents of n are divisible by certain numbers
2. Mathematical induction is used to show that a statement is true. A counter example is used to show that a statement is false.
3. Sample answer: 3n 1
4. Step 1: When n 1, the left side of the given equation is 1(1 1)
1. The right side is or 2 1, so the equation is true for n 1. Step 2: Assume 1 2 3 p k
k(k 1) 2
for
some positive integer k. Step 3: 1 2 3 p k 1k 12
©Glencoe/McGraw-Hill
305
k(k 1) (k 1) 2 k(k 1) 2(k 1) 2 (k 1) (k 2) 2
Algebra 2
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The last expression is the right side of the equation to be proved, where n k 1. Thus, the equation is true for n k 1. Therefore, 1 2 3 p n
n(n 1) 2
for all
positive integers n. 5. Step 1: When n 1, the left side of the given equation is 1 . 2 1 , 2
The right side is 1
1 2
6. Step 1: 41 1 3, which is divisible by 3. The statement is true for n 1. Step 2: Assume that 4k 1 is divisible by 3 for some positive integer k. This means that 4k 1 3r for some whole number r. Step 3: 4k 1 3r 4k 3r 1 4k1 12r 4 4k1 1 12r 3 4k1 1 314r 12 Since r is a whole number, 4r 1 is a whole number. Thus, 4k1 1 is divisible by 3, so the statement is true for n k 1. Therefore, 4n 1 is divisible by 3 for all positive integers n.
or
so the equation is true for
n 1. 1 1 2 2 2 1 1 1 p k 1 k for some 3 2 2 2
Step 2: Assume
positive integer k. Step 3:
1 1 2 2 2 1 1 1 p k k1 3 2 2 2 1 1 1 k k1 2 2 2 1 1 k1 k1 2 2 1 1 k1 2
The last expression is the right side of the equation to be proved, where n k 1. Thus, the equation is true for n k 1. Therefore,
1
1 2n
1 1 1 1 2 3 p n 2 2 2 2
for all positive
integers n.
©Glencoe/McGraw-Hill
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Algebra 2
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8. Sample answer: n 2
7. Step 1: 51 3 8, which is divisible by 4. The statement is true for n 1. Step 2: Assume that 5k 3 is divisible by 4 for some positive integer k. This means that 5k 3 4r for some positive integer r. Step 3: 5k 3 4r 5k 4r 3 5k1 20r 15 5k1 3 20r 12 5k1 3 415r 32 Since r is a positive integer, 5r 3 is a positive integer. Thus, 5k1 3 is divisible by 4, so the statement is true for n k 1. Therefore, 5n 3 is divisible by 4 for all positive integers n. 9. Sample answer: n 3
10. Step 1: After the first guest has arrived, no handshakes have taken place.
1(1 1) 2
0,
so the formula is correct for n 1. Step 2: Assume that after k guests have arrived, a total of
k(k 1) 2
handshakes have
take place, for some positive integer k. Step 3: When the (k 1)st guest arrives, he or she shakes hands with the k guests already there, so the total number of handshakes that have then taken place is k(k 1) 2
©Glencoe/McGraw-Hill
307
k.
Algebra 2
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k(k 1) k(k 1) 2k k 2 2 k [(k 1) 2] 2 (k 1)k k (k 1) or 2 2
The last expression is the formula to be proved, where n k 1. Thus, the formula is true for n k 1. Therefore, the total number of handshakes is
for
all positive integers n. 12. Step 1: When n 1, the left side of the given equation is
11. Step 1: When n 1, the left side of the given equation is 1. The right side is 1[2(1) 1] or 1, so the equation is true for n 1. Step 2: Assume 1 5 9 p (4k 3) k (2k 1) for some positive integer k. Step 3: 1 5 9 p (4k 3) [4(k 1) 3] k (2k 1) [4(k 1) 3] 2k 2 k 4k 4 3 2k 2 3k 1 (k 1)(2k 1) (k 1)[2(k 1) 1] The last expression is the right side of the equation to be proved, where n k 1. Thus, the equation is true for n k 1. Therefore, 1 5 9 p (4n 3) n(2n 1) for all positive integers n.
©Glencoe/McGraw-Hill
n(n 1) 2
1[3(1) 1]
2. The right side is 2 or 2, so the equation is true for n 1. Step 2: Assume 2 5 8 p (3k 1)
k(3k 1) 2
for some positive integer k. Step 3: 2 5 8 p (3k 1) [3(k 1) 1] k(3k 1) [3(k 1) 1] 2 k(3k 1) 2[3(k 1) 1] 2 2 3k k 6k 6 2 2
3k 2 7k 4 2 (k 1)(3k 4) 2 (k 1)[3(k 1) 1] 2
308
Algebra 2
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The last expression is the right side of the equation to be proved, where n k 1. Thus, the equation is true for n k 1. Therefore, 2 5 8 p n(3n 1)
(3n 1) for all 2 positive integers n. 14. Step 1: When n 1, the left side of the given equation is 12 or 1. The right side is
13. Step 1: When n 1, the left side of the given equation is 13 or 1. The right side is 12(1 1)2 4
1[2(1) 1][2(1) 1] 3
or 1, so the equation is true for n 1. Step 2: Assume 13 23 33 p k 3
k 2 1k 12 2 4
the equation is true for n 1. Step 2: Assume 12 32 52 p (2k 1)2
for
some positive integer k. Step 3: 13 2 3 33 p k 3 (k 1)3
k (2k 1)(2k 1) 3
for some
positive integer k. Step 3: 12 32 52 p (2k 1)2 [2(k 1) 1]2
k 2(k 1)2 (k 1)3 4 k 2(k 1)2 4(k 1)3 4 2 2 (k 1) [k 4(k 1)] 4 2 2 (k 1) (k 4k 4) 4 2 (k 1) (k 2)2 4 2 (k 1) [(k 1) 1]2 4
k(2k 1)(2k 1) 3
[2(k 1) 1]2
The last expression is the right side of the equation to be proved, where n k 1. Thus, the equation is true for n k 1.
©Glencoe/McGraw-Hill
or 1, so
309
k(2k 1)(2k 1) 3(2k 1)2 3 (2k 1)[k (2k 1) 3(2k 1)] 3 2 (2k 1)(2k k 6k 3) 3 2 (2k 1)(2k 5k 3) 3 (2k 1)(k 1)(2k 3) 3 (k 1)[2(k 1) 1][2(k 1) 1] 3
Algebra 2
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Therefore, 13 23 33 p n (n 1) 4 2
n3
The last expression is the right side of the equation to be proved, where n k 1. Thus, the equation is true for n k 1. Therefore, 12 32 52 p
2
for all
positive integers n.
(2n 1)2
n(2n 1)(2n 1) 3
for all positive integers n. 16. Step 1: When n 1, the left side of the given equation is
15. Step 1: When n 1, the left side of the given equation is 1 1 . The right side is a1 2 3 1 or , so the equation is 3
1 . 4
true
or , so the equation is true
1 3
p
1 a1 2
3 3 1 kb for 3
1 3
1 32
1 33
3
1 43
some
1 1 1 a2 kb k 1 2 3 3 1 1 1 k k 1 2 23 3 3k 1 3 2 2 3k 1 3k 1 1 2 3k 1 1 3k 1 1 a b 2 3k 1
p
3k 1
1 2
©Glencoe/McGraw-Hill
p
Step 3:
1
a1
4 4 1 1 1 a1 kb k 4 3 4
for some positive integer k. 1 4k
1 3k
1 4
for n 1. 1 1 Step 2: Assume 2
positive integer k. Step 3:
1 3
1 4
for n 1. 1 1 1 Step 2: Assume 2 2 1 3k
The right side is a1 b
b
1 4
1 42
1 43
p
1 k 1
4
1 1 1 kb k 1 3 4 4 1 1 1 k k 1 3 34 4 4k 1 4 3 3 4k 1 4k 1 1 3 4k 1 1 4k 1 1 a b 3 4k 1
a1
a1 1 3
b 3k 1 1
310
k 1b
1
4
Algebra 2
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The last expression is the right side of the equation to be proved, where n k 1. Thus, the equation is true for n k 1. 1 1 1 1 Therefore, 2 3 p n
The last expression is the right side of the equation to be proved, where n k 1. Thus, the equation is true for n k 1. Therefore,
1 a1 2
1 1 1 1 2 3p n 3 3 3 3 1 b for all positive 3n
4
a1 1 3
1 b 4n
4
4
4
for all positive
integers n.
integers n. 17. Step 1: 81 1 7, which is divisible by 7. The statement is true for n 1. Step 2: Assume that 8k 1 is divisible by 7 for some positive integer k. This means that 8k 1 7r for some whole number r. Step 3: 8k 1 7r 8k 7r 1 8k 1 56r 8 8k1 1 56r 7 8k 1 1 7(8r 1) Since r is a whole number, 8r 1 is a whole number. Thus, 8k1 1 is divisible by 7, so the statement is true for n k 1. Therefore, 8n 1 is divisible by 7 for all positive integers n.
18. Step 1: 91 1 8, which is divisible by 8. The statement is true for n 1. Step 2: Assume that 9k 1 is divisible by 8 for some positive integer k. This means that 9k 1 8r for some whole number r. Step 3: 9k 1 8r 9k 8r 1 9k1 72r 9 9k1 1 72r 8 9k1 1 8(9r 1) Since r is a whole number, 9r 1 is a whole number. Thus, 9k 1 1 is divisible by 8, so the statement is true for n k 1. Therefore, 9n 1 is divisible by 8 for all positive integers n.
19. Step 1: 121 10 22, which is divisible by 11. The statement is true for n 1. Step 2: Assume that 12k 10 is divisible by 11 for some positive integer k. This means that 12k 10 11r for some positive integer r.
20. Step 1: 131 11 24, which is divisible by 12. The statement is true for n 1. Step 2: Assume that 13k 11 is divisible by 12 for some positive integer k. This means that 13k 11 12r for some positive integer r.
©Glencoe/McGraw-Hill
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Step 3: 12k 10 11r 12k 11r 10 12k1 132r 120 12k1 10 132r 110 12k1 10 11(12r 10) Since r is a positive integer, 12r 10 is a positive integer. Thus, 12k 1 10 is divisible by 11, so the statement is true for n k 1. Therefore, 12n 10 is divisible by 11 for all positive integers n.
Step 3: 13k 11 12r 13k 12r 11 13k 1 156r 143 13k 1 11 156r 132 13k 1 11 12(13r 11) Since r is a positive integer, 13r 11 is a positive integer. Thus, 13k1 11 is divisible by 12, so the statement is true for n k 1. Therefore, 13n 11 is divisible by 12 for all positive integers n.
21. Step 1: There are 6 bricks in the top row, and 12 5(1) 6, so the formula is true for n 1. Step 2: Assume that there are k 2 5k bricks in the top k rows for some positive integer k. Step 3: Since each row has 2 more bricks than the one above, the numbers of bricks in the rows form an arithmetic sequence. The number of bricks in the (k 1)st row is 6 [(k 1) 1](2) or 2k 6. Then the number of bricks in the top k 1 rows is k 2 5k (2k 6) or k 2 7k 6. k 2 7k 6 (k 1)2 5(k 1), which is the formula to be proved, where n k 1. Thus, the formula is true for n k 1.
22. Step 1: When n 1, the left side of the given equation is 1 a1. The right side is a1(1 r )
©Glencoe/McGraw-Hill
1r
or a1, so the equation is true for n 1. Step 2: Assume a1 a1r a1r 2 p a1r k1 a1(1 r k ) 1r
for some positive
integer k. Step 3: a1 a1r a1r 2 p a1r k1 a1r k
a1(1 r k) a1r k 1r a1(1 r k ) (1 r )a1r k 1r k a1 a1r a1r k a1r k 1 1r k a1(1 r 1) 1r
The last expression is the right side of the equation to be proved, where n k 1.
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Thus, the equation is true for n k 1. Therefore, a1 a1r a1r 2 p a1r n1
Therefore, the number of bricks in the top n rows is n 2 5n for all positive integers n.
a1(1 r n ) 1r
for all positive integers n. 23. Step 1: When n 1, the left side of the given equation is
24. Step 1: The figure shows how to cover a 21 by 21 board, so the statement is true for n 1.
1 2
a1. The right side is [2a1 (1 1)d ] or a1, so the equation is true for n 1. Step 2: Assume a1 (a1 d ) (a1 2d 2 p [a1 (k 1)d ]
k [2a1 2
(k 1)d ] for
Step 2: Assume that a 2k by 2k board can be covered for some positive integer k.
some positive integer k. Step 3: a1 (a1 d ) (a1 2d ) p [a1 (k 1)d ] [a1 (k 1 1)d ]
k [2a1 2
(k 1)d ]
k [2a1 2
(k 1)d ]
[a1 (k 1 1)d ]
a1 kd
k [2a1 (k 1)d ] 2(a1 kd ) 2 2 k 2a1 (k k)d 2a1 2kd 2 (k 1)2a1 (k 2 k 2k)d 2 (k 1)2a1 k(k 1)d 2 k1 (2a1 kd ) 2 k1 [2a1 (k 1 1)d ] 2
©Glencoe/McGraw-Hill
Step 3: Divide a 2k1 by 2k1 board into four quadrants. By the inductive hypothesis, the first quadrant can be covered. Rotate the design that covers Quadrant I 90 clockwise and use it to cover Quadrant II. Use the design that covers Quadrant I to cover Quadrant III.
313
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The last expression is the right side of the formula to be proved, where n k 1. Thus, the formula is true for n k 1. Therefore, a1 (a1 d ) (a1 2d ) p [a1 (n 1)d ]
Rotate the design that covers Quadrant I 90
counterclockwise and use it to cover Quadrant IV. This leaves three empty squares near the center of the board, as shown. Use one more L-shaped tile to cover these 3 squares. Thus, a 2k1 by 2k1 board can be covered. The statement is true for n k 1. Therefore, a 2n by 2n checkerboard with the top right square missing can be covered for all positive integers n.
n [2a1 (n 1)d ] 2
for all positive integers n.
25. Sample answer: n 3
26. Sample answer: n 4
27. Sample answer: n 2
28. Sample answer: n 3
29. Sample answer: n 11
30. Sample answer: n 41
31. Write 7n as (6 1)n. Then use the Binomial Theorem. 7n 1 (6 1)n 1
32. An analogy can be made between mathematical induction and a ladder with the positive integers on the steps. Answers should include the following. • Showing that the statement is true for n 1 (Step 1). • Assuming that the statement is true for some positive integer k and showing that it is true for k 1 (Steps 2 and 3).
6n n 6n1
n(n 1) 2
6n2 p n 6 1 1 6n n 6n1
n(n 1) 2
6n2 p n 6 Since each term in the last expression is divisible by 6, the whole expression is divisible by 6. Thus, 7n 1 is divisible by 6. 33. C
34. A
35. x 6 6x 5y 15x 4y 2 20x 3y 3 15x 2y 4 6xy 5 y 6
36. a7 7a6b 21a5b 2 35a4b3 35a3b4 21a 2b 5 7ab6 b7
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37. 256x 8 1024x 7y 1792x 6y 2 1792x 5y 3 1120x 4y 4 448x 3y 5 112x 2y 6 16xy 7 y 8
38. 4, 10, 28
39. 2, 14, 782
40. 12 h
41. 0, 1
42. 14
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Chapter 12 Probability and Statistics Lesson 12-1 The Counting Principle Pages 634–637 1. HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
2. Sample answer: buying a shirt that comes in 3 sizes and 6 colors
3. The available colors for the car could be different from those for the truck.
4. independent
5. dependent
6. 30
7. 256
8. 20 10. dependent
9. D 11. independent
12. independent
13. dependent
14. 6
15. 16
16. 6
17. 30
18. 48
19. 1024
20. 240
21. 10,080
22. 151,200
23. 362,880
24. 17
25. 27,216
26. 160
27. 800
28. See students’ work.
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29. The maximum number of license plates is a product with factors of 26s and 10s, depending on how many letters are used and how many digits are used. Answers should include the following.
30. A
• There are 26 choices for the first letter, 26 for the second, and 26 for the third. There are 10 choices for the first number, 10 for the second, and 10 for the third. By the Fundamental Counting Principle, there are 263 103 or 17,576,000 possible license plates. • Replace positions containing numbers with letters. 31. C
32. 45
33. 20 mi
34. Step 1: When n 1, the left side of the given equation is 4. The right side is
1[3(1) 5] 2
or 4, so the equation is true for n 1. Step 2: Assume 4 7 10 p (3k 1)
k(3k 5) 2
for some positive integer k. Step 3: 4 7 10 p (3k 1) [3(k 1) 1]
©Glencoe/McGraw-Hill
317
k(3k 5) [3(k 1) 1] 2 k(3k 5) 2[3(k 1) 1] 2 2 3k 5k 6k 6 2 2 2 3k 11k 8 2
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(k 1)(3k 8) 2 (k 1)[3(k 1) 5] 2
The last expression is the right side of the equation to be proved, where n k 1. Thus, the equation is true for n k 1. Therefore, 4 7 10 p n(3n 5)
(3n 1) 2 for all positive integers n. 35. 28x 6 y 2
36. 280a3b4
37. 7
38. 5
39.
1 2
40. 1
41.
x x 5y
42. 36 mi
43. 1, 2
44. 0, 2
45. y (x 3)2 2
46. y 2(x 1)2 4
1 2
47. y x 2 8
48. 4
49. 3
50. 4
1 R 3 53. no inverse exists 51.
2 3
55. y x
1 1 B 6 2
5 R 4 54. y 2x 2
1 1 B 7 4
52.
1 3
56. 60
57. 30
58. 840
59. 720
60. 6
61. 15
62. 56
63. 1
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Permutations and Combinations Pages 641–643 2. C (n, n r)
1. Sample answer: There are six people in a contest. How many ways can the first, second, and third prizes be awarded?
n! [n (n r )]!(n r )! n! r !(n r )! n! (n r )!r !
C(n, r ) 3. Sometimes; the statement is only true when r 1.
4. 60
5. 120
6. 6
7. 6
8. combination; 15 10. permutation; 90,720
9. permutation; 5040 11. 84
12. 56
13. 9 15. 665,280
14. 2520 16. 10
17. 70
18. 792
19. 210
20. 27,720
21. 1260
22. permutation; 5040
23. combination; 28
24. permutation; 2520
25. permutation; 120
26. combination; 220
27. permutation; 3360
28. combination; 45
29. combination; 455
30. 11,880
31. 60
32. 75,287,520
33. 111,540
34. 267,696
35. 80,089,128
36. 528
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37. C (n 1,r ) C (n 1, r 1)
(n 1)! (n 1 r )!r ! (n 1)! [n 1 (r 1)] !(r 1)! (n 1)! (n r 1 )!r ! (n
38. Permutations and combinations can be used to find the number of different lineups. Answers should include the following. • There are 9! different 9-person lineups available: 9 choices for the first player, 8 choices for the second player, 7 for the third player, and so on. So, there are 362,880 different lineups. • There are C(16, 9) ways to choose 9 players from
(n (n (n (n
(n 1)! r )! (r 1)! (n 1)! nr r 1)!r ! n r (n 1)! r r )!(r 1)! r 1)!(n r ) (n 1)!r (n r )!r ! (n r )!r ! 1)!(n r r ) (n r )!r !
16: C (16, 9) 11,440.
16! 7!9!
or
(n 1)!n (n r )!r ! n! (n r )!r !
C(n, r ) 39.
D
40. A
41. 24
42. 6
43. 120
44. 8
45. 80
46. Sample answer: n 3
47. Sample answer: n 2
48. 1.0986
49. x 0.8047
50. 21.0855
51. 20 days 53.
(y 4)2 9
52. (x 4) 2 4
1
x2 16
y2 9
1
7 53 2 2
54. ;
55. 4; 128
56. {4, 4}
57. {2, 5}
58. e 3, f
59. 822
60. 0 x 3 0 y 3 23
61. 425
62. (1, 3)
©Glencoe/McGraw-Hill
1 3
320
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4 3
64.
63. (0, 2) 65.
6 7
66. 0
67. {7, 15}
68.
69.
3 5
70.
1 2
71.
1 5
72.
1 3
Lesson 12-3 Probability Pages 647–650 1. Sample answer: The event July comes before June has a probability of 0. The event June comes before July has a probability of 1.
2.
3 5
3. There are 6 6 or 36 possible outcomes for the two dice. Only 1 outcome, 1 and 1, results in a sum of 2,
4.
1 7
6.
4 7
1
so P(2) . There are 2 36 outcomes, 1 and 2 as well as 2 and 1, that result in a sum 2 1 of 3, so P(3) or . 36
5.
18
2 7
7. 8:1
8. 1:5
9. 2:7
10.
6 11
11.
10 11
12.
2 7
13.
1 8
14.
3 8
©Glencoe/McGraw-Hill
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15.
1 10
16.
21 50
17.
2 25
18.
1 50
19.
6 55
20.
21 55
21.
28 55
22.
14 575
23.
11 115
24.
7 115
25.
6 115
26.
132 575
27.
24 115
28.
6 115
29. 0
30.
1 22,957,480
31. 0.007
32. 0.623
33. 0.109
34. 1:1
35. 3:5
36. 11:1
37. 5:3
38. 4:3
39. 1:4
40. 4:7
41. 3:1
42.
6 7
43.
3 10
44.
5 11
45.
4 9
46.
9 17
47.
1 9
48.
7 16
49.
3 5
50.
1 10
51. 2:23
52. 1:999
53. 1:4
54.
©Glencoe/McGraw-Hill
322
540 1771
Algebra 2
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55.
1 20
56.
9 20
57.
9 20
58.
1 20
59.
9 20
60.
9 20
61.
1 120
62.
1
64. C
63. Probability and odds are good tools for assessing risk. Answers should include the following. • P(struck by lightning) 1 s , so sf
750,000
odds 1:(750,000 1) or 1:749,999. P(surviving a lightning strike) s sf
3 4
, so odds
3:(4 3) or 3 :1. • In this case, success is being struck by lightning or surviving the lightning strike. Failure is not being struck by lightning or not surviving the lightning strike. 1 36
65. D
66. theoretical;
67. experimental; about 0.307
68. experimental;
69. theoretical;
1 17
1 5
70. permutation; 120
71. permutation; 1260
72. combination; 35
73. 16
74. 24
75. direct variation
76. square root
77. (4, 4)
78. (1, 3)
©Glencoe/McGraw-Hill
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Algebra 2
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79.
6 35
80.
3 14
81.
1 4
82.
2 21
83.
9 20
Chapter 12 Practice Quiz 1 Page 650 1. 24
2. 756
3. 18,720
4. 1320
5. 56
6. permutation; 40,320
7. combination; 20,358,520
8.
1 221
10.
8 663
9.
13 102
Lesson 12-4 Multiplying Probabilities Pages 654–657 1. Sample answer: putting on your socks, and then your shoes
2. P(A, B, C, and D) P(A) P(B) P(C) P(D)
3. Mario; the probabilities of rolling a 4 and rolling a 2 are
4.
1 36
1 6
both . 5.
1 4
6.
1 17
7.
4 663
8.
7 30
9.
1 4
10.
1 16
©Glencoe/McGraw-Hill
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21 220
12. independent;
13.
1 12
14.
1 36
15.
25 36
16.
1 36
17.
1 6
18.
1 6
19.
5 6
20.
1 42
21.
1 49
22.
25 49
23.
10 21
24. 0
25. 0
26.
1 15 1 10
27.
2 15
28.
29.
2 15
30. dependent;
31. independent;
25 81
1 36
3 28
32. independent;
168 4913 1 32
33. dependent;
1 21
34. independent;
35. dependent;
81 2401
36.
1 9
First Spin
Second Spin R
R P(R,B)
B 1 3
1 3
or
1 9
B
Y R B Y R B
Y
Y
©Glencoe/McGraw-Hill
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Algebra 2
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First Spin Blue Yellow Red 1 1 1 3 3 3 Blue 1 3 Second Spin
BB 1 9
BY 1 9
BR 1 9
Yellow YB 1 1 3 9
YY 1 9
YR 1 9
RY 1 9
RR 1 9
Red 1 3
RB 1 9
38.
1 3
39.
1 9
40.
1 635,013,559,600
41.
19 1,160,054
42.
1 158,753,389,900
43.
6327 20,825
44. a
99 4 b 100 1 1320
45. about 4.87%
46.
47. no
48. no
49. Sample answer: As the number of trials increases, the results become more reliable. However, you cannot be absolutely certain that there are no black marbles in the bag without looking at all of the marbles.
50. 21
51. Probability can be used to analyze the chances of a player making 0, 1, or 2 free throws when he or she goes to the foul line to shoot 2 free throws. Answers should include the following.
52. D
©Glencoe/McGraw-Hill
326
or about 0.96
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• One of the decimals in the table could be used as the value of p, the probability that a player makes a given free throw. The probability that a player misses both free throws is (1 p)(1 p) or (1 p) 2. The probability that a player makes both free throws is p p or p 2. Since the sum of the probabilities of all the possible outcomes is 1, the probability that a player makes exactly 1 of the 2 free throws is 1 (1 p) 2 p 2 or 2p (1 p). • The result of the first free throw could affect the player’s confidence on the second free throw. For example, if the player makes the first free throw, the probability of making the second free throw might increase. Or, if the player misses the first free throw, the probability of making the second free throw might decrease. 53. C 55.
3 340
57. 1440 ways
1 204
56.
1 119
58. 6
59. 36
©Glencoe/McGraw-Hill
54.
60. x 2 4x 2
327
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61. x, x 4
62.
y
x
O
y x2 x 2
63.
64.
y
O
y
x
O
x
y x 2 3x
y x2 4
65. 153
66. 9
69. (1, 2)
70. (13, 9)
71. (2, 4)
72.
2 3 5 4
67. 0 b 0
68. 5a 4 0 b 3 0
73.
5 6
74.
75.
11 12
76. 1
1 6
5 12
77. 1
©Glencoe/McGraw-Hill
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Lesson 12-5 Adding Probabilities Pages 660–663 1. Sample answer: mutually exclusive events: tossing a coin and rolling a die; inclusive events: drawing a 7 and a diamond from a standard deck of cards
2.
3. The events are not mutually exclusive, so the chance of rain is less than 100%.
4.
1 3
French and Algebra
French 150
5.
1 3
6.
1 3
7.
1 2
8.
5 6
9.
2 3
11. inclusive;
4 13
12.
13 16
14.
1 6
15.
25 42
16.
37 42
17.
35 143
18.
105 143
19.
3 143
20.
84 143
21.
38 143
22.
32 39
23. mutually exclusive;
©Glencoe/McGraw-Hill
Algebra 300
10. mutually exclusive;
13. 1
25. inclusive;
400
7 9
24. inclusive;
21 34
1 2
26. mutually exclusive;
329
2 13
4 13
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27.
4 13
28.
2 3
29.
55 221
30.
11 221
31.
188 663
32.
63 221
33.
1 8
34.
1 8
35.
1 4
36.
3 4
37.
1 780
38.
145 156
39.
9 130
40.
1 26
41.
11 780
42.
1 78
43.
3 5
44.
53 108
45.
17 27
46.
17 162
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47. Subtracting P(A and B) from each side and adding P (A or B) to each side results in the equation P (A or B) P(A) P(B) P(A and B). This is the equation for the probability of inclusive events. If A and B are mutually exclusive, then P (A and B) 0, so the equation simplifies to P(A or B) P(A) P(B), which is the equation for the probability of mutually exclusive events. Therefore, the equation is correct in either case.
48. Probability can be used to estimate the percents of people who do the same things before going to bed. Answers should include the following. • The events are inclusive because some people brush their teeth and set their alarm. Also, you know that the events are inclusive because the sum of the percents is not 100%. • According to the information in the text and the table, P (read book) 38 and P (brush teeth) 100 81 . Since the events 100
are inclusive, P (read book and brush teeth) P (read book) P (brush teeth) P (read book and brush 38 81 teeth) 100 1200 59 . 2000 100
49. C
100
50. A
51.
1 216
52.
125 216
53.
1 216
54.
1 8
55. 4:1
56. 1:8
57. 2:5
58. 5:3
59. 254
60. 24
61. (8, 10)
62. (12, 5)
63. (x 1)2(x 1)(x 2 1)
64. min: (0, 5); max: (1.33, 3.81)
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65. min: (0.42, 0.62); max: (1.58, 1.38)
66.
y
x
O
(0, 2), (2, 0), (0, 2); max: f(2, 0) 6; min: f (0, 2) 2 67.
68. d 12.79t
y
x
O
(1, 3), (1, 1), (3, 3), (3, 5); max: f(3, 5) 23; min: f (1, 1) 3 69. direct variation
70. 323.4, 298, no mode, 143
71. 35.4, 34, no mode, 72
72. 3.6, 3.45, 2.1, 3.6
73. 63.75, 65, 50 and 65, 30
74. 79.83, 89, 89, 57
75. 12.98, 12.9, no mode, 4.7
Lesson 12-6 Statistical Measures Pages 666–670 2. Sample answer: The variance of the set {0, 1} is 0.25 and the standard deviation is 0.5.
1. Sample answer: {10, 10, 10, 10, 10, 10} n
3.
1 a (xi B n i1
x )2
4. 40, 6.3
5. 8.2, 2.9
©Glencoe/McGraw-Hill
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7. $7300.50, $5335.25
8. The mean is more representative for the southwest central states because the data for the Pacific states contains the most extreme value, $10,650.
9. 2500, 50
10. 1.6, 1.3
11. 3.1, 1.7
12. 4.8, 2.2
13. 37,691.2, 194.1
14. 569.4, 23.9
15. 82.9, 9.1
16. 43.6, 6.6
17. 114.5, 105, 23
18. The mean and median both seem to represent the center of the data.
19. Mean; it is highest.
20. Mode; it is lower and is what most employees make. It reflects the most representative worker.
21. $1047.88, $1049.50, $695
22. Mode; it is the least expensive price.
23. Mean or median; they are nearly equal and are more representative of the prices than the mode.
24. 2,290,403; 2,150,000; 2,000,000
25. Mode; it is lowest.
26. Mean; it is highest.
27. 19.3
28. 28.9
29. 19.5
30. Washington; see students’ work.
31. 59.8, 7.7
32. 64%
33. 100%
34. Different scales are used on the vertical axes.
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35. Sample answer: The first graph might be used by a sales manager to show a salesperson that he or she does not deserve a big raise. It appears that sales are steady but not increasing fast enough to warrant a big raise.
36. Sample answer: The second graph might be shown by the company owner to a prospective buyer of the company. It looks like there is a dramatic rise in sales.
37. A: 2.5, 2.5, 0.7, 0.8; B: 2.5, 2.5, 1.1, 1.0
38. The first histogram is lower in the middle and higher on the ends, so it represents data that are more spread out. Since set B has the greater standard deviation, set B corresponds to the first histogram and set A corresponds to the second.
39. The statistic(s) that best represent a set of test scores depends on the distribution of the particular set of scores. Answers should include the following. • mean, 73.9; median, 76.5; mode, 94 • The mode is not representative at all because it is the highest score. The median is more representative than the mean because it is influenced less than the mean by the two very low scores of 34 and 19.
40. A
41. D
42. 3
43. 1.9
44. The mean deviations would be greater for the greater standard deviation and lower for the groups of data that have the smaller standard deviation.
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45. inclusive;
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4 13
46. mutually exclusive;
47.
1 169
48.
4 663
49.
13 204
50.
1 16
51. 10, 92; 10, 21062;
52. 3
53. 17
54. 2
9 5
55. 12 cm3
56. 14, 62
57. (1, 5)
58. (3, 5)
59. 136
60. 340
61. 380
62. 475
63. 396
64. 495
3 7
Chapter 12 Practice Quiz 2 Page 670 1.
3 20
2.
1 6
3.
2 9
4.
1 4
5.
1 6
6.
2 3
7.
3 4
8. 6.6, 2.6
9. 23.6, 4.9
©Glencoe/McGraw-Hill
10. 134.0, 11.6
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Lesson 12-7 The Normal Distribution Pages 673–675 1. Sample answer:
2. The mean of the three graphs is the same, but the standard deviations are different. The first graph has the least standard deviation, the standard deviation of the middle graph is slightly greater, and the standard deviation of the last graph is greatest.
the use of cassettes since CDs were introduced 3. Since 99% of the data is within 3 standard deviations of the mean, 1% of the data is more than 3 standard deviations from the mean. By symmetry, half of this, or 0.5%, is more than 3 standard deviations above the mean.
4. normally distributed
5. 68%
6. 13.5%
7. 95%
8. 6800
9. 250
10. 1600
11. 81.5%
12. positively skewed
13. normally distributed
14. Negatively skewed; the histogram is high at the right and has a tail to the left.
15. 68%
16. 34%
17. 0.5%
18. 16%
19. 50%
20. 50%
21. 95%
22. 50%
23. 815
24. 25
25. 16%
26. 652
27. The mean would increase by 25; the standard deviation would not change; and the
28. If a large enough group of athletes is studied, some of the characteristics may be
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graph would be translated 25 units to the right.
normally distributed; others may have skewed distributions. Answers should include the following. • 10
Frequency
8 6 4 2 0 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Height (in.)
• Since the histogram has two peaks, the data may not be normally distributed. This may be due to players who play certain positions tending to be of similar large sizes while players who play the other positions tend to be of similar smaller sizes. 29. A
30. D
31. 17.5, 4.2
32. 42.5, 6.5
33.
2 13
34.
35.
4 13
36. 5, 0, 1 38. 1, 1
37. 3, 2, 4 39.
1 , 4
4 13
1
40.
y 50 2
1
1
O
2t
50 100
y 216t 2 53
about 45 min 41. 0.76 h
42. 21a 5b 2
43. 56c 5d 3
44. 126x 5y 4
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Lesson 12-8 Binomial Experiments Pages 678–680 1. Sample answer: In a 5-card hand, what is the probability that at least 2 cards are hearts?
2. RRRWW, RRWRW, RRWWR, RWRRW, RWRWR, RWWRR, WRRRW, WRRWR, WRWRR, WWRRR
3a. Each trial has more than two possible outcomes. 3b. The number of trials is not fixed. 3c. The trials are not independent.
4.
3 8
5.
1 8
6.
7 8
7.
1 28,561
8.
48 28,561
9.
27,648 28,561
10. about 0.05
11. about 0.37
12.
1 16
13.
1 16
14.
3 8
15.
1 4
16.
5 16
17.
11 16
18.
3125 7776
19.
125 3888
20.
625 648
21.
23 648
22.
243 1024
23.
1 1024
24.
15 1024
25.
135 512
26.
459 512
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27.
53 512
28.
105 512
29.
105 512
30.
319 512
31.
319 512
32. about 0.02
33. about 0.44
34.
560 2187
35. about 0.32
36.
1 3 15 5 15 3 1 , , , , , , 64 32 64 16 64 32 64
37.
1 4
38. C (n, m)p m(1 p)nm 40. 2
39. Getting a right answer and a wrong answer are the outcomes of a binomial experiment. The probability is far greater that guessing will result in a low grade than in a high grade. Answers should include the following. • Use (r w)5 r 5 5r 4w 10r 3w 2 10r 2w 3 5rw 4 w 5 and the chart on page 676 to determine the probabilities of each combination of right and wrong. 1 5 4
• P(5 right): r 5 P a b
1 1024
or about 0.098%; P (4 right, 1 wrong):
15 1024
or about
1.5%; P (3 right, 2 wrong): 1 3 3 2 4 4
10r 2w 3 10 a b a b
45 512
or about 8.8%; P (3 wrong, 2 right): 10r 2w 3 1 2 3 3 4 4
10 a b a b
135 512
or about
26.4%; P (4 wrong, 1 right):
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5r w 4 5 a b a b 1 4
405 1024
or
about 39.6%; P (5 wrong): 5
w5 a b 3 4
243 1024
or about
23.7%. 41. B
42. See students’ work.
43. normal distribution
44. 68%
45. 10
46. 16%
47. Mean; it is highest.
48.
y x 3 x
O
49.
y
50.
xy4
y y |5x |
x O
O
51. 0.1
52. 0.05
53. 0.039
54. 0.027
55. 0.041
56. 0.031
©Glencoe/McGraw-Hill
340
x
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Lesson 12-9 Sampling and Error Pages 683–685 1. Sample answer: If a sample is not random, the results of a survey may not be valid.
2. Sample answer for good sample: doing a random telephone poll to rate the mayor’s performance; sample answer for bad sample: conducting a survey on how much the average person reads at a bookstore
3. The margin of sampling error decreases when the size of the sample n increases. As n p (1 p) increases, decreases.
4. Yes; last digits of social security numbers are random.
5. No; these students probably study more than average.
6. about 9%
7. about 4%
8. about 4%
n
9. The probability is 0.95 that the percent of Americans ages 12 and older who listen to the radio every day is between 72% and 82%.
10. about 283
11. No; you would tend to point toward the middle of the page.
12. Yes; all seniors would have the same chance of being selected.
13. Yes; a wide variety of people would be called since almost everyone has a phone.
14. No; freshmen are more likely than older students to be still growing, so a sample of freshmen would not give representative heights for the whole school.
15. about 8%
16. about 4%
17. about 4%
18. about 3%
19. about 3%
20. about 2%
21. about 4%
22. about 2%
23. about 3%
24. about 2%
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25. about 2%
26. about 3%
27. about 983
28. 36 or 64
29. A political candidate can use the statistics from an opinion poll to analyze his or her standing and to help plan the rest of the campaign. Answers should include the following. • The candidate could decide to skip areas where he or she is way ahead or way behind, and concentrate on areas where the polls indicate the race is close. • about 3.5% • The margin of error indicates that with a probability of 0.95 the percent of the Florida population that favored Bush was between 43.5% and 50.5%. The margin of error for Gore was also about 3.5%, so with probability 0.95 the percent that favored Gore was between 40.5% and 47.5%. Therefore, it was possible that the percent of the Florida population that favored Bush was less than the percent that favored Gore.
30. A
31. C
32.
1 32
5 32
34.
1 2
33.
35. 95%
36. 210
37. 97.5%
38. x 2, x 3
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Chapter 13 Trigonometry Lesson 13-1 Right Triangle Trigonometry Pages 706–708 2.
1. Trigonometry is the study of the relationships between the angles and sides of a right triangle.
hypotenuse
adjacent
opposite
8 15 ; cos ; 17 17 8 17 tan ; csc ; 15 8 17 15 sec ; cot 15 8
4. sin
3. Given only the measures of the angles of a right triangle, you cannot find the measures of its sides.
5. sin
285 ; 11
cos
tan
285 ; 6
csc
sec
11 ; 6
7. cos 23
cot
32 ; x
6 ; 11
5 6
6. sin ; cos
11 285 ; 85
6 285 85
x 34.8
tan
5 211 ; 11
csc ;
sec
6 211 ; 11
cot
8. tan x
9. B 45, a 6, c 8.5
211 ; 6
15 ; 21
6 5
211 5
x 36
10. A 34, a 8.9, b 13.3
11. a 16.6, A 67, B 23
12. c 19.1, A 47, B 43
13. 1660 ft
14. B
15. sin
4 ; 11
cos
tan
4 2105 ; csc 105
sec
112105 ; 105
Glencoe/McGraw-Hill
2105 ; 11
3 5 3 tan ; csc 4 5 sec ; cot 4
11 ; 4
cot
4 5 5 ; 3 4 3
16. sin ; cos ;
2105 4
343
Algebra 2
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17. sin
27 ; 4
cos ;
tan
27 ; 3
csc
3 4
4 3
sec ; cot
19. sin tan
18. sin
4 27 ; 7
3 27 7
25 2 25 ; ; cos 5 5 1 ; csc 25; 2 25 ; 2
sec
cot 2
21. tan 30
x , 10
23. sin 54
17.8 , x
25. cos x
15 , 36
27a. sin 30
opp hyp x 2x
sin 30
cos 30
©Glencoe/McGraw-Hill
csc
2106 ; 9
sec
2106 ; 5
9 5
tan ;
cot
5 9
20. sin
215 ; 8
cos ;
tan
215 ; 7
csc
7 8
8 215 ; 15
7 215 15
3 x
22. cos 60 , x 6
x 22.0
24. tan 17.5
x 65
26. sin x
16 , 22
Replace opp with x and hyp with 2x.
x ; 23.7
sin 45 sin 45
1 12
12 2 adj hyp
sin 45 28b. cos 45
Simplify.
344
x 7.5
x 47
opp hyp x 12x
28a. sin 45
sine ratio
1 Simplify. 2 adj cosine ratio hyp 23x Replace adj with 13x and 2x hyp with 2x.
cos 30
5 2106 ; 106
8 7
x 5.8
23 2
cos
sec ; cot
sin 30 27b. cos 30
9 2106 ; 106
sine ratio Replace opp with x and hyp with 12x. Simplify. Rationalize the denominator. cosine ratio
cos 45
x 12x
Replace adj with x and hyp with 12x.
cos 45
1 12
Simplify.
cos 45
12 2
Rationalize the denominator.
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opp hyp
sine ratio
sin 60
23x 2x
Replace opp with 13x and hyp with 2x.
sin 60
23 2
Simplify.
27c. sin 60
28c. tan 45 tan 45
opp adj
tangent ratio
x x
Replace opp with x and adj with x.
tan 45 1
Simplify.
29. B 74, a 3.9, b 13.5
30. A 63, a 13.7, c 15.4
31. B 56, b 14.8, c 17.9
32. A 75, a 24.1, b 6.5
33. A 60, a 19.1, c 22
34. B 45, a 7, b 7
35. A 72, b 1.3, c 4.1
36. B 80, a 2.6, c 15.2
37. A 63, B 27, a 11.5
38. A 26, B 64, b 8.1
39. A 49, B 41, a 8, c 10.6
40. A 19, B 71, b 14.1, c 15
41. about 300 ft
42. about 142.8 ft
43. about 6
44. about 3.2 in.
45. 93.54 units2
46. about 1.72 km high
47. The sine and cosine ratios of acute angles of right triangles each have the longest measure of the triangle, the hypotenuse, as their denominator. A fraction whose denominator is greater than its numerator is less than 1. The tangent ratio of an acute angle of a right triangle does not involve the measure of the hypotenuse,
48. When construction involves right triangles, including building ramps, designing buildings, or surveying land before building, trigonometry is likely to be used. Answers should include the following. • If you view the ramp from the side then the vertical rise is opposite the angle that the ramp makes with the horizontal. Similarly, the horizontal run is the adjacent side. So the tangent of the angle is the ratio of the rise to the run or the slope of the ramp. • Given the ratio of the slope of ramp, you can find the angle of inclination by calculating tan–1 of this ratio.
opp . adj
If the measure of the opposite side is greater than the measure of the adjacent side, the tangent ratio is greater than 1. If the measure of the opposite side is less than the measure of the adjacent side, the tangent ratio is less than 1. ©Glencoe/McGraw-Hill
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49. C
50. 7.7
51. No; band members may be more likely to like the same kinds of music. 3 53. 8 15 55. 16 57. {2, 1, 0, 1, 2}
52. Yes; this sample is random since different kinds of people go to the post office. 1 54. 16
59. 20 qt
60. 35,904 ft
61. 12 m2
62. 48 L
Lesson 13-2
56. {222, 2i 22} 58. {121}
Angles and Angle Measure Pages 712–715
1. reals
2. In a circle of radius r units, one radian is the measure of an angle whose rays intercept an arc length of r units.
3.
4.
y
y
290˚
70˚ x
O
x
O
⫺70˚
5.
6.
y
y
300˚
570˚ O
©Glencoe/McGraw-Hill
x
O
346
x
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7.
y
O
⫺45˚
8.
13 18
10.
97 36
x
18
9.
11. 135
12. 30
13. 1140
14. 420, 300
15. 785, 295
16.
17. 21 h
18. 2 h
19.
7 , 3
5 3
20.
y 235˚
y 270˚
x
O
21.
22.
y
380˚
24.
y
x
O
y
O
⫺150˚
©Glencoe/McGraw-Hill
x
O
790˚
23.
y
x
O
x
O
347
⫺50˚
x
Algebra 2
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y
25.
26.
y
O
x
x
O 2 ⫺ 3
27.
2 3
28.
12
3 5 4
29.
30.
31.
11 3
32.
19 6
33.
79 90
34.
13 9
35. 150
36. 495
37. 45
38. 60
39. 1305
40. 510
41.
1620
515.7
42.
540
171.9
43. Sample answer: 585, 135
44. Sample answer: 390, 330
45. Sample answer: 345, 375
46. Sample answer: 220, 500
47. Sample answer: 8, 352
48. Sample answer: 400, 320
49. Sample answer:
11 , 4
51. Sample answer:
3 , 4
53. Sample answer:
13 , 2
5 4
50. Sample answer:
19 , 6
13 4
52. Sample answer:
4 , 3
3 2
54. Sample answer:
25 , 4
5 6
8 3
7 4
55. 2689 per second; 47 radians per second
56. 209.4 in2
57. about 188.5 m2
58. number 17
59. about 640.88 in2
60a. a 2 (b)2 a 2 b 2 1 60b. b 2 a 2 a 2 b 2 1 60c. b 2 (a)2 a 2 b 2 1
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61. Student answers should include the following. • An angle with a measure of more than 180 gives an indication of motion in a circular path that ended at a point more than halfway around the circle from where it started. • Negative angles convey the same meaning as positive angles, but in an opposite direction. The standard convention is that negative angles represent rotations in a clockwise direction. • Rates over 360 per minute indicate that an object is rotating or revolving more than one revolution per minute.
62. C
63. D
64. a 3.4, c 6.0, B 56
65. A 22, a 5.9, c 15.9
66. A 35, a 9.2, b 13.1
67. c 0.8, A 30, B 60
68. about 8.98%
69. about 7.07%
70. permutation, 17,100,720
71. combination, 35
72. [g h](x) 6x 8, [h g](x) 6x 4
73. [g h](x) 4x 2 6x 23, [h g](x) 8x 2 34x 44
74. 1041.8
75. 1418.2 or about 1418; the number of sports radio stations in 2008
76.
2 23 3
77.
3 25 5
78.
2 26 3
79.
210 2
80.
214 2
81.
210 4
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Chapter 13 Practice Quiz 1 Page 715 1. A 42, a 13.3, c 17.9
2. A 59, B 31, b 10.8
3.
4. sin
y
O
5.
⫺60˚
x
19 18
6.
10 2149 ; 149
cos
7 2149 ; 149
tan
10 ; 7
sec
2149 ; 7
csc
2149 ; 10
cot
7 10
5 2
8. 396
7. 210 9. 305; 415
10.
5 ; 3
3
Lesson 13-3 Trigonometric Functions of General Angles Pages 722–724 1. False; sec 0 tan 0
0 r
r r
or 1 and
2. Sample answer: 190
or 0. 8 15 , cos , 17 17 8 17 tan , csc , 15 8 17 15 sec , cot 15 8
4. sin
3. To find the value of a trigonometric function of , where is greater than 90, find the value of the trigonometric function for , then use the quadrant in which the terminal side of lies to determine the sign of the trigonometric function value of .
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5. sin 0, cos 1, tan 0, csc undefined, sec 1, cot undefined
6. sin
7. 55
8.
22 , 2
cos
22 , 2
tan 1, csc 22, sec 22, cot 1
y
4
y 7 4
235˚ O
O
x
'
'
9. 60
10.
y
x
23 2
' O
x
⫺240˚
11. 1
12. 23
2 23 3
13.
14. sin csc
23 , 2
2 23 , 3
cot 15. sin
26 , 3
cos
23 , 3
tan 22, cos
tan 23, sec 2,
23 3
16. about 12.4 ft
26 , 2
sec 23 24 , 25 24 tan , 7 25 sec , 7
17. sin
©Glencoe/McGraw-Hill
7 , 25 25 csc , 24 7 cot 24
cos
25 , cos 5 1 , csc 2
18. sin tan
sin
351
25 , 5
2 25 , 5
25,
cos 2,
Algebra 2
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19. cos
8 289 , 89
cos
5 289 , 89 8 5
tan , csc sec
289 , 5
3 4 5 5 3 5 tan , csc , 4 3 5 4 sec , cot 4 3
20. sin , cos ,
289 , 8 5 8
cot
22. sin 0, cos 1, tan 0, csc undefined, sec 1, cot undefined
21. sin 1, cos 0, tan undefined, csc 1, sec undefined, cot 0 23. sin
22 , 2
cos
22 , 2
tan 1, csc 22, sec 22, cot 1
25. 45
26 , 3
24. sin
23 , 3
tan 22, csc
26 , 2
sec 23, cot
22 2
26. 60
y
cos
y 240˚
315˚ O
27. 30
'
x
x
O '
y
28. 55
y
' O
x
'
⫺210˚
©Glencoe/McGraw-Hill
352
x
O ⫺125˚
Algebra 2
Chapter 13
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29.
4
y
30.
6
5 6
5 4
'
x
O
'
31.
y
7
x
O
y
32.
3
y
13 7
O
33.
'
x
23 2
36. 23
37. undefined
38.
1 2
39. 23
40.
22 2
41. undefined
42. 2
23 2
44. 1
45. 0.2, 0, 0.2, 0, 0.2, 0, and 0.2; or about 11.5, 0, 11.5, 0, 11.5, 0, and 11.5 4 5 5 csc , 4 3 cot 4
46. 6092.5 ft
4 3 5 , 3
47. sin , tan ,
©Glencoe/McGraw-Hill
2 ⫺ 3
34. 2
35. 23
43.
x
O
'
sec
48. sin
226 , 26
5 226 , 26
cos
csc 226, sec
226 , 5
cot 5
353
Algebra 2
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2 22 , 3
49. cos
tan
22 , 4
2 25 , 5
50. sin
csc 3, sec
cos
25 , 5
tan 2,
cot 222
csc
25 , 2
sec 25
3 22 , 4
3 210 , 10
1 5
51. sin
52. sin , cos
cos
210 , 10
tan 3,
csc
210 , 3
cot
tan
26 , 12
2 26 , 5
sec
5 26 , 12
cot 226
1 3
53. about 173.2 ft
54. 45; 2 45 or 90 yields the greatest value for sin 2.
55. 9 meters
56. I, II
57. II
58. III
59. Answers should include the following. • The cosine of any angle is x defined as , where x is r the x-coordinate of any point on the terminal ray of the angle and r is the distance from the origin to that point. This means that for angles with terminal sides to the left of the y-axis, the cosine is negative, and those with terminal sides to the right of the y-axis, the cosine is positive. Therefore, the cosine function can be used to model real-world data that oscillate between being positive and negative.
60. C
©Glencoe/McGraw-Hill
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• If we knew the length of the cable we could find the vertical distance from the top of the tower to the rider. Then if we knew the height of the tower we could subtract from it the vertical distance calculated previously. This will leave the height of the rider from the ground. 61. a ,
62.
2
63. 300
64.
900
5 2
5 23 b 2
65. sin 28 67. sin x
x , 12
5 , 13
286.5
66. cos 43
5.6
23
x , 83
60.7
68. 635
69. (7, 2)
70. (4, 3)
71. (5, 4)
72. 4.7
73. 15.1
74. 2.7
75. 32.9
76. 20.6
77. 39.6
Lesson 13-4 Law of Sines Pages 729–732 2. Sample answer: A 42, a 2.6 cm, b 3.2 cm
1. Sometimes; only when A is acute, a b sin A, or a b and when A is obtuse, a b.
C 3.2 cm
A
2.6 cm
3.9 cm
B C
3.2 cm 2.6 cm
A 0.9 cm ©Glencoe/McGraw-Hill
355
B Algebra 2
Chapter 13
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3. Gabe; the information given is of two sides and an angle, but the angle is not between the two sides, therefore the area formula involving sine cannot be used.
4. 57.5 in2
5. 6.4 cm2
6. C 30, a 2.9, c 1.5
7. B 80, a 32.0, b 32.6
8. B 20, A 20, a 20.2 10. two; B 42, C 108, c 5.7; B 138, C 12, c 1.2
9. no solution
11. one; B 24, C 101, c 12.0
12. one; B 19, C 16, c 8.7
13. 5.5 m
14. 43.1 m2
15. 19.5 yd2
16. 572.8 ft2
17. 62.4 cm2
18. 4.2 m2
19. 14.6 mi2
20. B 101, c 3.0, b 3.4
21. C 73, a 55.6, b 48.2
22. B 21, C 37, b 13.1
23. B 47, C 68, c 5.1
24. C 97, a 5.5, b 14.4
25. A 40, B 65, b 2.8
26. C 67, B 63, b 2.9
27. A 20, a 22.1, c 39.8
28. no
29. one; B 36, C 45, c 1.8
30. two; B 72, C 75, c 3.5; B 108, C 39, c 2.3
31. no
32. one; B 90, C 60, c 24.2
33. one; B 18, C 101, c 25.8
34. two; B 56, C 72, c 229.3; B 124, C 4, c 16.8
35. two; B 85, C 15, c 2.4; B 95, C 5, c 0.8
36. one; B 23, C 129, c 14.1
37. two; B 65, C 68, c 84.9; B 115, C 18, c 28.3
38. 4.6 and 8.5 mi
©Glencoe/McGraw-Hill
356
Algebra 2
Chapter 13
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39. 7.5 mi from Ranger B, 10.9 mi from Ranger A
40. 690 ft
41. 107 mph
42a. 14.63 b 20 42b. b 14.63 or b 20 42c. b 14.63
43. Answers should include the following. • If the height of the triangle is not given, but the measure of two sides and their included angle are given, then the formula for the area of a triangle using the sine function should be used. • You might use this formula to find the area of a triangular piece of land, since it might be easier to measure two sides and use surveying equipment to measure the included angle than to measure the perpendicular distance from one vertex to its opposite side. 1 • The area of ABC is ah.
44. D
2
C b
a h
A
sin B Area Area
©Glencoe/McGraw-Hill
B
c h or h c sin c 1 ah or 2 1 a (c sin B) 2
B
357
Algebra 2
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45. B 78, a 50.1, c 56.1 47.
46.
23 3
48. 22
49. 660, 60 51.
17 , 6
53.
55 221
23 2
50. 407, 313
7 6
52.
3 68
54. 780 ft
55. 5.6
56. 7.8
57. 39.4
58. 136.0
Lesson 13-5 Law of Cosines Pages 735–738 1. Mateo; the angle given is not between the two sides, therefore the Law of Sines should be used.
©Glencoe/McGraw-Hill
2a. Use the Law of Cosines to find the measure of one angle. Then use the Law of Sines or the Law of Cosines to find the measure of a second angle. Finally, subtract the sum of these two angles from 180 to find the measure of the third angle. 2b. Use the Law of Cosines to find the measure of the third side. Then use the Law of Sines or the Law of Cosines to find the measure of a second angle. Finally, subtract the sum of these two angles from 180 to find the measure of the third angle.
358
Algebra 2
Chapter 13
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4. cosines; A 76, B 69, c 6.5
3. Sample answer:
15
9
13
5. sines; B 70, a 9.6, b 14
6. sines; C 101, B 37, c 92.5
7. cosines; A 23, B 67, C 90
8. 19.5 m 10. sines; A 60, b 14.3, c 11.2
9. 94.3 11. cosines; A 48, B 62, C 70
12. cosines; A 46, B 74, C 59.6
13. sines; B 102, C 44, b 21.0
14. cosines; A 56.8, B 82, c 11.5
15. sines; A 80, a 10.9, c 5.4
16. cosines; A 55, C 78, b 17.9
17. cosines; A 30, B 110, C 40
18. no
19. sines; C 77, b 31.7, c 31.6
20. cosines; A 103, B 49, C 28
21. no
22. cosines; A 15, B 131, C 34
23. cosines; A 52, C 109, b 21.0
24. sines; C 102, b 5.5, c 14.4
25. cosines; A 24, B 125, C 31
26. cosines; A 107, B 35, c 13.8
27. cosines; B 82, C 58, a 4.5
28. about 159.7
29. about 100.1
30. Since the step angle for the carnivore is closer to 180, it appears as though the carnivore made more forward progress with each step than the herbivore did.
©Glencoe/McGraw-Hill
359
Algebra 2
Chapter 13
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31. 4.4 cm, 9.0 cm
32. about 1362 ft; about 81,919 ft 2
33. 91.6
34. Since cos 90 0, a 2 b 2 c 2 2bc cos A becomes a 2 b 2 c 2.
35. Answers should include the following. • The Law of Cosines can be used when you know all three sides of a triangle or when you know two sides and the included angle. It can even be used with two sides and the nonincluded angle. This set of conditions leaves a quadratic equation to be solved. It may have one, two, or no solution just like the SSA case with the Law of Sines. • Given the latitude of a point on the surface of Earth, you can use the radius of the Earth and the orbiting height of a satellite in geosynchronous orbit to create a triangle. This triangle will have two known sides and the measure of the included angle. Find the third side using the Law of Cosines and then use the Law of Sines to determine the angles of the triangle. Subtract 90 degrees from the angle with its vertex on Earth’s surface to find the angle at which to aim the receiver dish.
36. B
37. A
38. 100.0
©Glencoe/McGraw-Hill
360
Algebra 2
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39. Sample answer: 100.2
40. By finding the measure of angle C in one step using the Law of Cosines, only the given information was used. By finding this angle measure using the Law of Cosines and then the Law of Sines, a calculated value that was not exact was introduced; 100.0.
41. one; B 46, C 79, c 9.6
42. no solution
12 5 , cos , 13 13 12 13 tan , csc 5 12 13 5 sec , cot 5 12
44. sin
43. sin
cos csc cot
45. sin
26 , 4
cos
tan
215 , 5
sec
2 210 , 5
210 , 4
csc
4 265 , 65 265 , 7 4 7
48. 4.3891
51. 540, 180
52.
5 , 2
54.
10 , 3
5 6
©Glencoe/McGraw-Hill
265 , 4
215 3
50. 390, 330
19 , 6
sec
46. 1.3863
49. 405, 315
53.
7 4
tan ,
2 26 , 3
cot
47. {x 0 x 0.6931}
7 265 , 65
361
3 2
2 3
Algebra 2
Chapter 13
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Chapter 13 Practice Quiz 2 Page 738 1. sin
3 213 , 13
2 23 3
2.
2 213 , 13
cos
3 2
tan , csc sec
213 , 2
213 , 3 2 3
cot
4. two; B 27; C 131; c 30.2; B 153; C 5; c 3.5
3. 27.7 m2
5. cosines; c 15.9, A 59, B 43
Lesson 13-6 Circular Functions Pages 742–745 1. The terminal side of the angle in standard position must intersect the unit circle at P (x, y).
2. Sample answer: the motion of the minute hand on a clock; 60 s
3. Sample answer: The graphs have the same shape, but cross the x-axis at different points.
4. sin , cos
5. sin
22 ; 2
cos
12 13
22 2
6.
1 2
7.
©Glencoe/McGraw-Hill
5 13
23 2
8. 720
362
Algebra 2
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10.
9. 2 s
h 3
O
2
1
3
4
t
⫺3
4 5
5 13
3 5
13. sin
15 ; 17
15. sin
23 ; 2
cos
8 17
1 2
14. sin ; cos 1 2
cos
1 2
18.
23 2
19. 1
20.
23 2
21. 1
22.
22 2
9 4
23.
1 4
24.
25.
1 23 2
26. 23
27. 323
28. 1
29. 6
30. 9
31. 2
32. 8
1 440
13 2
16. sin 0.8; cos 0.6
17.
33.
12 13
12. sin ; cos
11. sin ; cos
s
34. y 1
O ⫺1
©Glencoe/McGraw-Hill
363
1 440
1 220
Algebra 2
x
Chapter 13
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1 13 1 13 b, a , b, 2 2 2 2
35. a ,
(1, 0), a , 1 2
1 a , 2
37.
36. The population is around 425 near the 60th day of the year. It rises to around 625 in May/June. It falls to around 425 again by August/September. It continues to drop to around 225 in November/December.
13 b, 2
13 b 2
y x
38. tan x y
39.
40. cot
41. 23
42.
43. sine: D {all reals}, R {1 y 1}; cosine: D {all reals}, R {1 y 1}
44. Answers should include the following. • Over the course of one period both the sine and cosine function attain their maximum value once and their minimum value once. From the maximum to the minimum the functions decrease slowly at first, then decrease more quickly and return to a slow rate of change as they come into the minimum. Similarly, the functions rise slowly from their minimum. They begin to increase more rapidly as they pass the halfway point, and then begin to rise more slowly as they increase into the maximum. Annual temperature fluctuations behave in exactly the same manner.
©Glencoe/McGraw-Hill
364
23 3
Algebra 2
Chapter 13
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• The maximum value of the sine function is 1 so the maximum temperature would be 50 25(1) or 75 F. Similarly, the minimum value would be 50 25(1) or 25 F. The average temperature over this time period occurs when the sine function takes on a value of 0. In this case that would be 50 F. 23 3
45. A
46.
47. cosines; c 12.4, B 59, A 76
48. cosines; A 34, B 62, C 84
49. 27.0 in2
50. 12.5 m2
51. 6800
52. 9500
53. 5000
54. 5000
55. 250
56. 50
57. does not exist
58.
59. 8
60. 4x 5
61. 2x 9
62. 5y 2 4y 4
63. 2y 7
5 y3
64 3
11 y1
64. 20
65. 110
66. 73
67. 80
68. 56
69. 89
©Glencoe/McGraw-Hill
365
Algebra 2
Chapter 13
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Lesson 13-7
Inverse Trigonometric Functions Pages 749–751
1. Restricted domains are denoted with a capital letter.
2. Sample answer: Cos 45 Cos1
22 2
22 ; 2
45
3. They are inverses of each other.
4. Arctan x
5. Arccos 0.5
6. 45
7. 0
8. 0.52
6
9. 3.14
10. 0.22
11. 0.75
12. 0.66
13. 0.58
14. 30
15. Arcsin
16. a Arctan b
17. y Arccos x
18. 30 Arcsin
19. Arccos y 45
20. Arctan a b x
21. 60
22. 30
23. 45
24. 30
25. 45
26. 90
27. 2.09
28. does not exist
29. 0.52
30. 0.52
31. 0.5
32. 0.66
33. 0.60
34. 0.5
35. 0.8
36. 0.81
37. 0.5
38. 3
39. 0.5
40. 1.57
41. 0.71
42. does not exist
43. 0.96
44. 0.87
©Glencoe/McGraw-Hill
1 2
4 3
366
Algebra 2
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45. 60 south of west
46. 83
47. No; with this point on the terminal side of the throwing angle , the measure of is found by solving the equation
48. 60
tan
17 . Thus 18
tan1
17 18
or about 43.4, which is greater than the 40 requirement. 49. 31
50. 102
51. Suppose P (x 1, y 1) and Q (x 2, y 2) lie on the line y mx b. Then m y2 y1 . The tangent of
52. Trigonometry is used to determine proper banking angles. Answers should include the following. • Knowing the velocity of the cars to be traveling on a road and the radius of the curve to be built, then the banking angle can be determined. First find the ratio of the square of the velocity to the product of the acceleration due to gravity and the radius of the curve. Then determine the angle that had this ratio as its tangent. This will be the banking angle for the turn. • If the speed limit were increased and the banking angle remained the same, then in order to maintain a safe road the curvature would have to be decreased. That is, the radius of the curve would also have to increase, which would make the road less curved.
x2 x1
the angle the line makes with the positive x-axis is opp equal to the ratio or adj
y2 y1 . x2 x 1
Thus tan m.
y
Q (x 2, y 2) P (x , y 1)
x2 ⫺ x1
O
y2 ⫺ y1 x
y ⫽ mx m ⫹b
53. 37 ©Glencoe/McGraw-Hill
54. D 367
Algebra 2
Chapter 13
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56. Sin1 x Cos1 x for 2 all values of x.
55. 23 1 22 23 1 22 1 1 2 2 2 2 2 2 y 2 2 2 2 2 2 2 2 2 x 0
23 2
57. From a right triangle perspective, if an acute angle has a given sine x, then the complementary angle has that same value 2 as its cosine. This can be verified by looking at a right triangle. Therefore, the sum of the angle whose sine is x and the angle whose cosine is x should be .
58.
59. 1
60. 1
61. sines; B 69, C 81, c 6.1 or B 111, C 39, C 3.9
62. cosines; A 13, B 77, C 90
63. 46, 39
64. 22, 57
65. 11, 109
66. 2.5 s
2
©Glencoe/McGraw-Hill
368
Algebra 2
Chapter 13
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2:03 PM
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Chapter 14 Trigonetmetric Graphs and Identities Lesson 14-1 Graphing Trigonometric Functions Pages 766–768 1. Sample answer: Amplitude is half the difference between the maximum and minimum values of a graph; y tan has no maximum or minimum value.
2. Sample answer: The graph repeats itself every 180.
3. Jamile; The amplitude is 3 and the period is 3.
4. amplitude: ; period 360 or 2
1 2
y 2.5 2 1.5 1 0.5 270˚ 180˚ 90˚ 1 1.5 2 2.5
90˚ 180˚ 270˚
6. amplitude: ; period 360 or 2
y
y
5 4 3 2 1
©Glencoe/McGraw-Hill
O
2 3
5. amplitude: 2; period: 360 or 2
270˚ 180˚ 90˚ 2 3 4 5
1
y 2 sin
2 1.5 1 0.5
y 2 sin
O
90˚ 180˚ 270˚
0.5 1 1.5 2
369
y
O
2 cos 3
90˚ 180˚ 270˚ 360˚
Algebra 2
Chapter 14
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2:03 PM
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8. amplitude: does not exist; period: 180 or
7. amplitude: does not exist; period: 180 or
y
y
2 1.5 1 0.5
2 1.5 1 0.5 270˚ 180˚ 90˚ 1 1 1.5 y 4 tan 2
O
90˚ 180˚ 270˚
O 270˚ 180˚ 90˚ 1 1.5 y csc 2 2
9. amplitude: 4; period: 180 or
10. amplitude: 4; period: 480 or
y 5 4 3 2 1
5 4 3 2 1
90˚ 180˚ 270˚ 360˚
y 1.25 1 0.75 0.5 0.25
30˚
60˚
O 90˚ 0.5 0.75 1 1.25
90˚ 120˚ 150˚ 1
y 2 sec 3
13. 12 months; Sample answer: The pattern in the population will repeat itself every 12 months.
©Glencoe/McGraw-Hill
3 4
2 3
2 1.5 1 0.5 60˚ 30˚ 1 1.5 2
90˚ 180˚ 270˚ 360˚ 450˚
12. amplitude: ; period: 720 or 4
y
O
3 4
y 4 cos
O 1 2 3 4 5
11. amplitude: does not exist; period: 120 or
8 2
y
y 4 sin 2
O 1 2 3 4 5
90˚ 180˚ 270˚
3
1
y 4 cos 2
90˚ 180˚ 270˚ 360˚ 450˚
14. 4250; June 1
370
Algebra 2
Chapter 14
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15. amplitude: 3; period: 360 or 2
16. amplitude: 5; period: 360 or 2
y 5 4 3 2 1 270˚ 180˚ 90˚ 2 3 4 5
y 5 4 3 2 1
y 3 sin
O
90˚ 180˚ 270˚
O 270˚ 180˚ 90˚ 2 3 4 5
y
y 10 8 6 4 2
5 4 3 2 1 O 270˚180˚90˚ 2 y 2 csc 3 4 5
90˚ 180˚ 270˚
270˚ 180˚ 90˚ 4 6 8 y 2 tan 10
1 5
y 1 0.8 0.6 0.4 0.2
90˚ 180˚ 270˚
y 10 8 6 4 2
1 y 5 sin
O
O
20. amplitude: does not exist; period: 360 or 2
19. amplitude: ; period: 360 or 2
©Glencoe/McGraw-Hill
90˚ 180˚ 270˚
18. amplitude: does not exist; period: 180 or
17. amplitude: does not exist; period: 360 or 2
270˚ 180˚ 90˚ 0.4 0.6 0.8 1
y 5 cos
90˚ 180˚ 270˚
O 270˚180˚90˚ 4 6 1 y 3 sec 8 10
371
90˚ 180˚ 270˚
Algebra 2
Chapter 14
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21. amplitude: 1; period 90 or
2
22. amplitude: 1; period: 180 or y
y 5 4 3 2 1 270˚ 180˚ 90˚ 2 3 4 5
5 4 3 2 1
y sin 4
O
90˚ 180˚ 270˚
period: 36 or
O 60˚ 30˚ 2 3 4 5
5
y
30˚
60˚
O 72˚ 54˚ 36˚ 18˚ 2 3 4 y cot 5 5
25. amplitude: does not exist; period: 540 or 3
18˚
36˚
54˚
72˚
26. amplitude: does not exist; period: 360 or 2 y
y
10 8 6 4 2
10 8 6 4 2
©Glencoe/McGraw-Hill
5 4 3 2 1
5 4 3 2 1
810˚ 540˚270˚ 4 6 1 y 4 tan 3 8 10
90˚ 180˚ 270˚
24. amplitude: does not exist;
2 3
y
y sec 3
O
270˚ 180˚ 90˚ 2 3 4 5
23. amplitude: does not exist; period: 120 or
y sin 2
O
270˚ 540˚ 810˚
O 540˚ 360˚ 180˚ 4 6 1 y 2 cot 2 8 10
372
180˚ 360˚ 540˚
Algebra 2
Chapter 14
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27. amplitude: 6; period: 540 or 3
28. amplitude: 3; period: 720 or 4
y 10 8 6 4 2 270˚ 180˚ 90˚ 4 6 8 10
y 2 y 6 sin 3
O
90˚ 180˚ 270˚
10 8 6 4 2
O 540˚360˚180˚ 4 6 8 10
29. amplitude: does not exist; period: 720 or 4
2
y
10 8 6 4 2 O 540˚360˚180˚ 4 1 y 3 csc 2 6 8 10
180˚ 360˚ 540˚
10 8 6 4 2
135˚ 90˚ 45˚ 4 6 1 y 2 cot 2 8 10
O
45˚
90˚ 135˚
8 9
31. amplitude: does not exist; period: 180 or
32. amplitude: ; period: 600 or 10 3
y 10 8 6 4 2
©Glencoe/McGraw-Hill
180˚ 360˚ 540˚
30. amplitude: does not exist; period: 90 or
y
O 270˚180˚ 90˚ 4 6 2y tan 8 10
1
y 3 cos 2
y
90˚ 180˚ 270˚
5 4 3 2 1
540˚360˚180˚ 2 3 4 5
373
3 2 3 y 3 sin 5 4
O
180˚ 360˚ 540˚
Algebra 2
Chapter 14
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33.
34. y
y 5 4 3 2 1 135˚ 90˚ 45˚ 2 3 4 5
y 35.
3 5
5 4 3 2 1
3
y 5 sin 4
O
45˚
90˚ 135˚
7
y 8 cos 5
O
135˚ 90˚ 45˚ 2 3 4 5
y
sin 4
1 107
7 8
45˚
90˚ 135˚
cos 5
36. y 0.25 sin 128t, y 0.25 sin 512t, y 0.25 sin 1024t 38. f (x ) cos x and f (x ) sec x
37. Sample answer: The amplitudes are the same. As the frequency increases, the period decreases.
f (x ) 5 4 3 2 1
f (x ) cos x f (x ) cos (x )
O 270˚ 180˚ 90˚ 2 3 4 5
x
90˚ 180˚ 270˚
f (x ) f (x ) sec x f (x ) sec (x )
5 4 3 2 1
O 270˚180˚90˚ 2 3 4 5
©Glencoe/McGraw-Hill
374
90˚ 180˚ 270˚
Algebra 2
x
Chapter 14
PQ245-6457F-P14[369-410].qxd
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39. y 2 sin
5
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t
40. y
2.5 2 1.5 1 0.5 0.5 1 1.5 2 2.5
y 2 sin 5 t
O
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
41. about 1.9 ft
42. Sample answer: Tides display periodic behavior. This means that their pattern repeats at regular intervals. Answers should include the following information. • Tides rise and fall in a periodic manner, similar to the sine function. • In f (x ) a sin bx, the amplitude is the absolute value of a.
43. A
44. C
45. 90
46. 90
47. 45
48.
1 2
49.
22 2
50.
22 2
51.
13 16
52. 3, 11, 27, 59, 123
©Glencoe/McGraw-Hill
375
Algebra 2
t
Chapter 14
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53.
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54.
y 15 13 11 9 7 5 3 1 8
4
y x
15 13 11 9 7 5 3 1
2
y 3x 2
O
4
8
x
8
3 5
55.
8
4
15 13 11 9 7 5 3 1 O
y 2x 2
4
8
x
8
O
4
3 5
x
8
y 3x 2 4
4
15 13 11 9 7 5 3 1 O
y (x 3)2 2 4
x
8
3 5
Translations of Trigonometric Graphs Pages 774–776
1. vertical shift: 15; amplitude: 3; period: 180; phase shift: 45
©Glencoe/McGraw-Hill
y 3x 2
y
y x2 2
3 5
Lesson 14-2
4
56.
y
y 2(x 1)2
y
2. The midline of a trigonometric function is the line about which the graph of the function oscillates after a vertical shift.
376
Algebra 2
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3. Sample answer: y sin ( 45)
4. 1; 2;
2 y
(
y sin 2
5. no amplitude; 180; 60
3 2
)
2
5 4 3 2 1 O
2
2 3 4 5
3 2
6. 1; 360; 45 y
y 5 4 3 2 1
0.75
y cos ( 45˚)
0.5
O 270˚180˚ 90˚ 2 3 y tan ( 60˚) 4 5
90˚ 180˚ 270˚
0.25
O 45˚ 0.25
45˚ 90˚ 135˚ 180˚ 225˚
0.5 0.75
3
7. no amplitude; 2; (
y sec 3
3 2
)
1 ; 4
1 4
y ; 1; 360 y
y 4 3 2 1
O 1 2 2 3 4
©Glencoe/McGraw-Hill
8.
2
3 2
5 4 3 2 1
O 270˚ 180˚ 90˚ 2 3 4 5
377
1
y cos 4
90˚ 180˚ 270˚
Algebra 2
Chapter 14
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9. 5; y 5; no amplitude; 360
10. 4; y 4; no amplitude; 180 y 7
y
6
10 8 6 4 2
5 4
O 270˚180˚90˚ 4 6 8 y sec 5 10
1 135˚ 90˚ 45˚
y sin 0.25
0.5
90˚
180˚
270˚
45˚
y 3 sin [2( 30˚)] 10 14 12 10 8 6 4 2
1
0.5
O 1
90˚
135˚
12. 10; 3; 180; 30
y
O
2
90˚ 180˚ 270˚
11. 0.25; y 0.25; 1; 360 1.5
3
y tan 4
360˚
y
O 270˚ 180˚ 90˚ 4
1
90˚ 180˚ 270˚
1.5
2 4
13. 6; no amplitude; 60; 45
14. 1; no amplitude; ;
y
y
1 45˚
©Glencoe/McGraw-Hill
O 2 3 4 5 6 7 8 9 10 11
45˚
4 3 2 1
3 8
4
1O 8
2 3 4 1 y 2 sec 4 4 1
y 2 cot (3 135˚) 6
[(
378
8
4
3 8
)]
Algebra 2
Chapter 14
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6
2 3
15. 2; ; 4;
16. 4; 1; 4 s
y 1 O 3 2 1
2
3
2 3
2
[ 1(
y 3 cos 2 6
)] 2
17. h 4 cos t or 2 h 4 cos 90t
18.
h
h 4 cos 2 t
6 5 4 3 2 1 O 1 2 3 4
19. 1; 360; 90
1
2
O 270˚ 180˚ 90˚ 2 3 4 5
©Glencoe/McGraw-Hill
4
t
20. no amplitude; 180; 30 y
y 5 4 3 2 1
3
5 4 3 2 1
y cos ( 90˚)
90˚ 180˚ 270˚
O 135˚ 90˚ 45˚ 2 3 4 5
379
45˚
90˚ 135˚
y cot ( 30˚)
Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
21. 1; 2;
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3
4
22. 1; 2;
y
y 5 4 3 2 1
3 2
(
y sin 4
5 4 3 2 1
)
O 2 2 3 4 5
2
3 2
23. no amplitude; 180; 22.5
3 2
O 2 2 3 4 5
5 4 3 2 1 45˚
90˚ 135˚
3 2
y 3 sin ( 75˚)
O 270˚ 180˚ 90˚ 2 3 4 5
1 y 4 tan ( 22.5˚)
90˚ 180˚ 270˚
26. 2; y 2; no amplitude; 360
25. 1; y 1; 1; 360 y
y
5 4 3 2 1
©Glencoe/McGraw-Hill
2
)
y
5 4 3 2 1
O 270˚ 180˚ 90˚ 2 3 y sin 1 4 5
24. 3; 360; 75
y
O 135˚ 90˚ 45˚ 2 3 4 5
(
y cos 3
5 4 3 2 1
O 270˚180˚90˚ 2 3 y sec 2 4 5
90˚ 180˚ 270˚
380
90˚ 180˚ 270˚
Algebra 2
Chapter 14
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3 4
y
360
2 1
y
O 270˚ 180˚ 90˚ 2 3 4 5 6 7 8
29.
1 ; 2
3 4
28. ; y ; no amplitude;
27. 5; y 5; 1; 360
90˚ 180˚ 270˚
5 4 3 2 1
O 270˚180˚90˚ 2 3 3 y csc 4 4 5
y cos 5
1 1 2 2
30. 1.5; y 1.5; 6; 360
y ; ; 360
y y 5 4 3 2 1
1
1
90˚ 180˚ 270˚
y 6 cos 1.5
10 8 6 4 2
y 2 sin 2
O 270˚ 180˚ 90˚ 2 3 4 5
O 270˚ 180˚ 90˚ 4 6 8 10
32.
31. 18 16 14 12 10 8 6 4 2 3 2 4
4
O
(
y 5 tan 4
4
2
3 4
O 270˚ 180˚ 90˚ 2 3 4 5
)
2 y 3 cos ( 50˚) 2
90˚ 180˚ 270˚
translation 50 right and 2 units up with an amplitude 2 of unit
3
translation units left and 4 5 units up
©Glencoe/McGraw-Hill
90˚ 180˚ 270˚
y 5 4 3 2 1
y
90˚ 180˚ 270˚
381
Algebra 2
Chapter 14
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34. 5; 4; 180; 30
33. 1; 2; 120; 45 y 5 4 3 2 1
y 10 8 6 4 2
y 2 sin [3( 45˚)] 1
O
270˚ 180˚ 90˚ 2 3 4 5
270˚ 180˚ 90˚ 4 6 8 10
90˚ 180˚ 270˚
35. 3.5; does not exist; 720; 60
90˚ 180˚ 270˚
y 4 cos [2( 30˚)] 5
36. 0.75; does not exist; 270; 90 y 20 16 12 8 4
y 8 6 4 2 O 270˚180˚90˚ 4 6 8 10 12
O
90˚ 180˚ 270˚
[1 (
270˚180˚90˚ 8 12 16 20
)]
y 3 csc 2 60˚ 3.5
O
90˚ 180˚ 270˚
[2(
)]
y 6 cot 3 90˚ 0.75
1 4
38. 4; does not exist; 30;22.5
37. 1; ; 180; 75
y y 5 4 3 2 1
O 270˚ 180˚ 90˚ 2 3 4 5
2 1
1 y 4 cos (2 150˚) 1
22.5˚ 90˚ 180˚ 270˚
O
22.5˚
2 3 4 5 6 7 8
2 y 5 tan (6 135˚) 4
©Glencoe/McGraw-Hill
382
Algebra 2
Chapter 14
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4
2 3
40. 4; does not exist; 6;
39. 3; 2; ;
y
y 8 7 6 5 4 3 2 1
3 2
[(
O
y 3 2 sin 2 4
2
2 2
3 2
16 14 12 10 8 6 4 2
)]
O 4 2 4
2
[1(
4 2
y 4 sec 3 3
41.
42.
1
5 4 3 2 1
3 2
2
y y 3 2 cos 1 y 3 cos ( )
2
2 3 4 5
3 2
)]
y 5 4 3 2 1
2
O
O 4 2 2 3 4 5
The graphs are identical.
[ 1 ( )] 1 3 y cos [ 4 ( 2 )] y sin 4 2
2
4
The graphs are identical. 43. c
44. 180; 5 yr
45. 300; 14.5 yr
46. Sample answer: When the prey (mouse) population is at its greatest the predator will consume more and the predator population will grow while the prey population falls. 9
47. h 9 6 sin c (t 1.5)d
48. a 1, b 1, h
49. Sample answer: You can use changes in amplitude and period along with vertical and horizontal shifts to show an animal population’s starting point and display changes to that population over a period of
50. B
©Glencoe/McGraw-Hill
383
2
Algebra 2
Chapter 14
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7/24/02
2:04 PM
Page 384 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
time. Answers should include the following information. • The equation shows a rabbit population that begins at 1200, increases to a maximum of 1450, then decreases to a minimum of 950 over a period of 4 years. • Relative to y a cos bx, y a cos bx k would have a vertical shift of k units, while y a cos [b (x h)] has a horizontal shift of h units. 52. amplitude: does not exist; period: 360 or 2
51. D
y 5 4 3 2 1 O 270˚180˚90˚ 2 3 y 3 csc 4 5
period: 270 or
y
3 2
y
5 4 3 2 1
10 8 6 4 2
y sin 2
O
90˚ 180˚ 270˚
360˚ 180˚ 4 6 2 y 3 tan 3 8 10
O
180˚ 360˚
56. 0.57
55. 0.75 ©Glencoe/McGraw-Hill
54. amplitude: does not exist;
53. amplitude: 1; period: 720 or 4
270˚ 180˚ 90˚ 2 3 4 5
90˚ 180˚ 270˚
384
Algebra 2
Chapter 14
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2:04 PM
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57. 0.83
58. 0.8
59. 35
60. 2.29
61. 0.66
62. 0.66
63.
5a 13 (a 2)(a 3)
64.
65.
3y 2 10y 5 2(y 5)(y 3)
66.
1 4
67. 1
23 2
68. 1
69.
1 2
70. 0
71.
23 3
72.
22 2
73. 1
Lesson 14-3 Trigonometric Identities Pages 779–781 1. Sample answer: The sine function is negative in the third and fourth quadrants. Therefore, the terminal side of the angle must lie in one of those two quadrants.
2. Sample answer: Pythagorean identities are derived by applying the Pythagorean Theorem to trigonometric concepts.
3. Sample answer: Simplifying a trigonometric expression means writing the expression as a numerical value or in terms of a single trigonometric function, if possible.
4.
5 4
3 5
5.
6.
7. 22
8. 1 10. sec
9. tan2
12. sin cos
11. csc
©Glencoe/McGraw-Hill
23 3
385
v2 gR
Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
13.
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1 2
14.
15. 25
25 3
16. 2 22 3 5
17.
5 4
18.
19.
23 2
20.
21.
3 4
22.
3 25 5
4 5 4 217 17
4 27 7
23.
24.
25. cot
26. 1
27. cos
28. sin
29. 2
30. 3
31. cot2
32. tan
33. 1
34. cot2
35. csc2
36. 1
37. about 11.5
38. about 4 m/s
39. about 9.4
40. E
I tan cos E I sin simplifies to E . R2
42. P I 2R sin2 2ft
41. No; R 2
43. P I 2R
I 2R . 1 tan2 2ft
44.
45. Sample answer: You can use equations to find the height and the horizontal distance of a baseball after it has been hit. The equations involve using the initial angle the ball makes with the ground with the sine function. Answers should include the following information.
©Glencoe/McGraw-Hill
I cos R2
9 16
46. B
386
Algebra 2
Chapter 14
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7/24/02
2:04 PM
Page 387 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
• Both equations are quadratic in nature with a leading negative coefficient. Thus, both are inverted parabolas which model the path of a baseball. • model rockets, hitting a golf ball, kicking a rock 48. 1; y 1; 1; 360
47. A
y 5 4 3 2 1 O 270˚ 180˚ 90˚ 2 3 4 5
49. 12; y 12; no amplitude; 180 20
y sin 1
y 5 4 3 2 1
15 10 5
©Glencoe/McGraw-Hill
50. amplitude: does not exist; period: 180 or
y
O 270˚ 180˚ 90˚ 5 y tan 12
90˚ 180˚ 270˚
90˚ 180˚ 270˚
O 135˚ 90˚ 45˚ 2 3 y csc 2 4 5
387
45˚
90˚ 135˚
Algebra 2
Chapter 14
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2:04 PM
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51. amplitude: 1; period: 120 or
2 3
52. amplitude: does not exist; period: 36 or
y 5 4 3 2 1
5
y 5 4 3 2 1
y cos 3
O 135˚ 90˚ 45˚ 2 3 4 5
45˚
90˚ 135˚
1
y 3 cot 5
O
22.5˚
22.5˚
2 3 4 5
1 6
53. 93
54. y (x 11)2
55. Symmetric ()
56. Substitution ()
57. Multiplication ()
58. Substitution ()
1 2
Chapter 14 Practice Quiz 1 Page 781 1.
3 , 4
2. 5, 2, 8,
720 or 4 y 5 4 3 2 1
270˚ 180˚ 90˚ 2 3 4 5
y
3
6 4 2
1
y 4 sin 2
O
90˚ 180˚ 270˚
O 4 2 4 6 8 10 12 14
3 5
3. 5.
4
4.
2
4
[1 (
y 2 cos 4 4
)] 5
213 3
25 2
©Glencoe/McGraw-Hill
388
Algebra 2
Chapter 14
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7/24/02
2:04 PM
Lesson 14-4
Page 389 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
Verifying Trigonometric Identities Pages 784–785
?
1. sin tan sec cos ?
sin tan
1 cos
2. Sample answer: Use various identities, multiply or divide terms to form an equivalent expression, factor, and simplify rational expressions.
cos
sec
1 cos
1 cos2 , cos cos Multiply by the LCD, cos . ? 1 cos2 tan cos Subtract. 1 cos 2 ? sin2 tan sin2 cos sin ? tan sin cos Factor. ?
sin tan sin sin sin
sin tan sin tan
sin tan cos ?
3. Sample answer: sin2 1 cos2 ; it is not an identity because sin 2 1 cos2 . 5.
4. tan (cot tan ) sec 2 ?
1 tan2 sec2 sec2 sec2
?
tan2 cos2 1 cos2 sin2 cos2
6.
?
cos2 sin2 sin2 sin2
cos2 1 sin 1 sin2 1 sin (1 sin )(1 sin ) 1 sin
?
1 sin ?
1 sin ?
1 sin
1 sin 1 sin
©Glencoe/McGraw-Hill
389
Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
7.
7/24/02
2:04 PM
1 tan2 csc2
tan2
sec2 csc2
tan2
Page 390 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
?
8.
?
cos
1 cos 2 ? 1 ?
tan2
sin cos
sin sec sin sec sin sec
sin tan 2
2
tan2 tan2
9.
sec 1 tan
?
sec 1 tan sec 1 tan sec 1 tan sec 1 tan
11.
?
?
?
tan sec 1
sin2 cos2
?
12. cot (cot tan ) csc2
?
cot2 cot tan csc2
?
cos2 1 2
cot 2
?
cos sin
?
csc 2 ?
?
14.
?
?
sin sec cot 1 sin
sin2 sec2
1 cos
cos sin
?
1
11
?
1 tan2 sec2 sec2 sec2
©Glencoe/McGraw-Hill
sin cos
cot2 1 csc2 csc2 csc2
1 sec2 sin2 sec2 1 cos2
?
cos sin 1 11
1
?
sin cos sin2 cos2 sin cos 1 sin sec
tan sec 1 sec 1 sec 1 tan (sec 1) sec2 1 tan (sec 1) tan2 sec 1 tan
2
13.
?
10. D
cos2 tan2 cos2 1 cos2
cos
sin ? 1 sin2 cos2 sec
sin 2
1 cos2
1 sin ? sec tan cot sin ? 1 sin2 sin sec
390
Algebra 2
Chapter 14
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7/24/02
2:04 PM
Page 391 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
1 2 cos2 ? tan cot sin cos (1 cos2 ) cos2 ? tan cot sin cos sin2 cos2 ? tan cot sin cos sin2 cos2 ? tan cot sin cos sin cos sin cos ? tan cot cos sin tan cot tan cot
16.
15. 1 cos ? (csc cot )2 1 cos 1 cos ? csc2 2 cot csc 1 cos cot2 1 cos ? 1 cos 2 2 1 cos sin sin 1 cos2 sin sin2 cos2 1 cos ? 1 2cos 1 cos sin2 sin2 sin2 1 cos ? 1 2 cos cos2 1 cos sin2 1 cos ? (1 cos )(1 cos ) 1 cos 1 cos2 1 cos ? (1 cos )(1 cos ) 1 cos (1 cos )(1 cos ) 1 cos 1 cos 1 cos 1 cos
17. ?
cot csc
cos ? sin
cot csc
cot csc
cot csc
18.
cot csc sin tan
?
sin cos
1
sin
1 ?
sin cos
sin sin cos cos 1 sin ? sin cos sin cos cos 1 sin ? sin (cos 1) cos ?
cot csc
?
cot csc
1 tan sec
sin cos
sin
cos
1 cos sin cos cos ? 1 cos ?
sin cos
sin cos cos
cos
sin cos sin cos
cos 1 sin cos sin (cos 1) cos sin
1 sin
cot csc cot csc ©Glencoe/McGraw-Hill
391
Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
19.
sec sin 1 cos sin
7/24/02
sin cos
2:04 PM
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?
cot
sin 1 cos ? 2csc 1 cos sin sin sin 1 cos 1 cos ? 2csc sin 1 cos 1 cos sin
20.
sin ?
cos cot
1 sin cos
sin2 sin cos 1 sin2 sin cos cos2 sin cos cos sin
1 2cos cos2 ? sin2 2csc sin (1 cos ) sin (1 cos )
?
sin2 cos2 1 2cos ? 2csc sin (1 cos )
?
2 2cos ? 2csc sin (1 cos )
cot cot
2(1 cos ) ? 2csc sin (1 cos )
?
cot
2 ? 2csc sin
?
cot
?
2csc 2csc
cot cot
21.
23.
1 sin sin 1 sin sin 1 sin sin 1 sin sin 1 sin sin 1 sin sin 1 sin sin 1 sec2
?
?
?
?
cot2 csc 1 cot2 csc 1 csc 1 csc 1 cot2 (csc 1) csc2 1 cot2 (csc 1) cot2
1
?
?
sin cos cos
1 sin sin sin 1 sin sin 1
csc2
?
sin
cos
?
sin cos sin cos
cos sin sin cos cos ? sin cos sin cos sin
1
csc 1
?
24. 1
?
1
1
cos2 sin2 1 11
1 1 1 1
©Glencoe/McGraw-Hill
1 tan 1 cot
22.
392
sin sin cos sin cos
?
1 cos 1 cos 1 cos
1 cos 1 cos
sec 1
?
?
sin cos sin cos
tan2 sec 1 tan2 sec 1 sec 1 sec 1 tan2 (sec 1) sec 2 1 2 tan (sec 1)
1 cos ?
?
tan2
?
1
1 cos
Algebra 2
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25. 1 tan4
26.
?
cos4 sin4 cos2 sin2 (cos2 sin2 )(cos2 sin2 )
?
cos2 sin2 (cos2 sin2 ) 1
?
cos2 sin2 cos2 sin2 cos2 sin2
?
2 sec2 sec4 (1 tan2 )(1 tan2 )
?
sec2 (2 sec2 ) [1 (sec2 1)](sec2 )
?
(2 sec2 )(sec2 ) (2 sec2 )(sec2 ) (2 sec2 )(sec2 ) 1 cos sin
1 cos 1 cos
27. 1 cos sin
1 cos2 sin (1 cos ) sin2 sin (1 cos ) sin 1 cos
29.
?
?
?
?
sin 1 cos
cos ? cos 2sec 1 sin 1 sin 1 sin cos 1 sin ? cos 2sec 1 sin 1 sin 1 sin 1 sin
28.
sin 1 cos
cos 11 sin 2 cos 11 sin 2 11 sin 211 sin 2
sin 1 cos
cos sin cos cos sin cos ? 2 sec 1 sin 2 2cos ? 2sec cos2 2 ? 2sec cos2 2sec 2sec
sin 1 cos sin 1 cos ?
tan sin cos csc2 1 sin cos
sin cos
1 2
sin2 1 cos
30.
sin2 1 cos 1 cos 1 cos sin2 (1 cos ) 1 cos2 sin2 (1 cos )
?
sin
?
2sec
1
11
sin2
?
1 cos ?
1 cos ?
1 cos ?
1 cos
1 cos 1 cos 31.
v 02
tan2
2g sec2
©Glencoe/McGraw-Hill
sin2 cos2 1 2g cos2
v 02
v 02
sin2 cos2
2g v 20 sin2
32. 598.7 m
cos 2 1
2g
393
Algebra 2
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34. Sample answer: Trigonometric identities are verified in a similar manner to proving theorems in geometry before using them. Answers should include the following. • The expressions have not yet been shown to be equal, so you could not use the properties of equality on them. • To show two expressions you must transform one, or both independently. • Graphing two expressions could result in identical graphs for a set interval, that are different elsewhere.
33. Sample answer: Consider a right triangle ABC with right angle at C. If an angle A has a sine of x, then angle B must have a cosine of x. Since A and B are both in a right triangle and neither is the right angle, their sum 2
must be .
35. D
36. B
37.
38.
[360, 360] scl: 90 by [5, 5] scl: 1
[360, 360] scl: 90 by [5, 5] scl: 1
is not
may be 40.
39.
[360, 360] scl: 90 by [5, 5] scl: 1
[360, 360] scl: 90 by [5, 5] scl: 1
may be
may be
©Glencoe/McGraw-Hill
394
Algebra 2
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41.
42.
[360, 360] scl: 90 by [5, 5] scl: 1
[360, 360] scl: 90 by [5, 5] scl: 1
may be
is not
43.
25 2
44.
25 3
45.
2193 12
46.
27 4
48. 1: 360; 45
47. 1: 360; 30 y 5 4 3 2 1
y 5 4 3 2 1
y cos ( 30˚)
O
270˚ 180˚ 90˚ 2 3 4 5
90˚ 180˚ 270˚
O
270˚ 180˚ 90˚ 2 3 4 5
2
49. 3; 2;
50.
5 6
52.
22 4
54.
2 23 4
y sin ( 45˚) 90˚ 180˚ 270˚
y 5 4 3 2 1
3 2
51.
26 4
53.
26 4
O 2 2 3 4 5
(
y 3 cos 2
2
3 2
)
22 2
©Glencoe/McGraw-Hill
395
Algebra 2
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Lesson 14-5
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Sum and Difference of Angles Formulas Pages 788–790 2. Use the formula sin( ) sin cos cos sin . Since sin 105 sin(60 45), replace with 60 and with 45 to get sin 60 cos 45 cos 60 sin 45. By finding the sum of the products of the values,
1. sin ( ) sin sin sin cos cos sin sin
the result is
26 22 4
or about 0.9659. 3. Sometimes; sample answer: The cosine function can equal 1.
4.
26 22 4
5.
26 22 4
6.
22 26 4
7.
23 2
8.
23 2
1 2
10. cos (270 )
9.
?
cos 270 cos sin 270 sin ?
0 (1 sin ) sin 2
sin a b cos
11. sin cos
2
cos sin
2
?
12.
sin( 30) cos( 60)
?
sin cos 30 cos cos 30 cos cos 60 sin sin 60
?
cos
?
?
sin 0 cos 1 cos cos cos
23 sin 2
1 cos 2 ?
1 2
cos
23 sin 2
1 2
1 2
cos cos cos
13.
5 23 1 5 23
©Glencoe/McGraw-Hill
14.
396
22 2
Algebra 2
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15.
22 26 4
16.
26 22 4
17.
26 22 4
18.
26 22 4
19.
26 22 4
20.
21.
22 2
22.
22 2
22 26 4 23 2
23.
22 2
24.
25.
22 26 4
26.
27.
26 22 4
28. sin (270 )
22 2
?
sin 270 cos cos 270 sin ?
1 cos 0 cos 30. cos (90 )
29. cos (90 )
?
?
cos 90 cos sin 90 sin
?
0 cos 1 sin sin
cos 90 cos sin 90 sin
?
0 1 sin sin 31.
?
sin(90 ) cos sin 90 cos cos 90 sin
32.
sin cos
?
cos
3 2
3 2 2
?
cos
cos sin
3 2
?
?
cos
?
sin 0 cos (1) cos
1 cos 0 sin cos
?
cos 0 cos
?
0 (cos ) cos
cos cos
©Glencoe/McGraw-Hill
sin 1
cos cos
397
Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
33.
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?
cos ( ) cos cos cos sin sin
34.
?
cos(2 ) cos cos 2 cos [sin 2 sin ]
?
cos
?
cos
?
1 cos 0 sin cos cos cos
?
1 cos [0 sin ] cos ?
1 cos 0 cos cos cos ?
36. sin (60 ) sin (60 )
sin( ) sin
35.
?
sin 60 cos cos 60 sin sin 60 cos cos 60 sin
?
sin cos [cos sin ] sin ?
0 cos [1 sin ] sin
?
?
0 [sin ] sin sin sin
23 2
23 2
cos
cos
1 2
1 2
sin
sin
23 cos 3
6
37. sin a b cos a b ?
3 cos 6
38. sin ( ) sin ( ) sin2 sin2
3 sin 6
sin cos cos sin cos
sin
23 sin cos 2 1 cos sin 2 1 ? 1 sin sin 2 2 1 2 ?
?
(sin cos cos sin ) (sin cos cos sin ) ?
23 2
sin2 cos2 cos2 sin2 ?
sin2 (1 sin2 ) (1 sin2 ) sin2 ?
sin2 sin2 sin2
sin
sin2 sin2 sin2 sin2 sin2
©Glencoe/McGraw-Hill
398
Algebra 2
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39.
y
40. ?
cos ( ) ?
cos ( )
4
1 tan tan sec sec 1
sin
y 10 sin (2t 30°) 10 cos (2t 60°)
sin
cos cos 1 cos
180° 90°
1 cos
cos ( ) sin
1 cos
sin cos
1 cos
90°
180° t
2 4
?
1 cos
O
cos cos cos cos
?
cos ( ) cos cos sin sin 1
cos ( ) cos ( ) 41. Destructive; the resulting graph has a smaller amplitude than the two initial graphs.
42. 0.3681 E
43. 0.4179 E
44. 0.6157 E
45. 0.5563 E
46. tan ( )
©Glencoe/McGraw-Hill
399
sin( ) cos( ) sin cos cos sin cos cos sin sin
sin cos cos cos cos cos cos cos
cos sin
tan tan 1 tan tan
cos cos sin sin
cos cos
Algebra 2
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tan ( )
sin( ) cos( ) sin cos cos sin cos cos sin sin
sin cos cos cos cos cos cos cos
cos sin
tan tan 1 tan tan
cos cos
47. Sample answer: To determine communication interference, you need to determine the sine or cosine of the sum or difference of two angles. Answers should include the following information. • Interference occurs when waves pass through the same space at the same time. When the combined waves have a greater amplitude, constructive interference results and when the combined waves have a smaller amplitude, destructive interference results.
48. A
49. C
50. cot sec ?
?
?
sin sin
cos cos
cos2 sin sin cos cos2 sin sin cos sin cos cos sin
1 cos
cot sec
©Glencoe/McGraw-Hill
400
Algebra 2
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51. sin2 tan2
52.
?
(1 cos2 ) ?
sin2 ?
sin2 ?
sin2
sec2 csc2 1 cos2 sin2 cos2
csc
sec2
?
sin (sin csc ) 2 cos2
2
?
sin2 1 2 cos2
?
1 cos2 1 2 cos2 2 cos2 2 cos2
1 sin2
sin2 tan2 53. 1
cos 1 cos
sec tan sin cos cos sin 1 sin
?
csc
54. 1
?
csc ?
csc ?
csc
csc csc 55. 4
56. sec
57. 2 sec
58. sin
3 234 , 34
cos
5 234 , 34
csc sec 59.
4 5
3 5
sin , cos , 4 3
5 3
234 , 5
5 4 3 4
61. 360
62. 3,991,680
63. 56
64. 210
65. about 228 mi
66.
2 25 2
5 3
cot
y2 34
x2 6
1
3 5
67.
©Glencoe/McGraw-Hill
234 , 3
60. sin 1, cos 0, tan undefined, csc 1, sec undefined, cot 0
tan , csc , sec , cot
3 5
tan ,
68. 401
Algebra 2
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3 4
69.
25 5
70.
71.
26 2
72.
73.
3 25 5
2 16 12 2
74.
Lesson 14-6
222 12 2
Double-Angle and Half-Angle Formulas Pages 794–797
1. Sample answer: If x is in the x third quadrant, then is 2 between 90 and 135. Use the half-angle formula for cosine knowing that the value is negative.
2. Sample answer: 45; cos 2(45) cos 90 or 0,
3. Sample answer: The identity used for cos 2 depends on whether you know the value of sin , cos , or both values.
4.
24 , 25
6.
23 1 22 13 , , , 2 2 2
5.
4 25 , 9
1 230 , 9 6
,
2 cos 45 2
26 6
12 2
or 22
7 25 2 25 , 25 5 5
,
22 13 2 3 27 , 8
7.
1 8
,
28 2 17 , 4
9.
8.
22 13 2
28 2 17 4
22 13 2
?
sin 2x 1 cos 2x
?
2 sin x cos x 1 (1 2 sin2 x)
?
2 sin x cos x 2 sin2 x
?
cos x sin x
10. cot x
cot x ©Glencoe/McGraw-Hill
402
Algebra 2
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Page 403 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
?
11. cos2 2x 4 sin2 x cos2 x 1
12. 1.64
?
cos2 2x sin2 2x 1 11 120 119 5 226 226 , , , 169 169 26 26
4 26 , 25
13. 15.
4 22 , 9
7 26 , 9 3
,
23 3
16.
3 255 23 28 155 , , , 32 32 4
17. 19.
24 7 3 210 , , , 25 25 10
18.
17 215 221 , 18 6 6
,
4 22 7 23 16 4 13 , , 9 9 6
21.
23.
,
215 , 8
215 5
210 10
7 210 26 , 8 4 4
,
28 155 4
235 , 18
23 210 , 25 5
14.
20.
120 119 5 226 , , , 169 169 26
22.
215 7 , , 8 8
23 16 4 13 6
4 25 , 9
1 26 230 , 9 6 6
226 26
28 2 115 , 4
28 2 115 4
4 221 17 , , 5 25
,
24.
25 110 1210 , 10 25 110 1210 10
25.
22 13 2
26.
27.
22 12 2
28.
22 13 2
30.
22 13 2
29. 31.
22 12 2 ?
sin 2x 2 cot x sin2 x cos x ? 2 sin x cos x 2 sin2 x sin x 2 sin x cos x 2 sin x cos x
22 12 2
2 cos2
32. 2a
3
x 2
?
1 cos x
2
1 cos x b 2
1 cos x b 2
2a
?
1 cos x ?
1 cos x
1 cos x 1 cos x
©Glencoe/McGraw-Hill
403
Algebra 2
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?
1 2
?
1 2
?
1 2
34. sin2 x (1 cos2 x)
33. ?
sin4 x cos4 x 2 sin2 x 1 (sin2 x cos2 x)(sin2 x cos2 x)
sin2 x [1 (1 2 sin2 x)]
?
2 sin2 x 1 (sin2 x cos2 x) 1
sin2 x (2 sin2 x) sin2 x sin2 x
?
2 sin2 x 1 [sin2 x (1 sin2 x)] 1 ?
2 sin2 x 1 sin2 x 1 sin2 x ?
2 sin2 x 1 2 sin2 x 1 2 sin2 x 1 tan2
35.
x 2
?
1 cos x 1 cos x
36.
x
sin2 2 cos a
3
a
3
2 x
?
1 cos x 1 cos x
2
1 cos x 2 b 2 ? 1 cos x 1 cos x 2 b 2
1 cos x 1 cos x
1 cos x sin x cos x sin x 1 cos2 x sin x cos x sin2 x sin x cos x sin x cos x
?
tan x ?
tan x ?
tan x
tan x tan x
1 cos x 1 cos x 1 cos x
1 37. 46.3
38. 1
39. 2 23
40.
2 g
3
1 cos L 1 cos L
?
2 g
v 2 tan (1 sin2 )
?
2 g
v 2 tan cos2
?
2 2 v sin g v 2 sin 2 g
404
3
1 cos L 1 cos L
v 2 (tan tan sin2 )
©Glencoe/McGraw-Hill
?
tan x
cos
Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
41.
1 tan 4
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y
42. 2.5 2 1.5 1 0.5
3 2
2
y sin2 x
x
O
2
1
1 1.5 y 2 cos 2x 2
3 2
y cos2 x
2.5
Sample answer: They all have the same shape and are vertical translations of each other. 2
43. The maxima occur at x and
3 . The 2
44.
y 2.5 2 1.5 1 0.5
minima occur
at x 0, and 2.
270˚ 180˚ 90˚ 1 1.5 2 2.5
y sin 2x
x O
90˚ 180˚ 270˚
45. The graph of f(x) crosses the x-axis at the points specified in Exercise 41.
46. c 1 and d 0.5
47. Sample answer: The sound waves associated with music can be modeled using trigonometric functions. Answers should include the following information. • In moving from one harmonic to the next, the number of vibrations that appear as sine waves increase by 1.
48. D
©Glencoe/McGraw-Hill
405
Algebra 2
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• The period of the function as you move from the nth harmonic to the (n 1)th harmonic decreases from 2 2 . to n1
n
51.
26 22 4
53. 55.
26 22 4
50.
49. B
52.
23 2
26 22 4
54.
1 2
22 2
56. ?
cot2 sin2
cos2 csc2 sin2 sin2 csc2 1
?
cot sin 2
2
cos2 sin2 sin2 1
sin2 sin2
cot2 sin2 1 cot2 sin2 cot2 sin2 ?
cot2 sin2
57. cos (cos cot )
58. 101 or 10
?
cot cos 1sin 12 ?
cos cos sin cot cos sin
?
cos2 cot cos cos (cos cot ) 59. 102.5 or about 316 times greater
60. 6, 5
61. 1, 1
62. 0, 2
63.
5 , 2
1 1 2 2
64. ,
2 1 2
65. 0,
©Glencoe/McGraw-Hill
406
Algebra 2
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Chapter 14 Practice Quiz 2 Page 797 ?
1. sin sec tan sin
1 cos sin cos
?
2.
sec cos sin tan cos ? 1 cos sin tan cos cos cos2 ? 1 sin tan cos cos 1 cos2 ? sin tan cos sin2 ? sin tan cos sin ? sin tan sin cos sin tan sin tan
4.
sin (90 ) cos
?
tan ?
tan ?
tan tan
sin (cos 1) cos sin cos sin ? sin tan cos sin ? sin cos sin tan cos cos sin tan sin tan
? 3. sin tan
?
sin 90 cos cos 90 sin cos ?
cos 0 cos cos cos
3 ? b sin 2 3 3 ? cos cos sin sin sin 2 2 cos a
5.
?
6. sin ( 30) cos ( 60) ?
(sin cos 30 cos sin 30) (cos cos 60 sin sin 60)
?
0 (1 sin ) sin
a ?
sin sin
13 2
sin
1 a cos 2 ?
1 2
1 2
cos b
13 2
sin b
1 2
cos cos cos 7.
23 2
9.
22 13 2
©Glencoe/McGraw-Hill
9 282 82
8. 10.
407
22 12 2
Algebra 2
Chapter 14
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Lesson 14-7
Page 408 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
Solving Trigonometric Equations Pages 802–804
1. Sample answer: If sec 0 1 then 0. Since no value cos
of makes
1 cos
2. Sample answer: The function is periodic with two solutions in each of its infinite number of periods.
0. there
are no solutions. 3. Sample answer: sin 2
4. 60, 120, 240, 300
5. 135, 225
6.
7.
6
5 3 , , , 6 2 6 2
8. 0
9. 0 k
2k 3
10. 90 k 360, 180 k 360
11. 60 k 360, 300 k 360
12.
7 6
11 6
2k,
2k or
210 k 360, 330 k 360 13.
6
2k,
5 6
2k,
2
2k
14. 31.3
or 30 k 360, 150 360, 90 k 360 15. 60, 300
16. 240, 300
17. 210, 330
18. 30, 150, 210, 330
19.
5 3 , , 6 6 2
20.
2
21.
7 11 , 6 6
22.
3 2 4 , , , 2 2 3 3
23.
3
2k,
5 3
2k
24. 2k, 5 3
25.
2 3
27.
3
2k,
2k,
©Glencoe/McGraw-Hill
4 3
5 3
2k
3
2k,
2k
26. 0 2k
2k
28. 0 k, 408
6
2k,
5 6
2k
Algebra 2
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29. 45 k 180
30. 0 k 180
31. 270 k 360
32. 30 k 360, 150 k 360
33. 0 k 180, 60 k 180
34. 120 k 360, 240 k 360
3 2k, 2k 2 2
35. 0 2k,
36.
7 6
2k,
11 6
2k or
210 k 360, 330 k 360
or 0 k 360, 90 k 360, 270 k 360 37. 0 k or 0 k 180
38.
2
k,
2 3
2k,
4 3
2k
or 90 k 180, 120 k 360, 240 k 360 39. 0 2k,
3
2k,
5 3
2,
40.
2
4k or 90 k 720
or 0 k 360, 60 k 360, 300 k 360 41. S
352 tan
43. y
3 2
or S 352 cot
3 2
42. about 32 44. 10
sin (t)
y 4 3.5 3 2.5 2 1.5 1 0.5 O 1 1
3
3
y 2 2 sin ( t )
1 2 3 4 5 6 7 8 9
t
45. (4.964, 0.598)
©Glencoe/McGraw-Hill
46. Sample answer: Temperatures are cyclic and can be modeled by trigonometric functions. Answers should include the following information.
409
Algebra 2
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• A temperature could occur twice in a given period such as when the temperature rises in the spring and falls in autumn. 47. D
48. B
49.
24 7 210 3 210 , , , 25 25 10 10
50.
1 1 23 23 , , , 2 2 2 2
51.
5 211 7 23 233 , , , 18 18 6 6
52.
7 25 2 25 24 , , , 25 25 5 5
54.
22 2
53.
23 2
55. b 11.0, c 12.2, mC 78
©Glencoe/McGraw-Hill
410
Algebra 2
Chapter 14
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