Answer KeyTransparencies
Provides transparencies withanswers for each lesson in the Student Edition
ISBN 0-07-828001-X
9
780078 280016
90000
1. First, find the sum of c andd. Divide this sum by e.Multiply the quotient by b.Finally, add a.
3. b; The sum of the cost ofadult and children ticketsshould be subtracted from50. Therefore, parenthesesneed to be inserted aroundthis sum to insure that thisaddition is done beforesubtraction.
5. 6
7. 1
9. 119
11. !23
13. $432
15. $1162.50
17. 3
19. 25
21. !34
23. 5
25. !31
27. 14
29. !3
31. 162
33. 2.56
35.
37. 31.25 drops per min
39. 2
25
13
2. Sample answer:
4. 72
6. 23
8. !2
10. 0
12. 18
14. $1875
16. 20
18. 29
20. 54
22. 19
24. 11
26. 7
28. !15
30. !52
32. 15.3
34. !7
36. about 1.8 lb
38. 3.4
40. 45
14 ! 45
Chapter 1 Solving Equations and InequalitiesLesson 1-1 Expressions and Formulas
Pages 8–10
©Glencoe/McGraw-Hill 1 Algebra 2 Chapter 1
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©Glencoe/McGraw-Hill 2 Algebra 2 Chapter 1
41. !4.2
43. !4
45. 1.4
47. !8
49.
51. !16
53. $8266.03
55. Sample answer:
57. C
59. 3
61. 10
63. !2
65. 23
144 ! 42 " 4 # 104 $ 4 $ 4 " 4 # 914 $ 42 % 14 " 42 # 844 " 4 ! 4 # 714 $ 42 " 4 $ 4 # 614 % 4 $ 42 " 4 # 54 % 14 ! 42 $ 4 # 414 $ 4 $ 42 " 4 # 34 " 4 $ 4 " 4 # 24 ! 4 $ 4 " 4 # 1
2
16
42. 5.3
44. 75
46. !4
48. 36.01
50. &
52. 30
54. 400 ft
56. Nurses use formulas tocalculate a drug dosagegiven a supply dosage and adoctor’s drug order. Theyalso use formulas tocalculate IV flow rates.Answers should include thefollowing.• A table of IV flow rates is
limited to those situationslisted, while a formula canbe used to find any IV flowrate.
• If a formula used in anursing setting is appliedincorrectly, a patient coulddie.
58. D
60. 4
62. 13
64. !5
66. 67
ay $ 52b2
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©Glencoe/McGraw-Hill 3 Algebra 2 Chapter 1
1a. Sample answer: 21b. Sample answer: $51c. Sample answer: !111d. Sample answer: 1.31e. Sample answer:1f. Sample answer: !1.3
3. 0; Zero does not have a multiplicative inverse since is undefined.
5. N, W, Z, Q, R
7. Multiplicative Identity
9. Additive Identity
11. 3
13.
15.
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)
19. W, Z, Q, R
21. N, W, Z, Q, R
23. I, R
25. N, W, Z, Q, R
27. Q, R; 2.4, 2.49,
29. Associative Property (%)
31. Associative Property ($)
33. Multiplicative Inverse
35. Multiplicative Identity
37. !m; Additive Inverse
2.92.49,2.49,
3c $ 18d
!2x $ 4y
!13,
10
12
2. A rational number is the ratioof two integers. Since isnot an integer, is not arational number.
4. Z, Q, R
6. Q, R
8. Associative Property ($)
10. 8,
12. !1.5,
14. 13p
16.
18. $175.50
20. Q, R
22. Q, R
24. Z, Q, R
26. I, R
28. Additive Inverse
30. Additive Identity
32. Commutative Property ($)
34. Distributive
36. 0
38. Multiplicative Inverse1m
;
!17a ! 1
23
!18
132
13
Lesson 1-2 Properties of Real NumbersPages 14–18
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©Glencoe/McGraw-Hill 4 Algebra 2 Chapter 1
39. 1
41. units
43. 10;
45. 0.125; !8
47.
49.
51.
53.
55.
57.
59. true
61. false; 6
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)
!8 $ 9y
!3.4m $ 1.8n
!12r $ 4t
40x ! 7y
3a ! 2b
!43,
34
!110
12
40. natural numbers
42. The square root of 2 isirrational and thereforecannot be described by anatural number.
44. !2.5; 0.4
46.
48.
50.
52.
54.
56.
58.
60. false; !3
62. true
64. 3.6; $327.60
910
x !196
y
4.4p ! 2.9q
32c ! 46d
11m $ 10a
10x $ 2y
4
35, !
523
58; !
85
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©Glencoe/McGraw-Hill 5 Algebra 2 Chapter 1
65.
Def. of a mixed number
Distributive
Multiply.
Comm. ($)
Add.
Assoc. ($)
Add.
67. 4700 ft2
69. $62.15
71. Answers should include thefollowing.• Instead of doubling each
coupon value and thenadding these valuestogether, the DistributiveProperty could be appliedallowing you to add thecoupon values first andthen double the sum.
# 8 $ 1 or 9
# 8 $ a34
$14b
# 8 $34
$14
# 6 $ 2 $34
$14
# 6 $34
$ 2 $14
# 3 122 $ 3 a14b $ 2 112 $ 2 a1
8b
# 3 a2 $14b $ 2 a1 $
18b
3 a2
14b $ 2 a11
8b 66. 50(47 $ 47); 50(47) $ 50(47)
68. $113(0.36 $ 0.19);$113(0.36) $ $113(0.19)
70. Yes; 7;dividingby a number is the same asmultiplying by its reciprocal.
72. B
6 $ 82
#62
$82
#
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©Glencoe/McGraw-Hill 6 Algebra 2 Chapter 1
• If a store had a 25% offsale on all merchandise,the Distributive Propertycould be used to calculatethese savings. Forexample, the savings on a$15 shirt, $40 pair ofjeans, and $25 pair ofslacks could be calculatedas 0.25(15) $ 0.25(40) $0.25(25) or as 0.25(15 $40 $ 25) using theDistributive Property.
73. C
75. False; 0 ! 1 # !1, which isnot a whole number.
77. False; which is not
a whole number.
79. 6
81. !2.75
83. !11
85. !4.3
23
,2 " 3 #
74. true
76. true
78. 9
80. !5
82. 358 in2
84.
86. 36
710
Chapter 1Practice Quiz 1
Page 18
1. 14
3. 6
5. 2 amperes
7. N, W, Z, Q, R
9. !67,
76
2. !9
4. !1
6. Q, R
8. Additive Inverse
10. 50x ! 64y
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©Glencoe/McGraw-Hill 7 Algebra 2 Chapter 1
Lesson 1-3 Solving EquationsPages 24–27
1. Sample answer: 2x # !14
3. Jamal; his method can beconfirmed by solving theequation using an alternativemethod.
5.
7. Sample answer: 5 plus 3 times the square of anumber is twice that number.
9. Addition Property of Equality
11. 14
13. !4.8
15. 16
17.
19.
21.
23.
25. a n4b 2
519 $ n2n2 ! 4
5 $ 3n
p #Ir t
2n ! n3
95C $ 32 # F
95
cC $59
1322d # F
C $59
1322 #59F
C #59F !
59
1322 C #
59
1F ! 322
2. Sometimes true; only whenthe expression you aredividing by does not equalzero.
4.
6. Sample answer: 9 times anumber decreased by 3 is 6.
8. Reflexive Property ofEquality
10. !21
12. !4
14. 1.5
16.
18. D
20.
22.
24.
26. 1n ! 72321n $ 82!6n3
10n $ 7
y #9 $ 2n
4
5 $ 4n
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©Glencoe/McGraw-Hill 8 Algebra 2 Chapter 1
27.
29. Sample answer: 5 less thana number is 12.
31. Sample answer: A numbersquared is equal to 4 timesthe number.
33. Sample answer: A numberdivided by 4 is equal to twicethe sum of that number and 1.
35. Substitution Property (#)
37. Transitive Property (#)
39. Symmetric Property (#)
41. 7
43. 3.2
45.
47.
49.
51. 1
53.
55.
57.
59.
61.
63. n # number of games;2(1.50) $ n(2.50) # 16.75; 5
65. x # cost of gasoline per mile;972 $ 114 $ 105 $ 7600x #1837; 8.5¢/mi
b #x 1c ! 32
a$ 2
3V&r 2 # h
dt
# r
!552
14
!7
!8
112
2&rh $ 2&r 2 28.
30. Sample answer: Twice anumber plus 3 is !1.
32. Sample answer: Three timesthe cube of a number isequal to the number plus 4.
34. Sample answer: 7 minus halfa number is equal to 3divided by the square of x.
36. Subtraction Property (#)
38. Addition Property (#)
40. Multiplication Property (#)
42. 8
44. 2.5
46.
48.
50.
52.
54. 19
56.
58.
60.
62.
64. s # length of a side; 8s #124, 15.5 in.
66. n # number of students thatcan attend each meeting;2n $ 3 # 83; 40 students
4x1 ! x
# y
2Ah
! a # b
a #!b2x
1017
!12
23
!11
!34
2&r 1h $ r 2
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©Glencoe/McGraw-Hill 9 Algebra 2 Chapter 1
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
69. n # number of lamps broken;12(125) ! 45n # 1365;3 lamps
71. 15.1 mi/month
73. The Central Pacific had tolay their track through theRocky Mountains, while theUnion Pacific mainly builttrack over flat prairie.
75. the product of 3 and thedifference of a number and 5added to the product of fourtimes the number and thesum of the number and 1
68. c # cost per student;
70. h # height of can A;
8 units
72. Central: 690 mi.; Union:1085 mi
74. $295
76. To find the most effectivelevel of intensity for yourworkout, you need to useyour age and 10-secondpulse count. You must alsobe able to solve the formulagiven for A. Answers shouldinclude the following.• Substitute 0.80 for I and
27 for P in the formulaand
solve for A. To solve thisequation, divide theproduct of 6 and 28 by 0.8.Then subtract 220 anddivide by The result is17.5. This means that thisperson is 17 years old.1
2
!1.
I # 6 % P " 1220 ! A2
13
&11.222h # &12223;
1800; $31452 #
505
50130 ! c2 $
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©Glencoe/McGraw-Hill 10 Algebra 2 Chapter 1
77. B
79.
81. 6.6
83. 105 cm2
85. 3
87.
89. !5 $ 6y
!14
!6x $ 8y $ 4z
• To find the intensity level fordifferent values of A and Pwould require solving a newequation but using thesame steps as describedabove. Solving for A wouldmean that for futurecalculations of A you wouldonly need to simplify anexpression, 220 ! , ratherthan solve an equation.
78. D
80.
82. 7.44
84.
86.
88. 3x
!2.5
!5
11a $ 8b
6PI
Lesson 1-4 Solving Absolute Value EquationsPages 30–32
1. when a is anegative number and thenegative of a negativenumber is positive.
3. Always; since the opposite of0 is still 0, this equation hasonly one case,
The solution is
5. 8
7.
9.
11. 5!32, 3665!18, !126!17
!ba
.
ax $ b # 0.
0a 0 # !a 2a.2b.
4. Sample answer:
6. 9
8.
10.
12. '
5!11, 2965!21, 136
04 ! 6 0 ; 2
0x ! 6 0 # 20x 0 # 4
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©Glencoe/McGraw-Hill 11 Algebra 2 Chapter 1
13.
15. least: 158(F; greatest: 162(F
17. 15
19. 0
21. 3
23.
25.
27. 55
29. {8, 42}
31.
33.
35.
37.
39. '
41. {!5, 11}
43.
45. {8}
47. maximum:205(F; minimum: 195(F
49. maximum:18 km, minimum: 8 km
51. sometimes; true only if c ) 0
0x ! 13 0 # 5;
0x ! 200 0 # 5;
e!113
, !3f
e2, 92f
e32f
5!2, 1665!45, 216
!9.4
!4
586 14.
16. 162(F; This would ensure aminimum internaltemperature of 160(F.
18. 24
20. 4
22. 13
24.
26. 5
28.
30.
32.
34.
36. '
38.
40. '
42. {3, 15}
44.
46.
48. heaviest:16.3 oz, lightest: 15.7 oz
50. sometimes; true only if and or if and
52. Answers should include thefollowing.• This equation needs to
show that the difference ofthe estimate E from theoriginally stated magnitudeof 6.1 could be plus 0.3 or
b * 0a * 0b ) 0
a ) 0
0x ! 16 0 # 0.3;
5!46e3,
53f
5!4, !16
e2, !163f
5!28, 206512, !306!22
!7.8
0x ! 160 0 # 2
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©Glencoe/McGraw-Hill 12 Algebra 2 Chapter 1
53. B
55.
57.
59.
61.
63. 14
65. Distributive Property
67. Additive Identity
69. true
71. false; 1.2
73. 364 ft2
75. 8
77.
79. !34
23
163
21n ! 1125!1.56
0x $ 1 0 $ 2 # !1x $ 420x $ 1 0 $ 2 # x $ 4;
minus 0.3, as shown inthe graph below. Insteadof writing two equations,
andabsolute
value symbols can beused to account for bothpossibilities,
• Using an originalmagnitude of 5.9, theequation to represent theestimated extremes wouldbe
54. A
56. x $ 1 $ 2 # x $ 4;!x ! 1 $ 2 # x $ 4;x $ 1 $ 2 # !x ! 4;!x ! 1 $ 2 # !x ! 4
58. 8
60.
62.
64. Commutative Property ($)
66. Multiplicative Inverse
68. false;
70. true
72.
74. 2
76.
78. 6
!2
12
1x $ 32 1x $ 52
23
!2
5n2
0E ! 5.9 0 # 0.3.
6.76.66.56.46.36.26.16.05.95.85.75.6
0.3 units 0.3 units
0E ! 6.1 0 # 0.3.
E ! 6.1 # !0.3,E ! 6.1 # 0.3
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©Glencoe/McGraw-Hill 13 Algebra 2 Chapter 1
1. Dividing by a number is thesame as multiplying by itsinverse.
3. Sample answer:
5.
7.
9.
11. all real numbers or
13.
15.
17.
19.
21.
!6 !5 !4 !3 !2!7
5k 0k ) !3.56 or 3!3.5, $+224 2620 22 28 30
5g 0g * 276 or 1!+, 27 43 4 6 72 5 8 9 10!1 10
5x 0 x , 76 or 1!+, 72!14 !12 !10 !6!8 !4
5n 0n ) !116 or 3!11, $ +22n ! 3 * 5; n * 4
!6 !4 !2 20 4
1!+, $+216 191514 181710 1398 1211
5p 0p - 156 or 115, $+2!1 0 1 3 4 6 72 5 8 9 10
5y 0y - 66 or 16, $+20 1 2 3
ex `x *53f or a!+,
53d f
x $ 2 , x $ 1
2. Sample answer:
4.
6.
8.
10.
12.
14. at least 92
16.
18.
20.
22.
2 4 6!4 !2 0
5y 0y , 56 or 1!+, 52!6 !4 !2 0 2 4
5p 0p * !36 or 1!+, !3 4!6 !4 !2!10 !8 0
5d 0d - !86 or 1!8, $+211 13 14 16 1712 15 18 19 20 2110
5b 0b * 186 or 1!+, 18 4
12n - 36; n - 3
!30 !28 !26 !24 !22 !20
5n 0n * !246 or 1!+, !24 4!6 !4 !2!10 !8 0
5w 0w , !76 or 1!+, !72!2 0 4 62 8
5c 0c ) 36 or 33, $+2!2 !1 0 21 3
5a 0a , 1.56 or 1!+, 1.52
!2n - !6
Lesson 1-5 Solving InequalitiesPages 37–39
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©Glencoe/McGraw-Hill 14 Algebra 2 Chapter 1
23.
25.
27.
29.
31.
33.
35.
37. '
39. at least 25 h
41.
43.
45.
47. at least 2 child-care staff members
1714
, 217m2 ) 17; m )
21n $ 52 * 3n $ 11; n ) !1
12n ! 7 ) 5; n ) 24
n $ 8 - 2; n - !6
2 4!6 !4 !2 0
135
15
35
15
ey `y ,15f or a!+,
15b
!6 !4 !2 0 2 4
5g 0g , 26 or 1!+, 22!6 !4 !2!8 0 2
5d 0d ) !56 or 3!5, $+2!286 !284 !282 !280 !278 !276
5x 0x , !2796 or 1!+, !27920 0.5 1 1.5 2 2.5
5n 0n ) 1.756 or 31.75, $+2!4 !2 0 2 4 6
5t 0 t * 06 or 1!+, 0 42 4!6 !4 !2 0
5m 0m - !46 or 1!4, $+2 24.
26.
28. or
30.
32.
34.
36.
38.
40. no more than 14 rides
42.
44.
46.
48.40,000$24,000 $ 0.015130,500n2 )
n ! 9 *n2; n * 18
!3n $ 1 , 16; n - !5
!4n ) 35; n * !8.75
!4 !3 !2 !1 10
en `n * !32f or a!+, !
32d
2 4 6!4 !2 0!6
5p 0p - 06 or 10, $+20 1 11
7107
97
87
67
57
47
37
27
17
ea `a )57f or c5
7, $+b
2.0 2.2 2.4 2.6 2.8 3.0
5z 0z - 2.66 or 12.6, $ +2!20 !18 !16 !14 !12 !10
5c 0c - !186 or 1!18, $+2720! 3
20! 120! 1
20320
14!
a! 120
, $+bew `w - !120f
2 4 6 8!2 0
5r 0 r * 66 or 1!+, 6 4!1 1 20
eb `b )23f or c2
3 $+b
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©Glencoe/McGraw-Hill 15 Algebra 2 Chapter 1
49. She must sell atleast 35 cars.
51. Ahmik must score atleast 91 on her next test tohave an A test average.
53. Answers should include thefollowing.•• Let n equal the number of
minutes used. Write anexpression representing thecost of Plan 1 and for Plan2 for n minutes. The costfor Plan 1 would include a$35 monthly access feeplus 40¢ for each minuteover 150 minutes or
Thecost for Plan 2 for 400minutes or less would be$55. To find where Plan 2would cost less than Plan1 solve
for n. Thesolution set is which means that for morethan 200 minutes of calls,Plan 2 is cheaper.
55. D
57.
59.
61. '
63. N, W, Z, Q, R
5!14, 206x ) !2
5n 0n - 2006,0.41n ! 150255 , 35 $
35 $ 0.41n ! 1502.
150 , 400
s ) 91;
n ) 34.97; 50.
52a. It holds only for 52b.52c. For all real numbers a, b,
and c, if and then
54. D
56.
58.
60.
62. b # online browsers eachyear; 6b $ 19.2 # 106.6;about 14.6 million onlinebrowsers each year
64. Q, R
e!54,
114f
x ) !1
x - !3
a , c.b , ca , b
1 , 2 but 2 . 12 ! 2.* or );
) 9085 $ 91 $ 89 $ 94 $ s5
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©Glencoe/McGraw-Hill 16 Algebra 2 Chapter 1
65. I, R
67.
69.
71. 5!11, !16e4, !
45f
5!7, 7666. 4.25(5.5 $ 8); 4.25(5.5) $
4.25(8)
68.
70.
72. 5!18, 106511, 256513, !236
Chapter 1Practice Quiz 2
Page 39
1. 0.5
3. 14
5.
10 89
23
29! 4
929
em `m -49f or a4
9, $ +b
2.
4. e!193
, 5f
2st 2 # g
Lesson 1-6 Solving Compound and Absolute Value Inequalities
Pages 43–46
1.
3. Sabrina; an absolute valueinequality of the form should be rewritten as an orcompound inequality, or
5.
7. 0n 0 , 2
42!6 !4 !2 0
0n 0 - 3
a , b.a - b
0a 0 - b
5 * c * 15 2. Sample answer: and
4.
6.
8. or
4 62!4 !2 0
y , !165y 0y - 4
0n 0 ) 4
!4!12 !8 0 4 8
0n 0 , 8
x - 2x , !3
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©Glencoe/McGraw-Hill 17 Algebra 2 Chapter 1
9.
11.
13. all real numbers
15.
17.
19.
21.
23.
25.
27. or
29.
31.
33.
35. '
4 62!4 !2 0
8 124!8 !4 0
5g 0!9 * g * 96!10 !8 !6 !4 !2 0
5f 0!7 , f , !564 62!4 !2 0
5x 0!2 , x , 468 124!8 !4 0
p ) 865p 0 p * 2
0n $ 1 0 - 1
0n 0 ) 1.5
0n 0 - 1
8 124!8 !4 0
0n 0 - 8
4 62!4 !2 0
0n 0 , 4
8 124!8 !4 0
0n 0 ) 5
4 62!4 !2 0
4!8!12!16 !4 0
5g 0!13 * g * 564 62!4 !2 0
5d 0!2 , d , 36 10. or
12.
14. 55 * * 60; 343.75 * c *
375; between $343.75 and$375
16.
18.
20.
22.
24.
26.
28.
30. or
32. all real numbers
34. or
36.
8 124!8 !4 0
5y 0!7 , y , 764 62!4 !2 0
m * !465m 0m ) 4
4 62!4 !2 0
4 62!4 !2 0
c ) 165c 0c , !2
4 6 82!2 0
5t 01 , t , 360n ! 1 0 * 3
0n 0 , 6
0n 0 * 5
!1.4 !1.2 0 1.2 1.4 1.6
0n 0 * 1.2
8 124!8 !4 0
0n 0 ) 6
8 124!8 !4 0
0n 0 , 7
c6.25
42!2 860
5k 0!3 , k , 764 8 12!4!8 0
a * !565a 0a ) 5
PQ245-6457F-PO1[001-019].qxd 7/24/02 12:19 PM Page 17 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-01:
©Glencoe/McGraw-Hill 18 Algebra 2 Chapter 1
37. or
39.
41. all real numbers
43.
45.
47.
49.
51.
53a.
53b.
53c.
53d. can berewritten as and
The solution ofis or
The solution ofis 6.
Therefore, the union of thesetwo sets is or x , !521x - 1
!10 * x *0x $ 2 0 ) 8x , !5.
x - 10x $ 2 0 - 30x $ 2 0 * 8.
0x $ 2 0 - 33 , 0x $ 2 0 * 8
4 62!4 !2 0
4 62!4 !2 0
4 62!4 !2 0
b $ c - aa $ b - c, a $ c - b,
108 in. , L $ D * 130 in.
45 * s * 55
6.8 , x , 7.4
32 540 1
en `n #72f
4 62!4 !2 0
10!2 !1
ew `!73
* w * 1f8 12 164!4 0
b , !265b 0b - 10 38.
40. '
42.
44.
46.
48.or
50. .
52.
54. Compound inequalities canbe used to describe theacceptable time frame for thefasting state before a glucosetolerance test is administeredto a patient suspected ofhaving diabetes. Answersshould include the following.• Use the word and when
both inequalities must besatisfied. Use the word orwhen only one or theother of the inequalitiesmust be satisfied.
• 10 * h * 16
a ! b , c , a $ b
84 in. , L * 106 in
b , 90.660 t ! 98.6 0 ) 8; 5b 0b - 106.6
45 * s * 65
32!2 !1 0 1
5n 0n - 1.564 62!4 !2 0
5n 0n ) 064 62!4 !2 0
4 62!4 !2 0
5r 0!3 , r , 46
PQ245-6457F-PO1[001-019].qxd 7/24/02 12:19 PM Page 18 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-01:
©Glencoe/McGraw-Hill 19 Algebra 2 Chapter 1
and 6). The unionof the graph of or
and the graph ofis shown
below. From this we can seethat solution can be rewrittenas or
55. or
57.
59. oror
61. or
63. or
65. {!10, 16}
67. '
69. Symmetric Property (#)
71.
73. 2
75. !7
3a $ 7b
4 62!4 !2 0
1!+, !12n , !1
2!4!6!8 !2 0
3!6, !+2d ) !6
x * !1615x $ 2 * !32; 5x 0x ) 0.215x $ 2 ) 32
x , !6x - !5
!4!12 !8 0 4 8
11 , x * 62.1!10 * x , !52
!10 * x * 6x , !5
x - 11!10 * x * • 12 hours would be an
acceptable fasting state forthis test since it is part ofthe solution set of
as indicated on thegraph below.
56. D
58.
60.or
62. or
64. highest: 592keys, lowest: 582 keys
66.
68. Addition Property of Equality
70. Transitive Property of Equality
72.
74. 92
!2m ! 7n ! 18
5!11, 46
0x ! 587 0 # 5;
4 62!4 !2 0
1!+, 42x , 4
x - 865x " x , !2abs12x ! 62 - 10;
2 , x , 3
1514 16 17 18 1998 10 11 12 13
* 16,10 * h
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©Glencoe/McGraw-Hill 20 Algebra 2 Chapter 2
1. Sample answer: {(!4, 3),(!2, 3), (1, 5), (!2, 1)}
3. Molly; to find g(2a), replace x with 2a. Teisha found 2g(a), not g(2a).
5. yes
7. D " {7}, R " {!1, 2, 5, 8}, no
9. D " all reals, R " all reals, yes
11. 10
y
O x
y ! "2x # 1
(7, 2)
(7, "1)
(7, 5)
(7, 8)y
O x
2. Sample answer:
4. yes
6. no
8. D " {3, 4, 6}, R " {2.5}, yes
10.R " all reals, no
12. !7
x ! y 2
xO
y
D " 5x 0 x # 06,
(6, 2.5)(3, 2.5)
(4, 2.5)
y
O x
y
xO
Chapter 2 Linear Relations and FunctionsLesson 2-1 Relations and Functions
Pages 60–62
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©Glencoe/McGraw-Hill 21 Algebra 2 Chapter 2
13. D " {70, 72, 88}, R " {95, 97, 105, 114}
15.
17. yes
19. no
21. yes
23. D " {!3, 1, 2}, R " {0, 1, 5}; yes
25. D " {!2, 3}, R " {5, 7, 8}; noy
O x
(3, 7)
("2, 5)
("2, 8)
y
O x
(2, 1)
(1, 5)
("3, 0)
July
95
100
105
110
115
January700 80 90
Record High Temperatures
14. {(88, 97), (70, 114), (88, 95),(72, 105)}
16. No; the domain value 88 ispaired with two range values.
18. no
20. yes
22. no
24. D " {3, 4, 6}, R " {5}; yes
26. D " {3, 4, 5, 6}, R " {3, 4, 5, 6}; yes
y
O x
(4, 3)
(3, 4)(6, 5)
(5, 6)
y
O x
(3, 5) (4, 5)
(6, 5)
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©Glencoe/McGraw-Hill 22 Algebra 2 Chapter 2
27. D " {!3.6, 0, 1.4, 2}, R " {!3, !1.1, 2, 8}; yes
29. D " all reals, R " all reals; yes
31. D " all reals, R " all reals; yes
y
O x
y ! 3x " 4
y
O x
y ! "5x
y
O x
(1.4, 2)
(2, "3)(0, "1.1)
("3.6, 8)
28. D " {!2.5, !1, 0}, R " {!1, 1}; no
30. D " all reals, R " all reals; yes
32. D " all reals, R " all reals; yes
y
O x
y ! 7x " 6
y
O x
y ! 3x
y
O x
("1, "1)
("2.5, 1) ("1, 1)(0, 1)
PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 22 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
©Glencoe/McGraw-Hill 23 Algebra 2 Chapter 2
33. D " all reals, yes
35.
37. No; the domain value 56 ispaired with two differentrange values.
39.
41. Yes; each domain value ispaired with only one rangevalue.
Pri
ce (
$)
70
60
50
40
30
20
10
0
Year19981996 2000 2002 2004
Stock Price
RB
I
170
165
160
155
150
145
140
HR480 50 52 54 56
American League Leaders
y
O x
y ! x 2
R " 5y 0 y # 06, 34.R " all reals; no
36. D " {47, 48, 52, 56}, R " {145, 147, 148, 157, 165}
38. {(1997, 39), (1998, 43),(1999, 48), (2000, 55),(2001, 61), (2002, 52)}
40. D " {1997, 1998, 1999, 2000, 2001, 2002},
R " {39, 43, 48, 52, 55, 61}
42. {(1987, 12), (1989, 13),(1991, 11), (1993, 12), (1995, 9), (1997, 6), (1999, 3)}
y
O x
x ! 2y 2 " 3
D " 5x 0 x # !36,
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©Glencoe/McGraw-Hill 24 Algebra 2 Chapter 2
44. D " {1987, 1989, 1991,1993, 1995, 1997, 1999},
R " {3, 6, 9, 11, 12, 13}
46. !14
48.
50. 3a ! 5
52. !4
54. 39
56. Relations and functions canbe used to representbiological data. Answersshould include the following.• If the data are written as
ordered pairs, then thoseordered pairs are arelation.
• The maximum lifetime ofan animal is not a functionof its average lifetime.
58. C
60. continuous
62. continuous
64.
66. $2.85
5m 0 4 $ m $ 66
!29
43.
45. Yes; no; each domain valueis paired with only one rangevalue so the relation is afunction, but the range value12 is paired with two domainvalues so the function is notone-to-one.
47. 6
49. !3
51. 25n2 ! 5n
53. 11
55. f(x) " 4x ! 3
57. B
59. discrete
61. discrete
63.
65. 5x 0 x $ 5.165y 0!8 $ y $ 66
Rep
rese
nta
tives
14
12
10
8
6
4
2
0
Year’87 ’91 ’95 ’99
30+ Years of Service
PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 24 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
©Glencoe/McGraw-Hill 25 Algebra 2 Chapter 2
1. The function can be written asf(x) " so it is of the form f(x) " mx % b, where m " and b " 1.
3. Sample answer: x % y " 2
5. yes
7. 2x ! 5y " 3; 2, !5, 3
9.
11. 2, 3y
O x
3x # 2y ! 6
y
O x
y ! "3x " 5
!53, !5
12
x % 1,12
2. 5, !2
4. No, the variables have anexponent other than 1.
6. 3x ! y " 5; 3, !1, 5
8. 2x ! 3y " !3; 2, !3, !3
10. 2, !2
12.y
Ox
4x # 8y ! 12
3, 32
y
O x
x " y " 2 ! 0
Lesson 2-2 Linear EquationsPages 65–67
67. $29.82
69. 31a % 10b
71. 2
73. 15
68. 43
70. !1
72. 6
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©Glencoe/McGraw-Hill 26 Algebra 2 Chapter 2
13. $177.62
15. yes
17. No; y is inside a square root.
19. No; x appears in adenominator.
21. No; x has an exponent otherthan 1.
23. x2 % 5y " 0
25. 7200 m
27. 3x % y " 4; 3, 1, 4
29. x ! 4y " !5; 1, !4, !5
31. 2x ! y " 5; 2, !1, 5
33. x % y " 12; 1, 1, 12
35. x " 6; 1, 0, 6
37. 25x % 2y " 9; 25, 2, 9
39. 3, 5y
Ox
5x # 3y ! 15
14. 563.00 euros
16. No; x appears in adenominator.
18. No; x has exponents otherthan 1.
20. yes
22. No; x is inside a square root.
24. h(x ) " x3 ! x2 % 3x
26. Sound travels only 1715 m in5 seconds in air, so it travelsfaster underwater.
28. 12x ! y " 0; 12, !1, 0
30. x ! 7y " 2; 1, !7, 2
32. x ! 2y " !3; 1, !2, !3
34. x ! y " !6; 1, !1, !6
36. y " 40; 0, 1, 40
38. 5x ! 4y " 2; 5, !4, 2
40. 6, !2y
O x
2x " 6y ! 12
PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 26 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
©Glencoe/McGraw-Hill 27 Algebra 2 Chapter 2
41.
43. 0, 0
45. none, !2
47. 8, none
x ! 8
2 4"4"6"8 6
8642
"2"4"6"8
"2x
y
O
y
O x
y ! "2
y
Ox
y ! x
y
O x
3x " 4y " 10 ! 0
103
, !52
42. 5, 2
44.
46. none, 4
48. 1, noney
Ox
x ! 1
y
O x
y ! 4
y
O x
y ! 4x " 2
12, !2
y
O x
2x # 5y " 10 ! 0
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©Glencoe/McGraw-Hill 28 Algebra 2 Chapter 2
49.
51.
The lines are parallel buthave different y-intercepts.
53. 90&C
55.
57. c
100 200 400
35030025020015010050
b0
1.75b # 1.5c ! 525
T(d )
O1 2"2"3"4 3 4
1601208040
"40"80
"120"160
d
T(d ) ! 35d # 20
y
Ox
x # y ! 5
x # y ! "5
x # y ! 0
f (x)
O x
f (x) ! 4x " 1
14, !1 50. 6, !3
52. Sample answer: x % y " 2
54. 4 km
56. 1.75b % 1.5c " 525
58. Yes; the graph passes thevertical line test.
g(x)
O x
g(x) ! 0.5x " 3
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©Glencoe/McGraw-Hill 29 Algebra 2 Chapter 2
59. no
61. A linear equation can beused to relate the amountsof time that a student spendson each of two subjects ifthe total amount of time isfixed. Answers should includethe following.• x and y must be
nonnegative becauseLolita cannot spend anegative amount of timestudying a subject.
• The intercepts representLolita spending all of hertime on one subject. Thex-intercept represents herspending all of her time onmath, and the y-interceptrepresents her spendingall of her time on chemistry.
63. B
65. D " {0, 1, 2}, R " {!1, 0, 2, 3}; no
y
O x(1, 0)
(1, 3)(0, 2)
(2, "1)
60. units2
62. B
64. D " {!1, 1, 2, 4}, R " {!4, 3, 5}; yes
66. 5x 0 !1 $ x $ 26
y
O x
(1, 3)("1, 5)
(2, "4)
(4, 3)
212
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©Glencoe/McGraw-Hill 30 Algebra 2 Chapter 2
2. Sometimes; the slope of avertical line is undefined.
4. 0
6. 1
8. y
O x
1. Sample answer: y " 1
3. Luisa; Mark did not subtractin a consistent manner whenusing the slope formula. If y2 " 5 and y1 " 4, then x2must be !1 and x1 must be2, not vice versa.
5.
7. y
O x
!12
67.
69. 3s % 14
71.
73. 2
75. !5
77. 0.4
13
5x 0 x $ !6 or x ' !26 68. $7.95
70. 4
72.
74.
76.
78. !0.8
415
!32
!14
Lesson 2-3 SlopePages 71–74
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©Glencoe/McGraw-Hill 31 Algebra 2 Chapter 2
9.
11.
13. 1.25&/hr
15.
17.
19. 0
21. 8
23. !4
25. undefined
27. 1
29. about 0.6
31. y
O x
35
!52
y
O x
y
O x
10.
12. 5.5&/hr
14. 2:00 P.M.–4:00 P.M.
16. 13
18. 4
20. !1
22. undefined
24.
26. 0
28. 9
30. about 1.3
32. y
O x
!54
y
O x
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©Glencoe/McGraw-Hill 32 Algebra 2 Chapter 2
33.
35.
37. about 68 million per year
39. The number of cassettetapes shipped has beendecreasing.
41. 45 mph
43.
45. y
O x
y
O x
y
O x
y
O x
34.
36.
38. about !32 million per year
40. 55 mph
42. speed or velocity
44.
46. y
O x
y
O x
y
O x
y
O x
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©Glencoe/McGraw-Hill 33 Algebra 2 Chapter 2
47.
49.
51. Yes; slopes show thatadjacent sides areperpendicular.
53. The grade or steepness of aroad can be interpretedmathematically as a slope.Answers should include thefollowing.• Think of the diagram at the
beginning of the lesson asbeing in a coordinateplane. Then the rise is achange in y-coordinatesand the horizontal distance is a change in x-coordinates. Thus, thegrade is a slope expressedas a percent.
y
Ox
y
O x
48.
50.
52. !1
54. D
y
O x
y
O x
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©Glencoe/McGraw-Hill 34 Algebra 2 Chapter 2
•
55. D
57. The graphs have the same y-intercept. As the slopesbecome more negative, thelines get steeper.
59.
61. !7
63.
65. 5x 0 !1 $ x $ 36!
52
y
O x
4x " 3y # 8 ! 0
!2, 83
y
O
xy ! 0.08x
56. The graphs have the same y-intercept. As the slopesincrease, the lines getsteeper.
58. !10, 4
60. 0, 0
62. 5
64. 3a !4
66. 5z 0 z # 7356
y
Ox
y ! 7x
y
O2 4"4"6"8"10
8642
"2"4"6"8
"2 x
"2x # 5y ! 20
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©Glencoe/McGraw-Hill 35 Algebra 2 Chapter 2
1. D " {!7, !3, 0, 2}, R " {!2, 1, 2, 4, 5}
3. 6x % y " 4
5. y
O x
2. 375
4. 10, 6y
Ox
3x # 5y ! 30
Chapter 2Practice Quiz 1
Page 74
67. at least 8
69. 9
71. y " !4x % 2
73.
75. y " !23x %
113
y "52x !
12
68. 17a !b
70. y " 9 ! x
72. y " !3x % 7
74. y "35x %
45
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©Glencoe/McGraw-Hill 36 Algebra 2 Chapter 2
Lesson 2-4 Writing Linear EquationsPages 78–80
1. Sample answer: y " 3x % 2
3. Solve the equation for y to getThe slope of this
line is The slope of a parallel line is the same.
5.
7.
9.
11.
13.
15.
17. undefined, none
19. y " 0.8x
21. y " !4
23. y " 3x ! 6
25.
27. y " !0.5x ! 2
29.
31. y " 0
33. y " x % 4
y " !45x %
175
y " !12x %
72
12, !
52
!23, !4
y "54x % 7
y " !35x %
165
y " !34x % 2
!32, 5
35
.
"35
x !25
.y
2. 6, 0
4. 2, !5
6. y " 0.5x % 1
8.
10. y " !x ! 2
12. B
14.
16.
18. !c, d
20.
22. y " 2
24. y " 0.25x % 4
26.
28. y " 4x
30. no slope-intercept form forx " 7
32.
34. y "34x !
14
y "32x
y "32x %
172
y " !53x %
293
!35, 6
34, 0
y " !52x % 16
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©Glencoe/McGraw-Hill 37 Algebra 2 Chapter 2
35.
37.
39. y " 3x ! 2
41. d " 180c ! 360
43. 540&
45. 10 mi
47. 68&F
49. y " 0.35x % 1.25
51. y " 2x % 4
53. C
55.
57. !2
x52
!y5
" 1
y " !115
x !235
y "23x %
103
36. y " !4x % 3
38. y " !x ! 4
40. y " !2x % 6
42. 180, !360
44. y " 75x % 6000
46.
48. !40&
50. $11.75
52. A linear equation cansometimes be used to relatea company’s cost to thenumber they produce of aproduct. Answers shouldinclude the following.• The y-intercept, 5400, is
the cost the companymust pay if they produce0 units, so it is the fixedcost. The slope, 1.37,means that it costs $1.37to produce each unit. Thevariable cost is 1.37x.
• $6770
54. A
56.
58. 3
52, !5
y
xO10 20"30 30
806040
"40"20
"10
y ! x # 3295
y "95x % 32
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©Glencoe/McGraw-Hill 38 Algebra 2 Chapter 2
Lesson 2-5 Modeling Real-World Data:Using Scatter Plots
Pages 83–86
1. d
3. Sample answer using(4, 130.0) and (6, 140.0):y " 5x % 110
5a. Cable Television
100
203040
Hou
seho
lds
(mill
ions
)
50607080
Year’88 ’90 ’92 ’94 ’96 ’98 ’00
2. D " {!1, 1, 2, 4}, R " {0, 2, 3}; Sampleanswer using (!1, 0) and (2, 2): 4
4a.
4b. Sample answer using (2000, 11.0) and (3000, 9.1):y " !0.0019x % 14.8
4c. Sample answer: 5.3&C
6a. Lives Saved byMinimum Drinking Age
50
101520
Live
s (t
hous
ands
) 25
Year’94 ’95 ’96 ’97 ’98 ’99 ’00
Atmospheric Temperature
2
0
468
Tem
pera
ture
(˚C
)
10121416
Altitude (ft)1000 2000 3000 4000 5000
59. 0
61. (
63.
65. 6.5
67. 5.85
5r 0 r # 6660. 0.55 s
62.
64. 3
66. 323.5
5x 0 x # !66
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©Glencoe/McGraw-Hill 39 Algebra 2 Chapter 2
5b. Sample answer using (1992, 57) and (1998, 67):y " 1.67x ! 3269.64
5c. Sample answer: about 87 million
7a.
7b. Sample answer using (4, 5)and (32, 37):y " 1.14x % 0.44
7c. Sample answer: about 13
9a.
9b. Sample answer using(1, 499) and (3, 588):y " 44.5x % 454.5, where xis the number of seasonssince 1995–1996
9c. Sample answer: about$1078 million or $1.1 billion
11. Sample answer: $1091
BroadwayPlay Revenue
100
0
200300400
Reve
nue
($ m
illio
ns)
500600700
Seasons Since ’95–’961 2 3 4
2000–2001Detroit Red Wings
10
0
203040
Ass
ists
5060
Goals10 20 30 40
6b. Sample answer using (1996, 16.5) and (1998, 18.2):y " 0.85x ! 1680.1
6c. Sample answer: 28,400
8a.
8b. Sample answer using (1993,9.4) and (1996, 12.5):y " 1.03x ! 2043.39
8c. Sample answer: about 26.9 gal
10. Sample answer using(1990, 563) and (1995, 739):y " 35.2x ! 69,485
12. The value predicted by theequation is somewhat lowerthan the one given in thegraph.
Bottled Water Consumption
20
468
Gal
lons
101214
Year’91 ’93 ’95 ’97 ’99
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©Glencoe/McGraw-Hill 40 Algebra 2 Chapter 2
13. Sample answer: Using thedata for August andNovember, a predictionequation for Company 1 isy " !0.86x % 25.13, wherex is the number of monthssince August. The negativeslope suggests that the valueof Company 1’s stock isgoing down. Using the datafor October and November, aprediction equation forCompany 2 is y " 0.38x %31.3, where x is the numberof months since August. Thepositive slope suggests thatthe value of Company 2’sstock is going up. Since thevalue of Company 1’s stockappears to be going down,and the value of Company2’s stock appears to be goingup, Della should buyCompany 2.
15.
17. Sample answer: about 23 in.
World Cities
5
0
101520
Prec
ipit
atio
n (in
.)
25303540
Elevation (ft)200 400 600
14. No. Past performance is noguarantee of the futureperformance of a stock.Other factors that should beconsidered include thecompanies’ earnings dataand how much debt theyhave.
16. Sample answer using(213, 26) and (298, 23):y " !0.04x ! 34.52
18. Sample answer: Thepredicted value differs fromthe actual value by morethan 20%, possibly becauseno line fits the data very well.
PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 40 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
©Glencoe/McGraw-Hill 41 Algebra 2 Chapter 2
19. Sample answer using (1975,62.5) and (1995, 81.7): 96.1%
21. See students’ work.
23. D
25. 1988, 1993, 1998; 247,360.5, 461
27. 354
29. y " 21.4x ! 42,294.03
31. y " 4x % 6
33. 3
20. Sample answer: Thepredicted percent is almostcertainly too high. Since thepercent cannot exceed100%, it cannot continue toincrease indefinitely at alinear rate.
22. Data can be used to write alinear equation thatapproximates the number ofCalories burned per hour interms of the speed that aperson runs. Answers shouldinclude the following.•
• Sample answer using(5, 508) and (8, 858):y " 116.67x ! 75.35
• about 975 calories;Sample answer: Thepredicted value differsfrom the actual value byonly about 2%.
24. A
26. y " 21.4x ! 42,296.2
28. about (1993, 356.17)
30. about 613, about 720
32.
34. 7
y " !37x !
67
Calories BurnedWhile Running
200
0
400600800
Calo
ries
1000
Speed (mph)5 6 7 8 9
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©Glencoe/McGraw-Hill 42 Algebra 2 Chapter 2
35.
37.
39. 11
41. 23
5x 0 x $ !7 or x ' !16293
36.
38. 3
40. 0
42. 1.5
373
Lesson 2-6 Special FunctionsPages 92–95
1. Sample answer: [[1.9]] " 1
3. Sample answer: f(x) " 0x ! 105. S
7. D " all reals, R " all integers
xO
g (x) ! !2x"
g(x)
2. !1
4. A
6. D " all reals, R " all integers
8. D " all reals, R " all nonnegative reals
xO
h (x) ! |x " 4|
h(x)
PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 42 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
©Glencoe/McGraw-Hill 43 Algebra 2 Chapter 2
9. D " all reals, R " all nonnegative reals
11. D " all reals, R " all reals
13.
15. C
17. S
19. A
21.
1
60 180 300
2345
xO
y
Time (hr)0
10. D " all reals,
12. step function
14. $6
16. A
18. S
20. P
22.
C C
0.10
1 2 3 4 5 6 7 8 9
0.200.300.400.500.600.700.800.901.00
Minutes
0
xO
g(x)R " 5y 0 y ) 26
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©Glencoe/McGraw-Hill 44 Algebra 2 Chapter 2
23. $1.00
25. D " all reals, R " all integers
27. D " all reals, R " {3a 0a is an integer.}
29. D " all reals, R " all integers
xO
f (x) ! !x" " 1
f(x)
x
h (x) ! "3!x"6
1"3"6"9
"12
"2"3"4 2 3 4
912
O
h(x)
"1
xO
g (x) ! !x " 2"
g(x)
24. D " all reals, R " all integers
26. D " all reals, R " all even integers
28. D " all reals, R " all integers
30. D " all reals, R " all nonnegative reals
xO
g (x) ! !x" # 3
g(x)
xO
f (x) ! 2!x"
f(x)
xO
f (x) ! !x # 3"
f(x)
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©Glencoe/McGraw-Hill 45 Algebra 2 Chapter 2
31. D " all reals, R " all nonnegative reals
33. D " all reals, R " {y 0 y # !4}
35. D " all reals, R " all nonnegative reals
37. D " all reals, R " all nonnegative reals
xOf (x) ! |x # |1
2
f(x)
xO
f (x) ! |x # 2|
f(x)
xO
g (x) ! |x | " 4
g(x)
xO
h (x) ! |"x |
h(x)
32. D " all reals, R " {y 0 y # 3}
34. D " all reals, R " all nonnegative reals
36. D " all reals, R " all nonnegative reals
38. D " all reals, R " {y 0 y # !3}
xO
f(x)
xOf (x) ! |x " |1
4
f(x)
xOh (x) ! |x # 3|
h(x)
xO
g (x) ! |x | # 3
g(x)
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©Glencoe/McGraw-Hill 46 Algebra 2 Chapter 2
39. D " {x 0 x $ !2 or x ' 2}, R " {!1, 1}
41. D " all reals, R " {y 0 y $ 2}
43. D " all reals, R " all nonnegative whole
numbers
45. f (x) " 0x ! 2 047.
xO
g (x) ! |!x"|
g(x)
xO
g(x)
xO
h(x)
40. D " all reals, R " {y 0 y ) 0 or y " 2}
42. D " all reals, R " all nonnegative whole
numbers
44.f(x)
46. {x 0 x # 0}
48. f (x) " e0 if 0 ) x ) 3000.8 (x ! 300) if x ' 300
" •2 if x $ !12x if !1 ) x ) 1!x if x ' 1
xO
f (x) ! !|x|"
f(x)
xO
f(x)
PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 46 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
©Glencoe/McGraw-Hill 47 Algebra 2 Chapter 2
49.
51. B
53.
55. Sample answer: 78.7 yr
Life Expectancy
7072747678
Years Since 1950
xO
y|x| # |y| ! 3
50. A step function can be usedto model the cost of a letterin terms of its weight.Answers should include thefollowing.• Since the cost of a letter
must be one of the values$0.34, $0.55, $0.76, $0.97,and so on, a step functionis the best model for thecost of mailing a letter. Thegas mileage of a car canbe any real number in aninterval of real numbers,so it cannot be modeled bya step function. In otherwords, gas mileage is acontinuous function of time.
•
52. D
54. Sample answer using (10, 69.7) and (47, 76.5):y " 0.18x % 67.9
56. y " 3x % 10
0.30
1 2 3 4 5 6 7
0.600.901.201.501.802.10
Weight (oz)
Cost
($)
0
PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 47 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
©Glencoe/McGraw-Hill 48 Algebra 2 Chapter 2
57. y " x ! 2
59.
61. no
63. yes
65. yes
3210"1"3 "2
ey `y '56f
58. {x 0 x # 3}
60. yes
62. no
64. no
654321"1 0
Chapter 2Practice Quiz 2
Page 95
1.
3. Sample answer using (66, 138) and (74, 178):y " 5x ! 192
5. D " all reals, R " nonnegative reals
xO
f (x) ! |x " 1|
f(x)
y " !23x %
113
2.
4. Sample answer: 168 Ib
Houston Comets
50
0
100150200250
Height (in.)65 70 75 80
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©Glencoe/McGraw-Hill 49 Algebra 2 Chapter 2
2. Substitute the coordinates ofa point not on the boundaryinto the inequality. If theinequality is satisfied, shadethe region containing thepoint. If the inequality is notsatisfied, shade the regionthat does not contain thepoint.
4.
6.
8.
xO
y
y ! |2x|
xO
y
x " y ! 0
xO
y
y ! 2
Lesson 2-7 Graphing InequalitiesPages 98–99
1. y ) !3x % 4
3. Sample answer: y # 0 x 0
5.
7.
xO
y
x " 2y ! 5
PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 49 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
©Glencoe/McGraw-Hill 50 Algebra 2 Chapter 2
10. 10c % 13d ) 40
12. No; (3, 2) is not in theshaded region.
14.
16.
18.
xO
y
y " 2 ! 3x
xO
y
3 ! x " 3y
9.
11.
13.
15.
17.
xO
y
y ! "4x # 3
xO
y
y ! 6x " 2
xO
y
x # y ! "5
cO
d
10c # 13d ! 40
xO
y
y ! 3|x| " 1
PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 50 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
©Glencoe/McGraw-Hill 51 Algebra 2 Chapter 2
19.
21.
23.
25.
xO
y
y ! |x|
xO
y
y ! x # 513
xO
y
4x " 5y " 10 ! 0
xO
y
y ! 1
20.
22.
24.
26.
xO
y
y ! |4x|
xO
y
y ! x " 512
xO
y
x " 6y # 3 ! 0
xO
y
y # 1 ! 4
PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 51 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
©Glencoe/McGraw-Hill 52 Algebra 2 Chapter 2
27.
29.
31. x $ !2
33.
50 150 250 350 xO
y
50
150
250
350
0.4x # 0.6y ! 90
x
y
O
x ! "2
xO
y
x # y ! 1
x # y ! "1
xO
y
y # |x| ! 3
28.
30.
32. y $ 3x ! 5
34. yes
xO
y
y ! 3x " 5
x
y
O
y ! "|x|
y ! |x|
xO
y
y ! |x " 1| " 2
PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 52 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
©Glencoe/McGraw-Hill 53 Algebra 2 Chapter 2
35. 4a % 3s # 2000
37. yes
39. yes
41. Linear inequalities can beused to track theperformance of players infantasy football leagues.Answers should include thefollowing.
36.
38. 1.2a %1.8b # 9000
40.
42. A
xO
y
|y| ! x
2000 4000 6000 8000a
O
b
2000
4000
6000 1.2a # 1.8b ! 9000
200 400 600 800a
O
s
200
400
600
800
4a # 3s ! 2000
PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 53 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
©Glencoe/McGraw-Hill 54 Algebra 2 Chapter 2
• Let x be the number ofreceiving yards and let ybe the number oftouchdowns. The numberof points Dana gets fromreceiving yards is 5x andthe number of points hegets from touchdowns is100y. His total number ofpoints is 5x % 100y. Hewants at least 1000 points,so the inequality 5x %100y # 1000 representsthe situation.
•
• the first one
43. B
45.
["10, 10] scl: 1 by ["10, 10] scl: 1
100"50 200 300 xO
y
2
4
6
8
10
12
5x # 100y ! 1000
44.
46.
["10, 10] scl: 1 by ["10, 10] scl: 1
["10, 10] scl: 1 by ["10, 10] scl: 1
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©Glencoe/McGraw-Hill 55 Algebra 2 Chapter 2
47.
49. D " all reals, R " {y 0 y # !1}
51.
53. Sample answer: $10,000
55. 3
Sales vs. Experience
2000
0
400060008000
Sale
s ($
)
10,000
Years1 2 3 4 5 6 7
xO
g (x) ! |x | " 1
g(x)
["10, 10] scl: 1 by ["10, 10] scl: 1
48. D " all reals, R " all integers
50. D " all reals, R " all nonnegative reals
52. Sample answer using (4, 6000) and (6, 8000):y " 1000x % 2000
54. 8
56. 12
xO h (x) ! |x " 3|
h(x)
xO
f (x)
f (x) ! !x" " 4
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1. Two lines cannot intersect inexactly two points.
3. A graph is used to estimatethe solution. To determinethat the point lies on bothlines, you must check that itsatisfies both equations.
5.
7. consistent and independent
x
y
O
y ! x " 4
y ! 6 # x
x
y
O
(2, 2)
2x " 3y ! 10
3x " 2y ! 10
2. Sample answer:
4.
6.
8. inconsistent
x
y
O
x " 2y ! 2
2x " 4y ! 8
x
y
O
(4, #3)6x " 9y ! #3
4x # 2y ! 22
x
y
O
(#2, 5)
y ! 2x " 9
y ! #x " 3
x ! y " 4, x # y " 2
Chapter 3 Systems of Equations and InequalitiesLesson 3-1 Solving Systems of Equations by Graphing
Pages 112–115
©Glencoe/McGraw-Hill 56 Algebra 2 Chapter 3
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9. consistent and dependent
11. The cost is $5.60 for bothstores to develop 30 prints.
13.
15.
17.
x
y
O
(5, 3)3x # 7y ! #6
x " 2y ! 11
x
y
O(4, 1)
x " 2y ! 6
2x " y ! 9
x
y
O
(1, #2)y ! 2x # 4
y ! #3x " 1
x
y
O
x # 2y ! 8
x # y ! 412
10. ,
12. You should use SpecialtyPhotos if you are developingless than 30 prints, and youshould use The Photo Lab ifyou are developing more than30 prints.
14.
16.
18.
x
y
O
(7, 6)
5x # 11 ! 4y
7x # 1 ! 8y
x
y
O
(3, 2)2x " 3y ! 12
2x # y ! 4
xy
O
(0, #8)
y ! 3x # 8
y ! x # 8
y " 0.10x ! 2.6y " 0.08x ! 3.2
©Glencoe/McGraw-Hill 57 Algebra 2 Chapter 3
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©Glencoe/McGraw-Hill 58 Algebra 2 Chapter 3
19.
21.
23.
25. inconsistent
x
y
O
y ! x # 4
y ! x " 4
x
y
O
(#4, #2)
x " y ! #214
12
x # y ! 012
x
y
O
(4, 2)
2x # y ! 6
x " 2y ! 514
y
Ox
(3.5, 0)2x " 3y ! 7
2x # 3y ! 7
20.
22.
24.
26. consistent and independent
x
y
O
y ! 2x " 6
y ! x " 3
x
y
O
(3, #5)x # y ! 52
335
x " y ! 343
15
x
y
O
(#9, 3)
y # x ! 613
x " y ! #323
x
y
O
(1.5, 5)4x # 2y ! #4
8x # 3y ! #3
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©Glencoe/McGraw-Hill 59 Algebra 2 Chapter 3
27. consistent and independent
29. inconsistent
31. consistent and independent
33. consistent and independent
x
y
O
1.2x " 2.5y ! 4
0.8x # 1.5y ! #10
x
y
O
2y ! x
8y ! 2x " 1
x
y
O
2y # 2x ! 8
y # x ! 5
x
y
O#4x " y ! 9
x " y ! 4
28. consistent and dependent
30. consistent and dependent
32. inconsistent
34. consistent and dependent
x
y
O
1.6y ! 0.4x " 1
0.4y ! 0.1x " 0.25
x
y
O
2y ! 5 # x
6y ! 7 # 3x
x
y
O
4x # 2y ! 6
6x # 3y ! 9
x
y
O
3x " y ! 3
6x " 2y ! 6
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©Glencoe/McGraw-Hill 60 Algebra 2 Chapter 3
35. inconsistent
37. (#3, 1)
39.
41. Deluxe Plan
43. Supply, 300,000; demand,200,000; prices will tend tofall.
y " 52 ! 0.23x, y " 80
x
y
O
3y # x ! #2
y # x ! 213
36. consistent and independent
38. (1, 3), (2, #1), (#2, #3)
40. (120, 80)
42. Supply, 200,000; demand,300,000; prices will tend torise.
44. 250,000; $10.00
Cost
($)
0
40
80
120
Miles40 80 120 160
y ! 52 " 0.23x
y ! 80
x
y
O2y # x ! #4
y # 2x ! 14x " y ! 7
x
y
O
2y # 4x ! 3
x # y ! #243
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©Glencoe/McGraw-Hill 61 Algebra 2 Chapter 3
45. y " 304x !15,982, y " 98.6x !18,976
47. FL will probably be rankedthird by 2020. The graphsintersect in the year 2015, soNY will still have a higherpopulation in 2010, but FLwill have a higher populationin 2020.
49. You can use a system ofequations to track sales andmake predictions aboutfuture growth based on pastperformance and trends inthe graphs. Answers shouldinclude the following.• The coordinates (6, 54)
represent that 6 yearsafter 1999 both the in-store sales and onlinesales will be $54,000.
• The in-store sales and theonline sales will never beequal and in-store saleswill continue to be higherthan online sales.
51. C
53. (#5.56, #12)
55. no solution
46. 2015
48a.
48b.
48c.
50. A
52. (3.40, #2.58)
54. (4, 3.42)
56. (#9, 3.75)
ab
"de,
cb
$fe
ab
$de
ab
"de,
cb
"fe
Popu
lati
on (T
hous
ands
)
0
16,000
12,000
20,000
24,000
Years After 19994 8 12 16 20
y ! 304x " 15,982
y ! 98.6x " 18,976
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©Glencoe/McGraw-Hill 62 Algebra 2 Chapter 3
57. (2.64, 42.43)
59.
61. A
63. S
65. {#15, 9}
67. {#2, 3}
69. {9}
71.
73.
75.
77.
79. x ! 4y
12x ! 18y # 6
9y ! 1
z3
! 1
x 2 # 6
y
xO
2x " y ! #4
58.
60.
62. C
64. {#13, 13}
66. %
68.
70.
72.
74.
76.
78. 15x ! 10y ! 10
#3x ! 6y
x ! 2
41a ! 528 ! 2n
e#5, 72f
y
xO
2y # 1 ! x
y
xO
y ! 5 " 3x
Lesson 3-2 Solving Systems of Equations Algebraically
Pages 119–122
1. See students’ work; oneequation should have avariable with a coefficient of 1.
2. There are infinitely manysolutions.
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3. Vincent; Juanita subtractedthe two equations incorrectly;
not 0.
5. (1, 3)
7. (5, 2)
9. (6, #20)
11.
13. (9, 5)
15. (3, #2)
17. no solution
19. (4, 3)
21. (2, 0)
23. (10, #1)
25. (4, #3)
27. (#8, #3)
29. no solution
31.
33. (#6, 11)
35. (1.5, 0.5)
37. 8, 6
39. ,
41. 4 2-bedroom, 2 3-bedroom
43. ,
45.
47. Yes; they should finish thetest within 40 minutes.
2x ! 4y " 100, y " 2x
700x ! 200y " 15,000x ! y " 30
16x ! 19y " 478x ! y " 28
a#12,
32b
a3
13, 2
23b
#y # y "#2y,
4. (4, 8)
6. (4, #1)
8. (9, 7)
10. no solution
12. C
14. (2, 7)
16. (#6, 8)
18. (1, 1)
20. (#1, 8)
22. (3, #1)
24. (#7, 9)
26. (6, 5)
28. (7, #1)
30. (#5, 8)
32.
34. infinitely many
36. (2, 4)
38. 2, 12
40. 18 members rented skis and 10 members rentedsnowboards.
42. (#5, #2), (4, 4), (#2, #8), (1, 10)
44. 18 printers, 12 monitors
46. 10 true/false, 20 multiple-choice
48. a ! s " 40, 11a ! 4s " 335
a13, 2b
©Glencoe/McGraw-Hill 63 Algebra 2 Chapter 3
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49. 25 min of step aerobics, 15 min of stretching
51. You can use a system ofequations to find the monthlyfee and rate per minutecharged during the monthsof January and February.Answers should include thefollowing.• The coordinates of the
point of intersection are(0.08, 3.5).
• Currently, Yolanda ispaying a monthly fee of$3.50 and an additional 8¢ per minute. If shegraphs y " 0.08x ! 3.5 (torepresent what she ispaying currently) and y " 0.10x ! 3 (to representthe other long-distanceplan) and finds theintersection, she canidentify which plan wouldbe better for a person withher level of usage.
53. A
50. (4, 6)
52. C
54. inconsistent
x
y
O
y ! x " 2
y ! x # 1
©Glencoe/McGraw-Hill 64 Algebra 2 Chapter 3
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©Glencoe/McGraw-Hill 65 Algebra 2 Chapter 3
55. consistent and dependent
57.
59.
61.
63.
65.
67. yes
69. no
3x ! 2y " 21; 3, 2, 21
2x # y " #3; 2, #1, #3
x # y " 0; 1, #1, 0
y
xO
3x " 9y ! #15
y
xO
x " y ! 3
x
y
O
4y # 2x ! 4
y # x ! 112
56. consistent and independent
58.
60.
62.
64.
66. 0.6 ampere
68. no
70. yes
x # 2y " 6; 1, #2, 6
3x ! 5y " 2; 3, 5, 2
7x # y " #4; 7, #1, #4
y
xO
5y # 4x ! #20
x
y
O
3x " y ! 1
y ! 2x # 4
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©Glencoe/McGraw-Hill 66 Algebra 2 Chapter 3
Lesson 3-3 Solving Systems of Inequalities by GraphingPages 125–127
Chapter 3Practice Quiz 1
Page 122
1.
3. (2, 7)
5. Hartsfield, 78 million; O’Hare, 72.5 million
x
y
O
(#1, 7)
y ! #x " 6
y ! 3x " 10
2.
4. (4, #1)
x
y
O
(3, 2)2x " 3y ! 12
2x # y ! 4
1. Sample answer:
3a. 43b. 23c. 13d. 3
5. y
xO
y ! x # 2
y ! #2x " 4
y & x ! 3, y ' x # 22. true
4.
6. y
xO
x " y ! 2
x ! #1x ! 3
y
xO
y ! 2
x ! 4
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7.
9. (#4, 3), (1, #2), (2, 9), (7, 4)
11. Sample answer: 3 packages of bagels, 4 packages ofmuffins; 4 packages ofbagels, 4 packages ofmuffins; 3 packages ofbagels, 5 packages ofmuffins
13. y
xO
y ! #4x ! #1
y
xO x " 2y ! #3
y ! 2x " 1x ! 1
8. (#3, #3), (2, 2), (5, #3)
10.
12.
14. y
xO
y ! x " 4
y ! 2 # x
y
xO
y ! 3
x ! 2
m
bM
uffi
ns
2
0
4
6
8
10
Bagels2 4 6 8 10 12
2.5b " 3.5m ! 28
m ! 3
b ! 2
©Glencoe/McGraw-Hill 67 Algebra 2 Chapter 3
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©Glencoe/McGraw-Hill 68 Algebra 2 Chapter 3
15.
17.
19. no solution
21.
23. y
xO
2x " 4y ! #7
x # 3y ! 2
2x # y ! 4
y
xO
x " 3y ! 6
x ! #4
x ! 2
y
xO
2y # x ! #6
4x # 3y ! 7
y
xO y ! #2
y ! 2
y ! x # 3
16.
18.
20.
22. no solution
24. (0, 0), (0, 4), (8, 0)
y
xO
y ! #1x ! #3
y ! 1 x ! 3
y
xO
y ! 2x # 3
y ! x " 112
y
xO
4x # y ! 2
3x " 2y ! 6
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25. (#3, #4), (5, #4), (1, 4)
27. (#6, #9), (2, 7), (10, #1)
29. (#4, 3), (#2, 7),
(4, #1),
31. 64 units2
33. ,
Stor
m S
urge
(ft)
0
8
10
12
14
16
Wind Speed (mph)80 100 120 140 160
s ! 111
s ! 130
h ! 12
h ! 9
h
s
h ( 9, h ) 12s ( 111, s ) 130
a713, 2
13b
26. (0, 4), (3, 0), (3, 5)
28. (#11, #3), (#1, #3),
(6, 4),
30. 16 units2
32.
34. category 4; 13-18 ft
Hou
rs R
akin
g Le
aves
2
0
4
6
8
10
12
14
16
Hours Cutting Grass2 4 6 8 10 12 14 16
x " y ! 15
10x " 12y ! 120
y
x
a6, 512b
©Glencoe/McGraw-Hill 69 Algebra 2 Chapter 3
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©Glencoe/McGraw-Hill 70 Algebra 2 Chapter 3
35.
37. 6 pumpkin, 8 soda
39. The range for normal bloodpressure satisfies fourinequalities that can begraphed to find theirintersection. Answers shouldinclude the following.• Graph the blood pressure
as an ordered pair; if thepoint lies in the shadedregion, it is in the normalrange.
• High systolic pressure isrepresented by the regionto the right of andhigh diastolic pressure isrepresented by the regionabove
41. Sample answer:
43. (6, 5)
45. y
xO
y ! 2x " 1
y ! # x # 412
(#2, #3)
y ) 6, y ( 2, x ) 5, x ( 1
y " 90.
x " 140
Swed
ish
Soda
2
0
4
6
8
10
12
14
Pumpkin2 4 6 8 10 12 14
x " 2.5y ! 26
2x " 1.5y ! 24y
x
36. Sample answer: 2 pumpkin, 8 soda; 4 pumpkin, 6 soda;8 pumpkin, 4 soda
38. 42 units2
40. B
42. (#3, 8)
44. (8, #5)
46. infinitely manyy
xO
2x " y ! #3
6x " 3y ! #9
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1. sometimes
3.
vertices: (1, 2), (1, 4), (5, 2);max: f (5, 2) " 4, min: f (1, 4) " #10
x
y
O
(1, 4)
(1, 2)
(5, 2)
2. Sample answer:
4.
vertices: (#3, 1), Q , 1R;min: f (#3, 1) " #17;no maximum
53
x
y
O
(#3, 1) ( , 1)53
y ( #x, y ( x # 5, y ) 0
©Glencoe/McGraw-Hill 71 Algebra 2 Chapter 3
Lesson 3-4 Linear ProgrammingPages 132–135
47.
49. #5
51. 8
53. 5
x
y
O
#x " 8y ! 12
2x # y ! 6
(4, 2)
48.
50. #12
52. 27
54. #8.25
y "12x ! 6
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5.
vertices: (0, 1), (1, 3), (6, 3),(10, 1); max: f (10,1) " 31, min: f (0,1) " 1
7.
vertices: (#2, 4), (#2, #3),(2, #3), (4, 1); max: f(2, #3) "5; min: f (#2, 4) " #6
9.
11. (0, 0), (26, 0), (20, 12),
13. 20 canvas tote bags and 12 leather tote bags
a0, 18
23b
4c ! 2l) 104c ( 0, l( 0, c ! 3l) 56,
x
y
O
(#2, 4)
(4, 1)
(2, #3)
(#2, #3)
x
y
O
(1, 3) (6, 3)
(0, 1) (10, 1)
6.
vertices: (2, 0), (2, 6), (7, 8.5),(7, #5); max: f(7, 8.5) " 81.5,min: f (2, 0) " 16
8.
vertices: (#3, #1), (#1, 2),(2, 3), (3,#2); max: f(3, #2) "5, min: f(#1, 2) " #3
10.
12. f(c, l) " 20c ! 35l
14. $820
c
!
Leat
her
Tote
Bag
s
4
0
8
12
16
20
24
28
Canvas Tote Bags4 8 12 16 20 24 28
(0, 0)
(20, 12)
(26, 0)
(0, 18 )23
x
y
O
(#3, #1)(3, #2)
(#1, 2) (2, 3)
x
y
O
(7, 8.5)
(7, #5)
(2, 0)
(2, 6)
©Glencoe/McGraw-Hill 72 Algebra 2 Chapter 3
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©Glencoe/McGraw-Hill 73 Algebra 2 Chapter 3
15.
vertices: (0, 1), (6, 1), (6, 13);max: f (6, 13) " 19;min: f (0, 1) " 1
17.
vertices: (1, 4), (5, 8), (5, 2), (1, 2); max: f (5, 2) "11, min: f (1, 4) " #5
19.
vertices: (#3, #1), (3, 5);min: f (#3, #1) " #9;no maximum
x
y
O(#3, #1)
(3, 5)
x
y
O
(1, 2)
(1, 4)
(5, 2)
(5, 8)
x
y
O
(6, 13)
(6, 1)(0, 1)
16.
vertices: (0, #4), (3, 5), (3, #4); max: f (3, #4) " 7, min: f (3, 5) " #2
18.
vertices: (2, 1), (2, 3), (4, 4),(4, 1); max: f (4, 4) " 16;min: f (2, 1) " 5
20.
vertices: (2, 2), (2, 8), (6, 12),(6, #6); max: f(6, 12) " 30, min: f(6, #6) " #24
x
y
O
(6, #6)
(2, 8)
(6, 12)
(2, 2)
4 8#4
#4
#8
4
8
12
x
y
O
(4, 1)(2, 1)
(4, 4)(2, 3)
x
y
O
(3, 5)
(3, #4)(0, #4)
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©Glencoe/McGraw-Hill 74 Algebra 2 Chapter 3
21.
vertices: (0, 0), (0, 2), (2, 1),(3, 0); max: f (0, 2) " 6;min: f (3, 0) " #12
23.
vertices: (3, 0), (0, #3);min: f (0, #3) " #12;no maximum
25.
vertices: (0, 2), (4, 3),max: f (4, 3) " 25,
min: f (0, 2) " 6
a73
, #13b;
x
y
O
( , )13#7
3
(4, 3)(0, 2)
x
y
O(3, 0)
(0, #3)
x
y
O(3, 0)(0, 0)
(0, 2) (2, 1)
22.
vertices: (0, 0), (0, 7), (4, 3),(2, 0); max: f (4, 3) " 14;min: f (0, 7) " #14
24.
vertices: (0, 4), (4, 0), (8, 6);max: f (4, 0) " 4; min:f (0, 4) " #8
26.
vertices: (0, 2), (4, 3), (2, 0);max: f (4, 3) " 13, min: f (2, 0) " 2
x
y
O
(4, 3)(0, 2)
(2, 0)
x
y
O
(8, 6)
(4, 0)
(0, 4)
x
y
O (2, 0)
(0, 0)
(0, 7)
(4, 3)
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©Glencoe/McGraw-Hill 75 Algebra 2 Chapter 3
27.
vertices: (2, 5), (3, 0);min: f (3, 0) " 3, no maximum
29.
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
33. (0, 0), (0, 42.5), (30, 20),(40, 0)
35. 30 graphing calculators, 20 CAS calculators
37. See students’ work.
x
y
O
(2, 3)
(2, 1) (4, 1)
(4, 4)
(5, 3)
x
y
O
(3, 0)
(2, 5)
28.
vertices: (0, 0), (0, 1), (2, 2),(4, 1), (5, 0); max: f (5, 0) "15, min: f (0, 1) " #5
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
32.
34. f (g, c) " 50g ! 65c
36. $2800
38. c ( 0, s ( 0, c ! s ) 4500, 4c ! 5s ) 20,000
g
cGra
phin
g Ca
lcul
ator
s
10
0
20
30
40
50
CAS Calculators10 20 30 40 50
(0, 0) (40, 0)
(30, 20)(0, 42.5)
x
y
O(0, 0)
(0, 1)
(2, 2)
(4, 1)
(5, 0)
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©Glencoe/McGraw-Hill 76 Algebra 2 Chapter 3
39. (0, 0), (0, 4000), (2500, 2000),(4000, 0)
41. 4000 acres corn, 0 acressoybeans; $130,500
43. There are many variables inscheduling tasks. Linearprogramming can help makesure that all the requirementsare met. Answers shouldinclude the following.• Let x " the number of
buoy replacements and lety " the number of buoyrepairs. Then, x ( 0, y ( 0,x ) 8 and x ! 2.5y ) 24.The captain would want tomaximize the number ofbuoys that a crew couldrepair and replace, so f (x, y) " x ! y.
• Graph the inequalities andfind the vertices of theintersection of the graphs.The coordinate (0, 24)maximizes the function. Sothe crew can service themaximum number ofbuoys if they replace 0and repair 24 buoys.
S
c0
1000
2000
2000 4000
3000
4000
(0, 0)
(2500, 2000)(0, 4000)
(4500, 0)
40. 2500 acres of corn, 2000 acres soybeans;$125,000
42. 3 chocolate chip, 9 peanutbutter
44. A
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©Glencoe/McGraw-Hill 77 Algebra 2 Chapter 3
45. C
47. %
49. (2, 3)51. c " average cost each year;
15c ! 3479 " 748953. Additive Inverse55. Multiplicative Inverse57. 959. 1661. 8
x
y
O
3x # 2y ! #6
y ! x # 132
46.
48. (#5, 8)
50. (5, 1)52. about $267 per year
54. Associative Property (*)56. Distributive Property58. 560. #362. #4
x
y
O
2y " x ! 4
y ! x # 4
Chapter 3Practice Quiz 2
Page 135
1.
x
y
O
y " x ! 4
y # x ! 0 2.
x
y
O
y ! 3x # 4
y ! x " 3
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©Glencoe/McGraw-Hill 78 Algebra 2 Chapter 3
3.
5.
vertices: (1, #3), (#1, 3), (5, 6), (5, 1); max: f (5, 1) "17, min: f (#1, 3) " #13
x
y
O
(5, 1)
(5, 6)
(#1, 3)
(1, #3)
x
y
O
x " 3y ! 15
4x " y ! 16 4.
vertices: (0, 0), (0, 4), (1, 6),(3, 0); max: f (1, 6) " 8, min: f (0, 0) " 0
x
y
O
(0, 4)
(3, 0)
(1, 6)
(0, 0)
Lesson 3-5 Solving Systems of Equations in Three Variables
Pages 142–144
1. You can use elimination orsubstitution to eliminate oneof the variables. Then youcan solve two equations intwo variables.
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
2. No; the first two equations dorepresent the same plane,however they do not intersectthe third plane, so there isno solution of this system.
4. (6, 3, #4)
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©Glencoe/McGraw-Hill 79 Algebra 2 Chapter 3
5. (#1, #3, 7)
7. (5, 2, #1)
9. (4, 0, 8)
11. 4 lb chicken, 3 lb sausage, 6 lb rice
13. (#2, 1, 5)
15. (4, 0, #1)
17. (1, 5, 7)
19. infinitely many
21.
23. (#5, 9, 4)
25. 8, 1, 3
27. enchilada, $2.50; taco, $1.95;burrito, $2.65
29. x ! y ! z " 355, x ! 2y !3z " 646, y " z ! 27
31.
y "43x 2 !
13x ! 3
a "43, b "
13, c " 3;
a13, #
12,
14b
12
6. infinitely many
8. no solution
10. 6c ! 3s ! r " 42,
c ! s ! r " , r " 2s
12. (3, 4, 7)
14. (2, #3, 6)
16. no solution
18. (1, 2, #1)
20.
22. (8, 3, #6)
24. 3, 12, 5
26. 1-$100, 3-$50, and 6-$20 checks
28. $7.80
30. 88 3-point goals, 115 2-pointgoals, 152 1-point free throws
32. You can write a system ofthree equations in threevariables to find the number ofeach type of medal. Answersshould include the following.• You can substitute b ! 6
for g and b # 8 for s in theequation g ! s ! b " 97.This equation is now interms of b. Once you findb, you can substitute againto find g and s. The U.S.Olympians won 39 goldmedals, 25 silver medals,and 33 bronze medals.
a12,
32,
92b
13
12
PQ245-6457F-P03[056-080].qxd 7/24/02 12:23 PM Page 79 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-03:
©Glencoe/McGraw-Hill 80 Algebra 2 Chapter 3
33. D
35. 120 units of notebook paperand 80 units of newsprint
37.
39. Sample answer using (7, 15)and (14, 22):
41.
43. 9s ! 4t
x ! 3y
y " x ! 8
x
y
O
4y # 2x ! 4
3x " y ! 3
• Another situation involvingthree variables is winningtimes of the first, second,and third place finishers ofa race.
34. A
36.
38.
40. Sample answer: about 47¢
42.
44. 18a ! 16b
#2z ! 8
y
xO
3x " y ! 1
2y # x ! #4
x
y
O
y ! 7 # 2x
y ! x " 2
PQ245-6457F-P03[056-080].qxd 7/24/02 12:23 PM Page 80 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-03:
1. The matrices must have thesame dimensions and eachelement of one matrix mustbe equal to the correspondingelement of the other matrix.
3. Corresponding elements areelements in the same rowand column positions.
5.
7. (3, 3)
9.
11.
13.
15.
17.
19. (3, !5, 6)
21. (4, !3)
23. (14, 15)
25. (5, 3, 2)
a3, !13b3 " 2
3 " 3
3 " 1
2 " 5
3 " 4
Chapter 4 MatricesLesson 4-1 Introduction to Matrices
Pages 156–158
©Glencoe/McGraw-Hill 81 Algebra 2 Chapter 4
2. Sample answers:row matrix, , 1 " 3;
column matrix, , 2 " 1;
square matrix, , 2 " 2;
zero matrix, , 2 " 2
4.
6. (5, 6)
8. Fri Sat Sun Mon Tue
10.
12.
14.
16. (2.5, 1, 3)
18. (5, 3)
20. (2, !5)
22. (1.5, 3)
24. (!2, 7)
26. Evening Matinee Twilight
Adult
Child
Senior
C7.50 5.50 3.754.50 4.50 3.755.50 5.50 3.75
S
2 " 5
4 " 3
2 " 3
B88 88 90 86 8554 54 56 53 52
RHigh
Low
1 " 5
B0 00 0
RB1 23 4
RB12R[1 2 3]
PQ245-6457F-P04[081-109].qxd 7/31/02 9:41 AM Page 81 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-04:
27.
29. Sample answer: Mason’sSteakhouse; it was given thehighest rating possible forservice and atmosphere,location was given one of thehighest ratings, and it ismoderately priced.
31. Single Double Suite
33. row 6, column 9
35. B
37. (7, 5, 4)
B6079
7089
7595RWeekday
Weekend
3 " 3 28. Cost Service Atmosphere Location
Catalina Grill ** * * *Oyster Club *** ** * **Casa di Pasta **** *** *** ***Mason’s ** **** **** ***Steakhouse
30. Weekday Weekend
32.
34. Matrices are used toorganize information so itcan be read and comparedmore easily. Answers shouldinclude the following.• If you want the least
expensive vehicle, thecompact SUV has the bestprice; the large SUV hasthe most horsepower,towing capacity and cargospace, and the standardSUV has the best fueleconomy.
• Sample answer: Matricesare used to report stockprices in the newspaper.
36. C
38. (7, 3, !9)
G 1 3 6 10 15 212 5 9 14 20 274 8 13 19 26 347 12 18 25 33 42
11 17 24 32 41 5116 23 31 40 50 6122 30 39 49 60 72
WC60 79
70 8975 95
SSingle
Double
Suite
©Glencoe/McGraw-Hill 82 Algebra 2 Chapter 4
! ¥
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39.
41.
vertices: (3, 1),
.max:
min:
43.
Cost
($)
1
0
23456
Hours1 2 3 4 5
f a32,
172b # !1
f a152
, 52b # 35,
a32,
172b;a15
2,
52b,
y
a!43,
35, !11b 40.
vertices: (1, 3), (6, 3),
max:
min: f (1, 3)#11
42.
vertices: (2, 1), (6, 3);min: f(2, 1) # 1, no maximum
44. step function
! 123
f a133
, 193b #
833
,
a133
, 193b;
y
xO
y " #2x ! 15y " x ! 2
y " 3
©Glencoe/McGraw-Hill 83 Algebra 2 Chapter 4
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45. $4.50
47. 2
49. 20
51. !10
53. !18
55. !3
57. 32
46. 2
48. 0
50. 3
52. 6.2
54. 17
56. 75
©Glencoe/McGraw-Hill 84 Algebra 2 Chapter 4
Lesson 4-2 Operations with MatricesPages 163–166
1. They must have the samedimensions.
3.
5.
7.
9.
11.
Females # E16,439 456,87314,545 405,16312,679 340,480
7931 257,5865450 133,235
UMales # E16,763 549,499
14,620 477,96014,486 455,305
9041 321,4165234 83,411
U,B!21 29
12 !22RB!22 8
3 24RB 1 10
!7 5RC4 4
4 44 4
S 2. Sample answer: [!3 1], [3 !1]
4. impossible
6.
8.
10.
12. E1,006,372883,123795,785579,002216,646
UB!3 30
26 11RB 10 6
!1 7RB18 !3 15 6
21 9 !6 24R
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13. No; many schools offer thesame sport for males andfemales, so those schoolswould be counted twice.
15. impossible
17.
19.
21.
23.
25.
27.
29. D 2 4
23
1 5
6 !1
TC38 4
32 !618 42
SC!2 !14 !1
!7 !4S
C!512
3 9
1023
123
!212
SB1.5 34.5 9
RC!13!323SC !4 8 !2
6 !10 !16!14 !12 4
S14.
16.
18. [15 !29 65 !2]
20.
22.
24.
26.
28.
30.
Saturday: C112 87 56 7484 65 39 7088 98 43 60
SFriday: C120 97 64 7580 59 36 6072 84 29 48
S ,
C!12 !133 !8
13 37SC 0 16
!8 2028 !4
SC13 10
4 77 !5
SC!4 !1532
!2S
C 1.8 9.083.18 31.04
10.41 56.56S
B15 0 40 13 !5
RC 10
!45S
©Glencoe/McGraw-Hill 85 Algebra 2 Chapter 4
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31.
33.
35. 1996, floods; 1997, floods;1998, floods; 1999,tornadoes; 2000, lightning
37.
39.
41. You can use matrices to trackdietary requirements and addthem to find the total eachday or each week. Answersshould include the following.
• Breakfast # C566 18 7482 12 17530 10 11
S ,
B1.00 1.001.50 1.50
RB1.50 2.251.00 1.75
R
E245228319227117
UC232 184 120 149
164 124 75 130160 182 72 108
S 32.
34.
36. Residents:Child Adult
Nonresidents:Child Adult
38. Before 6:00:Child Adult
After 6:00:Child Adult
40.
42. D
2B0.5 0.75 31 4 0.1
R# B1 1.5 62 8 0.2
RB2.00 3.503.00 5.25
RResidentsNonresidents
B3.00 4.504.50 6.75
RResidentsNonresidents
B4.50 6.753.00 5.25
RBefore 6After 6
B3.00 4.502.00 3.50
RBefore 6After 6
E1541352751
UC!8 !10 !8 !1
4 6 3 1016 14 14 12
S
©Glencoe/McGraw-Hill 86 Algebra 2 Chapter 4
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• Add the three matrices:
43. A
45.
47.
49.
51. (5, 3, 7)
53. (2, 5)
55. (6, !1)
57.
59. Multiplicative Inverse
61. Distributive Property
4 " 3
3 " 3
1 " 4
£2608 80 522091 67 822620 65 61
§ .Dinner # C1257 40 26
987 32 451380 29 38
SLunch # C785 22 19622 23 20710 26 12
S ,
44. 2 " 2
46. 2 " 4
48. 3 " 2
50. (3, !4, 0)
52.
54. (!3, 1)
56.
58. No, it would cost $6.30.
60. Associative Prop. ($)
62. Commutative Prop. (")
0.30p $ 0.15s % 6
a14, 6, !
16b
©Glencoe/McGraw-Hill 87 Algebra 2 Chapter 4
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1. Sample answer:
3. The Right DistributiveProperty says that
butsince
the Commutative Propertydoes not hold for matrixmultiplication in most cases.
5. undefined
7.
9. B2441RB15 !5 20
24 !8 32R
AC $ BC & CA $ CB1A $ B2C # AC $ BC,
C1 23 45 6
S ! B7 89 10
R 2. Never; the inner dimensionswill never be equal.
4. 3 " 2
6. [19 15]
8. not possible
10. yesA(BC)
(AB)C
# B!50 !2881 62
R# B!16 228 3
R ! B 3 2!1 2
R# ¢B2 !13 5
R ! B!4 18 0R ! ! B 3 2
!1 2R
# B!50 !2881 62
R# B2 !13 5
R ! B!13 !624 16
R# B2 !13 5
R ! ¢ B!4 18 0R ! B 3 2
!1 2R !
©Glencoe/McGraw-Hill 88 Algebra 2 Chapter 4
Lesson 4-3 Multiplying MatricesPages 171–174
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11.
13.
15. undefined
17. undefined
19. [6]
21. not possible
23.
25.
27. yes
$
# B!20 !452 !16
R# B!4 08 6R ! B5 1
2 !4R# ¢ B1 !2
4 3R $ B!5 2
4 3R ! ! B5 1
2 !4R(A $ B)C
# B!20 !452 !16
R# B 1 926 !8
R $ B!21 !1326 !8
RB!5 24 3
R ! B5 12 !4
RB5 12 !4
R# B1 !24 3
R !
AC $ BC
C 24 16!32 !5!48 !11
SB 1 !25 229 1 !30
R4 " 2
[45 55 65], C350 280320 165180 120
S 12. $74,525
14.
16.
18.
20.
22.
24. not possible
26.
28. yes
# B!39 !12!24 51
R# B1 !24 3
R B!15 612 9
R! ¢3 B!5 24 3
R !# B1 !24 3
RA(cB)
# B!39 !12!24 51
R# 3 B!13 !4!8 17
R! B!5 24 3
R !# 3 ¢ B1 !24 3
Rc (AB)
C 0 64 !409 11 !11
!3 39 !23S
B!3918RB 8 !11
22 12R3 " 5
1 " 5
2 " 2
©Glencoe/McGraw-Hill 89 Algebra 2 Chapter 4
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30. noABC
CBA
32.
34. $31,850
36.
38. Juniors
40. $24,900
42. $1460
D72 4968 6390 5686 62
T, B1.000.50R
C222518S
# B 31 81!58 28
R# B!21 13!26 !8
R ! B1 !24 3
R# B5 12 !4
R ! B!5 24 3R ! B1 !2
4 3R
# B!73 3!6 !76
R# B!13 !4!8 17
R ! B5 12 !4
R# B1 !24 3
R ! B!5 24 3R ! B5 1
2 !4R29. no
31.
33.
35. any two matrices and
where
and
37.
39. $431
41. $26,360
D96.5099.50
118117
Te # h
bg # cf, a # d,Be fg h
R Ba bc dR
C14,28513,270
4295SC290 165 210
175 240 190110 75 0
S# B!20 !452 !16
R# B 1 926 !8
R $ B!21 !1326 !8
RB!5 24 3
R ! B5 12 !4
R# B1 !24 3
R ! B5 12 !4
R $
AC $ BC
# B!12 6!40 !24
R# B5 12 !4
R ! B!4 08 6
R#B5 12 !4
R ! ¢ B1 !24 3
R$ B!5 24 3R !C (A $ B)
©Glencoe/McGraw-Hill 90 Algebra 2 Chapter 4
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44. Sports statistics are oftenlisted in columns andmatrices. In this case, youcan find the total number ofpoints scored by multiplyingthe point matrix, whichdoesn’t change, by therecord matrix, which changesfor each season. Answersshould include the following.• [479]• Basketball and wrestling
use different point valuesin scoring.
46. A
48. impossible
50. (7, !4)
52. (2, !5, !7)
54.
56. 2; !5
32; 3
P ! R #
43.the original matrix
45. B
47.
49.
51. (5, !9)
53. $2.50; $1.50
55. 8; !16
B!20 2!28 12
RB12 !6!3 21
R
a # 1, b # 0, c # 0, d # 1;
©Glencoe/McGraw-Hill 91 Algebra 2 Chapter 4
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57.
59.
©Glencoe/McGraw-Hill 92 Algebra 2 Chapter 4
58.
60.
1. (6, 3)
3. (1, 3, 5)
5.
7.
9. not possible
B4 31 3
RB232 159 120 149134 200 159 103
R2. (5, !1)
4.
6.
8.
10. B 15 !8 !10!7 23 16
RB!10 20 250 !20 35
RB!3 53 13
RB112 79 56 7469 95 82 50
RB120 80 64 7565 105 77 53
R,Chapter 4
Practice Quiz 1Page 174
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©Glencoe/McGraw-Hill 93 Algebra 2 Chapter 4
Lesson 4-4 Transformations with MatricesPages 178–181
1.
3. Sample answer:
5.
7.
9.
11. B
13.
15. B02
1.5!1.5
!2.50R
F¿(!10, !4) D¿(!3, 6), E¿(!2, !3),
C¿(5, 0), D¿(0, 0)A¿(0, !4), B¿(5, !4),
B04
54
50
00R
C¿(!3, !7)A¿(4, 3), B¿(5, !6),
B!41
!41
!41R
Transformation Shape Size Isometry
reflection same same yes
rotation p same p same p yes
translation same same yes
dilation changes same no
2.
4.
6.
8.
10.
12.
14.
16.C¿(!7.5, 0)A¿(0, 6), B¿(4.5, !4.5),
B!42
!42
!42RC¿(!5, 0), D¿(0, 0)
A¿(0, !4), B¿(!5, !4),
C¿(15, 0), D¿(0, 0)A¿(0, 12), B¿(15, 12),
B 3!1
3!1
3!1R
B!3!2
!3!2
!3!2R
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17.
19.
21.
23.
25. J(!5, 3), K(7, 2), L(4, !1)
y
xOD'
E'
F
ED
G
F'
G'
B24
54
41
11R
Z¿(!1, 7)X¿(!1, 1), Y¿(!4, 2),
18.
20.
22.
24.
26. B!2!3
!43
!25
32RB2
34
!32
!5!3!2R ! (!1) #
E¿(6, !2), F¿(8, !9)
F¿(1, !4), G¿(1, !1)D¿(4, !2), E¿(4, !5),
B 1!1
2!4
7!1R
©Glencoe/McGraw-Hill 94 Algebra 2 Chapter 4
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27.
29.
31.
33.
35.
37. and
39. Multiply the coordinates
by , then add the
result to .
41. (17, !2), (23, 2)
B60RB1
00
!1R
(8, !7)(!8, 7), (!7, !8),
B34R(!6, !3.75), (!3, !3.75)
(!1.5, !1.5), (!4.5, !1.5),
B 4!4
44
!44
!4!4RB 4
!4!4!4
!44
44R
28. 180' rotation
30.
32. The figures in Exercise 29and Exercise 30 have thesame coordinates, but thefigure in Exercise 31 hasdifferent coordinates.
34.
36. (6.5, 6.25)
38. The object is reflected overthe x-axis, then translated6 units to the right.
40. No; since the translationdoes not change the y-coordinate, it does notmatter whether you do thetranslation or the reflectionover the x-axis first. However,if the translation did changethe y-coordinate, then orderwould be important.
42. There is no single matrix toachieve this. However, youcould reflect the object overthe y-axis and then translateit 2(3) or 6 units to the right.
(!3.75, !2.625)
B 4!4
!4!4
!44
44R
©Glencoe/McGraw-Hill 95 Algebra 2 Chapter 4
PQ245-6457F-P04[081-109].qxd 7/31/02 9:41 AM Page 95 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-04:
43. Transformations are used incomputer graphics to createspecial effects. You cansimulate the movement of anobject, like in space, whichyou wouldn’t be able torecreate otherwise. Answersshould 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
matrix and
multiplied on the left by the2 " 2 rotation matrix.
• The object would getsmaller and appear to bemoving away from you.
45. A
47. undefined
49.
51.
yesD # 53, 4, 56, R # 5!4, 5, 66;
C111833
24!13
!8
!78
21S
Bab
cd
ef
gh
ijR
44. B
46.
48.
50.
52.
D # {all real numbers}, R # {all real numbers}; yes
C 2031
!10
10!46
3
!24!9
7S2 " 5
2 " 2
©Glencoe/McGraw-Hill 96 Algebra 2 Chapter 4
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53.
no
55.
57.
59. 6
61. 28
63. 94
0x ! 1 0 ( 1
0x 0 ( 2.8
R # 5all real numbers6;D # 5x 0 x ) 06,
54.
56.
58.
60. 5
62.
64. 53
103
513
23 mi
0x $ 1 0 * 2
0x 0 ) 4
©Glencoe/McGraw-Hill 97 Algebra 2 Chapter 4
Lesson 4-5 DeterminantsPages 185–188
1. Sample answer:
3. It is not a square matrix.
5. Cross out the column androw that contains 6. Theminor is the remaining 2 " 2matrix.
B28
14R 2. Khalid; the value of the
determinant is the difference ofthe products of the diagonals.
4. Sample answer:
,
6.
# 213# 60 $ 108 $ 45# !2(!30) ! 3(!36) $ 5(9)
5(0 ! (!9))3(0 ! 36) $# !2(!2 ! 28) !
!2 `!17
42` ! 3 ` 0
942` $ 5 ` 0
9!1
7`
† !209
3!1
7
542† #
B41
33RB3
615R
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7. !38
9. !40
11. !43
13. 45
15. 20
17. !22
19. !29
21. 63
23. 32
25. 32
27. !58
29. 62
31. 172
33. !22
35. !5
37. !141
39. !6
41. 14.5 units2
43. about 26 ft2
4 108 0
!45 !56 0
8. 0
10. !28
12. 0
14. 26 units2
16. !22
18. 0
20. !14
22. !6
24. !37
26. 11.3
28. 0
30. 60
32. !265
34. 21
36. 49
38. !123
40.
42. 12
44. 2875 mi2
53, !1
(!56) ! 0 # 2134 $ 108 $ 0 ! (!45) !
†!2 3 50 !1 49 7 2
†!2 30 !19 7
†!2 3 50 !1 49 7 2
†!2 30 !19 7
©Glencoe/McGraw-Hill 98 Algebra 2 Chapter 4
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45. Sample answer:
47. If you know the coordinatesof the vertices of a triangle,you can use a determinant tofind the area. This isconvenient since you don’tneed to know any additionalinformation such as themeasure of the angles.Answers should include thefollowing.• You could place a
coordinate grid over a mapof the Bermuda Trianglewith one vertex at theorigin. By using the scaleof the map, you coulddetermine coordinates torepresent the other twovertices and use adeterminant to estimatethe area.
• The determinant method isadvantageous since youdon’t need to physicallymeasure the lengths ofeach side or the measureof the angles between thevertices.
49. C
51. !36.9
53. !493
55. !3252
†111
111
111† 46. Multiply each member in the
top row by its minor andposition sign. In this case theminor is a 3 " 3 matrix.Evaluate the 3 " 3 matrixusing expansion by minorsagain.
48. C
50. 63.25
52. !25.21
54. 0
56. B!21
12
2!3R
©Glencoe/McGraw-Hill 99 Algebra 2 Chapter 4
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©Glencoe/McGraw-Hill 100 Algebra 2 Chapter 4
Lesson 4-6 Cramer’s RulePages 192–194
1. The determinant of thecoefficient matrix cannot bezero.
3.
5. (0.75, 0.5)
7. no solution
9. a6, !12, 2b
3x $ 5y # !6, 4x ! 2y # 30
2. Sample answer:and
4. (5, 1)
6. (!6, !8)
8.
10.0.065s $ 0.08d # 297.50s $ d # 4000,
a!5, 23, !
12b
6x $ 3y # 82x $ y # 5
57.
59. [!4]
61. undefined
63. [14 !8]
65. 138,435 ft
67.
69.
71. (1, 9)
73. (!1, 1)
75. (4, 7)
y #12x $ 5
y # !43x
C¿(5, !7.5)A¿(!5, 2.5), B¿(2.5, 5), 58.
60.
62. undefined
64.
66.
68.
70. (0, !3)
72. (2, 1)
74. (2, 5)
y # 2x $ 1
y # x ! 2
B 7!5
6916RB 2
!926
!12R
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11. savings account, $1500;certificate of deposit, $2500
13. (!12, 4)
15. (6, 3)
17. (!0.75, 3)
19. (!8.5625, !19.0625)
21. (4, !8)
23.
25. (3, !4)
27. (2, !1, 3)
29.
31.
33. race car, 5 plays; snowboard,3 plays
35. silk, $34.99; cotton, $24.99
37. peanuts, 2 lb; raisins, 1 lb;pretzels, 2 lb
a!15528
,
14370
,
673140b
a14129
, !10229
,
24429b
a23,
56b
12. (2, !1)
14. (3, 5)
16. (2.3, 1.4)
18. (!0.75, 0.625)
20.
22. (3, 10)
24. (!1.5, 2)
26. (!1, 3, 4)
28.
30. (11, !17, 14)
32.
34.
36.
38. If the determinant is zero,there is no unique solution tothe system. There is eitherno solution or there areinfinitely many solutions.Sample answer:and has a det #0; there are infinitely manysolutions of this system.
andhas a det # 0;
there are no solutions of thissystem.
4x $ 2y # 102x $ y # 4
4x $ 2y # 82x $ y # 4
3.2p $ 2.4r $ 4c # 16.8p $ r $ c # 5, 2r ! p # 0,
512s $ 14c # 542.30
8s $ 13c # 604.79,
r $ s # 8, 7r $ 5s # 50
a!1119
, 3919
, !1419b
a23, !1b
©Glencoe/McGraw-Hill 101 Algebra 2 Chapter 4
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39. Cramer’s Rule is a formulafor the variables x and ywhere (x, y) is a solution fora system of equations.Answers should include thefollowing.• Cramer’s Rule uses
determinants composed ofthe coefficients andconstants in a system oflinear equations to solvethe system.
• Cramer’s Rule isconvenient whencoefficients are large orinvolve fractions ordecimals. Finding thevalue of the determinant issometimes easier thantrying to find a greatestcommon factor if you aresolving by usingelimination or substitutingcomplicated numbers.
41. 111', 69'
43. 40
45.
47.
B1 1 13 3 3
R
40. B
42. 16
44. !53
46.
48.
(!2, !1)
y
xO
y ! 3x " 5
y ! #2x # 5(#2, #1)
C¿(!1, !1)A¿(1, 5), B¿(!2, 2),
©Glencoe/McGraw-Hill 102 Algebra 2 Chapter 4
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©Glencoe/McGraw-Hill 103 Algebra 2 Chapter 4
49.
(4, 3)
51.
53. B7266
9!23Rc # 10h $ 35
y
xO
(4, 3)
x " y ! 7
x #y ! #1 12
50.
no solution
52. [!4 32]
54. B2143R
y
xO2x # 4y ! 12
x # 2y ! 10
Chapter 4Practice Quiz 2
Page 194
1.
3.
5. !58
7. 26
9. (4, !5)
B12
4!1
1!4
!2!1R 2.
4. 22
6. !105
8. (1, !2)
10. (1, 2, 1)
C¿(!1, !4), D¿(2, !1)A¿(!1, 2), B¿(!4, !1),
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©Glencoe/McGraw-Hill 104 Algebra 2 Chapter 4
1.
3. Sample answer:
5. yes
7. no inverse exists
9. See students’ work.
11. yes
13. no
15. yes
17. true
19. false
21. no inverse exists
23.
25. 14
B!6!2
!7!3R1
7 B1
4!1
3R
B33
33R
D1000
0100
0010
0001
T 2. Exchange the values for aand d in the first diagonal inthe matrix. Multiply thevalues for b and c by !1 inthe second diagonal in thematrix. Find the determinantof the original matrix. Multiplythe negative reciprocal of thedeterminant by the matrixwith the above mentionedchanges.
4. no
6.
8.
10. yes
12. no
14. yes
16. true
18. true
20.
22.
24. no inverse exists
26. 134
B 7!2
34R
!13
B 1!2
!21R1
5 B1
005R
!127
B 4!7
!1!5RB2
358R
Lesson 4-7 Identity and Inverse MatricesPages 198–201
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27.
29.
31.
33a. yes33b. Sample answer:
35.
37. dilation by a scale factor of 12
B00
!44
412
88R
10 C 34
!15
!58
310
S132
B 1!6
52R!
112
B 6!5
0!2R 28. no inverse exists
30.
32a. no32b. Sample answer:
34.
36. dilation by a scale factor of 2
38. the graph of the
inverse transformation is theoriginal figure.
B!1D120
0
12
T;
B00
!22
26
44R
y
xA'A''
AB
C
C'
B'
C''
B''O
4 C 14
!16
34
12
S
©Glencoe/McGraw-Hill 105 Algebra 2 Chapter 4
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39. MEET_IN_THE_LIBRARY
41. BRING_YOUR_BOOK
43.
45. A
47.
49.
51.
53. (2, !4)
F!1
!13
73
1
23
!83
13
0
!13
VC35
15
!15
!25
SB!5!6
!9!11R
a # +1, d # +1, b # c # 0
40. AT_SIX_THIRTY
42. See students’ work.
44. A matrix can be used tocode a message. The key tothe message is the inverseof the matrix. Answersshould include the following.• The inverse matrix undoes
the work of the matrix. Soif you multiply a numericmessage by a matrix itchanges the message.When you multiply thechanged message by theinverse matrix, the result isthe original numericmessage.
• You must consider thedimensions of the codingmatrix so that you canwrite the numericmessage in a matrix withdimensions that can bemultiplied by the codingmatrix.
46. D
48. no inverse exists
50.
52.
54. (0, 7)
F 516
!14
532
!18
0
316
!116
14
!132
VC 3
5
312
25
212
S
©Glencoe/McGraw-Hill 106 Algebra 2 Chapter 4
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55. (!5, 4, 1)
57. !14
59. 1
61. !5
63.
65. 7.82 tons/in2
67.
69. 3
71. 300
73. !2
75. 4
77. !34
512
52
56. 52
58. 0
60. !3
62.
64.
66. 27
68.
70. 296
72. !1
74. 6
76. !27
!12
!38
13
©Glencoe/McGraw-Hill 107 Algebra 2 Chapter 4
1. 2r ! 3s # 4, r $ 4s # !2
3. Tommy; a 2 " 1 matrixcannot be multiplied by a 2 " 2 matrix.
5.
7. (5, !2)
9. (!3, 5)
B 2!4
3!7R ! Bg
hR # B 8
!5R
2. Sample answer:and
4.
6.
8. (1.5, !4)
10. (1, 1.75)
# C 93
12SC3 !5 2
4 7 12 0 !1
S ! CabcS
B11
!13R ! Bx
yR # B!3
5R2x $ 6y # 16
x $ 3y # 8
Lesson 4-8 Using Matrices to Solve Systems of EquationsPages 205–207
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11.
13.
15.
17.
19.
21. (3, 4)
23. (6, 1)
25.
27. (!2, !2)
29. (0, 9)
31.
33. 2010
a32,
13b
a!13, 4b
# C 2115
!7SC 3
11!5
!5!12
8
616!3S ! C r
stS# C 9
11!1SC3
14
!5!70
23
!3S ! Cx
yzS
B36
!75R ! Bm
nR # B!43
!10RB4
3!7 5R ! Bx
yR # B2
9Rh # 1, c # 12 12.
14.
16.
18.
20. (5, !2)
22. (!2, 3)
24.
26. (2, !3)
28. (7, 3)
30.
32. 27 h of flight instruction and23 h in the simulator
34. 80 mL of the 60% solution, and120 mL of the 40% solution
a!1, 92b
a12, !3b
# C 8!27
54SC 1 !1 0
!2 !5 !69 10 !1
S ! CxyzS# C 1
7!5SC2 3 !5
7 0 33 !6 1
S ! CabcSB5
3!6
2R ! Ba
bR # B!47
!17RB3
1!1
2R ! Bx
yR # B 0
!21R
©Glencoe/McGraw-Hill 108 Algebra 2 Chapter 4
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35. The solution set is the emptyset or infinite solutions.
37. D
39. (!6, 2, 5)
41. (0, !1, 3)
43.
45. (4, !2)
47. (!6, !8)
49. {!4, 10}
51. {2, 7}
B 4!7
!59R
36. The food and territory that twospecies of birds require forma system of equations. Anyindependent system ofequations can be solved usinga matrix equation. Answersshould include the following.• Let a represent the
number of nesting pairs ofSpecies A and let brepresent the number ofnesting pairs of Species B.Then, 20,000 and
•
so thearea can support 70 pairsof Species A and 85 pairsof Species B.
38. 17 small, 24 medium, 11 large
40. (1, !3, 2)
42.
44. no inverse exists
46. (4.27, !5.11)
48. about 114.3 ft
50. {!5, 1}
C 34
!12
!1
1S
a # 70 and b # 85,
!1
4000B 400 !120!500 140
R ! B20,00069,000
R;BabR #
69,000.500a $ 400b #
140a $ 120b #
©Glencoe/McGraw-Hill 109 Algebra 2 Chapter 4
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©Glencoe/McGraw-Hill 110 Algebra 2 Chapter 5
Chapter 5 PolynomialsLesson 5-1 Monomials
Pages 226–228
1. Sample answer:(2x2)3 ! 8x6 since (2x2)3 ! 2x2 2x2 2x2 !2x x 2x x 2x x ! 8x6
3. Alejandra; when Kyle used thePower of a Product Propertyin his first step, he forgot to put an exponent of "2 on a.Also, in his second step,("2)"2 should be , not 4.
5. 16b4
7. "6y 2
9. 9p2q3
11.
13.
15. 3.762 # 103
17. about 1.28 s
19. b4
21. z10
23. "8c3
25. "y3z2
27. "21b5c3
29. "24r7s5
31. 90a4b4
4.21 # 105
9c
2d
2
14
!!!!!!!
2. Sometimes; in general, xy xz ! xy$z, so xy xz ! xyz when
such as whenand
4. x10
6. 1
8.
10.
12.
14.
16. 5 # 100
18. a8
20. n16
22. 16x 4
24. an
26.
28. ab
30. 24x4y4
32. "1
4y 4
28x 4
y 2
8.62 # 10"4
14x
6
1w12z
6
"ab
4
9
z ! 2.y ! 2y $ z ! yz,
!!
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©Glencoe/McGraw-Hill 111 Algebra 2 Chapter 5
33.
35.
37.
39.
41.
43. 7
45. 4.32 # 104
47. 6.81 # 10"3
49. 6.754 # 108
51. 6.02 # 10"5
53. 6.2 # 1010
55. 1.681 # 10"7
57. 2 # 10"7 m
59. about 330,000 times
61. Definition of an exponent
2x 3y 2
5z 7
1v 3w 6
8y 3
x 6
"m 4n 9
3
a2c2
3b4 34.
36.
38.
40.
42. 6
44. 4.623 # 102
46. 1.843 # 10"4
48. 5.0202 # 108
50. 1.245 # 1010
52. 4.5 # 102
54. 4.225 # 109
56. 6.08 # 109
58. 1.67 # 1025
60. 10010 ! (102)10 or 1020, and10100 % 1020, so 10100 % 10010.
62. (ab)m
!
!
! ambm
m factors64748b ! b ! p ! b
m factors64748a ! a ! p ! a !
m factors6447448
ab ! ab ! p ! ab
a4b2
2
1x 2y 2
a4
16b4
cd 4
5
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©Glencoe/McGraw-Hill 112 Algebra 2 Chapter 5
63. Economics often involveslarge amounts of money.Answers should include thefollowing.• The national debt in 2000
was five trillion, six hundredseventy-four billion, twohundred million or 5.6742 #1012 dollars. The populationwas two hundred eighty-onemillion or 2.81 # 108.
• Divide the national debt bythe population.
orabout $20,193 per person.
65. B
67. ("3, 3)
69.
71. 7
73. (2, 0, 4)
75. Sample answer using (0, 4.9)and (28, 8.3): y ! 0.12x $ 4.9
77. 7
79. 2x $ 2y
C"12
1
32
"2 S
$2.0193 # 104
!5.6742 # 1012
2.81 # 108
64. D
66. (1, 2)
68.
70. "6
72. (2, 3, "1)
74.
76. Sample answer: 9.7 yr
78. "3
80. 3x " 3z
4
00
5
6
7
Med
ian
Age
(yr) 8
Years Since 197010 20 30
y
x
Median Age of Vehicles
c"21
"52d
PQ245-6457F-P05[110-134] 7/24/02 1:30 PM Page 112 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-05:
©Glencoe/McGraw-Hill 113 Algebra 2 Chapter 5
1. Sample answer:
3.
5. yes, 3
7.
9.
11.
13.
15.
17. yes, 3
19. no
21. yes, 7
23.
25.
27.
29.
31.
33.
35.
37.
39.
41. 6x 2 $ 34x $ 48
b2 " 25
p2 $ 2p " 24
"0.001x2 $ 5x " 500
2a4 " 3a3b $ 4a4b4
6x 2y 4 " 8x 2y 2 $ 4xy 5
12a3 $ 4ab
7x 2 " 8xy $ 4y 2
10m2 $ 5m " 15
"3y " 3y2
7.5x2 $ 12.5x ft 2
4z2 " 1
y2 " 3y " 70
6xy $ 18x
10a " 2b
x
x
x
x
x
x
x 2x
x x x
x 2 x 2
2
x5 $ x4 $ x3 2. 4
4. yes, 1
6. no
8.
10.
12.
14.
16. yes, 2
18. no
20. yes, 6
22.
24.
26.
28.
30.
32.
34.
36. $5327.50
38.
40. 36 " z2
42. 8y 2 $ 16y " 42
a2 $ 9a $ 18
46.75 " 0.018x
6x3 $ 9x2y " 12x3y2
15a3b3 " 30a4b3 $ 15a5b6
4b2c " 4bdz
4x2 " 3xy " 6y2
r 2 " r $ 6
4x2 $ 3x " 7
4m2 " 12mn $ 9n2
x2 $ 9x $ 18
10p3q2 " 6p5q3 $ 8p3q5
"3x2 " 7x $ 8
Lesson 5-2 PolynomialsPage 231–232
81.
83. "5x $ 10y
4x $ 8 82.
84. 3y " 15
"6x $ 10
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©Glencoe/McGraw-Hill 114 Algebra 2 Chapter 5
43.
45.
47.
49.
51.
53.
55. The expression for how muchan amount of money will growto is a polynomial in terms ofthe interest rate. Answersshould include the following.• If an amount A grows by r
percent for n years, theamount will be A(1 $ r )n
after n years. When thisexpression is expanded, apolynomial results.
• 13,872(1 $ r )3, 13,872r3 $41,616r 2 $ 41,616r $13,872
• Evaluate one of theexpressions when r ! 0.04.For example, 13,872(1 $ r)3 !13,872(1.04)3 or $15,604.11to the nearest cent. Thevalue given in the table is$15,604 rounded to thenearest dollar.
57. B
59. 20r 3t 4
61. b 2
4a 2
R2 $ 2RW $ W 2
9c2 " 12cd $ 7d2
27b3 " 27b2c $ 9bc2 " c3
d 2 " 2 $1
d 4
x 2 " 6xy $ 9y 2
a6 " b2 44.
46.
48.
50.
52.
54. 14; Sample answer:(x8 $ 1)(x6 $ 1) !
56. D
58. "64d 6
60.
62. (1, 4)
xz 2
y 2
x14 $ x8 $ x6 $ 1
"18x 2 $ 27x " 10
x3 " y3
xy3 $ y $1x
1 $ 8c $ 16c2
2m4 " 7m2 " 15
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©Glencoe/McGraw-Hill 115 Algebra 2 Chapter 5
1. Sample answer:(x2 $ x $ 5) & (x $ 1)
3. Jorge; Shelly is subtracting inthe columns instead of adding.
5.
7.
9.
11.
13.
15.
17. 2c2 " 3d $ 4d2
3ab " 6b2
3b $ 5
b3 $ b " 1
x2 " xy $ y2
3a3 " 9a2 $ 7a " 6
5b " 4 $ 7a
2. The divisor contains an x2
term.
4.
6.
8.
10.
12.
14. B
16.
18. 4n2 $ 3mn " 5m
5y "6y 2
x$ 3xy2
2y $ 5
x2 $ 11x " 34 $60
x $ 2
z4 $ 2z3 $ 4z2 $ 5z $ 10
x " 12
6y " 3 $ 2x
Lesson 5-3 Dividing PolynomialsPages 236–238
63.
65.
67. 2y3
69. 3a2
y
xO
2x ! y " 1
y
xO
y " # x ! 2 13
64.
66. x2
68. xy2
y
xO
x ! y $ #2
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©Glencoe/McGraw-Hill 116 Algebra 2 Chapter 5
19.
21.
23.
25.
27. x2
29.
31.
33.
19 "
35. g $ 5
37.
39.
41.
43.
45. x " 3
47. x $ 2
49. x2 " x $ 3
x3 " x "6
2x $ 3
3d2 $ 2d $ 3 "2
3d " 2
3t2 " 2t $ 3
t4 $ 2t3 $ 4t2 $ 5t $ 10
56x $ 3
x4 " 3x3 $ 2x2 " 6x $
a3 " 6a2 " 7a $ 7 $3
a $ 1
y2 " y " 1
39x $ 2
x3 " 5x2 $ 11x " 22 $
n2 " 2n $ 3
b2 $ 10b
2y2 $ 4yz " 8y3z4 20.
22.
24.
26.
28. x2 $ 3x $ 9
30.
32.
34.
36.
38.
40.
42.
44.
46.
48. x " 3
50. 2y2 " 3y $ 1
x2 " 1 $"3x $ 7x 2 $ 2
2x3 $ x2 " 1 $2
3x $ 1
x2 $ x " 1
h2 " 4h $ 17 "51
2h $ 3
y4 " 2y3 $ 4y2 " 8y $ 16
2b2 " b " 1 $4
b $ 1
9 "13
c $ 2
3c4 " c3 $ 2c2 " 4c $
1 $5
m " 3
2m3 $ m2 $ 3m "
m2 " 7
30w $ 606w4 $ 12w3 $ 24w2 $
2c2 $ c $ 5 $6
c " 2
x " 15
"a2b $ a "2b
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©Glencoe/McGraw-Hill 117 Algebra 2 Chapter 5
51.
53.
55.
57. ft /s
59. Division of polynomials canbe used to solve for unknownquantities in geometricformulas that apply to manufacturing situations.Answers should include thefollowing.• 8x in. by 4x $ s in.• The area of a rectangle is
equal to the length timesthe width. That is, .
• Substitute 32x2 $ x for A,8x for /, and 4x $ s for w.Solving for s involvesdividing 32x2 $ x by 8x.
32x2 $ x ! 8x (4x $ s)
! 4x $ s
4x $ ! 4x $ s
! s
The seam is inch.
61. D
18
18
18
32x2 $ x8x
A ! /w
A ! /w
x2 $ 3x $ 12
x 3 $ x 2 $ 6x " 24 ft
170 "170
t 2 $ 1
$0.03x $ 4 $1000
x52. Let x be the number.
Multiplying by 3 results in 3x.The sum of the number, 8,and the result of themultiplication is x $ 8 $ 3xor 4x $ 8. Dividing by thesum of the number and2 gives or 4. The end
result is always 4.
54. 85 people
56. x " 2 s
58. Sample answer: r3 " 9r2 $27r " 28 and r " 3
60. A
62. "x2 " 4x $ 14
4x $ 8x $ 2
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©Glencoe/McGraw-Hill 118 Algebra 2 Chapter 5
63.
65.
67.
69. 9
71. 4
73. 6
y ! "x $ 2
a2 " 2ab $ b2
y4z4 " y3z3 $ 3y2z
1. 6.53 # 108
3. "108x8y3
5.
7.
9. m2 " 3 "19
m " 4
3t 2 $ 2t " 8
x 2
z 6
64.
66. 5 # 102 s or 8 min 20 s
68.
70. 12
72. 3
74. 5
y !23x "
43
y2 $ 2y " 15
2. 7.2 # 10"3
4.
6.
8.
10. d2 $ d " 3
n3 " n2 " 5n $ 2
2x $ 5y
a3
b4c3
Chapter 5Practice Quiz 1
Page 238
Lesson 5-4 Factoring PolynomialsPages 242–244
1. Sample answer:
3. sometimes
5. a(a $ 5 $ b)
7. (y $ 2)(y " 4)
9. 3(b " 4)(b $ 4)
11. (h $ 20)(h2 " 20h $ 400)
x2 $ 2x $ 1 2. Sample answer: If a ! 1 andb ! 1, then but
4. "6x(2x $ 1)
6. (x $ 7)(3 " y)
8. (z " 6)(z $ 2)
10. (4w $ 13)(4w " 13)
12. x " 4x " 7
1a $ b22 ! 4.a2 $ b2 ! 2
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©Glencoe/McGraw-Hill 119 Algebra 2 Chapter 5
13.
15. 2x(y3 " 5)
17. 2cd2(6d " 4c $ 5c4d )
19. (2z " 3)(4y " 3)
21. (x $ 1)(x $ 6)
23. (2a $ 1)(a $ 1)
25. (2c $ 3)(3c $ 2)
27. 3(n $ 8)(n " 1)
29. (x $ 6)2
31. prime
33.
35.
37.
39.
41.
43.
45. 30 ft by 40 ft
47.
49.
51. x $ 2
53. 16x $ 16 ft /s
x " 4x 2 $ 2x $ 4
x $ 5x " 6
(3x " 2)(x $ 1)
(a " b)(5ax $ 4by $ 3cz)
(7a $ 2b)(c $ d )(c " d )
(p2 $ 1)(p $ 1)(p " 1)
(z $ 5)(z2 " 5z $ 25)
(y2 $ z)(y2 " z)
2yy " 4
14. x $ y cm
16.
18. prime
20. (3a $ 1)(x " 5)
22. (y " 1)(y " 4)
24. (2b " 1)(b $ 7)
26. (3m $ 2)(4m " 3)
28. 3(z $ 3)(z $ 5)
30. (x " 3)2
32. 3(m $ n)(m " n)
34. 3(x $ 3y)(x " 3y)
36.
38.
40.
42.
44.
46.
48.
50. x
52. x " 1 s
54. x " 8 cm
x " 5x " 2
x $ 1x " 4
(2y $ 1)(y $ 4)
(a $ 3b)(3a $ 5)(a " 1)
(8x $ 3)(x $ y $ z)
(x2 $ 9)(x $ 3)(x " 3)
(t " 2)(t2 $ 2t $ 4)
6ab2(a $ 3b)
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©Glencoe/McGraw-Hill 120 Algebra 2 Chapter 5
55. (8pn $ 1)2
57. B
59. yes
61. no; (2x $ 1)(x " 3)
63. t2 " 2t $ 1
65. x2 $ 2
67.
69. ["2]
71. 15 in. by 28 in.
73. no
75. Associative Property ($)
77. irrational
79. rational
81. rational
4x2 $ 3xy " 3y2
56. Factoring can be used to findpossible dimensions of ageometric figure, given thearea. Answers should includethe following.• Since the area of the
rectangle is the product ofits length and its width, thelength and width are factorsof the area. One set ofpossible dimensions is4x " 2 by x $ 3.
• The complete factorizationof the area is 2(2x " 1)(x $ 3), so the factor of 2 could be placed witheither 2x " 1 or x $ 3when assigning thedimensions.
58. C
60. no;
62. yes
64. y $ 3
66.
68.
70.
72. yes
74. Distributive Property
76. rational
78. rational
80. irrational
c"3618
74d
14x2 $ 26x " 4
x3 $ x2 " 2x $ 2 $1
3x " 2
(x $ 2)(x2 " 2x $ 4)
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©Glencoe/McGraw-Hill 121 Algebra 2 Chapter 5
2. If all of the powers in theresult of an even root haveeven exponents, the result isnonnegative without takingabsolute value.
4. 8.775
6. 2.632
8. 2
10. not a real number
12.
14.
16. 11.358
18. 0.933
20. 3.893
22. 4.953
24. 4.004
26. 26.889
28. 15
30. not a real number
32. "3
34.
36. 0.5
38. z2
40.
42. 3r
44. 25g2
46. 5x2 0y3 0
7 0m3 0
14
04x $ 3y 00y 0
1. Sample answer: 64
3. Sometimes; it is true when x % 0.
5. "2.668
7. 4
9. "3
11. x
13.
15. about 3.01 mi
17. "12.124
19. 2.066
21. "7.830
23. 3.890
25. 4.647
27. 59.161
29. '13
31. 18
33. "2
35.
37. "0.4
39.
41. 8a4
43. "c2
45. 4z2
" 0x 015
6 0a 0b2
Lesson 5-5 Roots of Real NumbersPages 247–249
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©Glencoe/McGraw-Hill 122 Algebra 2 Chapter 5
47. 6x2z2
49.
51. "3c3d4
53. p $ q
55.
57. not a real number
59. "5
61. about 1.35 m
63. x ! 0 and y ( 0, or y ! 0and x ( 0
65. B
67. 7xy2(y " 2xy3 $ 4x2)
69. (2x $ 5)(x $ 5)
71.
73.
75. (1, "3)
77.
79.
81. x2 " 9y2
a2 " 7a " 18
x2 $ 11x $ 24
c 8101418
23202504
d4x2 $ x $ 5 $
8x " 2
0z $ 4 0
3p6 0q3 0 48. 13x4y2
50. 2ab
52.
54.
56.
58. 2
60. about 127.28 ft
62. about 11,200 m)s
64. The speed and length of awave are related by anexpression containing asquare root. Answers shouldinclude the following.• about 1.90 knots, about
3.00 knots, and 4.24 knots• As the value of / increases,
the value of s increases.
66. D
68. (a $ 3)(b " 5)
70. (c " 6)(c2 $ 6c $ 36)
72.
74. ("2, 2)
76. (9, 4)
78.
80.
82. 6w2 " 7wz " 5z2
a2 $ 3ab $ 2b2
y2 $ 3y " 10
x3 " x2 $ x
02a $ 1 0" 0x $ 2 004x " y 0
PQ245-6457F-P05[110-134] 7/24/02 1:30 PM Page 122 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-05:
©Glencoe/McGraw-Hill 123 Algebra 2 Chapter 5
1. Sometimes; only
when 3. The product of two
conjugates yields adifference of two squares.Each square produces arational number and thedifference of two rationalnumbers is a rationalnumber.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
35.
37. 723 " 222
323
262
3627
a22bb2
23 62
13
c 0d 024 c
6y2z23 7
3 0x 0y 22y
5x222
323 2
923
2 $ 25
2223 2
2a2b223
"24235
2x 0y 024 x
a ! 1.
! 1n a11n a2. Sample answer:
4.
6.
8.
10.
12.
14. about 49 mph
16.
18.
20.
22.
24.
26.
28.
30.
32.
34.
36.
38. 425 $ 2326
522
2105
"60230
2r 42tt 5
24 543
12wz 25 wz2
4mn23 3mn2
2ab2210a
2y23 2
224 6
622
3 $ 323 " 25 " 215
523 $ 324 3
23 25
214y4y
1527
22 $ 23 $ 22
Lesson 5-6 Radical ExpressionsPages 254–256
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©Glencoe/McGraw-Hill 124 Algebra 2 Chapter 5
39.
41.
43.
45.
47.
49.
51. 0 ft /s
53. about 18.18 m
55. x and y are nonnegative.
57. B
59. 12z4
61.
63. x $ 1x $ 4
0y $ 2 0
6 $ 1622 yd, 24 $ 622 yd2
2x 2 " 1x " 1
"1 " 232
28 $ 72313
13 " 2222
25 " 522 $ 526 " 223 40.
42.
44.
46.
48.
50. The square root of adifference is not thedifference of the squareroots.
52.
54. 80 ft /s or about 55 mph
56. The formula for the time ittakes an object to fall acertain distance can bewritten in various formsinvolving radicals. Answersshould include the following.• By the Quotient Property
of Radicals, t !
Multiply by to
rationalize the denominator.
The result is h !
• about 1.12 s
58. D
60. 6ab3
62.
64. £"293
"415
"5§
x $ 7x " 4
22dgg
.
2g2g
22d2g.
d ! v 24.9h
4.9
2x $ 1
12 $ 72223
526 " 32222
8 " 2215
6 $ 326 $ 227 $ 242
PQ245-6457F-P05[110-134] 7/24/02 1:30 PM Page 124 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-05:
©Glencoe/McGraw-Hill 125 Algebra 2 Chapter 5
65.
67. consistent and independent
69. "5
71. "2, 4
73.
75.
77.
79.
81. 38
1324
56
14
5x 0x % 66
c 1"5
4"4d
1.
3.
5.
7.
9. "1 "17
02n $ 3 06 0x 0 0y3 0a(x $ 3)2
x2y (3x $ y $ 1) 2. prime
4. 8(r " 2s2)(r2 $ 2rs2 $ 4s4)6.
8.
10. 8 " 3222
x 22yy 2
"4a2b3
66. 16, "15
68. $4.20
70. 2
72.
74.
76.
78.
80.
82. "512
1930
1312
12
5x 0x ( "76"
73, 1
Chapter 5Practice Quiz 2
Page 256
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©Glencoe/McGraw-Hill 126 Algebra 2 Chapter 5
1. Sample answer: 64
3. In exponential form isequal to . By the Powerof a Power Property,
. But, is also equal to by the Power ofa Power Property. This lastexpression is equal to Thus,
5. or
7.
9.
11. 2
13.
15.
17.
19.
21.
23.
25.
27.
29. 2
31. 15
2 z
12
2312
25 c2 or (25 c)2
25 6
23
z (x " 2y)x " 2y
12
a
32 b
23
x
23
13
613x
53y
73
(23 x)223 x2
n2bm ! (n2b)m.
( n2b)m .
(b1n)m
bmn(bm)
1n ! b
mn
(bm)1n
2n bm
2. In radical form, the expressionwould be which isnot a real number becausethe index is even and theradicand is negative.
4.
6.
8. 5
10. 9
12.
14.
16.
18.
20. $5.11
22.
24.
26.
28.
30. 6
32. 127
513 x
23 y
13
6213
x223 x2
23 4
23x
m n mn
23
13
z 2z
23
a1112
2614
23 7
2"16,
Lesson 5-7 Rational ExponentsPages 260–262
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©Glencoe/McGraw-Hill 127 Algebra 2 Chapter 5
33.
35. 81
37.
39.
41. y 4
43.
45.
47.
49.
51.
53.
55.
57.
59.
61.
63.
65.
67. 880 vibrations per second
232 $ 3
12
216 " 5
12
xy1zz
24 5x2y2
1726 17
25
y 2 " 2y y " 4
32
a 6a
512
t 14
w
w
15
b15
43
23
19
34.
36. 4096
38. 27
40.
42. x3
44.
46.
48.
50.
52.
54.
56.
58.
60.
62.
64.
66.
68. about 262 vibrations persecond
2 ! 3
13
x " x
13z
23
23 6
ab23 c 2
c
b23 9a2b
526 55
23
x $ 3x $ 2x " 1
12
2c c
1516
r12
x x
56
a19
12
"18
PQ245-6457F-P05[110-134] 7/24/02 1:30 PM Page 127 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-05:
©Glencoe/McGraw-Hill 128 Algebra 2 Chapter 5
69. about 336
71. The equation that determinesthe size of the region around aplanet where the planet’s gravityis stronger than the Sun’s canbe written in terms of afractional exponent. Answersshould include the following.• The radical form of the
equation is r ! D or
r ! D Multiply the
fraction under the radical by .
r ! D
! D
! D
!
The simplified radical form is
r !
• If and are constant,then r increases as Dincreases because r is alinear function of D withpositive slope.
MsMp
D25 Mp2M3
s
Ms.
D25 Mp2M3
s
Ms
25 Mp2M3
s 25 Ms5
B5 M2p Ms
3
Ms5
B5 Mp2
Ms2 !
Ms3
Ms3
M3s
M3s
B5 Mp2
Ms2.
B5 aMp
Msb2
70. Rewrite the equation so thatthe bases are the same oneach side.
Since the bases are thesame and this is an equation,the exponents must be equal.Solve 2x ! x $ The result
is x !
72. C
12
.
12
.
32x ! 3x$12
(32)x ! 3x$12
9x ! 3x$12
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©Glencoe/McGraw-Hill 129 Algebra 2 Chapter 5
73. C
75.
77. 8
79.
81.
83. x $ 22x $ 1
x " 2
12x2
3622
74.
76.
78.
80. 1440
82.
84. 4x " 122x $ 9
2x " 3
4 0x " 5 0222
2x 0y 02x
Lesson 5-8 Radical Equations and InequalitiesPages 265–267
1. Since x is not under theradical, the equation is alinear equation, not a radicalequation. The solution is
x !
3. Sample answer:
5. "9
7. 15
9. 31
11.
13. 16
15. no solution
17. 9
19. "1
21. "20
0 * b + 4
2x $ 2x $ 3 ! 3
23 " 12
.
2. The trinomial is a perfectsquare in terms of .
so the equation can bewritten as Take the square root of eachside to get Usethe Addition Property ofEquality to add 3 to eachside, then square each sideto get x ! 9.
4. 2
6. no solution
8. 18
10.
12. about 13.42 cm
14. 49
16. no solution
18. 5
20.
22. 5
272
"32
* x * 39
1x " 3 ! 0.
(1x " 3)2 ! 0.
(1x " 3)2,x " 61x $ 9 !1x
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©Glencoe/McGraw-Hill 130 Algebra 2 Chapter 5
23. no solution
25. x % 1
27. x * "11
29. no solution
31. 3
33. 0 * x * 2
35. b ( 5
37. 3
39. 1152 lb
41. 34 ft
43. Since andthe left side of
the equation is nonnegative.Therefore, the left side of theequation cannot equal "1.Thus, the equation has nosolution.
12x " 3 ( 0,1x $ 2 ( 0
24. 9
26. "2 * x * 1
28. y % 4
30. 4
32. no solution
34. 0 * a + 3
36.
38. 16
40.
42. 21.125 kg
44. If a company’s cost andnumber of units manufacturedare related by an equationinvolving radicals or rationalexponents, then theproduction level associatedwith a given cost can befound by solving a radicalequation. Answers shouldinclude the following.•
•
Round down so that the costdoes not exceed $10,000.The company can make atmost 24,781 chips.
24,781.55 ! n
85032 ! n
850 ! n23
8500 ! 10n23
10,000 ! 10n23 $ 1500
C ! 1023 n2 $ 1500
t ! B 4,2r 3
GM
c % "7916
C ! 10,000
Subtract 1500from each side.
Divide eachside by 10.
Raise eachside to the
power.
Use acalculator.
32
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©Glencoe/McGraw-Hill 131 Algebra 2 Chapter 5
45. D
47.
49.
51.
53.
55. 1 " y
57. "11
59. "3 " 10x " 8x2
y
xO
(2, 5)
x ! y " 7
30x ! 20y " 160
(2, 5)x $ y ! 7, 30x $ 20y ! 160;
23 10010
(x2 $ 1)23
537
46. C
48.
50.
52.
54. 4 $ x
56. 2 $ 4x
58. 4 $ 6z $ 2z2
28 " 1023
6 0x3 0y22y
(x $ 7)12
Lesson 5-9 Complex NumbersPages 273–275
1a. true1b. true
3. Sample answer: 1 $ 3i and 1 " 3i
5.
7.
9. 6 $ 3i
11.
13.
15. 3, "3
17. 10 $ 3j amps
'2i22
717
"1117
i
"18023
5i 0xy 022
2. all of them
4. 6i
6. 12
8. i
10. 42 " 2i
12. '3i
14.
16. 5, 4
18. 12i
'i25
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©Glencoe/McGraw-Hill 132 Algebra 2 Chapter 5
19. 9i
21.
23. "12
25. "75i
27. 1
29. "i
31. 6
33. 4 " 5i
35. 6 " 7i
37. "8 $ 4i
39.
41.
43. 20 $ 15i
45.
47.
49. '4i
51.
53.
55.
57. 4, "3
59.
61.
63. 13 $ 18j volts
6711
, 1911
53, 4
'252
i
'2i210
'2i23
(5 " 2i )x2 $ ("1 $ i )x $ 7 $ i
"13
"222
3i
25
$15i
1017
"617
i
10a2 0b 0 i 20. 8x2i
22.
24. "48i
26. i
28. "1
30. 9 $ 2i
32. 2
34. 25
36. 8 $ 4i
38.
40.
42. "163 " 16i
44.
46.
48. 'i
50.
52.
54.
56. 4, 5
58.
60. 3, 1
62. 5 " 2j ohms
64. 4 $ 2j amps
"72, "3
'3i25
'i23
'i26
(j $ 4)x2 $ (3 " i )x $ 2 " 4i
1114
"523
14i
3917
$1417
i
25
$65i
"1322
PQ245-6457F-P05[110-134] 7/24/02 1:30 PM Page 132 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-05:
©Glencoe/McGraw-Hill 133 Algebra 2 Chapter 5
65. Case 1: i % 0Multiply each side by i to geti2 % 0 i or "1 % 0. This is acontradiction.Case 2: i + 0Since you are assuming i isnegative in this case, youmust change the inequalitysymbol when you multiplyeach side by i. The result isagain i2 % 0 i or "1 % 0, a contradiction. Since bothpossible cases result incontradictions, the orderrelation “+” cannot beapplied to the complexnumbers.
67. C
69. "1, "i, 1, i, "1, "i, 1, i, "1
71. 12
73. 4
75.
77. c23
1"2
"21d
y
13
!
!
66. Some polynomial equationshave complex solutions.Answers should include thefollowing.• a and c must have the
same sign.• 'i
68. C
70. Examine the remainder whenthe exponent is divided by 4.If the remainder is 0, theresult is 1. If the remainder is1, the result is i. If theremainder is 2, the result is"1. And if the remainder is3, the result is "i.
72. 11
74.
76.
78. c10
0"1d
aa
14
x
715
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©Glencoe/McGraw-Hill 134 Algebra 2 Chapter 5
79.
81. sofa: $1200, love seat: $600,coffee table: $250
83.
85. 0
y
xO
x ! y " 1
x # 2y " 4
c 2"3
12
"2"1d 80.
82.
84. 110
y
xO
y " #2x # 2
y " x ! 1
y
xO
A'
C'
B'
PQ245-6457F-P05[110-134] 7/24/02 1:30 PM Page 134 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-05:
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.
5a.
5b.
5c. f(x)
xO(!1, !1)f(x) " x 2 # 2x
x f(x)#3 3#2 p 0#1 #1
0 01 3
0; x ! #1; #1
Chapter 6 Quadratic Functions and InequalitiesLesson 6-1 Graphing Quadratic Functions
Pages 290–293
©Glencoe/McGraw-Hill 135 Algebra 2 Chapter 6
2a.2b.
4a. 0; x ! 0; 0
4b.
4c.
6a.
6b.
6c.
xO
f(x)
(2, 3) f(x) " !x 2 # 4x ! 1
x f(x)0 #11 p 22 33 24 #1
#1; x ! 2; 2
xO (0, 0)
f(x) " !4x 2
f(x)
x f(x)#1 #4
0 p 01 #4
(#3, #2); x ! #3(2, 1); x ! 2
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7a.
7b.
7c.
9a.
9b.
9c.
11. min.; #254
!4 !2 2
4
!4
!8
!12
xO
f(x)
f (x) " 3x 2 # 10x(! , ! )53
253
0; x ! #53; #
53
xO
f(x)
!4!8!10
!8
!4
!12f(x) " x 2 # 8x # 3
(!4, !13)
x f(x)#6 #9#5 p #12#4 #13#3 #12#2 #9
3; x ! #4; #4 8a.
8b.
8c.
10. max.; 7
12. min.; 0
xO
f(x)
(1, !1)
f (x) " 2x 2 ! 4x # 1
x f(x)#1 7
0 p 11 #12 13 7
1; x ! 1; 1
©Glencoe/McGraw-Hill 136 Algebra 2 Chapter 6
x f(x)#3 #3#2 #8
#1 #70 0
#253
#53
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13. $8.75
15a. 0; x ! 0; 0
15b.
15c.
17a. #9; x ! 0; 0
17b. x f(x)#2 #5#1 p #8
0 #91 #82 #5
(0, 0) xO
f (x) " !5x 2
f(x)
x f(x)#2 #20#1 p #5
0 01 #52 #20
14a. 0; x ! 0; 0
14b.
14c.
16a. 4; x ! 0; 0
16b.
16c.
18a. #4; x ! 0; 0
18b. x f(x)#2 4#1 p #2
0 #41 #22 4
xO
f (x) " x 2 # 4
!2!4 2 4
8
12
(0, 4)
f(x)
x f(x)#2 8#1 p 5
0 41 52 8
xO
(0, 0)
f(x)
f (x) " 2x 2
x f(x)#2 8#1 p 2
0 01 22 8
©Glencoe/McGraw-Hill 137 Algebra 2 Chapter 6
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18c.
20a. 4; x ! 2; 2
20b.
20c.
22a. #5; x ! 2; 2
22b.
22c.xO
f(x)
(2, !9)f (x) " x 2 ! 4x ! 5
x f(x)0 #51 p #82 #93 #84 #5
xO
f(x)
(2, 0)
f (x) " x 2 ! 4x # 4
x f(x)0 41 p 12 03 14 4
xO
(0, !4)f (x) " 2x 2 ! 4
f(x)17c.
19a. 191; x ! 0; 0
19b.
19c.
21a. 9; x ! 4.5; 4.5
21b.
21c.2
4 8 12
!4
!8
!12 (4 , !11 )12
14
xO
f(x)
f (x) " x 2 ! 9x # 9
x f(x)3 #94 p #114.5 #11.255 #116 #9
f (x)
xO
(0, 1)f (x) " 3x 2 # 1
x f(x)#2 13#1 p 4
0 11 42 13
xO!2!4 2 4
!4
4
(0, !9) f (x) " x 2 ! 9
f(x)
©Glencoe/McGraw-Hill 138 Algebra 2 Chapter 6
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23a. 36; x ! #6; #6
23b.
23c.
25a. #3; x ! 2, 2
25b.
25c.
27a. 0; x ! #54; #
54
xO
f(x) (2, 5)
f (x) " !2x 2 # 8x ! 3
x f(x)0 #31 p 32 53 34 #3
xO
f(x)
2
4
6
!8 !4!12!16(!6, 0)
f (x) " x 2 # 12x # 36
x f(x)#8 4#7 p 1#6 0#5 1#4 4
24a. #1; x ! #1; #1
24b.
24c.
26a.
26b.
26c.
28a. #1; x ! 0; 0
xO
f(x)(! , )2
343
f (x) " !3x 2 ! 4x
0; x ! #23, #
23
xO
f(x)
(!1, !4)
f (x) " 3x2 # 6x ! 1
x f(x)#3 8#2 p #1#1 #4
0 #11 8
©Glencoe/McGraw-Hill 139 Algebra 2 Chapter 6
x f(x)#2 #4#1 1
0 01 #7
43
#23
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28b.
28c.
30a.
30b.
30c.
xO
f(x)
(!3, 0)
f (x) " x 2 # 3x # 12
92
x f(x)#5 2#4 p 0.5#3 0#2 0.5#1 2
92; x ! #3, #3
xO
f(x)
(0, !1)
f (x) " 0.5x 2 ! 1
27b.
27c.
29a. 0; x ! #6; #6
29b.
29c.
xO
f(x)
!4
8
4
!4!8
(!6, 9)
f (x) " !0.25x2 ! 3x
x f(x)#8 8#7 p 8.75#6 9#5 8.75#4 8
xO
f(x)
(! , ! )54
258
f (x) " 2x 2 # 5x
©Glencoe/McGraw-Hill 140 Algebra 2 Chapter 6
x f(x)#3 3#2 #2
#1 #30 0
#258
#54
x f(x)#2 1
#1 p0 #1
1 p2 1
#12
#12
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31a.
31b.
31c.
33. max.; #9
35. min.; #11
37. max.; 12
39. max.;
41. min.; #11
43. min.;
45. 40 m
47. The y-intercept is the initialheight of the object.
49. 60 ft by 30 ft
51. $11.50
#10
13
#78
xO
f(x)
( , !1)13
f (x) " x 2 ! x !23
89
#89; x !
13;
13
32. min.; 0
34. min.; #14
36. max.; 5
38. min.;
40. max.; 5
42. max.; 5
44. x ! 40; (40, 40)
46. 300 ft, 2.5 s
48. 120 # 2x
50. 1800 ft2
52. $2645
92
©Glencoe/McGraw-Hill 141 Algebra 2 Chapter 6
x f(x)
#1
0
#1
1
2 179
#59
13
#89
79
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53. 5 in. by 4 in.
55. If a quadratic function can beused to model ticket priceversus profit, then by findingthe x-coordinate of the vertexof the parabola you candetermine the price per ticketthat should be charged toachieve maximum profit.Answers should include thefollowing.• If the price of a ticket is too
low, then you won’t makeenough money to coveryour costs, but if the ticketprice is too high fewerpeople will buy them.
• You can locate the vertexof the parabola on thegraph of the function. Itoccurs when x ! 40.Algebraically, this is found
by calculating x ! #
which, for this case, is x ! or 40. Thus the
ticket price should be setat $40 each to achievemaximum profit.
57. C
59. 3.20
61. 3.38
63. 1.56
65. #1 " 3i
#40002(#50)
b2a
54. c; The x-coordinate of thevertex of or 0, so the y-coordinate ofthe vertex, the minimum ofthe function, is a(0)2 " c orc ; #12.5
56. C
58. #2.08
60. 0.88
62. 0.43
64. #1
66. 9 # 5i
#0
2ay ! ax2 " c is
©Glencoe/McGraw-Hill 142 Algebra 2 Chapter 6
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67. 23
69. 4
71. [5 #13 8]
73.
75. 5
77. #2
C 6 0 #24
14 #23
#8S
68. #13
70. [10 #4 5]
72.
74.
(#1, 3); consistent andindependent
76. 8
78. #1
y
xO
(!1, 3)
y " !3x
y ! x " 4
c#28 20 #448 #16 36
d
©Glencoe/McGraw-Hill 143 Algebra 2 Chapter 6
1a. The solution is the value thatsatisfies an equation.
1b. A root is a solution of anequation.
1c. A zero is the x value of afunction that makes thefunction equal to 0.
1d. An x-intercept is the point atwhich a graph crosses the x-axis. The solutions, or roots,of a quadratic equation arethe zeros of the relatedquadratic function. You canfind the zeros of a quadraticfunction by finding the x-intercepts of its graph.
2. Sample answer: f (x) ! 3x 2 "2x # 1; 3x 2 " 2x # 1 ! 0
Lesson 6-2 Solving Quadratic Equations by GraphingPages 297–299
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3. The x-intercepts of therelated function are thesolutions to the equation. Youcan estimate the solutions bystating the consecutiveintegers between which thex-intercepts are located.
5. #2, 1
7. #7, 0
9. #7, 4
11. between #2 and #1, 3
13. #2, 7
15. 3
17. 0
19. no real solutions
21. 0, 4
23. between #1 and 0; between2 and 3
25. 3, 6
27. 6
29.
31.
33. between 0 and 1; between 3and 4
35. between #3 and #2;between 2 and 3
37. no real solutions
#2˛
12, 3
#12, 2˛
12
4. #4, 1
6. #4
8. #4, 6
10. #5
12. between #1 and 0; between1 and 2
14. 0, 6
16.
18.
20. 0, 3
22. between #5 and #4;between 0 and 1
24. #4, 5
26. #7
28.
30.
32. between #4 and #3;between 0 and 1
34. between #1 and 0, between2 and 3
36. no real solutions
38. #8, #9
#4, 112
#112, 3
#12, 3
#2, 112
©Glencoe/McGraw-Hill 144 Algebra 2 Chapter 6
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39. Let x be the first number.Then, 7 # x is the othernumber.
x(7 # x) ! 14
Since the graph of therelated function does notintersect the x-axis, thisequation has no realsolutions. Therefore, no suchnumbers exist.
41. #2, 14
43. 3 s
45. about 35 mph
y
xO
y " !x2 # 7x ! 14
#x 2 " 7x # 14 ! 0
40. Let x be the first number.Then, is the othernumber.
x(#9 # x) ! 24
Since the graph of therelated function does notintersect the x-axis, thisequation has no realsolutions. Therefore, no suchnumbers exist.
42. 4 s
44. about 12 s
46. about 8 s
y
xO
y " !x 2 ! 9x ! 24
#x 2 # 9x # 24 ! 0
#9 # x
©Glencoe/McGraw-Hill 145 Algebra 2 Chapter 6
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48. Answers should include thefollowing.•
• Locate the positive x-intercept at about 3.4.This represents the timewhen the height of the rideis 0. Thus, if the ride wereallowed to fall to theground, it would take about3.4 seconds.
50. B
52. $3
54. #9, 1
56. no real solutions
58. #1; x ! 1; 1
xO
f (x) " !4x 2 # 8x ! 1
(1, 3)
f(x)
h(t)
t0 321 54
4060
20
80100120140160180 h(t) " !16t 2 # 185
©Glencoe/McGraw-Hill 146 Algebra 2 Chapter 6
47. #4 and #2; The value of thefunction changes fromnegative to positive, thereforethe value of the function iszero between these twonumbers.
49. A
51. #1
53. 3, 5
55. $1.33
57. 4, x ! 3; 3
(3, !5)
xO
f (x) " x 2 ! 6x # 4
f(x)
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2. Sample answer: roots 6 and #5;x 2 # x # 30 ! 0
4. {0, 11}
6. {#7, 7}
8.
10. x 2 " 3x # 28 ! 0
12. 15x 2 " 14x " 3 ! 0
14. {#8, 3}
16. {#5, 5}
18. {#6, 3}
e#34, 4f
59. 4; x ! #6; #6
61.
63. 24
65. #60
67. x(x " 5)
69. (x # 7)(x # 4)
71. (3x " 2)(x " 2)
1013
"213
i
f(x)
xO
!12 !8 !4
8
4
!4
14f (x) " x2 # 3x # 4
(!6, !5)
60.
62.
64. #8
66. $500
68. (x # 10)(x " 10)
70. (x # 9)2
72. 2(3x " 2)(x # 3)
113
"513
i
15
"35
i
©Glencoe/McGraw-Hill 147 Algebra 2 Chapter 6
Lesson 6-3 Solving Quadratic Equations by FactoringPages 303–305
1. Sample answer: If theproduct of two factors iszero, then at least one of thefactors must be zero.
3. Kristin; the Zero ProductProperty applies only whenone side of the equation is 0.
5. {#8, 2}
7. {3}
9. {#3, 4}
11. 6x 2 # 11x " 4 ! 0
13. D
15. {#4, 7}
17. {#9, 9}
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19. {#3, 7}
21.
23. {8}
25.
27.
29.
31. {#3, 1}
33. 0, #3, 3
35. x 2 # 5x # 14 ! 0
37. x 2 " 14x " 48 ! 0
39. 3x 2 # 16x " 5 ! 0
41. 10x 2 " 23x " 12 ! 0
43. #14, #16
45. B ! D 2 # 8D " 16
47. y ! (x # p)(x # q)
y ! x2 # (p " q)x # pqa ! 1, b ! #(p " q),
axis of symmetry:
x !p " q
2
x ! ##(p " q)
2(1)
x ! #b
2a
c ! #pq
y ! x2 # px # qx # pq
e34,
94f
e#23, #
32f
e14, 4f
e0, #34f
20.
22. {6}
24.
26.
28.
30. {2, 4}
32. 0, #6, 5
34. x 2 # 9x " 20 ! 0
36. x 2 " x # 20 ! 0
38. 2x 2 # 7x " 3 ! 0
40. 12x 2 # x # 6 ! 0
42.
44. 12 cm by 16 cm
46. 4; The logs must have adiameter greater than 4 in.for the rule to producepositive board feet values.
48. #1
14 s
e#83, #
23f
e#12, #
32f
e#2,
14f
e0,
53f
©Glencoe/McGraw-Hill 148 Algebra 2 Chapter 6
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The axis of symmetry is theaverage of the x-intercepts.Therefore the axis ofsymmetry is located halfwaybetween the x-intercepts.
49. #6
51. D
53. #5, 1
55. between #1 and 0; between3 and 4
57.
59.
61. (3, #5)
63.
65.
67. 2i˛23
323
222
33 " 2022
322 # 223
50. Answers should include thefollowing.• Subtract 24 from each side
of x 2 " 5x ! 24 so thatthe equation becomes x 2 " 5x # 24 ! 0. Factorthe left side as (x # 3)(x " 8). Set each factorequal to zero. Solving eachequation for x. Thesolutions to the equationare 3 and #8. Since lengthcannot be negative, thewidth of the rectangle is 3 inches, and the length is3 " 5 or 8 inches.
• To use the Zero ProductProperty, one side of theequation must equal zero.
52. B
54.
56. min.; #19
58.
60. (#4, #4)
62.
64.
66.
68. 4i23
5i˛22
225
a13, 2b
523
#12
©Glencoe/McGraw-Hill 149 Algebra 2 Chapter 6
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©Glencoe/McGraw-Hill 150 Algebra 2 Chapter 6
Chapter 6Practice Quiz 1
Page 305
1. 4; x ! 2; 2
3.
5. 3x 2 " 11x # 4 ! 0
112, 4
!4
!8
8 124
4
x
f(x)
(2, !8)
f (x) " 3x2 ! 12x # 4
O
1. Completing the squareallows you to rewrite oneside of a quadratic equationin the form of a perfectsquare. Once in this form,the equation is solved byusing the Square RootProperty.
3. Tia; before completing thesquare, you must first checkto see that the coefficient ofthe quadratic term is 1. If it isnot, you must first divide theequation by that coefficient.
5.
7.
9. 54 $ 25694, ax #
32b2
e4 $ 223
f
2. max.; or
4. e#5, 12f
9
14
374
2. Never; the value of c thatmakes ax 2 " bx " c aperfect square trinomial is
the square of and the square of a number cannever be negative.
4. {#10, #4}
6. 36; (x # 6)2
8. {#6, 3}
10. 5#1 $ i 256
b2
Lesson 6-4 Completing the SquarePages 310–312
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11.
13. Earth: 4.5 s, Jupiter: 2.9 s
15. {#2, 12}
17.
19.
21. {#1.6, 0.2}
23. about 8.56 s
25. 81; (x # 9)2
27.
29. 1.44; (x # 1.2)2
31.
33. {#12, 10}
35.
37. {#3 $ 2i}
39.
41.
43.
45. {0.7, 4}
47.
49. x1,
1x # 1
e34 $ 22f
e#5 $ i 1236
fe2 $ 210
3f
e12, 1f
52 $ 236
2516
, ax "54b2
494
; ax #72b2
e#5 $ 2113
f53 $ 2226
e3 $2334
f 12. Jupiter
14. {3, #7}
16.
18.
20.
22. 25 ft
24. 64; (x " 8)2
26.
28. 0.09; (x " 0.3)2
30.
32. {3, 5}
34.
36.
38.
40.
42.
44. {#2, 0.6}
46.
48. in. by in.
50. 1 " 252
5
12
5
12
e13 $ 23f
e7 $ i2474
fe5 $ 213
6f
e#52, 1f
52 $ i 65#1 $ 276
169
; ax # 43b2
2254
; ax #152b2
e#54,
14f
e7 $ 252
f5#4 $ 27, #4 # 276
©Glencoe/McGraw-Hill 151 Algebra 2 Chapter 6
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51. Sample answers: The goldenrectangle is found in much ofancient Greek architecture,such as the Parthenon, aswell as in modernarchitecture, such as in thewindows of the UnitedNations building. Many songshave their climax at a pointoccurring 61.8% of the waythrough the piece, with 0.618being about the reciprocal ofthe golden ratio. Thereciprocal of the golden ratiois also used in the design ofsome violins.
53. 18 ft by 32 ft or 64 ft by 9 ft
55. D
52a. n ! 052b. n % 052c. n & 0
54. To find the distance traveledby the accelerating race carin the given situation, youmust solve the equation t 2 " 22t " 121 ! 246 or t 2 " 22t # 125 ! 0.Answers should include thefollowing.• Since the expression
t 2 " 22t # 125 is prime,the solutions of t 2 " 22t "121 ! 246 cannot beobtained by factoring.
• Rewrite t 2 " 22t " 121 as(t " 11)2. Solve (t " 11)2 !246 by applying theSquare Root Property.Then, subtract 11 fromeach side. Using acalculator, the twosolutions are about 4.7 and#26.7. Since time cannotbe negative, the drivertakes about 4.7 seconds toreach the finish line.
56. D
©Glencoe/McGraw-Hill 152 Algebra 2 Chapter 6
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58. x 2 # 6x # 27 ! 0
60. 12x 2 " 13x " 3 ! 0
62. 6, 8
64.
66.
68. greatest: #255'C; least:#259'C
70. #16
72. 0
a4321
, 67b
537
2. The square root of anegative number is acomplex number.
©Glencoe/McGraw-Hill 153 Algebra 2 Chapter 6
57. x 2 # 3x " 2 ! 0
59. 3x 2 # 19x " 6 ! 0
61. between #4 and #3;between 0 and 1
63.
65. (2, #5)
67. 0x # (#257) 0 ! 2
69. 37
71. 121
#4, #112
1a. Sample answer:
1b. Sample answer:y
xO
y
xO
Lesson 6-5 The Quadratic Formula and the DiscriminantPages 317–319
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1c. Sample answer:
3. must equal 0.
5a. 85b. 2 irrational
5c.
7a.7b. two complex
7c.
9.
11.
13. No; the discriminant ofis
indicating that the equationhas no real solutions.
15a. 24015b. 2 irrational
15c.
17a.17b. 2 complex
17c.
19a. 4919b. 2 rational
1 $ i2232
#23
8 $ 2215
#455,#16t 2" 85t ! 120
#5 $ i222
#3, #2
#3 $ i232
#3
2 $ 222
b2 # 4ac
y
xO
4a. 4844b. 2 rational
4c.
6a. 06b. one rational
6c.
8.
10.
12. at about 0.7 s and again atabout 4.6 s
14a. 2114b. 2 irrational
14c.
16a.16b. 2 complex16c.
18a. 12118b. 2 rational
18c.
20a. 2020b. 2 irrational
#14,
23
1 $ 2i
#16
#3 $ 2212
1 $ 23
0, #8
#12
14, #
52
©Glencoe/McGraw-Hill 154 Algebra 2 Chapter 6
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19c.
21a. 2421b. 2 irrational
21c.
23a. 023b. one rational
23c.
25a.
25b. 2 complex
25c.
27a. 1.4827b. 2 irrational
27c.
29.
31.
33.
35.
37.
39.
41. This means that the cablesdo not touch the floor of thebridge, since the graph doesnot intersect the x-axis andthe roots are imaginary.
43. 1998
#2, 6
0, #310
5 $ 2463
92
#3 $ 2152
$i2217
#1 $ 220.370.8
#1 $ i2154
#135
#52
#1 $ 26
#2, 13
20c.
22a. 022b. one rational
22c.
24a.24b. 2 complex
24c.
26a.
26b. 2 irrational
26c.
28.
30.
32.
34.
36.
38.
40.
42. domain: range:
44. about 40.2 mph
73.7 ( A(t ) ( 1201.20 ( t ( 25,
! #0.00288
3 $ 222
4 $ 27
#3 $ i27
$22
2 $ i23
#2, 32
2 $ 4279
289
9 $ i2318
#31
13
#2 $ 25
©Glencoe/McGraw-Hill 155 Algebra 2 Chapter 6
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45a.45b. or 45c.
47. D
49.
51.
53.
55.
57.
59.
61. no
63. yes; (2x " 3)2
65. no
y
xO 2 4 6 8
8642
!6!4
!4!6x ! y " 3
x # y " 9
y ! x " 4
4b 2c 2
a4b10
#2, 7
1 $ 2222
#14, #4
#6 & k & 6k % 6k & #6
k ! $6 46. The person’s age can besubstituted for A in theappropriate formula,depending upon their gender,and their average bloodpressure calculated. Seestudent’s work.• If a woman’s blood pressure
is given to be 118, thensolve the equation 118 !0.01A2 " 0.05A " 107 tofind the value of A. Use theQuadratic Formula,substituting 0.01 for a, 0.05for b, and #11 for c. Thisgives solutions of about#35.8 or 30.8. Since agecannot be negative, the onlyvalid solution for A is 30.8.
48. C
50.
52.
54.
56.
58.
60.
62. yes; (x # 7)2
64. yes; (5x " 2)2
66. yes; (6x # 5)2
y
xO
y " !1
x " 1
y " x
7.98 ) 106
10p6 0q 023, 5
#2, 0
4 $ 27
©Glencoe/McGraw-Hill 156 Algebra 2 Chapter 6
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©Glencoe/McGraw-Hill 157 Algebra 2 Chapter 6
1a. y ! 2(x " 1)2 " 51b. y ! 2(x " 1)2
1c. y ! 2(x " 3)2 " 31d. y ! 2(x # 2)2 " 31e. Sample answer:
y ! 4(x " 1)2 " 31f. Sample answer:
y ! (x " 1)2 " 31g. y ! #2(x " 1)2 " 3
3. Sample answer:y ! 2(x # 2)2 #1
5. (#3, #1); x ! #3; up
7. y ! #3(x # 3)2 " 38;(#3, 38); x ! #3; down
9.
11. y ! 4(x # 2)2
y
xO
y " (x ! 1)2 # 313
2. Substitute the x-coordinate ofthe vertex for h and the y -coordinate of the vertex for kin the equation y ! a(x # h)2 "k. Then substitute the x-coordinate of the other pointfor x and the y-coordinate fory into this equation and solvefor a. Replace a with thisvalue in the equation youwrote with h and k.
4. Jenny; when completing thesquare is used to write aquadratic function in vertexform, the quantity added isthen subtracted from thesame side of the equation tomaintain equality.
6. y ! (x " 4)2 # 19, (#4, #19); x ! #4; up
8.
10.
12. y ! #(x " 3)2 " 6
y
xO
y " ! 2x 2 # 16x ! 31
xO
y
y " 3(x # 3)2
Lesson 6-6 Analyzing Graphs of Quadratic FunctionsPages 325–328
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13.
15. (#3, 0); x ! #3 down
17. (0,#6); x ! 0 up
19. y ! #(x " 2)2 " 12;(#2, 12);
21. y ! #3(x # 2)2 " 12;(2, 12);
23. y ! 4(x " 1)2 # 7; (#1, #7);
25.
up
27.
29.
31. y
xO
y " x 2 # 6x # 2
y
xO
y " (x ! 2)2 # 414
y
xOy " 4(x # 3)2 # 1
a#12, #
74b; x ! #
12;
y ! 3 ax "12b2
# 74;
x ! #1; up
x ! 2; down
x ! #2; down
y ! #12
(x " 2)2 # 3 14. h(d ) ! #2d 2 " 4d " 6
16. (1, 2); x ! 1; up
18. (0, 3); x ! 0; down
20. y ! (x # 3)2 # 8;(3, #8); x ! 3; up
22. y ! 4(x " 3)2 # 36;(#3, #36); x ! #3; up
24. y ! #2(x # 5)2 " 15;(5, 15); x ! 5; down
26.
up
28.
30.
32. y
xO
y " x 2 ! 8x # 18
y
x
y " (x ! 3)2 ! 512
O
y
xOy " !(x ! 5)2 ! 3
a32, #20b; x !
32,
y ! 4 ax #32b2# 20;
©Glencoe/McGraw-Hill 158 Algebra 2 Chapter 6
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33.
35.
37. Sample answer: the graph ofy ! 0.4(x " 3)2 " 1 isnarrower than the graph of y ! 0.2(x " 3)2 " 1.
39. y ! 9(x # 6)2 " 1
41.
43.
45.
47. 34,000 feet; 32.5 s after theaircraft begins its parabolicflight
49.
51. Angle A; the graph of theequation for angle A ishigher than the other twosince 3.27 is greater than2.39 or 1.53.
d (t ) ! #16t 2 " 8t " 50
y ! #2x 2
y !13 x 2 " 5
y ! #23 (x # 3)2
y
xO
12
272y " ! x 2 # 5x !
y
xO
y " !4x 2 # 16x ! 1134.
36.
38. Sample answer: the graphshave the same shape, but thegraph of y ! 2(x # 4)2 " 1 is1 unit to the left and 5 unitsbelow the graph of y !2(x # 5)2 # 4
40. y ! 3(x " 4)2 " 3
42. y ! #3(x # 5)2 " 4
44.
46.
48. about 1.6 s
50. about 2.0 s
52. Angle B; the vertex of theequation for angle B isfarther to the right than theother two since 3.57 isgreater than 3.09 or 3.22.
y !43(x " 3)2 # 4
y !52
(x " 3)2 # 2
y
xO 13
y " x 2 ! 4x # 15
y
xO
y " !5x 2 ! 40x ! 80
©Glencoe/McGraw-Hill 159 Algebra 2 Chapter 6
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53.
The axis of symmetry is
or .
55. D
#b
2ax ! h
y ! a ax "b2ab2 " c #
b 2
4a
c # a a b2ab2
y ! a cx 2 "ba
x " a b2ab2d "
y ! a ax2 "ba
xb " c
y ! ax2 " bx " c 54. All quadratic equations are atransformation of the parentgraph By identifyingthese transformations whena quadratic function is writtenin vertex form, you canredraw the graph of Answers should include thefollowing.• In the equation
translated the graph ofunits to the right
when h is positive and hunits to the left when h isnegative. The graph of
is translated k unitsup when k is positive andk units down when k isnegative. When a ispositive, the graph opensupward and when a isnegative, the graph opensdownward. If the absolutevalue of a is less than 1,the graph will be narrowerthan the graph of and if the absolute value ofa is greater than 1, thegraph will be wider thanthe graph of
• Sample answer:is the
graph of translated2 units left and 3 unitsdown. The graph opensupward, but is narrowerthan the graph of
56. B
y ! x2.
y ! x2y ! 2(x " 2)2 # 3
y ! x2.
y ! x2,
y ! x2
y ! x 2 h
y ! a(x # h)2 " k, h
y ! x2.
y ! x2.
©Glencoe/McGraw-Hill 160 Algebra 2 Chapter 6
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57. 12; 2 irrational
59. 2 complex
61.
63.
65.
67a. Sample answer using (1994,76,302) and (1997, 99,448):y ! 7715x # 15,307,408
67b. 161,167
69. no
71. no
n3 # 3n2 # 15n # 21
2t2 " 2t #3
t # 1
53 $ 3i 6#23;
58. 225; 2 rational
60.
62.
64.
66.
68. yes
70. yes
y3 " 1 # 4
y " 3
t 2 # 2t " 1
e#2 $ 2132
f5#5 $ 2226
©Glencoe/McGraw-Hill 161 Algebra 2 Chapter 6
Chapter 6Practice Quiz 2
Page 328
1.
3. complex
5.
7.
9.downy ! #1x # 622; 16, 02, x ! 6;
y !23
1x # 222 # 5
e#9 $ 5252
f#11; 2
5#7 $ 2236 2.
4. 100; 2 rational
6.
8. y ! (x " 4)2 " 2; (#4, 2), x ! #4; up
10. y ! 2(x " 3)2 # 5; (#3, #5), x ! #3; up
e2 $ 2i223
f
e1 $ 3i2f
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©Glencoe/McGraw-Hill 162 Algebra 2 Chapter 6
1.
3a.3b. or 3c.
5.
7.
9.
11. *
5x 0#1 & x & 76O!2 2 4 6
12
8
4
y
x
y " !x 2 # 5x # 6
!4!8
!12
!20
!2!4
1284
2 4
y
xO
y " x 2 ! 16
#1 ( x ( 5x + 5x ( #1
x ! #1, 5
y + 1x # 322 # 1 2. Sample answer: one numberless than one numberbetween and 5, and onenumber greater than 5
4.
6.
8. or
10. or
12. 5x 0#23 ( x ( 236x % 465x 0x & #3
x % 5x & 1
y
xO
y " !2x2 ! 4x # 3
y
xOy " x2 ! 10x # 25
#3#3,
Lesson 6-7 Graphing and Solving Quadratic InequalitiesPages 332–335
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13. about 6.1 s
15.
17.
19. y
xO
y " x2 # 6x # 5
y
xO
y " x 2 # 4x
84!4
4
8
12
y
xO
y " !x 2 # 7x # 8
14.
16.
18.
20. y
xO!2 2!6 !4
14
10
6
2
y " !x2 ! 3x # 10
y
xO 84!8 !4
5
!10
!20
!30y " x 2 ! 36
y
xOy " x 2 # 4x # 4
15
5
!5!8 !4 4 8
!15
!25
y
xO
y " x 2 # 3x ! 18
©Glencoe/McGraw-Hill 163 Algebra 2 Chapter 6
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21.
23.
25.
27.
29. or
31.
33. or
35. or
37. all reals
39.
41. *
43. 0 to 10 ft or 24 to 34 ft
5x 0x ! 76x + 165x 0x ( #7
x + 465x 0x ( #6
5x 0#7 & x & 46x % #3x & #7
#2 ( x ( 6
y
xO
y " 2x 2 # x ! 3
1062
!8
6
2
!4
y
xO
y " !x 2 # 13x ! 36
!4 4!12 !8
20
12
4
!4
y
xO
y " !x2 ! 7x # 10 22.
24.
26. 5
28. or
30. or
32.
34.
36.
38. *
40. all reals
42. or
44a. 0.98, 4.81; The owner willbreak even if he charges$0.98 or $4.81 per squarefoot.
x % 365x 0#4 & x & 1
ex `x ! 13f
5x 0#4 ( x ( 365x 0#1 ( x ( 56
x % 665x 0x & #3
x % 3x & #3
42!2
4
!4
!8
y
xO
y " 2x2 # 3x ! 5
y
xO
y " !x 2 # 10x ! 23
©Glencoe/McGraw-Hill 164 Algebra 2 Chapter 6
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45. The width should be greaterthan 12 cm and the lengthshould be greater than 18 cm
47. 6
49. y
xO
y " !x 2 # 4
y " x 2 ! 4
44b. The ownerwill make a profit if the rent isbetween $0.98 and $4.81.
44c. If rent is setbetween $1.34 and $4.45per sq ft, the profit will begreater than $10,000
44d. or If rent isset between $0 and $1.34 orabove $4.45 per sq ft, the profitwill be less than $10,000.
46. P(n) ! n[15 " 1.5(60 # n)] #525 or
48. $1312.50; 35 passengers
50. Answers should include thefollowing.•• One method of solving this
inequality is to graph therelated quadratic functionh(t ) ! #16t 2 " 42t "3.75 # 10. The interval(s) at which the graph is abovethe x-axis represents thetimes when the trampolinistis above 10 feet. A secondmethod of solving thisinequality would be to findthe roots of the relatedquadratic equation #16t 2 "42t " 3.75 # 10 ! 0 andthen test points in the threeintervals determined bythese roots to see if theysatisfy the inequality. Theinterval(s) at which theinequality is satisfiedrepresent the times whenthe trampolinist is above 10 feet.
#16t2 " 42t " 3.75 %10
#1.5n2 " 105n # 525
r % 4.45;r & 1.34
1.34 & r & 4.45;
0.98 & r & 4.81;
©Glencoe/McGraw-Hill 165 Algebra 2 Chapter 6
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51. C
53. all reals,
55. or
57.
59. y ! (x # 1)2 " 8; (1, 8), x ! 1; up
61.
up
63.
65. 4a 2b 2 " 2a 2b " 4ab 2 " 12a #7b
67.
69.
71. 0x # 0.008 0 ( 0.002;0.078 ( x ( 0.082
B#21#13
4822R
xy 3 " y "1x
#5 $ i232
x ! #6;
y !12(x " 6)2; (#6, 0),
5x 0#1.2 ( x ( #0.46x % 365x 0x & #9
x , 265x 052. A
54.
56. or
58. no real solutions
60.down
62.
64.
66.
68.
70. 3#54 6 4#15a2 " 14a # 3
#6x 3 # 4x 2y " 13xy 2
#3 $ 2263
#4, #8
y ! #21x # 422; 14, 02, x ! 4;
x + #2.565x 0x ( #3.5
5x 0#7 & x & 76
©Glencoe/McGraw-Hill 166 Algebra 2 Chapter 6
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1. 4 ! 4x0; x ! x1
3. Sample answer given.
5. 6; 5
7. "21; 3
9. 2a9 # 6a3"12
11. 6a3 " 5a2 # 8a " 45
13a. as as
13b. even13c. 0
15. 109 lumens
17. 3; 1
19. 4; 6
x S"$f(x )S#$x S#$,f(x ) S# $
xO
f(x)
Chapter 7 Polynomial FunctionsLesson 7-1 Polynomial Functions
Pages 350–352
©Glencoe/McGraw-Hill 167 Algebra 2 Chapter 7
2. Sample answer: Even-degree polynomial functionswith positive leadingcoefficients have graphs inwhich as and as Odd-degreepolynomial functions withpositive leading coefficientshave graphs in which
as andas
4. Sometimes; a polynomialfunction with 4 real rootsmay be a sixth-degreepolynomial function with2 imaginary roots. Apolynomial function that has4 real roots is at least afourth-degree polynomial.
6. 5; "3
8. 4; 12
10. 100a2 # 20
12a. as as
12b. odd12c. 3
14a. as as
14b. odd14c. 1
16. 1; "1
18. No, the polynomial containstwo variables, a and b.
20. 3; "5
x S"$f (x )S"$x S# $,f (x )S# $
xS"$f (x )S# $x S# $,f (x )S"$
x S"$.f (x ) S"$x S# $f (x ) S# $
xS"$.x S# $f (x ) S# $
PQ245-6457F-P07[167-202].qxd 7/24/02 12:02 PM Page 167 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
21. No, this is not a polynomial
because the term cannot be written in the form xn,where n is a nonnegativeinteger.
23. 12; 18
25. 1008; "36
27. 86; 56
29. 7; 4
31. 12a2 " 8a # 20
33. 12a6 " 4a3 # 5
35. 3x4 # 16x2 # 26
37. "x6 # x3 # 2x2 # 4x # 2
39a. as as
39b. odd39c. 3
41a. as as
41b. even41c. 0
43a. as as
43b. odd43c. 1
45. 5.832 units
47. as as
49. 12
x S"$f(x )S"$x S#$;f (x )S"$
x S"$f (x)S"$x S#$,f (x)S#$
x S"$f (x)S"$xS#$,f (x)S#$
xS"$f (x )S"$x S#$,f (x )S#$
1c
22. "2; 4
24. 125; "37
26. "166; 50
28. 100; 4
30. 27a3 # 3a # 1
32. 3a4 " 2a2 # 5
34. x3 # 3x2 # 4x # 3
36. 6x2 # 44x # 90
38. 9x4 " 12x2 " 8x # 50
40a. as as
40b. even40c. 4
42a. as as
42b. odd42c. 5
44a. as as
44b. even44c. 2
46. even
48. Sample answer: Decrease;the graph appears to beturning at x ! 30 indicating arelative maximum at thatpoint. So attendance willdecrease after 2000.
50. "1, 0, 4
x S"$f (x )S"$x S#$,f (x )S"$
x S"$f (x )S#$x S#$,f (x )S#$
x S"$f (x)S#$x S#$,f (x)S#$
©Glencoe/McGraw-Hill 168 Algebra 2 Chapter 7
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52.
54. 16 regions
56. Many relationships in naturecan be modeled bypolynomial functions; forexample, the pattern in ahoneycomb or the rings in atree trunk. Answers shouldinclude the following.• You can use the equation
to find the number ofhexagons in a honeycombwith 10 rings and thenumber of hexagons in ahoneycomb with 9 rings.The difference is thenumber of hexagons in the tenth ring.
• Other examples of patternsfound in nature includepinecones, pineapples, andflower petals.
58. C
60. 5x 0 x % "9 or x & 76
xO!4
!4
!8
2
4
8
!2
f(x)
f (x) " x 3 ! x 2! 2x12
32
©Glencoe/McGraw-Hill 169 Algebra 2 Chapter 7
51.
53. 4
55. 8 points
57. C
59. 5x 0 2 ' x ' 66
f(x) !12x3 "
32x2 " 2x
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61.
63.
65.
67. 23,450(1 # p);23,450(1 # p)3
69. y
xO
y " !x2 # 6x ! 5
54 ( 3226
y
xO!12 !8
!2
!4
2
y " (x # 5)2 ! 113
ex `"1 % x %45f 62.
64.
66.
68.
70. y
xO!8 !4
!4
4 8
4
8
y " x 2 # 2x ! 612
y
xO
y " x2 # 4
e"76,
56f
y
xO
y " x 2 # x # 12
32
y
xO
y " !2(x ! 2)2 # 3
©Glencoe/McGraw-Hill 170 Algebra 2 Chapter 7
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©Glencoe/McGraw-Hill 171 Algebra 2 Chapter 7
Lesson 7-2 Graphing Polynomial FunctionsPages 356–358
1. There must be at least onereal zero between two pointson a graph when one of thepoints lies below the x-axisand the other point liesabove the x-axis.
3.
5.
7. between "2 and "1, between"1 and 0, between 0 and 1,and between 1 and 2
xO
f(x)
f (x) " x 4 ! 4x 2 # 2
xO
f(x)
!4 !2
!4
!8
42
4
8
f (x) " x 4 ! 7x 2 # x # 5
x f(x )
"3 20"2 p "9"1 "2
0 51 02 "53 26
xO
f(x)
f (x) " x 3 ! x 2 ! 4x # 4
2. 4
4.
6. between "1 and 0
8.
Sample answer: rel. max. atx ! "2, rel. min.: at x ! 0.5
xO!4 !2
!4
!8
42
4
8f(x)
f (x) " x 3 # 2x 2 ! 3x ! 5
xO
f(x)
f (x) " x 3 ! x 2 # 1
xO
f(x)
!4 !2!2
!4
42
4
8
f(x) " x 3 ! x 2 ! 4x # 4
x f(x )
"3 "20
"2 p 0
"1 6
0 4
1 0
2 0
3 10
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9.
Sample answer: rel. max. atx ! 0, rel. min. at x ! "2and at x ! 2
11. rel. max. between x ! 15and x ! 16, and no rel. min.;
as as
13a.
13b. at x ! "4 and x ! 013c. Sample answer: rel. max. at
x ! 0, rel. min. at x ! "3
xO
!2
!4
!8
42
4
f (x)
f (x) " !x 3 ! 4x 2
x f (x )
"5 25"4 p 0"3 "9"2 "8"1 "3
0 01 "52 "24
x S#$.f(x )S"$
x S"$,f(x )S"$
xO!4 !2
!4
4 2
4
8
f(x)
f (x) " x 4 ! 8x 2 # 10
10.
12. The number of cable TVsystems rose steadily from1985 to 2000. Then thenumber began to decline.
14a.
14b. between "2 and "114c. Sample answer: rel. max. at
x ! 0, rel. max. at x ! 1
xO!4 !2
!4
2 4
8
4
f (x)
f(x) " x 3 ! 2x 2# 6
x f (x )
"2 "10"1 p 3
0 61 52 63 154 38
C(t)
tO
2000
4000
6000
8000
10000
12000
161284
Cabl
e TV
Sys
tem
s
Years Since 1985
C(t) " !43.2t 2 # 1343t # 790
©Glencoe/McGraw-Hill 172 Algebra 2 Chapter 7
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15a.
15b. at x ! 1, between "1 and 0,and between 2 and 3
15c. Sample answer: rel. max. atx ! 0, rel. min. at x ! 2
17a.
17b. between 0 and 1, at x ! 2,and at x ! 4
17c. Sample answer: rel. max. atx ! 3, rel. min. at x ! 1
xO!4
!8
!4
!2 2 4
4
f(x) " !3x 3 # 20x 2 ! 36x # 16
f (x)x f (x )
"1 750 p 161 "32 03 74 05 "39
xO
f (x)
f(x) " x 3 ! 3x 2# 2
x f (x )
"2 "18"1 p "2
0 21 02 "23 24 18
16a.
16b. between "5 and "4,between "2 and "1, andbetween 1 and 2
16c. Sample answer: rel. max. atx ! "3, rel. min. at x ! 0
18a.
18b. between 3 and 418c. Sample answer: rel. max. at
x ! 0.5, rel. min. at x ! 2.5
xO
f (x)
f (x ) " x 3 ! 4x 2# 2x ! 1
x f (x )
"2 "29"1 p "8
0 "11 "22 "53 "44 75 34
xO!4!8 84
4
!8
!4
f (x)
f (x ) " x 3 # 5x 2 ! 9
x f (x )
"5 "9"4 p 7"3 9"2 3"1 "5
0 "91 "32 19
©Glencoe/McGraw-Hill 173 Algebra 2 Chapter 7
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19a.
19b. between "2 and "1 andbetween 1 and 2
19c. Sample answer: no rel. max.,rel. min. at x ! 0
21a.
21b. between "3 and "2, between"1 and 0, between 0 and 1,and between 1 and 2
21c. Sample answer: rel. max. atx ! "2 and at x ! 1.5, rel.min. at x ! 0
xO!4 !2
!4
!8
42
8
4
f (x)
f (x ) " !x 4 # 5x 2! 2x ! 1
x f (x )
"4 "169"3 p "31"2 7"1 5
0 "11 12 "13 "43
xO!4 !2
!4
!8
2
4
f (x)
f(x) " x 4 ! 8
x f (x )
"3 73"2 p 8"1 "7
0 "81 "72 83 73
20a.
20b. at x ! "3, x ! "1, x ! 1,and x ! 3
20c. Sample answer: rel. max. atx ! 0, rel. min. at x ! "2and x ! 2
22a.
22b. between "3 and "2, between"1 and 0, between 0 and 1,and between 3 and 4
22c. Sample answer: rel. max. atx ! "1.5 and at x ! 2.5, rel.min. at x ! 0.
xO!4 !2
!8
2 4
16
24
8
f (x)
f(x) " !x 4 # x 3# 8x 2! 3
x f (x )
"3 "39"2 p 5"1 3
0 "31 52 213 154 "67
xO!4 !2
!8
!16
2
16
8
f (x)
f (x ) " x 4 ! 10x 2# 9
x f (x )
"3 0"2 p"15"1 0
0 91 02 "153 04 105
©Glencoe/McGraw-Hill 174 Algebra 2 Chapter 7
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24a.
24b. between "2 and "1, andbetween 2 and 3
24c. Sample answer: rel. max. atx ! 0.5; rel. min. at x !"0.5 and at x ! 1.5
26a.
26b. between "2 and "1, between"1 and 0, between 0 and 1,between 2 and 3, andbetween 4 and 5
26c. Sample answer: rel. max. atx ! "1 and at x ! 2, rel.min. at x ! 0 and at x ! 3.5
O!4 !2
!20
!40
2 4
20
40
x
f (x)
f(x) " x 5 ! 6x 4 # 4x 3 # 17x 2 ! 5x ! 6
x f (x )
"2 "88
"1 p 5
0 "6
1 5
2 20
3 "3
4 "10
5 269
xO
f (x)
!2
!4
!8
2 4 6
4
f (x) " 2x 4 ! 4x 3 ! 2x 2 # 3x ! 5
x f (x )
"2 45"1 p "4
0 "51 "62 "73 40
23a.
23b. between 0 and 1, between 1and 2, between 2 and 3, andbetween 4 and 5
23c. Sample answer: rel. max. atx ! 2, rel. min. at x ! 0.5and at x ! 4
25a.
25b. between "4 and "3, between"2 and "1, between "1 and0, between 0 and 1, andbetween 1 and 2
25c. Sample answer: rel. max. atx ! "3 and at x ! 0, rel.min. at x ! "1 and at x ! 1
!4 !2 2 4
8
16
24
xO
f (x)
f(x) " x 5 # 4x 4 ! x 3 ! 9x 2 # 3
x f (x )
"4 "77"3 p 30"2 7"1 "2
0 31 "22 55
xO
f (x)
!2
!4
!8
2 4
4
f(x) " x 4 ! 9x 3 # 25x 2 ! 24x # 6
x f (x )
"1 650 p 61 "12 23 "34 "105 11
©Glencoe/McGraw-Hill 175 Algebra 2 Chapter 7
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27. highest: 1982; lowest: 2000
29. 5
31. x 0 2 4 6 8 10B(x) 25 34 40 45 50 54G(x) 26 33 39 44 49 53
x 12 14 16 18 20B(x) 59 64 68 71 71G(x) 56 59 61 61 60
33. 0 and between 5 and 6
35. 3 s
y
x0
2520
303540455055606570
14 16 18128 1062 4
G (x )
B (x )
Ave
rage
Hei
ght
(in.)
Age (yrs)
©Glencoe/McGraw-Hill 176 Algebra 2 Chapter 7
28. Rel. max. between 1980 and1985 and between 1990 and1995, rel. min. between 1975and 1980 and between 1985and 1990; as the number ofyears increases, the percentof the labor force that isunemployed decreases.
30. Sample answer: increase,based on the pastfluctuations of the graph
32. The growth rate for bothboys and girls increasessteadily until age 18 andthen begins to level off, withboys averaging a height of71 in. and girls a height of 60 in.
34. 5 s
36.
O
y
x
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37.
39.
41. D
43. "1.90; 1.23
45. 0; "1.22, 1.22
47. 24a3 " 4a2 " 2
49. 8a4 " 10a2 # 4
51. 2x4 # 11x2 # 16
O
y
x
O
y
x
38.
40. The turning points of apolynomial function thatmodels a set of data canindicate fluctuations that mayrepeat. Answers shouldinclude the following.• To determine when the
percentage of foreign-borncitizens was at its highest,look for the rel. max. of thegraph, which is at t ! 5.The lowest percentage isfound at t ! 75, the rel.min. of the graph.
• Polynomial equations bestmodel data that containturning points, rather thana constant increase ordecrease like linearequations.
42. B
44. 3.41; 0.59
46. 0.52; "0.39, 1.62
48. 10c2 " 25c # 20
50. 3x3 " 10x2 # 11x " 6
52. 4x4 " 9x3 # 28x2 " 33x # 20
O
y
x
©Glencoe/McGraw-Hill 177 Algebra 2 Chapter 7
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53.
55.
57. ("3, "2)
59. (1, 3)
61. (x # 5)(x " 6)
63. (3a # 1)(2a # 5)
65. (t " 3)(t 2 # 3t # 9)
y
xO
y " x 2 ! 2x
y
xO
y " x 2 ! 4x # 6
54.
56. (7, "4)
58. (4, "2)
60. 1 ft
62. (2b " 1)(b " 4)
64. (2m # 3)(2m " 3)
66. (r 2 # 1)(r # 1)(r " 1)
y
xO
y " !x 2 # 6x ! 3
©Glencoe/McGraw-Hill 178 Algebra 2 Chapter 7
Lesson 7-3 Solving Equations Using Quadratic Techniques
Pages 362–364
1. Sample answer:16x4 " 12x2 ! 0;4[4(x2)2 " 3x2] ! 0
3. Factor out an x and write theequation in quadratic form soyou have x[(x2)2 " 2(x2) #1] ! 0. Factor the trinomialand solve for x using theZero Product Property. Thesolutions are "1, 0, and 1.
2. The solutions of a polynomialequation are the points atwhich the graph intersectsthe x-axis.
4. not possible
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©Glencoe/McGraw-Hill 179 Algebra 2 Chapter 7
5. 84(n2)2 " 62(n2)
7. "4, "1, 4, 1
9. 64
11. 2(x 2)2 # 6(x 2) " 10
13. 11(n3)2 # 44(n3)
15. not possible
17. 0, "4, "3
19.
21.
23.
25. 81, 625
27. 225, 16
29. 1, "1, 4
31. w ! 4 cm, / ! 8 cm, h ! 2 cm
33. 3 ) 3 in.
35. h2 # 4, 3h # 2, h # 3
37. Write the equation inquadratic form, u2 " 9x #8 ! 0, where u ! |a " 3|.Then factor and use the ZeroProduct Property to solve for a; 11, 4, 2, and "5.
9 " 9i132
"9, 9 # 9i13
2,
2, "2, 212, "212
"13, 13, "i 13, i 13
6. 0, "5, "4
8.
10. 8 feet
12. not possible
14. b[7(b2)2 " 4(b2) # 2)]
16.
18. 0, "1, "5
20. 0, "4, 4, "4i, 4i
22.
24.
26. "343, "64
28. 400
30.
32. x4 " 7x2 # 9 ! 27
34. 6 ) 6 in.
36. The height increased by 3,the width increased by 2, andthe length increased by 4.
38. Answers should include thefollowing.• Solve the cubic equation
4x3 # ("164x2) #1600x ! 3600 in order todetermine the dimensionsof the cut square if thedesired volume is 3600 in3.Solutions are 10 in. and
in.• There can be more than
one square cut to producethe same volume becausethe height of the box is notspecified and 3600 has avariety of different factors.
31"26012
8, i 23, "i 23
8, "4 # 4i13, "4 " 4i13
12, "12, 3, "3
6 (x 15 )2 " 4 (x 1
5 ) " 16 ! 0
6, "3 # 3i13, "3 " 3i13
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39. D
41.
43. 17; 27
45.
47.
49. x2 # 5x " 4
51. x3 # 3x2 " 2
C¿(1, 3)A¿("1, "2), B¿(3, "3),
17153
; 135
xO
f(x) " x 3 ! 4x 2 # x # 5
f (x)x f (x )
"2 "21"1 p "1
0 51 32 "13 "14 95 35
40.
42.
44. 262; 2
46.
48.
50.
52. x 3 # 2x 2 " 10x #15 "21
x # 1
4x 2 " 16x # 27 "64
x # 2
y
xO
B'
C'
B
A'
A
C
B"2 "3 31 "3 "1
R
xO
f (x) " x 4 ! 6x 3 # 10x 2 ! x ! 3
f (x)x f (x )
"1 150 p "31 12 33 34 25
118
©Glencoe/McGraw-Hill 180 Algebra 2 Chapter 7
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1. 2a3 " 6a2 # 5a " 1
3. Sample answer: maximum atx ! "2, minimum at x ! 0.5
5. "3, 3, "i13, i13
xO
f (x) " x 3 # 2x 2 ! 4x ! 6
f (x)
!2!4
!4
!8
2 4
4
8
2. as as ; odd; 3
4.36(23 x)2 # 18(23 x) # 5(6x
13)2 # 3(6x 13) # 5 or
x S "$f(x ) S #$x S #$,f(x ) S "$
©Glencoe/McGraw-Hill 181 Algebra 2 Chapter 7
Chapter 7Practice Quiz 1
Page 364
Lesson 7-4 The Remainder and Factor TheoremsPages 368–370
2. 4
4. 7, "91
6. x # 1, x " 3
8. 2x # 1, x " 4
10. $2.894 billion
12. Sample answer: Directsubstitution, because it canbe done quickly with acalculator.
14. 37, "19
16. 55, 272
1. Sample answer:f(x ) ! x2 " 2x " 3
3. dividend: x3 # 6x # 32;divisor: x # 2; quotient:x2 " 2x # 10; remainder: 12
5. 353, 1186
7. x " 1, x # 2
9. x " 2, x2 # 2x # 4
11. $2.894 billion
13. "9, 54
15. 14, "42
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©Glencoe/McGraw-Hill 182 Algebra 2 Chapter 7
17. "19, "243
19. 450, "1559
21. x # 1, x # 2
23. x " 4, x # 1
25. or 2x " 1
27. x # 7, x " 4
29. x " 1, x2 # 2x # 3
31. x " 2, x # 2, x2 # 1
33. 3
35. 1, 4
37. 5 1 "14 69 "140 1005 "45 120 "100
1 "9 24 "20 0
39. 7.5 ft/s, 8 ft/s, 7.5 ft/s
41. By the Remainder Theorem,the remainder when f(x ) isdivided by x " 1 is equivalentto f(1), or a # b # c #d # e. Since a # b # c #d # e ! 0, the remainderwhen f (x ) is divided by x " 1is 0. Therefore, x " 1 is afactor of f(x ).
43. $16.70
x # 3, x "12
18. 267, 680
20. 422, 3110
22. x " 4, x # 2
24. x " 3, x " 1
26. or 3x # 4
28. x " 1, x # 6
30. 2x " 3, 2x # 3, 4x2 # 9
32. 8 1 "4 "29 "248 32 24;
1 4 3 0(x # 3)(x # 1)
34. 8
36. "3
38. Yes; 2 ft lengths. Thebinomial x " 2 is a factor ofthe polynomial since f (2) ! 0.
40. 0; The elevator is stopped.
42. $31.36
44. B(x) ! 2000x5 " 340(x4 #x3 # x2 # x # 1)
x " 1, x #43
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45. No, he will still owe $4.40.
47. D
49. (x2)2 " 8(x2) # 4
51. not possible
53. Sample answer: maximum atx ! "1, rel. max. at x ! 1.5,rel. min. at x ! 1
xO
f (x)
!2!4
!4
2 4
4
8
f (x) " !x 4 # 2x 3 # 3x 2 ! 7x # 4
46. Using the RemainderTheorem you can evaluate apolynomial for a value a bydividing the polynomial by x " a using syntheticdivision. Answers shouldinclude the following.• It is easier to use the
Remainder Theorem whenyou have polynomials ofdegree 2 and lower orwhen you have access toa calculator.
• The estimated number ofinternational travelers to theU.S. in 2006 is 65.9 million.
48. x " 2, x # 2, x # 1, x2 # 1
50. 9(d3)2 # 5(d3) " 2
52. Sample answer: rel. max, atx ! 0.5, rel. min. at x ! 3.5
54. T !2*2mrFc
Fc
xO
f (x)
!2
!8
!16
2 4
8
16f(x) " x 3 ! 6x 2 # 4x # 3
©Glencoe/McGraw-Hill 183 Algebra 2 Chapter 7
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©Glencoe/McGraw-Hill 184 Algebra 2 Chapter 7
56. ("3, "1)
58. C
60.
62. "3 ( i174
"7 ( 1172
55. (4, "2)
57. A
59. S
61. 9 ( 1576
Lesson 7-5 Roots and ZerosPages 375–377
1. Sample answer: p(x ) ! x3 "6x2 # x # 1; p(x ) has either2 or 0 positive real zeros, 1negative real zero, and 2 or0 imaginary zeros.
3. 6
5. "7, 0, and 3; 3 real
7. 2 or 0; 1; 2 or 4
9. 2, 1 # i, 1 " i
11. 2 # 3i, 2 " 3i, "1
13. real
15. 0, 3i, "3i; 1 real, 2 imaginary
17. 2, "2, 2i, and "2i; 2 real, 2 imaginary
19. 2 or 0; 1; 2 or 0
21. 3 or 1; 0; 2 or 0
23. 4, 2, or 0; 1; 4, 2, or 0
25. "2, "2 # 3i, "2 " 3i
"83; 1
2. An odd-degree functionapproaches positive infinity inone direction and negativeinfinity in the other direction,so the graph must cross thex-axis at least once, giving itat least one real root.
4. 2i, "2i; 2 imaginary
6. 2 or 0; 1; 2 or 0
8. "4, 1 # 2i, 1 " 2i
10. 2i, "2i, 3
12. f (x ) ! x3 " 2x2 # 16x " 32
14. 2 imaginary
16. 3i, 3i, "3i, and "3i; 4imaginary
18. "2, "2, 0, 2, and 2, 5 real
20. 2 or 0; 1; 2 or 0
22. 1; 3 or 1; 2 or 0
24. 5, 3, or 1; 5, 3, or 1; 0, 2, 4,6, or 8
26. 4, 1 # i, 1 " i
5 ( i2714
;
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©Glencoe/McGraw-Hill 185 Algebra 2 Chapter 7
27.
29.
31. 4 " i, 4 # i, "3
33. 3 " 2i, 3 # 2i, "1, 1
35. f(x ) ! x3 " 2x2 " 19x # 20
37. f(x ) ! x4 # 7x2 " 144
39. f(x ) ! x3 " 11x2 # 23x " 45
41a.
41b.
41c.
O
f (x )
x
O
f (x )
x
O
f (x )
x
"32, 1 # 4i, 1 " 4i
2i, "2i, i2, "
i2
28. 5i, "5i, 7
30.
32. 3 " i, 3 # i, 4, "1
34. 5 " i, 5 # i, "1, 6
36. f(x ) ! x4 " 10x3 # 20x2 #40x " 96
38. f(x ) ! x5 " x4 # 13x3 "13x2 # 36x " 36
40. f(x) ! x3 " 10x2 # 32x " 48
42. (3 " x)(4 " x)(5 " x) ! 24
12, 4 # 5i, 4 " 5i
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©Glencoe/McGraw-Hill 186 Algebra 2 Chapter 7
43. 1 ft
45. radius ! 4 m, height ! 21 m
47. "24.1, "4.0, 0, and 3.1
49. Sample answer: f(x) ! x3 "6x2 # 5x # 12 and g(x) !2x3 " 12x2 # 10x # 24each have zeros at x ! 4, x ! "2, and x ! 3.
[!30, 10] scl: 5 by [!20, 20] scl: 5
44. V(r ) ! *r 3 # 17*r 2
46. 1; 2 or 0; 2 or 0
48. Nonnegative roots representtimes when there is noconcentration of dyeregistering on the monitor.
50. One root is a double root.Sample graph:
xO
f (x)
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51. If the equation models thelevel of a medication in apatient’s bloodstream, adoctor can use the roots ofthe equation to determinehow often the patient shouldtake the medication tomaintain the necessaryconcentration in the body.Answers should include thefollowing.• A graph of this equation
reveals that only the firstpositive real root of theequation, 5, has meaningfor this situation, since thenext positive real rootoccurs after themedication level in thebloodstream has droppedbelow 0 mg. Thusaccording to this model,after 5 hours there is nosignificant amount ofmedicine left in thebloodstream.
• The patient should not gomore than 5 hours beforetaking their next dose ofmedication.
53. C
55. "254, 915
57. min.; "13
59. min.; "7
61. (6p " 5)(2p " 9)
63. C"3 23 "4
"2 9S
52. A
54. "127, 41
56. 36 in.
58. max.; 32
60. 5ab2(3a " c2)
62. 4y (y # 3)2
64. C11 57 04 "5
S
©Glencoe/McGraw-Hill 187 Algebra 2 Chapter 7
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65.
67.
69. (
19, (
13, (1, (3
(12, (1, (
52, (5
£29 "88 9
16 "16§ 66.
68.
70. (2, (4(116
, (18, (
14, (
12, (1,
(
114
, (17, (
27, (
12, (1, (2
y & "23x " 1
©Glencoe/McGraw-Hill 188 Algebra 2 Chapter 7
Lesson 7-6 Rational Zero TheoremPages 380–382
1. Sample answer: You limit thenumber of possible solutions.
3. Luis; Lauren found numbersin the form not as Luis did according to the Rational Zero Theorem.
5.
7. "2, "4, 7
9. "2, 2,
11. 10 cm ) 11 cm ) 13 cm
13. (1, (2, (3, (6
15. (1, (2, (3, (6, (9, (18
17. (1, , (3, (9, (27
19. "1, "1, 2
21. 0, 9
23. 0, 2, "2
25. "2, "4
(13, (
19
72
(1, (2, (12, (
13, (
16, (
23
pq
qp
,
2. Sample answer:
4. (1, (2, (5, (10
6. "4, 2, 7
8. 2, "2, 3, "3
10.
12. (1, (2
14. (1, (3, (5, (15,
16. (1, , (3
18. "6, "5, 10
20. 1, "1
22. , "1, 1
24. 0, 3
26. "7, 1, 3
12
(13
(13, (
53
23,
"3 ( 2174
2x2 # x # 3
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©Glencoe/McGraw-Hill 189 Algebra 2 Chapter 7
28. , 2
30.
32.
34.
36. r ! 2 in., h ! 6 in.
38. ! 36 in., w ! 48 in., h ! 32 in.
40.
42. k ! "3; "3, "6, 5
44. D
46. "6, "3, 5
6300 !13/3 " 3/2
/
V !13
˛*r 3 #43
˛*r 2
3, 23, "
23,
"3 ( 2132
"2, 43,
"3 ( i2
"12,
15
27. , "2
29.
31.
33. "1, "2, 5, i, "i
35.
37.
39.
41. ! 30 in., w ! 30 in., h ! 21 in.
43. The Rational Zero Theoremhelps factor large numbersby eliminating some possiblezeros because it is notpractical to test all of themusing synthetic substitution.Answers should include thefollowing.• The polynomial equation
that represents the volumeof the compartment is
• Reasonable measures ofthe width of thecompartment are, ininches, 1, 2, 3, 4, 6, 7, 9,12, 14, 18, 21, 22, 28, 33,36, 42, 44, 63, 66, 77, and84. The solution showsthat w ! 14 in., ! 22in., and d ! 9 in.
45. Sample answer:
106x " 120x5 " x4 " 27x3 # 41x2 #
/
V ! w3 # 3w 2 " 40w.
/
V !13/3 " 3/2
V ! 2h3 " 8h2 " 64h
2, "2 ( i13; 2
45, 0,
5 ( 132
i
"12,
13,
12,
34
12, "
13
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47.
49. "7, 5 # 2i, 5 " 2i
51.
53.
55. 6 cm, 8 cm, 10 cm
57.
59. 106x2 "85x # 25
61. x2 # x " 4 #5
x # 1
x5 " 7x4 " 8x3 #
4x2 " 8x # 3
(3xy22x
x " 4, 3x2 # 2
"4, 2 # i, 2 " i 48. "5, 3i, "3i
50.
52.
54.
56.
58.
60. x " 9 #33
x # 7
x3 # 5x2 # x " 10
x3 # 4x 2 " 6
04x " 5 0725
4x # 3, 5x " 1
©Glencoe/McGraw-Hill 190 Algebra 2 Chapter 7
Chapter 7Practice Quiz 2
Page 382
1. "930, "145
3.
5. "32
22x # 24 ! 0x4 " 4x3 " 7x2 #
2. 0, "180
4. 45
Lesson 7-7 Operations on FunctionsPages 386–389
1. Sometimes; sample answer:If f(x ) ! x " 2, g(x ) ! x # 8,
and
3. Danette; [g ! f ](x ) ! g [f (x )]means to evaluate the f function first and then the g function. Marquanevaluated the functions in the wrong order.
g ! f ! x # 6.then f ! g ! x # 6
2. Sample answer: g(x ) !{("2, 1), ("1, 2), (4, 3)}, f (x ) ! {(1, 7),(2, 9), (3, 3)}
4.
x + "5
3x # 4x # 5
,3x2 # 19x # 20;4x # 9; 2x " 1;
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©Glencoe/McGraw-Hill 191 Algebra 2 Chapter 7
5.;
7. {(2, "7)}; {(1, 0), (2, 10)}
9.
11. 11
13.
15. $33.74; price of CD whencoupon is subtracted andthen 25% discount is taken
17.
19.
21.
23. {(1, "3), ("3, 1), (2, 1)};{(1, 0), (0, 1)}
25. {(0, 0), (8, 3), (3, 3)};{(3, 6), (4, 4), (6, 6), (7, 8)}
27. {(5, 1), (8, 9)}; {(2, "4)}
x 3 # x 2 " x # 1x
, x + 0
x + "1; x2 " x ; x + "1
x 3 # x2 " 2x " 1x # 1
,
x 3 # x 2 " 1x # 1
, x + "1
"2x3 # 16x2; 2x2
8 " x, x + 8
2x2 " x # 8; 2x2 # x " 8;
2x ; 18; x2 " 81; x # 9x " 9
, x + 9
p(x ) !34
x; c (x ) ! x " 5
x2 # 11; x2 # 10x # 31
x2 # 3x " 4
, x + 4
x3 " 4x2 # 3x " 12x2 # x " 1; x2 " x # 7; 6. {("5, 7), (4, 9)}; {(4, 12)}
8.
10. 30
12. 1
14. $32.49; price of CD when25% discount is taken andthen the coupon issubtracted
16. Discount first, then coupon;sample answer: 25% of49.99 is greater than 25% of44.99.
18.
20.
22.
24. {(2, 4), (4, 4)}; {(1, 5), (3, 3),(5, 3)}
26. {(4, 5), (2, 5), (6, 12), (8, 12)};does not exist
28. {(2, 3), (2, 2)}; {("5, 6), (8, 6),("9, "5)}
x2 # 4x # 4, x + "2, 3x2 " 6x # 9, x + 2;
x 3 # x2 " 9x " 9x # 2
, x + "2;
x 3 # x 2 " 7x " 15x # 2
, x + "2;
x # 3
2, x + "3
2x3 # 18x2 # 54x # 54;x 2 # 8x # 15; x2 # 4x # 3;
2x " 34x # 9
, x +"94
8x2 # 6x " 27;6x # 6; "2x " 12;
6x " 8; 6x " 4
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©Glencoe/McGraw-Hill 192 Algebra 2 Chapter 7
29.
31.
33.
35. "12
37. 39
39. 25
41. 2
43. 79
45. 226
47. P(x ) ! "50x # 1939
49. p(x ) ! 0.70x; s(x ) ! 1.0575x
51. $110.30
53. 373 K; 273 K
55. $700, $661.20, $621.78,$581.73, $541.04
57. Answers should include thefollowing.• Using the revenue and
cost functions, a newfunction that representsthe profit is p(x ) ! r(c(x )).
• The benefit of combiningtwo functions into onefunction is that there arefewer steps to computeand it is less confusing tothe general population ofpeople reading theformulas.
59. C
8x3 # 4x2 # 2x # 12x3 # 2x2 # 2x # 2;
x2 # 2; x2 # 4x # 4
8x " 4; 8x " 1 30.
32.
34.
36. 50
38. 68
40. "48
42.
44. 104
46. 36
48. 939,000
50. s[p(x )]; The 30% would betaken off first, and then thesales tax would be calculatedon this price.
52. [K ! C](F ) ! (F " 32) # 273
54. 309.67 K
56. 244
58. A
60. (1, (2, (4, (8
59
112
2x2 " 5x # 9; 2x2 " x # 5
3x2 " 4; 3x2 " 24x # 48
15x " 5; 15x # 1
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61.
63.
65.
67.
69.
71.
73.
75.
77.
79.
81. m !Fr 2
GM
t !I
pr
y !1 " 4x 2
"5x
"116
B"5"3
"2 2R
"12 B"1
"3"2"4R
c 3 "2"1 1
d10 # 2j
x3 " 9x2 # 31x " 39
6x3 " 13x2 # 9x " 2
x3 " 4x2 " 17x # 60
(34, (6
(2, (3, (32,(1, (
12, (
14, 62.
64.
66.
68.
70.
72. does not exist
74. does not exist
76.
78.
80. F !95C # 32
x !"2
3 # 7y
x !6 # 3y
2
"12 B 5
"7"6 8R
x4 # x3 " 14x2 # 26x " 20
x3 " 6x2 # 4x " 24
x3 " 3x2 " 34x " 48
(1, (13, (
19
©Glencoe/McGraw-Hill 193 Algebra 2 Chapter 7
Lesson 7-8 Inverse Functions and RelationsPages 393–394
1. no
3. Sample answer:f (x) ! 2x, f "1(x) ! 0.5x;
5. {(4, 2), (1, "3), (8, 2)}
f [f "1(x )] ! f "1[f (x )] ! x
2. Switch x and y in theequation and solve for y.
4. n is an odd whole number.
6. {(3, 1), ("1, 1), ("3, 1), (1, 1)}
PQ245-6457F-P07[167-202].qxd 7/24/02 12:02 PM Page 193 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
7. f "1(x ) ! "x
9.
11. no
13. 15.24 m/s2
15. {(8, 3), ("2, 4), ("3, 5)}
17. {("2, "1), ("2, "3), ("4, "1), (6, 0)}
19. {(8, 2), (5, "6), (2, 8), ("6, 5)}
x
y
O 4 8 12
8
12
4
!4 y !1 " 2x !10
y " x # 512
y ! 2x " 10
xO 2 4
4
2
!4
!2
!2!4
f (x)
f !1(x ) " !xf (x) " !x
8. g"1(x ) !
10. yes
12. 32.2 ft/s2
14. {(6, 2), (5, 4), ("1, "3)}
16. {("4, 7), (5, 3), (4, "1), (5, 7)}
18. {(11, 6), (7, "2), (3, 0), (3, "5)}
20.
xO 2 4
4
2
!4
!2
!2!4
y
y " !3
x " !3
x ! "3
xO 2 4
2
!2
!2!4
g(x)
g!1(x) " x ! 13
13
g(x ) " 3x # 1
13
x "13
©Glencoe/McGraw-Hill 194 Algebra 2 Chapter 7
PQ245-6457F-P07[167-202].qxd 7/24/02 12:02 PM Page 194 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
©Glencoe/McGraw-Hill 195 Algebra 2 Chapter 7
21.
23.
25.
xO 2 4
4
!4
!2
!2!4
y
y !1 " ! x ! 12
12
y " !2x ! 1
y ! "12x "
12
xO 2 4
4
2
!4
!2
!2!4
g(x)
g!1(x) " x ! 4
g(x ) " x # 4
g"1(x ) ! x " 4
xO 2 4
4
2
!2
!2!4
g(x)
g(x ) " !2x
g!1(x ) " ! x12
g"1(x ) ! "12x 22.
24.
26.
xO 2 4
4
2
!4
!2
!2!4
y
y !1 " 3x
y " x13
y ! 3x
xO 2 4
4
2
!4
!2
!2!4
f (x) " 3x # 3
f (x)
f !1(x) " x !113
f "1(x ) !13x " 1
xO 2 4
4
2
!4
!2
!2!4
f (x)
f !1(x) " x # 5
f (x) " x ! 5
f"1(x ) ! x # 5
PQ245-6457F-P07[167-202].qxd 7/24/02 12:02 PM Page 195 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
©Glencoe/McGraw-Hill 196 Algebra 2 Chapter 7
27.
29.
31.
33. no
35. yes
37. yes
39. y !12x "
112
xO 2 4
4
2
!4
!2
!2!4
f !1(x) " x # 87
47
f (x)
f (x) " 7x ! 48
f "1(x ) !87x #
47
f (x)x
f (x) " x ! 745
f (x)!1 " x # 54
354
!30 !20 !10
!10
!20
!30
!40
!40 O
f "1(x ) !54x #
354
xO 2 4
4
2
!4
!2
!2!4
f (x)
f !1(x) " x85
f (x) " x58
f "1(x ) !85x 28.
30.
32. yes
34. no
36. yes
38.
40. 12
y !4(x # 7) " 6
2
xO 2 4
4
2
!4
!2
!2!4
g(x)
g(x) " 2x # 36
g!1(x) " 3x ! 32
g"1(x ) ! 3x "32
x21 3 4 5 6 7 8
78
65
1
32
f (x) " x # 413
f (x)
O
f !1(x ) " 3x ! 12
f "1(x ) ! 3x " 12
PQ245-6457F-P07[167-202].qxd 7/24/02 12:02 PM Page 196 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
41.
43. It can be used to convertCelsius to Fahrenheit.
45. Inverses are used to convertbetween two units ofmeasurement. Answersshould include the following.• Even if it is not necessary,
it is helpful to know theimperial units when giventhe metric units becausemost measurements in theU.S. are given in imperialunits so it is easier tounderstand the quantitiesusing our system.
• To convert the speed oflight from meters persecond to miles per hour,
47. B
49.
51. "7, "2, 3
53. 64
55. 3
57. 117
h[g(x )] ! 6xg[h(x )] ! 6x " 10;
675,000,000 mi/hr
3600 seconds1 hour
!1mile
1600 meters!
! 3.0 ) 108 meters1 second
!f (x)
I(m) ! 320 # 0.04m; $4500 42.
44. Sample answer:
or
46. A
48.
50.
52.
54. 32
56. 4
58. 196
"14,
43,
52
h[g(x)] ! x2 # 5x " 24g [h(x)] ! x2 " 3x " 24;
h[g(x)] ! 4x # 5g [h(x)] ! 4x # 20;
f(x ) ! "x and f "1(x ) ! "x
f(x ) ! x and f "1(x ) ! x
C [C"1(x )] ! C"1[C (x )] ! x
C"1(x ) !95x # 32;
©Glencoe/McGraw-Hill 197 Algebra 2 Chapter 7
PQ245-6457F-P07[167-202].qxd 7/24/02 12:02 PM Page 197 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
59. "7
61. 254
60. "
©Glencoe/McGraw-Hill 198 Algebra 2 Chapter 7
Lesson 7-9 Square Root Functions and InequalitiesPages 397–399
1. In order for it to be a squareroot function, only thenonnegative range can beconsidered.
3. Sample answer:
5.
7.
D: x & 1; R: y & 3
y
O x
87654321
1 2 3 4 5 6 7 8
y " "x ! 1 # 3
D: x & 0; R: y & 0
y
O x
87654321
1 2 3 4 5 6 7 8
y " "4x
y ! 22x " 4
2. Both have the shape of thegraph of but
is shifted down4 units, and isshifted to the right 4 units.
4.
6.
8. y
O x
87654321
1 2 3 4 5 6 7 8
y ""x ! 4 # 1
D: x & 0; R: y % 3
!2
4
2
4 8 12
y " 3 ! "x
y
Ox
D: x & 0, R: y & 2
y
O x
87654321
1 2 3 4 5 6 7 8
y " "x # 2
y ! 2x " 4y ! 2x " 4
y ! 2x,
PQ245-6457F-P07[167-202].qxd 7/24/02 12:02 PM Page 198 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
©Glencoe/McGraw-Hill 199 Algebra 2 Chapter 7
9.
11.
13. Yes; sample answer: Theadvertised pump will reach amaximum height of 87.9 ft.
15.
D: x & 0, R: y % 0
1 2 3 4 5 6 7 8
y " !"5x
!1!2!3!4!5!6!7!8
yO x
y
O x
4321
1 2
!2!3!4
3 4 5 6!2
y " "x # 2 ! 1
2
4
6
8
2!2 4 6
y " "2x # 4
y
O x
10.
12.
14.
16.
D: x & 0, R: y % 0
!1!2!3!4!5!6!7!8
1 2 3 4 5 6 7 8
y " !4"x
yO x
D: x & 0, R: y & 0
y
O x
87654321
1 2 3 4 5 6 7 8
y " "3x
v ! 22gh
y
O x
4321
1 2 3 4 5 6 7!1
y " 3 ! "5x # 1
!2!3!4
PQ245-6457F-P07[167-202].qxd 7/24/02 12:02 PM Page 199 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
©Glencoe/McGraw-Hill 200 Algebra 2 Chapter 7
17.
19.
21.
23.
D: x & "4, R: y % 5
y
Ox
8
6
4
2
2 4!2!4
y " 5 !"x # 4
D: x & 0.6, R: y & 0
y
O x
87654321
1 2 3 4 5 6 7 8
y " "5x ! 3
D: x & 7, R: y & 0
y
O x
87654321
1 2 3 4 5 6 7 8
y " "x ! 7
D: x & 0, R: y & 0
y
O x
87654321
1 2 3 4 5 6 7 8
y " "x 12
18.
20.
22.
24.
D: x & 2, R: y & 4
y
O x
87654321
1 2 3 4 5 6 7 8
y " "3x ! 6 # 4
D: x & "6, R: y & "3
!2
!4
4
2
2!6 !4 !2
y " "x # 6 ! 3
y
O x
D: x & "0.5, R: y % 0
!1!2!3!4!5!6!7!8
1 2 3 4 5 6 7 8
y " !"2x # 1
yO x
D: x & "2, R: y & 0
8
6
4
2
2!2 4 6
y " "x # 2
y
O x
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25.
27.
29.
31.
33. 317.29 mi
35. See students’ work.
y
O x
87654321
1 2 3 4 5 6 7 8
y " "6x ! 2 # 1
y
O x
87654321
1 2 3 4 5 6 7 8
y " "5x ! 8
y
Ox
8
6
4
2
2!4 !2
y " "x # 5
D: x % 0.75, R: y & 3
y
O x
8
6
4
2
!3 !2 !1
y " 2"3 ! 4x # 3
26.
28.
30.
32. 125 ft
34. 119 lb
36. If a is negative, the graph isreflected over the x-axis. Thelarger the value of a, the lesssteep the graph. If h ispositive, the origin is
y
O x
87654321
1 2 3 4 5 6 7 8
y " "x ! 3 # 4
y
Ox
8
6
4
2
2 4!4 !2
y " "2x # 8
1 2 3 4 5 6 7 8
y " !6"x
!2!4!6!8!10!12!14!16
yO x
©Glencoe/McGraw-Hill 201 Algebra 2 Chapter 7
PQ245-6457F-P07[167-202].qxd 7/24/02 12:02 PM Page 201 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
translated to the right, and ifh is negative, the origin istranslated to the left. When kis positive, the origin istranslated up, and when k isnegative, the origin istranslated down.
38. C
40. yes
42. yes
44.
46. 4; If x is your number, youcan write the expression
which equals 4
after dividing the numeratorand denominator by the GCF,
48. 6p 2 " 2p " 20
x # 2.
3x # x # 8x # 2
,
10x2 " 40x # 40; 10; x + 211x " 22; 9x"18;
©Glencoe/McGraw-Hill 202 Algebra 2 Chapter 7
37. Square root functions areused in bridge designbecause the engineers mustdetermine what diameter ofsteel cable needs to be usedto support a bridge based onits weight. Answers shouldinclude the following.• Sample answer: When the
weight to be supported isless than 8 tons.
• 13,608 tons
39. D
41. no
43.
45.
47.
49. a3 " 1
2x2 " 4x " 16
x + "32
8x3 # 12x3 " 18x " 27,
x + "32; 2x " 3, x + "
32;
8x 3 # 12x 2 " 18x " 282x # 3
,
x + "32;
8x 3 # 12x 2 " 18x " 262x # 3
,
x # 5x " 3
, x + 3
2x # 2; 8; x2 # 2x " 15;
PQ245-6457F-P07[167-202].qxd 7/24/02 12:02 PM Page 202 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
©Glencoe/McGraw-Hill 203 Algebra 2 Chapter 8
1. Since the sum of the x-coordinates of the givenpoints is negative, the x-coordinate of the midpointis negative. Since the sum ofthe y-coordinates of thegiven points is positive, they-coordinate of the midpointis positive. Therefore, themidpoint is in Quadrant II.
3. Sample answer: (0, 0) and(5, 2)
5. (2.5, 2.25)
7. units
9. D
11. (!4, !2)
13.
15. (3.1, 2.7)
17.
19. (7, 11)
21. Sample answer: Drawseveral line segments acrossthe U.S. One should go fromthe NE corner to the SWcorner; another should gofrom the SW corner to theNW corner; another shouldgo across the middle (east towest); and so on. Find themidpoints of these segments.Locate a point to representall of these midpoints.
a 124
,
58b
a172
,
272b
1122
Chapter 8 Conic SectionsLesson 8-1 Midpoint and Distance Formulas
Pages 414–416
2. all of the points on theperpendicular bisector of thesegment
4.
6. 10 units
8. units
10. (12, 5)
12. (2, !6)
14. (0.075, 3.2)
16.
18.
20. around 8th St. and 10th Ave.
22. near Lebanon, Kansas
a12,
132b, a5
2, 2b, a5,
12b
a 512
, !524b
12.61
a!2,
132b
23. See students’ work.
25. 25 units
27. units
29. units
31. 1 unit
33. units
35. units, 10 units2
37. units
39. about 0.9 h
41. The slope of the line through
(x1, y1) and (x2, y2) is and the point-slope form of the equation of the line is
y ! y1 " (x ! x1).
Substitute
into this equation. The left
side is ! y1 or .
The right side is
or Therefore, the
point with coordinates
lies on the
line through (x1, y1) and (x2, y2). The distance from
to (x1, y1) is
or . TheB¢x1 ! x2
2!2 # ¢y1 ! y2
2!2
B¢x1 !x1 # x2
2!2 # ¢y1 !
y1 # y2
2!2
¢x1 # x2
2,
y1 # y2
2!
¢x1 # x2
2,
y1 # y2
2!
y2 ! y1
2.
y2 ! y1
x2 ! x1 ¢x2 ! x1
2!¢x1 # x2
2! x1! "
y2 ! y1
x2 ! x1
y2 ! y1
2y1 # y2
2
¢x1 # x2
2,
y1 # y2
2!
y2 ! y1
x2 ! x1
y2 ! y1
x2 ! x1
1130
712 # 158
181312
170.25
3117
24. 13 units
26. units
28. 0.75 unit
30. units
32. units
34. units, units2
36.units
38. about 85 mi
40. 14 in.
42. The formulas can be used todecide from which locationan emergency squad shouldbe dispatched. Answersshould include the following.• Most maps have a
superimposed grid. Thinkof the grid as a coordinatesystem and assignapproximate coordinatesto the two cities. Then usethe Distance Formula tofind the distance betweenthe points with thosecoordinates.
• Suppose the bottom left ofthe grid is the origin. Thenthe coordinates of Lincolnare about (0.7, 0.3); thecoordinates of Omaha areabout (4.6, 3.3); and thecoordinates of Fremontare about (1.5, 4.6). Thedistance from Omaha toFremont is about
or 34 miles. The distancefrom Lincoln to Fremont is
102(1.5 ! 4.6)2 # (4.6 !3.3)2
1277165 # 212 # 1122 #
90$6110$
1271
165
12
©Glencoe/McGraw-Hill 204 Algebra 2 Chapter 8
distance from
to (x2, y2) is
or
.
Therefore, the point with
coordinates
is equidistant from (x1, y1)and (x2, y2).
43. C
45. on the line with equation y " x
47.
49.
51. !1 # 13i
53. 4 ! 3i
y
xO
y ! 2!x " 1
R " 5y 0y % 16D " 5x 0x % 06,
y
xO
y ! !x # 2
R " 5y 0y % 06D " 5x 0x % 26,
¢x1 # x2
2,
y1 # y2
2!
B¢x1 ! x2
2!2 # ¢y1 ! y2
2!2
B¢x2 ! x1
2!2 # ¢y2 ! y1
2!2
B¢x1 !x1 #x2
2b2 # ¢y2 !
y1 #y2
2!2 "
¢x1 # x2
2,
y1 # y2
2! about
or 44 miles. Since Omahais closer than Lincoln, thehelicopter should bedispatched from Omaha.
44. B
46. !1; is perpendicular tothe line with equation y " x,which has slope 1.
48.
50. no
52. 6 ! 2i
54. y " (x # 3)2
y
xO
y ! !x # 1
R " 5y 0 % !16D " 5x 0x % 06,
AA!" ¿
102(1.5 ! 0.7)2 # (4.6 !0.3)2
©Glencoe/McGraw-Hill 205 Algebra 2 Chapter 8
55. y " (x ! 2)2 ! 3
57. y " 3(x ! 1)2 # 2
59. y " !3(x # 3)2 # 17
56. y " 2(x # 5)2
58. y " !(x # 2)2 # 10
©Glencoe/McGraw-Hill 206 Algebra 2 Chapter 8
1. (3, !7), 3, !6 x " 3,
y "!7
3. When she added 9 tocomplete the square, she forgotto also subtract 9. Thestandard form is y " (x # 3)2!9 # 4 or y " (x # 3)2 ! 5.
5. (3, !4), 3, !3 x " 3,
y " !4 upward, 1 unity
xO
y ! (x # 3) 2 # 4
14,
b,34
a
116
b,1516
a 2. Sample answer: x "!y2
4. y " 2(x ! 3)2 ! 12
6. (!7, 3), x " !7,
y " 2 upward, unit
y
xy ! 2(x " 7)2 " 3
#2 2#4#6#8#10#12#14
2468
10121416
12
78,
a!7, 318b,
Lesson 8-2 ParabolasPages 423–425
8.
right, units
10.
12. y " (x ! 3)2 # 2
14.
16. (0, 0), , x " 0,
downward, 6 unitsy
xO
#6y ! x 2
y "32,a0, !
32b
y "12
(x # 12)2 ! 80
y
xOx ! # (y " 1) 2 " 51
8
x " !18
(y # 1)2 # 5
y
xO
x ! y 2 # 6y " 1223
32
x " !158
,
y "92,a!9
8,
92b,a!3
2,
92b,
©Glencoe/McGraw-Hill 207 Algebra 2 Chapter 8
7.
downward, unit
9.
11.
13. x " (y # 7)2 ! 29
15. x " 3 ay #56b2 ! 11
112
x "124
y2 ! 6
y
xO
y ! (x # 3) 2 " 618
y "18
(x ! 3)2 # 6
y
xO
y ! #3x2 # 8x # 6
13
y " !712
,
x " !43,a!4
3, !
34b,a!4
3, !
23b,
18. (!6, 3), x " !6,
upward, 3 units
20. (2, !3), (3, !3), y " !3,x " 1, right, 4 units
22. (6, !16), x " 6,
upward, 1 unit
#2#4#6#8
#10#12#14#16
#2 2 4 6 8 10 12 14y
xO
y ! x 2 # 12x " 20
y " !1614,
a6, !1534b,
y
xO
4(x # 2) ! (y " 3)2
y
xO
3(y # 3) ! (x " 6)2
y " 214,
a!6, 334b,
©Glencoe/McGraw-Hill 208 Algebra 2 Chapter 8
17. (0, 0), y " 0,
right, 2 units
19. (1, 4), x " 1,
downward, 2 units
21. (4, 8), (3, 8), y " 8, x " 5,left, 4 units
161412108642
#4 #3#2#1 1 2 3 4
y
xO
(y # 8)2 ! #4(x # 4)
y
x
O
#2(y # 4) ! (x # 1)2
y " 412,a1, 3
12b,
y
xO
y 2 ! 2x
x " !12,a1
2, 0b,
24.
right, unit
26.
downward, unit
28. (3, 5), x " 3,
upward, 2 unitsy
xO
y ! x 2 # 3x " 192
12
y " 4
12,a3, 5
12b,
#4
#8
#12
#16
#2 2 4y
xO
x ! #2x 2 " 5x # 10
12
y " !274
,
x "54,a5
4, !7b,a5
4, !
558b,
O
21
x
#2#3#4#5#6
10 20 30 40 50 60 70 80
yx ! 5y 2 " 25y " 60
15
x "28710
,
y " !52,a144
5, !
52b,a115
4, !
52b,
©Glencoe/McGraw-Hill 209 Algebra 2 Chapter 8
23. (!24, 7), y " 7,
right, 1 unit
25. (4, 2), x " 4,
upward, unit
27.
left, unit
y
xO
x ! #4y 2 " 6y " 2
14
x "6916
,
y "34,a67
16,
34b,a17
4,
34b,
y
xO
y ! 3x 2 # 24x " 50
13
y " 11112
,
a4, 2
112b,
24
16
8
#8
#24 #16 #8 8
y
xO
x ! y 2 # 14y " 25
x " !2414,
a!2334, 7b,
29. (123, !18),
y " !18, x " 123 left, 3 units
31. 1
33.
35. 0.75 cm
37.
141210
1 2 3 4 5 6 7 8#2
8642
y
xO
x ! # (y # 6)2" 8124
x " !124
(y ! 6)2 # 8
y " !23
20
#20
#40
#60
#120 #60 60 120
y
xO
x ! # y 2 # 12y " 1513
34,
a122
14, !18b, 30.
32. !1 and !
34.
36.
38.
y
xO
x ! (y " 2)2# 618
x "18
(y # 2)2 ! 6
y
xO
y ! x 2 " 1116
y "116
x2 # 1
a!13, !
23b
13
y
xO
x ! 3y 2 " 4y " 1
©Glencoe/McGraw-Hill 210 Algebra 2 Chapter 8
39.
41.
43. about y " !0.00046x2 # 325
45.
47. A parabolic reflector can beused to make a car headlightmore effective. Answersshould include the following.• Reflected rays are focused
at that point.• The light from an
unreflected bulb wouldshine in all directions. Witha parabolic reflector, mostof the light can be directedforward toward the road.
49. A
51. 10 units
y " !1
26,200x2 # 6550
y
xO
x ! (y # 3)2" 414
x "14
(y ! 3)2 # 4
#2#4#6#8
#4 #3#2#1 1 2 3 4 5 6
8642
y
xO
y ! (x # 1)2" 7116
y "116
(x ! 1)2 # 7 40.
42.
44.
46. x " (y ! 3)2 # 4
48. B
50. 13 units
52. units234
y " !1
100 (x ! 50)2 # 25
y "29x2 ! 2
XBox.XBox.
#12 #8 #4
#4
#8
8
4
y
xO
y ! # (x " 7)2" 416
y " !16(x # 7)2 # 4
©Glencoe/McGraw-Hill 211 Algebra 2 Chapter 8
53.
55. 4
57. 9
59.
61. 423
223
y
xO y !!x " 1
54. 2.016 & 105
56. 5
58. 12
60.
62. 622
322
©Glencoe/McGraw-Hill 212 Algebra 2 Chapter 8
2. (x # 3)2 # (y ! 1)2 " 64; left3 units, up 1 unit
4. (x ! 3)2 # (y # 1)2 " 9
6. x2 # (y # 2)2 " 25
8. (4, 1), 3 unitsy
xO
(x # 4)2 " (y # 1)2 ! 9
Lesson 8-3 CirclesPages 428–431
1. Sample answer: (x ! 6)2 #(y # 2)2 " 16
3. Lucy; 36 is the square of theradius, so the radius is 6units.
5. (x # 1)2 # (y # 5)2 " 4
7. (x ! 3)2 # (y # 7)2 " 9
9. (0, 14), units
11. unit
13. (!2, 0), units
15.y
x
Earth
Satellite
6400km 42,200
km
35,800km
y
xO
(x " 2)2 " y 2 ! 12
223
y
xO
(x " )2 " (y # )2
!23
12
89
2123
b!23
, 12
a
O#16 #8 8 16
#8
24
16
8
y
x
x 2 " (y # 14)2 ! 34
234 10. (4, 0), unit
12. (!4, 3), 5 units
14. x2 # y 2 " 42,2002
16. (x # 1)2 # (y ! 1)2 " 16
y
xO
(x " 4)2 " (y # 3)2 ! 25
y
xO
(x # 4)2 " y 2 ! 1625
45
©Glencoe/McGraw-Hill 213 Algebra 2 Chapter 8
17. (x ! 2)2 # (y # 1)2 " 4
19. (x # 8)2 # (y ! 7)2 "
21. (x # 1)2 #
23.1777
25. (x ! 4)2 # (y ! 2)2 " 4
27. (x # 5)2 # (y ! 4)2 " 25
29. (x # 2.5)2 # (y # 2.8)2 " 1600
31. (0, 0), 12 units
8 16#16 #8
16
8
#8
#16
y
xO
x2 " y 2 ! 144
(x # 213)2 # (y ! 42)2 "
ay #12b2 "
19454
14
18. x2 # (y ! 3)2 " 49
20. (x # 1)2 # (y ! 4)2 " 20
22. (x ! 8)2 # (y # 9)2 " 1130
24. (x # 8)2 # (y # 7)2 " 64
26. (x ! 1)2 # (y ! 4)2 " 16
28. x2 # y2 " 18
30. (0, !2), 2 units
32. (3, 1), 5 unitsy
xO
(x # 3)2 " (y # 1)2 ! 25
y
xO
x 2 " (y " 2)2 ! 4
©Glencoe/McGraw-Hill 214 Algebra 2 Chapter 8
33. (!3, !7), 9 units
35. (3, !7), units
37. unitsy
xO
229(!2, 23),
2
#6#4#2 2 4 6 8 10#2#4#6#8
#10#12#14
y
xO
(x # 3)2 " (y " 7)2 ! 50
522
42
#12#10#8#6#4#2 2 4 6 8#2#4#6#8
#10#12#14#16
O
y
x
(x " 3)2 " (y " 7)2 ! 81
34. (3, 0), 4 units
36. , 5 units
38. (!7, !3), unitsy
xO
222
y
xO
(!25, 4)
y
xO
(x # 3)2 " y 2 ! 16
©Glencoe/McGraw-Hill 215 Algebra 2 Chapter 8
40. (!1, 0), units
42. units
44. (6, 8), 4 units
161412108642
#2#2 2 4 6 8 10 12 14
y
xO
y
xO
21292
a!92, 4b,
y
xO
21139. (0, 3), 5 units
41. (9, 9), units
43. units
42
#2#4#6#8
#10#12
#6#4#2 2 4 6 8 10
y
xO
a32, !4b, 3217
2
18161412108642
#2#2 2 4 6 8 10 12 14 16 18
y
xO
2109
y
xO
©Glencoe/McGraw-Hill 216 Algebra 2 Chapter 8
45. (!1, !2), units
47. units
49. (x # 1)2 # (y # 2)2 " 5
51. A
53. y " '216 ! (x # 3)2
yxO
a0, !92b, 219
y
xO
214 46. (!2, 1), units
48. about 109 mi
50. A circle can be used torepresent the limit at whichplanes can be detected byradar. Answers shouldinclude the following.• x2 # y2 " 2500• The region whose
boundary is modeled by x2 # y2 " 4900 is larger,so there would be moreplanes to track.
52. D
54.
y " !216 ! (x # 3)2
y " 216 ! (x # 3)2,
y
xO
22
©Glencoe/McGraw-Hill 217 Algebra 2 Chapter 8
55.
57. (1, 0), y " 0,
left, unit
59. (!2, !4), x "!2,
upward, 1 unit
61. (!1, !2)
63. !4, !2, 1
65. 28 in. by 15 in.
67. 6
69. 25
71. 222
y
xO
y ! x 2 " 4x
y " !414,
a!2, !3
34b,
y
xO
x ! #3y 2 " 1
13
x " 1112
,a1112
, 0b,[#10, 10] scl:1 by [#10, 10] scl:1
56.The equations with the #symbol and ! symbolrepresent the right and lefthalves of the circle,respectively.
58. (3, !2), x " 3,
downward, 1 unit
60. (4, !4)
62.
64.
66. 12
68. 4
70. 225
!12, 2, 3
a32, 6b
y
xOy " 2 ! #(x # 3)2
y " !134,
a3, !214b,
x " !3 ' 216 ! y2;
©Glencoe/McGraw-Hill 218 Algebra 2 Chapter 8
1. 13 units
3. (0, 0), , y " 0,
right, 6 units
5. (0, 4), 7 units
12108642
#2#4
#8#6#4#2 2 4 6 8
y
xO
x 2 " (y # 4)2 ! 49
y
xO
y 2 ! 6x
x " !112,
a112, 0b
2. units
4. (!4, 4), x " !4,
upward, 1 unit
y
xO
y ! x 2 " 8x " 20
y " 334,
a!4, 414b,
2226
©Glencoe/McGraw-Hill 219 Algebra 2 Chapter 8
Chapter 8Practice Quiz 1
Page 431
1. x " !1, y " 2
3. Sample answer:
5.
7. (0, 0): (0, '3); 6
9. (0, 0); ('2, 0); 4y
xO
4x 2 " 8y 2 ! 32
412;
y
xO
y 2
18 x 2
9 " ! 1
612;
(y # 4)2
36#
(x ! 2)2
4" 1
(x ! 2)2
4#
(y # 5)2
1" 1
2. Let the equation of a circlebe (x ! h)2 # (y ! k)2 " r2.Divide each side by r2 to get
"1. This
is the equation of an ellipsewith a and b both equal to r. Inother words, a circle is anellipse whose major and minoraxes are both diameters.
4.
6.
8. (1, !2); (5, !2), (!3, !2);4
10. (4, !2); 10; 2y
xO
(4 ' 216, !2);
y
xO
(x # 1)2 20
(y " 2)2 4" ! 1
415;
y 2
100#
x 2
36" 1
x 2
36#
y 2
20" 1
(x ! h)2
r 2 #(y ! k)2
r 2
©Glencoe/McGraw-Hill 220 Algebra 2 Chapter 8
Lesson 8-4 EllipsesPages 437–440
11. about
13.
15.
17.
19.
21.
23. about
25.
27. (0, 0);y
xO
y 2
10 x 2
5 " ! 1
2152110;(0, '15);
y 2
20#
x 2
4" 1
" 1
y 2
2.00 & 1016x 2
2.02 & 1016 #
x 2
169#
y 2
25" 1
(x ! 5)2
64#
(y ! 4)2
814
" 1
(y ! 4)2
64#
(x ! 2)2
4" 1
y 2
16#
(x # 2)2
4" 1
x 2
16#
y 2
7" 1
" 1
y2
1.27 & 1015x 2
1.32 & 1015 # 12.
14.
16.
18.
20.
22.
24.
26.
28. (0, 0); ('4, 0); 10; 6y
xO
x 2
25 y 2
9 " ! 1
#(y # 1)2
5" 1
(x ! 1)2
30
#y 2
279,312.25" 1
x 2
193,600
x 2
324#
y 2
196" 1
(x ! 1)2
81#
(y ! 2)2
56" 1
(y ! 2)2
100#
(x ! 4229
" 1
(x # 2)2
81#
(y ! 5)2
16" 1
(x ! 5)2
64#
(y ! 4)2
9" 1
y2
64#
x2
39" 1
©Glencoe/McGraw-Hill 221 Algebra 2 Chapter 8
29. (!8, 2); 24; 18
31. (0, 0); 6;
33. (0, 0); 8; 6
35. (!3, 1); (!3, 5), (!3, !3);
y
xO
412416;
y
xO
16x 2 " 9y 2 ! 144
(0, '17);
y
xO
3x 2 " 9y 2 ! 27
213('16, 0);
8#8#16#24
16
8
#8
#16
y
xO
(x " 8)2 144
(y # 2)2 81 " ! 1
(!8 ' 317, 2); 30. (5, !11);24; 22
32. (0, 0); 6;
34. (0, 0); 18; 12
36. (!2, 7);
12
8
4
#4
4#4#8
y
xO
4124110 ;(!2 ' 412, 7);
y
xO
36x 2 " 81y 2 ! 2916
8642
#2#4#6#8
2 4 6 8#2#4#6#8
('315, 0);
y
xO
27x 2 " 9y 2 ! 81
213(0, '16);
y
xO
(y " 11)2 144
(x # 5)2 121 " ! 1
4 8 12 16 20#4#8#12
4
#4#8
#12#16#20#24#28
(5, !11 ' 123);
©Glencoe/McGraw-Hill 222 Algebra 2 Chapter 8
37. (2, 2); (2, 4), (2, 0);
39.
41. C
43. about
45. (x ! 4)2 # (y ! 1)2 " 101
47. (x ! 4)2 # (y # 1)2 " 16
" 1
y 2
1.26 & 1019#x
2
1.35&1019
x2
12#
y 2
9" 1
y
xO
213217; 38. (!1, 3); (2, 3), (!4, 3); 10; 8
40. Knowledge of the orbit ofEarth can be used inpredicting the seasons and inspace exploration. Answersshould include the following.• Knowledge of the path of
another planet would beneeded if we wanted tosend a spacecraft to thatplanet.
• 1.55 million miles
42. B
44. (x ! 3)2 # (y # 2)2 " 25
46. (x # 1)2 # y 2 " 45
48. (x ! 3)2 # 1y
xO y ! (x # 3)2 " 112
y "12
y
xO
©Glencoe/McGraw-Hill 223 Algebra 2 Chapter 8
49.
51. Sample answer: 128,600,000
53.
55. y
xO
y ! x12
y
xOy ! #2x
Peop
le (m
illio
ns)
00 2 4 6 8 10 12 14 16 18 20
120118116114112110108106104
0000
Married Americans 50. Sample answer using (0, 104.6) and (10, 112.6):y " 0.8x # 104.6
52.
54.
56. y
xO
y " 2 ! 2(x # 1)
y
xO
y ! # x12
y
xOy ! 2x
©Glencoe/McGraw-Hill 224 Algebra 2 Chapter 8
1. sometimes
3. Sample answer:
5. x 2
1!
y 2
15" 1
x 2
4!
y 2
9" 1
2. As k increases, the branchesof the hyperbola becomewider.
4.
6.
y
x
#4#2
#6#8
2 6 84
42
68
#8#6#4#2 O
! 1 y 2
18 #x 2
20
x311010
y " '
(0, '238);(0, '322);
y 2
4!
x 2
21" 1
©Glencoe/McGraw-Hill 225 Algebra 2 Chapter 8
Lesson 8-5 HyperbolasPages 445–448
57. y
xOy " 2 ! #2(x # 1)
7.
9.
11.
13.
15.
17.
19. x 2
16!
y 2
9" 1
(x ! 2)2
49!
(y # 3)2
4" 1
x 2
25!
y 2
36" 1
ay # b 2
112
254
!x 2
6" 1
x 2
4!
y 2
12" 1
x
y
#8#4
#12#16
4 12 16 208
84
1216
#12#8#4 O
y # 2 " '252
(x ! 4)
(4 ' 325, !2);(4 ' 225, !2);
yxO
! 1 (y " 6)2
20 #(x # 1)2
25
y # 6 " '225
5 (x ! 1)
(1, !6 ' 325);(1, !6 ' 225); 8. ('6, 0);
10. (0, '15); (0, '25);
12.
14.
16.
18.
20. y 2
36!
x 2
4" 1
(y ! 5)2
16!
(x # 4)2
81" 1
y 2
16!
x 2
49" 1
(x ! 3)2
4!
(y # 5)2
9" 1
(y ! 3)2
1!
(x # 2)2
4" 1
x
y
5 15 2010
105
1520
#10#15#20 O
#10#15#20
#5#5
! 1 y 2
225x 2
400#
y " '34x
x
y
#8
#16
168
8
16
#16 O#8
x 2 # 36y 2 ! 36
y " '16x
('237, 0);
©Glencoe/McGraw-Hill 226 Algebra 2 Chapter 8
21. ('9, 0);
23. (0, '4);
25.
x
y
O
x 2 # 2y 2 ! 2
y " '222
x
('23, 0);(' 22, 0);
x
y
#4#2
#6#8
2 6 84
42
68
#2#4#6#8 O
! 1 y 2
16 #x 2
25
y " '45x(0, '241);
x
y
#8#4
#12#16
4 12 168
84
1216
#4#8#12#16 O
! 1 x 2
81 #y 2
49
y " '79x
(' 2130, 0); 22. (0, '6); y "'3x
24. ('3, 0);
26. ('2, 0);
x
y
O
x 2 # y 2 ! 4
('222, 0); y " 'x
x
y
O
! 1 x 2
9 #y 2
25
( '234, 0); y " '53x
x
y
#4#2
#6#8
1 3 42
42
68
#1#2#3#4 O
! 1 y 2
36 #x 2
4
(0, '2210);
©Glencoe/McGraw-Hill 227 Algebra 2 Chapter 8
27. (0, '6); y " '2x
29. (!2, 0), (!2, 8); (!2, !1),
(!2, 9);
31. (!3, !3), (1, !3);
x
y
O
! 1 (x " 1)2
4 #(y
" 3)2
9 ! 1 #
y # 3 " '32
(x # 1)
(!1 ' 213, !3);
4
8
4
12
#4
#4
#8O
x
y
! 1 (y # 4)2
16(x
" 2)2
9 ! 1 #
y ! 4 " '43
(x # 2)
#4
#12
168
8
16
#8#16 O x
y
y 2 ! 36 " 4x 2
(0, '325); 28.
30. (2, !2), (2, 8);
32. (!12, !3), (0, !3);
x
y
#4#2
#6#8
#10
2
642
#2#4#6#8#10#12#14O
! 1 #(y
" 3)2
9 ! 1 #(x " 6)2
36
y # 3 " '12
(x # 6)
(!6 ' 325, !3);
x
y
O
#4#2
#6
2 6 8 104
108642
#2#4#6
! 1 #(x
# 2)2
16 ! 1 #(y # 3)2
25
y ! 3 " '54
(x ! 2)
(2, 3 ' 241);
x
y
O
6y 2 ! 2x 2 " 12
y " '233
x
(0, '222);(0, '22);
©Glencoe/McGraw-Hill 228 Algebra 2 Chapter 8
33.
35.
37. 120 cm, 100 cm
x2
1.1025!
y2
7.8975" 1
x
y
#4#2
#6#8
#10
2 4 6 8
642
#2#4#6#8 O
y2 # 3x2 " 6y " 6x # 18 ! 0
y # 3 " '23(x ! 1)
(1, !3 ' 422);
(1, !3 ' 226); 34. (!4, 0), (6, 0);
36.
38. (x ! 2)2
4!
(y ! 3)2
4" 1
O x
y
Station Station
x
y
#4#2
#6#8
2 4 6 8
8642
#2#4#6#8 O
4x 2 # 25y 2 # 8x # 96! 0
y " '25
(x ! 1)
(1 ' 229, 0);
©Glencoe/McGraw-Hill 229 Algebra 2 Chapter 8
39. about 47.32 ft
41. C
43.
45.
47. (x ! 5)2
16#
(y ! 2)2
1" 1
y
xO
xy ! #2
x
y
O
xy ! 2
40. Hyperbolas and parabolashave different graphs anddifferent reflective properties.Answers should include thefollowing.• Hyperbolas have two
branches, two foci, andtwo vertices. Parabolashave only one branch, onefocus, and one vertex.Hyperbolas haveasymptotes, but parabolasdo not.
• Hyperbolas reflect raysdirected at one focustoward the other focus.Parabolas reflect parallelincoming rays toward theonly focus.
42. B
44.
46. The graph of xy " !2 canbe obtained by reflecting thegraph of xy " 2 over the x-axis or over the y-axis. Thegraph of xy " !2 can alsobe obtained by rotating thegraph of xy " 2 by 90(.
48. (y ! 1)2
16#
(x # 3)2
9" 1
(22, 22), (!22, !22)
©Glencoe/McGraw-Hill 230 Algebra 2 Chapter 8
49.
51. !4, !2
53.
55. about 5,330,000 subscribersper year
57. 2x # 17y
59. 1, !2, 9
61. 5, 0, !2
63. 0, 1, 0
C!7 05 20
S
(x ! 1)2
25#
(y ! 4)2
9" 1 50. (5, !1), 2 units
52. !7,
54. [13 !8 1]
56. !5, 4
58. 2, 3, !5
60. !3, 1, 2
62. 1, 0, 0
32
x
y
O
©Glencoe/McGraw-Hill 231 Algebra 2 Chapter 8
Chapter 8Practice Quiz 2
Page 448
1. (y ! 1)2
81#
(x ! 3)2
32" 1 2. (4, !2); 6; 2
x
y
O
! 1 (x # 4)2
9 "(y
" 2)2
1
(4 ' 222, !2);
3. (!1, 1); 8;
5. (x ! 2)2
16!
(y ! 2)2
5" 1
x
y
O
225
(!1, 1 ' 211); 4. x2
9!
y 2
16" 1
©Glencoe/McGraw-Hill 232 Algebra 2 Chapter 8
Lesson 8-6 Conic SectionsPages 450–452
1. Sample answer:
3. The standard form of theequation is
This is anequation of a circle centeredat (2, !1) with radius 0. Inother words, (2, !1) is theonly point that satisfies theequation.
5. hyperbola
x
y
O2 4 6 8#8#6#4#2
8642
#2#4#6#8
y 2
16!
x 2
8" 1,
(y # 1)2 " 0.(x ! 2)2 #
2x2 # 2y2 ! 1 " 02.
4. parabola
6. circle
y
xO
ax !12b2 # y 2 "
94,
x
y
O
y " ax #32b2 !
54,
2x2 ! 4x # 7y # 1 " 0
7.
ellipse
9. ellipse
11.
13. ellipse
y
xO
y2
4#
x2
2" 1,
y
xO2 4 6 8 10 12 14
108642
#2#4#6
y
xO
(x # 1)2
4#
(y ! 3)2
1" 1, 8. parabola
10. hyperbola
12. circle
14. parabola
y
xO
y "18x2,
4 8#8 #4
8
4
#4
#8
4 8
y
xO
x2 # y2 " 27,
©Glencoe/McGraw-Hill 233 Algebra 2 Chapter 8
15. hyperbola
17. parabola
19.circle
y
xO
(x # 2)2 # (y ! 3)2 " 9,
y
xO
y " (x ! 2)2 ! 4,
y
xO
x2
4!
y 2
1" 1, 16.
hyperbola
18. , parabola
20. circle
4 8#8 #4
#4
#8
4
y
xO
x2 # (y # 3)2 " 36,
y
xO
x "19(y ! 4)2 # 4
4 8 12#8 #4
12
8
4
#4
y
xO
(x ! 1)2
36!
(y ! 4)2
4" 1,
©Glencoe/McGraw-Hill 234 Algebra 2 Chapter 8
21. hyperbola
23. circle
25. ellipse
y
xO
x2
4#
(y # 1)2
3" 1,
y
xO
x2 # (y ! 4)2 " 5,
y
xO 2 4#8#6#12#10 #4#2
8642
#2#4#6#8
(x # 4)2
32!
y 2
32" 1, 22. ellipse
24. hyperbola
26. ellipse
y
xO
(x # 1)2
16#
(y ! 1)2
4" 1,
2 4 6 8#8#6#4#2
8642
#2#4#6#8
y
xO
(y # 1)2
25!
x 2
9" 1,
y
xO
(x ! 1)2
9#
y 2
92
" 1,
©Glencoe/McGraw-Hill 235 Algebra 2 Chapter 8
27. parabola
29. ellipse
31. hyperbola
33. circle
35. parabola
37. ellipse
39. parabola
41. b
43. c
y
xO
(x ! 3)2
25#
(y ! 1)2
9" 1,
4#8#12 #4
#4
#8
#12
#16
yxO
y " !(x # 4)2 ! 7, 28.
hyperbola
30. parabolas and hyperbolas
32.
34. hyperbola
36. ellipse
38. circle
40. hyperbola
42. a
44. 2 intersecting lines
y
xO
y
xO
(x ! 2)2
5!
(y # 1)2
6" 1,
©Glencoe/McGraw-Hill 236 Algebra 2 Chapter 8
45. The plane should be verticaland contain the axis of thedouble cone.
47. D
49.
51.
53.
55.
57. (2, 6)
59. (0, 2)
x7
y 4
x12
(x ! 3)2
9!
(y # 6)2
4" 1
0 ) e ) 1, e * 1
46. If you point a flashlight at aflat surface, you can makedifferent conic sections byvarying the angle at whichyou point the flashlight.Answers should include thefollowing.• Point the flashlight directly
at a ceiling or wall. Thelight from the flashlight isin the shape of a coneand the ceiling or wall actsas a plane perpendicularto the axis of the cone.
• Hold the flashlight close toa wall and point it directlyvertically toward the ceiling.A branch of a hyperbola willappear on the wall. In thiscase, the wall acts as aplane parallel to the axis ofthe cone.
48. C
50.
52.
54.
56. 196 beats per min
58. (3, 2)
m12n
yxO
(3, !4); (3 ' 25, !4); 6; 4
(y ! 4)2
36!
(x ! 5)2
16" 1
©Glencoe/McGraw-Hill 237 Algebra 2 Chapter 8
1a. (!3, !4), (3, 4)
1b. ('1, 4)
3. Sample answer:
5. (!4, !3), (3, 4)
7. (1, '5), (!1, '5)
x2 # y2 " 40, y " x2 # x
y
O x
y ! 5 # x 2
y ! 2x 2 " 2
y
Ox
4x # 3y ! 0
x 2 " y 2 ! 25
2. The vertex of the parabola ison the ellipse. The parabolaopens toward the interior ofthe ellipse and is narrowenough to intersect theellipse in two other points.Thus, there are exactly threepoints of intersection.
4. ('4, 5)
6. no solution
8.
x
y
O#6#8#10 8 1042 6
42
68
101214
#2#4#2
#4#6
©Glencoe/McGraw-Hill 238 Algebra 2 Chapter 8
Lesson 8-7 Solving Quadratic SystemsPages 458–460
9.
11. (2, 4), (!1, 1)
13.
15.
17. (5, 0), (!4, '6)
19. ('8, 0)
21. no solution
23. (!5, 5), (!5, 1), (3, 3)
25. (1, 3)
27. 0.5 s
29. a40 ! 24255
, 45 ! 12255
b
a!53, !
73b,
(25, 25), (!25, !25)
(!1 ! 217, 1 ! 217)
(!1 # 217, 1 # 217),
y
O x
10. (40, 30)
12. (!1, 2)
14. no solution
16. no solution
18. (0, 3),
20. (0, '5)
22. (4, '3), (!4, '3)
24. (6, 3), (6, 1), (!4, 4), (!4, 0)
26. (3, '4), (!3, '4)
28. Sample answer:
30. (39.2, '4.4)
x 2
2#
y 2
16" 1
x2
16#
(y ! 2)2
4" 1,
x 2
36#
y 2
16" 1,
b1232
, !114
a'
a32, 9
2b,
©Glencoe/McGraw-Hill 239 Algebra 2 Chapter 8
31. No; the comet and Pluto maynot be at either point ofintersection at the sametime.
33.
35.
37.
39. none
41. none
y
xO
y
O x
y
O x
32.
34.
36.
38.
40.
42. !3 ) k ) !2 or 2 ) k ) 3
k " '2 or k " '3
k ) !3, !2 ) k ) 2, or k * 3
#6#8 86
42
68
#4#2
#4
#8
#2
#6
42
y
xO
y
O x
y
O x
©Glencoe/McGraw-Hill 240 Algebra 2 Chapter 8
43. Systems of equations can beused to represent thelocations and/or paths ofobjects on the screen.Answers should include thefollowing.•• The y-intercept of the
graph of the equation is 0, so the path of
the spaceship contains theorigin.
• orabout (!15.81, !47.43)
45. B
47. Sample answer:
49. Sample answer:
51. impossible
x2
4#
y 2
100" 1
x2 # y2 " 81,
(x # 2)2
16!
y 2
4" 1
x2 # y 2 " 36,
(!5210, !15210)
y " 3x
y " 3x, x2 # y2 " 2500
44. A
46. Sample answer:
48. Sample answer:
50. Sample answer: ,
52. 11, circle
y
xO
(x # 2)2 # (y # 1)2 "
" 1x2
64!
y2
16
x2
64#
y 2
16" 1
x2
16#
y2
4" 1x2 # y2 " 100,
x " (y ! 2)2y " x2,
©Glencoe/McGraw-Hill 241 Algebra 2 Chapter 8
53. ellipse
55. !7, 0
57. !7, 3
59.
61a. 4061b. two real, irrational
61c.
63. 2 # 9i
65.
67. 6
69. !51
71. y " 3x ! 2
85
!15i
'210
5
!43
y
xO
(y ! 3)2
9#
x 2
4" 1, 54. (0, '2); (0, '4);
56. 0, 3
58. 7, !5
60.
62a. !4862b. two imaginary
62c.
64. 29 ! 28i
66. about 1830 times
68. !2
70. (5, 3, 7)
72. y " !53x !
4
3
1 '2i23
3
!34, 1
2
y
xO
6y 2 # 2x 2 ! 24
y " '233
x
©Glencoe/McGraw-Hill 242 Algebra 2 Chapter 8
Glencoe/McGraw-Hill 243 Algebra 2 Chapter 9Glencoe/McGraw-Hill 243 Algebra 2 Chapter 9
1. Sample answer:
3. Never; solving the equationusing cross products leads to15 ! 10, which is never true.
5.
7.
9.
11.
13. D
15.
17.
19.
21. a " 12a " 1
12
s3
#n2
7m
cd2x
65
3c20b
1a # b
46,
4(x " 2)6(x " 2)
Chapter 9 Rational Expressions and EquationsLesson 9-1 Multiplying and Dividing Rational Expressions
Pages 476–478
2. To multiply rational numbersor rational expressions, youmultiply the numerators andmultiply the denominators. Todivide rational numbers orrational expressions, youmultiply by the reciprocal ofthe divisor. In either case, youcan reduce your answer bydividing the numerator andthe denominator of the resultsby any common factors.
4.
6.
8.
10.
12.
14.
16.
18.
20.
22. 3x2
2y
y " 23y # 1
5t " 1
#3x 4y
5c2b
2y(y # 2)3(y " 2)
p " 5p " 1
512x
3y 2
y " 4
9m4n4
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23.
25. #2p2
27.
29.
31. 1
33.
35.
37. #2p
39.
41.
43. a ! #b or b
45.
47. (2x2 " x # 15)m2
49. A rational expression can beused to express the fractionof a nut mixture that ispeanuts. Answers shouldinclude the following.• The rational expression
is in simplest form
because the numeratorand the denominator haveno common factors.
8 " x13 " x
6827 " m13,129 " a
43
2x " y2x # y
2(a " 5)(a # 2)(a " 2)
w # 3w # 4
43
b3
x2y 2
#4bc27a
24. #f
26.
28. 3
30.
32.
34.
36.
38.
40. y " 1
42. d ! #2, #1 or 2
44.
46. 2x " 1 units
48.
50. C
1a # 2
682713,129
m " nm2 " n2
#3nm
3(r " 4)r " 3
5(x # 3)2(x " 1)
23
xz8y
©Glencoe/McGraw-Hill 244 Algebra 2 Chapter 9
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• Sample answer:could be used to representthe fraction that is peanutsif x pounds of peanuts andy pounds of cashews wereadded to the originalmixture.
51. A
53.
55.
hyperbola
57. odd; 3
59. #1, 4
61. 0, 5
63. $
65. #119
4 8 12
8
4
x
AA C A BLACK
y
O
! 1 " (y " 2)2
1(x " 7)2
9"4
"8
(x # 7)2
9 (y # 2)2
1! 1;
(%217, %222)
8 " x13 " x " y
52. (#1, %4), (5, %2)
54. x (y " 3)2 " 1; parabola
56. even; 2
58. even; 0
60.
62. 4.99 & 102 s or about 8 min19 s
64.
66. #11124
32,
1916
#16,
13
x
y
O
x ! 1 3 (y # 3)2 # 1
!13
©Glencoe/McGraw-Hill 245 Algebra 2 Chapter 9
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67.
69. #1118
1415
68.
70. 16
#1116
©Glencoe/McGraw-Hill 246 Algebra 2 Chapter 9
Lesson 9-2 Adding and Subtracting Rational EpressionsPages 481–484
1. Catalina; you need acommon denominator, not acommon numerator, tosubtract two rationalexpressions.
3a. Always; since a, b, and c arefactors of abc, abc is alwaysa common denominator of
3b.Sometimes; if a, b, and chave no common factors,then abc is the LCD of
3c. Sometimes; if a and b haveno common factors and c isa factor of ab, then ab is theLCD of
3d. Sometimes; if a and c arefactors of b, then b is the LCD of
3e. Always; since
the sum is
always
5. 80ab3c
7. 2 # x3
x 2y
bc " ac " ababc
.
bcabc
"acabc
"ababc
,
1a
"1b
"1c
!
1a
"1b
"1c
.
1a
"1b
"1c
.
1a
"1b
"1c
.
1a
"1b
"1c
.
2. Sample answer: d 2 # d, d " 1
4. 12x2y2
6. x(x # 2)(x " 2)
8. 42a2 " 5b2
90ab2
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9.
11.
13. units
15. 180x2yz
17. 36p3q4
19. x2(x # y)(x " y)
21. (n # 4)(n # 3)(n " 2)
23.
25.
27.
29.
31.
33.
35.
37.
39.
41. #1
43.
45. 12 ohms
a " 7a " 2
2y 2 " y # 4(y # 1)(y # 2)
x2 # 6(x " 2)2(x " 3)
#8d " 20(d # 4)(d " 4)(d # 2)
y (y # 9)(y " 3)(y # 3)
a " 3a # 4
110w # 42390w
25b # 7a3
5a2b 2
2x " 15y3y
3112v
13x2 " 4x # 92x (x # 1)(x " 1)
3a # 10(a # 5)(a " 4)
3742m
10.
12.
14. 70s2t2
16. 420a3b3c3
18. 4(w # 3)
20. (2t " 3)(t # 1)(t " 1)
22.
24.
26.
28.
30.
32.
34.
36.
38. 0
40.
42.
44.
46. 24x
h
3x # 42x (x # 2)
2s # 12s " 1
1b " 1
#4h " 15(h # 4)(h # 5)2
7x " 382(x # 7)(x " 4)
5m # 43(m " 2)(m # 2)
13y # 8
#3
20q
9x 2 # 2y 3
12x 2y
5 " 7rr
6 " 8bab
85
5d " 16(d " 2)2
©Glencoe/McGraw-Hill 247 Algebra 2 Chapter 9
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47.
49.
51. Subtraction of rationalexpressions can be used todetermine the distancebetween the lens and thefilm if the focal length of thelens and the distancebetween the lens and theobject are known. Answersshould include the following.• To subtract rational
expressions, first find acommon denominator.Then, write each fractionas an equivalent fractionwith the commondenominator. Subtract thenumerators and place thedifference over thecommon denominator. Ifpossible, reduce theanswer.
• could be used
to determine the distancebetween the lens and thefilm if the focal length ofthe lens is 10 cm and thedistance between the lensand the object is 60 cm.
53. C
1q
!1
10#
160
2md(d 2 # L2)2
2md(d # L)2(d " L)2 or
24x # 4
h 48.
50. Sample answer:
52. B
54. 415xyz 2
1x " 1
, 1
x # 2
48(x # 2)x(x # 4)
h
©Glencoe/McGraw-Hill 248 Algebra 2 Chapter 9
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55.
57.
59.
61. y
xO
2
2468
"8
"4
4 6"2"6
10
" ! 116 25
(x # 2)2 (y " 5)2
x
y
8
2
6
"6
O
"2"8
" ! 1 x 2 y 2
2016
x
y
O
x 2 ! y # 4
(y " 3)2 ! x # 2
a(a " 2)a " 1
56.
58. 2.5 ft
60.
x
y
5"5 10"10
5
10
15
O
"5
"10
"15" ! 1 y 2
49 x 2
25
y
xO 8"8
9x 2 # y 2 ! 81
x 2 # y 2 ! 16
8
"8
©Glencoe/McGraw-Hill 249 Algebra 2 Chapter 9
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2.
4.
6. x # 1
8.
10. 14
6ax " 20by15a2b3
72
c6b 2
©Glencoe/McGraw-Hill 250 Algebra 2 Chapter 9
Chapter 9Practice Quiz 1
Page 484
1.
3.
5. (w " 4)(3w " 4)
7.
9. n # 29(n " 6)(n # 1)
4a " 1a " b
#y 2
32
t " 2t # 3
1. Sample answer:
3. x ! 2 and y ! 0 areasymptotes of the graph. They-intercept is 0.5 and there isno x-intercept because y ! 0is an asymptote.
5. asymptote: x ! #5; hole:x ! 1
f(x) !1
(x " 5)(x # 2)
2. Each of the graphs is astraight line passing through(#5, 0) and (0, 5). However,
the graph of
has a hole at (1, 6), and thegraph of doesnot have a hole.
4. asymptote: x ! 2
6.
xO
f (x)
f (x) ! xx # 1
g(x ) ! x " 5
f (x) !(x # 1)(x " 5)
x # 1
Lesson 9-3 Graphing Rational FunctionsPages 488–490
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©Glencoe/McGraw-Hill 251 Algebra 2 Chapter 9
7.
9.
11.
13.
15. y ' 0 and 0 ( C ( 1
17. asymptotes: x ! #4, x ! 2
19. asymptotes: x ! #1, hole:x ! 5
yO
10
6
2
"8"16
"4
8 16
C ! y
y # 12
C
xO
f (x)
f(x) ! x # 2x2 " x " 6
C C
xO
f (x)
f (x) ! x " 5x # 1
2
4
4 8"4
"2
"4
"8
xO
f (x)
2
4
4 8"4
"2
"4
"8
f (x) ! 6(x " 2)(x # 3)
8.
10.
12. 100 mg
14. y ! #12, C ! 1; 0; 0
16. asymptotes: x ! 2, hole:x ! 3
18. asymptotes: x ! #4, hole:x ! #3
20. hole: x ! 4
xO
f (x)
f (x) ! 4(x " 1)2
xO
10
6
2
2"4
"4
6 10
f (x) ! x2 " 25x " 5
f (x)
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21. hole: x ! 1
23.
25.
27.
29.
xO
f (x)
f (x) ! 1(x # 3)2
xO
f (x)
f (x) ! 5xx # 1
4
8
4 8"4
"4
"8
xO
f (x)
f (x) ! " 5x # 1
2
6
4 8"4
"4
"8
"8
xO
f (x)
f (x) ! 3x
22.
24.
26.
28.
30.
xO
f (x)
f (x) ! x # 4x " 1
2
6
4 8"4
"4
"8
"8
xO
f (x) f (x) ! "3(x " 2)2
C C
xO
f (x)
f (x) ! xx " 3
xO
f (x)
f (x) ! 1x # 2
xO
f (x)
f (x) ! 1x
©Glencoe/McGraw-Hill 252 Algebra 2 Chapter 9
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31.
33.
35.
37.
xO
f (x)
f (x) ! x " 1x2 " 4
xO
f (x)
f (x) ! "1(x # 2)(x " 3)
xO
f (x)
f (x) ! x2 " 1x " 1
xO
f (x)
f (x) ! x " 1x " 3
32.
34.
36.
38.
xO
f (x)
f (x) ! 6(x " 6)2
xO
f (x)
f (x) ! xx2 " 1
xO
f (x)
f (x) ! 3(x " 1)(x # 5)
xO
f (x)
f (x) ! x2 " 36x # 6
4
4"4
"4
"8
"12
"8
©Glencoe/McGraw-Hill 253 Algebra 2 Chapter 9
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Glencoe/McGraw-Hill 254 Algebra 2 Chapter 9
39.
41. The graph is bell-shapedwith a horizontal asymptoteat f(x) ! 0.
43.
45. about #0.83 m/s
47.
O
P(x)
x
4
8
4"12 "4
"4
"8
"8
P(x ) ! 6 # x10 # x
O
Vf
m1
Vf ! 5m1 " 7
m1 # 7
4
8"16
12
20
"8 "4
xO
f (x)
f (x) ! 1(x # 2)2
40.
42. Since
the graph of
would be a reflection of the
graph of over
the x-axis.
44. m1 ! #7; 7; #5
46. Sample answers:
48. the part in the first quadrant
f (x) !5(x " 2)
(x " 2)(x # 3)
f (x) !2(x " 2)
(x " 2)(x # 3),
f(x) !x " 2
(x " 2)(x # 3),
f (x) !64
x 2 " 16
f (x) !#64
x 2 " 16
#64x 2 " 16
! #a 64x 2 " 16
b,
xO
f (x)
f (x) ! 64x2 # 16
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Glencoe/McGraw-Hill 255 Algebra 2 Chapter 9
49. It represents her original free-throw percentage of 60%.
51. A rational function can beused to determine how mucheach person owes if the costof the gift is known and thenumber of people sharing thecost is s. Answers shouldinclude the following.•
• Only the portion in the firstquadrant is significant inthe real world becausethere cannot be a negativenumber of people nor anegative amount of moneyowed for the gift.
53. B
55.
57. (6, 2); 5
xO
y
(x " 6)2 # (y " 2)2! 25
3x # 16(x " 3)(x # 2)
sO
c ! 150s50
50 100
100
"50
"50
"100
"100
c ! 0
s ! 0
c
50. y ! 1; This represents 100%,which she cannot achievebecause she has alreadymissed 4 free throws.
52. A
54.
56.
58.y
xO
x 2 # y 2 # 4x ! 9
(#2, 0); 113
5(w # 2)(w " 3)2
3m " 4m " n
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Glencoe/McGraw-Hill 256 Algebra 2 Chapter 9
1a. inverse1b. direct
3. Sample answers: wages andhours worked, total cost andnumber of pounds of apples;distances traveled andamount of gas remaining inthe tank, distance of anobject and the size itappears
5. direct; #0.5
7. 24
9. #8
11. 25.8 psi
13. Depth(ft) Pressure(psi)0 01 0.432 p 0.863 1.294 1.72
2. Both are examples of directvariation. For y ! 5x, yincreases as x increases. Fory ! #5x, y decreases as xincreases.
4. inverse; 20
6. joint;
8. #45
10. P ! 0.43d
12. about 150 ft
14. direct; 1.5
12
Lesson 9-4 Direct, Joint, and Inverse VariationPages 495–498
59. $65,892
61. #12, 10
63. 4.5
65. 20
60. #4 % 2i
62.
64. 1.4
66. 12
#7 % 32132
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Glencoe/McGraw-Hill 257 Algebra 2 Chapter 9
15. joint; 5
17. direct; 3
19. direct; #7
21. inverse; 2.5
23. V ! kt
25. 118.5 km
27. 20
29. 64
31. 4
33. 9.6
35. 0.83
37.
39. 100.8 cm3
41. m ! 20sd
43. 1860 lb
45. joint
47. I !k
d 2
16
P
dO
P ! 0.43d
16. inverse; #18
18. inverse; 12
20. joint;
22.
24. directly; 2)
26. 60
28. 216
30. 25
32. 1.25
34. #12.6
36.
38. 30 mph
40. See students’ work.
42. joint
44. !15md
46. See students’ work.
48. I
dO
I ! 16d 2
/
214
V !kp
13
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49. The sound will be heard asintensely.
51. about 127,572 calls
53. no;
55. A direct variation can beused to determine the totalcost when the cost per unitis known. Answers shouldinclude the following.• Since the total cost T is
the cost per unit u timesthe number of units n or C ! un, the relationship isa direct variation. In thisequation u is the constantof variation.
• Sample answer: The schoolstore sells pencils for 20¢each. John wants to buy5 pencils. What is the totalcost of the pencils? ($1.00)
57. C
59. asymptotes: x ! #4, x ! 3
61.
63.
65. 0.4; 1.2
67.
69. A
#35; 3
m (m " 1)m " 5
xy # x
d * 0
14 50.
52. about 601 mi
54. Sample answer: If the averagestudent spends $2.50 for lunchin the school cafeteria, writean equation to represent theamount s students will spendfor lunch in d days. How muchwill 30 students spend in aweek? a ! 2.50sd ; $375
56. D
58. asymptote: x ! 1; hole x ! #1
60. hole: x ! #3
62.
64. 9.3 & 107
66. 3; 7
68. C
70. S
t 2 # 2t # 2(t " 2)(t # 2)
0.02; C !0.02P
1P
2
d 2
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©Glencoe/McGraw-Hill 259 Algebra 2 Chapter 9
1.
3. 49
5. 112
xO
f (x )
f (x) ! x " 1x " 4
2.
4. 4.4
xO
f (x )
f (x) ! "2x2
" 6x # 9
Chapter 9Practice Quiz 2
Page 498
Lesson 9-5 Classes of FunctionsPages 501–504
1. Sample answer:
This graph is a rationalfunction. It has an asymptoteat x ! #1.
3. The equation is a greatestinteger function. The graphlooks like a series of steps.
5. inverse variation or rational
P
dO
2. constant (y ! 1),direct variation (y ! 2x),identity (y ! x)
4. greatest integer
6. constant
71. P
73. C
72. A
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7. c
9. identity or direct variation
11. absolute value
13. absolute value
15. rational
17. quadratic
19. b
21. g
23. constanty
xO
y ! "1.5
y
xO
y ! x # 2
y
xO
y ! x
8. b
10. quadratic
12. A ! )r2; quadratic; the graph is a parabola
14. square root
16. direct variation
18. constant
20. e
22. a
24. direct variationy
xO
y ! 2.5x
y
xO
y ! "x 2 # 2
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25. square root
27. rational
29. absolute value
31. C ! 4.5 m
33. a line slanting to the right andpassing through the origin
y
xO
y ! 2x
y
xO
y ! x2 " 1
x " 1
y
xOAA C A BLACK
y ! !9x
26. inverse variation or rational
28. greatest integer
30. quadratic
32. direct variation
34. similar to a parabola
y
xO
y ! 2x 2
y
xO
y ! 3[x ]
y
xO
y ! 4x
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36. The graph is similar to thegraph of the greatest integerfunction because both graphslook like a series of steps. Inthe graph of the postagerates, the solid dots are onthe right and the circles areon the left. However, in thegreatest integer function, thecircles are on the right andthe solid dots are on the left.
38. A graph of the function thatrelates a person’s weight onEarth with his or her weighton a different planet can beused to determine a person’sweight on the other planet byfinding the point on thegraph that corresponds withthe weight on Earth anddetermining the value on theother planet’s axis. Answersshould include the following.• The graph comparing
weight on Earth and Marsrepresents a direct variationfunction because it is astraight line passing throughthe origin and is neitherhorizontal nor vertical.
• The equation V ! 0.9Ecompares a person’sweight on Earth with his orher weight on Venus.
V
E
Ven
us
20
0
40
60
80
Earth20 40 60 80
35.
37a. absolute value37b. quadratic37c. greatest integer37d. square root
y
x
Cost
(cen
ts)
40
0
80
120
160
Ounces2 4 6 8 10
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39. C
41. 22
43.
45.
up; unit
y
xO
2 4 6 10 12
8101214
642
"2"2
(y # 1) ! (x " 8)212
12
y ! #118;
x ! 8; (8,#1); a8,#78b;
f (x )
xO
f (x) ! 8(x " 1)(x # 3)
40. D
42.
44.
46.
right; 4 units
y
xO
x ! y 2 " y " 314
12
y ! 1; x ! #4
14;
a#3
14, 1b ; a#2
14, 1b;
f (x )
xO
f (x) ! x2 " 5x # 4
x " 4
f (x )
xO
f (x) ! 3x # 2
©Glencoe/McGraw-Hill 263 Algebra 2 Chapter 9
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1. Sample answer:
3. Jeff; when Dustin multipliedby 3a, he forgot to multiplythe 2 by 3a.
5. 2, 6
7. #6, #2
15
"2
a " 2! 1
2. 2(x " 4); #4
4. 3
6.
8. #2 ( c ( 2
23
©Glencoe/McGraw-Hill 264 Algebra 2 Chapter 9
Lesson 9-6 Solving Rational Equations and InequalitiesPages 509–511
47.
right; 3 units
49. impossible
51.
53. 1
55.
57. 45x3y3
59. 3(x # y)(x " y)
61. (t # 5)(t " 6)(2t " 1)
#176
a13, 2b
y
xO
3x " y 2 ! 8y # 31
x ! 4
14;
y ! #4;(5, #4); a5
34, #4b; 48.
50. (7, #5)
52. (2, #2)
54. 12
56. 60a3b2c2
58. 15(d # 2)
60. (a # 3)(a " 1)(a " 2)
c#25 23 #5466 #26 57
d
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9.
11. 2
13. #6, 1
15. #1 ( a ( 0
17. 11
19. t ( 0 or t ' 3
21. 0 ( y ( 2
23. 14
25. $
27. 7
29.
31. 32
33. band, 80 members; chorale,50 members
35. 24 cm
37. 5 mL
39. 6.15
41. If something has a generalfee and cost per unit, rationalequations can be used todetermine how many units aperson must buy in order forthe actual unit price to be agiven number. Answersshould include the following.
• To solve ! 6,
multiply each side of theequation by x to eliminatethe rational expression.
500 " 5xx
#3 % 3222
v ( 0 or v ' 116
10.
12.
14. #3, 2
16. #1 ( m ( 1
18. 3
20. 0 ( b ( 1
22. p ( 0 or p ' 2
24.
26. $
28.
30.
32. 2 or 4
34. 4.8 cm/g
36. 15 km/h
38. 5
40.
42. B
bbc " 1
1 % 21454
73
32
12
#43
2
29 h
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Then subtract 5x from eachside. Therefore, 500 ! x.A person would need tomake 500 minutes of longdistance calls to make the actual unit price 6¢.
• Since the cost is 5¢ perminute plus $5.00 permonth, the actual cost perminute could never be 5¢or less.
43. C
45. square root
47. 36
49.
51.
53. 5x 0 0 + x + 462137
22130
y
O x
y ! 2 x!
44. quadratic
46. direct variation
48. 33.75
50.
52.
54. eb `#112
( b ( 2f5x 0 x ( #11 or x ' 36225
y
O xy ! 0.8x
y
O x
y ! 2x 2 # 1
©Glencoe/McGraw-Hill 266 Algebra 2 Chapter 9
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©Glencoe/McGraw-Hill 267 Algebra 2 Chapter 10
1. Sample answer: 0.8
3. c
5. b
7. D ! {x 0x is all real numbers.},R ! {y 0y " 0}
9. decay
11.
13. or
15. or
17. x # 0
19. y ! 65,000(6.20)x
27223322
4272227
y ! 3a12bx
y
xO
y ! 2( )x13
Chapter 10 Esponential and Logarithmic RelationsLesson 10-1 Exponential Functions
Pages 527–530
2a. quadratic2b. exponential2c. linear2d. exponential
4. a
6. D ! {x 0x is all real numbers.},R ! {y 0y " 0}
8. growth
10. growth
12.
14. a4$
16. %9
18. 2
20. 22,890,495,000
y ! %18132x
y
xO
y ! 3(4)x
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21. D ! {x 0x is all real numbers.},R ! {y 0y " 0}
23. D ! {x 0x is all real numbers.},R ! {y 0y " 0}
25. D ! {x 0x is all real numbers.},R ! {y 0y & 0}
27. growth
29. decay
31. decay
33.
35. y ! 7(3)x
37. y ! 0.2(4)x
y ! %2a14bx
yx
O
y ! "( )x15
y
xO
y ! 0.5(4)x
y
xO
y ! 2(3)x
22. D ! {x 0x is all real numbers.},R ! {y 0y " 0}
24. D ! {x 0x is all real numbers.},R ! {y 0y " 0}
26. D ! {x 0x is all real numbers.},R ! {y 0y & 0}
28. growth
30. growth
32. decay
34. y ! 3(5)x
36.
38. y ! %0.3(2)x
y ! %5a13bx
y
xO
y ! "2.5(5)x
y
xO
y ! 4( )x13
y
xO
y ! 5(2)x
©Glencoe/McGraw-Hill 268 Algebra 2 Chapter 10
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39. 54 or 625
41.
43. n2'$
45. n ! 5
47. 1
49.
51. n & 3
53. %3
55. 10
57. y ! 100(6.32)x
59. y ! 3.93(1.35)x
61. 2144.97 million; 281.42million; No, the growth ratehas slowed considerably.The population in 2000 wasmuch smaller than theequation predicts it would be.
63. A(t ) ! 1000(1.01)4t
65. s . 4x
67. Sometimes; true when b " 1,but false when b & 1.
%83
7412
40.
42.
44. 25$
46.
48. n # %2
50. 0
52.
54. p ( %2
56. %3, 5
58. about 1,008,290
60. 9.67 million; 17.62 million;32.12 million; These answersare in close agreement withthe actual populations inthose years.
62. Exponential; the base, 1 ' ,is fixed, but the exponent, nt,is variable since the time tcan vary.
64. $2216.72
66. 1.5 three-year periods or 4.5 yr
68. The number of teams y thatcould compete in atournament with x roundscan be expressed as y ! 2x.The 2 teams that make it tothe final round got there as aresult of winning gamesplayed with 2 other teams,for a total of 2 ) 2 ! 22 or 4games played in the previousrounds. Answers shouldinclude the following.
rn
53
23
y 213
x115
©Glencoe/McGraw-Hill 269 Algebra 2 Chapter 10
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69. A
71.
The graphs have the sameshape. The graph of y ! 2x ' 3 is the graph of y ! 2x translated three unitsup. The asymptote for thegraph of y ! 2x is the line y ! 0 and for y ! 2x ' 3 isthe line y ! 3. The graphshave the same domain, allreal numbers, but the rangeof y ! 2x is y " 0 and therange of y ! 2x ' 3 is y " 3.The y-intercept of the graphof y ! 2x is 1 and for thegraph of y ! 2x ' 3 is 4.
["5, 5] scl: 1 by ["1, 9] scl: 1
• Rewrite 128 as a power of2, 27. Substitute 27 for y inthe equation y ! 2x. Then,using the Property ofEquality for Exponents, xmust be 7. Therefore, 128teams would need to play7 rounds of tournamentplay.
• Sample answer: 52 wouldbe an inappropriatenumber of teams to play inthis type of tournamentbecause 52 is not a powerof 2.
70. 780.25
72.
The graphs have the sameshape. The graph of y ! 3x'1 is the graph of y ! 3x translated one unit tothe left. The asymptote forthe graph of y ! 3x and for y ! 3x'1 is the line y ! 0.The graphs have the samedomain, all real numbers,and range, y " 0. The y-intercept of the graph of y ! 3x is 1 and for the graphof y ! 3x'1 is 3.
["5, 5] scl: 1 by ["1, 9] scl: 1
©Glencoe/McGraw-Hill 270 Algebra 2 Chapter 10
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73.
The graphs have the sameshape. The graph of
y ! is the graph of
y ! translated two units
to the right. The asymptote
for the graph of y ! and
for y ! is the line
y ! 0. The graphs have thesame domain, all realnumbers, and range, y " 0.The y-intercept of the graph
of y ! is 1 and for the
graph of y ! is 25.
75. For h " 0, the graph of y ! 2x is translated 0h 0 unitsto the right. For h & 0, thegraph of y ! 2x is translated|h| units to the left. For k "0, the graph of y ! 2x istranslated 0k 0 units up. For k & 0, the graph of y ! 2x istranslated 0k 0 units down.
a15bx%2
a15bx
a15bx%2
a15bx
a15bx
a15bx%2
["5, 5] scl: 1 by ["1, 9] scl: 1
74.
The graphs have the sameshape. The graph of
y ! %1 is the graph of
y ! translated one
unit down. The asymptote
for the graph of y ! is
the line y ! 0 and
for the graph of y ! % 1
is the line y ! %1. Thegraphs have the samedomain, all real numbers,
but the range of y !
is y " 0 and of y ! % 1
is y " %1. The y-intercept
of the graph of y ! is 1
and for the graph of
y ! % 1 is 0.
76. 1, 15
a14bx
a14bx
a14bxa1
4bx
a14bx
a14bx
a14bx
a14bx
["5, 5] scl: 1 by ["3, 7] scl: 1
©Glencoe/McGraw-Hill 271 Algebra 2 Chapter 10
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77. 1, 6
79. 0 & x & 3 or x " 6
81. greatest integer
83.
85.
87. g [h(x)] ! 2x % 6;h [g(x)] ! 2x % 11
89. g [h(x)] ! %2x % 2;h [g(x)] ! %2x ' 11
151
B 311
%6%5R
B10
01R
78. 3
80. square root
82. constant
84. does not exist
86. about 23.94 cm
88. g [h(x)] ! x 2 ' 6x ' 9;h [g(x)] ! x2 ' 3
y
xO
y ! 8
%133
,
©Glencoe/McGraw-Hill 272 Algebra 2 Chapter 10
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1. Sample answer: x ! 5y andy ! log5 x
3. Scott; the value of alogarithmic equation, 9, isthe exponent of theequivalent exponentialequation, and the base ofthe logarithmic expression, 3,is the base of the exponentialequation. Thus x ! 39 or19,683.
5. log7
7.
9. %3
11. %1
13. 1000
15. 1
17. 3
19. 107.5
21. log8 512 ! 3
23.
25. log100
27. 53 ! 125
29.
31.
33. 4
823 ! 4
4%1 !14
10 !12
log5 1
125! %3
12,
3612 ! 6
149
! %2
2. They are inverses.
4. log5 625 ! 4
6. 34 ! 81
8. 4
10. 21
12. 27
14.
16. x " 6
18. 1013
20. 105.5 or about 316,228 times
22. log3 27 ! 3
24. 9 ! %2
26. log2401
28. 132 ! 169
30.
32.
34. 2
a15b%2
! 25
100%12 !
110
7 !14
log13
12
# x # 5
©Glencoe/McGraw-Hill 273 Algebra 2 Chapter 10
Lesson 10-2 Logarithms and Logarithmic FunctionsPages 535–538
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35.
37. %5
39. 7
41. n % 5
43. %3
45. 1018.8
47. 81
49. 0 & y # 8
51. 7
53. x ( 24
55. 4
57. 2
59. 5
61. a " 3
63.
InverseProp. ofExp. and Logarithms
2 ! 2! Simplify.
2 !? 2(1)
log5 52 !? 2 log5 51
log5 25 !? 2 log5 5
12
36.
38. %4
40. 45
42. 3x ' 2
44. 2x
46. 1010.67
48. c " 256
50. 125
52. 0 & p & 1
54. *3
56. 11
58.
60. y ( 3
62. *8
64.
Inverse Prop.of Exp. andLogarithms
1 ! 1!
14
(4) !? 1
log16 1614 ! log2 24 !
? 1
log16 2 ! log2 16 !? 1
25
52
©Glencoe/McGraw-Hill 274 Algebra 2 Chapter 10
Originalequation
25 ! 52
and 5 ! 51
Originalequation
and 16 ! 242 ! 16
14
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65.
Originalequation
8 ! 23
InverseProp. ofExp. andLogarithms
3 ! 31
InverseProp. ofExp. andLogarithms
0 ! 0! InverseProp. ofExp. andLogarithms
67a.
67b. The graph of y ! log2 x ' 3is the graph of y ! log2 xtranslated 3 units up. The graphof y ! log2 x % 4 is the graphof y ! log2 x translated 4 unitsdown. The graph of log2 (x % 1)is the graph of y ! log2 xtranslated 1 unit to the right.The graph of log2 (x ' 2) isthe graph of y ! log2 xtranslated 2 units to the left.
69. 101.4 or about 25 times as great
y
xO
y ! log2x # 3
y ! log2(x # 2)
y ! log2(x " 1)
y ! log2x " 4
1 ! 70log7 70 !? 0
log7 1 !? 0
log7 (log3 31) !? 0
log7 (log3 3) !? 0
log7 [log3 (log2 23)] !? 0
log7 [log3 (log2 8)] !? 0
66a.
66b. The graphs are reflections ofeach other over the line y ! x.
68. 103 or 1000 times as great
70. 101.7 or about 50 times
y
xO
y ! ( )x12
y ! log x 12
©Glencoe/McGraw-Hill 275 Algebra 2 Chapter 10
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72. All powers of 1 are 1, so theinverse of y ! 1x is not afunction.
74. B
76.
78. +
80.
82.
84. $2400, CD; $1600, savings
86. y 24
88. an6
90. 1
4330y
*73
x 216
71. 2 and 3; Sample answer: 5 is between 22 and 23.
73. A logarithmic scale illustratesthat values next to each othervary by a factor of 10. Answersshould include the following.• Pin drop: 1 , 100;
Whisper: 1 , 102; Normalconversation: 1 , 106;Kitchen noise: 1 , 1010;Jet engine: 1 , 1012
•
• On the scale shown above,the sound of a pin drop andthe sound of normalconversation appear not todiffer by much at all, when infact they do differ in terms ofthe loudness we perceive.The first scale shows thisdifference more clearly.
75. D
77. b12
79. %3,
81.
83.
85. x10
87. 8a6b3
89. x 3
y 2z 3
6x % 58(x % 3)(x ' 3)(x ' 7)
5 * 2734
145
2 $ 1011 4 $ 1011 6 $ 1011 8 $ 1011 1 $ 10120
Pindrop
Whisper(4 feet)
Normalconversation
Jetengine
Kitchennoise
©Glencoe/McGraw-Hill 276 Algebra 2 Chapter 10
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2. Sample answer:2log3 x ' log3 5; log3 5x 2
4. 1.1402
6. %0.3690
8. 2
10. 4
12. 20:1
14. 1.2921
16. 0.2519
18. 2.1133
20. 0.0655
©Glencoe/McGraw-Hill 277 Algebra 2 Chapter 10
1. properties of exponents
3. Umeko; Clemente incorrectlyapplied the product andquotient properties oflogarithms.log7 6 ' log7 3 ! log7 (6 ) 3)or log7 18 Product Property of
Logarithms
log7 18 % log7 2 !log7 (18 - 2) or log7 9
Quotient Prop. of Logarithms
5. 2.6310
7. 6
9. 3
11. pH ! 6.1 ' log10
13. 1.3652
15. %0.2519
17. 2.4307
19. %0.4307
BC
Lesson 10-3 Properties of LogarithmsPages 544–546
Chapter 10Practice Quiz 1
Page–538
1. growth
3. log4 4096 ! 6
5.
7.
9. x " 26
35
43
2. y ! 2(4)x
4.
6. 15
8. n # %1
10. 3
932 ! 27
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21. 2
23. 4
25. 14
27. 2
29. +
31. 10
33.
35. False; log2 (22 ' 23) ! log2 12,log2 22 ' log2 23 ! 2 ' 3 or 5,and
37. 2
39. about 0.4214 kilocalories pergram
41. 3
43. About 95 decibels;L ! 10 log10 R, where L is theloudness of the sound indecibels and R is the relativeintensity of the sound. Sincethe crowd increased by afactor of 3, we assume thatthe intensity also increases bya factor of 3. Thus, we needto find the loudness of 3R.
log2 12 Z 5, since 25 Z 12.
x 3
4
22. 3
24. %2
26. 12
28. *4
30. 6
32. 12
34.
36.
Power Prop. of Logarithms
Product Prop. of Logarithms
Product of Powers Prop.!
Power Prop. of Logarithms
38. E ! 1.4 log
40. about 0.8429 kilocalories pergram
42. 3
44. 5
C2
C1
(n ' m)logb x ! (n ' m)logb
logb (xn'm) !? (n ' m)logb x
logb (xn ! xm) !? (n ' m)logb x
logb x n ' logb
x m !? (n ' m)logb x
n logb x ' m logb x !? (n ' m)logb x
12
1x % 12
©Glencoe/McGraw-Hill 278 Algebra 2 Chapter 10
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©Glencoe/McGraw-Hill 279 Algebra 2 Chapter 10
L ! 10 log10 3RL ! 10 (log10 3 ' log10 R )L ! 10 log103 ' 10 log10 RL ! 10(0.4771) ' 90L ! 4.771 ' 90 or about 95
45. 7.5
47. Let bx ! m and by ! n. Thenlogb m ! x and logb n ! y.
!
bx%y ! Quotient Prop.
logb bx %y ! logb
mn
logb m % logb n ! logb
mn
x % y ! logb
mn
mn
mn
bx
b y
46. about 22
48. Since logarithms areexponents, the properties oflogarithms are similar to theproperties of exponents. TheProduct Property states thatto multiply two powers thathave the same base, add theexponents. Similarly, thelogarithm of a product is thesum of the logarithms of itsfactors. The Quotient Propertystates that to divide twopowers that have the samebase, subtract theirexponents. Similarly, thelogarithm of a quotient is thedifference of the logarithms ofthe numerator and thedenominator. The PowerProperty states that to findthe power of a power, multiplythe exponents. Similarly, thelogarithm of a power is theproduct of the logarithm andthe exponent. Answers shouldinclude the following.• Quotient Property:
log2 ! log2
! 5 % 3 or 2
! log2 2(5%3)
a25
23b a328b
Prop. of Equality forLogarithmicEquations
InverseProp. ofExp. andLogarithms
Replace xwith logbmand y withlogbn.
Replace 32with 25 and 8with 23.Quotient ofPowersInverse Prop.of Exp. andLogarithms
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49. A
51. 4
53. 2x
55. %8
57. odd; 3
log2 32 % log2 8 ! log2 25 % log2 23
Replace 32 with25 and 8 with 23.
! 5 % 3 or 2Inverse Prop. ofExp. and Logarithms
So, log2 ! log2 32 % log2 8
Power Property:log3 94 ! log3 (32)4 Replace 9 with 32.
Power of a Power
! or 8
4 log3 9 !
!
! or 8
So, log3 94 ! 4 log3 9.• The Product of Powers
Property and ProductProperty of Logarithmsboth involve the addition ofexponents, sincelogarithms are exponents.
50. Let bx ! m, then logb m ! x.(bx)p ! mp
bxp ! mp Product of Powerslogb b xp ! logb mp
xp ! logb mp
plogb m ! logb mp
52. %3
54. 6
56. d & 4
58. even; 4
2 ! 4
(log3 32) ! 4
(log3 9) ! 4
2 ! 4! log3 3(2!4)
a328b
©Glencoe/McGraw-Hill 280 Algebra 2 Chapter 10
Inverse Prop.of Exp. and Logarithms
Comm (,)
Replace 9 with 32.Inverse Prop.of Exp. andLogarithms
Prop. of Equalityfor LogarithmicEquations
Inverse Prop.of Exp. andLogarithms
Replace xwith logb m.
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2. Sample answer:5x ! 2; x ! 0.4307
4. 0.6021
6. %0.3010
8. {n 0n " 0.4907}
10. *1.1615
12. {p 0 p # 4.8188}
14. 3.4022
16. at most 0.00003 mole perliter
18. 1.0792
20. 0.3617
22. %1.5229
24. 2.2
26. 3.5
28. 2.4550
30. 0.5537
32. 4.8362
34. 8.0086
log 42log 3
;
1. 10; common logarithms
3. A calculator is notprogrammed to find base 2 logarithms.
5. 1.3617
7. 1.7325
9. 4.9824
11. 11.5665
13. 0.8271
15. 3.1699
17. 0.6990
19. 0.8573
21. %0.0969
23. 11
25. 2.1
27. {x 0x ( 2.0860}
29. {a 0a & 1.1590}
31. 0.4341
33. 4.7820
log 9log 2
;
log 5log 7
;
©Glencoe/McGraw-Hill 281 Algebra 2 Chapter 10
Lesson 10-4 Common LogarithmsPages 549–551
59.
61.
63. 1
65. x "53
53x
3ba
60. 2
62. 3.06 s
64. 5
66. %34
& x & 2
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35. *1.1909
37. {n 0n " %1.0178}
39. 3.7162
41. 0.5873
43. %7.6377
45.
47.
49.
51. between 0.000000001 and0.000001 mole per liter
53. Sirius
55. Vega
57. about 3.75 yr or 3 yr 9 mo
2log 1.6log 4
" 0.6781
log 3log 7
" 0.5646
log 13log 2
" 3.7004
36. *2.6281
38. 1.0890
40. {p 0p # 1.9803}
42. 4.7095
44. 2.7674
46.
48.
50.
52. 8
54. Sirius: 1.45, Vega: 0.58
56a. 3;
56b.
56c. conjecture: loga b !
proof:
Original statement
Change of BaseFormula
!Inverse Prop. ofExponents andLogarithms
58. about 11.64 yr or 11 yr, 8 mo
1logb a
!1
logb a
logb blogb a
!? 1
logb a
loga b !
? 1log b a
1logba
;
32;
23
13
0.5 log 5log 6
" 0.4491
log 8log 3
" 1.8928
log 20log 5
" 1.8614
©Glencoe/McGraw-Hill 282 Algebra 2 Chapter 10
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©Glencoe/McGraw-Hill 283 Algebra 2 Chapter 10
59. Comparisons betweensubstances of differentacidities are more easilydistinguished on alogarithmic scale. Answersshould include the following.• Sample answer:
Tomatoes: 6.3 , 10%5 moleper literMilk: 3.98 , 10%7 mole perliterEggs: 1.58 , 10%8 moleper liter
• Those measurementscorrespond to pHmeasurements of 5 and 4,indicating a weak acid and astronger acid. On thelogarithmic scale we cansee the difference in theseacids, whereas on a normalscale, these hydrogen ionconcentrations would appearnearly the same. Forsomeone who has to watchthe acidity of the foods theyeat, this could be thedifference between anenjoyable meal and heartburn.
61. C
63. 1.6938
65. 64
67. 62
69. (d ' 2)(3d % 4)
71. prime
73. 32 ! x
75. log5 45 ! x
77. logb x ! y
60. A
62. 1.4248
64. 1.8416
66.
68. %22
70. (7p ' 3)(6q % 5)
72. 2x ! 3
74. 53 ! 125
76. log7 x ! 3
z #164
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©Glencoe/McGraw-Hill 284 Algebra 2 Chapter 10
1. the number e
3. Elsu; Colby tried to writeeach side as a power of 10.Since the base of the naturallogarithmic function is e, heshould have written eachside as a power of e;10ln 4x Z 4x.
5. 0.0334
7. %2.3026
9. e0 ! 1
11. 5x
13. 1.0986
15. 0 & x & 403.4288
17. *90.0171
19. about 15,066 ft
21. 148.4132
23. 1.6487
25. 2.3026
27. %3.5066
29. about 49.5 cm
31. 2 ! ln 6x
33. ex ! 5.2
35. y
37. 45
39. %0.6931
41. x " 0.4700
43. 0.5973
45. x ( %0.9730
2. ex ! 8
4. 403.4288
6. 0.1823
8. x ! ln 4
10. 3
12. x " 3.4012
14. %0.8047
16. 2.4630
18.
20. 54.5982
22. 0.3012
24. 1.0986
26. 1.6901
28. $183.21
30. %x ! ln 5
32. e1 ! e
34. 0.2
36. %4x
38. 0.2877
40. x & 1.5041
42. 0.2747
44. x ( 0.6438
46. 27.2991
h ! %26200 ln P
101.3
Lesson 10-5 Base e and Natural LogarithmsPages 557–559
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47. 49.4711
49. 14.3891
51. 45.0086
53. 1
55.
57.
59. about 55 yr
61. about 21 min
63. The number e is used in theformula for continuouslycompounded interest, A ! Pert. Although no banksactually pay interestcompounded continually, theequation is so accurate incomputing the amount ofmoney for quarterlycompounding or dailycompounding, that it is oftenused for this purpose.Answers should include thefollowing.
t !110
r
t !100 ln 2
r
48. 1.7183
50. 232.9197
52. 2, 6
54. about 19.8 yr
56. 100 ln 2 " 70
58. about 7.33 billion
60. about 32 students
62. always;
Originalstatement
Change ofBase Formula
Multiply
by the
reciprocal of
.
Simplify.
64. B
log xlog y
!? log x
log y
log ylog e
log xlog e
log xlog y
!? log x
log e!
log elog y
log xlog elog ylog e
logxlogy
!?
log xlog y
!? In x
In y
©Glencoe/McGraw-Hill 285 Algebra 2 Chapter 10
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• If you know the annualinterest rate r and theprincipal P, the value ofthe account after t years iscalculated by multiplying Ptimes e raised to the rtimes t power. Use acalculator to find the valueof ert.
• If you know the value Ayou wish the account toachieve, the principal P,and the annual interestrate r, the time t needed toachieve this value is foundby first taking the naturallogarithm of A minus thenatural logarithm of P.Then, divide this quantityby r.
65. 1946, 1981, 2015; It takesbetween 34 and 35 years forthe population to double.
67.
69. 5
71. inverse; 4
73. direct; %7
75. 3.32
77. 1.43
79. 13.43
log 0.047log 6
! %1.7065
66.
68.
70. 4
72. joint; 1
74.
76. 1.54
78. 323.49
80. 9.32
x !120
y 2 % 5
log 23log 50
! 0.8015
log 68log 4
! 3.0437
©Glencoe/McGraw-Hill 286 Algebra 2 Chapter 10
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1.
3. 3
5. 1.3863
log 5log 4
; 1.1610 2. e2 ! 3x
4. x " 5.3219
©Glencoe/McGraw-Hill 287 Algebra 2 Chapter 10
Chapter 10Practice Quiz 2
Page 559
Lesson 10-6 Exponential Growth and DecayPages 563–565
1. y ! a(1 ' r)t, where r " 0represents exponentialgrowth, and r & 0 representsexponential decay.
3. Sample answer: money in abank
5. about 33.5 watts
7. y ! 212,000e0.025t
9. C
11. at most $108,484.93
13. No; the bone is only about21,000 years old, anddinosaurs died out63,000,000 years ago.
15. about 0.0347
17. $12,565 billion
19. after the year 2182
2. Take the common logarithmof each side, use the PowerProperty to write log (1 ' r)t
as t log(1 ' r), and thendivide each side by thequantity log(1 ' r).
4. Decay; the exponent isnegative.
6. about 402 days
8. about 349,529 people
10. $1600
12. about 8.1 days
14. more than 44,000 years ago
16. y ! ae0.0347t
18. about 2025
20. 4.7%
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©Glencoe/McGraw-Hill 288 Algebra 2 Chapter 10
22. Answers should include thefollowing.• Find the absolute value of
the difference between theprice of the car for twoconsecutive years. Thendivide this difference bythe price of the car for theearlier year.
• Find 1 minus the rate ofdecrease in the value ofthe car as a decimal.Raise this value to thenumber of years it hasbeen since the car waspurchased, and thenmultiply by the originalvalue of the car.
24. D
26. ln 29 ! 4n % 2
28. 1.5323
30. 9
32.
34. hyperbola
36. parabola
38. 2.06 , 108
40. about 38.8%
p60
21. Never; theoretically, theamount left will always behalf of the previous amount.
23. about 19.5 yr
25. ln y ! 3
27. 4x2 ! e8
29. p " 3.3219
31.
33.
35. ellipse
37. circle
39. 8 , 107
p150
0.5 (0.08 p)6
'0.5 (0.08 p)
4
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Glencoe/McGraw-Hill 289 Algebra 2 Chapter 11
1. The differences between theterms are not constant.
3. Sample answer: 1, !4, !9,!14, . . .
5. !3, !5, !7, !9
7. 14, 12, 10, 8, 6
9. !112
11. 15
13. 56, 68, 80
15. 30, 37, 44, 51
17. 6, 10, 14, 18
19. 3,
21. 5.5, 5.1, 4.7, 4.3
23. 2, 15, 28, 41, 54
25. 6, 2, !2, !6, !10
27. 1,
29. 28
31. 94
33. 335
35.
37. 27
39. 61
41. 37.5 in.
43. 30th
45. 82nd
47. an " !7n # 25
263
23,
13, 0
43,
113
, 133
73,
Chapter 11 Sequences and SeriesLesson 11-1 Arithmetic Sequences
Pages 580–582
2. 95
4. 24, 28, 32, 36
6. 5, 8, 11, 14, 17
8. 43
10. 79
12. an " 11n ! 37
14. $12,000
16. 10, 3, !4, !11
18. 1, 4, 7, 10
20. 2,
22. 8.8, 11.3, 13.8, 16.3
24. 41, 46, 51, 56, 61
26. 12, 9, 6, 3, 0
28. 1,
30. !49
32. !175
34. 340
36.
38. !47
40. 173
42. 304 ft
44. 19th
46. an " 9n ! 2
48. an " !2n ! 1
!252
118
, 74,
178
58,
85,
65
125
,
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49. 13, 17, 21
51. Yes; it corresponds ton " 100.
53. 4, !2
55. 7, 11, 15, 19, 23
57. Arithmetic sequences can beused to model the numbersof shingles in the rows on asection of roof. Answersshould include the following.• One additional shingle is
needed in eachsuccessive row.
• One method is tosuccessively add 1 to theterms of the sequence: a8 "9 # 1 or 10, a9 " 10 # 1or 11, a10 " 11 # 1 or 12,a11 " 12 # 1 or 13, a12 "13 # 1 or 14, a13 "14 # 1 or 15, a14 " 15 # 1or 16, a15 " 16 # 1 or 17.Another method is to usethe formula for the nthterm:a15 " 3 # (15 ! 1)1 or 17.
59. B
61. !0.4055
50. pn " 4n ! 3
52. 70, 85, 100
54. !5, !2, 1, 4
56. z " 2y ! x
58. B
60. about 26.7%
62. 0.4621
©Glencoe/McGraw-Hill 290 Algebra 2 Chapter 11
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64. 15
66. 5, 4, 3, 2
63. 146.4132
65. 2, 5, 8, 11
67. 11, 15, 19, 23, 27
©Glencoe/McGraw-Hill 291 Algebra 2 Chapter 11
Lesson 11-2 Arithmetic SeriesPages 586–587
1. In a series, the terms areadded. In a sequence, theyare not.
3. Sample answer:
5. 230
7. 552
9. 260
11. 95
13. !6, 0, 6
15. 344
17. 1501
19. !9
21. 104
23. !714
25. 14
27. 10 rows
29. 721
31. 162
33. 108
35. !195
37. 315,150
39. 1,001,000
41. 17, 26, 35
a4
n"1(3n # 4)
2. Sample answer: 0 # 1 # 2 #3 # 4
4. 1300
6. 1932
8. 800
10. 63
12. 11, 20, 29
14. 28
16. 663
18. 2646
20. !88
22. 182
24. 225
26.
28. 8 days
30. 735
32. !204
34. !35
36. 510
38. 24,300
40. 166,833
42. !13, !8, !3
!2456
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43. !12, !9, !6
45. 265 ft
47. False; for example, 7 # 10 #13 # 16 " 46, but 7 # 10 #13 # 16 # 19 # 22 # 25 #28 " 140.
49. C
51. 5555
53. 6683
55. !135
57.
59.
61.
63. 16
65. 227
26221
3 $ 2892
!92
©Glencoe/McGraw-Hill 292 Algebra 2 Chapter 11
44. 13, 18, 23
46. True; for any series, 2a1 #2a2 # 2a3 # # 2an "2(a1 # a2 # a3 # # an).
48. Arithmetic series can beused to find the seatingcapacity of an amphitheater.Answers should include thefollowing.• The sequence represents
the numbers of seats inthe rows. The sum of thefirst n terms of the seriesis the seating capacity ofthe first n rows.
• One method is to write outthe terms and add them:18 # 22 # 26 # 30 # 34 #38 # 42 # 46 # 50 # 54 "360. Another method is touse the formulaSn " [2a1 # (n ! 1)d ]:
S10 " [2(18) #
(10 ! 1)4] or 360.
50. C
52. 3649
54. 111
56. about 3.82 days
58.
60.
62.
64. !54
222
23
!163
102
n2
pp
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2. Sample answer:
1,
4. 67.5, 101.25
6. !2, !6, !18, !54, !162
8. 56
10. an " 4 % 2n!1
12. A
14. 192, 256
16. 48, 32
18.
20. !21.875, 54.6875
22. 1, 4, 16, 64, 256
24. 576, !288, 144, !72, 36
26. 2592
28. 1024
30.
32. 192
34. 2
36.
38. $46,794.34
40.
42. an " 4(!3)n !1
an " 64 a14b
n!1
572
14
12524
, 62548
p23,
49,
827
,
©Glencoe/McGraw-Hill 293 Algebra 2 Chapter 11
1a. Geometric; the terms havea common ratio of !2.
1b. Arithmetic; the terms have acommon difference of !3.
3. Marika; Lori divided in thewrong order when finding r.
5. 2, !4
7.
9. !4
11. 3, 9
13. 15, 5
15. 54, 81
17.
19. !2.16, 2.592
21. 2, !6, 18, !54, 162
23. 243, 81, 27, 9, 3
25.
27. 729
29. 243
31. 1
33. 78,125
35. !8748
37. 655.36 lb
39.
41. an " !2(!5)n !1
an " 36 a13b
n!1
316
2027
, 4081
1564
Lesson 11-3 Geometric SequencesPages 590–592
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43. $18, 36, $72
45. 16, 8, 4, 2
47. 8 days
49. False; the sequence 1, 4, 9,16, for example, is neither arithmetic norgeometric.
51. The heights of the bouncesof a ball and the heightsfrom which a bouncing ballfalls each form geometricsequences. Answers shouldinclude the following.• 3, 1.8, 1.08, 0.648, 0.3888• The common ratios are the
same, but the first termsare different. Thesequence of heights fromwhich the ball falls is thesequence of heights of thebounces with the term 3inserted at the beginning.
53. C
55. 203
57. !12, !16, !20
59. 127
61. 6181
p,
©Glencoe/McGraw-Hill 294 Algebra 2 Chapter 11
44. $12, 36, $108
46. 6, 12, 24, 48
48. 5 mg
50. False, the sequence 1, 1, 1,1, for example, isarithmetic (d " 0) andgeometric (r " 1).
52. A
54. 632.5
56. 19, 23
58. units
60. 6332
522 # 3210
p,
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1. 46
3. 187
5. 1
2.
4. 816
112
©Glencoe/McGraw-Hill 295 Algebra 2 Chapter 11
Chapter 11Practice Quiz 1
Page 592
1. Sample answer:
3. Sample answer: The firstterm is . Divide thesecond term by the first tofind that the common ratio is
. Therefore, the nth termof the series is given by
. There are fiveterms, so the series can be
written as
5. 39,063
7. 165
9. 129
11.
13. 3
15. 728
17. 1111
10939
a5
n"12 ! 6n!1.
2 ! 6n!1
r " 6
a1 " 2
4 # 2 # 1 #12
2. The polynomial is ageometric series with firstterm 1, common ratio x, and4 terms. The sum is
.
4. 732
6. 81,915
8.
10.
12. 3
14. 93 in. or 7 ft 9 in.
16. 765
18. 300
314
13309
1(1 ! x 4)
1 ! x"
x 4 ! 1
x ! 1
Lesson 11-4 Geometric SeriesPages 596–598
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19. 244
21. 2101
23.
25. 1040.984
27. 6564
29. 1,747,625
31. 3641
33.
35. 2555
37.
39. 3,145,725
41. 243
43. 2
45. 80
47. about 7.13 in.
49. If the number of people thateach person sends the joketo is constant, then the totalnumber of people who haveseen the joke is the sum of ageometric series. Answersshould include the following.• The common ratio would
change from 3 to 4.• Increase the number of
days the joke circulates sothat it is inconvenient tofind and add all the termsof the series.
3874
546116
7283
20. 1,328,600
22. 1441
24.
26. 7.96875
28.
30. $10,737,418.23
32. 206,668
34.
36.
38.
40. 86,093,440
42. 1024
44. 6
46. 8
48. If the first term and commonratio of a geometric seriesare integers, then all theterms of the series areintegers. Therefore, the sumof the series is an integer.
50. A
58,975256
!364
!1829
!118,096
2154
©Glencoe/McGraw-Hill 296 Algebra 2 Chapter 11
PQ245-6457F-P11[289-315].qxd 7/31/02 9:56 AM Page 296 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-11:
51. C
53. 3.99987793
55.
57. 232
59.
61. Sample answer: 294
63. 2
65.
67. 0.6
23
Drive-In Movie Screens
Scre
ens
1000
900
800
700
600
0
Years Since 19950 1 2 3 4 5 6
$14,
32, $9
52.
54. 6.24999936
56. !3,
58. 192
60. Sample answer using(1, 826) and (3, 750):
62. 2
64.
66. !2
14
y " !38x # 864
!818
!274
,!92,
!1,048,575
©Glencoe/McGraw-Hill 297 Algebra 2 Chapter 11
Lesson 11-5 Infinite Geometric SeriesPages 602–604
1. Sample answer: a&
n"1 a1
2b
n
2. 0.999999 . . . can be writtenas the infinite geometricseries The first term of this series is
and the common ratio is
so the sum is or 1.910
1 ! 110
110
,
910
p .910
#9
100#
91000
#
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3. Beth; the common ratio forthe infinite geometric seriesis Since ' 1, the
series does not have a sumand the formula S "
does not apply.
5. does not exist
7.
9. 100
11.
13. 96 cm
15. does not exist
17. 45
19. !16
21.
23. does not exist
25. 1
27.
29.
31. 2
33.
35. 900 ft
37. 75, 30, 12
39. !8, !315, !1
725
, !64125
40 # 2022 # 20 # p
32
23
545
7399
34
a1
1 ! r
`!43`!
43
.
4. 108
6. does not exist
8.
10.
12.
14. 14
16. 7.5
18. 64
20. does not exist
22. 3
24. 1
26. 7.5
28. 144
30. 6
32. 30 ft
34. or about136.6 cm
36. 27, 18, 12
38.
40. 79
24, 1612, 11
1132
, 7409512
80 # 4022
175999
59
307
©Glencoe/McGraw-Hill 298 Algebra 2 Chapter 11
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41.
43.
45.
47.
49. The total distance that a ballbounces, both up and down,can be found by adding thesums of two infinitegeometric series. Answersshould include the following.•
or S "
• The total distance the ballfalls is given by the infinitegeometric series 3 #
Thesum of this series is
or 7.5. The total distance the ball bouncesup is given by the infinitegeometric series 3(0.6) #
The sum of this series is
or 4.5. Thus, the total distance the ball travelsis 7.5 # 4.5 or 12 feet.
51. C
53.
55. 3
874481
3(0.6)1 ! 0.6
3(0.6)3 # p .3(0.6)2 #
31 ! 0.6
3(0.6) # 3(0.6)2 # p .
a1
1 ! ra1(1 ! r n)
1 ! r,
an " a1 ! rn!1, Sn "
229990
427999
8299
19
42.
44.
46.
48.
50. D
52. !182
54. 32.768%
56. !32
S "a1
1 ! r
S(1 ! r )" a1
# 0 # pS ! rS " a1 # 0 # 0 # 0(!)rS " a1r # a1r 2 # a1r 3 # a1r 4 # p
S " a1 # a1r # a1r 2 # a1r 3 # p
511
82333
411
©Glencoe/McGraw-Hill 299 Algebra 2 Chapter 11
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57.
59.
61.
63.
65.
67.
69. The number of visitors wasdecreasing.
71. 3
73.
75. !4
12
x2 ! 10x # 24 " 0
x2 ! 36 " 0
!12,
32,
72
(x ! 2)2 # (y ! 4)2 " 36
!x # 7(x ! 3)(x # 1)
x ' 5 58.
60.
62.
64.
66.
68. about !180,724 visitors peryear
70. 2
72. 2
74. 4
x2 # 9x # 14 " 0
!12, !
13, 0,
12
(x ! 3)2 # (y # 1)2 " 32
3x # 7(x # 4)(x # 2)
!2a # 5ba2b
©Glencoe/McGraw-Hill 300 Algebra 2 Chapter 11
Lesson 11-6 Recursion and Special SequencesPages 608–610
1.
3. Sometimes; if andthen or 4, soBut, if then
5. !3, !2, 0, 3, 7
7. 1, 2, 5, 14, 41
9. 1, 3, !1
11.
13. !6, !3, 0, 3, 6
15. 2, 1, !1, !4, !8
17. 9, 14, 24, 44, 84
19. !1, 5, 4, 9, 13
bn " 1.05bn!1 ! 10
x2 " 1, so x2 " x1.x1 " 1,x2 ( x1.
x2 " 22x1 " 2,f(x) " x2
an " an!1 # d; an " r ! an!1 2. Sample answer:
4. 12, 9, 6, 3, 0
6. 0, !4, 4, !12, 20
8. 5, 11, 29
10. 3, 11, 123
12. $1172.41
14. 13, 18, 23, 28, 33
16. 6, 10, 15, 21, 28
18. 4, 6, 12, 30, 84
20. 4, !3, 5, !1, 9
an " 2an!1 # an!2
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21.
23. 67
25. 1, 1, 2, 3, 5, . . .
27. $99,921.21, $99,841.95,$99,762.21, $99,681.99,$99,601.29, $99,520.11,$99,438.44, $99,356.28
29.
31. 16, 142, 1276
33. !7, !16, !43
35. !3, 13, 333
37.
39. $75.77
41. Under certain conditions, theFibonacci sequence can beused to model the number ofshoots on a plant. Answersshould include the following.• The 13th term of the
sequence is 233, so thereare 233 shoots on the plantduring the 13th month.
• The Fibonacci sequence isnot arithmetic because thedifferences (0, 1, 1, 2, . . .)of the terms are notconstant. The Fibonaccisequence is not geometricbecause the ratiosQ1, 2, . . .R of the terms
are not constant.
32
,
52
, 372
, 1445
2
tn " tn!1 # n
72
, 74
, 76
, 78
, 710
22.
24. !2.1
26. the Fibonacci sequence
28. 1, 3, 6, 10, 15
30. 20,100
32. 5, 17, 65
34. !4, !19, !94
36. !1, !1, !1
38.
40. No; according to the first twoiterates, f(4) " 4. Accordingto the second and thirditerates, f(4) " 7. Since f(x)is a function, it cannot havetwo values when x " 4.
42. D
43
, 103
, 763
34
, 32
, 154
, 252
, 4258
©Glencoe/McGraw-Hill 301 Algebra 2 Chapter 11
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43. C
45.
47. !5208
49. units
51. 5040
53. 20
55. 210
3x # 7
16
44. 27
46.
48.
50. 120
52. 6
54. 126
1093243
125
©Glencoe/McGraw-Hill 302 Algebra 2 Chapter 11
Lesson 11-7 The Binomial TheoremPages 615–617
1. 1, 8, 28, 56, 70, 56, 28, 8, 1
3. Sample answer:
5. 17,160
7.
9.
11.
13. 362,880
15. 72
17. 495
19.
21.
23.270x
2 # 405x # 243x
5 # 15x 4 # 90x
3 #
56r 3s 5 # 28r 2s
6 # 8rs 7 # s
856r 5s
3 # 70r 4s 4 #
r 8 # 8r 7s # 28r 6s 2 #
a 3 ! 3a
2b # 3ab 2 ! b
3
1,088,640a6b4
108xy3 # 81y4x4 ! 12x3y # 54x2y2 !
10p2q3 # 5pq4 # q5p5 # 5p4q # 10p3q2 #
(5x # y)4
2. n
4. 40,320
6. 66
8.
10.
12. 10
14. 6,227,020,800
16. 210
18. 2002
20.
22.
24. a4 ! 8a3 # 24a2 ! 32a # 16
10m 2a
3 # 5ma 4 ! a
5m
5 ! 5m 4a # 10m
3a 2 !
4mn3 # n4m4 # 4m3n # 6m2n2 #
56a5b3
160t3 # 240t2 # 192t # 64t6 # 12t5 # 60t4 #
PQ245-6457F-P11[289-315].qxd 7/31/02 9:56 AM Page 302 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-11:
©Glencoe/McGraw-Hill 303 Algebra 2 Chapter 11
25.
27.
29.
31. 27x3 # 54x2 # 36x # 8 cm3
33. 45
35.
37.
39.
41.
43. The coefficients in a binomialexpansion give the numbersof sequences of birthsresulting in given numbers ofboys and girls. Answersshould include the following.• (b # g)5 " b5 # 5b4g #
10b3g2 # 10b2g3 # 5bg4 # g5;
!638
x5
145,152x6y3
5670a4
924x6y6
20a2 # 40a # 32
a5
32#
5a4
8# 5a3 #
! 720x2y3 # 240xy 4 ! 32y 5
243x5 ! 810x4y # 1080x3y2
24b 2x
2 ! 8bx 3 # x
416b4 ! 32b3x # 26.
28.
30.
32. 1, 4, 6, 4, 1
34.
36.
38.
40.
42. and represent the 6th
and 7th entries in the row for in Pascal’s triangle.
represents the seventh
entry in the row for Since is below and
in Pascal’s triangle,
44. D
12!7!5!
#12!6!6!
"13!7!6!
.
12!6!6!
12!7!5!
13!7!6!
n " 13.
13!7!6!
n " 12
12!6!6!
12!7!5!
3527
x4
1,088,640a6b4
280x4
!126x4y5
10m3
3#
5m4
27#
m5
243
243 # 135m # 30m2 #
216x2y2 # 96xy3 # 16y481x4 # 216x3y #
12ab5 # b6160a3b3 # 60a2b4 #
64a6 # 192a5b # 240a4b2 #
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There is one sequence ofbirths with all 5 boys, fivesequences with 4 boys and1 girl, ten sequences with3 boys and 2 girls, tensequences with 2 boys and3 girls, five sequences with1 boy and 4 girls, and onesequence with all 5 girls.
• The number of sequencesof births that have exactlyk girls in a family of nchildren is the coefficientof in the expansionof According tothe Binomial Theorem, thiscoefficient is
45. C
47. 3, 5, 9, 17, 33
49. 2.3219
51. 1.2920
53. asymptotes:
55. hyperbola
57. yes
59. True; or 1.
61. True; or 1.12(1 # 1)2
4"
1(4)4
1(1 # 1)2
"1(2)
2
x " !4, x " 1
log 8log 5
;
log 5log 2
;
n!(n ! k)!k!
.
(b # g)n.bn!kgk
46. 7, 5, 3, 1, !1
48. 125 cm
50. 2.0959
52. asymptotes:
54. hole:
56. parabola
58. no
60. False;
or 3.
62. True; which iseven.
31 ! 1 " 2,
2(3)2
(1 # 1)(2 ! 1 # 1)2
"
x " !3
x " !2, x " !3
1log 3
;
©Glencoe/McGraw-Hill 304 Algebra 2 Chapter 11
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1. 1,328,600
3. 24
5. 1, 5, 13, 29, 61
7. 5, !13, 41
9.160a3 # 240a2 # 192a # 64a6 # 12a5 # 60a4 #
2. !364
4.
6. 2, 4, 8, 14, 22
8.
10. 4032a5b4
90x2y3 # 15xy 4 # y 5243x5 # 405x 4y # 270x3y 2 #
254
©Glencoe/McGraw-Hill 305 Algebra 2 Chapter 11
Chapter 11Practice Quiz 2
Page 617
Lesson 11–8 Proof and Mathematical InductionPages 619–621
1. Sample answers: formulasfor the sums of powers of thefirst n positive integers andstatements that expressionsinvolving exponents of n aredivisible by certain numbers
3. Sample answer: 3n ! 1
2. Mathematical induction isused to show that astatement is true. A counterexample is used to show thata statement is false.
4. Step 1: When the leftside of the given equation is
1. The right side is or 1, so the equation is true for
Step 2: Assume
3 # # k " for
some positive integer k.Step 3:
# (k # 1)
"(k # 1) # (k # 2)
2
"k(k # 1) # 2(k # 1)
2
"k(k # 1)
2
k # 1k # 121 # 2 # 3 # p #
k(k # 1)2
p1 # 2 #
n " 1.
1(1 # 1)2
n " 1,
PQ245-6457F-P11[289-315].qxd 7/31/02 9:56 AM Page 305 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-11:
5. Step 1: When the leftside of the given equation is
The right side is or
so the equation is true for
Step 2: Assume
" for some
positive integer k.
Step 3:
" 1
" 1
" 1
The last expression is theright side of the equation tobe proved, where Thus, the equation is true for
Therefore,
" 1 for all positive integers n.
!12n
123 # p #
12n
12
#122 #
n " k # 1.
n " k # 1.
!1
2k#1
!2
2k#1 # 1
2k#1
!12k #
12k#1
123 # p #
12k #
12k#1
12
#122 #
12k1!
123 # p #
12k
12
#122 #
n " 1.
12
,
12
1 !12
.
n " 1,
The last expression is theright side of the equation tobe proved, where Thus, the equation is true for
Therefore,
n " for all
positive integers n.
6. Step 1: which isdivisible by 3. The statementis true for Step 2: Assume that is divisible by 3 for somepositive integer k. Thismeans that forsome whole number r.Step 3:
Since r is a whole number,is a whole number.
Thus, is divisible by3, so the statement is truefor Therefore,
is divisible by 3 for allpositive integers n.4n ! 1
n " k # 1.
4k#1 ! 14r # 1
4k#1 ! 1 " 314r # 124k#1 ! 1 " 12r # 3
4k#1 " 12r # 44k " 3r # 1
4k ! 1 " 3r
4k ! 1 " 3r
4k ! 1n " 1.
41 ! 1 " 3,
n(n # 1)2
1 # 2 # 3 # p #n " k # 1.
n " k # 1.
©Glencoe/McGraw-Hill 306 Algebra 2 Chapter 11
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7. Step 1: which isdivisible by 4. The statementis true for Step 2: Assume that is divisible by 4 for somepositive integer k. Thismeans that forsome positive integer r.Step 3:
Since r is a positive integer,is a positive integer.
Thus, is divisible by4, so the statement is truefor Therefore, is divisibleby 4 for all positive integers n.
9. Sample answer: n " 3
5n # 3n " k # 1.
5k#1 # 35r ! 3
5k#1 # 3 " 415r ! 325k#1 # 3 " 20r ! 125k#1 " 20r ! 15
5k " 4r ! 35k # 3 " 4r
5k # 3 " 4r
5k # 3n " 1.
51 # 3 " 8, 8. Sample answer:
10. Step 1: After the first guesthas arrived, no handshakes
have taken place. ,so the formula is correct for
Step 2: Assume that after kguests have arrived, a total
of handshakes have
take place, for some positiveinteger k.Step 3: When the stguest arrives, he or sheshakes hands with the kguests already there, so thetotal number of handshakesthat have then taken place is
.# kk(k ! 1)2
(k # 1)
k(k ! 1)2
n " 1.
" 01(1 ! 1)2
n " 2
©Glencoe/McGraw-Hill 307 Algebra 2 Chapter 11
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11. Step 1: When the leftside of the given equation is1. The right side is 1[2(1) ! 1]or 1, so the equation is truefor Step 2: Assume
for some positiveinteger k.Step 3:
The last expression is theright side of the equation tobe proved, where Thus, the equation is true for
Therefore, for all
positive integers n.(4n ! 3) " n(2n ! 1)
1 # 5 # 9 # p #n " k # 1.
n " k # 1.
" (k # 1)[2(k # 1) ! 1]" (k # 1)(2k # 1)" 2k2 # 3k # 1" 2k2 ! k # 4k # 4 ! 3
[4(k # 1) ! 3]" k (2k ! 1) #(4k ! 3) # [4(k # 1) ! 3]
1 # 5 # 9 # p #
k (2k ! 1)# p # (4k ! 3) "
1 # 5 # 9n " 1.
n " 1,
The last expression is theformula to be proved, where
Thus, the formulais true for Therefore, the total number
of handshakes is for
all positive integers n.
12. Step 1: When the leftside of the given equation is
2. The right side is or 2, so the equation is truefor Step 2: Assume
for some positive integer k.Step 3:
"(k # 1)[3(k # 1) # 1]
2
"(k # 1)(3k # 4)
2
"3k 2 # 7k # 4
2
"3k 2 # k # 6k # 6 ! 2
2
"k(3k # 1) # 2[3(k # 1) ! 1]
2
# [3(k # 1) ! 1]"k(3k # 1)
2
(3k ! 1) # [3(k # 1) ! 1]2 # 5 # 8 # p #
k(3k #1)2
8 # p # (3k ! 1) "
2 # 5 #n " 1.
1[3(1) # 1]2
n " 1,
n(n ! 1)2
n " k # 1.n " k # 1.
"k (k # 1)
2 or
(k # 1)k2
"k [(k ! 1) # 2]
2
k(k ! 1)2
# k "k(k ! 1) # 2k
2
©Glencoe/McGraw-Hill 308 Algebra 2 Chapter 11
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13. Step 1: When the leftside of the given equation is13 or 1. The right side is
or 1, so theequation is true for Step 2: Assume
for some positive integer k.Step 3:
"
"
"
"
"
The last expression is theright side of the equation tobe proved, where Thus, the equation is true forn " k # 1.
n " k # 1.
(k # 1)2[(k # 1) # 1]2
4
(k # 1)2(k # 2)2
4
(k # 1)2(k 2 # 4k # 4)4
(k # 1)2[k 2 # 4(k # 1)]4
k 2(k # 1)2 # 4(k # 1)3
4
"k
2(k # 1)2
4# (k # 1)3
k3 # (k # 1)3# p ## 33# 2 313
"k2 1k # 122
433 # p # k
3
13 # 23 #n " 1.
12(1 # 1)2
4
n " 1,
The last expression is theright side of the equation tobe proved, where Thus, the equation is true for
Therefore,
for allpositive integers n.
14. Step 1: When the leftside of the given equation is12 or 1. The right side is
or 1, so
the equation is true for
Step 2: Assume
for some
positive integer k.Step 3:
"
"
"
"
"
"
"(k #1)[2(k #1) ! 1][2(k # 1) #1]
3
(2k # 1)(k # 1)(2k # 3)3
(2k # 1)(2k 2 # 5k # 3)
3
(2k # 1)(2k 2 ! k # 6k # 3)
3
(2k # 1)[k (2k ! 1) # 3(2k # 1)]3
k(2k ! 1)(2k # 1) # 3(2k # 1)2
3
[2(k # 1) ! 1]2
k(2k ! 1)(2k # 1)3
#
(2k ! 1)2 # [2(k # 1) ! 1]212 # 32 # 52 # p #
k (2k ! 1)(2k # 1)3
12#32 #52 # p #(2k !1)2 "
n " 1.
1[2(1) ! 1][2(1) # 1]3
n " 1,
n(3n # 1)2
(3n ! 1) "
2 # 5 # 8 # p #n " k # 1.
n " k # 1.
©Glencoe/McGraw-Hill 309 Algebra 2 Chapter 11
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Therefore,
for all
positive integers n.
15. Step 1: When the leftside of the given equation is
. The right side is
or , so the equation is true
for Step 2: Assume
for some
positive integer k.
Step 3:
"12a1 !
13k#1b
"12
a3k #1 ! 13k #1 b
"3k#1 ! 12 ! 3k #1
"3k#1 ! 3 # 2
2 ! 3k #1
"12
!1
2 ! 3k #1
3k #1
"12 a2 !
13kb #
13k#1
13k #
13k#1
13
#132 #
133 # p #
p #13k "
12 a1 !
13kb
13
#132 #
132 #
n " 1.
13
12a1 !
13b1
3
n " 1,
n2(n # 1)2
4n3 "
13 # 23 # 33 # p # The last expression is theright side of the equation tobe proved, where Thus, the equation is true for
Therefore,
for all positive integers n.
16. Step 1: When the leftside of the given equation is
The right side is
or so the equation is true
for Step 2: Assume
for some positive integer k.
Step 3:
"13a1 !
14k#1b
"13
a4k #1 ! 14k #1 b
"4k #1 ! 13 ! 4k #1
"4k #1 ! 4 # 3
3 ! 4k #1
"13
!1
3 ! 4k #1
4k #1
"13a1 !
14kb #
14k#1
14k #
14k #1
14
#142 #
143 # p #
a1 ! 14kb1
4k "13
# p #143
14
#142 #
n " 1.
14
,
13a1 !
14b1
4.
n " 1,
n(2n ! 1)(2n # 1)3
(2n ! 1)2 "
12 # 32 # 52 # p #n " k # 1.
n " k # 1.
©Glencoe/McGraw-Hill 310 Algebra 2 Chapter 11
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The last expression is theright side of the equation tobe proved, where Thus, the equation is true for
Therefore,
for all positive
integers n.17. Step 1: which is
divisible by 7. The statementis true for Step 2: Assume that is divisible by 7 for somepositive integer k. Thismeans that forsome whole number r.Step 3:
Since r is a whole number,is a whole number.
Thus, is divisible by7, so the statement is truefor Therefore, is divisibleby 7 for all positive integers n.
19. Step 1:which is divisible by 11. Thestatement is true for Step 2: Assume that is divisible by 11 for somepositive integer k. Thismeans that for some positive integer r.
12k # 10 " 11r
12k # 10n " 1.
121 # 10 " 22,
8n ! 1n " k # 1.
8k#1 ! 18r # 1
8k #1 ! 1 " 7(8r # 1)8k#1 ! 1 " 56r # 7
8k #1 " 56r # 88k " 7r # 1
8k ! 1 " 7r
8k ! 1 " 7r
8k ! 1n " 1.
81 ! 1 " 7,
"12a1 !
13nb
# 13np1
3#
132 #
133 #
n " k # 1.
n " k # 1.
The last expression is theright side of the equation tobe proved, where Thus, the equation is true for
Therefore,
for all positive
integers n.
18. Step 1: which isdivisible by 8. The statementis true for Step 2: Assume that is divisible by 8 for somepositive integer k. Thismeans that forsome whole number r.Step 3:
Since r is a whole number,is a whole number.
Thus, is divisible by 8, so the statement is truefor Therefore, is divisibleby 8 for all positive integers n.
20. Step 1:which is divisible by 12. Thestatement is true for Step 2: Assume that is divisible by 12 for somepositive integer k. Thismeans that for some positive integer r.
13k # 11 " 12r
13k #11n " 1.
131 # 11 " 24,
9n ! 1n " k # 1.
9k #1 ! 19r # 1
9k#1 ! 1 " 8(9r # 1)9k#1 ! 1 " 72r # 8
9k#1 " 72r # 99k " 8r # 1
9k ! 1 " 8r
9k ! 1 " 8r
9k ! 1n " 1.
91 ! 1 " 8,
"13
a1 !14nb
# 14np1
4#
142 #
143 #
n " k # 1.
n " k # 1.
©Glencoe/McGraw-Hill 311 Algebra 2 Chapter 11
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Step 3:
Since r is a positive integer,is a positive
integer. Thus, isdivisible by 11, so thestatement is true for
Therefore, isdivisible by 11 for all positiveintegers n.
21. Step 1: There are 6 bricks inthe top row, and 6, so the formula is true for
Step 2: Assume that thereare bricks in the topk rows for some positiveinteger k.Step 3: Since each row has2 more bricks than the oneabove, the numbers of bricksin the rows form anarithmetic sequence. Thenumber of bricks in the
st row isor
Then the number ofbricks in the top rowsis or
which is theformula to be proved, where
Thus, the formulais true for n " k # 1.n " k # 1.
5(k # 1),k 2 # 7k # 6 " (k # 1)2 #k 2 # 7k # 6.
k 2 # 5k # (2k # 6)k # 1
2k # 6.6 # [(k # 1) ! 1](2)(k # 1)
k 2 # 5k
n " 1.
12 # 5(1) "
12n #10n " k # 1.
12k #1 # 1012r ! 10
12k#1 # 10 " 11(12r ! 10)12k#1 # 10 " 132r ! 110
12k#1 " 132r ! 12012k " 11r ! 10
12k # 10 " 11rStep 3:
Since r is a positive integer,is a positive
integer. Thus, isdivisible by 12, so thestatement is true for
Therefore, isdivisible by 12 for all positiveintegers n.
22. Step 1: When the left side of the given equation is
The right side is
or so the equation is truefor Step 2: Assume
for some positive
integer k.Step 3:
The last expression is theright side of the equation tobe proved, where n " k # 1.
"a1(1 ! r k#1)
1 ! r
"a1 ! a1r k # a1r k ! a1r k#1
1 ! r
"a1(1 ! r k ) # (1 ! r )a1r k
1 ! r
# a1r k"a1(1 ! r k)
1 ! r
# a1r ka1r 2 # p # a1r k!1
a1 # a1r #
a1(1 ! r k )1 ! r
a1r 2 # p # a1r k!1 "
a1 # a1r #n " 1.a1,
a1(1 ! r1)1 ! r
a1.
n " 1,
13n # 11n " k # 1.
13k#1 # 1113r ! 11
13k #1 # 11 " 12(13r ! 11)13k #1 # 11 " 156r ! 132
13k #1 " 156r ! 14313k " 12r ! 11
13k # 11 " 12r
©Glencoe/McGraw-Hill 312 Algebra 2 Chapter 11
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Therefore, the number ofbricks in the top n rows is
for all positiveintegers n.
23. Step 1: When the leftside of the given equation is
a1. The right side is
or so theequation is true for Step 2: Assume a1#
for
some positive integer k.Step 3:
[2a1 # (k # 1 ! 1)d ]k # 1
2"
(2a1 # kd )k # 12
"
(k # 1)2a1 # k(k # 1)d2
"
(k # 1)2a1 # (k 2 ! k # 2k)d2
"
k ! 2a1 # (k 2 ! k)d # 2a1 # 2kd2
"
k [2a1 # (k ! 1)d ] # 2(a1 # kd )2
"
# a1 # kd
[2a1 # (k ! 1)d ]k2
"
[a1 # (k # 1 ! 1)d ]
[2a1 # (k ! 1)d ] #k2
"
[a1 # (k # 1 ! 1)d ](k ! 1)d ] #
p # [a1 #(a1 # 2d ) #a1 # (a1 # d ) #
[2a1 # (k ! 1)d ]k2
"
[a1 # (k ! 1)d ](a1 # d ) # (a1 # 2d 2 # p #
n " 1.a1,(1 ! 1)d ]
[2a1 #12
n " 1,
n2 # 5n
Thus, the equation is true for
Therefore,
for all positive integers n.
24. Step 1: The figure showshow to cover a 21 by 21
board, so the statement istrue for
Step 2: Assume that a byboard can be covered for
some positive integer k.
Step 3: Divide a byboard into four
quadrants. By the inductivehypothesis, the first quadrantcan be covered. Rotate thedesign that covers QuadrantI 90) clockwise and use it tocover Quadrant II. Use thedesign that covers QuadrantI to cover Quadrant III.
2k#12k#1
2k2k
n " 1.
a1(1 ! r n )1 ! r
a1r 2 # p # a1rn!1 "a1 # a1r #
n " k # 1.
©Glencoe/McGraw-Hill 313 Algebra 2 Chapter 11
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The last expression is theright side of the formula tobe proved, where Thus, the formula is true for
Therefore,
for all positive integers n.
25. Sample answer:
27. Sample answer:
29. Sample answer:
31. Write as Thenuse the Binomial Theorem.
Since each term in the lastexpression is divisible by 6,the whole expression isdivisible by 6. Thus, is divisible by 6.
33. C
35.20x 3y 3 # 15x 2y 4 # 6xy 5 # y 6
x 6 # 6x 5y # 15x 4y 2 #
7n ! 1
6n!2 # p # n ! 6
n(n ! 1)2
!" 6n # n ! 6n!1 #
6n!2 # p # n ! 6 # 1 ! 1
n(n ! 1)2
!" 6n # n ! 6n!1 #
7n ! 1 " (6 # 1)n ! 1
(6 # 1)n.7n
n " 11
n " 2
n " 3
[2a1#(n ! 1)d ]n2
(n ! 1)d ] "
[a1 #(a1# 2d ) # p ##a1 # (a1 # d )
n " k # 1.
n " k # 1.
Rotate the design that coversQuadrant I 90)counterclockwise and use itto cover Quadrant IV. Thisleaves three empty squaresnear the center of the board,as shown. Use one more L-shaped tile to cover these3 squares. Thus, a by
board can be covered.The statement is true for
Therefore, a by checkerboard with the topright square missing can becovered for all positiveintegers n.
26. Sample answer:
28. Sample answer:
30. Sample answer:
32. An analogy can be madebetween mathematicalinduction and a ladder withthe positive integers on thesteps. Answers shouldinclude the following.• Showing that the
statement is true for(Step 1).
• Assuming that thestatement is true for somepositive integer k andshowing that it is true for
(Steps 2 and 3).
34. A
36.
21a2b5 # 7ab6 ! b735a4b3 # 35a3b4 !
a7 ! 7a6b # 21a5b 2 !
k # 1
n " 1
n " 41
n " 3
n " 4
2n2nn " k # 1.
2k#12k#1
©Glencoe/McGraw-Hill 314 Algebra 2 Chapter 11
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37.
39. 2, 14, 782
41. 0, 1
112x 2y 6 # 16xy 7 # y 81120x 4y 4 # 448x 3y 5 #1792x 6y 2 # 1792x 5y 3 #
256x8 # 1024x7y # 38. 4, 10, 28
40. 12 h
42. !14
©Glencoe/McGraw-Hill 315 Algebra 2 Chapter 11
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1. HHH, HHT, HTH, HTT, THH,THT, TTH, TTT
3. The available colors for thecar could be different fromthose for the truck.
5. dependent
7. 256
9. D
11. independent
13. dependent
15. 16
17. 30
19. 1024
21. 10,080
23. 362,880
25. 27,216
27. 800
Chapter 12 Probability and StatisticsLesson 12-1 The Counting Principle
Pages 634–637
©Glencoe/McGraw-Hill 316 Algebra 2 Chapter 12
2. Sample answer: buying ashirt that comes in 3 sizesand 6 colors
4. independent
6. 30
8. 20
10. dependent
12. independent
14. 6
16. 6
18. 48
20. 240
22. 151,200
24. 17
26. 160
28. See students’ work.
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29. The maximum number oflicense plates is a productwith factors of 26s and 10s,depending on how manyletters are used and howmany digits are used.Answers should include thefollowing.
• There are 26 choices forthe first letter, 26 for thesecond, and 26 for thethird. There are 10 choicesfor the first number, 10 forthe second, and 10 for thethird. By the FundamentalCounting Principle, thereare or 17,576,000possible license plates.
• Replace positions containingnumbers with letters.
31. C
33. 20 mi
263 ! 103
30. A
32. 45
34. Step 1: When , the leftside of the given equation is
4. The right side is
or 4, so the equation is truefor .Step 2: Assume
!
for some positive integer k.Step 3:
!
!
!
!3k 2 " 11k " 8
2
3k 2 " 5k " 6k " 6 " 22
k(3k " 5) " 2[3(k " 1) " 1]2
" [3(k " 1) " 1]k(3k " 5)2
(3k " 1) " [3(k " 1) " 1]4 " 7 " 10 "p"
k(3k " 5)2
" p " (3k " 1)
4 " 7 " 10n ! 1
1[3(1) " 5]2
n ! 1
©Glencoe/McGraw-Hill 317 Algebra 2 Chapter 12
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35. 28x6y2
37. 7
39.
41.
43. #1, #2
45. y ! (x $ 3)2 " 2
47.
49. 3
51.
53. no inverse exists
55.
57. 30
59. 720
61. 15
63. 1
y !23x "
13
17B1 $14 3
R
y ! $12x2 " 8
$x
x " 5y
12
!
!
The last expression is theright side of the equation tobe proved, where .Thus, the equation is true for
.Therefore,
!
for all positive integers n.
36. 280a3b4
38. 5
40. $1
42. 36 mi
44. 0, $2
46. y ! $2(x " 1)2 " 4
48. 4
50. 4
52.
54. y ! $2x $ 2
56. 60
58. 840
60. 6
62. 56
16B$1 5$2 4
R
n(3n " 5)2
" (3n " 1)
4 " 7 " 10 " pn ! k " 1
n ! k " 1
(k " 1)[3(k " 1) " 5]2
(k " 1)(3k " 8)2
©Glencoe/McGraw-Hill 318 Algebra 2 Chapter 12
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1. Sample answer: There aresix people in a contest. Howmany ways can the first,second, and third prizes beawarded?
3. Sometimes; the statement isonly true when r ! 1.
5. 120
7. 6
9. permutation; 5040
11. 84
13. 915. 665,280
17. 70
19. 210
21. 1260
23. combination; 28
25. permutation; 120
27. permutation; 3360
29. combination; 455
31. 60
33. 111,540
35. 80,089,128
©Glencoe/McGraw-Hill 319 Algebra 2 Chapter 12
Lesson 12-2 Permutations and CombinationsPages 641–643
2.
4. 60
6. 6
8. combination; 15
10. permutation; 90,720
12. 56
14. 252016. 10
18. 792
20. 27,720
22. permutation; 5040
24. permutation; 2520
26. combination; 220
28. combination; 45
30. 11,880
32. 75,287,520
34. 267,696
36. 528
! C(n, r )
!n!
(n $ r )!r !
!n!
r !(n $ r )!
!n !
[n $ (n $ r )]!(n $ r )!
C (n, n $ r)
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37.
39. D
41. 24
43. 120
45. 80
47. Sample answer: n ! 2
49. x % 0.8047
51. 20 days
53.
55. $4; 128
57. {$2, 5}
59.
61. 425
822
(y $ 4)2
9"
(x $ 4)2
4! 1
! C (n, r )
!n!
(n $ r )!r !
!(n $ 1)!n(n $ r )!r !
!(n $ 1)!(n $ r " r )
(n $ r )!r !
!(n $ 1)!(n $ r )
(n $ r )!r !"
(n $ 1)!r(n $ r )!r !
(n $ 1)!(n $ r )!(r $ 1)!
!rr
!(n $ 1)!
(n $ r $ 1)!r !!
n $ rn $ r
"
(n $ 1)!(n $ r )! (r $ 1)!
!(n $ 1)!
(n $ r $ 1 )!r !"
(n $ 1)![n $ 1 $ (r $ 1)] !(r $ 1)!
!(n $ 1)!
(n $ 1 $ r )!r !"
C (n $ 1,r ) "C (n $ 1, r $ 1) 38. Permutations andcombinations can be used tofind the number of differentlineups. Answers shouldinclude the following.• There are 9! different
9-person lineups available:9 choices for the firstplayer, 8 choices for thesecond player, 7 for thethird player, and so on.So, there are 362,880different lineups.
• There are C(16, 9) ways tochoose 9 players from
16: C (16, 9) ! or 11,440.
40. A
42. 6
44. 8
46. Sample answer: n ! 3
48. $1.0986
50. 21.0855
52.
54.
56. {$4, 4}
58.
60.
62. ($1, 3)
0x3 0y323
e$3, 13f
$72;
532
x2
16"
y2
9! 1
16!7!9!
©Glencoe/McGraw-Hill 320 Algebra 2 Chapter 12
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63. (0, 2)
65.
67. {$7, 15}
69.
71. 15
35
$67
64.
66. 0
68. &
70.
72. 13
12
$43
©Glencoe/McGraw-Hill 321 Algebra 2 Chapter 12
Lesson 12-3 ProbabilityPages 647–650
1. Sample answer: The eventJuly comes before June hasa probability of 0. The eventJune comes before July hasa probability of 1.
3. There are or 36possible outcomes for thetwo dice. Only 1 outcome, 1and 1, results in a sum of 2,
so P(2) ! There are 2outcomes, 1 and 2 as well as2 and 1, that result in a sumof 3, so or
5.
7. 8:1
9. 2:7
11.
13. 18
1011
27
118
.2
36P(3) !
136
.
6 ! 6
2.
4.
6.
8. 1:5
10.
12.
14. 38
27
611
47
17
35
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15.
17.
19.
21.
23.
25.
27.
29. 0
31. 0.007
33. 0.109
35. 3:5
37. 5:3
39. 1:4
41. 3:1
43.
45.
47.
49.
51. 2:23
53. 1:4
35
19
49
310
24115
6115
11115
2855
655
225
110
16.
18.
20.
22.
24.
26.
28.
30.
32. 0.623
34. 1:1
36. 11:1
38. 4:3
40. 4:7
42.
44.
46.
48.
50.
52. 1:999
54. 5401771
110
716
917
511
67
122,957,480
6115
132575
7115
14575
2155
150
2150
©Glencoe/McGraw-Hill 322 Algebra 2 Chapter 12
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55.
57.
59.
61.
63. Probability and odds aregood tools for assessing risk.Answers should include thefollowing.• P(struck by lightning) !
! so
odds ! 1:(750,000 $ 1) or1:749,999. P(surviving alightning strike) !
! , so odds !
3:(4 $ 3) or 3 :1.• In this case, success is
being struck by lightningor surviving the lightningstrike. Failure is not beingstruck by lightning or notsurviving the lightningstrike.
65. D
67. experimental; about 0.307
69. theoretical;
71. permutation; 1260
73. 16
75. direct variation
77. (4, 4)
117
34
ss " f
1750,000
,s
s " f
1120
920
920
120
56.
58.
60.
62.
64. C
66. theoretical;
68. experimental;
70. permutation; 120
72. combination; 35
74. 24
76. square root
78. (1, 3)
15
136
' $ 1'
920
120
920
©Glencoe/McGraw-Hill 323 Algebra 2 Chapter 12
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1. 24
3. 18,720
5. 56
7. combination; 20,358,520
9. 13102
2. 756
4. 1320
6. permutation; 40,320
8.
10. 8663
1221
©Glencoe/McGraw-Hill 324 Algebra 2 Chapter 12
Chapter 12Practice Quiz 1
Page 650
1. Sample answer: putting onyour socks, and then yourshoes
3. Mario; the probabilities ofrolling a 4 and rolling a 2 are
both
5.
7.
9. 14
4663
14
16.
2. P(A, B, C, and D) !
4.
6.
8.
10. 116
730
117
136
P(A) ! P(B) ! P(C) ! P(D)
Lesson 12-4 Multiplying ProbabilitiesPages 654–657
79.
81.
83. 920
14
635
80.
82. 221
314
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©Glencoe/McGraw-Hill 325 Algebra 2 Chapter 12
11. dependent;
13.
15.
17.
19.
21.
23.
25. 0
27.
29.
31. independent;
33. dependent;
35. dependent;81
2401
121
2581
215
215
1021
149
56
16
2536
112
21220
12. independent;
14.
16.
18.
20.
22.
24. 0
26.
28.
30. dependent;
32. independent;
34. independent;
36. 19
132
1684913
328
110
115
2549
142
16
136
136
136
R
B
Y
R
B
Y
R
B
Y
R
B
Y
First Spin Second Spin
P(R,B) ! " or 19
13
13
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38.
40.
42.
44. or about 0.96
46.
48. no
50. 21
52. D
11320
a 99100b4
1158,753,389,900
1635,013,559,600
13
©Glencoe/McGraw-Hill 326 Algebra 2 Chapter 12
37.
39.
41.
43.
45. about 4.87%
47. no
49. Sample answer: As thenumber of trials increases,the results become morereliable. However, youcannot be absolutely certainthat there are no blackmarbles in the bag withoutlooking at all of the marbles.
51. Probability can be used toanalyze the chances of aplayer making 0, 1, or 2 freethrows when he or she goesto the foul line to shoot 2free throws. Answers shouldinclude the following.
632720,825
191,160,054
19
First SpinBlue Yellow Red
Blue BB BY BR
SecondYellow YB YY YR
Spin
Red RB RY RR19
19
19
13
19
19
19
13
19
19
19
13
13
13
13
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• One of the decimals in thetable could be used as thevalue of p, the probabilitythat a player makes agiven free throw. Theprobability that a playermisses both free throws is
or The probability that aplayer makes both freethrows is or p2.Since the sum of theprobabilities of all thepossible outcomes is 1, theprobability that a playermakes exactly 1 of the 2free throws is
or
• The result of the first freethrow could affect theplayer’s confidence on thesecond free throw. Forexample, if the playermakes the first free throw,the probability of makingthe second free throwmight increase. Or, if theplayer misses the first freethrow, the probability ofmaking the second freethrow might decrease.
53. C
55.
57. 1440 ways
59. 36
3340
2p (1 $ p).1 $ (1 $ p)2 $ p2
p ! p
(1 $ p)2.(1 $ p)(1 $ p)
54.
56.
58. 6
60. x2 $ 4x " 2
1119
1204
©Glencoe/McGraw-Hill 327 Algebra 2 Chapter 12
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61.
63.
65. 153
67.
69. (1, 2)
71. ($2, 4)
73.
75.
77. 1512
1112
56
0b 0
y
xO
y ! x2 # 4
x, x $ 4 62.
64.
66.
68.
70. (13, 9)
72.
74.
76. 116
54
23
5a4 0b3 0$9
y
xO
y ! x2 # 3x
y
xO
y ! x2 $ x # 2
©Glencoe/McGraw-Hill 328 Algebra 2 Chapter 12
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1. Sample answer: mutuallyexclusive events: tossing acoin and rolling a die;inclusive events: drawing a 7and a diamond from astandard deck of cards
3. The events are not mutuallyexclusive, so the chance ofrain is less than 100%.
5.
7.
9.
11. inclusive;
13. 1
15.
17.
19.
21.
23. mutually exclusive;
25. inclusive;2134
79
38143
3143
35143
2542
413
23
12
13
2.
4.
6.
8.
10. mutually exclusive;
12.
14.
16.
18.
20.
22.
24. inclusive;
26. mutually exclusive;413
12
3239
84143
105143
3742
16
1316
213
56
13
13
AlgebraFrench
French and Algebra
150 300400
©Glencoe/McGraw-Hill 329 Algebra 2 Chapter 12
Lesson 12-5 Adding ProbabilitiesPages 660–663
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27.
29.
31.
33.
35.
37.
39.
41.
43.
45. 1727
35
11780
9130
1780
14
18
188663
55221
413
28.
30.
32.
34.
36.
38.
40.
42.
44.
46. 17162
53108
178
126
145156
34
18
63221
11221
23
©Glencoe/McGraw-Hill 330 Algebra 2 Chapter 12
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47. Subtracting P(A and B) fromeach side and adding P (A orB) to each side results in theequation P (A or B) ! P(A) "P(B) $ P(A and B). This isthe equation for theprobability of inclusiveevents. If A and B aremutually exclusive, then P (Aand B) ! 0, so the equationsimplifies to P(A or B) !P(A) " P(B), which is theequation for the probability ofmutually exclusive events.Therefore, the equation iscorrect in either case.
49. C
51.
53.
55. 4:1
57. 2:5
59. 254
61.
63. (x " 1)2(x $ 1)(x 2 " 1)
(#8, $10)
1216
1216
48. Probability can be used toestimate the percents ofpeople who do the samethings before going to bed.Answers should include thefollowing.• The events are inclusive
because some peoplebrush their teeth and settheir alarm. Also, you knowthat the events areinclusive because the sumof the percents is not100%.
• According to theinformation in the text andthe table, P (read book) !
and P (brush teeth) !
Since the events
are inclusive, P (read bookand brush teeth) ! P (readbook) " P (brush teeth) $P (read book and brush teeth) !
50. A
52.
54.
56. 1:8
58. 5:3
60. 24
62.
64. min:max: ($1.33, $3.81)
(0, $5);
(#12, #5)
18
125216
12002000
!59
100.
38100
"81
100$
81100
.
38100
©Glencoe/McGraw-Hill 331 Algebra 2 Chapter 12
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65. min:max:
67.
(1, 3), (3, 3), (3, 5);max: f(3, 5) ! 23;min:
69. direct variation
71. 35.4, 34, no mode, 72
73. 63.75, 65, 50 and 65, 30
75. 12.98, 12.9, no mode, 4.7
f (1, $1) ! $3
(1, $1),
y
xO
($1.58, 1.38)($0.42, 0.62); 66.
(0, 2), (2, 0), max: f(2, 0) ! 6;min:
68. d ! 12.79t
70. 323.4, 298, no mode, 143
72. 3.6, 3.45, 2.1, 3.6
74. 79.83, 89, 89, 57
f (0, $2) ! $2
(0, $2);
y
xO
©Glencoe/McGraw-Hill 332 Algebra 2 Chapter 12
Lesson 12-6 Statistical MeasuresPages 666–670
1. Sample answer:{10, 10, 10, 10, 10, 10}
3.
5. 8.2, 2.9
( ! B1na
n
i!1(xi $ x )2
2. Sample answer: The varianceof the set {0, 1} is 0.25 andthe standard deviation is 0.5.
4. 40, 6.3
6. 424.3, 20.6
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7. $7300.50, $5335.25
9. 2500, 50
11. 3.1, 1.7
13. 37,691.2, 194.1
15. 82.9, 9.1
17. 114.5, 105, 23
19. Mean; it is highest.
21. $1047.88, $1049.50, $695
23. Mean or median; they arenearly equal and are morerepresentative of the pricesthan the mode.
25. Mode; it is lowest.
27. 19.3
29. 19.5
31. 59.8, 7.7
33. 100%
8. The mean is morerepresentative for thesouthwest central statesbecause the data for thePacific states contains themost extreme value,$10,650.
10. 1.6, 1.3
12. 4.8, 2.2
14. 569.4, 23.9
16. 43.6, 6.6
18. The mean and median bothseem to represent the centerof the data.
20. Mode; it is lower and is whatmost employees make. Itreflects the mostrepresentative worker.
22. Mode; it is the leastexpensive price.
24. 2,290,403; 2,150,000;2,000,000
26. Mean; it is highest.
28. 28.9
30. Washington; see students’work.
32. 64%
34. Different scales are used onthe vertical axes.
©Glencoe/McGraw-Hill 333 Algebra 2 Chapter 12
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35. Sample answer: The firstgraph might be used by asales manager to show asalesperson that he or shedoes not deserve a big raise.It appears that sales aresteady but not increasing fastenough to warrant a bigraise.
37. A: 2.5, 2.5, 0.7, 0.8; B: 2.5,2.5, 1.1, 1.0
39. The statistic(s) that bestrepresent a set of test scoresdepends on the distribution ofthe particular set of scores.Answers should include thefollowing.• mean, 73.9; median, 76.5;
mode, 94• The mode is not
representative at allbecause it is the highestscore. The median is morerepresentative than themean because it isinfluenced less than themean by the two very lowscores of 34 and 19.
41. D
43. 1.9
36. Sample answer: The secondgraph might be shown by thecompany owner to aprospective buyer of thecompany. It looks like thereis a dramatic rise in sales.
38. The first histogram is lowerin the middle and higher onthe ends, so it representsdata that are more spreadout. Since set B has thegreater standard deviation,set B corresponds to the firsthistogram and set Acorresponds to the second.
40. A
42. 3
44. The mean deviations wouldbe greater for the greaterstandard deviation and lowerfor the groups of data thathave the smaller standarddeviation.
©Glencoe/McGraw-Hill 334 Algebra 2 Chapter 12
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45. inclusive;
47.
49.
51.
53. 17
55. 12 cm3
57. (1, 5)
59. 136
61. 380
63. 396
10, #92; 10, #21062; #95
13204
1169
413
46. mutually exclusive;
48.
50.
52.
54.
56.
58. (3, 5)
60. 340
62. 475
64. 495
1$4, 62$2
$3
116
4663
37
©Glencoe/McGraw-Hill 335 Algebra 2 Chapter 12
Chapter 12Practice Quiz 2
Page 670
1.
3.
5.
7.
9. 23.6, 4.9
34
16
29
320
2.
4.
6.
8. 6.6, 2.6
10. 134.0, 11.6
23
14
16
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1. Sample answer:
the use of cassettes sinceCDs were introduced
3. Since 99% of the data iswithin 3 standard deviationsof the mean, 1% of the datais more than 3 standarddeviations from the mean. Bysymmetry, half of this, or0.5%, is more than 3 standarddeviations above the mean.
5. 68%
7. 95%
9. 250
11. 81.5%
13. normally distributed
15. 68%
17. 0.5%
19. 50%
21. 95%
23. 815
25. 16%
27. The mean would increase by25; the standard deviationwould not change; and the
2. The mean of the threegraphs is the same, but thestandard deviations aredifferent. The first graph hasthe least standard deviation,the standard deviation of themiddle graph is slightlygreater, and the standarddeviation of the last graph isgreatest.
4. normally distributed
6. 13.5%
8. 6800
10. 1600
12. positively skewed
14. Negatively skewed; thehistogram is high at the rightand has a tail to the left.
16. 34%
18. 16%
20. 50%
22. 50%
24. 25
26. 652
28. If a large enough group ofathletes is studied, some ofthe characteristics may be
©Glencoe/McGraw-Hill 336 Algebra 2 Chapter 12
Lesson 12-7 The Normal DistributionPages 673–675
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graph would be translated 25units to the right.
29. A
31. 17.5, 4.2
33.
35.
37.
39.
41. 0.76 h
43. 56c 5d
3
14, 1
$3, 2, 4
413
213
normally distributed; othersmay have skeweddistributions. Answers shouldinclude the following.•
• Since the histogram has twopeaks, the data may not benormally distributed. Thismay be due to players whoplay certain positionstending to be of similar largesizes while players who playthe other positions tend tobe of similar smaller sizes.
30. D
32. 42.5, 6.5
34.
36.
38.
40.
about 45 min
42.
44. 126x 5y
4
21a 5b
2
y
tO
y ! 216t2 # 53
#2 #1
#50
50
#100
1 2
1, $1
$5, 0, 1
413
70 71 72 73 74 75 76 77 78 79 80696867
4
6
2
0
8
Freq
uenc
y
10
Height (in.)
©Glencoe/McGraw-Hill 337 Algebra 2 Chapter 12
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1. Sample answer: In a 5-cardhand, what is the probabilitythat at least 2 cards arehearts?
3a. Each trial has more than twopossible outcomes.
3b. The number of trials is notfixed.
3c. The trials are notindependent.
5.
7.
9.
11. about 0.37
13.
15.
17.
19.
21.
23.
25. 135512
11024
23648
1253888
1116
14
116
27,64828,561
128,561
18
2. RRRWW, RRWRW,RRWWR, RWRRW,RWRWR, RWWRR,WRRRW, WRRWR,WRWRR, WWRRR
4.
6.
8.
10. about 0.05
12.
14.
16.
18.
20.
22.
24.
26. 459512
151024
2431024
625648
31257776
516
38
116
4828,561
78
38
©Glencoe/McGraw-Hill 338 Algebra 2 Chapter 12
Lesson 12-8 Binomial ExperimentsPages 678–680
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27.
29.
31.
33. about 0.44
35. about 0.32
37.
39. Getting a right answer and awrong answer are theoutcomes of a binomialexperiment. The probability isfar greater that guessing willresult in a low grade than ina high grade. Answersshould include the following.• Use
andthe chart on page 676 todetermine the probabilitiesof each combination ofright and wrong.
• P(5 right): r 5 ! P
or about 0.098%; P (4 right,
1 wrong): or about
1.5%; P (3 right, 2 wrong):
10r 2w 3 ! 10
or about 8.8%; P (3 wrong,2 right): 10r 2w 3 !
10 ! or about
26.4%; P (4 wrong, 1 right):
135512
a14b2 a3
4b3
a14b3 a3
4b2 !
45512
151024
a14b5
!
11024
10r 2w
3 " 5rw 4 " w
5r
5 " 5r 4w " 10r
3w 2 "
(r " w)5 !
14
319512
105512
53512
28.
30.
32. about 0.02
34.
36.
38.
40. 2
C (n, m)pm(1 $ p)n$m
164
332
,1564
,516
,1564
,332
,164
,
5602187
319512
105512
©Glencoe/McGraw-Hill 339 Algebra 2 Chapter 12
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5rw 4 ! 5 or
about 39.6%; P (5 wrong):
w 5 ! ! or about
23.7%.
41. B
43. normal distribution
45. 10
47. Mean; it is highest.
49.
51. 0.1
53. 0.039
55. 0.041
y
xO
x $ y ! 4
2431024
a34b5
a14b a3
4b4 !
4051024
42. See students’ work.
44. 68%
46. 16%
48.
50.
52. 0.05
54. 0.027
56. 0.031
y
xO
y ! |5x |
y
xO
x ! #3
©Glencoe/McGraw-Hill 340 Algebra 2 Chapter 12
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1. Sample answer: If a sampleis not random, the results ofa survey may not be valid.
3. The margin of sampling errordecreases when the size ofthe sample n increases. As nincreases, decreases.
5. No; these students probablystudy more than average.
7. about 4%
9. The probability is 0.95 thatthe percent of Americansages 12 and older who listento the radio every day isbetween 72% and 82%.
11. No; you would tend to pointtoward the middle of thepage.
13. Yes; a wide variety of peoplewould be called since almosteveryone has a phone.
15. about 8%
17. about 4%
19. about 3%
21. about 4%
23. about 3%
p (1 $ p)n
2. Sample answer for goodsample: doing a randomtelephone poll to rate themayor’s performance; sampleanswer for bad sample:conducting a survey on howmuch the average personreads at a bookstore
4. Yes; last digits of socialsecurity numbers arerandom.
6. about 9%
8. about 4%
10. about 283
12. Yes; all seniors would havethe same chance of beingselected.
14. No; freshmen are more likelythan older students to be stillgrowing, so a sample offreshmen would not giverepresentative heights for thewhole school.
16. about 4%
18. about 3%
20. about 2%
22. about 2%
24. about 2%
©Glencoe/McGraw-Hill 341 Algebra 2 Chapter 12
Lesson 12-9 Sampling and ErrorPages 683–685
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25. about 2%
27. about 983
29. A political candidate can usethe statistics from an opinionpoll to analyze his or herstanding and to help plan therest of the campaign.Answers should include thefollowing.• The candidate could
decide to skip areas wherehe or she is way ahead orway behind, andconcentrate on areaswhere the polls indicatethe race is close.
• about 3.5%• The margin of error
indicates that with aprobability of 0.95 thepercent of the Floridapopulation that favoredBush was between 43.5%and 50.5%. The margin oferror for Gore was alsoabout 3.5%, so withprobability 0.95 the percentthat favored Gore wasbetween 40.5% and 47.5%.Therefore, it was possiblethat the percent of theFlorida population thatfavored Bush was less thanthe percent that favoredGore.
31. C
33.
35. 95%
37. 97.5%
532
26. about 3%
28. 36 or 64
30. A
32.
34.
36. 210
38. x $ 2, x $ 3
12
132
©Glencoe/McGraw-Hill 342 Algebra 2 Chapter 12
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Glencoe/McGraw-Hill 343 Algebra 2 Chapter 13
1. Trigonometry is the study ofthe relationships between theangles and sides of a righttriangle.
3. Given only the measures ofthe angles of a right triangle,you cannot find themeasures of its sides.
5. ;
7.
9.
11.
13. 1660 ft
15. ;
cot ! !2105
4sec ! !
112105105
;
csc ! ! 114
; tan ! !42105
105;
sin ! !411
; cos ! !2105
11
a ! 16.6, A ! 67", B ! 23"
B ! 45", a ! 6, c ! 8.5
cos 23" !32x
; x ! 34.8
cot ! !6285
85sec ! !
116
;
tan ! !285
6; csc !!
1128585
;
sin ! !28511
; cos ! !611
Chapter 13 TrigonometryLesson 13-1 Right Triangle Trigonometry
Pages 706–708
2.
4. ;
;
;
6. ;
;
;
8.
10.
12.
14. B
16.
sec ! !54; cot ! !
43
tan ! !34; csc ! !
53;
sin ! !35; cos ! !
45;
c ! 19.1, A ! 47", B ! 43"
A ! 34", a ! 8.9, b ! 13.3
tan x " !1521
; x ! 36
cot ! !211
5sec ! !
621111
csc ! !65
tan ! !5211
11;
sin ! !56; cos ! !
2116
cot ! !158
sec ! !1715
csc ! !178
tan ! !815
;
sin ! !817
; cos ! !1517
adjacent
opposite
hypotenuse!
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17. ;
19. ;
;
21.
23.
25.
27a. sine ratio
Replace opp withx and hyp with2x.
Simplify.
27b. cosine ratio
Replace adjwith andhyp with 2x.
Simplify.cos 30" !232
13xcos 30" !23x2x
cos 30" !adjhyp
sin 30" !12
sin 30" !x
2x
sin 30" !opphyp
cos x " !1536
, x ! 65
sin 54" !17.8
x, x ! 22.0
tan 30" !x
10, x ! 5.8
cot ! ! 2sec ! !252
tan ! !12; csc ! ! 25;
sin ! !255
; cos ! !225
5
cot ! !327
7sec ! !
43;
csc ! !427
7;tan ! !
273
;
sin ! !274
; cos ! !34
18.
;
;
;
20.
;
22.
24.
26.
28a. sine ratio
Replace oppwith x and hypwith
Simplify.
Rationalize thedenominator.
28b. cosine ratio
Replace adjwith x and hypwith
Simplify.
Rationalize thedenominator.cos 45" !
122
cos 45" !112
12x.cos 45" !
x12x
cos 45" !adjhyp
sin 45" !122
sin 45" !112
12x.
sin 45" !x12x
sin 45" !opphyp
sin x " !1622
, x ! 47
tan 17.5" !x
23.7; x ! 7.5
cos 60" !3x, x ! 6
sec ! !87; cot ! !
721515
csc ! !8215
15tan ! !
2157
;
sin ! !215
8; cos ! !
78;
cot ! !59
sec ! !2106
5
csc ! !2106
9
tan ! !95;cos ! !
52106106
sin ! !92106
106;
©Glencoe/McGraw-Hill 344 Algebra 2 Chapter 13
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27c. sine ratio
Replace oppwith andhyp with 2x.
Simplify.
29.
31.
33.
35.
37.
39. ,
41. about 300 ft
43. about
45. 93.54 units2
47. The sine and cosine ratios ofacute angles of righttriangles each have thelongest measure of thetriangle, the hypotenuse, astheir denominator. A fractionwhose denominator isgreater than its numerator isless than 1. The tangent ratioof an acute angle of a righttriangle does not involve themeasure of the hypotenuse,
If the measure of theopposite side is greater thanthe measure of the adjacentside, the tangent ratio isgreater than 1. If the measureof the opposite side is lessthan the measure of theadjacent side, the tangentratio is less than 1.
oppadj
.
6"
c ! 10.6A ! 49", B ! 41", a ! 8
A ! 63", B ! 27", a ! 11.5
A ! 72", b ! 1.3, c ! 4.1
A ! 60", a ! 19.1, c ! 22
B ! 56", b ! 14.8, c ! 17.9
B ! 74", a ! 3.9, b ! 13.5
sin 60" !232
13xsin 60" !23x2x
sin 60" !opphyp
28c. tangent ratio
Replace oppwith x and adjwith x.Simplify.
30.
32.
34.
36.
38.
40.
42. about 142.8 ft
44. about 3.2 in.
46. about 1.72 km high
48. When construction involvesright triangles, includingbuilding ramps, designingbuildings, or surveying landbefore building, trigonometryis likely to be used. Answersshould include the following.• If you view the ramp from
the side then the verticalrise is opposite the anglethat the ramp makes withthe horizontal. Similarly,the horizontal run is theadjacent side. So thetangent of the angle is theratio of the rise to the runor the slope of the ramp.
• Given the ratio of theslope of ramp, you canfind the angle of inclinationby calculating tan–1 of thisratio.
c ! 15b ! 14.1,A ! 19", B ! 71",
A ! 26", B ! 64", b ! 8.1
B ! 80", a ! 2.6, c ! 15.2
B ! 45", a ! 7, b ! 7
A ! 75", a ! 24.1, b ! 6.5
A ! 63", a ! 13.7, c ! 15.4
tan 45" ! 1
tan 45" !xx
tan 45" !oppadj
©Glencoe/McGraw-Hill 345 Algebra 2 Chapter 13
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49. C
51. No; band members may bemore likely to like the samekinds of music.
53.
55.
57. {# }
59. 20 qt
61. 12 m2
2, #1, 0, 1, 2
1516
38
50. 7.7
52. Yes; this sample is randomsince different kinds ofpeople go to the post office.
54.
56. { }
58. {121}
60. 35,904 ft
62. 48 L
$222, 2i 22
116
©Glencoe/McGraw-Hill 346 Algebra 2 Chapter 13
Lesson 13-2 Angles and Angle MeasurePages 712–715
1. reals
3.
5.
O
y
x
300˚
O
y
x
!70˚
290˚
2. In a circle of radius r units,one radian is the measure ofan angle whose raysintercept an arc length of runits.
4.
6.
O
y
x
570˚
O
y
x
70˚
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©Glencoe/McGraw-Hill 347 Algebra 2 Chapter 13
7.
9.
11. 135"
13.
15.
17. 21 h
19.
21.
23.
O
y
x
!150˚
O
y
x790˚
O
O
y
x
235˚
785", #295"
1140"
#%
18
O
y
x!45˚
8.
10.
12.
14.
16.
18. 2 h
20.
22.
24.
O
y
x!50˚
O
y
x380˚
O
y
x
270˚
7%
3, #
5%
3
420", #300"
#30"
97%
36
13%
18
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25.
27.
29.
31.
33.
35. 150"
37.
39. 1305"
41.
43. Sample answer:
45. Sample answer:
47. Sample answer:
49. Sample answer:
51. Sample answer:
53. Sample answer:
55. 2689" per second; 47 radiansper second
57. about 188.5 m2
59. about 640.88 in2
13%
2, #
3%
2
3%
4, #
13%
4
11%
4, #
5%
4
8", #352"
345", #375"
585", #135"
1620%
! 515.7"
#45"
79%
90
11%
3
#%
12
2%
3
O
y
x
"
26.
28.
30.
32.
34.
36. 495"
38.
40. 510"
42.
44. Sample answer:
46. Sample answer:
48. Sample answer:
50. Sample answer:
52. Sample answer:
54. Sample answer:
56. 209.4 in2
58. number 17
60a.60b.60c. b2 & (#a)2 ! a2 & b2 ! 1
b2 & a2 ! a2 & b2 ! 1a2 & (#b)2 ! a2 & b2 ! 1
25%
4, #
7%
4
4%
3, #
8%
3
19%
6, #
5%
6
400", #320"
220", #500"
390", #330"
540%
! 171.9"
#60"
13%
9
19%
6
#5%
4
%
3
O
y
x2"3!
©Glencoe/McGraw-Hill 348 Algebra 2 Chapter 13
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61. Student answers shouldinclude the following.• An angle with a measure
of more than 180" givesan indication of motion ina circular path that endedat a point more thanhalfway around the circlefrom where it started.
• Negative angles convey thesame meaning as positiveangles, but in an oppositedirection. The standardconvention is that negativeangles represent rotationsin a clockwise direction.
• Rates over 360" perminute indicate that anobject is rotating orrevolving more than onerevolution per minute.
63. D
65.
67.
69. about 7.07%
71. combination, 35
73.
75. 1418.2 or about 1418; thenumber of sports radiostations in 2008
77.
79.
81. 2104
2102
3255
[h ! g](x) ! 8x2 & 34x & 44[g ! h](x) ! 4x2 # 6x & 23,
c ! 0.8, A ! 30", B ! 60"
A ! 22", a ! 5.9, c ! 15.9
62. C
64.
66.
68. about 8.98%
70. permutation, 17,100,720
72.
74. 1041.8
76.
78.
80. 2142
2263
2233
[h ! g](x) ! 6x # 4[g ! h](x) ! 6x # 8,
A ! 35", a ! 9.2, b ! 13.1
a ! 3.4, c ! 6.0, B ! 56"
©Glencoe/McGraw-Hill 349 Algebra 2 Chapter 13
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©Glencoe/McGraw-Hill 350 Algebra 2 Chapter 13
Chapter 13Practice Quiz 1
Page 715
1.
3.
5.
7. 210"
9. 305"; #415"
19%
18
O
y
x!60˚
A ! 42", a ! 13.3, c ! 17.9 2.
4.
tan ! !
6.
8.
10. 5%
3; #
%
3
#396"
5%
2
sec ! !2149
7; cot ! !
710
107
; csc ! !2149
10;
cos ! !72149
149;
sin ! !102149
149;
A ! 59", B ! 31", b ! 10.8
2. Sample answer: 190"
4.
cot ! ! #158
sec ! ! #1715
,
csc ! !178
,tan ! ! #815
,
sin ! !817
, cos ! ! #1517
,
Lesson 13-3 Trigonometric Functions of General Angles
Pages 722–724
1. False; sec or 1 and
or 0.
3. To find the value of atrigonometric function of !,where ! is greater than 90",find the value of thetrigonometric function for !',then use the quadrant inwhich the terminal side of !lies to determine the sign ofthe trigonometric functionvalue of !.
0r
tan 0" !
rr
0" !
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5.
7. 55"
9. 60"
11.
13.
15.
17.
sec ! !257
, cot ! !724
tan ! !247
, csc ! !2524
,
sin ! !2425
, cos ! !725
,
sec ! ! 23
cos ! ! #262
,tan ! ! #22,
sin ! ! #263
, cos ! !233
,
#223
3
#1
O
y
x
!240˚
!'
O
y
x
235˚
!'
sec ! ! #1, cot ! ! undefinedtan ! ! 0, csc ! ! undefined,sin ! ! 0, cos ! ! #1, 6.
8.
10.
12.
14.
16. about 12.4 ft
18.
sin ! ! 255
, cos ! ! 2,
csc ! ! 25,tan ! !12,
sin ! ! 255
, cos ! ! 225
5,
cot ! ! #233
sec ! ! #2,csc ! !223
3,
sin ! !232
, tan ! ! #23,
#23
#232
O
y
x
7"4
!'
%
4
cot ! ! 1sec ! ! 22,
csc ! ! 22,tan ! ! 1,
sin ! !222
, cos ! !222
,
©Glencoe/McGraw-Hill 351 Algebra 2 Chapter 13
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19.
21. sin ! ! #1, cos ! ! 0,tan ! ! undefined, csc ! ! #1,sec ! ! undefined, cot ! ! 0
23.
tan ! ! #1, cot ! ! #1
25. 45"
27. 30"
O
y
x
!210˚
!'
O
y
x
315˚
!'
sec ! ! 22,csc ! ! #22,
sin ! ! #222
, cos ! !222
,
sec ! !289
5, cot ! ! #
58
tan ! ! #85, csc ! ! #
2898
,
cos ! ! 5289
89,
cos ! ! 8289
89, 20.
22. sin ! ! 0, cos ! ! #1,tan ! ! 0, csc ! ! undefined,sec ! ! #1,cot ! ! undefined
24.
26. 60"
28. 55"
O
y
x
!125˚!'
O
y
x
240˚
!'
sec ! ! #23, cot ! ! 222
tan ! ! 22, csc ! ! #262
,
sin ! ! #263
, cos ! ! #233
,
sec ! !54, cot ! ! #
43
tan ! ! #34, csc ! ! #
53,
sin ! ! #35, cos ! !
45,
©Glencoe/McGraw-Hill 352 Algebra 2 Chapter 13
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29.
31.
33.
35.
37. undefined
39.
41. undefined
43.
45. 0.2, 0, 0, 0.2, 0, andor about 11.5", 0",
0", 11.5", 0", and
47.
cot ! ! #34
csc ! ! #54, sec ! !
53,
sin ! ! #45, tan ! ! #
43,
#11.5"#11.5",#0.2;
#0.2,
232
23
#23
#232
O
y
x
13"7
!'
%
7
O
y
x
5"4
!'
%
430.
32.
34.
36.
38.
40.
42. 2
44.
46. 6092.5 ft
48.
cot ! ! #5
sec ! ! #226
5,csc ! ! 226,
#5226
26,cos ! !sin ! !
22626
,
#1
222
12
#23
#2
O
y
x2"3!!'
%
3
O
y
x
5"6
!'
%
6
©Glencoe/McGraw-Hill 353 Algebra 2 Chapter 13
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50.
52.
54. or 90" yields thegreatest value for sin 2!.
56. I, II
58. III
60. C
45"; 2 ( 45"
cot ! ! #226
tan ! ! #2612
, sec ! !526
12,
sin ! ! #15, cos ! !
2265
,
sec ! ! #25csc ! ! #252
,
tan ! ! 2,cos ! ! #255
,
sin ! ! #225
5,49.
51.
53. about 173.2 ft
55. 9 meters
57. II
59. Answers should include thefollowing.• The cosine of any angle is
defined as where x is the x-coordinate of anypoint on the terminal rayof the angle and r is thedistance from the origin tothat point. This means thatfor angles with terminalsides to the left of the y-axis, the cosine isnegative, and those withterminal sides to the rightof the y-axis, the cosine ispositive. Therefore, thecosine function can beused to model real-worlddata that oscillate betweenbeing positive andnegative.
xr,
csc ! ! #210
3, cot ! !
13
cos ! ! #21010
, tan ! ! 3,
sin ! ! #3210
10,
cot ! ! #222
#322
4,csc ! ! 3, sec ! !
#224
,cos ! ! #222
3, tan ! !
©Glencoe/McGraw-Hill 354 Algebra 2 Chapter 13
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• If we knew the length ofthe cable we could find thevertical distance from thetop of the tower to the rider.Then if we knew the heightof the tower we couldsubtract from it the verticaldistance calculatedpreviously. This will leavethe height of the rider fromthe ground.
61.
63. 300"
65.
67.
69. (7, 2)
71.
73. 15.1
75. 32.9"
77. 39.6"
(5, #4)
sin x " !513
, 23
sin 28" !x
12, 5.6
a52, #
5232b 62.
64.
66.
68. 635
70.
72. 4.7
74. 2.7
76. 20.6"
(#4, 3)
cos 43" !x
83, 60.7
900"
%! 286.5"
%
2
©Glencoe/McGraw-Hill 355 Algebra 2 Chapter 13
Lesson 13-4 Law of SinesPages 729–732
2. Sample answer:
A
C
B
3.2 cm2.6 cm
0.9 cm
A
C
B
3.2 cm 2.6 cm
3.9 cm
b ! 3.2 cma ! 2.6 cm,A ! 42",1. Sometimes; only when A is
acute, a ! b sin A, or and when A is obtuse, a ) b.
a ) b
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©Glencoe/McGraw-Hill 356 Algebra 2 Chapter 13
3. Gabe; the information givenis of two sides and an angle,but the angle is not betweenthe two sides, therefore thearea formula involving sinecannot be used.
5. 6.4 cm2
7.
9. no solution
11.
13. 5.5 m
15. 19.5 yd2
17. 62.4 cm2
19. 14.6 mi2
21.
23.
25.
27.
29.
31. no
33.
35.
37.
c ! 28.3c ! 84.9; B ! 115", C ! 18",two; B ! 65", C ! 68",
c ! 0.8c ! 2.4; B ! 95", C ! 5",two; B ! 85", C ! 15",
c ! 25.8one; B ! 18", C ! 101",
c ! 1.8C ! 45",one; B ! 36",
A ! 20", a ! 22.1, c ! 39.8
A ! 40", B ! 65", b ! 2.8
B ! 47", C ! 68", c ! 5.1
C ! 73", a ! 55.6, b ! 48.2
c ! 12.0one; B ! 24", C ! 101",
B ! 80", a ! 32.0, b ! 32.6
4. 57.5 in2
6.
8.
10.
12.
14. 43.1 m2
16. 572.8 ft2
18. 4.2 m2
20.
22.
24.
26.
28. no
30.
32.
34.
36.
38. 4.6 and 8.5 mi
c ! 14.1one; B ! 23", C ! 129",
C ! 4", c ! 16.8c ! 229.3; B ! 124",two; B ! 56", C ! 72",
c ! 24.2one; B ! 90", C ! 60",
c ! 2.3c ! 3.5; B ! 108", C ! 39",two; B ! 72", C ! 75",
C ! 67", B ! 63", b ! 2.9
C ! 97", a ! 5.5, b ! 14.4
B ! 21", C ! 37", b ! 13.1
B ! 101", c ! 3.0, b ! 3.4
c ! 8.7one; B ! 19", C ! 16",
c ! 1.2c ! 5.7; B ! 138", C ! 12",two; B ! 42", C ! 108",
B ! 20", A ! 20", a ! 20.2
C ! 30", a ! 2.9, c ! 1.5
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39. 7.5 mi from Ranger B,10.9 mi from Ranger A
41. 107 mph
43. Answers should include thefollowing.• If the height of the triangle
is not given, but themeasure of two sides andtheir included angle aregiven, then the formula forthe area of a triangleusing the sine functionshould be used.
• You might use this formulato find the area of atriangular piece of land,since it might be easier tomeasure two sides anduse surveying equipmentto measure the includedangle than to measure theperpendicular distancefrom one vertex to itsopposite side.
• The area of " is
sin B ! or
Area ! ah or
Area ! a (c sin B)12
12
h ! c sin Bhc
A
C
B
a
h
b
c
ah.12
ABC
40. 690 ft
42a.42b. or 42c.
44. D
b * 14.63b + 20b ! 14.63
14.63 * b * 20
©Glencoe/McGraw-Hill 357 Algebra 2 Chapter 13
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45.
47.
49.
51.
53.
55. 5.6
57. 39.4"
55221
17%
6, #
7%
6
660", #60"
233
B ! 78", a ! 50.1, c ! 56.1 46.
48.
50.
52.
54. 780 ft
56. 7.8
58. 136.0"
368
407", #313"
22
232
©Glencoe/McGraw-Hill 358 Algebra 2 Chapter 13
Lesson 13-5 Law of CosinesPages 735–738
1. Mateo; the angle given is notbetween the two sides,therefore the Law of Sinesshould be used.
2a. Use the Law of Cosines tofind the measure of oneangle. Then use the Law ofSines or the Law of Cosinesto find the measure of asecond angle. Finally,subtract the sum of thesetwo angles from 180" to findthe measure of the thirdangle.
2b. Use the Law of Cosines tofind the measure of the thirdside. Then use the Law ofSines or the Law of Cosinesto find the measure of asecond angle. Finally,subtract the sum of thesetwo angles from 180" to findthe measure of the thirdangle.
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3. Sample answer:
5.
7.
9.
11.
13.
15.
17.
19.
21. no
23.
25.
27.
29. about 100.1"
a ! 4.5cosines; B ! 82", C ! 58",
C ! 31"cosines; A ! 24", B ! 125",
b ! 21.0cosines; A ! 52", C ! 109",
c ! 31.6sines; C ! 77", b ! 31.7,
C ! 40"cosines; A ! 30", B ! 110",
c ! 5.4sines; A ! 80", a ! 10.9,
b ! 21.0sines; B ! 102", C ! 44",
C ! 70"cosines; A ! 48", B ! 62",
94.3"
C ! 90"B ! 67",cosines; A ! 23",
b ! 14sines; B ! 70", a ! 9.6,
15
13
9
4.
6.
8. 19.5 m
10.
12.
14.
16.
18. no
20.
22.
24.
26.
28. about 159.7"
30. Since the step angle for thecarnivore is closer to 180", itappears as though thecarnivore made more forwardprogress with each step thanthe herbivore did.
c ! 13.8cosines; A ! 107", B ! 35",
c ! 14.4sines; C ! 102", b ! 5.5,
C ! 34"cosines; A ! 15", B ! 131",
C ! 28"cosines; A ! 103", B ! 49",
b ! 17.9cosines; A ! 55", C ! 78",
c ! 11.5cosines; A ! 56.8", B ! 82",
C ! 59.6cosines; A ! 46", B ! 74",
c ! 11.2sines; A ! 60", b ! 14.3,
c ! 92.5sines; C ! 101", B ! 37",
c ! 6.5cosines; A ! 76", B ! 69",
©Glencoe/McGraw-Hill 359 Algebra 2 Chapter 13
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31. 4.4 cm, 9.0 cm
33. 91.6"
35. Answers should include thefollowing.• The Law of Cosines can
be used when you knowall three sides of a triangleor when you know twosides and the includedangle. It can even be usedwith two sides and thenonincluded angle. Thisset of conditions leaves aquadratic equation to besolved. It may have one,two, or no solution just likethe SSA case with theLaw of Sines.
• Given the latitude of apoint on the surface ofEarth, you can use theradius of the Earth and theorbiting height of a satellitein geosynchronous orbit tocreate a triangle. Thistriangle will have twoknown sides and themeasure of the includedangle. Find the third sideusing the Law of Cosinesand then use the Law ofSines to determine theangles of the triangle.Subtract 90 degrees fromthe angle with its vertex onEarth’s surface to find theangle at which to aim thereceiver dish.
37. A
32. about 1362 ft; about 81,919 ft2
34. Since
becomes
36. B
38. 100.0"
a2 ! b2 & c2.a2 ! b2 & c2 # 2bc cos A
cos 90" ! 0,
©Glencoe/McGraw-Hill 360 Algebra 2 Chapter 13
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39. Sample answer: 100.2"
41.
43.
45.
47. { }
49.
51.
53. 19%
6, #
5%
6
540", #180"
405, #315"
x 0x ) #0.6931
sec ! !2210
5, cot ! !
2153
tan ! !215
5, csc ! !
2263
,
sin ! !264
, cos ! !210
4,
sec ! !135
, cot ! !512
tan ! !125
, csc ! !1312
sin ! !1213
, cos ! !513
,
c ! 9.6one; B ! 46", C ! 79",
40. By finding the measure ofangle C in one step usingthe Law of Cosines, only thegiven information was used.By finding this anglemeasure using the Law ofCosines and then the Law ofSines, a calculated value thatwas not exact wasintroduced; 100.0".
42. no solution
44.
46. 1.3863
48. 4.3891
50.
52.
54. 10%
3, #
2%
3
5%
2, #
3%
2
390", #330"
cot ! !47
sec ! !265
4,csc ! !
2657
,
tan ! !74,cos ! !
426565
,
sin ! !7265
65,
©Glencoe/McGraw-Hill 361 Algebra 2 Chapter 13
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1.
3. 27.7 m2
5. B ! 43"cosines; c ! 15.9, A ! 59",
cot ! ! #23
sec ! ! #213
2,
tan ! ! #32, csc ! !
2133
,
cos ! ! #2213
13,
sin ! !3213
13, 2.
4.
C ! 5"; c ! 3.5B ! 153";c ! 30.2;
C ! 131";two; B ! 27";
#223
3
1. The terminal side of theangle ! in standard positionmust intersect the unit circleat
3. Sample answer: The graphshave the same shape, butcross the x-axis at differentpoints.
5.
7. #12
sin ! !222
; cos ! !222
P (x, y).
2. Sample answer: the motionof the minute hand on aclock; 60 s
4.
6.
8. 720"
232
sin ! ! #1213
, cos ! !513
©Glencoe/McGraw-Hill 362 Algebra 2 Chapter 13
Chapter 13Practice Quiz 2
Page 738
Lesson 13-6 Circular FunctionsPages 742–745
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9. 2 s
11.
13.
15.
17.
19.
21. 1
23.
25.
27.
29. 6
31.
33. 1440
s
2%
#323
1 # 232
14
#1
#12
sin ! !232
; cos ! ! #12
sin ! !1517
; cos ! !817
sin ! !45; cos ! ! #
35
10.
12.
14.
16.
18.
20.
22.
24.
26.
28. 1
30. 9
32. 8
34.y
xO
!1
1
1440
1220
23
94
#222
232
232
sin ! ! 0.8; cos ! ! 0.6
sin ! ! #12; cos ! !
132
sin ! ! #513
; cos ! ! #1213
h
tO
!3
3
4321
©Glencoe/McGraw-Hill 363 Algebra 2 Chapter 13
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35.
37.
39.
41.
43. sine: D ! {all reals}, R ! { };cosine: D ! {all reals}, R ! { }#1 , y , 1
#1 , y , 1
23
#xy
yx
a12, #132b
(#1, 0), a#12, #132b,
a12, 132b, a#1
2, 132b, 36. The population is around 425
near the 60th day of the year.It rises to around 625 inMay/June. It falls to around 425again by August/September.It continues to drop to around225 in November/December.
38. tan !
40. #cot !
42.
44. Answers should include thefollowing.• Over the course of one
period both the sine andcosine function attain theirmaximum value once andtheir minimum value once.From the maximum to theminimum the functionsdecrease slowly at first,then decrease morequickly and return to aslow rate of change asthey come into theminimum. Similarly, thefunctions rise slowly fromtheir minimum. They beginto increase more rapidlyas they pass the halfwaypoint, and then begin torise more slowly as theyincrease into themaximum. Annualtemperature fluctuationsbehave in exactly thesame manner.
#233
©Glencoe/McGraw-Hill 364 Algebra 2 Chapter 13
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45. A
47.
49. 27.0 in2
51. 6800
53. 5000
55. 250
57. does not exist
59. 8
61.
63.
65. 110"
67. 80"
69. 89"
2y & 7 &5
y # 3
2x & 9
A ! 76"B ! 59",cosines; c ! 12.4,
• The maximum value of thesine function is 1 so themaximum temperaturewould be 50 & 25(1) or 75" F. Similarly, theminimum value would be
or 25" F. Theaverage temperature overthis time period occurswhen the sine functiontakes on a value of 0. Inthis case that would be 50" F.
46.
48.
50. 12.5 m2
52. 9500
54. 5000
56. 50
58.
60.
62.
64. 20"
66. 73"
68. 56"
5y2 # 4y & 4 #11
y & 1
4x # 5
643
C ! 84"B ! 62",cosines; A ! 34",
233
50 & 25(#1)
©Glencoe/McGraw-Hill 365 Algebra 2 Chapter 13
PQ245-6457F-P13[343-368].qxd 7/31/02 9:46 AM Page 365 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-13:
1. Restricted domains aredenoted with a capital letter.
3. They are inverses of eachother.
5. " ! Arccos 0.5
7. 0"
9.
11. 0.75
13. 0.58
15. # ! Arcsin "
17. y ! Arccos x
19. Arccos y ! 45"
21. 60"
23. 45"
25. 45"
27. 2.09
29. 0.52
31. 0.5
33. 0.60
35. 0.8
37. 0.5
39.
41. 0.71
43. 0.96
#0.5
% ! 3.14
2. Sample answer:
4. ! ! Arctan x
6. 45"
8.
10. 0.22
12. 0.66
14. 30"
16. a ! Arctan b
18.
20. Arctan
22. 30"
24. 30"
26. 90"
28. does not exist
30. 0.52
32. 0.66
34. 0.5
36. 0.81
38. 3
40. 1.57
42. does not exist
44. 0.87
a#43b ! x
30" ! Arcsin 12
#%
6! #0.52
Cos#1 222
! 45"
Cos 45" !222
;
©Glencoe/McGraw-Hill 366 Algebra 2 Chapter 13
Lesson 13-7 Inverse Trigonometric FunctionsPages 749–751
PQ245-6457F-P13[343-368].qxd 7/31/02 9:46 AM Page 366 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-13:
45. 60" south of west
47. No; with this point on theterminal side of the throwingangle !, the measure of ! isfound by solving the equationtan ! ! . Thus ! ! tan#1
or about 43.4", which isgreater than the 40"requirement.
49. 31"
51. Suppose andlie on the line
Then The tangent of
the angle ! the line makeswith the positive x-axis is equal to the ratio or
Thus
53. 37"
y
xO
P (x , y1)
Q (x2, y2)
!
x2 ! x1
y2 ! y1
y # mxmx m $ b
tan ! ! m.y2 # y1
x2 # x1.
oppadj
y2 # y1
x2 # x1.
m !y ! mx & b.Q (x2, y2)
P (x1, y1)
1718
1718
46. 83"
48. 60"
50. 102"
52. Trigonometry is used todetermine proper bankingangles. Answers shouldinclude the following.• Knowing the velocity of the
cars to be traveling on aroad and the radius of thecurve to be built, then thebanking angle can bedetermined. First find theratio of the square of thevelocity to the product ofthe acceleration due togravity and the radius ofthe curve. Then determinethe angle that had thisratio as its tangent. Thiswill be the banking anglefor the turn.
• If the speed limit wereincreased and the bankingangle remained the same,then in order to maintain asafe road the curvaturewould have to bedecreased. That is, theradius of the curve wouldalso have to increase,which would make theroad less curved.
54. D
©Glencoe/McGraw-Hill 367 Algebra 2 Chapter 13
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56. forall values of x.
58.
60. 1
62.
64.
66. 2.5 s
#22, #57
C ! 90"B ! 77",cosines; A ! 13",
232
%
2Sin#1 x & Cos#1 x !55.
57. From a right triangleperspective, if an acute angle! has a given sine x, thenthe complementary angle
# ! has that same valueas its cosine. This can beverified by looking at a righttriangle. Therefore, the sumof the angle whose sine is xand the angle whose cosine is x should be
59.
61.or
63. 46, 39
65. 11, 109
C ! 3.9C ! 39",B ! 111",c ! 6.1
C ! 81",sines; B ! 69",
#1
%
2.
%
2
©Glencoe/McGraw-Hill 368 Algebra 2 Chapter 13
x 0 1
y%
2%
2%
2%
2%
2%
2%
2%
2%
2
#1#232
#222
#12
232
222
12
PQ245-6457F-P13[343-368].qxd 7/31/02 9:46 AM Page 368 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-13:
©Glencoe/McGraw-Hill 369 Algebra 2 Chapter 14
1. Sample answer: Amplitude ishalf the difference betweenthe maximum and minimumvalues of a graph;has no maximum orminimum value.
3. Jamile; The amplitude is 3and the period is 3!.
5. amplitude: 2; period: 360" or 2!
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " 2 sin !
y # tan !
Chapter 14 Trigonetmetric Graphs and IdentitiesLesson 14-1 Graphing Trigonometric Functions
Pages 766–768
2. Sample answer: The graphrepeats itself every 180".
4. amplitude: period 360" or 2!
6. amplitude: period 360" or 2!
90˚ 180˚ 270˚ 360˚
1
2
0.5
1.5
!1!1.5
!0.5
!2
y
!O
y " cos !23
23;
90˚!90˚ 180˚!180˚ 270˚!270˚
1
2
0.5
1.5
2.5
!1!1.5
!2!2.5
y
!O
y " sin !12
12;
PQ245-6457F-P14[369-410].qxd 7/24/02 2:03 PM Page 369 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
7. amplitude: does not exist;period: 180" or !
9. amplitude: 4; period: 180" or !
11. amplitude: does not exist;
period: 120" or
13. 12 months; Sample answer:The pattern in the populationwill repeat itself every 12months.
30˚!60˚ !30˚ 60˚ 90˚ 120˚ 150˚
1
2
0.5
1.5
!1!1.5
!2
y
!
O
y " sec 3!12
2!
3
90˚ 180˚ 270˚ 360˚
2
4
1
3
5
!2!1
!3!4!5
y
!O
y " 4 sin 2!
90˚!90˚ 180˚!180˚ 270˚!270˚
1
2
0.5
1.5
!1!1.5
!2
y
!O
y " tan !14
8. amplitude: does not exist;period: 180" or !
10. amplitude: 4; period: 480" or
12. amplitude: period:720" or 4!
14. 4250; June 1
90˚!90˚ 180˚ 270˚ 360˚ 450˚
0.5
1
0.25
0.75
1.25
!0.5!0.75
!1!1.25
y
!O
y " cos !12
34
34;
90˚ 180˚ 270˚ 360˚ 450˚
2
4
1
3
5
!2!1
!3!4!5
y
!O
y " 4 cos !34
8!
2
90˚!90˚ 180˚!180˚ 270˚!270˚
1
2
0.5
1.5
!1!1.5
!2
y
!O
y " csc 2!
©Glencoe/McGraw-Hill 370 Algebra 2 Chapter 14
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15. amplitude: 3; period: 360" or 2!
17. amplitude: does not exist;period: 360" or 2!
19. amplitude: period: 360" or 2!
90˚!90˚ 180˚!180˚ 270˚!270˚
0.4
0.8
0.2
0.6
1
!0.4!0.6!0.8
!1
y
!O
y " sin !15
15
;
90˚!180˚!270˚ !90˚ 180˚ 270˚
2
45
1
3
!2!3!4!5
y
!O
y " 2 csc !
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " 3 sin !
16. amplitude: 5; period: 360" or 2!
18. amplitude: does not exist;period: 180" or !
20. amplitude: does not exist;period: 360" or 2!
90˚!180˚!270˚ !90˚ 180˚ 270˚
4
810
2
6
!4!6!8
!10
y
!O
y " sec !13
90˚!90˚ 180˚!180˚ 270˚!270˚
4
810
2
6
!4!6!8
!10
y
!O
y " 2 tan !
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " 5 cos !
©Glencoe/McGraw-Hill 371 Algebra 2 Chapter 14
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21. amplitude: 1; period 90" or
23. amplitude: does not exist;
period: 120" or
25. amplitude: does not exist;period: 540" or 3!
270˚!270˚ 540˚!540˚ 810˚!810˚
4
810
2
6
!4!6!8
!10
y
!O
y " 4 tan !13
30˚!60˚ !30˚ 60˚
2
45
1
3
!2!3!4!5
y
!O
y " sec 3!
2!
3
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " sin 4!
!
2 22. amplitude: 1; period: 180" or !
24. amplitude: does not exist;
period: 36" or
26. amplitude: does not exist;period: 360" or 2!
!180˚ 360˚180˚!360˚!540˚ 540˚
4
810
2
6
!4!6!8
!10
y
!O
y " 2 cot !12
18˚!18˚ 36˚!36˚ 54˚ 72˚!54˚!72˚
2
45
1
3
!2!3!4!5
y
!O
y " cot 5!
!
5
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " sin 2!
©Glencoe/McGraw-Hill 372 Algebra 2 Chapter 14
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27. amplitude: 6; period: 540" or 3!
29. amplitude: does not exist;period: 720" or 4!
31. amplitude: does not exist;period: 180" or !
!90˚ 180˚90˚!180˚!270˚ 270˚
4
810
2
6
!4!6!8
!10
y
!O
2y " tan !
180˚!360˚!540˚ !180˚ 360˚ 540˚
4
810
2
6
!4!6!8
!10
y
!O
y " 3 csc !12
90˚!90˚ 180˚!180˚ 270˚!270˚
4
8
2
6
10
!4!6!8
!10
y
!O
y " 6 sin !23
28. amplitude: 3; period: 720" or 4!
30. amplitude: does not exist;period: 90" or
32. amplitude: period: 600" or
180˚!180˚ 360˚!360̊ 540˚!540˚
2
4
1
3
5
!2!3!4!5
y
!O
y " sin !35
23
34
10!
3
89;
!45˚ 90˚45˚!90˚!135˚ 135˚
4
810
2
6
!4!6!8
!10
y
!O
y " cot 2!12
!
2
180˚!180˚ 360˚!360˚ 540˚!540˚
4
8
2
6
10
!4!6!8
!10
y
!O
y " 3 cos !12
©Glencoe/McGraw-Hill 373 Algebra 2 Chapter 14
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33.
35.
37. Sample answer: Theamplitudes are the same. Asthe frequency increases, theperiod decreases.
1107
y #35 sin 4!
45̊!45˚ 90˚!90̊ 135˚!135˚
2
4
1
3
5
!2!3!4!5
y
!O
y " sin 4!35
34.
36.
38. and
90˚!180˚!270˚ !90˚ 180˚ 270˚
2
45
1
3
!2!3!4!5
O
f (x)
x
f (x) " sec xf (x) " sec (!x)
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
f (x)
xO
f (x) " cos xf (x) " cos (!x)
f (x ) # sec xf (x ) # cos x
y # 0.25 sin 1024!ty # 0.25 sin 512!t,y # 0.25 sin 128!t,
y #78 cos 5!
45̊!45˚ 90˚!90̊ 135˚!135˚
2
4
1
3
5
!2!3!4!5
y
!
O
y " cos 5!78
©Glencoe/McGraw-Hill 374 Algebra 2 Chapter 14
PQ245-6457F-P14[369-410].qxd 7/24/02 2:03 PM Page 374 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
39.
41. about 1.9 ft
43. A
45. 90"
47. 45"
49.
51. 1316
222
y # 2 sin !
5 t 40.
42. Sample answer: Tidesdisplay periodic behavior.This means that their patternrepeats at regular intervals.Answers should include thefollowing information.• Tides rise and fall in a
periodic manner, similar tothe sine function.
• In theamplitude is the absolutevalue of a.
44. C
46. $90"
48.
50.
52. 3, 11, 27, 59, 123
222
12
f (x ) # a sin bx,
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1
2
0.5
1.5
2.5
!1!0.5
!1.5!2
!2.5
y
tO
y " 2 sin t#5
©Glencoe/McGraw-Hill 375 Algebra 2 Chapter 14
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54.
56.
!4!8 4 8
3
79
1
5
1315
11
!3!5
y
O x
y " x 2 $ 2
y " (x ! 3)2 $ 2
!4!8 4 8
3
79
1
5
1315
11
!3!5
y
O x
y " 3x 2
y " 3x 2 ! 4
©Glencoe/McGraw-Hill 376 Algebra 2 Chapter 14
53.
55.
!4!8 4 8
3
79
1
5
1315
11
!3!5
y
O x
y " 2x 2
y " 2(x $ 1)2
!4!8 4 8
3
79
1
5
1315
11
!3!5
y
O x
y " x 2
y " 3x 2
Lesson 14-2 Translations of Trigonometric GraphsPages 774–776
1. vertical shift: 15; amplitude: 3;period: 180"; phase shift: 45"
2. The midline of atrigonometric function is theline about which the graph ofthe function oscillates after avertical shift.
PQ245-6457F-P14[369-410].qxd 7/24/02 2:03 PM Page 376 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
3. Sample answer:
5. no amplitude; 180"; $60"
7. no amplitude; 2!; $
#!#
234
1
!1!2!3!4
y
O
y " sec (! $ )#3
#2! #
23#2! 3#
2!
!
3
90˚!90˚ 180˚!180˚ 270˚!270˚
2
45
1
3
!2!3!4!5
y
!O
y " tan (! $ 60 )̊
y # sin (! % 45")4. 1; 2!;
6. 1; 360"; 45"
8. 1; 360"
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " cos ! $ 14
14; y #
14;
90˚45˚!45˚
0.75
0.25
0.5
!0.25
!0.5
!0.75
180˚ 225˚135˚
y
!O
y " cos (! ! 45˚)
!!
2
4
1
3
5
!2!3!4!5
y
!O
y " sin (! ! )#2
#2
#2
3#2
3#2
#!#
!
2
©Glencoe/McGraw-Hill 377 Algebra 2 Chapter 14
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9. $5; y # $5; no amplitude;360"
11. 0.25; y # 0.25; 1; 360"
13. $6; no amplitude; 60"; $45"
45˚!45˚
1
!2!3!4!5!6!7!8!9
!10!11
y
!O
y " 2 cot (3! $ 135 )̊ ! 6
90˚
1
0.5
1.5
!0.5
!1
!1.5
180˚ 270˚ 360˚
y
!O
y " sin ! $ 0.25
90˚!180˚!270˚ !90˚ 180˚ 270˚
4
810
2
6
!4!6!8
!10
y
!O
y " sec ! ! 5
10. 4; y # 4; no amplitude; 180"
12. 10; 3; 180"; 30"
14. 1; no amplitude;
234
1
!1!2!3!4
y
O #8
#4
y " sec [4(! ! )] $ 1#4
12
! #8! #
43#8! 3#
8!
!
2;
!
4
90˚!90˚ 180˚!180˚ 270˚!270˚
2468
101214
!4
y
!O
y " 3 sin [2(! ! 30 )̊] $ 10
45˚!45˚
2
3
4
5
6
7
1
!1 90˚!90˚ 135˚!135˚
y
!O
y " tan ! $ 4
©Glencoe/McGraw-Hill 378 Algebra 2 Chapter 14
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15.
17. t or
19. 1; 360"; $90"
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " cos (! $ 90 )̊
h # 4 $ cos 90"t
!
2h # 4 $ cos
2#!2# #!#
1
!1
!2
!3
3#!3#
y
!O
y " cos [ (! $ )] ! 212
23
#6
$2; 23; 4!; $
!
616. 4; 1; 4 s
18.
20. no amplitude; 180"; 30"
!45̊ 90˚45˚!90˚!135˚ 135˚
2
45
1
3
!2!3!4!5
y
!O
y " cot (! ! 30 )̊
1 2 3 4
2
4
1
3
56
!2!1
!3!4
h
tO
h " 4 ! cos t#2
©Glencoe/McGraw-Hill 379 Algebra 2 Chapter 14
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21.
23. no amplitude; 180"; $22.5"
25. $1; y # $1; 1; 360"
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!
O
y " sin ! ! 1
!45̊ 90˚45˚!90˚!135˚ 135˚
2
45
1
3
!2!3!4!5
y
!O
y " tan (! $ 22.5˚)14
!!
y " sin (! ! )#4
#2
#2
3#2
3#2
#!#
2
4
1
3
5
!2!3!4!5
y
!
O
1; 2!; !
422.
24. 3; 360"; 75"
26. 2; y # 2; no amplitude; 360"
90˚!180˚!270˚ !90˚ 180˚ 270˚
2
45
1
3
!2!3!4!5
y
!O
y " sec ! $ 2
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " 3 sin (! ! 75˚)
!!
y " cos (! $ )#3
#2
#2
3#2
3#2
#!#
2
4
1
3
5
!2!3!4!5
y
!O
1; 2!; $!
3
©Glencoe/McGraw-Hill 380 Algebra 2 Chapter 14
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27. $5; 1; 360"
29. 360"
31.
translation units left and 5 units up
!
4
!! #4
#2
#4!
#2
3#4
3#4
12
1618
2468
10
14
y
!O
y " 5 $ tan (! $ )#4
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " sin ! $12
12
12; y #
12;
12;
90˚!90˚ 180˚!180˚ 270˚!270˚
21
!2!3!4!5!6!7!8
y
!O
y " cos ! ! 5
y # $5; 28. y no amplitude;
360"
30. 1.5; y # 1.5; 6; 360"
32.
translation 50" right and 2 units up with an amplitudeof unit2
3
90˚!90˚ 180˚ 270˚!180˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " cos (! ! 50 )̊ $ 223
90˚!90˚ 180˚!180˚ 270˚!270˚
4
8
2
6
10
!4!6!8
!10
y
!O
y " 6 cos ! $ 1.5
90˚!180˚!270˚ !90˚ 180˚ 270˚
2
45
1
3
!2!3!4!5
y
!O
y " csc ! !34
# $34;$
34;
©Glencoe/McGraw-Hill 381 Algebra 2 Chapter 14
PQ245-6457F-P14[369-410].qxd 7/24/02 2:04 PM Page 381 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
33. 1; 2; 120"; 45"
35. $3.5; does not exist; 720";$60"
37. 1; 180"; 75"
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " cos (2! ! 150 )̊ $ 114
14;
90˚!180˚!270˚ !90˚ 180˚ 270˚
2
68
4
!4!6!8
!10!12
y
!O
y " 3 csc [ (! $ 60 )̊] ! 3.512
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!
O
y " 2 sin [3(! ! 45 )̊] $ 1
34. $5; 4; 180"; $30"
36. 0.75; does not exist; 270"; 90"
38. $4; does not exist; 30";$22.5"
22.5˚!22.5˚
12
!2!3!4!5!6!7!8
y
!O
y " tan (6! $ 135 )̊ ! 425
!90˚ 180˚90˚!180˚!270˚ 270˚
8
1620
4
12
!8!12!16!20
y
!O
y " 6 cot [ (! ! 90 )̊] $ 0.7523
90˚!90˚ 180˚!180˚ 270˚!270˚
4
8
2
6
10
!4!6!8
!10
y
!O
y " 4 cos [2(! $ 30 )̊] ! 5
©Glencoe/McGraw-Hill 382 Algebra 2 Chapter 14
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39.
41.
The graphs are identical.
43. c
45. 300; 14.5 yr
47.
49. Sample answer: You can usechanges in amplitude andperiod along with vertical andhorizontal shifts to show ananimal population’s startingpoint and display changes tothat population over a period of
h # 9 % 6 sin c!9
(t $ 1.5)d
2
4
1
3
5
!2!3!4!5
y
!O
y " 3 ! cos !12
y " 3 $ cos (! $ #)12
!! #2
#2
!# # 3#2
3#2
5
7
4321
6
8
!2
y
!O!! #
2#2
!# # 3#2
3#2
y " 3 $ 2 sin [2(! $ )]#4
3; 2; !; $!
440. 4; does not exist; 6!;
42.
The graphs are identical.
44. 180; 5 yr
46. Sample answer: When theprey (mouse) population is atits greatest the predator willconsume more and thepredator population will growwhile the prey population falls.
48.
50. B
a # $1, b # 1, h #!
2
!2#!4# 2# 4#
2
4
1
3
5
!2!3!4!5
y
!O
y " !sin [ (! ! )]14
#2
y " cos [ (! $ )]14
3#2
2468
10121416
!4
y
!O!2#!4# 2# 4#
y " 4 $ sec [ (! $ )]2#3
13
$2!
3
©Glencoe/McGraw-Hill 383 Algebra 2 Chapter 14
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time. Answers should includethe following information.• The equation shows a
rabbit population thatbegins at 1200, increasesto a maximum of 1450,then decreases to aminimum of 950 over aperiod of 4 years.
• Relative to would
have a vertical shift of k units, while
has a horizontalshift of h units.
51. D
53. amplitude: 1; period: 720" or4!
55. 0.75
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " sin !2
[b (x $ h)]y # a cos
y # a cos bx % ky # a cos bx,
52. amplitude: does not exist;period: 360" or 2!
54. amplitude: does not exist;
period: 270" or
56. 0.57
180˚!180˚ 360˚!360˚
4
810
2
6
!4!6!8
!10
y
!O
y " 3 tan !23
3!
2
90˚!180˚!270˚ !90˚ 180˚ 270˚
2
45
1
3
!2!3!4!5
y
!O
y " 3 csc !
©Glencoe/McGraw-Hill 384 Algebra 2 Chapter 14
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58. 0.8
60. 2.29
62. 0.66
64.
66.
68. $1
70. 0
72. $222
$232
$14
57. 0.83
59. 35
61. 0.66
63.
65.
67. $1
69.
71.
73. 1
233
12
3y 2 % 10y % 52(y $ 5)(y % 3)
5a $ 13(a $ 2)(a $ 3)
©Glencoe/McGraw-Hill 385 Algebra 2 Chapter 14
Lesson 14-3 Trigonometric IdentitiesPages 779–781
1. Sample answer: The sinefunction is negative in thethird and fourth quadrants.Therefore, the terminal sideof the angle must lie in oneof those two quadrants.
3. Sample answer: Simplifying atrigonometric expressionmeans writing the expressionas a numerical value or interms of a single trigonometricfunction, if possible.
5.
7.
9.
11. csc !
tan2 !
22
$54
2. Sample answer: Pythagoreanidentities are derived byapplying the PythagoreanTheorem to trigonometricconcepts.
4.
6.
8. 1
10. sec !
12. sin ! # cos !
v 2
gR
35
$233
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13.
15.
17.
19.
21.
23.
25. cot !
27. cos !
29. 2
31. cot2 !
33. 1
35. csc2 !
37. about 11.5"
39. about 9.4"
41. No;
simplifies to .
43.
45. Sample answer: You can useequations to find the heightand the horizontal distanceof a baseball after it hasbeen hit. The equationsinvolve using the initial anglethe ball makes with theground with the sine function.Answers should include thefollowing information.
P # I2R $I2R
1 % tan2 2!ft.
E #I sin !
R 2
R2 #I tan ! cos !
E
$427
7
34
232
54
$25
12
14.
16.
18.
20.
22.
24.
26. 1
28. sin !
30. $3
32. tan !
34. cot2 !
36. 1
38. about 4 m/s
40.
42.
44.
46. B
916
P # I2R sin2 2!ft
E #I cos !
R 2
$4217
17
45
$325
5
35
222
253
©Glencoe/McGraw-Hill 386 Algebra 2 Chapter 14
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• Both equations arequadratic in nature with aleading negative coefficient.Thus, both are invertedparabolas which model thepath of a baseball.
• model rockets, hitting agolf ball, kicking a rock
47. A
49. 12; y # 12; no amplitude; 180"
90˚!90˚ 180˚!180˚ 270˚!270˚
10
15
20
5
!5
y
!O
y " tan ! $ 12
48. $1; y # $1; 1; 360"
50. amplitude: does not exist;period: 180" or !
45˚!90˚!135˚ !45˚ 90˚ 135˚
2
45
1
3
!2!3!4!5
y
!O
y " csc 2!
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " sin ! ! 1
©Glencoe/McGraw-Hill 387 Algebra 2 Chapter 14
PQ245-6457F-P14[369-410].qxd 7/24/02 2:04 PM Page 387 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
51. amplitude: 1; period: 120" or
53. 93
55. Symmetric (#)
57. Multiplication (#)
45˚!45˚ 90˚!90˚ 135˚!135˚
2
4
1
3
5
!2!3!4!5
y
!O
y " cos 3!
2!
352. amplitude: does not exist;
period: 36" or
54.
56. Substitution (#)
58. Substitution (#)
y # $16
(x $ 11)2 %12
22.5˚!22.5˚
1
5432
!2!3!4!5
y
!O
y " cot 5!13
!
5
©Glencoe/McGraw-Hill 388 Algebra 2 Chapter 14
Chapter 14Practice Quiz 1
Page 781
1. 720" or 4!
3.
5. 252
$35
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " sin !34
12
34, 2. $5, 2, 8!,
4. $213
3
42
6
!4
!10!8!6
!12!14
y
!O!2#!4# 2# 4#
y " 2 cos [ (! ! )] ! 5#4
14
!
4
PQ245-6457F-P14[369-410].qxd 7/24/02 2:04 PM Page 388 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
1.
Multiply by the LCD, cos !.
Subtract.
Factor.
3. Sample answer:it is not an identity
because
5.
sin2 ! # sin2 !
sin2 !cos2 !
! cos2 ! #? sin2 !
tan2 ! cos2 ! #? 1 $ cos2 !
sin2 ! # 1 $ cos2 !.cos2 !;
sin2 ! # 1 %
sin !cos !
# tan !
sin ! tan ! # sin ! tan !
sin ! tan ! #? sin ! !
sin !cos !
# sin2 !1 $ cos2 !
sin ! tan ! #? sin2 !
cos !
sin ! tan ! #? 1 $ cos2 !
cos !
sin ! tan ! #? 1
cos !$
cos2 !cos !
,
sec ! #1
cos !
sin ! tan ! #? 1
cos !$ cos !
sin ! tan ! #? sec ! $ cos ! 2. Sample answer: Use various
identities, multiply or divideterms to form an equivalentexpression, factor, andsimplify rational expressions.
4.
6.
1 % sin ! # 1 % sin !
(1 $ sin !)(1 % sin !)1 $ sin !
#? 1 % sin !
1 $ sin2 !1 $ sin !
#? 1 % sin !
cos2 !1 $ sin !
#? 1 % sin !
sec2 ! # sec2 !1 % tan2 ! #
? sec2 !
tan !(cot ! % tan !) #? sec2 !
©Glencoe/McGraw-Hill 389 Algebra 2 Chapter 14
Lesson 14-4 Verifying Trigonometric IdentitiesPages 784–785
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7.
9.
11.
1 # 1
13.
sec2 ! # sec2 !1 % tan2 ! #
? sec2 !
1 %1
cos2 !! sin2 ! #
? sec2 !
1 % sec2 ! sin2 ! #? sec2 !
cos2 ! % sin2 ! #? 1
cos2 ! %sin2 !cos2 !
! cos2 ! #? 1
cos2 ! % tan2 ! cos2 ! #? 1
sec ! % 1tan !
#sec ! % 1
tan !
sec ! % 1tan !
#? tan ! ! (sec ! % 1)
tan2 !
sec ! % 1tan !
#? tan ! ! (sec ! % 1)
sec2 ! $ 1
sec ! % 1tan !
#? tan !
sec ! $ 1!
sec ! % 1sec ! % 1
sec ! % 1tan !
#? tan !
sec ! $ 1
tan2 ! # tan2 !
1cos2 !
! sin2 ! #? tan2 !
#? tan2 !
1cos2 !
1sin2 !
sec2 !csc2 !
#? tan2 !
1 % tan2 !csc2 !
#? tan2 ! 8.
10. D
12.
14.
1 # 1
sin ! !1
cos !!
cos !sin !
#? 1
sin ! sec ! cot ! #? 1
csc2 ! # csc2 !cot2 ! % 1 #
? csc2 !
cot2 ! %sin !cos !
!cos !sin !
#? csc2 !
cot2 ! % cot ! tan ! #? csc2 !
cot ! (cot ! % tan !) #? csc2 !
sin !sec !
#sin !sec !
sin !sec !
#? sin ! cos !
1
sin !sec !
#? sin ! cos !
sin2 ! % cos2 !
#?1
sin2 ! % cos2 !sin ! cos !
sin !sec !
#?1
sin2 !cos !
%sin !cos !
sin !sec !
sin !sec !
#? 1
tan ! % cot !
©Glencoe/McGraw-Hill 390 Algebra 2 Chapter 14
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15.
17.
cot ! csc ! # cot ! csc !
cot ! csc ! #? cos !
sin !!
1sin !
cos !sin !(cos ! % 1)
cot ! csc ! #? cos ! % 1
sin !!
cot ! csc ! #?
cos ! % 1sin !
sin ! (cos ! % 1)cos !
cot ! csc ! #?
cos ! $ 1sin !
sin ! cos ! % sin !cos !
cot ! csc ! #?
cos !sin !
%1
sin !
sin ! %sin !cos !
cot ! csc ! #? cot ! % csc !
sin ! % tan !
1 $ cos !1 % cos !
#1 $ cos !1 % cos !
1 $ cos !1 % cos !
#?(1 $ cos !)(1 $ cos !)(1 $ cos !)(1 % cos !)
1 $ cos !1 % cos !
#?(1 $ cos !)(1 $ cos !)
1 $ cos2 !
1 $ cos !1 % cos !
#?1 $ 2 cos ! % cos2 !
sin2 !
cos2 !sin2 !
1 $ cos !1 % cos !
#?1
sin2 !$
2cos !sin2 !
%
1sin !
%cos2 !sin2 !
1 $ cos !1 % cos !
#?1
sin2 !$ 2 !
cos !sin !
!
% cot2 !
1 $ cos !1 % cos !
#? csc2 ! $ 2 cot ! csc !
1 $ cos !1 % cos !
#? (csc ! $ cot !)2
16.
18.
sin ! % cos ! # sin ! % cos !
sin ! % cos ! #? sin ! % cos !
cos !& cos !
sin ! % cos ! #?
sin ! % cos !cos !
1cos !
sin ! % cos ! #?
1 %sin !cos !
1cos !
sin ! % cos ! #? 1 % tan !
sec !
tan ! $ cot ! # tan ! $ cot !
sin !cos !
$cos !sin !
#? tan ! $ cot !
sin2 !sin ! cos !
$cos2 !
sin ! cos !#? tan ! $ cot !
sin2 ! $ cos2 !sin ! cos !
#? tan ! $ cot !
(1 $ cos2 !) $ cos2 !sin ! cos !
#? tan ! $ cot !
1 $ 2 cos2 !sin ! cos !
#? tan ! $ cot !
©Glencoe/McGraw-Hill 391 Algebra 2 Chapter 14
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19.
21.
23.
1 # 1cos2 ! % sin2 ! #? 1
1
sec2 !%
1
csc2 !#? 1
1 % sin !sin !
#1 % sin !
sin !
1 % sin !sin !
#?1
sin !%
sin !sin !
1 % sin !sin !
#? csc ! % 1
1 % sin !sin !
#? cot2 ! (csc ! % 1)
cot2 !
1 % sin !sin !
#? cot2 !(csc ! % 1)
csc2 ! $ 1
1 % sin !sin !
#? cot2 !
csc ! $ 1!
csc ! % 1csc ! % 1
1 % sin !sin !
#? cot2 !
csc ! $ 1
cot ! # cot !
cos !sin !
#? cot !
cos2 !sin ! cos !
#? cot !
1 $ sin2 !sin ! cos !
#? cot !
1sin ! cos !
$sin2 !
sin ! cos !#? cot !
1cos !sin !
$sin !cos !
#? cot !
sec !sin !
$sin !cos !
#? cot ! 20.
22.
24.
1 %1
cos !# 1 %
1cos !
1 %1
cos !#? sec ! % 1
1 %1
cos !#? tan2 !(sec ! % 1)
tan2 !
1 %1
cos !#? tan2 !(sec ! % 1)
sec2 ! $ 1
1 %1
cos !#? tan2 !
sec ! $ 1!
sec ! % 1sec ! % 1
1 %1
cos !#? tan2 !
sec ! $ 1
sin !cos !
#sin !cos !
sin ! % cos !cos !
!sin !
sin ! % cos !#? sin !
cos !
sin ! % cos !cos !
sin ! % cos !
sin !
#? sin !
cos !
1 %sin !cos !
1 %cos !sin !
#? sin !
cos !
1 % tan !1 % cot !
#? sin !
cos !
2csc ! #? 2csc !
2sin !
#? 2csc !
2(1 $ cos !)sin !(1 $ cos !)
#? 2csc !
2 $ 2cos !sin !(1 $ cos !)
#? 2csc !
sin2 ! % cos2 ! % 1 $ 2cos !sin !(1 $ cos !)
#? 2csc !
sin2 !sin !(1 $ cos !)
% 1 $ 2cos ! % cos2 !
sin ! (1 $ cos !)#? 2csc !
sin !sin !
!sin !
1 $ cos !%
1 $ cos !1 $ cos !
!1 $ cos !
sin !#? 2csc !
sin !1 $ cos !
%1 $ cos !
sin !#? 2csc !
©Glencoe/McGraw-Hill 392 Algebra 2 Chapter 14
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25.
27.
29.
1 # 1
31.
#v 2
0 sin2 !2g
#v 2
0
2g!
sin2 !cos2 !
!cos2 !
1
v 20 tan2 !
2g sec2 !#
v 20
sin2 !cos2 !
2g
1cos2 !
sin !cos !
! sin ! ! cos ! !1
sin2 !
#? 1
tan ! sin ! cos ! csc2 ! #? 1
sin !1 % cos !
#sin !
1 % cos !
sin2 !sin ! (1 % cos !)
#? sin !
1 % cos !
1 $ cos2 !sin ! (1 % cos !)
#? sin !
1 % cos !
1 $ cos !sin !
!1 % cos !1 % cos !
#? sin !
1 % cos !
1 $ cos !sin !
#? sin !
1 % cos !
# (2 $ sec2 !)(sec2 !)(2 $ sec2 !)(sec2 !)
#? (2 $ sec2 !)(sec2 !)
[1 $ (sec2 ! $ 1)](sec2 !)#? sec2 !(2 $ sec2 !)
(1 $ tan2 !)(1 % tan2 !)#? 2 sec2 ! $ sec4 !
1 $ tan4 ! 26.
28.
30.
32. 598.7 m
1 % cos ! # 1 % cos !
sin2 !(1 % cos !)
sin2 !#? 1 % cos !
sin2 ! (1 % cos !)1 $ cos2 !
#? 1 % cos !
sin2 !1 $ cos !
!1 % cos !1 % cos !
#? 1 % cos !
sin2 !1 $ cos !
#? 1 % cos !
2sec ! # 2sec !
2cos2 !
#? 2sec !
2cos !cos2 !
#? 2sec !
cos ! $ sin ! cos ! % cos ! % sin ! cos !
1 $ sin2 !
#? 2 sec !
cos !11 $ sin !2 % cos !11 % sin !211 % sin !2 11 $ sin !2 #
? 2sec !
cos !1 % sin !
!1 $ sin !1 $ sin !
%cos !
1 $ sin !!
1 % sin !1 % sin !
#? 2sec !
cos !1 % sin !
%cos !
1 $ sin !#? 2sec !
cos2 ! $ sin2 ! # cos2 ! $ sin2 !#? cos2 ! $ sin2 !
(cos2 ! $ sin2 !) ! 1#? cos2 ! $ sin2 !
(cos2 ! $ sin2 !)(cos2 ! % sin2 !)cos4 ! $ sin4 ! #
? cos2 ! $ sin2 !
©Glencoe/McGraw-Hill 393 Algebra 2 Chapter 14
PQ245-6457F-P14[369-410].qxd 7/24/02 2:04 PM Page 393 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
33. Sample answer: Consider aright triangle ABC with rightangle at C. If an angle A hasa sine of x, then angle Bmust have a cosine of x.Since A and B are both in aright triangle and neither isthe right angle, their sum
must be
35. D
37.
is not
39.
may be[!360, 360] scl: 90 by [!5, 5] scl: 1
[!360, 360] scl: 90 by [!5, 5] scl: 1
!
2.
34. Sample answer:Trigonometric identities areverified in a similar mannerto proving theorems ingeometry before using them.Answers should include thefollowing.• The expressions have not
yet been shown to beequal, so you could notuse the properties ofequality on them.
• To show two expressionsyou must transform one,or both independently.
• Graphing two expressionscould result in identicalgraphs for a set interval,that are differentelsewhere.
36. B
38.
may be
40.
may be[!360, 360] scl: 90 by [!5, 5] scl: 1
[!360, 360] scl: 90 by [!5, 5] scl: 1
©Glencoe/McGraw-Hill 394 Algebra 2 Chapter 14
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41.
may be
43.
45.
47. 1: 360"; 30"
49. 3; 2!; $
51.
53. 264
%222
264
!! #2
#2
!# # 3#2
3#2
2
4
1
3
5
!2!3!4!5
y
!O
y " 3 cos (! $ )#2
!
2
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!O
y " cos (! ! 30 )̊
219312
252
[!360, 360] scl: 90 by [!5, 5] scl: 1
42.
is not
44.
46.
48. 1: 360"; 45"
50.
52.
54. 2 $ 234
224
56
90˚!90˚ 180˚!180˚ 270˚!270˚
2
4
1
3
5
!2!3!4!5
y
!
O
y " sin (! ! 45 )̊
$274
$253
[!360, 360] scl: 90 by [!5, 5] scl: 1
©Glencoe/McGraw-Hill 395 Algebra 2 Chapter 14
PQ245-6457F-P14[369-410].qxd 7/24/02 2:04 PM Page 395 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
©Glencoe/McGraw-Hill 396 Algebra 2 Chapter 14
Lesson 14-5 Sum and Difference of Angles FormulasPages 788–790
1. sin (" % #) $ sin " % sin #sin " cos # % cos " sin # 'sin " % #
3. Sometimes; sample answer:The cosine function canequal 1.
5.
7.
9.
11.
cos ! # cos !
13. 5 $ 231 % 523
sin ! ! 0 % cos ! ! 1 #? cos !
cos ! sin !
2#? cos !sin ! cos
!
2%
sin a! %!
2b #
? cos !
$12
232
26 $ 224
2. Use the formula
Since sin 105" # replace "
with 60" and # with 45" toget sin 60" cos 45" % cos 60"sin 45". By finding the sum ofthe products of the values,
the result is
or about 0.9659.
4.
6.
8.
10. cos(270" $ !)
% sin 270" sin !
# $sin !
12. sin(! % 30") % cos(! % 60")
% cos ! cos 30" %
cos ! cos 60" $ sin ! sin 60"
%
# cos !
14. 222
#? 1
2 cos ! %
12
cos !
12
cos ! $232
sin !
#? 23
2sin ! %
12
cos !
#? sin ! cos 30"
#? 0 % ($1 sin !)
#? cos 270" cos !
232
22 $ 264
26 % 224
26 % 224
sin(60" % 45"),
cos " sin #.# sin " cos # %sin(" % #)
PQ245-6457F-P14[369-410].qxd 7/24/02 2:04 PM Page 396 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
©Glencoe/McGraw-Hill 397 Algebra 2 Chapter 14
15.
17.
19.
21.
23.
25.
27.
29. cos (90" % !)
# $sin !
31.
cos ! # cos !
cos ! $ 0 #? cos !
1 ! cos ! $ 0 ! sin ! #? cos !
#? cos !
sin 90" cos ! $ cos 90" sin !sin(90" $ !) #
? cos !
#? 0 $ 1 sin !
#? cos 90" cos ! $ sin 90" sin !
$26 $ 224
22 $ 264
222
$222
$26 $ 224
$26 $ 224
22 $ 264
16.
18.
20.
22.
24.
26.
28. sin (270" $ !)
# $cos !
30. cos (90" $ !)
# sin !
32.
$cos ! # $cos !
0 % ($cos !) #? $cos !
sin ! ! 0 % cos !($1) #?
$cos !
#?
$cos !
sin ! cos 3!
2% cos ! sin
3!
2
sin 1! %3!
22 #
?$cos !
#? 0 ! cos ! % 1 ! sin !
#? cos 90" cos ! $ sin 90" sin !
#?
$1 cos ! $ 0
sin !$ cos 270"#? sin 270" cos !
222
$232
22 $ 264
$222
$26 $ 224
$26 $ 224
PQ245-6457F-P14[369-410].qxd 7/24/02 2:04 PM Page 397 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
©Glencoe/McGraw-Hill 398 Algebra 2 Chapter 14
33.
$cos ! # $cos !
35.
sin ! # sin !
37.
$
# sin !
#?
12 sin ! %
12 sin !
cos ! %12 sin !
#? 1
2 sin ! %
232
cos ! $ 232
cos ! cos
!
6% sin ! sin
!
6
#? sin ! cos
!
3% cos ! sin
!
3
sin a! %!
3b $ cos a! %
!
6b
0 $ [$sin !] #? sin !
0 ! cos ! $ [$1 ! sin !] #? sin !
sin ! cos !$ [cos ! sin !] #? sin !
sin(! $ !) #? sin !
$1 ! cos ! % 0 ! sin ! #?
$cos !
#?
$cos !
cos ! cos ! % sin ! sin !cos (! $ !) #
?$cos ! 34.
36.
%
38. sin ( " % #) sin ( " $ #)# sin2 " $ sin2 #
(sin " cos # $ cos " sin #)
$
sin2 #
sin2 # % sin2 " sin2 #
# sin2 " $ sin2 #
#? sin2 " $ sin2 " sin2 # $
(1 $ sin2 ")#? sin2 "(1 $ sin2 #)
cos2 " sin2 ##? sin2 " cos2 # $
#? (sin " cos # % cos " sin #)
# 23 cos !
232
cos ! $12 sin !
#? 232
cos ! %12 sin !
$ cos 60" sin !% sin 60" cos !#?
sin 60" cos ! % cos 60" sin !
sin (60" % !) % sin (60" $ !)
cos ! # cos !1 ! cos ! $ 0 #
? cos !
1 ! cos ! $ [0 ! sin !] #? cos !
#? cos !
cos 2! cos ! $ [sin 2! sin !]cos(2! % !) #
? cos !
PQ245-6457F-P14[369-410].qxd 7/24/02 2:04 PM Page 398 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
39.
cos (" % #) # cos (" % #)
41. Destructive; the resultinggraph has a smalleramplitude than the two initialgraphs.
43. 0.4179 E
45. 0.5563 E
cos " cos # $ sin " sin #1
cos (" % #) #?
1 $sin "cos "
!sin #
cos #1
cos "!
1cos #
&cos " cos #cos " cos #
cos (" % #) #?
cos (" % #) #? 1 $
sin "cos "
!sin #cos #
1cos "
!1
cos #
cos (" % #) #? 1 $ tan " tan #
sec " sec #
40.
42. 0.3681 E
44. 0.6157 E
46. tan (" % #)
#tan " % tan #
1 $ tan " tan #
#
sin " cos #cos " cos #
%cos " sin #cos " cos #
cos " cos #cos " cos #
$sin " sin #cos " cos #
#sin " cos # % cos " sin #cos " cos # $ sin " sin#
#sin(" % #)cos(" % #)
O
y
t
4
!2
!4
y " 10 sin (2t ! 30°) $ 10 cos (2t $ 60°)
!180° !90° 180°90°
©Glencoe/McGraw-Hill 399 Algebra 2 Chapter 14
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47. Sample answer: Todetermine communicationinterference, you need todetermine the sine or cosineof the sum or difference oftwo angles. Answers shouldinclude the followinginformation.• Interference occurs when
waves pass through thesame space at the sametime. When the combinedwaves have a greateramplitude, constructiveinterference results andwhen the combined waveshave a smaller amplitude,destructive interferenceresults.
49. C
tan (" $ #)
48. #A
50.
# cot ! % sec !
#? cos !
sin !%
1cos !
%sin !
sin ! cos !#? cos2 !
sin ! cos !
#? cos2 ! % sin !
sin ! cos !
cot ! % sec !
#tan " $ tan #
1 % tan " tan #
#
sin " cos #cos " cos #
$cos " sin #cos " cos #
cos " cos #cos " cos #
%sin " sin #
cos " cos #
#sin " cos # $ cos " sin #cos " cos # % sin " sin #
#sin(" $ #)cos(" $ #)
©Glencoe/McGraw-Hill 400 Algebra 2 Chapter 14
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51.
53.
55. 4
57. 2 sec !
59.
61. 360
63. 56
65. about 228 mi
67. (225
2
sec ! # $53, cot ! #
34
tan ! #43, csc ! # $
54,
sin ! # $45, cos ! # $
35,
csc ! # csc !
1sin !
#? csc !
1cos !
!cos !sin !
#? csc !
1cos !
)sin !cos !
#? csc !
sec !tan !
#? csc !
# sin2 ! % tan2 !
#? sin2 ! %sin2 !cos2 !
1sin2 !
#? sin2 ! %
1cos2 !
)
#? sin2 ! %
sec2 !csc2 !
#? (1 $ cos2 !) %
sec2 !csc2 !
sin2 ! % tan2 ! 52.
54. 1
56. sec !
58.
60. sin ! # 1, cos ! # 0,tan ! # undefined,csc ! # 1, sec ! # undefined,cot ! # 0
62. 3,991,680
64. 210
66.
68. (35
y 2
34$
x 2
6# 1
cot ! # $53
sec ! #234
5,
csc ! # $234
3,
tan ! # $35,cos ! #
523434
,
sin ! # $3234
34,
2 $ cos2 ! # 2 $ cos2 !1 $ cos2 ! % 1 #
? 2 $ cos2 !
sin2 ! % 1 #? 2 $ cos2 !
sin ! (sin ! % csc !) #? 2 $ cos2 !
©Glencoe/McGraw-Hill 401 Algebra 2 Chapter 14
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©Glencoe/McGraw-Hill 402 Algebra 2 Chapter 14
1. Sample answer: If x is in thethird quadrant, then is between 90" and 135". Usethe half-angle formula forcosine knowing that thevalue is negative.
3. Sample answer: The identityused for cos 2! depends onwhether you know the valueof sin !, cos !, or both values.
5.
7.
9. 22 $ 132
28 $ 2174
, $28 % 217
4
$327
8, $
18,
4259
, $19, 230
6, $266
x2
2. Sample answer: 45";cos 2(45") # cos 90" or 0,
or
4.
6.
8.
10.
# cot x
#? cos x
sin x
#? 2 sin x cos x
2 sin2 x
#? 2 sin x cos x
1 $ (1 $ 2 sin2 x)
cot x #? sin 2x
1 $ cos 2x
$22 $ 13
2
22 % 132
232
, 12, 22 $ 13
2,
2425
, $725
, 255
, 225
5
22122
2 cos 45" # 2 !
Lesson 14-6 Double-Angle and Half-Angle FormulasPages 794–797
69.
71.
73. (216$12
2
(262
(255
70.
72.
74. (22$212
2
(325
5
(34
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©Glencoe/McGraw-Hill 403 Algebra 2 Chapter 14
11.
1 # 1
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
2 sin x cos x # 2 sin x cos x
2 sin x cos x #? 2
cos xsin x
! sin2 x
sin 2x #? 2 cot x sin2 x
22 $ 122
$22 % 12
2
$22 % 13
2
4259
, $19, 266
,
2306
$2316 % 413
6
$422
9,
79, 2316 $ 413
6
23518
, $1718
, 215
6, 221
6
$28 % 155
4
28 $ 1554
,$3255
32,
2332
,
4229
, $79, 263
, $233
$120169
, 119169
, 5226
26, 22626
cos2 2x % sin2 2x #? 1
cos2 2x % 4 sin2 x cos2 x #? 1 12. 1.64
14.
16.
18.
20.
22.
24.
26.
28.
30.
32.
1 % cos x # 1 % cos x
2a1 % cos x2
b #? 1 % cos x
2a(31 % cos x2
b2
#? 1 % cos x
2 cos2 x2
#? 1 % cos x
$22 $ 13
2
$22 $ 13
2
22 $ 122
25110 $ 121010
25110 % 121010
,
$4221
5,
1725
,
$28 $ 2115
4
2158
, 78, $28 % 2115
4,
120169
, 119169
, 5226
26, $22626
$215
8, $
78, 210
4, 264
2425
, 725
, 3210
10, $21010
$426
25, $
2325
, 210
5, $215
5
PQ245-6457F-P14[369-410].qxd 7/24/02 2:04 PM Page 403 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
33.
35.
37. 46.3"
39. 2 % 23
1 $ cos x1 % cos x
#1 $ cos x1 % cos x
a(31 $ cos x2
b2
a(31 % cos x2
b2#? 1 $ cos x
1 % cos x
sin2
x2
cos2 x2
#? 1 $ cos x
1 % cos x
tan2 x2
#? 1 $ cos x
1 % cos x
2 sin2 x $ 1 # 2 sin2 x $ 1#? 2 sin2 x $ 1
sin2 x $ 1 % sin2 x#? 2 sin2 x $ 1
[sin2 x $ (1 $ sin2 x)] ! 1#? 2 sin2 x $ 1
(sin2 x $ cos2 x) ! 1#? 2 sin2 x $ 1
(sin2 x $ cos2 x)(sin2 x % cos2 x)sin4 x $ cos4 x #
? 2 sin2 x $ 1
34.
36.
38.
40.
#v 2 sin 2!
g
#? 2
g v 2 sin ! cos !
#? 2
g v 2 tan ! cos2 !
#? 2
g v 2 tan !(1 $ sin2 !)
2g v 2 (tan ! $ tan ! sin2 !)
1 ( 31 $ cos L1 % cos L
1 * 31 $ cos L1 % cos L
tan x # tan x
sin xcos x
#? tan x
sin2 x
sin x cos x#? tan x
1 $ cos2 xsin x cos x
#? tan x
1sin x cos x
$cos xsin x
#? tan x
sin2 x # sin2 x
sin2 x #? 1
2 (2 sin2 x)
sin2 x #? 1
2 [1 $ (1 $ 2 sin2 x)]
sin2 x #? 1
2 (1 $ cos2 x)
©Glencoe/McGraw-Hill 404 Algebra 2 Chapter 14
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41.
43. The maxima occur at
and . The minima occur
at
45. The graph of crosses thex-axis at the points specifiedin Exercise 41.
47. Sample answer: The soundwaves associated with musiccan be modeled usingtrigonometric functions.Answers should include thefollowing information.• In moving from one
harmonic to the next, thenumber of vibrations thatappear as sine wavesincrease by 1.
f(x)
x # 0, (! and (2!.
(3!
2
(!
2x #
14tan ! 42.
Sample answer: They allhave the same shape andare vertical translations ofeach other.
44.
46.
48. D
c # 1 and d # 0.5
90˚!90˚ 180˚!180˚ 270˚!270˚
1
2
0.5
1.5
2.5
!1!1.5
!2!2.5
y
xO
y " sin 2x
1
2
0.5
1.5
2.5
!1!1.5
!2!2.5
y
xO
y " ! cos 2x12 y " !cos2 x
y " sin2 x
!! #2
#2
!# # 3#2
3#2
©Glencoe/McGraw-Hill 405 Algebra 2 Chapter 14
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• The period of the functionas you move from the nthharmonic to the (n % 1)thharmonic decreases from
to
49. B
51.
53.
55.
57.
59. 102.5 or about 316 timesgreater
61. 1, $1
63. $2
65. 0, $12
52,
# cos !(cos ! % cot !)#? cos2 ! % cot ! cos !
#? cos !
sin ! cos ! sin ! % cot ! cos !
#? cot ! cos !1sin ! % 12cos !(cos ! % cot !)
12
$232
26 $ 224
2!
n % 1.
2!
n
50.
52.
54.
56.
58. 101 or 10
60. $6, 5
62. 0, $2
64. $12,
12
cot2 ! $ sin2 ! # cot2 ! $ sin2 !
cot2 ! $ sin2 ! #?cot2 ! $ sin2 !
1
cos2 ! 1
sin2 !$ sin2 !
sin2 ! 1
sin2 !
cot2 ! $ sin2 ! #?
cos2 ! csc2 ! $ sin2 !sin2 ! csc2 !
cot2 ! $ sin2 ! #?
26 % 224
$222
26 % 224
©Glencoe/McGraw-Hill 406 Algebra 2 Chapter 14
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©Glencoe/McGraw-Hill 407 Algebra 2 Chapter 14
1.
3.
5.
7.
9. 22 $ 132
232
$sin ! # $sin !
0 % ($1 ! sin !) #?
$sin !
$sin !cos 3!
2 cos ! % sin
3!
2 sin ! #
?
cos a3!
2$ !b #
?$sin !
sin ! % tan ! # sin ! % tan !
sin ! cos !cos
%sin !cos !
sin ! % tan ! #?
sin ! cos ! % sin !cos !
sin ! % tan ! #?
sin !(cos ! % 1)cos !
sin ! % tan ! #?
tan ! #? tan !
sin !cos !
#? tan !
sin ! !1
cos ! #
? tan !
sin ! sec ! #? tan !
Chapter 14Practice Quiz 2
Page 797
2.
4.
6.
8.
10. 22 $ 122
$9282
82
# cos !
#? 1
2 cos ! %
12
cos !
a12
cos ! $132
sin !b#? a13
2 sin ! %
12 cos !b %
(cos ! cos 60" $ sin ! sin 60")cos ! sin 30") %#
? (sin ! cos 30" %
sin (! % 30") % cos (! % 60")
cos ! # cos !cos ! % 0 #
? cos !
cos !sin 90" cos ! % cos 90" sin ! #?
sin (90" % !) #? cos !
sin ! tan ! # sin ! tan !
sin ! sin !cos !
#? sin ! tan !
sin2 !cos !
#? sin ! tan !
1 $ cos2 !
cos !#? sin ! tan !
1
cos !$
cos2 !cos !
#? sin ! tan !
1cos !
$ cos ! !cos !cos !
#? sin ! tan !
sec ! $ cos ! #? sin ! tan !
PQ245-6457F-P14[369-410].qxd 7/24/02 2:04 PM Page 407 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
©Glencoe/McGraw-Hill 408 Algebra 2 Chapter 14
Lesson 14-7 Solving Trigonometric EquationsPages 802–804
1. Sample answer: If sec then # 0. Since no value
of makes # 0. there
are no solutions.
3. Sample answer: sin
5.
7.
9.
11.
13.
or
15.
17.
19.
21.
23.
25.
27. !
3% 2k!,
5!
3% 2k!
2!
3% 2k!,
4!
3% 2k!
!
3% 2k!,
5!
3% 2k!
7!
6,
11!
6
!
6,
5!
6,
3!
2
210", 330"
60", 300"
360", 90" % k ! 360"30" % k ! 360", 150" %
!
6% 2k!,
5!
6% 2k!,
!
2% 2k!
60" % k ! 360", 300" % k ! 360"
0 % k!
!
6
135", 225"
! # 2
1cos !
!
1cos !
! # 0 2. Sample answer: The functionis periodic with two solutionsin each of its infinite numberof periods.
4.
6.
8.
10.
12. or
%
14.
16.
18.
20.
22.
24.
26.
28. 0 % k!, !
6% 2k!,
5!
6% 2k!
0 % 2k!
5!
3% 2k!
! % 2k!, !
3% 2k!,
!
2,
3!
2,
2!
3,
4!
3
!
2
30", 150", 210", 330"
240", 300"
31.3"
k ! 360"210" % k ! 360", 330"
7!
6% 2k!,
11!
6% 2k!
k ! 360"90" % k ! 360", 180" %
0 %2k!
3
!
6,
!
2,
5!
6,
3!
2
60", 120", 240", 300"
PQ245-6457F-P14[369-410].qxd 7/24/02 2:04 PM Page 408 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
29.
31.
33.
35.
or
37. or
39.
or
41. or
43.
45. (4.964, $0.598)
1 2 3 4 5 6 7 8 9!1
2.5
3.5
21.5
10.5
3
4
!1
y
tO
y " $ sin (# t )32
32
y #32
%32 sin (!t)
S # 352 cot !S #352tan !
360", 300" % k ! 360"0" % k ! 360", 60" % k !
0 % 2k!, !
3% 2k!,
5!
3% 2!,
0" % k ! 180"0 % k!
270" % k ! 360"
0" % k ! 360", 90" % k ! 360",
0 % 2k!, !
2% 2k!,
3!
2% 2k!
0" % k ! 180", 60" % k ! 180"
270" % k ! 360"
45" % k ! 180" 30.
32.
34.
36. or
38.
or
40. or
42. about 32"
44. 10
46. Sample answer:Temperatures are cyclic andcan be modeled bytrigonometric functions.Answers should include thefollowing information.
90" % k ! 720"!
2% 4k!
360", 240" % k ! 360"
90" % k ! 180", 120" % k !
!
2% k!,
2!
3% 2k!,
4!
3% 2k!
330" % k ! 360"210" % k ! 360",
7!
6% 2k!,
11!
6% 2k!
240" % k ! 360"
120" % k ! 360",
150" % k ! 360"
30" % k ! 360",
0" % k ! 180"
©Glencoe/McGraw-Hill 409 Algebra 2 Chapter 14
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47. D
49.
51.
53.
55. b # 11.0, c # 12.2, m!C # 78
$232
521118
, 718
, 236
, 233
6
2425
, 725
, 21010
, 3210
10
• A temperature could occurtwice in a given periodsuch as when thetemperature rises in thespring and falls in autumn.
48. B
50.
52.
54. 222
2425
, $725
, 255
, 225
5
232
, $12,
12, 232
©Glencoe/McGraw-Hill 410 Algebra 2 Chapter 14
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