algebra 2 answer key

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

PQ245-6457F-PO1[001-019].qxd 7/24/02 12:18 PM Page 1 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-01:


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



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


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


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



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



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

#



• 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



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



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



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 $



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



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



53. B

55.

57.

59.

61.

63. 14

65. Distributive Property


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 ($)


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



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



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



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



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


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



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


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



37. or

39.


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



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



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

PQ245-6457F-P02[020-055].qxd 7/24/02 12:22 PM Page 20 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:


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)



27. D " {!3.6, 0, 1.4, 2}, R " {!3, !1.1, 2, 8}; yes

29. 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



y

O x

y ! 7x " 6

y

O x

y ! 3x

y

O x

("1, "1)

("2.5, 1) ("1, 1)(0, 1)



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,



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



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



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



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



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



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



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



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



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



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



•

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



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


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



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



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



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



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



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.



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



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


8. D " all reals, R " all nonnegative reals

xO

h (x) ! |x " 4|

h(x)




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



23. $1.00


27. D " all reals, R " {3a 0a is an integer.}


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)


26. D " all reals, R " all even integers



xO

g (x) ! !x" # 3

g(x)

xO

f (x) ! 2!x"

f(x)

xO

f (x) ! !x # 3"

f(x)




33. D " all reals, R " {y 0 y # !4}



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}



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)



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)



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



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


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



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



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



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



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



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



• 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



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



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


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



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




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.


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




29. inconsistent



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



32. inconsistent


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



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


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



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



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.


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



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





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


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



Lesson 3-3 Solving Systems of Inequalities by GraphingPages 125–127


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


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




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


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




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


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


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


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)




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)



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)



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


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)



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



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


Page 135

1.

x

y

O

y " x ! 4

y # x ! 0 2.

x

y

O

y ! 3x # 4

y ! x " 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)



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



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


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


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


! ¥


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



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


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


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



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



• Add the three matrices:

43. A

45.

47.

49.

51. (5, 3, 7)

53. (2, 5)

55. (6, !1)

57.



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



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 !


Lesson 4-3 Multiplying MatricesPages 171–174


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



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)



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;



57.

59.


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



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


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



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



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



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


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


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



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




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


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



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),




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


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),



1.

3. Sample answer:

5. yes

7. no inverse exists


11. yes

13. no

15. yes

17. true

19. false


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.


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


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



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


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


50.

52.

54. (0, 7)

F 516

!14

532

!18

0

316

!116

14

!132

VC 3

5

312

25

212

S



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


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


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



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.


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 #




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,

!!

PQ245-6457F-P05[110-134] 7/24/02 1:30 PM Page 110 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-05:


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



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



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



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



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



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



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



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


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



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)



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


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


76. rational

78. rational

80. irrational

c"3618

74d

14x2 $ 26x " 4

x3 $ x2 " 2x $ 2 $1

3x " 2

(x $ 2)(x2 " 2x $ 4)



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


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



47. 6x2z2

49.

51. "3c3d4

53. p $ q

55.


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



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



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



65.


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


Page 256




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



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



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



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



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



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



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



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



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'


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


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


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


x f(x)#3 #3#2 #8

#1 #70 0

#253

#53


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



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)



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


x f(x)#2 #4#1 1

0 01 #7

43

#23


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


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


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


x f(x)

#1

0

#1

1

2 179

#59

13

#89

79


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



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


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


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


#2˛

12, 3

#12, 2˛

12

4. #4, 1

6. #4

8. #4, 6

10. #5


14. 0, 6

16.

18.

20. 0, 3


24. #4, 5

26. #7

28.

30.


34. between #1 and 0, between2 and 3


38. #8, #9

#4, 112

#112, 3

#12, 3

#2, 112



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



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


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


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)


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


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}


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



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


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





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


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



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



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.


57. x 2 # 3x " 2 ! 0

59. 3x 2 # 19x " 6 ! 0


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


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.


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


4c.

6a. 06b. one rational

6c.

8.

10.

12. at about 0.7 s and again atabout 4.6 s


14c.

16a.16b. 2 complex16c.


18c.


#14,

23

1 $ 2i

#16

#3 $ 2212

1 $ 23

0, #8

#12

14, #

52



19c.


21c.


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.


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



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




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


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;



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



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.



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



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



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


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



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



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;



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


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



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


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# $


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#$,f (x)S#$



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


51.

53. 4

55. 8 points

57. C

59. 5x 0 2 ' x ' 66

f(x) !12x3 "

32x2 " 2x


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




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


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



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



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



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



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)


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


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



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


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



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


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



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 #$,f(x ) S "$



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



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


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




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

;



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



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)


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



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


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



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


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



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;



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



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


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


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)}


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




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



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


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;



59. "7

61. 254

60. "


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,



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



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


25.

27.

29.

31.

33. 317.29 mi


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



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;


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;



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


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


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


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


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,


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,


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,


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


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


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


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


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


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


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


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


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;


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



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


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


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);


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


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


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

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y " '

(0, '238);(0, '322);

y 2

4!

x 2

21" 1


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);


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);


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);


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);


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)


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



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


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,


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,


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,


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,


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


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


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,


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


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,


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 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


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



• 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



67.

69. #1118

1415

68.

70. 16

#1116


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


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



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



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



2.

4.

6. x # 1

8.

10. 14

6ax " 20by15a2b3

72

c6b 2



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



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)


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



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 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



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



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



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


42. joint

44. !15md


48. I

dO

I ! 16d 2

/

214

V !kp

13


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




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


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


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



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


34. similar to a parabola

y

xO

y ! 2x 2

y

xO

y ! 3[x ]

y

xO

y ! 4x



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



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



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


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


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



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


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




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


8. growth

10. growth

12.

14. a4$

16. %9

18. 2

20. 22,890,495,000

y ! %18132x

y

xO

y ! 3(4)x




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



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



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



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



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



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

,



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


Lesson 10-2 Logarithms and Logarithmic FunctionsPages 535–538


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


Originalequation

25 ! 52

and 5 ! 51

Originalequation

and 16 ! 242 ! 16

14


65.

Originalequation

8 ! 23

InverseProp. ofExp. andLogarithms

3 ! 31


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



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



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


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


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


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




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


Replace xwith logbmand y withlogbn.

Replace 32with 25 and 8with 23.Quotient ofPowersInverse Prop.of Exp. andLogarithms


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


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.


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

;


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


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




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



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


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



• 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



1.

3. 3

5. 1.3863

log 5log 4

; 1.1610 2. e2 ! 3x

4. x " 5.3219



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%



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



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

,


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



64. 15

66. 5, 4, 3, 2

63. 146.4132

65. 2, 5, 8, 11

67. 11, 15, 19, 23, 27


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


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


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


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

,


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


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,


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,


1. 46

3. 187

5. 1

2.

4. 816

112



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


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



51. C

53. 3.99987793

55.

57. 232

59.


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


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

#


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



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



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


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


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



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


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 #



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 #


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

;



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



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,



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,


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.



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



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:


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.


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



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


"

"

"

"

"

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,


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.



Therefore,

for all

positive integers n.


. 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.


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.




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:


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.


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:


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.



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



Therefore, the number ofbricks in the top n rows is

for all positiveintegers n.


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,


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.



The last expression is theright side of the formula tobe proved, where Thus, the formula is true for

Therefore,


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



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



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


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



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

!


!

!

!

!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



35. 28x6y2

37. 7

39.

41.

43. #1, #2

45. y ! (x $ 3)2 " 2

47.

49. 3

51.


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,

!


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



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



29. combination; 455

31. 60

33. 111,540

35. 80,089,128


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



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)


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!



63. (0, 2)

65.

67. {$7, 15}

69.

71. 15

35

$67

64.

66. 0

68. &

70.

72. 13

12

$43


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


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



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;


73. 16


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;


72. combination; 35

74. 24

76. square root

78. (1, 3)

15

136

' $ 1'

920

120

920



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



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



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


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


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


• 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



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



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.


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


Lesson 12-5 Adding ProbabilitiesPages 660–663


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



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



65. min:max:

67.

(1, 3), (3, 3), (3, 5);max: f(3, 5) ! 23;min:


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


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


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


28. 28.9

30. Washington; see students’work.

32. 64%

34. Different scales are used onthe vertical axes.



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.



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


48.

50.

52.

54.

56.

58. (3, 5)

60. 340

62. 475

64. 495

1$4, 62$2

$3

116

4663

37



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


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


Lesson 12-7 The Normal DistributionPages 673–675


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.)



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


Lesson 12-8 Binomial ExperimentsPages 678–680


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



5rw 4 ! 5 or

about 39.6%; P (5 wrong):

w 5 ! ! or about

23.7%.

41. B

43. normal distribution

45. 10


49.

51. 0.1

53. 0.039

55. 0.041

y

xO

x $ y ! 4

2431024

a34b5

a14b a3

4b4 !

4051024


44. 68%

46. 16%

48.

50.

52. 0.05

54. 0.027

56. 0.031

y

xO

y ! |5x |

y

xO

x ! #3



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%


Lesson 12-9 Sampling and ErrorPages 683–685


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




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!


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;



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



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


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˚



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


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!



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"





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" !


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

,



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,



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



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 ! !



• 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


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



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


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



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


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.


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",



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,



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,



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



Page 738

Lesson 13-6 Circular FunctionsPages 742–745


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



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



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)



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

;


Lesson 13-7 Inverse Trigonometric FunctionsPages 749–751


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



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


x 0 1

y%

2%

2%

2%

2%

2%

2%

2%

2%

2

#1#232

#222

#12

232

222

12



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;


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


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!




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 !




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 !



21. amplitude: 1; period 90" or


period: 120" or


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 !


period: 36" or


!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!






!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


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



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



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



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


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.


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



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



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



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



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;



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



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



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,



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 !



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)


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


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



• 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"


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



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!


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



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


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 !


Lesson 14-4 Verifying Trigonometric IdentitiesPages 784–785


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 !



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 !



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 !



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 !



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



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




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(" % #)



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



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 !


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°



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(" $ #)



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 !




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



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


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)



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



• 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




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 !


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 !



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"


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"



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



algebra 2 answer key

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