CCA2 SA Ch12 - Hudson City School District

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© 2013 CPM Educational Program. All rights reserved. Lesson 12.1.1 12-5. a: always b: never c: always d: True for x = π 4 + 2π n and x = 5π 4 + 2π n 12-6. a: ( x 2)( x + 2) b: ( y 9)( y + 9) c: (1 x )(1 + x ) d: (1 sin x )(1 + sin x ) 12-7. a: 80.86 b: 24.05 c: 15.50º trigonometric ratios, Law of Sines, Law of Cosines, Pythagorean theorem 12-8. The graphs of y = sin(2x) and y = 2sin(x) intersect at integer multiples of π. Solving the equation algebraically yields x = 0 + πn. 12-9. a: x 2, f ( x ) 2 b: f 1 ( x ) = ( x 2) 2 2 + 2 c: x 2, f 1 ( x ) 2 12-10. a: Stretched (amplitude = 3), shifted left π 2 , and shifted down 4 b: See graph at right. 12-11. The first process is fully in control. The second process is wildly out of control. The third process is out of control; beginning at the 9 th hour, there are 12 consecutive points above the centerline. 12-12. a: 90 b: 190 c: 35 d: 405,150 12-13. a: 12 P 5 = 95, 040 b: 12 C 5 = 792 12-14. Sample answers: h = π 2 , 5π 2 , 3π 2 12-15. a: negative b: negative c: positive d: negative 12-16. a 25 24 12-17. See graph at right. 12-18. 5 x + 4 12-19. a: x = 9 b: x = –9 12-20. a: 3π 5 b: 16π 9 c: 140º d: 285º e: 1530º f: 13π 9 12-21. a: 5 C 2 i 4 C 1 12 C 3 = 2 11 b: 4 C 3 12 C 3 = 1 55 c: 5 C 1 i 4 C 1 i 3 C 1 12 C 3 = 3 11 d: 5 C 3 + 4 C 3 + 3 C 3 12 C 3 = 3 44 e: 5 C 1 i 4 C 2 12 C 3 = 3 22 f: 1 3 11 + 3 44 ( ) = 29 44 12-22. a: a 3 + 3a 2 b + 3ab 2 + b 3 b: 8 m 3 + 60 m 2 + 150 m + 125 x y

Transcript of CCA2 SA Ch12 - Hudson City School District

Page 1: CCA2 SA Ch12 - Hudson City School District

© 2013 CPM Educational Program. All rights reserved.

Lesson 12.1.1 12-5. a: always b: never c: always d: True for x = π

4 + 2πn and x = 5π4 + 2πn

12-6. a: (x − 2)(x + 2) b: (y − 9)(y + 9) c: (1− x)(1+ x) d: (1− sin x)(1+ sin x) 12-7. a: ≈ 80.86 b: ≈ 24.05 c: ≈ 15.50º trigonometric ratios, Law of Sines, Law of Cosines, Pythagorean theorem 12-8. The graphs of y = sin(2x) and y = 2sin(x) intersect at integer multiples of π. Solving the

equation algebraically yields x = 0 + πn. 12-9. a: x ≥ 2, f (x) ≥ 2 b: f −1(x) = (x−2)2

2 + 2 c: x ≥ 2, f −1(x) ≥ 2 12-10. a: Stretched (amplitude = 3), shifted left π2 , and shifted down 4

b: See graph at right. 12-11. The first process is fully in control. The second process is wildly

out of control. The third process is out of control; beginning at the 9th hour, there are 12 consecutive points above the centerline.

12-12. a: 90 b: 190 c: 35 d: 405,150 12-13. a: 12P5 = 95, 040 b: 12C5 = 792 12-14. Sample answers: h = π

2 ,5π2 , −

3π2

12-15. a: negative b: negative c: positive d: negative 12-16. a ≤ 25

24 12-17. See graph at right. 12-18. 5

x+4 12-19. a: x = 9 b: x = –9 12-20. a: 3π5 b: 16π9 c: 140º d: 285º e: 1530º f: 13π9 12-21. a:

5C2 i4C112C3

= 211 b: 4C3

12C3= 155 c:

5C1i4C1i3C1

12C3= 311

d: 5C3+4C3+3C312C3

= 344 e:

5C1i4C212C3

= 322 f: 1− 3

11 +344( ) = 29

44 12-22. a: a3 + 3a2b + 3ab2 + b3 b: 8m3 + 60m2 +150m +125

x

y

Page 2: CCA2 SA Ch12 - Hudson City School District

2 © 2013 CPM Educational Program. All rights reserved. Core Connections Algebra 2

Lesson 12.1.2 12-29. a: x = π

6 ,5π6 b: x = 5π

6 ,7π6 c: x = π

4 ,3π4 d: x = 0

12-30. No, 52º and 308º have the same value for the cosine, while 128º

has the exact opposite cosine value. See diagram at right. 12-31. See graph at right. 12-32. f −1(x) = −x + 6 12-33. x

x+2 12-34. a: yes b: x4 + x3 + x2 + x +1 ; yes c: xn + xn−1 + xn−2 +…+ x +1

12-35. a: See possible diagrams at right and answers below. Cell A is the proportion of people correctly identified as drug users. Cell B is the proportion correctly identified as not drug users. Cell C is the proportion the test failed to identify as drug users but who are. Cell D is the proportion identified as drug users who are not. This can also be modeled as a tree diagram.

b: 0.02 are actually using; 0.01 are told they are using, but are actually not.

c: 0.00977/(0.00977 + 0.02277) ≈ 30% d: From part (b), about 1 out of 100 people

receiving assistance will lose their assistance because they have been falsely accused of using drugs. That seems high considering that only 2 out of 100 are actually using drugs. From part (c), 30% of the people identified as using drugs will be falsely accused and unfairly lose their money.

e: Yes, they are independent because the accuracy of the test stays the same whether or not a person uses drugs. To test, check whether P(A) ⋅P(B) = P(A and B) , for example, P(drug user) ⋅P(test correct) = P(drug user and test correct) .

12-36. a: 0.0253 b: 26C5

52C5 c: 0.000495 d: 13C5

52C5 e: 0.0019808

12-37. a: 2491 b: 2091

x –x

128º 52º

360º – 52º =308º x

y

Drug User

correct

correct Not Drug User

incorrect

incorrect

0.02277

0.00023

0.98703

0.00977

Recipient

Dru

g Te

st

Not Using

User

Correct

Incorrect

A B

D C

Page 3: CCA2 SA Ch12 - Hudson City School District

Selected Answers © 2013 CPM Educational Program. All rights reserved. 3

Lesson 12.1.3 (Day 1) 12-43. a: 30º, 50º or π6 ,

5π6 b: 120º, 140º or 2π3 ,

4π3

c: 45º, 225º or π4 ,5π4 d: ≈ 35.26º, 144.74º, 215.26º, 324.74º or 0.62, 2.53, 3.76, 5.67

12-44. a: domain: –3 ≤ x ≤ 3, range: –3 ≤ y ≤ 3, not a function b: domain: –3 ≤ x ≤ 4, range: –2 ≤ y ≤ 4, not a function c: domain: x ≤ 3, range: y ≤ 4, yes a function d: domain: −∞ < x < ∞ , range: y ≥ –2, yes a function 12-45. a: x = 2 b: no solution 12-46. a: b: 12-47. a: (4, 8) b: (0, –2, 3)

12-48. a: See possible diagrams at right and answers below. Cell A is the proportion correctly identified as

suspects. Cell B is the proportion correctly identified as not

being suspects. Cell C is the proportion the software failed to

identify but who actually are suspects. Cell D is the proportion the software identified as

suspects who are not. This can also be modeled as a tree diagram.

b: 0.000999995/(0.000004995 + 0.000999995) ≈ 99.5%

12-49. 12C5 +12 C4 +12 C3 +12 C2 +12 C1 +12 C0 = 1586 12-50. 2x4 − 2x + −1

x−3 12-51. a: See graph at right. b: The number of defects seems to be staying at or within

control limits, but there is a cycle apparent every eight hours. Perhaps as the inspectors work through their shift, they get tired and catch fewer errors.

x y

z

3 –2 –2

x y

z

Person

Faci

al ID

Sof

twar

e

Not Suspect

Suspect

Incorrect

A

B

D C

Correct

Suspect

correct

correct Not a Suspect

incorrect

incorrect

0.000004995

0.000000005

0.99899501

0.000999995

Page 4: CCA2 SA Ch12 - Hudson City School District

4 © 2013 CPM Educational Program. All rights reserved. Core Connections Algebra 2

Lesson 12.1.3 (Day 2) 12-52. The restrictions are needed so that the inverses will be functions. The domain of the sine

function is restricted to − π2 ≤ x ≤

π2 , the domain of the cosine function is restricted to

0 ≤ x ≤ π , and the domain of the tangent function is restricted to − π2 < x <

π2 .

12-53. a: x = 7π

6 + 2πn, 11π6 + 2πn b: x = π6 + 2πn,

11π6 + 2πn

c: x = 3π4 +πn d: x = πn

12-54. a: shifted up 1 unit b: shifted left π4 c: reflected over the x-axis d: vertically stretched by a factor of 4 12-55. a: b: c: g(x) = − f (x) 12-56. a: 56 b: 3x+8

2x2 c: x2+2x+3

(x+1)(x−1) d: sin2 θ+cosθsinθ cosθ 12-57. f (x) = 2(x −1)2 −1 ; domain: all real numbers; range: f (x) ≥ −1 ;

vertex: (1, –1); line of symmetry: x = 1; See graph at right. 12-58. a: x ≈ 1.356 b: x ≈ 2.112 c: x ≈ 1.792 12-59. a: a4 + 4a3b + 6a2b2 + 4ab3 + b4 b: 81m4 − 216m3 + 216m2 − 96m +16 12-60. a: 8C3 +8 C4 = 56 + 70 = 126 b: If mushrooms are a known topping, then choose the rest of the toppings from only

7 remaining toppings so 7C3+7C2126 = 56126 =

49

x

y

x

y

y

x

Page 5: CCA2 SA Ch12 - Hudson City School District

Selected Answers © 2013 CPM Educational Program. All rights reserved. 5

Lesson 12.1.4 12-67. a: ac b: cb c: cb d: ba 12-68. a and b: ± π

6 , ±5π6 , ±

7π6 , ±

11π6

12-69. The solution is equivalent to the solutions of cos(x) = 1

2 and sin(x) = 0 . 0°, 180°, 60°, 300°; or 0,π , π3 ,

5π3

12-70. 9.10 12-71. 1.083, 8.3% 12-72. a: a3 + b3 b: x3 − 8 c: y2 +125 d: x3 − y3 e: They consist only of two terms; they are sums or differences of cubes. 12-73. a: (x + y)(x2 − xy + y2 ) b: (x − 3)(x2 + 3x + 9) c: (2x − y)(4x2 + 2xy + y2 ) d: (x +1)(x2 − x +1) 12-74. y = 10

216 (x + 6)3 −10

12-75. a:

3C1i10C313C4

= 360715 ≈ 0.503 b: 11

13C4= 165 = 0.015

12-76. a: ca b: ab c: ba d: ca 12-77. The solution is equivalent to the solutions of sin(x) = − 1

2 and sin(x) = 0 . 90° + 180°n, 210° + 360°n, 330° + 360°n; or π2 +πn,

7π6 + 2πn, 11π6 + 2πn

12-78. 2x2 + 4x −1 12-79. a: x2(x + 2y)(x2 − 2xy + 4y2 ) b: (2y2 − 5x)(4y2 +10xy2 + 25x2 ) c: (x + y)(x2 − xy + y2 )(x − y)(x2 + xy + y2 ) 12-80. Possible equation: y = 1

8 (x − 3)3 + 3 Inverse: y = 8(x − 3)3 + 3

12-81. y = 4(0.4)x + 5 12-82. a: 126a5b4 b: 1120x4y4 12-83. x2 − 6x + 34 = 0 12-84. P(3 or 4 or 5) ≈ 0.99

Page 6: CCA2 SA Ch12 - Hudson City School District

6 © 2013 CPM Educational Program. All rights reserved. Core Connections Algebra 2

Lesson 12.2.1 12-90. a: 1 b: cos(4w) c: tan(θ ) 12-91. ≈ 75.52°, 75.52°, and 28.96° 12-92. a: x = 30º + 360ºn or x = 150º + 360ºn b: no solution 12-93. 3x2 − x + 2 12-94. x3 − 2x2 − 3x + 9 12-95. 1

10! 12-96. (5 – 2)! because 3! > 2! 12-97. a: 18 b: –12 c: –1 + 7i d: –14.5 e: x = 0, –7 12-98. a: p = 44, a = 20 b: y = 20 cos π

22 (x −15)( ) + 3 12-99. Possibilities include: sin2 x = 1− cos2 x , sin x = ± 1− cos2 x ,

sin2 x = (1− cos x)(1+ cos x) 12-100. − 4

5 12-101. a: sinθcosθ b: 1

sinθ c: cosθsinθ d: 1cosθ

12-102. a: y − 9 = 315

2 (x − 2) or y = 3152 x − 306 b: y = 0.25(6)x

12-103. They intersect at 1

2 , 0( ) and (3, 10). 12-104. a: 1

2(x−1) b: 4x3x2+10x+3

12-105. a: 41.41° b: 28.30° 12-106. roots: −0.4 ± 0.8i 6 ; vertex: (–0.4, 19.2); f (x) = 5(x + 0.4)2 +19.2 12-107. a: (0.9)5 ≈ 0.59 b: 10(0.9)2(0.1)3 + 5(0.9)(0.1)4 + (0.1)5 ≈ 0.00856

Page 7: CCA2 SA Ch12 - Hudson City School District

Selected Answers © 2013 CPM Educational Program. All rights reserved. 7

Lesson 12.2.2 (Day 1) 12-109. 90.21 feet or 4.71 feet 12-110. a: 2x4 − x2 + 3x + 5 = (x −1)(2x3 + 2x2 + x + 4)+ 9 b: x5 − 2x3 +1= (x − 3)(x4 + 3x3 + 7x2 + 21x + 63)+190 12-111. a: x = 3π

2 b: x = π3 ,

5π3 c: x = π

4 ,5π4 d: x = 7π

6 ,11π6

12-112. a: x ≈ 69.34 b: x ≈ 5.35 12-113. a: 10P5 = 30, 240 b: 10 i 94 = 65, 610 12-114. ± 3

5 ,45( )

12-115. (x + 4)2 + (y − 6)2 = 64 ; circle, center: (–4, 6), r = 8

See graph at right. 12-116. a: See graph at right below. b: The process has a lot of variability for the first 14 hours.

There are two out-of-control points (one upper and one lower). Apparently an adjustment was made at hour 14 because the process is much less variable, but now there are nine consecutive points above the centerline of 0.075. Apparently another adjustment was made at hour 22, but this apparently swung the process to the very low end.

12-117. a: 2x+13x−2 b: x2+2x+4x(4x+5)

x

y

Page 8: CCA2 SA Ch12 - Hudson City School District

8 © 2013 CPM Educational Program. All rights reserved. Core Connections Algebra 2

Lesson 12.2.2 (Day 2) 12-118. a: (sinθ + cosθ )2 =

sin2 θ + 2 sinθ cosθ + cos2 θ =1+ 2 sinθ cosθ

b: tanθ + cotθ =sinθcosθ + cosθ

sinθ =

sin2 θ+cos2 θsinθ cosθ =1

sinθ cosθ = cscθ secθ

c: (tanθ cosθ ) sin2 θ + 1sec2 θ( ) =

sinθcosθ cosθ( ) (sin2 θ + cos2 θ ) =

(sinθ )(1) = sinθ

12-119. See unit circle at right. θ = π

6 ,5π6 ,

7π6 ,

11π6

12-120. m∠B = 86.17º or 1.5 radians 12-121. The equation of the parabola is y = 1

8 (x − 3)2 + 3 .

It is a function; every input has only one output. 12-122. a: x ≈ 1.839 b: x ≈ –1.839 c: x ≈ 1.839 12-123. a: The two lines intersect at (8, 17).

b: No solution; the lines are parallel. 12-124. a: There is one way to choose all five. 5!

5!0! = 1 . In order to have the formula give a reasonable result for all situations, it is necessary to define 0! as equal to 1.

b: There is one way to choose nothing. 5!0!5! = 1

12-125. AC = 10 inches 12-126. Possible answer: f (x) = x3 − 5x2 + 8x − 6

Page 9: CCA2 SA Ch12 - Hudson City School District

Selected Answers © 2013 CPM Educational Program. All rights reserved. 9

Lesson 12.2.3 12-130. sin π

12 = sinπ3 −

π4( ) = 6− 2

4 ; cos π12 = cos

π3 −

π4( ) = 2+ 6

4 12-131. a: sin π

3 +π4( ) = 6+ 2

4 b: cos 3π4 + π

6( ) = cos 7π6 − π

4( ) = − 6+ 24

12-132. sin(x + x) = sin x cos x + cos x sin x = 2 sin x cos x

cos(x + x) = cos x cos x − sin x sin x = cos2 x − sin2 x 12-133. a: See graph at right.

b: Possible answer: f (x) = − sin x

c: cos x + π2( ) = cos x cos π2 − sin x sin π

2 = − sin x 12-134. sin x + cos x 12-135. Divide by x – 3, then solve the resulting quadratic; x = 1 ± i. 12-136. a: 2 b: a – 2 12-137. 3C1 1

4( )2 34( ) = 9

64 ≈ 0.141

x

y