Answers to Precalculus Unit 1 Practicepshs.psd202.org/documents/jfinley/1506287062.pdfA4 SpringBoard...
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A1© 2015 College Board. All rights reserved. SpringBoard Precalculus, Unit 1 Practice
LeSSon 1-1 1. a. Total number of Bracelets
Week n 1 2 3 4 5
Total number of Bracelets Made Bn
128 144 160 176 192
Answers to Precalculus Unit 1 Practice
b. 208, 224, 240, 256, 272
c.
n
Bn
100125150175200
Tota
l Num
ber
of B
race
lets
225250275300
1 2 3 4 5
Week 6 7 8 9 10
Total Number of BraceletsMade
d. Bn 5 112 1 16n or Bn 5 128 1 16(n 2 1)
2. C
3. a. an 5 22 1 4n or an 5 2 1 4(n 2 1)
b.
n
an
5101520253035404550
1 2 3 4 5 6 7 8 9 10
4. a. 2, 8, 18, 32, 50, 72, 98, 128, 162, 200
b.
n
An
20406080
100120140160180
220
1 2 3 4 5 6 7 8 9 10
200
5. The graph for {An 1 1} increases more quickly than for the same n-values of {an} because {An 1 1} is a sequence that adds the values of the terms in {an}.
LeSSon 1-2 6. $153,043.03
7. Sample answer: She will not have 245 pencils because the values for the sequence are the total number of pencils each week.
8. D
9. 3015
10. n # 6
LeSSon 1-3 11. no, because logically the temperature cannot
increase infinitely
12. a. 0, 214
, 229
, 2316
; sample answer: Yes, the next
three numbers are 2425
, 2536
, and 2649
,
which are all negative.
A2© 2015 College Board. All rights reserved. SpringBoard Precalculus, Unit 1 Practice
b. Sample answer: Yes, the formula f (n) 5 n12 2
n1
always subtracts a larger number from a smaller number, which will always be negative.
13. A
14. For n 5 1: 1 2 1 5 1(1 1)
22
, so Step 1 is verified
for n 5 1. Assume that k 2 1 5 k k( 1)
22
is true.
So k k( 1)( 1 1)
21 1 2
5 k k( 1)
21
.
15. Sample answer: No, concluding that a statement is true comes from logical induction of the conditions initially set.
LeSSon 2-1 16. D
17. a.
102030405060708090
100110120130
1 2 3 4 5 6 7 8 9 10n
an
b. Sample answer: No, he is not correct because the points are not linear.
18. Sample answer: No, it is not true because in 30 years the investment will be worth less than $1 million. It will take about 100 years to be worth $1 million.
19. n 5 3
20. 803
LeSSon 2-2
21. S4 5 858 and S5 5
34132
22. A
23. 68.2448
24. ∑65(0.2)kk 1
5
5
25. 111.111
LeSSon 2-3 26. a. convergent to 0
b. divergent
c. divergent
d. convergent to 0
27. A
28. 310
3100
31000
...1 1 1
29. 6.85810
30. 79
LeSSon 3-1 31. a. The population is increasing each year.
b. 7.8 billion
c. u0 5 7.3; un 5 1.0114un21
d. 62 years
32. C 33. u0 5 7; un 5 un21 1 5
34. 11,264
35. 64,766
LeSSon 3-2 36. an 5 7.3(1.0114)n
37. Sample answer: The explicit expression has the variable as an exponent.
38. an 5 23.5 1 11.5n
39. a55 5 20.2105
40. C
LeSSon 4-1 41. C
42. Year 2232
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43. a. M(t) 5 1000(1.1)t
b. Sample answer: No, it is not true because in 50 years the investment will be worth about $117,391. It will actually take 73 years for the mutual fund to be worth more than $1 million.
44. about 28.5
45. a. neither
b. both
c. both
d. neither
e. both
LeSSon 4-2 46. a. D
b. A2(t) 5 35,572(1.1169)t
47. $36,327; $980,702
48. 4739 years
49. Sample answer: Neither will reach the goal of $2 million. Instead of investing, I would spend the money on college since a bachelor’s degree holder will earn at least a third more than a high school diploma holder. [Source: http://education-portal.com/articles/How_Much_More_Do_College_Graduates_Earn_Than_Non-College_Graduates.html]
50. about 37 years
LeSSon 4-3 51. a. 3.47%
b. 3.5%
52. Sample answer: Investing in a home is more lucrative than investing in a savings account, but it is less rewarding than investing in the stock market. One other factor to consider is risk; a savings account is a lot less risky than the stock market.
53. A
54. $63.80
55. 20.00866
LeSSon 5-1 56. B 57. a.
x
y
023 22 21
21
22
23
3
2
1
1 2 3
b. domain: (0, ∞); range: (2∞, ∞); x-intercept: 1; asymptote: x 5 0; end behavior: as x → 0, y → 2∞ and x → ∞, y → ∞.
58. 428
59. 2.4
60. 0.0000316
LeSSon 5-2 61. 1.269
62. Sample explanation: The equation 75(1.05)t 5 150 can be used to determine the time t in years for the number of users to reach 150 million. Divide both sides of the equation by 75: 1.05t 5 2. Write the logarithmic form of the equation: t 5 log1.05 2. Apply
the Change of Base Formula: t 5 log 2
log 1.05 ≈ 14. So,
it will take about 14 years for the number of users to reach 150 million.
63. log x 1
216
3 2
64. y log x 1 log z
65. C
LeSSon 5-3 66. 2.46
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67. a. 10,000 5 22(22t)
b. t ≈ 4.41 hours
c. It will take about 4.4 hours for the population to reach 10,000.
d. In 4 hours, the population will have doubled 8 times, so the population will be 22(28) 5 5632. In 4.5 hours, the population will have doubled 9 times, so the population will be 22(29) 5 11,264. So, it is reasonable that the population will be 10,000 in close to 4.4 hours.
68. x 5 1.61
69. x ≈ 2.95
70. x 5 7 3 52
6
LeSSon 6-1 71. a. g (x) 5 3|x 2 2| 1 1
b.
x
g(x)
21
21
1
2
3
1 2 3 4 5
72. a. g(x) 5 216t2 1 20t 1 5
b.
x
g(x)
22
22
5
10
5
73. a. g(x) 5 ln 2x 1 1
b.
x
g(x)
21
22
23
25 24 23
1
2
3
22 121
74. f (x) is neither odd nor even since f (2x) fi f (x) and f (2x) fi f (x). There is symmetry about the y-axis between x 5 20.5 and x 5 0.5.
75. a. T(t) 5 216t2 1 17t 1 3.5
b. 3.5 feet; 17 feet per second
LeSSon 6-2
76. a. f (x) 5
x xx xx x
0.2 31,8660.4 6373 31866 150,0000.45 53,627 150,000
,
1 < <
1 .
b. f (x) 5
x xx xx x
0.22 31,8660.42 7011 31866 150,0000.47 56,628 150,000
,
1 < <
1 .
77. a. g (x) 5 x x
x x
2 if 0if 0
3
2
2 >
,
b.
x
g(x)
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78. g (x) 5 x x
x x
2 if 0if 0
3
2
1 <
2 .
x
g(x)
79. ( f 1 g)(x) 5 3x 2 1 2x 2 1; D :
( f 2 g)(x) 5 2x 2 1 2x 1 1; D :
( f g)(x) 5 2x 4 1 4x 3 2 x 2 2 2x; D :
fg
(x) 5 xx
22 2
2
21
2; D : x x 1
2≠
| 6
80. ( f 1 g)(x) 5 3x 1 2 1 32x; D :
( f 2 g)(x) 5 3x 1 2 2 32x; D :
( f g)(x) 5 3x 1 2; D :
fg
(x) 5 32x 1 2; D :
LeSSon 7-1
81.
x
y
200400600800
1000
Acc
iden
ts 12001400160018002000
10 20 30 40 50
Speed Limit (mph)60 70 80
Hwy. 123
82. a. Sample answer: The scatterplot shows that the data points do not appear to lie on or very close to a single line. In addition, a linear regression of the data has a correlation coefficient of 0.543, which does not indicate a strong linear correlation.
b. Take the logarithm to the base 10 of the speed limit data and the logarithm to the base 10 of the accident data.
c. y 5 3.085x 2 3.004
d. r 5 0.921; Sample answer: The correlation coefficient for the transformed data is much closer to 1 than the correlation coefficient for the original data. This result indicates that a power function is a better model of the original data than a linear function.
83. D
84. about 1,060 accidents
85. about 41 mph
LeSSon 7-2 86.
x
y
5
10
15
Acc
iden
ts
20
5 10
Speed Limit (mph)
15 20 25
Hwy. 456
87. C
88. Sample answer: f (25) ≈ 400 and f (50) ≈ 6,750 makes sense since the higher the speed of vehicles on a road, the more likely it is that accidents will happen. All of the other choices go against this basic principle.
89. domain: ; range: ; symmetry: origin; max/min: neither; end behavior: y → ∞ as x → ∞, y → 2∞ as x → 2∞; increasing
90. domain: ; range: y # 0; symmetry: y-axis; max/min: maximum at (0, 0); end behavior: y → 2∞ as x → ∞, y → 2∞ as x → 2∞; increasing then decreasing
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LeSSon 8-1 91. f ( g(x)) 5
x4164
2; {x | x fi 22, 2, 6i 4}
g ( f (x)) 5
x
4 4
216; {x | x fi 0}
92. f ( g(x)) 5 x4 22 ; {x | 22 # x # 2}
g ( f (x)) 5 6 2 x 2 16; {x ∈ }
93. C
94. g (x) 5 80x3; f (x) 5 0.67x; f (g(x)) 5 53.6x3; {x | x $ 0}; f (g (x)) represents the value in dollars of a cube of loam with a side length of x feet.
95. f (g(10)) 5 $53,600
LeSSon 8-2 96. B
97. f 21(x) 5 x3 33 2 1 1; D:
98. yes
f ( g(x)) 5
x
13
3 1
13
1
2 5 x
g ( f (x)) 5
x
3
3 1 13
1
2 1
5 x
99. No, g(x) includes only the negative side of the function.
100. a. f 21(x) 5 x3
53
2 . The domain of f is restricted
to {x | x $ 22}. The domain of f 21 is {x | x $ 21} and the range of f 21 is {y | y $ 22}.
b.
x
y
22
24
22 2
2
4
4 6 8 10