Chapter 6 48. The function is of the form y = a(1 − r)t, where 1 − r < 1.
So, it represents exponential decay. Use the decay factor 1 − r to find the rate of decay. 4 1−r=— 5 −1 −1 1 −r = −—, or −0.2 5 So, the rate of decay is 20%.
(
4 5
4 5
5 5
1 5
— − 1 = — − — = −—
)
54. f(t) = 0.4(1.16)t−1
(1.16)t = 0.4—1 (1.16) 0.4 = —(1.16)t 1.16 ≈ 0.34(1.16)t The function is of the form y = a(1 + r)t, where 1 + r > 1. So, it represents exponential growth. 55. b(t) = 4(0.55)t+3
49. y = (0.9)t−4
⋅ (0.55) = ⋅ (0.55) = 4(0.166375) ⋅ (0.55)
(0.9)t = —4 (0.9) (0.9)t =— 0.6561 1 = — (0.9)t 0.6561 ≈ 1.52(0.9)t
3
4(0.55)3
t
t
⋅
≈
The function is of the form y = a(1 − r)t, where 1 − r < 1. So, it represents exponential decay.
0.67(0.55)t
The function is of the form y = a(1 − r)t, where 1 − r < 1. So, it represents exponential decay. 56. r(t) = (0.88)4t
= (0.884)t
50. y = (1.4)t+8
≈ (0.60)t
⋅ (1.4) ≈ (1.4) ⋅ (14.8)
y = (1.4)t
= 4(0.55)t
8
The function is of the form y = a(1 − r)t, where 1 − r < 1. So, it represents exponential decay.
t
≈ 14.8(1.4)t
The function is of the form y = a(1 + r)t, where 1 + r > 1. So, it represents exponential growth. 51. y = 2(1.06)9t
= 2(1.069)t
r n
(
)
nt
)
0.05 4t y = 2000 1 + — 4 y = 2000(1 + 0.0125)4t y = 2000(1.0125)4t
≈ 2(1.69)t The function is of the form y = a(1 + r)t, where 1 + r > 1. So, it represents exponential growth. 52. y = 5(0.82)t∕5
= 5(0.82)(1∕5)⋅t
(
r n
58. y = P 1 + —
(
)
nt
)
0.10 2t y = 1400 1 + — 2 y = 1400(1 + 0.05)2t y = 1400(1.05)2t
(
= 5(0.821∕5)t
r n
59. y = P 1 + —
5— t = 5( √0.82 )
≈ 5(0.96)t The function is of the form y = a(1 − So, it represents exponential decay.
(
57. y = P 1 + —
r)t,
where 1 − r < 1.
53. x(t) = (1.45)t∕2
= (1.45)(1∕2)⋅t = (1.451∕2)t = ( √1.45 )
— t
≈ (1.20)t
(
)
nt
)
0.084 12t y = 6200 1 + — 12 y = 6200(1 + 0.007)12t y = 6200(1.007)12t
(
r n
60. y = P 1 + —
(
)
nt
)
0.092 4t y = 3500 1 + — 4 y = 3500(1 + 0.023)4t y = 3500(1.023)4t
The function is of the form y = a(1 + r)t, where 1 + r > 1. So, it represents exponential growth.
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Algebra 1 Worked-Out Solutions
333
Chapter 6 +1 ⤻ +1 ⤻ +1 ⤻ +1 ⤻
61. a. Tree A
Year, t Basal area, A
0
1
120
132
2
3
y = P(1 + r)t y = 300(1.08)t
4
145.2 159.7 175.7
⤻ ⤻⤻⤻ ×1.1 ×1.1 ×1.1 ×1.1
From the table, you know the initial basal area of Tree A is 120 square inches, and it is multiplied by a growth factor of 1.1 each year. So, P = 120 and 1 + r = 1.1. y = P(1 + r)t
So, the function that represents the balance of the investment account is y = 300(1.03)4t, and the function that represents the balance of the savings account is y = 300(1.08)t. b. Investment account:
Year, t
0
1
2
3
Balance (dollars), y 300 337.65 380.03 427.73
y = 120(1.1)t Accounts
y = P(1 + r)t y = 154(1 + 0.06)t y = 154(1.06)t So, the function that represents the basal area of Tree A after t years is y = 120(1.1)t and the basal area of Tree B after t years is y = 154(1.06)t.
Balance (dollars)
The initial basal area of Tree B is 154 square inches, and the rate of growth is 6%, or 0.06.
b. Tree B
y 800 700 600 500 400 300 200 100 0
y = 300(1.03)4t
y = 300(1.08)t 0 1 2 3 4 5 6 7 8 t
Year
Year, t
0
1
2
3
4
Basal area, A 154 163.24 173.03 183.42 194.42
Both accounts start with the same balance. The investment account balance is increasing at a faster rate, so it is greater than the savings account balance after the start.
Basal area (in.2)
Tree Basal Area A 200 175 150 125 100 75 50 25 0
63. a. The initial value is 25,000, and the rate of growth is 5.5%,
or 0.055. AB = 154(1.06)t AA = 120(1.1)t
0 1 2 3 4 t
Year
The basal area of Tree B is larger than the basal area of Tree A, but the difference between the basal areas is decreasing.
y = P(1 + r)t y = 25,000(1 + 0.055)t y = 25,000(1.055)t A function that represents the city’s population is y = 25,000(1.055)t. 1 b. Use the fact that t = —(12t) and the properties of 12 exponents to rewrite the function in a form that reveals the monthly rate of growth. y = 25,000(1.055)t y = 25,000(1.055)(1/12)(12t)
62. a. The principal of the investment account is $300, the
annual interest rate is 6%, or 0.06, and because the interest is compounded quarterly, n = 4.
(
r y=P 1+— n
(
)
nt
)
0.12 4t y = 300 1 + — 4 y = 300(1 + 0.03)4t
y = 25,000(1.055(1/12))(12t) y ≈ 25,000(1.00447)(12t) 1 + r ≈ 1.00447 −1
−1 r ≈ 0.00447
So, the monthly percent increase is about 0.45%.
y = 300(1.03)4t The graph crosses the y-axis at (0, 300). So, the principal of the savings account is $300. The points on the graph are approximately (1, 325), (2, 350), and (3, 375). 350 350 325 Because — ≈ 1.08, — ≈ 1.08, and — ≈ 1.07, the 300 325 325 balance of the savings account has a growth factor of about 1.08. So, P = 300 and 1 + r = 1.08.
334
Algebra 1 Worked-Out Solutions
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