Solutions Manual For Applied Calculus 3rd Edition By Hughes-Hallett Thrash

SOLUTIONS FOR THE PRETEST 517


PRETEST
Solutions for the Pretest

1. y = 500 + 200x
2. a = 20
3. a1 = 1
4
and a2 = 1
5
4. A: −2, B: −1, C: 0, D: 1.
P
5. P0 =
(1 + r/n)t
6. r = n(P 1/t − 1)
7. 9
8. x = −7
9. (c): 25 − (x − 3)2
10. (0, 8)
11. y


y = x3 + 8
Q
P
x



Figure A.1

(−2, 0)
12. 20
E2 + T
13. (a):
E
u+y
14. (c):
uy
x2 + y 2
15. (b):
xy
16. −2/3
17. x28
18. W 13/6
19. 10uv − 15
5
20. p =
3
5−a
21. x =
7
22. w = 36
15
23. n =
2
A
24. q =
C +D
25. x = 21 y − 2
9
26. x = −
2

,518 APPENDIX /SOLUTIONS

27. s = −3
28. 6x(2x − a2 )
29. Cab(Ca2 − b2 )
30. (b): (x − 3)(x + 5)
31. x = 2, x = −1, x = 5/2
32. 10
33. W = 0 and W = 6.
34. s = g 2 + 3g + 3
35. (θ + 1)2 = x4
36. 4z 2 + 4z
37. a = 2
38. a = 4
39. w = 1
40. 55◦ F
41. M = 0.42
42. 8 cm2
43. 2.5 cm2
44. 4.2%
45. $14,280
46. $36.45




Solutions on Trigonometry

1. 1
2. x = 0, x = π, x = 2π
3. 540◦
4. (2π, 5)

, 11.1 SOLUTIONS 485


CHAPTER ELEVEN
Solutions for Section 11.1

1. Adding the terms, we see that
3 + 3 · 2 + 3 · 22 = 3 + 6 + 12 = 21.
We can also find the sum using the formula for a finite geometric series with a = 3, r = 2, and n = 3:

3(1 − 23 )
3 + 3 · 2 + 3 · 22 = = 3(8 − 1) = 21.
1−2


2. Adding the terms, we see that

50 + 50(0.9) + 50(0.9)2 + 50(0.9)3 = 50 + 45 + 40.5 + 36.45 = 171.95.

We can also find the sum using the formula for a finite geometric series with a = 50, r = 0.9, and n = 4:

50(1 − (0.9)4 )
50 + 50(0.9) + 50(0.9)2 + 50(0.9)3 = = 171.95.
1 − 0.9


3. The sum can be rewritten as
2 + 2(2) + 2(22 ) + · · · + 2(29 ).
This is a finite geometric series with a = 2, r = 2, and n = 10. We have

2(1 − (2)10 )
Sum = = 2046.
1−2


4. We use the formula for the sum of a finite geometric series with a = 20, r = 1.4, and n = 9. We have

20(1 − (1.4)9 )
Sum = = 983.05.
1 − 1.4


5. This is an infinite geometric series with a = 1000 and r = 1.08. Since r > 1, the series diverges and the sum does not
exist.
6. We use the formula for the sum of a finite geometric series with a = 500, r = 0.6, and n = 16. We have

500(1 − (0.6)16 )
Sum = = 1249.65.
1 − 0.6


7. This is an infinite geometric series with a = 30 and r = 0.85. Since −1 < r < 1, the series converges and we have
30
Sum = = 200.
1 − 0.85


8. This is an infinite geometric series with a = 25 and r = 0.2. Since −1 < r < 1, the series converges and we have
25
Sum = = 31.25.
1 − 0.2

, 486 Chapter Eleven /SOLUTIONS

9. We use the formula for the sum of a finite geometric series with a = 1, r = 1/2, and n = 9. We have
1 − (0.5)9
Sum = = 1.9961.
1 − 0.5

10. This is an infinite geometric series with a = 1 and r = 1/3. Since −1 < r < 1, the series converges and we have
1
Sum = = 1.5.
1 − 1/3

11. This is a finite geometric series with a = 3, r = 1/2, and n − 1 = 10, so n = 11. Thus
3(1 − (1/2)11 ) (211 − 1) 3(211 − 1)
Sum = =3·2 11
= .
1 − 1/2 2 210

12. Each term in this series is 1.5 times the preceding term, so this is an infinite geometric series with a = 1000 and r = 1.5.
Since r > 1, the series diverges and the sum does not exist.
13. Each term in this series is half the preceding term, so this is an infinite geometric series with a = 200 and r = 0.5. Since
−1 < r < 1, the series converges and we have
200
Sum = = 400.
1 − 0.5

14. This is an infinite geometric series with a = −2 and r = −1/2. Since −1 < r < 1, the series converges and
−2 4
Sum = =− .
1 − (−1/2) 3

15. Since a = 10 and r = 0.75, we find the partial sums using the formula
10(1 − (0.75)n )
Sn = .
1 − 0.75
For n = 5, we have
10(1 − (0.75)5 )
S5 = = 30.51.
1 − 0.75
For n = 10, we have
10(1 − (0.75)10 )
S10 = = 37.75.
1 − 0.75
For n = 15, we have
10(1 − (0.75)15 )
S15 = = 39.47.
1 − 0.75
For n = 20, we have
10(1 − (0.75)20 )
S20 = = 39.87.
1 − 0.75
As n gets larger, the partial sums appear to be approaching 40, as we expect.
16. We find the partial sums using the formula
250(1 − (1.2)n )
Sn = .
1 − 1.2
For n = 5, we have
250(1 − (1.2)5 )
S5 = = 1,860.40.
1 − 1.2
For n = 10, we have
250(1 − (1.2)10 )
S10 = = 6,489.67.
1 − 1.2
For n = 15, we have
250(1 − (1.2)15 )
S15 = = 18,008.78.
1 − 1.2
For n = 20, we have
250(1 − (1.2)20 )
S20 = = 46,672.00.
1 − 1.2
As n gets larger, the partial sums appear to be growing without bound, as we expect, since r > 1.