Solving A = Pet for P, we obtain P = Ae which is the present value of the amount A due in t years if money earns interest at an annual nominal rate r compounded continuously. (A) Graph P=16,000e 0.051, 0sts 90. (B) Find lim 16,000 0.05t [Guess, using part (A).] [Conclusion: The longer the time until the amount A is due, the smaller is its present value, as we would expect.] (A) Choose the correct graph of P = 16,000 0.05, Ost≤ 90. ○ A. Q 16000- -0.05t, (B) lim 16,000e - O B. 16000 O D. 16000- Q 16000- Q

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Solving A = Pet for P, we obtain P = Aet which is the present value of the amount A due in t years if money earns interest at an annual nominal rate r compounded continuously.
(A) Graph P = 16,000ē-0.05t
, Ost≤90.
(B) Find lim→→
16,000e 0.05t
[Guess, using part (A).]
[Conclusion: The longer the time until the amount A is due, the smaller is its present value, as we would expect.]
(A) Choose the correct graph of P = 16,000e 0.05t, Ost≤ 90.
○ A.
16000-
90
(B) lim → 16,000e
-0.05t
=
B.
16000-
ос.
16000-
90
○ D.
16000-
90
Q
Q
☑
Transcribed Image Text:Solving A = Pet for P, we obtain P = Aet which is the present value of the amount A due in t years if money earns interest at an annual nominal rate r compounded continuously. (A) Graph P = 16,000ē-0.05t , Ost≤90. (B) Find lim→→ 16,000e 0.05t [Guess, using part (A).] [Conclusion: The longer the time until the amount A is due, the smaller is its present value, as we would expect.] (A) Choose the correct graph of P = 16,000e 0.05t, Ost≤ 90. ○ A. 16000- 90 (B) lim → 16,000e -0.05t = B. 16000- ос. 16000- 90 ○ D. 16000- 90 Q Q ☑
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