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If [tex]$f(5) = 288.9$[/tex] when [tex]$r = 0.05$[/tex] for the function [tex]$f(t) = P e^{rt}$[/tex], then what is the approximate value of [tex]$P$[/tex]?

A. 24
B. 3520
C. 371
D. 225

Answer :

We start with the equation given by the function evaluated at [tex]$t=5$[/tex]:

[tex]$$
f(5) = P \cdot e^{0.05 \times 5}.
$$[/tex]

Since we know that [tex]$f(5) = 288.9$[/tex], we can substitute:

[tex]$$
P \cdot e^{0.05 \times 5} = 288.9.
$$[/tex]

First, calculate the exponent:

[tex]$$
0.05 \times 5 = 0.25.
$$[/tex]

Thus, the equation becomes:

[tex]$$
P \cdot e^{0.25} = 288.9.
$$[/tex]

Now, to solve for [tex]$P$[/tex], divide both sides of the equation by [tex]$e^{0.25}$[/tex]:

[tex]$$
P = \frac{288.9}{e^{0.25}}.
$$[/tex]

Evaluating [tex]$e^{0.25}$[/tex] gives approximately [tex]$1.2840$[/tex]. Therefore:

[tex]$$
P \approx \frac{288.9}{1.2840} \approx 225.
$$[/tex]

Thus, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{225}$[/tex].

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