High School

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

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

Answer :

We are given the function

[tex]$$
f(t) = P \, e^{rt},
$$[/tex]

with [tex]$r = 0.05$[/tex] and [tex]$f(5) = 288.9$[/tex]. First, substitute [tex]$t = 5$[/tex] and [tex]$r = 0.05$[/tex] into the function:

[tex]$$
f(5) = P \, e^{0.05 \times 5} = P \, e^{0.25}.
$$[/tex]

We obtain the equation

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

To solve for [tex]$P$[/tex], we divide both sides by [tex]$e^{0.25}$[/tex]:

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

It is known that

[tex]$$
e^{0.25} \approx 1.2840254166877414.
$$[/tex]

Thus,

[tex]$$
P \approx \frac{288.9}{1.2840254166877414} \approx 224.99554622932885.
$$[/tex]

Rounded to the nearest whole number, [tex]$P$[/tex] is approximately [tex]$225$[/tex].

Therefore, the correct answer is [tex]$\boxed{225}$[/tex].