High School

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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. 371
C. 225
D. 3520

Answer :

We are given a function

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

with [tex]$r = 0.05$[/tex]. For [tex]$t = 5$[/tex], the function value is

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

### Step 1: Simplify the Exponent

First, compute the exponent for [tex]$t = 5$[/tex]:

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

So the equation becomes:

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

### Step 2: Solve for [tex]$P$[/tex]

Isolate [tex]$P$[/tex] by dividing both sides by [tex]$e^{0.25}$[/tex]:

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

### Step 3: Compute the Value of [tex]$e^{0.25}$[/tex]

The value of [tex]$e^{0.25}$[/tex] is approximately

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

### Step 4: Calculate [tex]$P$[/tex]

Substitute the value of [tex]$e^{0.25}$[/tex] into the equation:

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

Rounding this value to the nearest whole number, we get approximately [tex]$225$[/tex].

### Conclusion

The approximate value of [tex]$P$[/tex] is [tex]$225$[/tex], which corresponds to option C.

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