High School

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. 371
B. 3520
C. 225
D. 24

Answer :

To find the approximate value of [tex]\( P \)[/tex], we need to use the function [tex]\( f(t) = P \cdot e^{r \cdot t} \)[/tex]. We are given that [tex]\( f(5) = 288.9 \)[/tex] and [tex]\( r = 0.05 \)[/tex].

Here’s how we can find [tex]\( P \)[/tex]:

1. Set up the equation for [tex]\( f(t) \)[/tex]:
[tex]\[
288.9 = P \cdot e^{0.05 \cdot 5}
\][/tex]

2. Calculate the exponent:
[tex]\[
e^{0.05 \cdot 5} = e^{0.25}
\][/tex]

3. Calculate the approximate value of [tex]\( e^{0.25} \)[/tex]:
[tex]\[
e^{0.25} \approx 1.28
\][/tex]
(Note: This is an approximation based on typical values.)

4. Solve for [tex]\( P \)[/tex]:
[tex]\[
P = \frac{288.9}{e^{0.25}}
\][/tex]

5. Substitute the value of [tex]\( e^{0.25} \)[/tex] into the equation:
[tex]\[
P = \frac{288.9}{1.28}
\][/tex]

6. Calculate [tex]\( P \)[/tex]:
[tex]\[
P \approx 225
\][/tex]

Therefore, the approximate value of [tex]\( P \)[/tex] is 225, which corresponds to option C.