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. 3520
C. 371
D. 225

Answer :

To find the approximate value of [tex]\( P \)[/tex] in the function [tex]\( f(t) = P \cdot e^{r \cdot t} \)[/tex], when we know that [tex]\( f(5) = 288.9 \)[/tex] and [tex]\( r = 0.05 \)[/tex], we can follow these steps:

1. Substitute the known values into the function:
We have [tex]\( f(t) = P \cdot e^{r \cdot t} \)[/tex]. Given [tex]\( f(5) = 288.9 \)[/tex], replace [tex]\( t \)[/tex] with 5 and [tex]\( r \)[/tex] with 0.05:
[tex]\[
288.9 = P \cdot e^{0.05 \times 5}
\][/tex]

2. Calculate the exponent:
Compute the exponent part, [tex]\( e^{0.05 \times 5} \)[/tex]:
[tex]\[
e^{0.25} \approx 1.284
\][/tex]

3. Solve for [tex]\( P \)[/tex]:
Rearrange the equation to solve for [tex]\( P \)[/tex]:
[tex]\[
P = \frac{288.9}{1.284}
\][/tex]

4. Approximate the division:
Dividing [tex]\( 288.9 \)[/tex] by [tex]\( 1.284 \)[/tex] gives:
[tex]\[
P \approx 225
\][/tex]

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

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