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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^{0.05t}[/tex], then what is the approximate value of [tex]P[/tex]?

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

Answer :

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

1. Identify the function and variables:
- [tex]\( f(t) = Pe^{rt} \)[/tex]
- Given values: [tex]\( f(5) = 288.9 \)[/tex], [tex]\( r = 0.05 \)[/tex], [tex]\( t = 5 \)[/tex].

2. Set up the equation using the given values:
[tex]\[
288.9 = P \cdot e^{0.05 \times 5}
\][/tex]

3. Calculate the exponent:
- First, compute the exponent in the expression [tex]\( e^{0.05 \times 5} \)[/tex]:
[tex]\[
0.05 \times 5 = 0.25
\][/tex]

4. Calculate [tex]\( e^{0.25} \)[/tex]:
- Use the fact that [tex]\( e^{0.25} \approx 1.284 \)[/tex].

5. Solve for [tex]\( P \)[/tex]:
- Substitute [tex]\( e^{0.25} \approx 1.284 \)[/tex] into the equation:
[tex]\[
288.9 = P \cdot 1.284
\][/tex]
- Divide both sides by 1.284 to solve for [tex]\( P \)[/tex]:
[tex]\[
P = \frac{288.9}{1.284} \approx 225
\][/tex]

6. Answer:
- The approximate value of [tex]\( P \)[/tex] is 225.

Therefore, the correct choice is C. 225.

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