College

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

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

Answer :

Sure, let's solve the problem step-by-step using the information provided.

We are given the function:

[tex]\[ f(t) = P \cdot e^{r \cdot t} \][/tex]

where [tex]\( f(5) = 288.9 \)[/tex], [tex]\( r = 0.05 \)[/tex], and [tex]\( t = 5 \)[/tex].

We need to find the value of [tex]\( P \)[/tex].

1. First, plug the known values into the equation:

[tex]\[ 288.9 = P \cdot e^{0.05 \cdot 5} \][/tex]

2. Simplify the exponent:

[tex]\[ e^{0.25} \][/tex]

3. Solve for [tex]\( P \)[/tex] by rearranging the equation:

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

4. Calculate [tex]\( e^{0.25} \)[/tex]. The approximate value of [tex]\( e^{0.25} \)[/tex] is 1.2840.

5. Now, divide 288.9 by 1.2840 to find [tex]\( P \)[/tex]:

[tex]\[ P \approx \frac{288.9}{1.2840} \approx 225 \][/tex]

Therefore, the approximate value of [tex]\( P \)[/tex] is 225.

The correct answer is C. 225.