Answer :
To find the value of [tex]\( P \)[/tex] given the function [tex]\( f(t) = P e^{r t} \)[/tex], we can follow these steps:
1. Identifying Known Values:
- [tex]\( f(5) = 288.9 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( t = 5 \)[/tex]
2. Substituting Known Values into the Function:
The function is [tex]\( f(t) = P e^{r t} \)[/tex]. With [tex]\( t = 5 \)[/tex], the equation becomes:
[tex]\[
f(5) = P e^{0.05 \cdot 5}
\][/tex]
3. Calculating the Exponential Term:
The exponent is:
[tex]\[
0.05 \times 5 = 0.25
\][/tex]
So, [tex]\( e^{0.25} \)[/tex].
4. Finding [tex]\( e^{0.25} \)[/tex]:
The value of [tex]\( e^{0.25} \)[/tex] is approximately [tex]\( 1.2840254166877414 \)[/tex].
5. Rewriting the Equation:
Now, we have:
[tex]\[
288.9 = P \times 1.2840254166877414
\][/tex]
6. Solving for [tex]\( P \)[/tex]:
To isolate [tex]\( P \)[/tex], divide both sides of the equation by [tex]\( 1.2840254166877414 \)[/tex]:
[tex]\[
P = \frac{288.9}{1.2840254166877414} \approx 224.99554622932885
\][/tex]
Therefore, the approximate value of [tex]\( P \)[/tex] is [tex]\( 225 \)[/tex].
Answer:
C. 225
1. Identifying Known Values:
- [tex]\( f(5) = 288.9 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( t = 5 \)[/tex]
2. Substituting Known Values into the Function:
The function is [tex]\( f(t) = P e^{r t} \)[/tex]. With [tex]\( t = 5 \)[/tex], the equation becomes:
[tex]\[
f(5) = P e^{0.05 \cdot 5}
\][/tex]
3. Calculating the Exponential Term:
The exponent is:
[tex]\[
0.05 \times 5 = 0.25
\][/tex]
So, [tex]\( e^{0.25} \)[/tex].
4. Finding [tex]\( e^{0.25} \)[/tex]:
The value of [tex]\( e^{0.25} \)[/tex] is approximately [tex]\( 1.2840254166877414 \)[/tex].
5. Rewriting the Equation:
Now, we have:
[tex]\[
288.9 = P \times 1.2840254166877414
\][/tex]
6. Solving for [tex]\( P \)[/tex]:
To isolate [tex]\( P \)[/tex], divide both sides of the equation by [tex]\( 1.2840254166877414 \)[/tex]:
[tex]\[
P = \frac{288.9}{1.2840254166877414} \approx 224.99554622932885
\][/tex]
Therefore, the approximate value of [tex]\( P \)[/tex] is [tex]\( 225 \)[/tex].
Answer:
C. 225
