To solve for the function [tex]\( f(t) = P e^{r t} \)[/tex], given [tex]\( P = 7 \)[/tex] and [tex]\( r = 0.07 \)[/tex], we need to determine the value of [tex]\( f(7) \)[/tex] to the nearest tenth.
Let's break it down step by step:
1. Identify the given values:
- [tex]\( P = 7 \)[/tex]
- [tex]\( r = 0.07 \)[/tex]
- [tex]\( t = 7 \)[/tex]
2. Substitute the values into the function [tex]\( f(t) = P e^{r t} \)[/tex]:
[tex]\[
f(7) = 7 \cdot e^{0.07 \cdot 7}
\][/tex]
3. Calculate the exponent [tex]\( 0.07 \cdot 7 \)[/tex]:
[tex]\[
0.07 \cdot 7 = 0.49
\][/tex]
4. Find the value of [tex]\( e^{0.49} \)[/tex]:
We know the base of the natural logarithm, [tex]\( e \)[/tex], is approximately 2.71828. Evaluating [tex]\( e^{0.49} \)[/tex]:
[tex]\[
e^{0.49} \approx 1.63231
\][/tex]
5. Multiply this result by 7:
[tex]\[
7 \cdot 1.63231 \approx 11.426213539687653
\][/tex]
6. Round this value to the nearest tenth:
[tex]\[
11.426213539687653 \approx 11.4
\][/tex]
Therefore, the value of [tex]\( f(7) \)[/tex] to the nearest tenth is [tex]\( 11.4 \)[/tex].
The correct answer is:
D. 11.4
