To determine the value of [tex]\(\$1\)[/tex] growing at an annual growth rate of 3.5% over 75 years, we use the formula for compound interest. The compound interest formula is given by:
[tex]\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\][/tex]
where:
- [tex]\(A\)[/tex] is the amount of money accumulated after [tex]\(n\)[/tex] years, including interest.
- [tex]\(P\)[/tex] is the principal amount (\[tex]$1 in this case).
- \(r\) is the annual interest rate (3.5% or 0.035 as a decimal).
- \(n\) is the number of times interest is compounded per year (since it's compounded annually, \(n = 1\)).
- \(t\) is the number of years the money is invested or borrowed for (75 years).
Plugging the given values into the formula, we get:
\[
A = 1 \left(1 + \frac{0.035}{1}\right)^{1 \times 75}
\]
Simplifying inside the parenthesis first:
\[
A = 1 \left(1 + 0.035\right)^{75}
\]
\[
A = 1 \left(1.035\right)^{75}
\]
Calculating \(1.035^{75}\):
This results in approximately \(13.19855\).
So, \(\$[/tex]1\) growing at an annual rate of 3.5% over 75 years would be worth approximately [tex]\( \$13.20\)[/tex].
From the given choices, the closest value is:
[tex]\[
\boxed{13.20}
\][/tex]
