How much should be invested now at 4.65% compounded annually to have $42,000 in 11 years?

To determine how much should be invested now to have $42,000 in 11 years at an interest rate of 4.65% compounded annually, we use the formula for compound interest:

\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]

Where:
- $ A $ is the amount of money you want in the future ($42,000).
- $ P $ is the initial investment (the amount we need to find).
- $ r $ is the annual interest rate (4.65% or 0.0465).
- $ n $ is the number of times the interest is compounded per year (since it's compounded annually, $ n = 1 $).
- $ t $ is the time in years (11 years).

We need to solve for $ P $:

\[
P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}
\]

Substituting the given values:

\[
P = \frac{42,000}{\left(1 + \frac{0.0465}{1}\right)^{1 \times

Find out how much to invest now to achieve $42,000 in 11 years with an interest rate of 4.65% compounded annually. This problem covers key concepts in compound interest and exponential growth, suitable for high school students in Grades 10-12.

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