Set up the formula to find the balance after 8 years for a total of \( \$ 12,000 \) invested at an annual interest rate of \( 9 \% \) compounded continuously. \( A=12,000\left(1+\frac{9}{365}\right)^{(365)(8)} \) \( A=12,000(e)^{(0.09)(8)} \) \( A=12,000\left(1+\frac{0.09}{365}\right)^{(365)(8)} \) \( A=12,000(e)^{(9)(8)} \)

To find the balance after 8 years for a total of $12,000 invested at an annual interest rate of 9% compounded continuously, we can use the formula for continuous compounding: \[ A = P \times e^{r \times t} \] Where: - \( A \) is the balance after \( t \) years, - \( P \) is the principal amount (initial investment), - \( r \) is the annual interest rate, - \( t \) is the time in years. Given: - \( P = \$12,000 \) - \( r = 9\% = 0.09 \) - \( t = 8 \) years Substitute the values into the formula: \[ A = 12,000 \times e^{0.09 \times 8} \] \[ A = 12,000 \times e^{0.72} \] Now, we can calculate the balance after 8 years using the formula. Calculate the value by following steps: - step0: Calculate: \(12000e^{0.72}\) - step1: Convert the expressions: \(12000e^{\frac{18}{25}}\) - step2: Rewrite the expression: \(12000\sqrt[25]{e^{18}}\) The balance after 8 years for a total of $12,000 invested at an annual interest rate of 9% compounded continuously is approximately $24,653.20.