\( \$ 1000 \) are deposited in an account with \( 8.5 \% \) interest rate, compounded continuously. What is the balance after 5 years? \( \mathrm{P}=\$ 1000 \quad \mathrm{r}=0.085 \quad \mathrm{t}=5 \) \( \mathrm{~F}= \) Pe \( ^{r t} \quad \mathrm{~F}=\$[?] \)

To find the balance after 5 years with continuous compounding, we can use the formula for continuous compounding: \[ F = Pe^{rt} \] Given: - Principal amount \( P = \$1000 \) - Interest rate \( r = 0.085 \) (8.5%) - Time \( t = 5 \) years Substitute the given values into the formula: \[ F = 1000e^{0.085 \times 5} \] Now, we can calculate the balance after 5 years. Calculate the value by following steps: - step0: Calculate: \(1000\times \exp \left(0.085\times 5\right)\) - step1: Multiply the numbers: \(1000\times \exp \left(0.425\right)\) - step2: Calculate: \(1000e^{0.425}\) - step3: Simplify: \(1000e^{\frac{17}{40}}\) - step4: Rewrite the expression: \(1000\sqrt[40]{e^{17}}\) The balance after 5 years with continuous compounding is approximately $1529.59.