A high-yield savings account that compounds interest continuously was opened in 1991. The recursive equation \( f(x) \approx f(x-1) \cdot 1.375 \) shows the amount of money in the account rounded to the nearest cent for each year after 2000, in which \( f(0)=3,500 \). Use the recursive formula to create an exponential equation that models the account's balance, \( A \), after \( t \) years, assuming there were no withdrawals or deposits. What is the correct equation, with \( r \) rounded to three decimal places and \( P \) rounded to two decimal places? While solving, round values to three decimal places. (1 point) \( A=3,500 e^{0.375 t} \) \( A=3,500(1.375)^{t} \) \( A=3,500(1.318)^{t} \) \( A=3,500 e^{0.318 t} \)

To convert the recursive formula \( f(x) \approx f(x-1) \cdot 1.375 \) into an exponential equation, we recognize that \( 1.375 \) represents the growth factor each year. Therefore, we can express the account balance \( A \) after \( t \) years as: \[ A = 3,500(1.375)^t \] Now, we can provide some context: Continuous compounding is a powerful concept. Instead of compounding annually, it's like having your money earn interest every fraction of a moment! This leads to higher returns over time—like magic for your savings! Understanding how this compound interest works can really boost your financial savvy! For a practical tip: When creating or working with such equations, always keep an eye on the base of your exponential expression. In this case, \( 1.375 \) means for every year, your investment grows by 37.5%. If you ever need to adjust your growth rate or invest in something new, knowing how to manipulate the growth factor will come in handy!