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Questions: Algebra BusinessCalculus
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Suppose that $19083 is invested at an interest rate of 39% per year, compounded continuously. What is the account balance after 38 years?
Using \(\displaystyle P=P_0e^{kt}\) gives \(\displaystyle P=19083e^{\frac{39 t}{100}}\). Evaluating at \(\displaystyle t=38\) gives \(\displaystyle P=52106376369\) dollars.
\begin{question}Suppose that \$19083 is invested at an interest rate of 39\% per year, compounded continuously. What is the account balance after 38 years?
\soln{4.5cm}{Using $P=P_0e^{kt}$ gives $P=19083e^{\frac{39 t}{100}}$. Evaluating at $t=38$ gives $P=52106376369$ dollars.}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Suppose that $19083 is invested at an interest rate of 39% per year, compounded continuously. What is the account balance after 38 years? </p> </p>
<p> <p>Using <img class="equation_image" title=" \displaystyle P=P_0e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20P%3DP_0e%5E%7Bkt%7D%20" alt="LaTeX: \displaystyle P=P_0e^{kt} " data-equation-content=" \displaystyle P=P_0e^{kt} " /> gives <img class="equation_image" title=" \displaystyle P=19083e^{\frac{39 t}{100}} " src="/equation_images/%20%5Cdisplaystyle%20P%3D19083e%5E%7B%5Cfrac%7B39%20t%7D%7B100%7D%7D%20" alt="LaTeX: \displaystyle P=19083e^{\frac{39 t}{100}} " data-equation-content=" \displaystyle P=19083e^{\frac{39 t}{100}} " /> . Evaluating at <img class="equation_image" title=" \displaystyle t=38 " src="/equation_images/%20%5Cdisplaystyle%20t%3D38%20" alt="LaTeX: \displaystyle t=38 " data-equation-content=" \displaystyle t=38 " /> gives <img class="equation_image" title=" \displaystyle P=52106376369 " src="/equation_images/%20%5Cdisplaystyle%20P%3D52106376369%20" alt="LaTeX: \displaystyle P=52106376369 " data-equation-content=" \displaystyle P=52106376369 " /> dollars.</p> </p>