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Questions: Algebra BusinessCalculus
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Suppose that $14141 is invested at an interest rate of 18% per year, compounded continuously. What is the account balance after 30 years?
Using \(\displaystyle P=P_0e^{kt}\) gives \(\displaystyle P=14141e^{\frac{9 t}{50}}\). Evaluating at \(\displaystyle t=30\) gives \(\displaystyle P=3130908\) dollars.
\begin{question}Suppose that \$14141 is invested at an interest rate of 18\% per year, compounded continuously. What is the account balance after 30 years?
\soln{4.5cm}{Using $P=P_0e^{kt}$ gives $P=14141e^{\frac{9 t}{50}}$. Evaluating at $t=30$ gives $P=3130908$ 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 $14141 is invested at an interest rate of 18% per year, compounded continuously. What is the account balance after 30 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=14141e^{\frac{9 t}{50}} " src="/equation_images/%20%5Cdisplaystyle%20P%3D14141e%5E%7B%5Cfrac%7B9%20t%7D%7B50%7D%7D%20" alt="LaTeX: \displaystyle P=14141e^{\frac{9 t}{50}} " data-equation-content=" \displaystyle P=14141e^{\frac{9 t}{50}} " /> . Evaluating at <img class="equation_image" title=" \displaystyle t=30 " src="/equation_images/%20%5Cdisplaystyle%20t%3D30%20" alt="LaTeX: \displaystyle t=30 " data-equation-content=" \displaystyle t=30 " /> gives <img class="equation_image" title=" \displaystyle P=3130908 " src="/equation_images/%20%5Cdisplaystyle%20P%3D3130908%20" alt="LaTeX: \displaystyle P=3130908 " data-equation-content=" \displaystyle P=3130908 " /> dollars.</p> </p>