How much would $400 invested at 9% interest compounded continuously be worth after 3 years? Rounded by the nearest cent

1 answer:

4 0

$\bf \qquad \textit{Continuously Compounding Earned Amount}
\\\\
A=Pe^{rt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$400\\
r=rate\to 9\%\to \frac{9}{100}\to &0.09\\
t=years\to &3
\end{cases}
\\\\\\
A=400e^{0.09\cdot 3}$

notice, "e" is the Euler's constant, and yes, is on your calculator.

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An actor invested some money at 5% simple interest, and $41,000 more than 4 times the amount at 9%. The total annual interest ea

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The amount invested at 5%=$77,000
The amount invested at 9%=$349,000

Step-by-step explanation:

Let the amount invested at 5% simple interest =$x

He invested $41,000 more than 4 times the amount at 9%.

This amount is: $(4x+41000)

Total Annual Interest Earned = $35,260

Therefore, Time=1 year

Simple Interest $=\frac{Principal X Rate X Time}{100}$

Therefore, his total interest

=Interest from Investment 1 + Interest from Investment 2

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