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3 years ago

Given an exponential function for compounding interest, A(x)=P(1.02)^x, what is the rate of change?

Mathematics

2 answers:

Leviafan [203]3 years ago

7 0

Answer:Its 2% guys

Step-by-step explanation:

Rudiy273 years ago

5 0

$\bf \qquad \textit{Amount for Exponential Growth}\\\\
A=I(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\\
\end{cases}\\\\
-------------------------------\\\\
\stackrel{A}{A(x)}=\stackrel{I}{P}(1.02)^{\stackrel{t}{x}}\implies A=P(1+\stackrel{r}{0.02})^x\implies \cfrac{r}{100}=0.02
\\\\\\
r=100\cdot 0.02\implies r=\stackrel{\%}{2}$

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