Mathematics, 17.01.2020 09:31 AWFHayami
Given an exponential function for compounding interest, a(x) = p(1.04)x, what is the rate of change? (1 point) a. 4% b. 1.04% c. 104% d. 0.04%
Answers: 1
![\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\\ r=rate\to r\%\to \frac{r}{100}\\ n= \begin{array}{llll} \textit{times it compounds per year} \end{array}\\ t=years \end{cases} \\\\[-0.35em] ~\dotfill](/tpl/images/0459/2142/e794e.png)
![\bf A(x)=P(1.04)^x\implies A(x)=P(1+0.04)^x\implies A(x)=P\left(1+\frac{0.04}{1} \right)^x \\\\[-0.35em] ~\dotfill\\\\ r=0.04\implies \stackrel{\textit{to percentage}}{r\% = 0.04\cdot 100}\implies r=\stackrel{\%}{4}](/tpl/images/0459/2142/debb0.png)
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Given an exponential function for compounding interest, a(x) = p(1.04)x, what is the rate of change?...
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