Respuesta :
[tex]\bf \qquad \textit{Amount for Exponential Growth}\\\\
A=I(1 + r)^t\qquad
\begin{cases}
A=\textit{accumulated amount}\\
I=\stackrel{multiplier}{\textit{initial amount}}\\
r=\stackrel{increase~rate}{rate\to r\%\to \frac{r}{100}}\\
t=\textit{elapsed time}\\
\end{cases}\\\\
-------------------------------[/tex]
[tex]\bf \textit{we know that in }\stackrel{year~0}{2000}\textit{ there were }\stackrel{A}{\$25}\implies 25=I(1+r)^0 \\\\\\ 25=I(1)\implies 25=I\qquad \qquad \boxed{A=25(1+r)^t}\\\\ -------------------------------\\\\ \textit{we also know that in }\stackrel{year~10}{2010}\textit{ it turned to }\stackrel{A}{72}\implies 72=25(1+r)^{10}[/tex]
[tex]\bf \cfrac{72}{25}=(1+r)^{10}\implies \sqrt[10]{\frac{72}{25}}=1+r\implies \sqrt[10]{\frac{72}{25}}-1=r \\\\\\ 0.111576\approx r\implies r\%=100\cdot 0.111576\implies r\approx \stackrel{\%}{11.1576}[/tex]
[tex]\bf \textit{we know that in }\stackrel{year~0}{2000}\textit{ there were }\stackrel{A}{\$25}\implies 25=I(1+r)^0 \\\\\\ 25=I(1)\implies 25=I\qquad \qquad \boxed{A=25(1+r)^t}\\\\ -------------------------------\\\\ \textit{we also know that in }\stackrel{year~10}{2010}\textit{ it turned to }\stackrel{A}{72}\implies 72=25(1+r)^{10}[/tex]
[tex]\bf \cfrac{72}{25}=(1+r)^{10}\implies \sqrt[10]{\frac{72}{25}}=1+r\implies \sqrt[10]{\frac{72}{25}}-1=r \\\\\\ 0.111576\approx r\implies r\%=100\cdot 0.111576\implies r\approx \stackrel{\%}{11.1576}[/tex]
