Respuesta :
[tex]\bf \textit{Amount of Population Growth}\\\\
A=Ie^{rt}\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}\\\\
-------------------------------\\\\
\textit{now, it was 500, in increased by 25\% so }500+\stackrel{\textit{25\% of 500}}{125}\implies 625[/tex]
[tex]\bf \textit{in 10 years}\implies \begin{cases} A=625\\ I=500\\ t=10 \end{cases}\implies 600=500e^{r10}\implies \cfrac{600}{500}=e^{10r} \\\\\\ \cfrac{625}{500}=e^{10r}\implies ln\left(\frac{625}{500} \right)=ln(e^{10r})\implies ln\left(\frac{625}{500} \right)=10r \\\\\\ \cfrac{ln\left(\frac{625}{500} \right)}{10}=r\implies 0.022314\approx r[/tex]
[tex]\bf -------------------------------\\\\ A=500e^{0.0.022314t}\qquad \qquad \stackrel{t=50}{A}=500e^{0.022314\cdot 50}\implies A\approx 1526 [/tex]
how fast is the population growing? well, I'd think is "r", or the rate, which is 0.022314 * 100 or 2.2314%.
[tex]\bf \textit{in 10 years}\implies \begin{cases} A=625\\ I=500\\ t=10 \end{cases}\implies 600=500e^{r10}\implies \cfrac{600}{500}=e^{10r} \\\\\\ \cfrac{625}{500}=e^{10r}\implies ln\left(\frac{625}{500} \right)=ln(e^{10r})\implies ln\left(\frac{625}{500} \right)=10r \\\\\\ \cfrac{ln\left(\frac{625}{500} \right)}{10}=r\implies 0.022314\approx r[/tex]
[tex]\bf -------------------------------\\\\ A=500e^{0.0.022314t}\qquad \qquad \stackrel{t=50}{A}=500e^{0.022314\cdot 50}\implies A\approx 1526 [/tex]
how fast is the population growing? well, I'd think is "r", or the rate, which is 0.022314 * 100 or 2.2314%.
