Mathematics, 10.11.2019 07:31 itzzzabbi
Leaves were falling off a tree at an exponential rate beginning oct.1, on the 10th there were 5287 leaves. on the 21st there were 692 leaves
y=ae^kt
a) find k (constant of proportionality)
b) how many leaves were there on the tree before it started losing leaves?
c) on what date was there only one leaf?
any insight is appreciated
Answers: 1
![\bf \textit{Amount of Population Growth, \boxed{\textit{10th day}}} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&5287\\ P=\textit{initial amount}\dotfill &P\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &10\\ \end{cases} \\\\\\ A=5287e^{10r} \\\\[-0.35em] ~\dotfill](/tpl/images/0367/8399/11db5.png)
![\bf \textit{Amount of Population Growth, \boxed{\textit{21st day}}} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&692\\ P=\textit{initial amount}\dotfill &P\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &21\\ \end{cases} \\\\\\ 692=Pe^{21r} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} 5287=Pe^{10r}\\ 692=Pe^{21r} \end{cases}](/tpl/images/0367/8399/9f07f.png)
![\bf ln\left( \cfrac{692}{5287}\right)=ln\left( e^{11r} \right)\implies ln\left( \cfrac{692}{5287}\right)=11r\implies \cfrac{ln\left( \frac{692}{5287}\right)}{11}=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill -0.18486\approx r~\hfill](/tpl/images/0367/8399/05fd8.png)
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Leaves were falling off a tree at an exponential rate beginning oct.1, on the 10th there were 5287 l...
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