At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
we know that initially there were 120 colonies or namely that C = 120, well, what time was that? that was the hour 0, or namely t = 0, so then
[tex]P=Ce^{rt}\qquad \begin{cases} P=\textit{accumulated amount}\\ C=\textit{initial amount}\dotfill &120\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &0\\ \end{cases} \\\\\\ P=120e^{r\cdot 0}\implies P=120e^0\implies P=120\cdot 1\implies P=120[/tex]
we also know that when t = 4, C = 550
[tex]\textit{Amount of Population Growth} \\\\ P=Ce^{rt}\qquad \begin{cases} P=\textit{accumulated amount}\dotfill&120\\ C=\textit{initial amount}\dotfill &550\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &4\\ \end{cases} \\\\\\ 120=550e^{4r}\implies \cfrac{120}{550}=e^{4r}\implies \cfrac{12}{55}=e^{4r}~\hfill \stackrel{\textit{using this log cancellation rule}}{\log_a(a^x)=x}[/tex]
[tex]\log_e\left( \frac{12}{55}\right)=\log_e(e^{4r})\implies \log_e\left( \frac{12}{55}\right)=4r\implies \ln\left( \frac{12}{55}\right)=4r \\\\\\ \cfrac{\ln\left( \frac{12}{55}\right)}{4}=r\implies -0.381\approx r = k[/tex]
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
