IDNLearn.com: Your trusted platform for finding reliable answers. Our platform is designed to provide reliable and thorough answers to all your questions, no matter the topic.
Sagot :
so, the car depreciates by half every 5 years, or namely it depreciates by 50% every 5 years, so
[tex]\textit{Periodic/Cyclical Exponential Decay} \\\\ A=P(1 - r)^{\frac{t}{c}}\qquad \begin{cases} A=\textit{current amount}\dotfill &\$7000\\ P=\textit{initial amount}\dotfill &\$19000\\ r=rate\to 50\%\to \frac{50}{100}\dotfill &0.5\\ t=\textit{elapsed time}\\ c=period\dotfill &5 \end{cases}[/tex]
[tex]7000=19000(1-0.5)^{\frac{t}{5}}\implies \cfrac{7000}{19000}=0.5^{\frac{t}{5}}\implies \cfrac{7}{19}=0.5^{\frac{t}{5}} \\\\\\ \log\left( \cfrac{7}{19} \right)=\log\left( 0.5^{\frac{t}{5}} \right)\implies \log\left( \cfrac{7}{19} \right)=t\log\left( 0.5^{\frac{1}{5}} \right) \\\\\\ \log\left( \cfrac{7}{19} \right)=t\log( \sqrt[5]{0.5})\implies \cfrac{\log\left( \frac{7}{19} \right)}{\log( \sqrt[5]{0.5})}=t\implies 7.2\approx t[/tex]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Find clear and concise answers at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.
