Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
[tex]\textit{Amount for Exponential Decay using Half-Life} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{current amount}\dotfill &20\\ P=\textit{initial amount}\dotfill &200\\ t=\textit{elapsed time}\\ h=\textit{half-life}\dotfill &12 \end{cases} \\\\\\ 20=200\left( \frac{1}{2} \right)^{\frac{t}{12}}\implies \cfrac{20}{200}=\left( \frac{1}{2} \right)^{\frac{t}{12}}\implies \cfrac{1}{10}=\left( \frac{1}{2} \right)^{\frac{t}{12}}[/tex]
[tex]\log\left( \cfrac{1}{10} \right)=\log\left[ \left( \frac{1}{2} \right)^{\frac{t}{12}} \right]\implies \log\left( \cfrac{1}{10} \right)=t\log\left[ \left( \sqrt[12]{\frac{1}{2}} \right) \right] \\\\\\ \cfrac{\log\left( \frac{1}{10} \right)}{\log\left[ \left( \sqrt[12]{\frac{1}{2}} \right) \right]}=t\implies \implies \stackrel{mins}{39.9}\approx t[/tex]
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.