Respuesta :
[tex]\bf \textit{Amount for Exponential Decay using Half-Life}
\\\\
A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &520\\
t=\textit{elapsed time}\to &72\\
h=\textit{half-life}\to &12
\end{cases}
\\\\\\
A=520\left( \frac{1}{2} \right)^{\frac{72}{12}}\implies A=520\left( \frac{1}{2} \right)^6\implies A=8.125[/tex]
is it linear or exponential? well, look above.
is it linear or exponential? well, look above.