#### Answer

a) The decay model is $A\left( t \right)=900{{e}^{-0.017t}}$.
b) The amount is $759.298\,\,\text{grams}$.

#### Work Step by Step

(a)
Consider that at any time $t$, the amount of the radioactive substance present is $A\left( t \right)$, following the exponential decay model:
$A\left( t \right)={{A}_{0}}{{e}^{kt}}$
Initially there are 900 grams of the substance present,$A\left( 0 \right)=900$, hence
$\begin{align}
& 900={{A}_{0}}{{e}^{0\times k}} \\
& {{A}_{0}}=900
\end{align}$
Therefore
$A\left( t \right)=900{{e}^{kt}}$
The substance has a half-life of 40 days; hence for $t=40$, $A\left( 40 \right)=450$
Therefore:
$\begin{align}
& 450=900{{e}^{40k}} \\
& {{e}^{40k}}=\frac{1}{2} \\
& k=\frac{1}{40}{{\log }_{e}}\frac{1}{2} \\
& =-0.017
\end{align}$
Therefore, the decaying model for the substance is
$A\left( t \right)=900{{e}^{-0.017t}}$
(b)
The radioactive substance remains after 10 days, that is, for $t=10$
$\begin{align}
& A\left( 10 \right)=900{{e}^{-0.017\times 10}} \\
& =759.298
\end{align}$
Thus, the radioactive substance that remains after 10 days is $759.298\,\,\text{grams}$.