Given initial mass:

\(\displaystyle{m}_{{0}}={801}\)kg at t=0

Let

m(t)=Mass at time t

The exponential function,

\(\displaystyle{m}{\left({t}\right)}={m}_{{0}}{e}^{{{k}{t}}}\), where k =constant

Also given:

half life=85days

\(\displaystyle\Rightarrow{m}{\left({85}\right)}={\frac{{{m}_{{0}}}}{{{2}}}}\)

Now,

When t=85days:

\(\displaystyle{m}{\left({85}\right)}={m}_{{0}}{e}^{{{k}\times{85}}}\)

\(\displaystyle\Rightarrow{\frac{{{m}_{{0}}}}{{{2}}}}={m}_{{0}}{e}^{{{85}{k}}}\)

\(\displaystyle\Rightarrow{e}^{{{85}{k}}}={\frac{{{1}}}{{{2}}}}\)

\(\displaystyle\Rightarrow{e}^{{k}}={\left({\frac{{{1}}}{{{2}}}}\right)}^{{{\frac{{{1}}}{{{85}}}}}}\)

\(\displaystyle\Rightarrow{k}={{\ln{{\left({\frac{{{1}}}{{{2}}}}\right)}}}^{{{\frac{{{1}}}{{{85}}}}}}}\)

\(\displaystyle\Rightarrow{k}=-{\frac{{{1}}}{{{85}}}}{\ln{{2}}}\)

Therefore,

When t=10days:

\(\displaystyle{m}{\left({10}\right)}={801}{e}^{{{k}\times{10}}}\)

\(\displaystyle={801}{\left[{e}^{{{10}\times{\left(-{\frac{{{1}}}{{{85}}}}{\ln{{2}}}\right\rbrace}}}\right]}\)

\(\displaystyle={738.273}\)kg

Hence,

The radioactive material remains after 10 days =738.273kg

\(\displaystyle{m}_{{0}}={801}\)kg at t=0

Let

m(t)=Mass at time t

The exponential function,

\(\displaystyle{m}{\left({t}\right)}={m}_{{0}}{e}^{{{k}{t}}}\), where k =constant

Also given:

half life=85days

\(\displaystyle\Rightarrow{m}{\left({85}\right)}={\frac{{{m}_{{0}}}}{{{2}}}}\)

Now,

When t=85days:

\(\displaystyle{m}{\left({85}\right)}={m}_{{0}}{e}^{{{k}\times{85}}}\)

\(\displaystyle\Rightarrow{\frac{{{m}_{{0}}}}{{{2}}}}={m}_{{0}}{e}^{{{85}{k}}}\)

\(\displaystyle\Rightarrow{e}^{{{85}{k}}}={\frac{{{1}}}{{{2}}}}\)

\(\displaystyle\Rightarrow{e}^{{k}}={\left({\frac{{{1}}}{{{2}}}}\right)}^{{{\frac{{{1}}}{{{85}}}}}}\)

\(\displaystyle\Rightarrow{k}={{\ln{{\left({\frac{{{1}}}{{{2}}}}\right)}}}^{{{\frac{{{1}}}{{{85}}}}}}}\)

\(\displaystyle\Rightarrow{k}=-{\frac{{{1}}}{{{85}}}}{\ln{{2}}}\)

Therefore,

When t=10days:

\(\displaystyle{m}{\left({10}\right)}={801}{e}^{{{k}\times{10}}}\)

\(\displaystyle={801}{\left[{e}^{{{10}\times{\left(-{\frac{{{1}}}{{{85}}}}{\ln{{2}}}\right\rbrace}}}\right]}\)

\(\displaystyle={738.273}\)kg

Hence,

The radioactive material remains after 10 days =738.273kg