Since you have posted a question with multiple subparts, we will answer the first three subparts (a,b,c). If you want the remaining subpart to be answered, repost the question and mention the subpart you want us to answer in your message.

Given:

Scientists with a sample that has 60 mg of Cobalt-56, and after 31 days, the mass becomes 45.43mg.

The differential equation which models expontial decay is \(\displaystyle{\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}=-{k}{m}\) and the solution or differential equation is \(\displaystyle{m}{\left({t}\right)}={m}_{{0}}{e}^{{-{k}{t}}}\), where \(\displaystyle{m}_{{0}}\) is the initial mass and k is the relative decay rate.

Then, \(\displaystyle{m}_{{0}}={60}\)mg and m(31)=45.43 mg

a) Compute the relative decay rate k using the given information as follows.

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

\(\displaystyle{m}{\left({31}\right)}={m}_{{0}}{e}^{{-{k}{\left({31}\right)}}}\)

\(\displaystyle{45.43}={60}{e}^{{-{31}{k}}}\)

Solve for k as shown below.

\(\displaystyle{e}^{{-{31}{k}}}={\frac{{{45.43}}}{{{60}}}}\)

=0.75716667

Take natural logarithm on both sides,

\(\displaystyle{\ln{{\left({e}^{{-{31}{k}}}\right)}}}={\ln{{\left({0.75716667}\right)}}}\)

\(\displaystyle-{31}{k}={\ln{{\left({0.75716667}\right)}}}\)

\(\displaystyle{k}=-{\frac{{{\ln{{\left({0.75716667}\right)}}}}}{{{31}}}}\)

\(\displaystyle\approx{0.00897}\) mg\day

Therefore, the relative decay rate is approximately 0.00897 mg\day

b) The half life of Cobalt-56 is the time after which only half of the original material remains.

The original amount taken is 60 mg. Then half of it will be 30 mg.

Now obtain the half-life of Cobalt-56 as shown below.

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

\(\displaystyle{30}={60}{e}^{{-{k}{t}}}\)

\(\displaystyle{30}={60}{e}^{{-{0.00897}{t}}}\)

Solve for t as follows.

\(\displaystyle{e}^{{-{0.00897}{t}}}={\frac{{{30}}}{{{60}}}}\)

\(\displaystyle{\ln{{\left({e}^{{-{0.00897}{t}}}\right)}}}={\ln{{\left({\frac{{{1}}}{{{2}}}}\right)}}}\)

\(\displaystyle-{0.00897}{t}=-{\ln{{\left({2}\right)}}}\)

\(\displaystyle{t}={\frac{{{\ln{{\left({2}\right)}}}}}{{{0.00897}}}}\)

\(\displaystyle\approx{77.27}\) days

Therefore, the half-life of Cobalt-56 is 77.27 days.

c) Obtain the number of days taken for the initial sample of 60 mg of Cobalt-56 to decay to just 10 mg as follows.

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

\(\displaystyle{10}={60}{e}^{{-{k}{t}}}\)

\(\displaystyle{10}={60}{e}^{{-{0.00897}{t}}}\)

Solve for t as follows

\(\displaystyle{e}^{{-{0.00897}{t}}}={\frac{{{10}}}{{{60}}}}\)

\(\displaystyle{\ln{{\left({e}^{{-{0.00897}{t}}}\right)}}}={\ln{{\left({\frac{{{1}}}{{{6}}}}\right)}}}\)

\(\displaystyle-{0.00897}{t}=-{\ln{{\left({6}\right)}}}\)

\(\displaystyle{t}={\frac{{{\ln{{\left({6}\right)}}}}}{{{0.00897}}}}\)

\(\displaystyle={199.750219}\)

\(\displaystyle\approx{200}\) days

Given:

Scientists with a sample that has 60 mg of Cobalt-56, and after 31 days, the mass becomes 45.43mg.

The differential equation which models expontial decay is \(\displaystyle{\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}=-{k}{m}\) and the solution or differential equation is \(\displaystyle{m}{\left({t}\right)}={m}_{{0}}{e}^{{-{k}{t}}}\), where \(\displaystyle{m}_{{0}}\) is the initial mass and k is the relative decay rate.

Then, \(\displaystyle{m}_{{0}}={60}\)mg and m(31)=45.43 mg

a) Compute the relative decay rate k using the given information as follows.

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

\(\displaystyle{m}{\left({31}\right)}={m}_{{0}}{e}^{{-{k}{\left({31}\right)}}}\)

\(\displaystyle{45.43}={60}{e}^{{-{31}{k}}}\)

Solve for k as shown below.

\(\displaystyle{e}^{{-{31}{k}}}={\frac{{{45.43}}}{{{60}}}}\)

=0.75716667

Take natural logarithm on both sides,

\(\displaystyle{\ln{{\left({e}^{{-{31}{k}}}\right)}}}={\ln{{\left({0.75716667}\right)}}}\)

\(\displaystyle-{31}{k}={\ln{{\left({0.75716667}\right)}}}\)

\(\displaystyle{k}=-{\frac{{{\ln{{\left({0.75716667}\right)}}}}}{{{31}}}}\)

\(\displaystyle\approx{0.00897}\) mg\day

Therefore, the relative decay rate is approximately 0.00897 mg\day

b) The half life of Cobalt-56 is the time after which only half of the original material remains.

The original amount taken is 60 mg. Then half of it will be 30 mg.

Now obtain the half-life of Cobalt-56 as shown below.

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

\(\displaystyle{30}={60}{e}^{{-{k}{t}}}\)

\(\displaystyle{30}={60}{e}^{{-{0.00897}{t}}}\)

Solve for t as follows.

\(\displaystyle{e}^{{-{0.00897}{t}}}={\frac{{{30}}}{{{60}}}}\)

\(\displaystyle{\ln{{\left({e}^{{-{0.00897}{t}}}\right)}}}={\ln{{\left({\frac{{{1}}}{{{2}}}}\right)}}}\)

\(\displaystyle-{0.00897}{t}=-{\ln{{\left({2}\right)}}}\)

\(\displaystyle{t}={\frac{{{\ln{{\left({2}\right)}}}}}{{{0.00897}}}}\)

\(\displaystyle\approx{77.27}\) days

Therefore, the half-life of Cobalt-56 is 77.27 days.

c) Obtain the number of days taken for the initial sample of 60 mg of Cobalt-56 to decay to just 10 mg as follows.

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

\(\displaystyle{10}={60}{e}^{{-{k}{t}}}\)

\(\displaystyle{10}={60}{e}^{{-{0.00897}{t}}}\)

Solve for t as follows

\(\displaystyle{e}^{{-{0.00897}{t}}}={\frac{{{10}}}{{{60}}}}\)

\(\displaystyle{\ln{{\left({e}^{{-{0.00897}{t}}}\right)}}}={\ln{{\left({\frac{{{1}}}{{{6}}}}\right)}}}\)

\(\displaystyle-{0.00897}{t}=-{\ln{{\left({6}\right)}}}\)

\(\displaystyle{t}={\frac{{{\ln{{\left({6}\right)}}}}}{{{0.00897}}}}\)

\(\displaystyle={199.750219}\)

\(\displaystyle\approx{200}\) days