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The mass m milligrams of a radioactive substance at time t days, is given by:m(t) = Ae^-tkwhere A and k are constants.The initial mass of the substance is 120 milligrams.The mass of the substance after 10 days is 90 milligrams.a)Sketch the graph to show the relationship between tand mfor t≥ 0 days.

The mass m milligrams of a radioactive substance at time t days, is given by:m(t) = Ae-example-1
User Thymen
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1 Answer

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Since the initial mass of the substance is 120 milligrams, then:


m(0)=120.

Therefore, the y-intercept of the graph of the given function is (0,120), also, evaluating the given function at t=0 we get:


m(0)=Ae^(-0k)=Ae^0=A\text{.}

Therefore A=120.

Now, we know that after 10 days, the mass of the substance is 90 milligrams, then the graph of the function passes through the point (10,90), also, evaluating the given function at t=10 we get:


m(10)=120e^(-10k).

Therefore:


90=120e^(-10k)\text{.}

Dividing the above equation by 120 we get:


\begin{gathered} (90)/(120)=(120e^(-10k))/(120), \\ (3)/(4)=e^(-10k)\text{.} \end{gathered}

Applying the natural logarithm to the above equation we get:


\begin{gathered} \ln ((3)/(4))=\ln (e^(-10k)), \\ \ln ((3)/(4))=-10k\text{.} \end{gathered}

Dividing the above equation by -10 we get:


k=-(\ln ((3)/(4)))/(10)\text{.}

Therefore, the function that models this behavior is:


\begin{gathered} m(e)=120e^{-(-\ln ((3)/(4)))\cdot(t)/(10)}=120e^{\ln ((3)/(4))\cdot(t)/(10)}\text{.} \\ \text{for t}\ge0. \end{gathered}

And its graph is:

Answer:

The mass m milligrams of a radioactive substance at time t days, is given by:m(t) = Ae-example-1
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