a) Check the graph below, please
b) 120.002
c) 0.02876
d) 75.729g
e) Approximately on the 31st day
1) Considering that the mass is given by this exponential function:
[tex]m(t)=Ae^{-kt}[/tex]And plugging into that the initial mass, the period, and the final amount of mass:
[tex]90=120e^{-10k}[/tex]Notice that the initial mass A, is constant and m, is the final amount of mass.
a) Based on that we can plot the following equation:
In this graph, the x-axis represents the days and the y-axis the mass. Since the exponent is negative the graph is like it was reflected across the y-axis.
b) We can demonstrate that the initial value of the mass, A is 120 doing this:
[tex]\begin{gathered} 90=Ae^{-10\cdot0.02877} \\ \frac{90}{A}=\frac{Ae^{-10\cdot0.02877}}{A} \\ \frac{90}{A}=e^{-10\cdot0.02877} \\ 90=Ae^{-10\cdot0.02877} \\ A=\frac{90}{e^{-10\cdot0.02877}} \\ A=120.002\approx120 \end{gathered}[/tex]Note that we have plugged into that the value of k=0.002877 so that we could have just one variable.
c) Plugging into the formula all the variables but k, we have:
[tex]\begin{gathered} 90=120e^{-10k} \\ \frac{90}{120}=\frac{120e^{-10k}}{120} \\ \frac{3}{4}=e^{-10k} \\ \ln (\frac{3}{4})=\ln e^{-10k} \\ -0.2876=-10k \\ k=0.02876\approx0.02877 \end{gathered}[/tex]Remember the logarithmic properties so that we could find out k as 0.02877 approximately.
d) Now the point here is the mass "m" in grams for 16 days decaying:
[tex]\begin{gathered} m=120e^{-0.02877\cdot16} \\ m=75.729g \end{gathered}[/tex]Note that we have used the Euler number as 2.718...
e) Similarly let's find it for the time:
[tex]\begin{gathered} 50=120e^{-0.0287\cdot t} \\ \frac{50}{120}=\frac{120e^{-0.0287\cdot t}}{120} \\ \frac{5}{12}=e^{-0.0287\cdot t} \\ \ln (\frac{5}{12})=\ln e^{-0.0287\cdot t} \\ -0.8754=-0.0287t \\ t=30.504\approx31 \end{gathered}[/tex]Hence, the mass will reach 50 grams about the 30th to the 31st day
