I'll Be Solving Two Different Questions.
Formula For Both : \(N(t)=N_{o}E^{-kt}\)
1. A 40 gram sample of a substance thats used for drug research has a k-value of 0.1455
\(\left[\begin{array}{ccc}N(t)=40E^{-0.1455t}\end{array}\right]\)
2. A 40 gram sample of a substance thats used for drug research has a k-value of 0.1446.
\(\left[\begin{array}{ccc}N(t)=40E^{-0.1446t}\end{array}\right]\)
\(\left[\begin{array}{ccc}Number&1\end{array}\right]\)
We Determine That 20 Is Half Of 40.
With That Being Said
\(\left[\begin{array}{ccc}N(t)=40E^{-0.1455t}&=20\end{array}\right]\)
\(\left[\begin{array}{ccc}In(E^{-0.1455})=\frac{In\frac{1}{2} }{-0.1455} \end{array}\right]\)
\(\left[\begin{array}{ccc}T=\frac{In\frac{1}{2} }{-0.1455}&Or&T=\frac{-In(2)}{0.1455}\end{array}\right]\)
\(\left[\begin{array}{ccc}-0.5In/0.1455\\=4.76\end{array}\right]\)
Rounding It To Be [ 4.8 ]
\(\left[\begin{array}{ccc}Number&2\end{array}\right]\)
Again 20 Is Determined To Be Half Of 40
With That Being Said
\(\left[\begin{array}{ccc}N(t)=40E^{-0.1446t}&=20\end{array}\right]\)
\(\left[\begin{array}{ccc}In(E^{-0.1446})=\frac{In\frac{1}{2} }{-0.1446} \end{array}\right]\)
\(\left[\begin{array}{ccc}T=\frac{In\frac{1}{2} }{-0.1446}&Or&T=\frac{-In(2)}{0.1446}\end{array}\right]\)
\(\left[\begin{array}{ccc}-0.5In/0.1446\\=4.79\end{array}\right]\)
Rounding It To Be [ 4.8 ]
