[tex]\begin{gathered} g(n)\propto\frac{1}{n} \\ g(n)=\frac{k}{n},(\text{where k = constant) } \end{gathered}[/tex][tex]\begin{gathered} \text{when} \\ g(n)=12 \\ n=3 \\ g(n)=\frac{k}{n} \\ 12=\frac{k}{3} \\ k=12(3) \\ k=36 \end{gathered}[/tex][tex]\begin{gathered} g(n)=\frac{36}{n} \\ \text{constant of proportionality} \end{gathered}[/tex]
find the value of n when g(n) = 6
[tex]\begin{gathered} g(n)=6 \\ g(n)=\frac{36}{n} \\ n=\frac{36}{g(n)} \\ n=\frac{36}{6} \\ n=6.0 \end{gathered}[/tex]
Therefore the value of n = 6.0 (tenth
