We are given that y varies directly as x and inversely as the square of z
Mathematically, the relationship is given by
[tex]y=k\frac{x}{z^2}[/tex]Where k is the constant of proportionality.
Let us first find the value of k
It is given that y = 108 when x = 81 and z = 3
[tex]\begin{gathered} y=k\frac{x}{z^2} \\ 108=k\frac{81}{3^2} \\ 108=k\frac{81}{9}^{} \\ 108=k\cdot9^{} \\ k=\frac{108}{9} \\ k=12 \end{gathered}[/tex]So, the value of constant k is 12
Find y when x = 44 and z = 2
[tex]\begin{gathered} y=12\frac{x}{z^2} \\ y=12\frac{44}{2^2} \\ y=12\frac{44}{4}^{} \\ y=12(11) \\ y=132 \end{gathered}[/tex]Therefore, the value of y is 132
Option D is the correct answer.
