Solution:
Given that,
z varies directly with x and inversely with [tex]y^2[/tex]
Which means,
[tex]z \propto \frac{x}{y^2}\\\\z = k \times \frac{x}{y^2}[/tex]
Where, "k" is constant of proportionality
Substitute x = 2 and y = 5, z = 8 in above
[tex]8 = k \times \frac{2}{5^2}\\\\8 = \frac{2k}{25}\\\\k = 25 \times 4\\\\k = 100[/tex]
What is the value of z when x = 4 and y = 9?
Substitute k = 100 , x = 4 and y = 9 in above
[tex]z = 100 \times \frac{4}{9^2}\\\\z = 100 \times \frac{4}{81}\\\\z = 4.938[/tex]
Thus value of z is 4.938
