Question

1. z varies directly with $x^2$ and inversely with y. when x = 2 and y = 4, z = 3. what is the value of z when x = 4 and y = 9? options: \\(\frac{16}{3}\\), 24, \\(\frac{2}{3}\\), \\(\frac{8}{3}\\)

Explanation:

Step1: Write variation formula

$z = k\frac{x^2}{y}$

Step2: Find constant k

$3 = k\frac{2^2}{4} \implies 3 = k\frac{4}{4} \implies k=3$

Step3: Calculate z for x=4,y=9

$z = 3\frac{4^2}{9} = 3\frac{16}{9} = \frac{16}{3}$

Answer:

$\frac{16}{3}$