If y varies directly as square of x and the square root of z and y=3 when x=4 and z=9,find the value of y when x=3 and z=16
y = k x^2 √z
Or,
y/(x^2 √z) = k and is constant. So,
3/(4^2 √9) = y/(3^2 √16)
Now just solve for y.
Or,
y/(x^2 √z) = k and is constant. So,
3/(4^2 √9) = y/(3^2 √16)
Now just solve for y.