Solve this equation

Question

y varies directly as x and inversely as the square root of z. y=10, when. x=9 and z=3. find y when x=2 and z=6

Answer 1

y = kx/sqrt(z)

10 = k*9/sqrt(3)

10 * sqrt(3) / 9 = k

y = 10 * sqrt(3) *x/ (9*sqrt(z))

THEN

y = 10 * sqrt(3)*2 / (9*sqrt(6))

= 10 * sqrt(3)*sqrt(6)*2 / 9 * 6 <--- rationalizes the denominator

= 20 * 3*sqrt(2) / 54

=10 * 3 * sqrt(2) / 27

= 10*sqrt(2) / 9