Im stuck on two problems and would really appreciate the help.

Question

1. y varies jointly as a and b, and inversely as the square root of c. y=28 when a=2, b=7, and c=9. Find y when a=4, b=2, and c=4

2. y varies directly as x and inversely as the square of z. y=24 when x=27 and z=3. Find y when x=50 and z=5

Answer 1

y = kab/√c
So, (y√c)/ab = k is constant.
so, you want

(y√4)/(4*2) = (28√9)/(2*7)

Now just find y.

y=kx/z^2
so, you want
(y*25)/50 = (24*9)/27

Answer 2

@Steve Would these be correct?

1. y=64/9

2. y=16

Answer 3

ummm, no.
#1
(y√4)/(4*2) = (28√9)/(2*7)
2y/8 = 84/14
y = 84*4/14 = 24

#2 ok