Question 795285
1st problem : y varies directly as x and inversely as the square of z. y=30 when x=40 and z=2. find y when x=80 and z=4.
y = kx/z^2
To find k the constant of variation,insert values given. 
y = kx/z^2
30 = k*40/2^2
30 = k40/4
30 = k10
k = 30/10
k = 3.
Your formula:-
y = 3x/z^2
.....
y = 3*80/16
y = 240/16
y = 15
......................
2nd problem: if y varies as x and y=7 when x=6 find y when x=18
y = kx
To find k insert values:
7 = k6
k = 7/6
.......
Your formula:
y = 7/6 x

y = 7/6 * 18
y = 21
..............
Hope this helps.
:-)