SOLUTION: there are 2 problems i dont know how to do can u pls help 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

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Question 795285: there are 2 problems i dont know how to do can u pls help
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
2nd problem: if y varies as x and y=7 when x=6 find y when x=18
Found 2 solutions by Cromlix, stanbon:
Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
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.
:-)

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1st problem : y varies directly as x and inversely as the square of z.
y = k*x/z^2
------
Solve for "k" using "y=30 when x=40 and z=2{
30 = k*40/2^2
120 = 40k
k = 3
-----
Equation for this problem::
y = 3x/z^2
----------------------
find y when x=80 and z=4
y = 3*80/4^2 = 3*5 =15
==============================
2nd problem: if y varies as x and y=7 when x=6 find y when x=18
y/18 = 7/6
y = (18/6)7
y = 21
------------------------
Cheers,
Stan H.
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