SOLUTION: m varies jointly with the square of n and the square root of q. m = k*n^2*sqrtq ----------- ------- If m = 24 when n =2 and q =4, find m when n = 5 and q = 9 If m = 24 w

SOLUTION: m varies jointly with the square of n and the square root of q. m = kn^2sqrtq ----------- ------- If m = 24 when n =2 and q =4, find m when n = 5 and q = 9 If m = 24 w

Question 135635: m varies jointly with the square of n and the square root of q.
m = k*n^2*sqrtq
-------------------
If m = 24 when n =2 and q =4,
find m when n = 5 and q = 9

If m = 24 when n =2 and q =4, find m when n = 5 and q = 9
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
m varies jointly with the square of n and the square root of q.
m = k*n^2*sqrtq
-------------------
If m = 24 when n =2 and q =4,
solve for K:
24 = k*4*2
k = 3
So, m = 3*n^2*sqrt(q)
----------------------------
find m when n = 5 and q = 9
m = 3*25*3
m = 225
=============
Cheers,
Stan H.

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
F varies jointly as x and y means where k is the constant of variation.

For your problem:

You are given that when and , so:

=> , therefore

Now you can write:

Finding m for n = 5 and q = 9 is a simple matter of substitution and arithmetic.
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