SOLUTION: Please help me with the following question: Find an equation of variation where y varies jointly as x and z and inversely as the square of w and y=20 when x=-0.5, z=4, and w=5.

Algebra ->  Systems-of-equations -> SOLUTION: Please help me with the following question: Find an equation of variation where y varies jointly as x and z and inversely as the square of w and y=20 when x=-0.5, z=4, and w=5.      Log On


   



Question 166394: Please help me with the following question:
Find an equation of variation where y varies jointly as x and z and inversely as the square of w and y=20 when x=-0.5, z=4, and w=5.

Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
y varies jointly as x and z means y = k*x*z (equation 1)
y varies inversely as square of w means y = k/w^2 (equation 2)
y varies jointly as x and z, and inversely as w^2 means y = k*x*z/w^2 (equation 3)
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given y = 20 when x = -.5, z = 4, and w = 5
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plug these into equation 3
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20 = k*((-.5)*(4))/(5^2)
simplify:
20 = -2*k/25
multiply both sides of equation by 25
25*20 = -2*k
simplify
500 = -2*k
divide both sides of equation by 2
250 = -k
multiply botrh sides of equation by -1
-250 = k
which is the same as
k = -250
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now that you have k, plug it back into the equation to see that it is correct.
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20 = [-250*(-.5)*(4)]/25
20 = [-250*(-2)]/25
20 = 500/25
20 = 20
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this is how it works:
k = constant of proportionality.
if y varies directly with x, then y = k*x
if y varies directly ,and jointly, with x and z, then y = k*x*z
if y varies inversely with x, then y = k/x
if y varies inversely with x^2, then y = k/x^2
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