SOLUTION: The variable w varies jointly with y and the square of x and inversely with the square root of z.
Also w=-3 when x= 2sqrt3, y=-1 and z=1/4. Suppose that w=1/2x, y=2x and z=8y. Fin
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Rational-functions
-> SOLUTION: The variable w varies jointly with y and the square of x and inversely with the square root of z.
Also w=-3 when x= 2sqrt3, y=-1 and z=1/4. Suppose that w=1/2x, y=2x and z=8y. Fin
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Question 1090443: The variable w varies jointly with y and the square of x and inversely with the square root of z.
Also w=-3 when x= 2sqrt3, y=-1 and z=1/4. Suppose that w=1/2x, y=2x and z=8y. Find x.
please help
You can put this solution on YOUR website! w=k*yx^2/sqrt(z)
now substitute
-3=k*-1*(2 sqrt(3)^2)/sqrt(1/4)=k * -1*12/(1/2)=k*-24
k=1/8
now use k in the problem with a different missing variable.
w=k*yx^2/sqrt(z)
(1/2)x=(1/8)*2x*x^2/sqrt(16x); note, I am setting w=(1/2)x not (1/2x)
(1/2)x=2x^3/[8 *4 sqrt(x)]
multiply by 2
x=4x^3/(8*4 sqrt(x))
32x^(3/2)=4x^3
8=x^(3/2)
raise everything to the (2/3) power
8^2/3)=4=x ANSWER
x=4; therefore w=2,y=8 and z=64. See if this works
2=(1/8)*8*16/sqrt(64).
This is 2=16/8, which is true.
