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If z varies jointly with x and the square of y, and z=600 when x=2 and y=5, find the positive value of y when z= 4608.
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"z varies jointly with x and the square of y" means z = Cxy^2 for some C. Now use z=600 when x=2 and y=5 to find C
I already got C to equal 60, and when I solve for y I get the \[\sqrt{12.8}\], which not one of the answer choices. The answer choices are 4, 6, 8, and 10.
That's the wrong value for C
600=50C thus C=12 and we have z=12xy^2
Are we still assuming x=2 in solving for y when z=4608?
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Oh, I see. I wasn't sure if I had to square y or not. Thank you.
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