anonymous
  • anonymous
The value of y varies directly with x2, and y = 64 when x = 4. What is the value of y when x = 6?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
if that x2 stands for \(x^2\), then: \[y=Ax^2\] where \(A\) is some constant, if you plug \(y=64\) for \(x=4\) you can find the value of \(A\) and then plug \(x=6\) and see what you get for \(y\)...
anonymous
  • anonymous
6^2 would make it 36 right?
anonymous
  • anonymous

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anonymous
  • anonymous
that woul be the case \(A=1\)... you can et the value for \(A\) from: \[64=A\times 4^2\] find that and then calculate \(y=A\times 6^2\)
anonymous
  • anonymous
what is a
anonymous
  • anonymous
the problem says: "The value of y varies directly with x^2, and y = 64 when x = 4" so y must equal "something" times x^2, i called that something A
anonymous
  • anonymous
im sorry im the worst XD
anonymous
  • anonymous
576
anonymous
  • anonymous
no problem, just keep asking until you understand ! what did you get for A ?
anonymous
  • anonymous
well 64/4 is 16 so i multiplied 36 by 16 to get 576
anonymous
  • anonymous
note that \(A=\frac{64}{4^2}\) dont forget the square!
anonymous
  • anonymous
oh yeah
anonymous
  • anonymous
i dont remember how to do that
anonymous
  • anonymous
well \(4^2=16\) so \(A=\frac{64}{16}=4\)
anonymous
  • anonymous
so 4 times 36 ?
anonymous
  • anonymous
for 144?
anonymous
  • anonymous
:) that is what i get... good work!

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