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anonymous

  • one year ago

The value of y varies directly with x2, and y = 64 when x = 4. What is the value of y when x = 6?

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  1. anonymous
    • one year ago
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    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 \(y\)...

  2. anonymous
    • one year ago
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    6^2 would make it 36 right?

  3. anonymous
    • one year ago
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    @Greg_D

  4. anonymous
    • one year ago
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    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 \(y=A\times 6^2\)

  5. anonymous
    • one year ago
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    what is a

  6. anonymous
    • one year ago
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    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

  7. anonymous
    • one year ago
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    im sorry im the worst XD

  8. anonymous
    • one year ago
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    576

  9. anonymous
    • one year ago
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    no problem, just keep asking until you understand ! what did you get for A ?

  10. anonymous
    • one year ago
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    well 64/4 is 16 so i multiplied 36 by 16 to get 576

  11. anonymous
    • one year ago
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    note that \(A=\frac{64}{4^2}\) dont forget the square!

  12. anonymous
    • one year ago
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    oh yeah

  13. anonymous
    • one year ago
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    i dont remember how to do that

  14. anonymous
    • one year ago
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    well \(4^2=16\) so \(A=\frac{64}{16}=4\)

  15. anonymous
    • one year ago
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    so 4 times 36 ?

  16. anonymous
    • one year ago
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    for 144?

  17. anonymous
    • one year ago
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    :) that is what i get... good work!

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