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Math Help - How can I find the percentage decrease? (Algebra)

  1. #1
    Junior Member PythagorasNeophyte's Avatar
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    Question How can I find the percentage decrease? (Algebra)

    Hello everyone. I have an algebra question:

    Given that y varies directly as the cube of x, find the percentage decrease in y if x is halved.


    Since y is in direct proportion with x^3,
    let y=kx^3, where k is a constant.

    If x is halved, it would be \dfrac{x^3}{2}.

    But it doesn't solve the problem.

    I wonder how the percentage can be calculated when there are no values for x and y...
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  2. #2
    MHF Contributor
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    Actually, if x is halved then you need to replace x with \frac{x}{2}, which gives

    y = k\left(\frac{x}{2}\right)^3

     = \frac{kx^3}{8}.


    Since you end up with \frac{1}{8} or 12.5\% of what you started with, that means the percentage decrease is 87.5\%.
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  3. #3
    Junior Member PythagorasNeophyte's Avatar
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    Hello Prove It,

    many thanks for your reply, however, I still need clarifications about finding the percentage decrese.

    Do you mean that \dfrac{kx^2}{8} can be written as \dfrac{1}{8} ?

    How can \dfrac{kx^2}{8} be known as \dfrac{1}{8} ?
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  4. #4
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    Dividing by 8 is the same as finding \frac{1}{8} of that quantity.

    So \frac{kx^2}{8} = \frac{1}{8}kx^2.
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  5. #5
    Junior Member PythagorasNeophyte's Avatar
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    I see...so I was wrong when I wrote \dfrac{x^3}{2}, when it should be y = k\left(\frac{x}{2}\right)^2.
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  6. #6
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    Quote Originally Posted by PythagorasNeophyte View Post
    I see...so I was wrong when I wrote \dfrac{x^3}{2}, when it should be y = k\left(\frac{x}{2}\right)^2.
    Correct.
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