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    This should be simple...

    A = 0.5, yx^2/2a = 10, and yx/a = 50

    If I substitute a in first, I get a different answer to when I substitute it in at the end. Here's my logic: 2a = 1 so yx^2=10 and yx = 100, then divide one by the other and you get x = 1/10, 0.1

    however,

    if you divide one by the other with a, you get x/2 = 1/5, so x = 0.4

    What's going on?
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    (Original post by IL9)
    This should be simple...

    A = 0.5, yx^2/2a = 10, and yx/a = 50

    If I substitute a in first, I get a different answer to when I substitute it in at the end. Here's my logic: 2a = 1 so yx^2=10 and yx = 100, then divide one by the other and you get x = 1/10, 0.1

    however,

    if you divide one by the other with a, you get x/2 = 1/5, so x = 0.4

    What's going on?
    Could you use latex or at least brackets

    Oh, and is A the same as a?
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    (Original post by IL9)
    This should be simple...

    A = 0.5, yx^2/2a = 10, and yx/a = 50

    If I substitute a in first, I get a different answer to when I substitute it in at the end. Here's my logic: 2a = 1 so yx^2=10 and yx = 100, then divide one by the other and you get x = 1/10, 0.1

    however,

    if you divide one by the other with a, you get x/2 = 1/5, so x = 0.4

    What's going on?
    If I'm right you're saying that the 2nd method is to divide by a?
    This isn't true.
    \dfrac{1}{x} \div x \not= 1
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    (Original post by IL9)
    Here's my logic: 2a = 1 so yx^2=10 and yx = 100
    yx does not equal 100.

    yx/a=50 so yx = 50a = 25
    • Thread Starter
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    A is the same as a, it is y times x squared, to be clear. The second method I'm dividing. One equation by the other, just like the first, but I'm leaving a in the equations rather than putting its value of 0.5 in. Obviously they cancel when divided by each other, because the a is on the denominator for both. Now can you explain why there are different answers?
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    (Original post by ghostwalker)
    yx does not equal 100.

    yx/a=50 so yx = 50a = 25

    Got it, knew it was something stupid, thanks.
 
 
 
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