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Math Help - Reverse calculation foruma for simple lookup

  1. #1
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    Reverse calculation foruma for simple lookup

    I am building an app for a client and not exactly a math wiz. I've been able to get pretty far using formulas they've given me, but now I need to reverse find a value based on some constants I already have.

    first find for x:
    15 = (x/0.6)squared * 40

    then using x find y:
    x = (y*20*15)/5940
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  2. #2
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    Re: Reverse calculation foruma for simple lookup

    Quote Originally Posted by bdub01 View Post
    first find for x:
    15 = (x/0.6)squared * 40
    then using x find y: x = (y*20*15)/5940
    This a a bit hard to read.
    But if you know that 15 = 40\left( {\frac{x}{{0.6}}} \right)^2 then there are two solutions:
    x =  \pm \left( {0.6} \right)\sqrt {\frac{{15}}{{40}}}

    I suspect that you need only the positive solution here.
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  3. #3
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    Re: Reverse calculation foruma for simple lookup

    Quote Originally Posted by bdub01 View Post
    then using x find y:
    x = (y*20*15)/5940
    x = 300y / 5940
    x = 10y / 198
    10y = 198x
    y = 19.8x
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