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Math Help - Applications and problem solving@

  1. #1
    Junior Member mcdanielnc89's Avatar
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    Applications and problem solving@

    I do not know how to get the following answer

    reloading image.


    The answer that is tehre i think is the right one, but i have no clue...
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mcdanielnc89 View Post
    I do not know how to get the following answer




    The answer that is tehre i think is the right one, but i have no clue...
    You have the system \left\{\begin{array}{lcr}x-y&=&6\\x+y&=&10\end{array}\right.

    Use the method of elimination to solve the system.

    Adding the two equations together, we see that the y term canceled out.

    So we get 2x=16\implies x=\dots

    Then substitute this value of x into either equation in the system to find y.

    Can you take it from here?

    --Chris
    Last edited by Chris L T521; August 28th 2008 at 07:03 PM. Reason: Sorry...typo on my part... >_>
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  3. #3
    Junior Member mcdanielnc89's Avatar
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    I'm terribly sorry. I'm so ticked i put the wrong one. the electricity is ging on and off.. hehres the one i don't know...


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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mcdanielnc89 View Post
    I'm terribly sorry. I'm so ticked i put the wrong one. the electricity is ging on and off.. hehres the one i don't know...


    Determine the equation that represents the area of the picture:

    The dimensions are given to you: (39-2x)~cm by (26-2x)~cm

    So we see that area is (39-2x)(26-2x)~cm^2

    However, the area has a value of 440~cm^2

    Thus, (39-2x)(26-2x)=440

    Factor out the left side, and then the equation becomes

    4x^2-130x+1014=440\implies4x^2-130x+574=0

    Solve this quadratic equation.

    Can you take it from here?

    --Chris
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  5. #5
    Junior Member mcdanielnc89's Avatar
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    Thank you so much Chris!!!! the answer is 5.3
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