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Math Help - Need help with the end of this problem

  1. #1
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    Need help with the end of this problem

    In order to save space, both the length and width of an advertisement in a newspaper had ot be decreased by the same amount in order to cut its area in half. The original advertisement had a length of 10 centimeters and a width of 15 centimeters. What are the new dimensions of the advertisement?

    My work:
    (15-x)910-x)=75
    150-15x-10x+x^2=75
    x^2-25x+75=0
    Then set it up in a quadratic and got 21.51 and 3.49. Would that be my answers or would I need to subtract those by 15-x and 10-x to find my new dimensions? If so would it be 15-3.49 and 10-3.49?

    Thanks!
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  2. #2
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    Re: Need help with the end of this problem

    Everything looks right. 11.51 by 6.51 are your new dimensions.
    Thanks from skweres1
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  3. #3
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    Re: Need help with the end of this problem

    Hello, skweres1!

    In order to save space, both the length and width of an advertisement in a newspaper
    had to be decreased by the same amount in order to cut its area in half.
    The original advertisement had a length of 10 cm and a width of 15 cms.
    What are the new dimensions of the advertisement?

    My work:
    . . (15-x)(10-x)\:=\: 75 \quad\Rightarrow\quad x^2-25x+75\:=\:0 . Good!

    Then set it up in a quadratic and got 21.51 and 3.49. . Right!
    Would that be my answers? . No, x is the width of the strips trimmed off.
    . . or would I need to subtract those by 15-x and 10-x to find my new dimensions? . Yes!

    If so, would they be 15-3.49 and 10-3.49? . Correct!

    I assume you made a sketch.

    Code:
          : - - 10  - - :
        - *---*-------- * -
        : |///://///////| x
        : * - + - - - - * -
        : |///:         | :
        : |///:         | :
       15 |///:         | :
        : |///:         |15-x
        : |///:         | :
        : |///:         | :
        : |///:         | :
        - *---*---------* -
          : x :   10-x  :
    Thanks from skweres1
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