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Math Help - help please

  1. #1
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    Question help please

    the length of a rectangle L is 1 more than twice the width, and perimeter is 62.find the width and the width???



    polynomials

    (0.7x^2 + 0.2x-0.8)-(0.9x^2-1.1x-1.4)=


    (3x-6)(2x+4)



    4xy^2(7x^3 +3x^2y^2-9y^3)




    (4x^3-3x^2+3x-4)(6x-4)
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  2. #2
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    Quote Originally Posted by BeBeMala View Post
    the length of a rectangle L is 1 more than twice the width, and perimeter is 62.find the width and the width???


    Please post each question in a separate thread with an explanantion on what is required and your attempts at solution.

    P = 2W+2L

    Where L = 2W+1 and P = 62

    therefore 62 = 2W+2(2W+1)

    62 = 2W+4W+2

    62 = 6W+2

    Can you solve it from here?
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  3. #3
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    Quote Originally Posted by BeBeMala View Post

    polynomials

    (0.7x^2 + 0.2x-0.8)-(0.9x^2-1.1x-1.4)=
    group like terms here, expand the negative through the second expression first.

     (0.7x^2 + 0.2x-0.8)-(0.9x^2-1.1x-1.4)=0.7x^2 + 0.2x-0.8-0.9x^2+1.1x+1.4 =(0.7x^2 -0.9x^2)+(1.1x+ 0.2x)+(1.4-0.8)=\dots

    Can you finish it?
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  4. #4
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    polynomial:=0.2x^2 +1.3x +0.6????

    the width is 10?? how i find the length??
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  5. #5
    MHF Contributor
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    Talking

    You find the length by solving the "width" equation you were given, and plugging the solution value into the "length" formula you were given.
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