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i need help in solving these 2 problems...please help..thank you...

Your home is built on a square lot. To add more space to your yard, you purchase an additional 4 feet along the side of the property. The area of the lot is now 9600 square feet. What are the dimentions of the new lot??

The Garys have a triangular pennant of area 420in.^2 flying from the flagpole in their yard. The height of the triangle is 10 in. than 5 times the base of the triangle. What are the dimensions of the pennant??

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sheila

  1. For the first problem:
    x=length of side before the addition
    x+4=the length of the longer side after the the additional 4 feet is purchased.

    The new area of the property is:
    x(x+4)=9600
    x^2 + 4x =9600
    x^2 + 4x - 9600=0
    which is:
    (x-96)(x+100)=0
    Ignore the negative solution for x since a negative number would not make much sense for a lot length.
    x=96 and x+4=100

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    Quidditch

  2. x * ( x + 4 ) = 9600

    x^2 + 4 x - 9600 = 0
    solve quadratic

    area = (1/2) b h
    420 = (1/2) b (5b+10)
    840 = 5 b^2 + 10 b

    b^2 + 2 b - 168 = 0
    solve quadratic

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    Damon

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