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Math Help - HELP!!!! These are impossible.

  1. #1
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    HELP!!!! These are impossible.

    I need a formula for 1 and 3. Number 2 i just can't figure out.

    1. The width of a rectangle is 7 less than twice its length. If the area of the rectangle is 23 cm, what is the length of the diagonal?

    Note: Your answer must be a number. It may not contain any arithmetic operations.

    The length of the diagonal is _______ cm.

    2.Find all possible solutions. Give your answers in increasing order. Give your answers to the nearest thousandth.

    The smaller possible solution is ________. Is it a solution? (yes or no)

    The larger possible solution is ________. Is it a solution? (yes or no)

    3.Expand the expression using the Binomial Theorem:
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by badandy328 View Post
    1. The width of a rectangle is 7 less than twice its length. If the area of the rectangle is 23 cm, what is the length of the diagonal?

    Note: Your answer must be a number. It may not contain any arithmetic operations.

    The length of the diagonal is _______ cm.
    Call the length of the rectangle x. Then w = 2x - 7. The area is then
    A = x(2x - 7) = 2x^2 - 7x. The area is known to be 23 cm^2 (Note the unit change, you had it wrong in the problem statement.) So:
    23 = 2x^2 - 7x

    2x^2 - 7x - 23 = 0

    By the quadratic formula:
    x = [-(-7) (+/-) sqrt{(-7)^2 - 4 * 2 * (-23)}]/(2 * 2)

    x = [7 (+/-) \sqrt{233}]/4 (after a bit of work.)

    Note: x must be positive, so we can only have the "+" solution here.

    Now, we want the diagonal. This is going to be
    sqrt{x^2 + (2x - 7)^2} = 5x^2 - 28x + 49 (by the Pythagorean theorem.)

    sqrt{5 [ (7 + sqrt{233})/4 ]^2 - 28 [ (7 + sqrt{233})/4 ] + 49}

    sqrt{5 [ 141/6 + 7*sqrt{233}/8 ] - 28 [ (7 + sqrt{233})/4 ] + 49}

    sqrt{ (705 - 21*sqrt{233})/4

    -Dan
    Last edited by topsquark; December 10th 2006 at 05:47 PM.
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  3. #3
    Forum Admin topsquark's Avatar
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    [quote=badandy328;31174]2.Find all possible solutions. Give your answers in increasing order. Give your answers to the nearest thousandth.

    The smaller possible solution is ________. Is it a solution? (yes or no)

    The larger possible solution is ________. Is it a solution? (yes or no)

    sqrt{-4x + 4} + 1 = 3x

    2 sqrt{-x + 1} = 3x - 1

    Square both sides:
    4(-x + 1) = 9x^2 - 6x + 1

    9x^2 - 2x - 3 = 0

    x = [-(-2) (+/-) sqrt{(-2)^2 - 4 * 9 * (-3)}]/(2 * 9)

    x = [1 (+/-) 2*sqrt{7}]/9

    Upon checking these two solutions in the original equation I find that only the "+" solution works. Thus:
    x = [1 + 2*sqrt{7}]/9[/tex]

    -Dan
    Last edited by topsquark; December 10th 2006 at 05:49 PM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Ummm...I'm not going to do the last one until I find out what's wrong with the LaTeX.

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by badandy328 View Post
    3.Expand the expression using the Binomial Theorem:
    _____________________________________________________
    I'm going to define the symbol:
    (nCr) = n!/(r! * (n-r)!)

    Then the expansion:
    (ax + b)^n = (nC0)*(ax)^n*b^0 + (nC1)*(ax)^{n-1}b^1 + (nC2)*(ax)^{n-2}b^2 +... + (nC(n-1))*(ax)^1*b^{n-1} + (nCn)*(ax)^0*b^n

    In this case a = 4, b = 1, n = 5
    (5C0) = 5!/(0! * (5 - 0)!) = 1
    (5C1) = 5!/(1! * (5 - 1)!) = 5
    (5C2) = 5!/(2! * (5 - 2)!) = 10
    (5C3) = (5C2)
    (5C4) = (5C1)
    (5C5) = (5C0)

    As you can verify for yourself.

    -Dan
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