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Math Help - Square Roots

  1. #1
    Member realintegerz's Avatar
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    Square Roots

    How do I find Cx

    CD = sq. root{ [(sq. root (16 - (Cx-1)^2) + 2] - Cy }^2

    Cx is supposed to equal + or - [sq. root (16- (CD+1)^2 ] +1
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  2. #2
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    CD, Cx, Cy ... what do these mean?
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  3. #3
    Member realintegerz's Avatar
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    CD is a length

    Cx is the x coordinate of C
    and Cy is the y coordinate of C


    I just need help how to get Cx outta there....
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  4. #4
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    CD = sq. root{ [(sq. root (16 - (Cx-1)^2) + 2] - Cy }^2
    I can't interpret your syntax ... how did you determine this for segment CD?

    It might help to know what the original problem statement says in its entirety.
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  5. #5
    Member realintegerz's Avatar
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    Create a rule that allows you to predict the location of C when you know the length of CD,



    Ignore the sentence to the right, that was a different problem involving the circle...
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  6. #6
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    point D is on the circle. the coordinates for point D are

    (x, \sqrt{16-(x-1)^2} + 2)

    point C has coordinates (x,3)

    CD is the vertical distance between D and C ... the difference between the y-values.

    CD = [\sqrt{16-(x-1)^2} + 2] - 3

    CD = \sqrt{16-(x-1)^2} - 1

    CD + 1 = \sqrt{16 - (x-1)^2}

    (CD + 1)^2 = 16 - (x-1)^2

    (x-1)^2 = 16 - (CD + 1)^2

    x-1 = \pm \sqrt{16 - (CD + 1)^2}

    x = 1 \pm \sqrt{16 - (CD + 1)^2}
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