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Math Help - Parabolic arch

  1. #1
    Junior Member
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    Parabolic arch

    I have one last question for the parabolic arch.

    If a parabolic arch satisfies the equation y=16 - 0.25x^2 for y>0.

    How would I find the width [I]w/I] of the arch?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by gretchen View Post
    I have one last question for the parabolic arch.

    If a parabolic arch satisfies the equation y=16 - 0.25x^2 for y>0.

    How would I find the width [i]w/I] of the arch?
    lets assume that the ground level corresponds to y=0, then is we
    solve 16-0.24 y^2=0, we will find the two points that the arch touches
    ground and the width will be the difference between them, so:

    16-0.24y^2=0,

    or:

    0.25 y^2=16

    taking square roots:

    (1/2) y= +/- 4

    y=-8 or +8.

    So the width of the arch is 8-(-8)=16.

    RonL
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