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3 May, 00:44

The flight of a cannonball toward a hill is described by the parabola y = 2 + 0.12x - 0.002x 2. the hill slopes upward along a path given by y = 0.15x. where on the hill does the cannonball land?

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  1. 3 May, 04:29
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    To find where on the hill the canonball lands

    So 0.15x = 2 + 0.12x - 0.002x^2

    Taking the LHS expression to the right and rearranging we have - 0.002x^2 + 0.12x -

    0.15x + 2 = 0.

    So we have - 0.002x^2 - 0.03x + 2 = 0

    I'll multiply through by - 1 so we have

    0.002x^2 + 0.03x - 2 = 0.

    This is a quadratic eqn with two solutions x1 = 25 and x2 = - 40 since x cannot be negative x = 25. The second solution y = 0.15 * 25 = 3.75
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