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an arrow is shot into the air with an initial velocity of 96 feet per second. the height in feet of the arrow t seconds after it was shot into the air is given by the function h(x)=-16t^2+96t. find the maximum height of the arrow. what is the answer

  • algebra -

    h(x) = -16t^2 + 96t.

    The parabola opens downward. Therefore,
    the max. point is the vertex.

    t(V) = -b/2a = -96 / -32 = 3s = The value of t at the vertex.

    h=-16*3^2 + 96*3 = -144 + 288 = 144Ft







    ;

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