A ball is thrown in the air from a ledge. Its height in tesspresented by R -16(x^2 – 5x-6), where x is the number of seconds sin

Question

A ball is thrown in the air from a ledge. Its height in tesspresented by
R -16(x^2
– 5x-6), where x is the number of seconds since the ball has
been thrown. The height of the ball is 0 feet when it hits the ground.
How many seconds does it take the ball to reach the ground?

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Brielle 1 month 2021-10-20T22:02:39+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-10-20T22:03:57+00:00

    Answer:

    6 .74 seconds

    Step-by-step explanation:

    The height of the ball is given by the equation:

    R(x) =  - 16( {x}^{2}  - 5x - 6)

    When the ball hit the ground, then;

    R(x) =  0 \\ - 16( {x}^{2}  - 5x - 6) = 0 \\ {x}^{2}  - 5x - 6 = 0 \\ x = -0.74 \:  Or \:  6.74

    We discard x=-0.74 since time is not negative.

    Therefore the ball hit the ground after 6.74 seconds.

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45:7+7-4:2-5:5*4+35:2 =? ( )