Will give brainiest. a ball is thrown from an initial height of 3 meters with an initial upward velocity of 20/ms . the ball's height h (in meters) after t seconds is given by the following. h=3+20t-5t^2 find all values of t for which the ball's height is 8 meters. round your answer(s) to the nearest hundredth. (if there is more than one answer, use the "or" button.)

Using the given equation, set H to 8 meters and solve for t:

8 = 3 + 20t -5t^2

Subtract 8 from each side:

3 + 20t -5t^2 -8 = 0

Simplify:

20t - 5t^2-5 =0

Factor out -5:

-5(t^2-4t+1)=0

Divide both sides by -5:

t^2 - 4t +1 = 0

Using the quadratic formula solve for t:

t = 2 +/- √3

t = 3.73 or 0.27

t = 5.897 sec or

t = 2.103 sec

Step-by-step explanation:

initial height = 3 meters

initial velocity = 20m/s

height in meters after t second (h) = 3 + 20t -5t²

The ball height is assumed to be 8 meters i.e it traveled 5 meters during that time with the velocity of 20m/s

5 = 3 + 20t - 5t²

5t²-20t +2 =0

solving quadratically

t = 5.897 sec or

t = 2.103 sec

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."