Aball is thrown from a height of 139 feet with an initial downward velocity of 15 ft/s. the ball's height h (in feet) after t seconds is given by the following. h=139-15t-16t^2 how long after the ball is thrown does it hit the ground? round your answer(s) to the nearest hundredth. (if there is more than one answer, use the "or" button.) t =

t = 2.52 seconds

Step-by-step explanation:

h=139-15t-16t^2

We want to know when the ball hits the ground

That would be when h=0

0 = 139-15t-16t^2

We can use the quadratic formula to find t

t = -b ± sqrt(b^2-4ac)

          2a

where a = -16   b = -15  and c = 139

t = -(-15) ± sqrt((-15)^2-4(-16)139)

          2(-16)

t = (15) ± sqrt(225+8896)

          -32

t = (15) ± sqrt(9121)

          -32

t = 15+ sqrt(9121)                       t = 15- sqrt(9121)

        or                      

           -32                                               -32

        -3.453247707        or              2.515747707

Since time cannot be negative

 2.515747707

Round to the nearest hundredth

t = 2.52 seconds

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

439400

step-by-step explanation:

130^2 = 16900

16900*26 = 439400