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

hi there!

question 1 answer: p/q can be terminating when q = 5^m * 2^n

question 2 answer: -1 and -1

question 3 answer: -4

question 4 answer: 5/12

hope this : )

the revenue of the fitness center, y, is the product of two factors: the number of members and the fee paid by each member. there are 320 members now, but number of members goes down by 10 for each $5 increase in fees, so if x is the number of $5 price increases then members is equal to (320 - 10x). since the fee is currently $45, and the fee increases by $5 for every price bump x, the fee is given by (45 + 5x). so, revenue y is given by y = (320 - 10x)(45 + 5x).