Aball is thrown from an initial height of 1 meter with an initial upward velocity of 15 m/s. the ball's height h (in meters) after seconds is given by the following.
h= 1+15t-5t^2
find all values of t for which the ball's height is 9 meters.
round your answer(s) to the nearest hundredth.
(if there is more than one answer, use the "or" button.)

t = __ seconds

0.69s

2.31s

Step-by-step explanation:

Given in the question an equation

h = 1 + 15t - 5t²

The problem asks for a time when the ball's height will be 9 m. To find these times, I plug 9 in for h and solve for t:

8 = 1 + 15t - 5t²

-5t² + 15t + 1 - 9 = 0

-5t² + 15t - 8 = 0

Using the quadratic equation:

t = -15 ± √15²-4(-5)(-8) / 2(-5)

t = -15±√65 / -10

t = -15 + √65 / -10 or  t = -15 - √65 / -10

t = 0.69 seconds or  t = 2.31 seconds

So, the ball is at a height of 9m twice: once on the way up after 0.69 seconds and once on the way back down after 2.31 seconds.

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.

the correct answer is c

good luck love