Answer:
Therefore the ball strikes the ground 5.98 s after kick.
Step-by-step explanation:
Given that , a ball is kicked in the air. The height of the ball can be calculated using the equation
where h is height in feet and t is time in second.
When the ball touches the ground then the height will be zero.
i.e h(t)=0
Therefore


Since -2≠0 then 
[The solution of a quadratic equation ax²+bx+c=0 is
here a=8, b=-32 and c=-95]
Therefore t= 5.98 or -1.98
since time can not be negative So , t=5.98 s
Therefore the ball strikes the ground 5.98 s after kick.
Answer: Choice B. sqrt(2)
Draw out a right triangle in quadrant IV as you see in the attached image below. The horizontal and vertical legs are both 1 unit long. To ensure that the signs are properly set up, I am making the vertical leg BC have a label "-1" to mean this is below the x axis. Note how
tan(theta) = opposite/adjacent = BC/AB = -1/1 = -1
Use the pythagorean theorem to find that the hypotenuse AC is sqrt(2) units long
a^2 + b^2 = c^2
(1)^2 + (1)^2 = c^2
2 = c^2
c^2 = 2
c = sqrt(2)
The secant of theta is the ratio of the hypotenuse over the adjacent side, so we end up with
sec(theta) = hypotenuse/adjacent
sec(theta) = AC/AB
sec(theta) = sqrt(2)/1
sec(theta) = sqrt(2) which is why choice B is the answer
30,000 - 19,500 = 10,500 gallons left to fill the pool. So I am assuming the answer would be w = 10,500 gallons... unless there was an equation missing in the question
7/8 = 21/24
5/12 = 10/24
21/24-10/24 = 11/24
Answer:
y = 9x + (-7) should be the answer