A ball is thrown vertically upward with an initial velocity of 96 feet per second. the distances (in feet) of the ball from the ground after t seconds is s(t)=96t-16t^2. a.) At what time t will the ball strikes the ground?
b.) from what time t is the ball more than 138ft above the ground?

a) The ball strikes the ground at 6 seconds

b) The ball will be at more than 138 ft above the ground when the time is between 2.39 s and 3.61 s

Step-by-step explanation:

The distance of the ball from the ground after t seconds is modeled by the equation:

a) The ball is at ground level when s=0, thus:

Factoring:

There are two solutions:

t=0

96-16t=0

The first solution corresponds to the moment when the ball is thrown upward.

The second solution comes from:

96-16t=0

Solving:

t = 96/16 = 6 s

The ball strikes the ground at 6 seconds.

b)The ball will have a distance of more than 138 ft above the ground when:

Rearranging:

Factoring:

This inequality is satisfied when t lies in the interval:

(2.39,3.61)

The ball will be at more than 138 ft above the ground when the time is between 2.39 s and 3.61 s

80° is the measure of gh

step-by-step explanation:

circle f is congruent to circle j

we have two circle with center f and j.

both circles are congruent. the radius of the circle would be equal.

see the attachment for the circle.

we are given ∠efd ≅ ∠gjh

therefore, arc (ed) = arc (gh)     (∴ circle f is congruent to circle j)

but ∠dfe=80°=arc (ed)

therefore, arc (gh) = 80°

thus, 80° is the measure of gh

y=-2x+1500

step-by-step explanation:

+1500

2 over one