Ajuggler tosses a ball into the air . the balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40t+4 . suppose the juggler missed and ball hit the ground . find the maximum height of the ball and time it took to reach the ground.

29 feet

2.6 seconds

Step-by-step explanation:

h(t) is a downwards parabola, so the maximum is at the vertex.

t = -b / (2a)

t = -40 / (2×-16)

t = 1.25

h(1.25) = -16(1.25)² + 40(1.25) + 4

h(1.25) = 29

When the ball lands, h(t) = 0.

0 = -16t² + 40t + 4

0 = 4t² − 10t − 1

t = [ -(-10) ± √((-10)² − 4(4)(-1)) ] / 2(4)

t = (10 ± √116) / 8

t = (5 ± √29) / 4

t is positive, so:

t = (5 + √29) / 4

t ≈ 2.6

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !