The equation h(t) = 1.2 + 40t – 5t2 models the height, in meters of a ball t
seconds after it has been launched from a catapult. At what times is the
ball exactly 40 meters above the surface.

Step-by-step explanation:

To determine the time when the height is 40 meters above the ground, we would substitute 40 for h in the given equation. It becomes

1.2 + 4t0t - 5t² = 40

5t² - 40t + 40 - 1.2 = 0

5t² - 40t + 38.8 = 0

The general formula for solving quadratic equations is expressed as

x = [- b ± √(b² - 4ac)]/2a

From the equation given,

a = 5

b = - 40

c = 38.8

Therefore,

t = [- - 40 ± √(- 40² - 4 × 5 × 38.8)]/2 × 5

t = [40 ± √(1600 - 776)]/10

t = [40 ± √824]/10

t = (40 + 28.71)/10 or x = (40 - 28.71)/10

t = 6.871 seconds or t = 1.129 seconds

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632

step-by-step explanation:

what are the statements? and what do you mean by "216 2 1 6"?