a quarterback throws a ball in the air from an initial height of 5 feet. The football has an initial vertical velocity of 50 feet per second. Another player catches the ball when it is 2 feet above the ground. How long is the ball in the air?

The range of the function is from 2 feet to 12.25 feet

Step-by-step explanation:

In a function f(x) = y

x is the domain of the function

y is the range of the function

The height (h) of the ball at time (t) seconds can be represented by the equation h(t) = - 16 t² + 20 t + 6

∵ h(t) = - 16 t² + 20 t + 6

∴ The domain is t

∴ The range is h(t)

- To find the range of the quadratic function find the maximum or

minimum value of it

∵ The leading coefficient of the function is -16

∴ The function has a maximum value

To find the maximum value differentiate h(t) with respect to t and equate it by 0 to find the value of t for the maximum height

∵ h'(t) = -16(2) t + 20(1)

∴ h'(t) = -32 t + 20

- Equate it by 0

∵ h'(t) = 0

∴ -32 t + 20 = 0

- Subtract 20 from both sides

∴ -32 t = - 20

- Divide both sides by -32

∴ t = 0.625 seconds ⇒ time for the maximum height

Substitute the value of t in h(t) to find the maximum height

∵ h(0.625) = -16(0.625)² + 20(0.625) + 6

∴ h(t) = 12.25 feet

∴ The maximum height of the ball is 12.25 feet

∵ The ball is caught at 2 feet

∴ The range of the function is 2 ≤ h(t) ≤ 12.25

The range of the function is from 2 feet to 12.25 feet

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step-by-step explanation:

i wish i could u

step-by-step explanation:

de-gf

hj=3x+20

ji=15x-64