A ball is kicked from an initial height of 2.5 feet with an initial velocity of 45 feet per second. The function ƒ(x) = −16x^2 + 45x + 2.5 models its path, where x is the time (in seconds) the ball travels and ƒ(x) is the height (in feet) the ball is kicked. What is the height of the ball 2 seconds after it is kicked?

28.5 ft

Step-by-step explanation:

The height of a ball is given by the equation ƒ(x) = −16x² + 45x + 2.5 where x is the tome the ball travels. To find the height of the ball after a seconds, we substitute a in place of x, the equation becomes:

ƒ(a) = −16a² + 45a + 2.5.

Therefore if the height of the ball is needed to be found after 2 seconds, we find f(2).

ƒ(2) = −16(2)² + 45(2) + 2.5 = -64 + 90 + 2.5 = 28.5 ft

f(2) = 28.5 ft

-13

step-by-step explanation:

f(x)=2x-5 for f(-4)

f(-4)   means let x=-4

f(-4) = 2(-4) -5

f(-4) =-8-5

      = -13