Aprojectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. the path of the projectile is modeled using the equation h(t) = –16t2 48t 190. approximately when will the projectile hit the ground? 1.5 seconds 3.2 seconds 5.3 seconds 6.2 seconds

5.3 seconds

Step-by-step explanation:

We have the motion equation as follows

When the projectile touches the ground we have the height equal to 0

-16t^{2}+ 48t +190=0

we need to to solve this quadratic equation with:

In this case A=-16, B=48, C=190

The results are -2.25 s and 5.3 s, since time is a positive variable the final answer is 5.3 seconds

-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