A rocket is launched from atop a 74-foot cliff with an initial velocity of 113 ft/s. The height of the rocket above the ground at time t is given by h=16t^2+113t+74. When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.

9514 1404 393

  7.7 seconds

Step-by-step explanation:

The solution is given by the quadratic formula.

For ax² +bx +c = 0, the roots are ...

  x = (-b±√(b²-4ac))/(2a)

We want the positive root, so ...

  x = (-113 -√(113² -4(-16)(74)))/(2(-16)) = (113 +√17505)/32 ≈ 7.6658

The rocket will hit the ground about 7.7 seconds after it is launched.

a is the correct

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3