A rock is thrown from the top of a tall building. The distance, in feet, between the rock in the ground T seconds after it is thrown it is given by d=-16t-4t+462. How long after the rock is thrown is 450 feet from the ground

0.75 seconds

Step-by-step explanation:

We simply set the distance d to 450 feet

Thus, we have

450 = -16t^2 - 4t + 462

= 16t^2 + 4t - 462 + 450

0 = 16t^2 + 4t - 12

divide through by 4

4t^2 + t - 3 = 0

4t^2 + 4t - 3t - 3 = 0

4t(t + 1) -3( t + 1) = 0

(4t-3)(t + 1) = 0

t + 1 = 0 or 4t-3 = 0

t = -1 or 3/4

t = -1 or 0.75

Since time cannot be negative , we have 0.75 as our only answer

Thus, it will take the rock 0.75 seconds to reach a distance of 450 feet after it is thrown from the ground

e) none of the above because the measure of angle cba is 101°