An object is thrown straight up from the top of a building 150 feet tall with an initial velocity of 12 feet per second. The height of the object as a function of time can be modeled by the function y = –16x2 +12x + 150, where y is the height of the object (in feet) and x is the time in seconds after it is thrown.

will this work or help ?

Step-by-step explanation:

The given equation governs the height versus time of the object. h(0) = 64, so the ground is considered at h = 0. The question is at what time is h = 0? So, set h = 0 and solve for t.

-16t^2 + 48t + 64 = 0

We can divide both sides of this equation by -16 for simpler numbers. We get

t^2 - 3t - 4 = 0

We can factor this equation:

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

So, t = 4 and t = -1 are solutions.

But, time must be positive, so t = -1 makes no sense, so we discard it. This leaves

t = 4 sec.

40% off the regular price is the correct answer.

step-by-step explanation:

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: