The distance of an object thrown upward is modeled by the equation h=-2x^2+3t+4, where h is the height of the object in feet at any given time and t is the time in seconds. the object is visible whenever it is at h (greater than or equal to) 0. for how long is the object visible?

So for this, we will have to find the zeros (x-intercepts) of the equation. In this case, we will be using the quadratic formula, which is (a = x^2 coefficient, b = x coefficient, c = constant). Using our info, our equation is such:

Firstly, solve the multiplications and the exponents:

Next, do the addition:

Next, plug in the equations into the calculator and your answer will be:

Since we can't have negative time in this situation, it can't be -0.41 seconds. Which means that the object is visible for about 0.60 seconds, or C.

a

step-by-step explanation: