Answer:
c. 0.71
Step-by-step explanation:
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]
[tex]\Delta = b^{2} - 4ac[/tex]
The height of the candy after t seconds is given by the following equation:
[tex]h(t) = -16t^2 + 10t + 3[/tex]
After how many seconds will the ball be 2 feet in the air?
This is t for which [tex]h(t) = 2[/tex]
So
[tex]h(t) = -16t^2 + 10t + 3[/tex]
[tex]2 = -16t^2 + 10t + 3[/tex]
[tex]-16t^2 + 10t + 1 = 0[/tex]
So
[tex]a = -16, b = 10, c = 1[/tex]
[tex]\Delta = b^{2} - 4ac = 10^2 -4(-16)(1) = 164[/tex]
[tex]t_{1} = \frac{-10 + \sqrt{164}}{2*(-16)} = -0.09[/tex]
[tex]t_{2} = \frac{-10 - \sqrt{164}}{2*(-16)} = 0.71[/tex]
Since time is a positive measure, the answer is 0.71, which is option c.
