Question

select the correct inequality and solution set. determine the time when the rocket will be more 40 feet above the ground. solution set: 0, 1.5 ∪ 6, 7.6 inequality: -4t² + 30t + 44 > 0 inequality: -4t² + 30t - 36 > 0 inequality: -4t² + 30t - 44 ≥ 0 solution set: (1.5, 6) solution set: (0, 1.5) ∪ (6, 7.6) solution set: 1.5, 6

Explanation:

Step1: Assume height - time formula

Let the height of the rocket as a function of time $t$ be $h(t)=-4t^{2}+30t + k$. When the rocket is on the ground, $h(t) = 0$. We want to find when $h(t)>40$. So, $-4t^{2}+30t + k>40$. If we assume the initial - condition related setup gives us a simplified form. Let's set up the inequality directly for the height condition. We want to find when $-4t^{2}+30t + k-40>0$. If we assume the initial height and other factors are accounted for in a standard quadratic - motion model, we get the inequality $-4t^{2}+30t - 36>0$ (by setting up the height equation and moving 40 to the left - hand side and simplifying the constant terms).

Step2: Solve the quadratic inequality

First, rewrite the inequality $-4t^{2}+30t - 36>0$ as $4t^{2}-30t + 36<0$ (multiplying both sides by - 1 and reversing the inequality sign). Then, divide through by 2: $2t^{2}-15t + 18<0$. Factor the quadratic: $2t^{2}-15t + 18=(2t - 3)(t - 6)<0$.
The roots of the equation $(2t - 3)(t - 6)=0$ are $t = 1.5$ and $t = 6$.
For a quadratic function $y=(2t - 3)(t - 6)=2t^{2}-15t + 18$ which is a parabola opening upwards (since the coefficient of $t^{2}$ is positive), the solution of the inequality $(2t - 3)(t - 6)<0$ is the interval between the roots. So the solution set is $(1.5,6)$.

Answer:

Inequality: $-4t^{2}+30t - 36>0$
Solution Set: $(1.5,6)$