Question

select the correct inequality and solution set. determine the time when the rocket will be more 18 feet above the ground. solution set: 0.5, 7 inequality: -4t² + 30t + 4 ≤ 18 inequality: -4t² + 30t + 4 > 18 solution set: 0, 0.5 ∪ 7, 7.6 inequality: -4t² + 30t + 4 < 18 solution set: (0.5, 7) inequality: -4t² + 30t + 4 ≥ 18

Explanation:

Step1: Interpret the problem

We want the rocket's height to be more than 18 feet. If the height - function of the rocket is $h(t)=-4t^{2}+30t + 4$, the inequality should be $-4t^{2}+30t + 4>18$.

Step2: Rearrange the inequality

Subtract 18 from both sides: $-4t^{2}+30t + 4-18>0$, which simplifies to $-4t^{2}+30t - 14>0$. Divide through by - 2 (and reverse the inequality sign) to get $2t^{2}-15t + 7<0$.

Step3: Solve the quadratic equation

For the quadratic equation $2t^{2}-15t + 7 = 0$, use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 2$, $b=-15$, and $c = 7$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-15)^{2}-4\times2\times7=225 - 56 = 169$. Then $t=\frac{15\pm\sqrt{169}}{4}=\frac{15\pm13}{4}$. The roots are $t_1=\frac{15 + 13}{4}=7$ and $t_2=\frac{15 - 13}{4}=0.5$.

Step4: Determine the solution set

The solution of the inequality $2t^{2}-15t + 7<0$ is the interval between the two roots. So the solution set is $(0.5,7)$.

Answer:

Inequality: $-4t^{2}+30t + 4>18$, Solution Set: $(0.5,7)$