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To determine during what time interval the tennis ball will be at least 15 feet above the ground, we need to solve the inequality given by the height function [tex]\(h(x) = 40x - 16x^2\)[/tex].
1. We start by setting up the inequality:
[tex]\[ 40x - 16x^2 \geq 15 \][/tex]
2. To solve this, we first convert the inequality into an equality to find the critical points where the height is exactly 15 feet:
[tex]\[ 40x - 16x^2 = 15 \][/tex]
3. Rearrange the equation into standard quadratic form:
[tex]\[ -16x^2 + 40x - 15 = 0 \][/tex]
4. Next, solve the quadratic equation for [tex]\(x\)[/tex]. The quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] is used where [tex]\(a = -16\)[/tex], [tex]\(b = 40\)[/tex], and [tex]\(c = -15\)[/tex]:
[tex]\[ x = \frac{-40 \pm \sqrt{40^2 - 4(-16)(-15)}}{2(-16)} \][/tex]
Simplifying inside the square root:
[tex]\[ x = \frac{-40 \pm \sqrt{1600 - 960}}{-32} \][/tex]
[tex]\[ x = \frac{-40 \pm \sqrt{640}}{-32} \][/tex]
[tex]\[ x = \frac{-40 \pm 8\sqrt{10}}{-32} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{10}}{4} \][/tex]
5. The solutions to the quadratic equation are:
[tex]\[ x = \frac{5 - \sqrt{10}}{4} \quad \text{and} \quad x = \frac{5 + \sqrt{10}}{4} \][/tex]
6. To determine the interval during which the height of the tennis ball is at least 15 feet, we analyze the values of the quadratic function [tex]\(40x - 16x^2\)[/tex] at these points:
- Let [tex]\( x_1 = \frac{5 - \sqrt{10}}{4} \approx 0.5 \)[/tex]
- Let [tex]\( x_2 = \frac{5 + \sqrt{10}}{4} \approx 2.04 \)[/tex]
7. The height of the ball [tex]\(h(x)\)[/tex] is a downward-opening parabola (since the coefficient of [tex]\(x^2\)[/tex] is negative), meaning the ball reaches at least 15 feet above the ground between the roots [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex].
8. So, the time interval where the height is at least 15 feet is:
[tex]\[ \frac{5 - \sqrt{10}}{4} \leq x \leq \frac{5 + \sqrt{10}}{4} \][/tex]
Interpreted in decimal form, this means:
[tex]\[ 0.5 \leq x \leq 2.04 \][/tex]
Therefore, the correct answer is:
[tex]\[ x \geq 0.5 \text{ seconds and } x \leq 2.04 \text{ seconds} \][/tex]
So, the answer choice is:
[tex]\[ x \geq 0.5 \text{ seconds and } x \leq 2.04 \text{ seconds} \][/tex]
1. We start by setting up the inequality:
[tex]\[ 40x - 16x^2 \geq 15 \][/tex]
2. To solve this, we first convert the inequality into an equality to find the critical points where the height is exactly 15 feet:
[tex]\[ 40x - 16x^2 = 15 \][/tex]
3. Rearrange the equation into standard quadratic form:
[tex]\[ -16x^2 + 40x - 15 = 0 \][/tex]
4. Next, solve the quadratic equation for [tex]\(x\)[/tex]. The quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] is used where [tex]\(a = -16\)[/tex], [tex]\(b = 40\)[/tex], and [tex]\(c = -15\)[/tex]:
[tex]\[ x = \frac{-40 \pm \sqrt{40^2 - 4(-16)(-15)}}{2(-16)} \][/tex]
Simplifying inside the square root:
[tex]\[ x = \frac{-40 \pm \sqrt{1600 - 960}}{-32} \][/tex]
[tex]\[ x = \frac{-40 \pm \sqrt{640}}{-32} \][/tex]
[tex]\[ x = \frac{-40 \pm 8\sqrt{10}}{-32} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{10}}{4} \][/tex]
5. The solutions to the quadratic equation are:
[tex]\[ x = \frac{5 - \sqrt{10}}{4} \quad \text{and} \quad x = \frac{5 + \sqrt{10}}{4} \][/tex]
6. To determine the interval during which the height of the tennis ball is at least 15 feet, we analyze the values of the quadratic function [tex]\(40x - 16x^2\)[/tex] at these points:
- Let [tex]\( x_1 = \frac{5 - \sqrt{10}}{4} \approx 0.5 \)[/tex]
- Let [tex]\( x_2 = \frac{5 + \sqrt{10}}{4} \approx 2.04 \)[/tex]
7. The height of the ball [tex]\(h(x)\)[/tex] is a downward-opening parabola (since the coefficient of [tex]\(x^2\)[/tex] is negative), meaning the ball reaches at least 15 feet above the ground between the roots [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex].
8. So, the time interval where the height is at least 15 feet is:
[tex]\[ \frac{5 - \sqrt{10}}{4} \leq x \leq \frac{5 + \sqrt{10}}{4} \][/tex]
Interpreted in decimal form, this means:
[tex]\[ 0.5 \leq x \leq 2.04 \][/tex]
Therefore, the correct answer is:
[tex]\[ x \geq 0.5 \text{ seconds and } x \leq 2.04 \text{ seconds} \][/tex]
So, the answer choice is:
[tex]\[ x \geq 0.5 \text{ seconds and } x \leq 2.04 \text{ seconds} \][/tex]
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