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Answer :
Certainly! Let's solve the problem step by step.
The height of the toy rocket is given by the equation:
[tex]\[ h(t) = 128t - 16t^2 \][/tex]
We need to find the times when the height [tex]\( h(t) \)[/tex] is at least 200 feet:
[tex]\[ 128t - 16t^2 \geq 200 \][/tex]
First, rearrange this inequality:
[tex]\[ 128t - 16t^2 - 200 \geq 0 \][/tex]
Rewriting it in the standard quadratic form [tex]\( at^2 + bt + c = 0 \)[/tex], we have:
[tex]\[ -16t^2 + 128t - 200 \geq 0 \][/tex]
To make it easier to solve, we can divide the entire inequality by -16 (which will flip the inequality sign):
[tex]\[ t^2 - 8t + 12.5 \leq 0 \][/tex]
Now, we need to solve the quadratic equation:
[tex]\[ t^2 - 8t + 12.5 = 0 \][/tex]
Use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation, [tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], [tex]\( c = 12.5 \)[/tex].
Calculate the discriminant:
[tex]\[ b^2 - 4ac = (-8)^2 - 4 \times 1 \times 12.5 = 64 - 50 = 14 \][/tex]
Now plug these values into the quadratic formula:
[tex]\[ t = \frac{8 \pm \sqrt{14}}{2} \][/tex]
Calculate the two roots:
[tex]\[ t = \frac{8 + \sqrt{14}}{2} \][/tex]
[tex]\[ t = \frac{8 - \sqrt{14}}{2} \][/tex]
Board calculations (approximating):
[tex]\[ \sqrt{14} \approx 3.74 \][/tex]
So the roots are:
[tex]\[ t \approx \frac{8 + 3.74}{2} = \frac{11.74}{2} = 5.87 \][/tex]
[tex]\[ t \approx \frac{8 - 3.74}{2} = \frac{4.26}{2} = 2.13 \][/tex]
These are the critical points where the height is exactly 200 feet. To find when the height is at least 200 feet, we check the interval between these two times. Since it's a downward-opening parabola, the height will be at least 200 feet between [tex]\( t \approx 2.13 \)[/tex] and [tex]\( t \approx 5.87 \)[/tex].
Thus, the inequality for when the height of the rocket is at least 200 feet is:
[tex]\[ 2.13 \leq t \leq 5.87 \][/tex]
Rounding to the nearest hundredths, the solution is:
[tex]\[ t \in [2.13, 5.87] \][/tex]
The height of the toy rocket is given by the equation:
[tex]\[ h(t) = 128t - 16t^2 \][/tex]
We need to find the times when the height [tex]\( h(t) \)[/tex] is at least 200 feet:
[tex]\[ 128t - 16t^2 \geq 200 \][/tex]
First, rearrange this inequality:
[tex]\[ 128t - 16t^2 - 200 \geq 0 \][/tex]
Rewriting it in the standard quadratic form [tex]\( at^2 + bt + c = 0 \)[/tex], we have:
[tex]\[ -16t^2 + 128t - 200 \geq 0 \][/tex]
To make it easier to solve, we can divide the entire inequality by -16 (which will flip the inequality sign):
[tex]\[ t^2 - 8t + 12.5 \leq 0 \][/tex]
Now, we need to solve the quadratic equation:
[tex]\[ t^2 - 8t + 12.5 = 0 \][/tex]
Use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation, [tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], [tex]\( c = 12.5 \)[/tex].
Calculate the discriminant:
[tex]\[ b^2 - 4ac = (-8)^2 - 4 \times 1 \times 12.5 = 64 - 50 = 14 \][/tex]
Now plug these values into the quadratic formula:
[tex]\[ t = \frac{8 \pm \sqrt{14}}{2} \][/tex]
Calculate the two roots:
[tex]\[ t = \frac{8 + \sqrt{14}}{2} \][/tex]
[tex]\[ t = \frac{8 - \sqrt{14}}{2} \][/tex]
Board calculations (approximating):
[tex]\[ \sqrt{14} \approx 3.74 \][/tex]
So the roots are:
[tex]\[ t \approx \frac{8 + 3.74}{2} = \frac{11.74}{2} = 5.87 \][/tex]
[tex]\[ t \approx \frac{8 - 3.74}{2} = \frac{4.26}{2} = 2.13 \][/tex]
These are the critical points where the height is exactly 200 feet. To find when the height is at least 200 feet, we check the interval between these two times. Since it's a downward-opening parabola, the height will be at least 200 feet between [tex]\( t \approx 2.13 \)[/tex] and [tex]\( t \approx 5.87 \)[/tex].
Thus, the inequality for when the height of the rocket is at least 200 feet is:
[tex]\[ 2.13 \leq t \leq 5.87 \][/tex]
Rounding to the nearest hundredths, the solution is:
[tex]\[ t \in [2.13, 5.87] \][/tex]
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