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Answer :
To solve this problem, we want to find the time interval during which Jerald's height is less than 104 feet above the ground. His height is modeled by the equation [tex]\( h = -16t^2 + 729 \)[/tex], where [tex]\( h \)[/tex] is the height in feet and [tex]\( t \)[/tex] is the time in seconds.
1. Set Up the Inequality:
We need to find when his height is less than 104 feet:
[tex]\[
-16t^2 + 729 < 104
\][/tex]
2. Simplify the Inequality:
Subtract 104 from both sides:
[tex]\[
-16t^2 + 729 - 104 < 0
\][/tex]
Simplifying the numbers gives:
[tex]\[
-16t^2 + 625 < 0
\][/tex]
3. Find the Critical Points:
To find when this inequality holds, first solve the equation [tex]\(-16t^2 + 625 = 0\)[/tex] for critical points.
Rearrange to:
[tex]\[
-16t^2 = -625
\][/tex]
Divide both sides by -16:
[tex]\[
t^2 = \frac{625}{16}
\][/tex]
4. Solve for [tex]\( t \)[/tex]:
Take the square root of both sides to find the critical points:
[tex]\[
t = \pm \sqrt{\frac{625}{16}}
\][/tex]
Simplifying the square root:
[tex]\[
t = \pm 6.25
\][/tex]
5. Identify the Interval:
The critical points are [tex]\( t = -6.25 \)[/tex] and [tex]\( t = 6.25 \)[/tex]. However, since time cannot be negative, we discard [tex]\( -6.25 \)[/tex].
The parabola opens downward (since the coefficient of [tex]\( t^2 \)[/tex] is negative), which means the height is less than 104 feet during the time interval between these two values. Thus, the interval in question is:
[tex]\[
0 \leq t \leq 6.25
\][/tex]
Therefore, Jerald is less than 104 feet above the ground during the time interval from [tex]\( t = 0 \)[/tex] to [tex]\( t = 6.25 \)[/tex] seconds.
The correct answer is: [tex]\(0 \leq t \leq 6.25\)[/tex].
1. Set Up the Inequality:
We need to find when his height is less than 104 feet:
[tex]\[
-16t^2 + 729 < 104
\][/tex]
2. Simplify the Inequality:
Subtract 104 from both sides:
[tex]\[
-16t^2 + 729 - 104 < 0
\][/tex]
Simplifying the numbers gives:
[tex]\[
-16t^2 + 625 < 0
\][/tex]
3. Find the Critical Points:
To find when this inequality holds, first solve the equation [tex]\(-16t^2 + 625 = 0\)[/tex] for critical points.
Rearrange to:
[tex]\[
-16t^2 = -625
\][/tex]
Divide both sides by -16:
[tex]\[
t^2 = \frac{625}{16}
\][/tex]
4. Solve for [tex]\( t \)[/tex]:
Take the square root of both sides to find the critical points:
[tex]\[
t = \pm \sqrt{\frac{625}{16}}
\][/tex]
Simplifying the square root:
[tex]\[
t = \pm 6.25
\][/tex]
5. Identify the Interval:
The critical points are [tex]\( t = -6.25 \)[/tex] and [tex]\( t = 6.25 \)[/tex]. However, since time cannot be negative, we discard [tex]\( -6.25 \)[/tex].
The parabola opens downward (since the coefficient of [tex]\( t^2 \)[/tex] is negative), which means the height is less than 104 feet during the time interval between these two values. Thus, the interval in question is:
[tex]\[
0 \leq t \leq 6.25
\][/tex]
Therefore, Jerald is less than 104 feet above the ground during the time interval from [tex]\( t = 0 \)[/tex] to [tex]\( t = 6.25 \)[/tex] seconds.
The correct answer is: [tex]\(0 \leq t \leq 6.25\)[/tex].
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