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Answer :
To solve this problem, we need to determine the time interval during which Jerald is less than 104 feet above the ground. We are given the height equation:
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to find the time [tex]\( t \)[/tex] such that:
[tex]\[ h < 104 \][/tex]
Substituting the height equation, we have:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Now, let's solve this inequality step-by-step:
1. Subtract 729 from both sides to isolate the quadratic term:
[tex]\[ -16t^2 < 104 - 729 \][/tex]
[tex]\[ -16t^2 < -625 \][/tex]
2. Divide both sides by -16. Remember to flip the inequality sign when dividing by a negative number:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
3. Calculate the right side:
[tex]\[ \frac{625}{16} = 39.0625 \][/tex]
So the inequality becomes:
[tex]\[ t^2 > 39.0625 \][/tex]
4. The solution to [tex]\( t^2 > 39.0625 \)[/tex] is:
[tex]\[ t > \sqrt{39.0625} \][/tex]
This gives one boundary, but we need to consider the practical scenario where time [tex]\( t \)[/tex] starts from zero:
[tex]\[ t > \sqrt{39.0625} \approx 6.25 \][/tex]
Since we are considering the time when Jerald is less than 104 feet, the interval of time he is less than 104 feet above the ground starts right after this value.
So, the interval for which Jerald is less than 104 feet above the ground is:
[tex]\[ t > 6.25 \][/tex]
This matches choice [tex]\( t > 6.25 \)[/tex].
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to find the time [tex]\( t \)[/tex] such that:
[tex]\[ h < 104 \][/tex]
Substituting the height equation, we have:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Now, let's solve this inequality step-by-step:
1. Subtract 729 from both sides to isolate the quadratic term:
[tex]\[ -16t^2 < 104 - 729 \][/tex]
[tex]\[ -16t^2 < -625 \][/tex]
2. Divide both sides by -16. Remember to flip the inequality sign when dividing by a negative number:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
3. Calculate the right side:
[tex]\[ \frac{625}{16} = 39.0625 \][/tex]
So the inequality becomes:
[tex]\[ t^2 > 39.0625 \][/tex]
4. The solution to [tex]\( t^2 > 39.0625 \)[/tex] is:
[tex]\[ t > \sqrt{39.0625} \][/tex]
This gives one boundary, but we need to consider the practical scenario where time [tex]\( t \)[/tex] starts from zero:
[tex]\[ t > \sqrt{39.0625} \approx 6.25 \][/tex]
Since we are considering the time when Jerald is less than 104 feet, the interval of time he is less than 104 feet above the ground starts right after this value.
So, the interval for which Jerald is less than 104 feet above the ground is:
[tex]\[ t > 6.25 \][/tex]
This matches choice [tex]\( t > 6.25 \)[/tex].
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