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Answer :
Sure! Let's solve this problem step-by-step.
The equation given for Jerald's height, [tex]\( h \)[/tex], in feet, in terms of time, [tex]\( t \)[/tex], in seconds is:
[tex]\[ h = -16t^2 + 729 \][/tex]
We need to determine when Jerald's height is less than 104 feet above the ground. So, we set up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Next, we solve this inequality step-by-step.
1. Set up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
2. Move 104 to the other side:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
Simplifying, this becomes:
[tex]\[ -16t^2 + 625 < 0 \][/tex]
3. Isolate the [tex]\( t^2 \)[/tex] term:
First, subtract 625 from both sides:
[tex]\[ -16t^2 < -625 \][/tex]
Then, divide both sides by -16, and remember to reverse the inequality sign because we are dividing by a negative number:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
4. Simplify:
[tex]\[ t^2 > 39.0625 \][/tex]
5. Take the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ |t| > \sqrt{39.0625} \][/tex]
So,
[tex]\[ |t| > 6.25 \][/tex]
This means,
[tex]\[ t < -6.25 \quad \text{or} \quad t > 6.25 \][/tex]
But, since time [tex]\( t \)[/tex] must be positive (because it represents time elapsed since Jerald jumped):
Therefore, [tex]\( t > 6.25 \)[/tex].
So, the correct interval of time for which Jerald is less than 104 feet above the ground is:
[tex]\[ t > 6.25 \][/tex]
Thus, the answer is:
[tex]\[ t > 6.25 \][/tex]
The equation given for Jerald's height, [tex]\( h \)[/tex], in feet, in terms of time, [tex]\( t \)[/tex], in seconds is:
[tex]\[ h = -16t^2 + 729 \][/tex]
We need to determine when Jerald's height is less than 104 feet above the ground. So, we set up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Next, we solve this inequality step-by-step.
1. Set up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
2. Move 104 to the other side:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
Simplifying, this becomes:
[tex]\[ -16t^2 + 625 < 0 \][/tex]
3. Isolate the [tex]\( t^2 \)[/tex] term:
First, subtract 625 from both sides:
[tex]\[ -16t^2 < -625 \][/tex]
Then, divide both sides by -16, and remember to reverse the inequality sign because we are dividing by a negative number:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
4. Simplify:
[tex]\[ t^2 > 39.0625 \][/tex]
5. Take the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ |t| > \sqrt{39.0625} \][/tex]
So,
[tex]\[ |t| > 6.25 \][/tex]
This means,
[tex]\[ t < -6.25 \quad \text{or} \quad t > 6.25 \][/tex]
But, since time [tex]\( t \)[/tex] must be positive (because it represents time elapsed since Jerald jumped):
Therefore, [tex]\( t > 6.25 \)[/tex].
So, the correct interval of time for which Jerald is less than 104 feet above the ground is:
[tex]\[ t > 6.25 \][/tex]
Thus, the answer is:
[tex]\[ t > 6.25 \][/tex]
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