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Answer :
To determine the interval of time during which Jerald is less than 104 feet above the ground, we start with the given height equation for Jerald’s jump:
[tex]\[ h(t) = -16t^2 + 729 \][/tex]
We want to find the time interval where his height is less than 104 feet, so we set up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
First, we subtract 104 from both sides to simplify the inequality:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
[tex]\[ -16t^2 + 625 < 0 \][/tex]
Next, we solve this inequality for [tex]\( t \)[/tex]. Start by moving 625 to the other side:
[tex]\[ -16t^2 < -625 \][/tex]
Divide both sides by -16, remembering to flip the inequality sign because we are dividing by a negative number:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
Now, calculate [tex]\( \frac{625}{16} \)[/tex]:
[tex]\[ t^2 > \left(\frac{25}{4}\right)^2 \][/tex]
Taking the square root on both sides gives:
[tex]\[ |t| > \frac{25}{4} \][/tex]
This results in the two inequalities:
[tex]\[ t > \frac{25}{4} \quad \text{or} \quad t < -\frac{25}{4} \][/tex]
Since time [tex]\( t \)[/tex] must be greater than 0 (as it represents seconds after Jerald jumps), we discard the negative solution. Hence, the interval of time for which Jerald is less than 104 feet above the ground is:
[tex]\[ t > \frac{25}{4} \][/tex]
Converting [tex]\(\frac{25}{4}\)[/tex] to a decimal, we get:
[tex]\[ t > 6.25 \][/tex]
Thus, the correct interval is [tex]\( t > 6.25 \)[/tex].
[tex]\[ h(t) = -16t^2 + 729 \][/tex]
We want to find the time interval where his height is less than 104 feet, so we set up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
First, we subtract 104 from both sides to simplify the inequality:
[tex]\[ -16t^2 + 729 - 104 < 0 \][/tex]
[tex]\[ -16t^2 + 625 < 0 \][/tex]
Next, we solve this inequality for [tex]\( t \)[/tex]. Start by moving 625 to the other side:
[tex]\[ -16t^2 < -625 \][/tex]
Divide both sides by -16, remembering to flip the inequality sign because we are dividing by a negative number:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
Now, calculate [tex]\( \frac{625}{16} \)[/tex]:
[tex]\[ t^2 > \left(\frac{25}{4}\right)^2 \][/tex]
Taking the square root on both sides gives:
[tex]\[ |t| > \frac{25}{4} \][/tex]
This results in the two inequalities:
[tex]\[ t > \frac{25}{4} \quad \text{or} \quad t < -\frac{25}{4} \][/tex]
Since time [tex]\( t \)[/tex] must be greater than 0 (as it represents seconds after Jerald jumps), we discard the negative solution. Hence, the interval of time for which Jerald is less than 104 feet above the ground is:
[tex]\[ t > \frac{25}{4} \][/tex]
Converting [tex]\(\frac{25}{4}\)[/tex] to a decimal, we get:
[tex]\[ t > 6.25 \][/tex]
Thus, the correct interval is [tex]\( t > 6.25 \)[/tex].
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