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Answer :
To solve for the time interval when Jerald is less than 104 feet above the ground, we need to work through the equation given for his height:
The equation is:
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to find when his height [tex]\( h \)[/tex] is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Let's solve this step-by-step:
1. Rearrange the inequality:
Subtract 729 from both sides:
[tex]\[ -16t^2 < 104 - 729 \][/tex]
Simplify:
[tex]\[ -16t^2 < -625 \][/tex]
2. Divide by -16:
When dividing by a negative number, remember to flip the inequality sign:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
3. Calculate the square root:
Find the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ t > \sqrt{\frac{625}{16}} \][/tex]
[tex]\[ t > \frac{\sqrt{625}}{\sqrt{16}} \][/tex]
[tex]\[ t > \frac{25}{4} \][/tex]
Calculate:
[tex]\[ t > 6.25 \][/tex]
So, Jerald is less than 104 feet above the ground at times greater than 6.25 seconds.
Therefore, the correct interval is: [tex]\( t > 6.25 \)[/tex] seconds.
The equation is:
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to find when his height [tex]\( h \)[/tex] is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Let's solve this step-by-step:
1. Rearrange the inequality:
Subtract 729 from both sides:
[tex]\[ -16t^2 < 104 - 729 \][/tex]
Simplify:
[tex]\[ -16t^2 < -625 \][/tex]
2. Divide by -16:
When dividing by a negative number, remember to flip the inequality sign:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
3. Calculate the square root:
Find the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ t > \sqrt{\frac{625}{16}} \][/tex]
[tex]\[ t > \frac{\sqrt{625}}{\sqrt{16}} \][/tex]
[tex]\[ t > \frac{25}{4} \][/tex]
Calculate:
[tex]\[ t > 6.25 \][/tex]
So, Jerald is less than 104 feet above the ground at times greater than 6.25 seconds.
Therefore, the correct interval is: [tex]\( t > 6.25 \)[/tex] seconds.
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