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Answer :
To solve the problem of finding for which interval of time Jerald is less than 104 feet above the ground, we need to work with the given height equation:
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to find when this height, [tex]\( h \)[/tex], is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Let's solve this inequality step-by-step:
1. Rearrange the inequality:
Subtract 729 from both sides:
[tex]\[ -16t^2 < 104 - 729 \][/tex]
Simplify the right side:
[tex]\[ -16t^2 < -625 \][/tex]
2. Isolate the quadratic term:
Since the inequality is negative, let's divide both sides by -16. Remember, dividing by a negative number reverses the inequality:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
3. Solve for [tex]\( t \)[/tex]:
Now, solve the inequality [tex]\( t^2 > 39.0625 \)[/tex].
Take the square root of both sides:
[tex]\[ t > \sqrt{39.0625} \quad \text{or} \quad t < -\sqrt{39.0625} \][/tex]
4. Interpret the solution in terms of time, [tex]\( t \)[/tex]:
Since time, [tex]\( t \)[/tex], cannot be negative (as we're considering the time after Jerald jumps), we only consider the solution where [tex]\( t \)[/tex] is positive:
[tex]\[ t > 6.25 \][/tex]
Therefore, the interval of time for which Jerald is less than 104 feet above the ground is:
[tex]\( t > 6.25 \)[/tex]
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to find when this height, [tex]\( h \)[/tex], is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Let's solve this inequality step-by-step:
1. Rearrange the inequality:
Subtract 729 from both sides:
[tex]\[ -16t^2 < 104 - 729 \][/tex]
Simplify the right side:
[tex]\[ -16t^2 < -625 \][/tex]
2. Isolate the quadratic term:
Since the inequality is negative, let's divide both sides by -16. Remember, dividing by a negative number reverses the inequality:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
3. Solve for [tex]\( t \)[/tex]:
Now, solve the inequality [tex]\( t^2 > 39.0625 \)[/tex].
Take the square root of both sides:
[tex]\[ t > \sqrt{39.0625} \quad \text{or} \quad t < -\sqrt{39.0625} \][/tex]
4. Interpret the solution in terms of time, [tex]\( t \)[/tex]:
Since time, [tex]\( t \)[/tex], cannot be negative (as we're considering the time after Jerald jumps), we only consider the solution where [tex]\( t \)[/tex] is positive:
[tex]\[ t > 6.25 \][/tex]
Therefore, the interval of time for which Jerald is less than 104 feet above the ground is:
[tex]\( t > 6.25 \)[/tex]
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