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Answer :
To determine the interval of time during which Jerald is less than 104 feet above the ground, we need to solve the inequality based on the equation of the height:
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to find when the height [tex]\( h \)[/tex] is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
1. Rearrange the inequality:
Subtract 104 from both sides:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
[tex]\[ -16t^2 < 104 - 729 \][/tex]
Simplify:
[tex]\[ -16t^2 < -625 \][/tex]
2. Solve for [tex]\( t^2 \)[/tex]:
Divide both sides by -16. Remember, dividing or multiplying by a negative number flips the inequality sign:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
Simplify the fraction:
[tex]\[ t^2 > 39.0625 \][/tex]
3. Find the values of [tex]\( t \)[/tex]:
We solve the equation [tex]\( t^2 = 39.0625 \)[/tex] to find the critical points.
Taking the square root of both sides, we get:
[tex]\[ t = \pm \sqrt{39.0625} \][/tex]
Calculating the square roots:
[tex]\[ t = \pm 6.25 \][/tex]
4. Determine the interval:
Since [tex]\( t^2 > 39.0625 \)[/tex], Jerald's height is less than 104 feet when [tex]\( t \)[/tex] is outside the range between -6.25 and 6.25:
[tex]\( t < -6.25 \)[/tex] or [tex]\( t > 6.25 \)[/tex]
Given that time [tex]\( t \)[/tex] in this context is non-negative (since it represents seconds after the jump starts), we only consider positive values, resulting in:
[tex]\[ t > 6.25 \][/tex]
Thus, Jerald is less than 104 feet above the ground for [tex]\( t > 6.25 \)[/tex] seconds.
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to find when the height [tex]\( h \)[/tex] is less than 104 feet:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
1. Rearrange the inequality:
Subtract 104 from both sides:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
[tex]\[ -16t^2 < 104 - 729 \][/tex]
Simplify:
[tex]\[ -16t^2 < -625 \][/tex]
2. Solve for [tex]\( t^2 \)[/tex]:
Divide both sides by -16. Remember, dividing or multiplying by a negative number flips the inequality sign:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
Simplify the fraction:
[tex]\[ t^2 > 39.0625 \][/tex]
3. Find the values of [tex]\( t \)[/tex]:
We solve the equation [tex]\( t^2 = 39.0625 \)[/tex] to find the critical points.
Taking the square root of both sides, we get:
[tex]\[ t = \pm \sqrt{39.0625} \][/tex]
Calculating the square roots:
[tex]\[ t = \pm 6.25 \][/tex]
4. Determine the interval:
Since [tex]\( t^2 > 39.0625 \)[/tex], Jerald's height is less than 104 feet when [tex]\( t \)[/tex] is outside the range between -6.25 and 6.25:
[tex]\( t < -6.25 \)[/tex] or [tex]\( t > 6.25 \)[/tex]
Given that time [tex]\( t \)[/tex] in this context is non-negative (since it represents seconds after the jump starts), we only consider positive values, resulting in:
[tex]\[ t > 6.25 \][/tex]
Thus, Jerald is less than 104 feet above the ground for [tex]\( t > 6.25 \)[/tex] seconds.
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