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Answer :
To solve the problem of determining when Jerald is less than 104 feet above the ground, we start with his height equation:
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to find the time [tex]\( t \)[/tex] when his height is less than 104 feet. So, we set up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
First, let's solve for when his height is exactly 104 feet to find the critical points:
[tex]\[ -16t^2 + 729 = 104 \][/tex]
Next, solve the equation:
1. Subtract 104 from 729 to isolate terms with [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 729 = 104 \][/tex]
[tex]\[ -16t^2 = 104 - 729 \][/tex]
[tex]\[ -16t^2 = -625 \][/tex]
2. Divide both sides by -16:
[tex]\[ t^2 = \frac{-625}{-16} \][/tex]
[tex]\[ t^2 = \frac{625}{16} \][/tex]
[tex]\[ t = \pm \sqrt{\frac{625}{16}} \][/tex]
[tex]\[ t = \pm \frac{25}{4} \][/tex]
This gives us the critical points [tex]\( t = -\frac{25}{4} \)[/tex] and [tex]\( t = \frac{25}{4} \)[/tex].
Now, the interval for [tex]\( t \)[/tex] will be when the height is less than 104 feet, which happens between these two critical points:
- Since time, [tex]\( t \)[/tex], typically starts from 0 in real-life scenarios and increases, we consider the positive value for practical reasons.
Thus, the interval where the height is less than 104 feet is:
[tex]\[ 0 \leq t < \frac{25}{4} \][/tex]
Now let's approximate this value:
[tex]\[ \frac{25}{4} = 6.25 \][/tex]
Therefore, Jerald is less than 104 feet above the ground during the interval:
[tex]\[ 0 \leq t < 6.25 \][/tex]
This corresponds to the interval option:
[tex]\( 0 \leq t \leq 6.25 \)[/tex]
[tex]\[ h = -16t^2 + 729 \][/tex]
We want to find the time [tex]\( t \)[/tex] when his height is less than 104 feet. So, we set up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
First, let's solve for when his height is exactly 104 feet to find the critical points:
[tex]\[ -16t^2 + 729 = 104 \][/tex]
Next, solve the equation:
1. Subtract 104 from 729 to isolate terms with [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 729 = 104 \][/tex]
[tex]\[ -16t^2 = 104 - 729 \][/tex]
[tex]\[ -16t^2 = -625 \][/tex]
2. Divide both sides by -16:
[tex]\[ t^2 = \frac{-625}{-16} \][/tex]
[tex]\[ t^2 = \frac{625}{16} \][/tex]
[tex]\[ t = \pm \sqrt{\frac{625}{16}} \][/tex]
[tex]\[ t = \pm \frac{25}{4} \][/tex]
This gives us the critical points [tex]\( t = -\frac{25}{4} \)[/tex] and [tex]\( t = \frac{25}{4} \)[/tex].
Now, the interval for [tex]\( t \)[/tex] will be when the height is less than 104 feet, which happens between these two critical points:
- Since time, [tex]\( t \)[/tex], typically starts from 0 in real-life scenarios and increases, we consider the positive value for practical reasons.
Thus, the interval where the height is less than 104 feet is:
[tex]\[ 0 \leq t < \frac{25}{4} \][/tex]
Now let's approximate this value:
[tex]\[ \frac{25}{4} = 6.25 \][/tex]
Therefore, Jerald is less than 104 feet above the ground during the interval:
[tex]\[ 0 \leq t < 6.25 \][/tex]
This corresponds to the interval option:
[tex]\( 0 \leq t \leq 6.25 \)[/tex]
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