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Jerald jumped from a bungee tower. If the equation that models his height, in feet, is h = –16t2 + 729, where t is the time in seconds, for which interval of time is he within 104 feet above the ground? A t > 6.25
B –6.25 < t < 6.25
C t < 6.25
D, 0

Respuesta :

JeanaShupp

Answer :- t > 6.25 would be the interval of time is he within 104 feet above the ground.


Explanation:-

Given equation:-[tex]h=-16t^2+729[/tex], where t is the time in seconds.

For the interval that he is at height within 104 feet above the ground.[tex]0<-16t^2+729<104[/tex]

[tex]-16t^2+729<104\\\Rightarrow\ -16t^2<104-729........[\text{subtract 729 both sides}]\\\Rightarrow-16t^2<-625\\\Rightarrow\ 16t^2>625....[\text{multiply -1 on both sides}]\\\Rightarrow\ 4t>25\text{ or }4t<-25....[\text{take square root on both sides}]\\\Rightarrow\ t>6.25\text{ or }t>-6.25..[\text{divide 4 on both sides}][/tex]

And time cannot be negative

Thus at t>6.25 seconds,for height= 104 feet above the ground.

⇒The interval of time is he within 104 feet above the ground

would be t > 6.25 .

Ver imagen JeanaShupp

Answer:

t >6.25

Step-by-step explanation:

Given : [tex]h = -16t^{2} +729[/tex]

To Find: for which interval of time is he within 104 feet above the ground?

Solution ;

[tex]h = -16t^{2} +729[/tex]

Since we are required to find for which interval of time is he within 104 feet above the ground

So,

[tex]-16t^{2} +729<104[/tex]

[tex]-16t^{2} <104-729[/tex]

[tex]-16t^{2} < -625[/tex]

[tex]16t^{2}>625[/tex]

[tex]t^{2} >\frac{625}{16}[/tex]

[tex]t >\sqrt{39.0625}[/tex]

[tex]t >6.25[/tex]

Thus for t >6.25 he is  within 104 feet above the ground

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