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Mathematics

2 answers:

aleksandr82 [10.1K]3 years ago

7 0

-13

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I think I am late sorry- :(

prisoha [69]3 years ago

4 0

<h2>Answer:</h2>

Your answers are all on the attachments and the explanations as well

Step-by-step explanation:

Click the attachments pleaseee!

hope this helped :3!

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3 years ago

Read 2 more answers

The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 – 5t2, where t is the tim

Leno4ka [110]

<h2>Time Interval : 1 sec to 1.414 sec</h2>

Step-by-step explanation:

The height of the stuntperson jumping off a $20\text{ }m$ is modelled as $h\text{ }=\text{ }20\text{ }-\text{ }5t^{2}$ . Here $t$ is the time in seconds.

We have to record the person when he is between the heights $15\text{ }m\text{ and }10\text{ }m$ from the ground.

When $h\text{ }=\text{ }15$ ,

$15\text{ }=20\text{ }-\text{ }5t^{2}\\5t^{2}\text{ }=\text{ }5\\t^{2}\text{ }=\text{ }1\\t\text{ }=\text{ }1\text{ }sec$

When $h\text{ }=\text{ }10$ ,

$10\text{ }=20\text{ }-\text{ }5t^{2}\\5t^{2}\text{ }=\text{ }10\\t^{2}\text{ }=\text{ }2\\t\text{ }=\text{ }\sqrt{2}\text{ }=\text{ }1.414\text{ }sec$

The stuntperson is at height $15\text{ }m$ at $t\text{ }=\text{ }1\text{ }sec$ , and is at height $10\text{ }m$ at $t\text{ }=\text{ }1.414\text{ }sec$ .