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 $t\text{ }=\text{ }1.414\text{ }sec$ .

∴ The camera should film the person for the time interval $\textbf{1}\text{ }sec\text{ to }\textbf{1.414}\text{ }sec$ .

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3 years ago

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Suppose you have two six-sided dice where each side is equally likely to land face up when rolled. What is the probability that

Basile [38]

Answer:

Step-by-step explanation:

brainly.com/question/14291598?referrer=fromSearch

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3 years ago

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Write y=2x^2+8x+5 in vertex form. Please

zmey [24]

--> y=2x^2+8x+5
Step 1: Group the first 2 terms together, separating them from the constant term.
--> (2 x X^2+{8}x X)+{5}
Step 2: Factor out leading coefficient, for completing the square to work, the coefficient of x2 must be 1.
--> 2 x (x^2+{4} x X) +{5}
Step 3: Complete the square, Take half of x coefficient and square it. Notice to keep equation balanced you must add this number and subtract it making the net effect zero.
--> 2 x(X^2{4}x X +4-4)+ {5}
--> 2 x(X+{2}^2 x 4) + {5}
Step 4: Distribute and add constants.
--> 2 x(X+{2}^2 -8+5
--> 2 x(X+{2}^2 +{-3}
Now it is successfully in vertex form and can be easily graphed.
The vertex is at (-2,-3)
The parabola opens up and has a y-intercept at (0, 5)
Here is a graph of this parabola:

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3 years ago