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Vedmedyk [2.9K]
3 years ago
5

An object is launched from a platform. It's height in meters, x seconds after the launch is modeled by:

Mathematics
1 answer:
Contact [7]3 years ago
3 0

We are  given function: h(x)=-5(x-4)^2+180 for the height in meters, x seconds after the launch.

We need to find the time when it hit the ground.

The height of the object would be 0, when it hit the ground.

<em>So, we need to set the given function equal to 0 and solve for x.</em>

-5(x-4)^2+180 = 0.

Subtracting both sides by 180.

-5(x-4)^2+180-180 = 0-180.

-5(x-4)^2 = -180.

Dividing both sides by -5, we get

(x-4)^2 = 36.

Taking square root on both sides, we get

\sqrt{(x-4)^2} = \sqrt{36}

x-4 = + - 6

x-4 =6 and x-4 =-6

Adding 4 on both sides in both equations, we get

x-4+4 =6+4 and x-4+6 =-6+4

x=10 and x=-2.

<em>x represents time in seconds. So we can't take a negative value for time.</em>

<h3>Therefore,  10 seconds after being launched will the object hit the ground.</h3>
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<h2>SOLVING</h2>

\Large\maltese\underline{\textsf{A. What is Asked}}

What is the slope of the line passing through the point (1,2) and (5,4)

\Large\maltese\underline{\textsf{This problem has been solved!}}

Formula used, here  \bf{\dfrac{y2-y1}{x2-x1}

_______________________________________________________

\bf{\dfrac{4-2}{5-1} | simplify

\bf{\dfrac{2}{4} | reduce

\bf{\dfrac{1}{2}

\rule{300}{1.7}

\bf{Result:}

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\boxed{\bf{aesthetic\not101}}

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