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Olegator [25]
3 years ago
12

I need help plz aijahsg​

Mathematics
1 answer:
lara31 [8.8K]3 years ago
6 0

Answer:

0.95

Step-by-step explanation:

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A rocket is launched from the top of a 99-foot cliff with an initial velocity of 122 ft/s.
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Remember that c is the initial height. Since we the rocket is in a 99-foot cliff, c=99. Also, we know that the velocity of the rocket is 122 ft/s; therefore v=122
Lets replace the values into the the vertical motion formula to get:
0=-16 t^{2} +122t+99
Notice that the rocket hits the ground at the bottom of the cliff, which means that the final height is 99-foot bellow its original position; therefore, our final height will be h=-99
Lets replace this into our equation to get:
-99=-16 t^{2} +122t+99
-16 t^{2} +122+198=0

Now we can apply the quadratic formula t= \frac{-b+or- \sqrt{ b^{2} -4ac} }{2a} where a=-16, b=122, and c=198
t= \frac{-122+or- \sqrt{ 122^{2}-(4)(-16)(198) } }{(2)(-16)}
t= \frac{-122+ \sqrt{27556} }{-32} or t= \frac{-122- \sqrt{27556} }{-32}
t= \frac{-122+166}{-32} or t= \frac{-122-166}{-32}
t= \frac{-11}{8} or t=9

Since the time can't be negative, we can conclude that the rocket hits the ground after 9 seconds.
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4 years ago
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You need to finish the question
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I’m kinda stuck on here on how to do this .... how am I suppose to go down again if it’s not gonna be on the line.
mars1129 [50]

To graph the new shape, you must move each vertex down 3 units. This causes Q to become (1, 0), L to become (5, -2), and Q to become (1, 0). These values can be found by simply counting down by three for each point.

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