Answer:
its 23.76
Step-by-step explanation:
I answered in the comments, but now I answered legit (there was a virus earlier lol)
The nearest tenth of how fast a rover will hit Mars' surface after a bounce of 15 ft in height is 20.7ft/s.
<h3>What is the approximation about?</h3>
From the question:
Mars: F(x) = 2/3
Therefore, If x = 15
Then:
f (15) = 2/3 ![\sqrt[8]{15}](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7B15%7D)
= 16/3
= 20.7ft/s
Hence, The nearest tenth of how fast a rover will hit Mars' surface after a bounce of 15 ft in height is 20.7ft/s.
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Answer:
25%
Step-by-step explanation:
28-21=7
7/28=×/100
7×100=700
700÷28=25
This is a system of equation:
7y - 5x = 31
x = 4 - 2y
Plug in 4 - 2y for x
7y - 5(4 - 2y) = 31
Distribute -5 to both 4 and -2y
7y - 20 + 10y = 31
Simplify
17y - 20 = 31
Isolate the y. Note the equal sign. What you do to one side you do to the other. Do the opposite of PEMDAS.
Add 20 to both sides
17y - 20 (+20) = 31 (+20)
17y = 51
Divide 17 from both sides to isolate the y
17y/17 = 51/17
y = 51/17 or 3
y = 3
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Plug in 3 for y in one of the equations.
x=4−2y
x = 4 - 2(3)
Simplify.
x = 4 - 6
x = -2
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x = -2, y = 3 is your answer, or B)
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hope this helps
