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:
d =32
Step-by-step explanation:

Answer:
The Answer is D
Step-by-step explanation:
Answer:
B. 8.
D. log2 256.
Step-by-step explanation:
Generally:
loga b + loga b
= loga b^2.
Therefore log2 16 + log2 16
= log2 16^2
= log2 256.
From the above:
256 = 2^x where x is the log.
Now 2^8 = 256 so the original log2 16 + log2 16 = 8.