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:no cap
Step-by-step explanation:that looks complicated
5.40 divided by 45 = 0.12
The cafe charges 12 cents per minute.
R = 10, T = 20
OK. Calculate the length of segment RT:
|RT| = |20 - 10| = |10| = 10
Divide |RT| into a ratio of 2:3
2 + 3 = 5
10 : 5 = 2
Therefore we have
|RS| = 2 · 2 = 4 and |ST| = 3 · 2 = 6 (4 + 6 = 10 CORRECT)
T = R + 4 and T = T - 6
T = 10 + 4 = 14; T = 20 - 6 = 14 CORRECT
<h3>Your answer is T = 14.</h3>
Answer:
Step-by-step explanation:
- The lengths of the sides of a 30-60-90 triangle are in the ratio 1:2:√3.
- The lengths of the sides of a 45-45-90 triangle are in the ratio 1 : 1 : √2
- We can get the length of the longest side of the bottom triangle is 9×2÷√3 =
- And based on the theory shown on the 2nd step, we can get x (the length of the longest side of the upper triangle) is x√2
