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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The answers are...
B. (3 ^3) ^9 * (7 ^3) ^6
D. (3 ^3 + 3 ^9) * (7 ^6 + 7 ^3)
The correct answer to this question is this one: "D. 11.474"
your calculator probably doesn't do "secant". But sec is 1/cos
so do cos(85) (in degree mode, not radian mode)
then "invert it": 1 / ANS or x^-1 if you have that key
or type
1/cos(85 degrees)= <span>11.4737</span>
Should be C
4+16+6 = 26
14/26 chance of getting a running song
so it’s likely
Answer: In step 1, Andrew used the commutative property
Explanation:
In step 1 he used commutative, which is a + b = b + a
(- 5.7 + 2.2) = 2.2 + (- 5.7)
Step 2, he used associative property, not the distributive.
Step 3, he just added5.7 + 1.1 = 6.8
Step 4, he distributed the negative sign:
- (6.8) = - 6.8