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.
Learn more about Estimation from
brainly.com/question/24592593
#SPJ1
Answer:
4
Step-by-step explanation:
2x2x2x2 = 16
Answer:
972 x^16 y^24
Step-by-step explanation:
Simplify the following:
(-2 x^3 y^7)^2 (3 x^2 y^2)^5
Multiply each exponent in -2 x^3 y^7 by 2:
(-2)^2 x^(2×3) y^(2×7) (3 x^2 y^2)^5
2×7 = 14:
(-2)^2 x^(2×3) y^14 (3 x^2 y^2)^5
2×3 = 6:
(-2)^2 x^6 y^14 (3 x^2 y^2)^5
(-2)^2 = 4:
4 x^6 y^14 (3 x^2 y^2)^5
Multiply each exponent in 3 x^2 y^2 by 5:
4 x^6 y^14×3^5 x^(5×2) y^(5×2)
5×2 = 10:
4×3^5 x^6 y^14 x^(5×2) y^10
5×2 = 10:
4×3^5 x^6 y^14 x^10 y^10
3^5 = 3×3^4 = 3 (3^2)^2:
4×3 (3^2)^2 x^6 y^14 x^10 y^10
3^2 = 9:
4×3×9^2 x^6 y^14 x^10 y^10
9^2 = 81:
4×3×81 x^6 y^14 x^10 y^10
3×81 = 243:
4×243 x^6 y^14 x^10 y^10
4 x^6 y^14×243 x^10 y^10 = 4 x^(6 + 10) y^(14 + 10)×243:
4×243 x^(6 + 10) y^(14 + 10)
14 + 10 = 24:
4×243 x^(6 + 10) y^24
6 + 10 = 16:
4×243 x^16 y^24
4×243 = 972:
Answer: 972 x^16 y^24
Answer:
Step-by-step explanation:
first you need to put the equation in y=mx+b form
y-3=(-1/2)(x+4)
y-3= (-1/2x)+4
y=-1/2x+1
then you just graph
the y intercept is 1 and the slope is -1/2 so it’s going to be a negative graph and it starts from 1 on the y axis
Constant is 4. 7a has a variable so that can’t be a constant value. The only option left in the expression is 4
