Answer:
From 3 to 6 seconds, the rocket is falling 11 yards per second.
Step-by-step explanation:
Normal pace of progress is exactly the same thing as the slant. Since this is explanatory, we can't track down the specific pace of progress as we could if this were a straight capacity. Be that as it may, we can utilize a similar thought. At the point when t = 3, h(t) = 33, so the organize point is (3, 33). At the point when t = 6, h(t) = 0, so the facilitate is (6, 0). Attachment those qualities into the incline recipe:
m= 0-33/6-3 and m=-33/3 which is 11
From 3 to 6 seconds, the rocket is falling 11 yards each second.
Brainliest?
Area of a square with side s is
![\sf{s^2}](https://tex.z-dn.net/?f=%5Csf%7Bs%5E2%7D)
In your question, the side or s is:
![\sf{8x^7y^3}](https://tex.z-dn.net/?f=%5Csf%7B8x%5E7y%5E3%7D)
And so the area of a square with that side length would be:
![\sf{(8x^7y^3)^2}](https://tex.z-dn.net/?f=%5Csf%7B%288x%5E7y%5E3%29%5E2%7D)
And using this formula:
![\sf{(a^b)^c =a^{b\times c}}](https://tex.z-dn.net/?f=%5Csf%7B%28a%5Eb%29%5Ec%20%3Da%5E%7Bb%5Ctimes%20c%7D%7D)
We get that the area is:
![\sf{(8x^7y^3)^2 = 8^2 x^{7\times 2} y^{3\times 2}}](https://tex.z-dn.net/?f=%5Csf%7B%288x%5E7y%5E3%29%5E2%20%3D%208%5E2%20x%5E%7B7%5Ctimes%202%7D%20y%5E%7B3%5Ctimes%202%7D%7D)
And simplifying that we get the final answer as:
I have these same questions but the way i solved them i did
A: 1/3 x 3.14 x 2*2 x 8 = 32/3 = 2000 divided by 32 = 63 scoops
The formula is 1/3 x 3.14 x r*2 x h
B: 1/3 x 3.14 x 8*2 x 8 = 128/3 = 2000 divided by 128 = 16 scoops