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Oksanka [162]
3 years ago
8

GIVING OUT BRAINLIEST ANSWER!!! HELP ME OUT PLEASEEE

Mathematics
1 answer:
Goshia [24]3 years ago
7 0

Answer:

10

Step-by-step explanation:

Hope It Helps

BRAINLIEST PLEASE

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brainliest plz it would really help out

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Which statements accurately describe the function f(x) = 3√18^x? Check all that apply.
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A and C are correct.
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A month of the year is chosen at random. What is the probability that it has 31 days
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Suppose that there are three astronauts outside a spaceship and they decide to play catch. all the astronauts weigh the same on
Lena [83]
<span>The 3rd astronaut would catch the 2nd astronaut and throw the 2nd astronaut towards the 1st and the game would end there. The key thing to remember is conservation of momentum. Since all of the astronauts have the same mass and strength, I will be introducing a unit called "A" which represents the maximum momentum that one astronaut can produce while throwing another. So here's the game of catch, throw by throw. Before the game begins, I will assume all three astronauts are stationary and have 0 momentum. So Astronaut 1 = 0 A (Stationary, next to astronaut 2) Astronaut 2 = 0 A (Stationary, next to astronaut 1) Astronaut 3 = 0 A (Stationary) 1st astronaut grabs the 2nd astronaut and throws him towards the 3rd. Since every action has an equal and opposite reaction, what will happen is the 1st astronaut will be sent moving backwards with a momentum of -1/2A and the 2nd astronaut will be heading towards the 3rd with a momentum of +1/A. So we're left with Astronaut 1 = -1/2 A (Moving to the left) Astronaut 2 = +1/2 A (Moving to the right) Astronaut 3 = 0 A (Stationary) Now the 3rd astronaut catches the 2nd who was thrown at him. Both of them continue moving in the same direction as the 2nd astronaut was just prior to being caught, but at a reduced velocity, giving Astronaut 1 = -1/2 A (Moving to the left) Astronaut 2 = +1/4 A (Moving to the right, slowly) Astronaut 3 = +1/4 A (Moving to the right, slowly) Finally, Astronaut 3 throws astronaut 2 back towards Astronaut 1, giving Astronaut 1 = -1/2 A (Moving to the left) Astronaut 2 = +1/4 A -1/2A = -1/4A (Moving to the left, slowly) Astronaut 3 = +1/4 A +1/2A = +3/4A (Moving to the right, rapidly) So what you're left with is Astronaut 1 moving to the left faster than Astronaut 2, so those two astronauts will never catch each other. Meanwhile, Astronaut 3 is moving to the right and getting further and further away from the other 2 astronauts. So none of the astronauts will ever be able to catch or throw anyone ever again.</span>
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