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matrenka [14]
3 years ago
7

Simplify please (n^11)^2

Mathematics
2 answers:
Montano1993 [528]3 years ago
8 0

Answer:

n^22

Step-by-nstep explanation:

because you would multiply the exponents so 11 x 2 is n^22

Citrus2011 [14]3 years ago
3 0

Answer:

n^22 (I love your profile picture by the way)

Step-by-step explanation:

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Please help! <3
igomit [66]

Answer:

8 square units

Step-by-step explanation:

Area of a rectangle= lenght x width

locating the four points on a Cartesian plane

lenght = BC= AD = (0,4)->(4,4) = 4units

width = AB = CD = (0,2)-> (0,4) = 2units

Area of rectangle = 4units x 2units = 8 square units

3 0
3 years ago
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>
5 0
3 years ago
What is the solution for the inequality? -4x - 8 &gt; -20
melamori03 [73]
B. x<3     is the answer to the question you have asked
7 0
3 years ago
Read 2 more answers
Please look at the question in the picture it’s a statistics problem
Klio2033 [76]

B is the correct answer

5 0
3 years ago
in the figure below trianglePQM and triangleQRP are right triangles. the measure of lineQM is 6 and the measure of lineQP is 8.
Kipish [7]

Answer:

Option 4 is correct. The length of PR is 6.4 units.

Step-by-step explanation:

From the given figure it is noticed that the triangle PQR and triangle MQR.

Let the length of PR be x.

Pythagoras formula

hypotenuse^2=base^2+perpendicular^2

Use pythagoras formula for triangle PQM.

PM^2=QM^2+PQ^2

PM^2=(6)^2+(8)^2

PM^2=36+64

PM^2=100

PM=10

The value of PM is 10. The length of PR is x, so the length of MR is (10-x).

Use pythagoras formula for triangle PQR.

PQ^2=QR^2+PR^2

(8)^2=QR^2+x^2

64-x^2=QR^2                   .....(1)

Use pythagoras formula for triangle MQR.

MQ^2=QR^2+MR^2

(6)^2=QR^2+(10-x)^2

36=QR^2+x^2-20x+100

36-x^2+20x-100=QR^2        .... (2)

From equation (1) and (2) we get

36-x^2+20x-100=64-x^2

20x-64=64

20x=128

x=6.4

Therefore length of PR is 6.4 units and option 4 is correct.

3 0
3 years ago
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