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Bogdan [553]
3 years ago
6

What happens to the distance between each billiard ball during this rigid transformation?

Mathematics
1 answer:
Delicious77 [7]3 years ago
6 0

The question is incomplete. Here is the complete question.

To set up a game of billiards, the first player moves the balls contained within a triangular rack as shown. What happens to the distance between each billiard during this rigid transformation?

A. The distance remains constant throughout the transformation.

B. The distance decreases at the start and increases after all motion stops.

C. The distance stays the same at the start but decreasesexactly when motion ends.

D. The distance increases at the start and then decreases as the rack gets further from the player.

Answer: A. The distance remains constant throughout the transformation.

Step-by-step explanation: In a <u>rigid</u> <u>motion</u>, all moving points in the plane are moving in way such tha:

1)  relative distance between them stays the same and

2) relative position of the points stays the same

There are four types of rigid motions: translation, rotation, reflexion and glide reflection.

<u>Translation</u>: every point or object is moved by the same amount and in the same direction;

<u>Rotation</u>: the object rotates by the same amount around a fixed point;

<u>Reflexion</u>: the object exchange points from one side of a line with points on the other side of the line at the same distance from the line;

<u>Glide</u> <u>Reflection</u>: is a mirror reflection followed by a translation parallel to the mirror.

In the game of billiards, because all the balls are inside the triangular rack, the distance, and also the position, between them stays the same, limited by the rack. Since they are moving by the same amount in the same direction, the rigid transformation is a translation.

Therefore, the distance of the balls in the triangular rack remains constant throughout the transformation.

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According to the statement

we have to explain the green mathematics.

In mathematics, Actually there is a Green Function which was founded by a mathematician George Green.

In this function, a Green's function is the impulse response of an in homogeneous linear differential operator defined on a domain with specified initial conditions or boundary conditions.

The example of green function is the Green's function G is the solution of the equation LG = δ, where δ is Dirac's delta function; the solution of the initial-value problem Ly = f is the convolution (G ⁎ f), where G is the Green's function.

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In this there is a green theorem, which  relates a line integral around a simply closed plane curve C and a double integral over the region enclosed by C.

So, The green mathematics tells about the impulse response of an in homogeneous linear differential operator.

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2 years ago
About 2,000,000 people visited the aquarium last year. If this number were rounded to the nearest million, what was the greatest
notsponge [240]
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bezimeni [28]

from the question, (-2,2) and (1,2) have the same y value so you can use that as your base and easily find the perpendicular height using the y axis since it's parallel to the x axis.

the third point, (0,-6), to the base is your height

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