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irga5000 [103]
3 years ago
6

Which statement describes the polygon?

Mathematics
2 answers:
Karo-lina-s [1.5K]3 years ago
8 0
THIS answer is A. It is equilateral but not equiangular. Hope this helps!

I'm All Fired Up!

Aye Sir!
goblinko [34]3 years ago
3 0

Answer:

A

Step-by-step explanation:

A.

The polygon is equilateral but not equiangular.

B.

The polygon is equiangular but not equilateral.

C.

The polygon is equilateral and equiangular.

D.

The polygon is neither equilateral nor equiangular.

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1232 yards or 3696 feet or 44352 inches or 0.7 of a mile

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Prove that it is impossible to dissect a cube into finitely many cubes, no two of which are the same size.
solniwko [45]

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The sides of a cube are squares, and they are covered by the respective sides of the cubes covering that side of the big cube. If we can show that a sqaure cannot be descomposed in squares of different sides, then we are done.

We cover the bottom side of that square with the bottom side of smaller squares. Above each square there is at least one square. Those squares have different heights, and they can have more or less (but not equal) height than the square they have below.

There is one square, lets call it A, that has minimum height among the squares that cover the bottom line, a bigger sqaure cannot fit above A because it would overlap with A's neighbours, so the selected square, lets call it B, should have less height than A itself.

There should be a 'hole' between B and at least one of A's neighbours, this hole is a rectangle with height equal to B's height. Since we cant use squares of similar sizes, we need at least 2 squares covering the 'hole', or a big sqaure that will form another hole above B, making this problem inifnite. If we use 2 or more squares, those sqaures height's combined should be at least equal than the height of B. Lets call C the small square that is next to B and above A in the 'hole'. C has even less height than B (otherwise, C would form the 'hole' above B as we described before). There are 2 possibilities:

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If the second case would be true, next to C and above A there should be another 'hole', making this problem infinite. Assuming the first case is true, then C would fit perfectly above A and between B and A's neighborhood.  Leaving a small rectangle above it that was part of the original hole.

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