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ozzi
3 years ago
12

To create an image, Figure D is going to be reflected across the y-axis, rotated 180° around the point indicated in the figure,

and translated vertically across the x-axis. What must be true about Figure D and its image?
Mathematics
1 answer:
uranmaximum [27]3 years ago
8 0
The image and pre-image are congruent
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Please help me! I need it
Studentka2010 [4]

Answer:

About 0.5

Step-by-step explanation:

By the law of sines:

\dfrac{2}{\sin 105}=\dfrac{c}{\sin 15} \\\\\\c=\dfrac{2}{\sin 105}\cdot \sin 15\approx 0.5cm

Hope this helps!

3 0
3 years ago
What is the side length? Please help, thanks
azamat

Answer:

if Area = 169/225

then

S = √(169/225)

S = √169/√225

S = 13/15

5 0
3 years ago
Simplify (question is in the picture)
Sholpan [36]
Your answer is 26.9 so u could round it up to 27
3 0
3 years ago
Read 2 more answers
It is not possible to divide 5 by 0 because there is no number you can___ by 0 and get 5 .
elixir [45]

Answer:

multiply

Step-by-step explanation:

5 0
2 years ago
Prove that it is impossible to dissect a cube into finitely many cubes, no two of which are the same size.
solniwko [45]

explanation:

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:

  • C has similar size than the difference between A and B
  • C has smaller size than the difference between A and B

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.

That small rectangle has base length similar than the sides of C, so it cant be covered by a single square. The small sqaure you would use to cover that rectangle that is above to C and next to B, lets call it D, would leave another 'hole' above C and between D and A's neighborhood.

As you can see, this problem recursively forces you to use smaller and smaller squares, to a never end. You cant cover a sqaure with a finite number of squares and, as a result, you cant cover a cube with finite cubes.

3 0
3 years ago
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