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shepuryov [24]
3 years ago
12

What is the answer to 95-72

Mathematics
1 answer:
Anni [7]3 years ago
3 0
The answer to 96-72 is 23
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olasank [31]

Answer:

B

Step-by-step explanation:

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3 years ago
You buy a package of 12 pencils for $2.64 write this rate as a unit rate?
iogann1982 [59]
Unit rate means how much does it cost for 1 unit

2.64 for 12 pencils
find 1 pencil
divide both sides by 12
0.22 for 1 pencil

0.22 per pencil
3 0
3 years ago
Number 31<br> Any help please?
Alex73 [517]
2,200*14=30,800
2,000*13=26,000
14+13=27
30,800+26,000=56,000

Your answer is C, 13(:
5 0
3 years ago
1.5 more than the quotient of m and 5 is g
seropon [69]
You can do it! :) I believe in you
3 0
3 years ago
The nine squares of a 3-by-3 chessboard are to be colored red and blue. The chessboard is free to rotate but cannot be flipped o
nignag [31]

Answer:

a_n = 2^{\frac{n^2-1}{4} + 1} + \frac{2^{n^2} - \, 2^{\frac{n^2-1}{4} + 1}}{4}

For n = 3, there are 134 possibilities

Step-by-step explanation:

First, lets calculate the generating function.

For each square we have 2 possibilities: red and blue. The Possibilities between n² squares multiply one with each other, giving you a total of 2^n² possibilities to fill the chessboard with the colors blue or red.

However, rotations are to be considered, then we should divide the result by 4, because there are 4 ways to flip the chessboard (including not moving it), that means that each configuration is equivalent to three other ones, so we are counting each configuration 4 times, with the exception of configurations that doesnt change with rotations.

A chessboard that doesnt change with rotations should have, in each position different from the center, the same colors than the other three positions it could be rotated into. As a result, we can define a <em>symmetric by rotations chessboard</em> with only (n²-1)/4 + 1 squares (the quarter part of the total of squares excluding the center plus the center).

We cocnlude that the total of configurations of symmetrical boards is 2^{\frac{n^2-1}{4} + 1}

Since we have to divide by 4 the rest of configurations (because we are counted 4 times each one considering rotations), then the total number of configutations is

a_n = 2^{\frac{n^2-1}{4} + 1} + \frac{2^{n^2} - \, 2^{\frac{n^2-1}{4} + 1}}{4}

If n = 3, then the total amount of possibilities are

a_3 = 2^{\frac{3^2-1}{4} + 1} + \frac{2^{3^2} - \, 2^{\frac{3^2-1}{4} + 1}}{4} =  134

3 0
3 years ago
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