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zhannawk [14.2K]
2 years ago
5

Determine the value of x in each figure.

Mathematics
2 answers:
RSB [31]2 years ago
7 0

8x = 120

x = 120/8 = 15.

so ans ia 15

velikii [3]2 years ago
3 0

Answer:

x=15is your answer

Step-by-step explanation:

8x=120 vertically opposite angle

x=120/8=15

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Isabella spends $4.87 on paintbrushes and paint. How many of each item does she buy? Explain how you found your answer
Rina8888 [55]

Answer:Answer: Isabella buys 2 paintbrushes and 3 jars of paint.

Step-by-step explanation:

Isabella can buy together 5 paintbrushes ( p ) and jars of paint ( p ) : j + p = 5, She can`t buy 4 items ( because 4 * $0.99 = $3.96 < $4.87 ) and she also can`t buy 6 items ( 6 * $0.95 = $5.70 > $4.87 ). So: p = 5 - j. Another equation is: 0.95 * ( 5 - j ) + 0.99 j = 4.87, 4.75 - 0.95 j + 0.99 j = 4.87; 0.04 j = 0.12; j = 0.12 : 0.04; j = 3, p = 5 - 3 = 2. We can prove it: 0.95 * 2 + 0.99 * 3 = 1.90 + 2.97 = 4.87.

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5 0
3 years ago
Solve p/7 = 14 for p
Paul [167]

P would equal 98 i believe because since you are dividing p by 7 you would have to do the reverse to get p by itself which would be p=14*7 or p=98

3 0
3 years ago
Read 2 more answers
How to make 3/20 a decimal
Contact [7]

3/20 as decimal = 0.15

6 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
Please help me with number 11 and number 12
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For 11. Not all of the y values (outputs) are consistent with the pattern
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