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spayn [35]
3 years ago
12

can you please just help me so that i dont fail this one because everything i've been doing i've been failing and it's because i

didn't take a picture and be like hey is this right. can you please help

Mathematics
2 answers:
Llana [10]3 years ago
7 0

Answer:

1. r=5 2. x=2 3. x=-3 4. y=11 5. x=-11

Step-by-step explanation:

1. 7r-7=2r+18

combine

5r=25r

divide

r=5

5. 3(x-4)=5(x+2)

multiply

3x-12=5x+10

combine

-2x=22

x=-11

Julli [10]3 years ago
6 0
1. r=5
2. x=2
3. x=-3
4. y=11
5. x=-11
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First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).

So, we can think of our numbers as

\begin{array}{c|c}x&x\mod 3\\0&0\\1&1\\2&2\\3&0\\4&1\\5&2\\6&0\\7&1\\8&2\\9&0\end{array}

In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.

For example, we could arrange our numbers following the pattern

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We have 3!=6 possible patterns. Suppose for example that we choose the pattern

0,1,2,0,1,2,0,1,2

One possible way of following this pattern would be the arrangement

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In fact, we substituted every '0' with a multiple of 3 (3, 6 or 9), every '1' with a number 1 away from a multiple of 3 (1, 4 or 7) and every '2' with a number 2 away from a multiple of 3 (2, 5 or 8).

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