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lbvjy [14]
3 years ago
9

Compute each sum or difference 9/10+7/9 1/2-3/11 2/5-1/15

Mathematics
2 answers:
Galina-37 [17]3 years ago
8 0
1. 9/10 + 7/9 = 81/90 + 70/90 = 151/90 = 1 61/90
2. 1/2-3/11 = 11/22-6/22 = 5/22
3. 2/5-1/15 = 6/15-1/15 = 5/15 =1/3
Helga [31]3 years ago
6 0

Answer:

Step-by-step explanation:

1) 9/10+7/9

The first step is to take the lowest common factor of 10 and 9 in the denominator. It is 90. Therefore,

9/10+7/9 = (9 × 9 + 7 × 10)/90 =

(81 + 70)/90 = 151/90 = 1 61/90

2) 1/2-3/11

The first step is to take the lowest common factor of 2 and 11 in the denominator. It is 22. Therefore,

1/2-3/11 = (11 - 6)/22 = 5/22

3) 2/5-1/15

The first step is to take the lowest common factor of 5 and 15 in the denominator. It is 15. Therefore,

2/5-1/15 = (6 - 1)/15 = 5/15 = 1/3

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Write an equation that represents the following statement. Five less than the product of r and two is 3.
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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

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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).

This means that, once we fix a patter, we have 3 choices for the first 3 slots, 2 choices for the next 3 slots, and the final slot will be fixed. So, we have

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216\cdot 6 = 1296

possible arrangements.

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