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mrs_skeptik [129]
3 years ago
15

Could use help on this one! I'll send thanks and mark Brainiest at when finished to see who's correct.

Mathematics
1 answer:
sdas [7]3 years ago
8 0

sqrt (3/10)

sqrt(3)/sqrt(10)

no sqrt in the denominator, so multiply by sqrt(10)/sqrt(1)

sqrt(3) sqrt(10)

-------------------

sqrt(10) sqrt(10)

sqrt(30)/10

Choice D


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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).

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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.

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216\cdot 6 = 1296

possible arrangements.

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