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STatiana [176]
3 years ago
15

T2.T?

Mathematics
1 answer:
Setler [38]3 years ago
4 0

9514 1404 393

Answer:

  63

Step-by-step explanation:

Reversing the digits changes the value by a factor of 9 times the difference in the digit values. Here, this means the digits differ by 27/9 = 3. If one digit is double the other, the higher digit is 2×3 = 6.

The number is 63.

_____

If you like, you can work through the algebra of it. Let the original number have tens digit x and ones digit y.

  original number value = 10x +y

  reversed number value = 10y +x

  difference of values = 27 = (10x +y) -(10y +x) = 9(x -y)

The tens digit is twice the ones digit, so we have ...

  x = 2y

Substituting in the difference equation, we have ...

  27 = 9(x -y)

  3 = (2y -y) = y . . . . . divide by 9, substitute for x

  x = 2(3) = 6

The number is 10x+y = 63.

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How many subsets of {1, 2, 3, 4, 6, 8, 10, 15} are there for which the sum of the elements is 15?
stepladder [879]

Answer:

512

Step-by-step explanation:

Suppose we ask how many subsets of {1,2,3,4,5} add up to a number ≥8. The crucial idea is that we partition the set into two parts; these two parts are called complements of each other. Obviously, the sum of the two parts must add up to 15. Exactly one of those parts is therefore ≥8. There must be at least one such part, because of the pigeonhole principle (specifically, two 7's are sufficient only to add up to 14). And if one part has sum ≥8, the other part—its complement—must have sum ≤15−8=7

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For instance, if I divide the set into parts {1,2,4}

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.

Once one makes that observation, the rest of the proof is straightforward. There are 25=32

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4 years ago
Please help me i need help please
Oksanka [162]
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angle 4= 20
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