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exis [7]
4 years ago
10

HELP MEEEEE PLSSSSSS

Mathematics
1 answer:
ad-work [718]4 years ago
4 0

Answer:

1. t + 6, 2. 11y, 3. 8, 4. \frac{6}{t}, 5. 6, 6. 3s, 7. 11 - r, 8. 6x + 2s, 9. \frac{5}{2}, 10. 3x - 10, 11. 7, 12. 41

Step-by-step explanation:

1. The sum of 6 and t = t + 6.

2. The product of y and 11 = 11y.

3. The difference of 4 from 12 = 12 - 4 = 8.

4. The quotient of 6 divided by t = \frac{6}{t}.

5. 19 less than 25 = 25 - 19 = 6.

6. Increase s by 2s = s + 2s = 3s.

7. Decrease 11 by r = 11 - r.

8. Add 6 times x and 2 times s = 6x + 2s.

9. The quotient of 17 minus 12 and 2 = \frac{17 - 12}{2} = \frac{5}{2}.

10. The product of 3 and x less 10 = 3x - 10.

11. The product of 7 and 2 divided by 2 = \frac{7 \times 2}{2} = 7.

12. Decrease the product of 7 and 6 by a number = (7 × 6) - 1 = 41.

(Answer)

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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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