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Igoryamba
3 years ago
9

Find IJ. A. 17 B. 10.5 C. 10 D. 13

Mathematics
2 answers:
Zolol [24]3 years ago
8 0

Answer:

B:10.5

Step-by-step explanation:

half of 21 is 10.5

oksano4ka [1.4K]3 years ago
3 0
B would be the correct anwser , if you scale factor it by 2 .
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Lina20 [59]

Answer:

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3 years ago
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find the number of subsets of s = 1 2 3 . . . 10 that contain exactly 4 elements including 3 or 4 but not both.
Anna007 [38]

Answer:

112

Step-by-step explanation:

Let A be a subset of S that satisfies such condition.

If 3∈A, then the other three elements of A must be chosen from the set B={1,2,5,6,7,8,9,10} (because 3 cannot be chosen again and 4 can't be alongside 3). B has eight elements, then there are \binom{8}{3}=56 ways to select the remaining elements of A (the binomial coefficient counts this). The remaining elements determine A uniquely, then there are 56 subsets A.

If 4∉A, we have to choose the remaining elements of A from the set B={1,2,5,6,7,8,9,10}. B has eight elements, then there are \binom{8}{3}=56 ways to select the remaining elements of A. Thus, there are 56 choices for A.

By the sum rule, the total number of subsets is 56+56=112

4 0
4 years ago
When a polynomial p(x) is divided by x+3 the quotient is x^2-x+5 and the remainder is 2. find p(x)?
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As we solve we generate a succession of equivalent equations.

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3 years ago
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