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Veronika [31]
3 years ago
11

Simplify. 32 + 5 × 2 − 8 A) −84 B) 8 C) 11 D) 20

Mathematics
2 answers:
satela [25.4K]3 years ago
6 0

32+5x2-8

32+10-8 (multiply 5 times 2)

42-8 (add 32 and 10)

34 (subtract 8 from 42)

The answer should be 34

navik [9.2K]3 years ago
4 0

32 + 5 × 2 − 8

32+10-8

42-8= 34

The answer should be 34 but I don't see it up there

:/

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Find the midpoint of the segment with the following endpoints.<br> (8,4) and (2,7)
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Answer:

Step-by-step explanation:

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So, the Change In X is 2-8 = -6

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Once you divide both of the changes by 2, you get

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Y:  \frac{3}{2}

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Find the number of distinguishable permutations of:
777dan777 [17]

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Consider one such permutation:

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I write one of the Bs in the bold for emphasis. This permutation is not distinguishable (but for the temporary boldface) from

BBHHHOOOOOO

which is to say they both count as the same permutation. To avoid counting this twice, you would divide the total number of permutations by the number of ways you can permute the identical character. In the case of Bs, this can be done in 2! = 2 ways.

Similarly, the 3 Hs can be rearranged in 3! = 6 ways, and the 6 Os can be rearranged in 6! = 720 ways.

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As a formula: given a word of length N with n different characters that respectively occur k_n times, the total number of distinct permutations is

\dfrac{N!}{k_1! k_2! k_3! \cdots k_n!}

where k_1+k_2+k_3+\cdots+k_n=N.

Try this with some simple examples:

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This has length 6, with E occurring three times, and every other letter occurs once. The total number of distinct permutations of GREECE is

6! / (1! • 1! • 1! • 3!) = 120

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This also has length 6, with B occurring once, N twice, and A three times, so the total number of distinct permutations of BANANA is

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This has length 13, with one each of C and Y; two each of S, A, I, L; and three Ts. Then the number of distinct permutations is

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4 0
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