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3 years ago

16 teams qualify for a tournament. How many possible ways can the final game (2 teams) be played?

Mathematics

1 answer:

Firdavs [7]3 years ago

7 0

Answer:

120

Step-by-step explanation:

so we have 16 teams and we are chosing two

so this formula is called the chose formula

and the formula is 16*(16-1)/2!=16*15/2=120

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Find the term indecent of x in the expansion of (x^2-1/x)^6

Mars2501 [29]

By the binomial theorem,

$\displaystyle \left(x^2-\frac1x\right)^6 = \sum_{k=0}^6 \binom 6k (x^2)^{6-k} \left(-\frac1x\right)^k = \sum_{k=0}^6 \binom 6k (-1)^k x^{12-3k}$

I assume you meant to say "independent", not "indecent", meaning we're looking for the constant term in the expansion. This happens for k such that

12 - 3k = 0 ===> 3k = 12 ===> k = 4

which corresponds to the constant coefficient

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3 years ago

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Basile [38]

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3 years ago

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bezimeni [28]

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Step-by-step explanation:

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3 years ago

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Damm [24]

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Step-by-step explanation:

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3 years ago

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avanturin [10]

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4 years ago