If ∆HCN~∆BMD, solve for x.

Mathematics

1 answer:

tekilochka [14]2 years ago

8 0

Answer:

Step-by-step explanation:

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By definition of the binomial coefficient,

$C(n,k) = \dfrac{n!}{k!(n-k)!}$

so we have

$C(50,3) + C(50,4) = \dfrac{50!}{3!47!} + \dfrac{50!}{4!46!} \\\\ = \dfrac{50!}{3!46!} \left(\dfrac1{47} + \dfrac14\right) \\\\ = \dfrac{50!}{3!46!} \times \dfrac{51}{188} \\\\ = \dfrac{51!}{3!46!} \times \dfrac1{4\times47} \\\\ = \dfrac{51!}{4!47!} \\\\ = \dfrac{51!}{4!(51-4)!} = C(51,4)$

as required.

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Domain restrictions are what x values make the denominator zero.
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Answer:

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Step-by-step explanation:

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