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miss Akunina [59]
2 years ago
11

(5√3-√27)^3 how do you prove that this is an integer?

Mathematics
2 answers:
mr Goodwill [35]2 years ago
5 0

Keys:

  • \left(a\cdot \:b\right)^n=a^nb^n

Step-by-step explanation:

\left(5\sqrt{3}-\sqrt{27}\right)^3\\\5\sqrt{3}-\sqrt{27}^3=\left(2\sqrt{3}\right)^3\\= 2\sqrt{3}\\\left(2\sqrt{3}\right)^3=2^3\left(\sqrt{3}\right)^3\\=2^3\left(\sqrt{3}\right)^3\\2^3=8\\=8\left(\sqrt{3}\right)^3\\=8\cdot \:3\sqrt{3}\\8\cdot \:3=24\\=24\sqrt{3}

kirill [66]2 years ago
4 0

Answer:

It is not an integer.

Step-by-step explanation:

(5\sqrt{3} -\sqrt{27} )^3\\\\=(\sqrt{3} *(5-3))^3\\\\=8*3*\sqrt{3} \\\\=24\sqrt{3} \\

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                       =\binom{10}{0}0.44^{0}(1-0.44)^{12-0}+\binom{10}{1}0.44^{1}(1-0.44)^{10-1}+\binom{10}{2}0.44^{2}(1-0.44)^{10-2}

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