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ZanzabumX [31]
3 years ago
6

Can anyone help me wit this problem..

Mathematics
1 answer:
Ket [755]3 years ago
5 0
Use formula for the number of combinations \\C_k^n= \frac{n!}{(n-k)!k!} \\ \\n=12, k=2
\\ C_2^{12}= \frac{12!}{(12-2)!2!}= \frac{12!}{10!2!}= \frac{12\times11}{2\times1}=66
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

First, solve

6^3 x 2^6/2^3

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because both expressions are equal to 1728, this is an answer. this answer can also eliminate 2^3 x 3^3 because it has a lower exponent than the answer that is equal to 1728. therefore, the value of it must be smaller than 1728

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This answer is correct, because both values are equal to each other, meaning that 12^6 is larger than 1728  

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