32. If 0 < n < 1, which of the following gives the correct ordering of n,n,and n2? (A) n < n < n2 (D)n

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32. If 0 < n < 1, which of the following gives the correct ordering of n,n,and n2? (A) n < n < n2 (D)n

Mathematics
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look at question 32
here is a trick if you get confused since you are not given numbers, clearly it doesn't matter what they are. so pick one, find the answer, and it will be right no matter what
i dont get it

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i pick \(n=\frac{1}{4}\) so \(n^2=\left(\frac{1}{4}\right)^2=\frac{1}{16}\) and \(\sqrt{n}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
now we order them, and whatever answer we get will be the right one this is a good trick to know for standardized tests
i get \[\frac{1}{16}<\frac{1}{4}<\frac{1}{2}\] so \[n^2
btw i picked \(\frac{1}{4}\) because it was easy to take the square root
so it's E
i didn't look at the answers
ok

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