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anonymous
 4 years ago
32. If 0 < n < 1, which of the following gives the correct ordering of n,n,and n2?
(A) n < n < n2 (D)n<n2 < n
(B) n < n2 < n (C) n < n < n2 (E) n2 <n< n
anonymous
 4 years ago
32. If 0 < n < 1, which of the following gives the correct ordering of n,n,and n2? (A) n < n < n2 (D)n<n2 < n (B) n < n2 < n (C) n < n < n2 (E) n2 <n< n

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0here 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i pick \(n=\frac{1}{4}\) so \(n^2=\left(\frac{1}{4}\right)^2=\frac{1}{16}\) and \(\sqrt{n}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now we order them, and whatever answer we get will be the right one this is a good trick to know for standardized tests

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i get \[\frac{1}{16}<\frac{1}{4}<\frac{1}{2}\] so \[n^2<n<\sqrt{n}\] must be correct

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0btw i picked \(\frac{1}{4}\) because it was easy to take the square root

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i didn't look at the answers
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