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KNicksfan4life
 3 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
KNicksfan4life
 3 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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KNicksfan4life
 3 years ago
Best ResponseYou've already chosen the best response.0look at question 32

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1here 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1i pick \(n=\frac{1}{4}\) so \(n^2=\left(\frac{1}{4}\right)^2=\frac{1}{16}\) and \(\sqrt{n}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1now we order them, and whatever answer we get will be the right one this is a good trick to know for standardized tests

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1i get \[\frac{1}{16}<\frac{1}{4}<\frac{1}{2}\] so \[n^2<n<\sqrt{n}\] must be correct

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1btw i picked \(\frac{1}{4}\) because it was easy to take the square root

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1i didn't look at the answers
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