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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<n2 < n
(B) n < n2 < n (C) n < n < n2 (E) n2 <n< n
 one year ago
 one year 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
 one year ago
 one year ago

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KNicksfan4lifeBest ResponseYou've already chosen the best response.0
look at question 32
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i pick \(n=\frac{1}{4}\) so \(n^2=\left(\frac{1}{4}\right)^2=\frac{1}{16}\) and \(\sqrt{n}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
now we order them, and whatever answer we get will be the right one this is a good trick to know for standardized tests
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i get \[\frac{1}{16}<\frac{1}{4}<\frac{1}{2}\] so \[n^2<n<\sqrt{n}\] must be correct
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
btw i picked \(\frac{1}{4}\) because it was easy to take the square root
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i didn't look at the answers
 one year ago
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