A community for students.
Here's the question you clicked on:
 0 viewing
 2 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
 2 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

This Question is Closed

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

satellite73
 2 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
 2 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
 2 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
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1i didn't look at the answers
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.