SOLUTION: If 0 < n < 1, which of the following gives the correct ordering of √n, n, and, n^2 ? A. √n < n < n^2 b √n < n^2 < n c n <√n < n^2 d n < n^2

SOLUTION: If 0 < n < 1, which of the following gives the correct ordering of √n, n, and, n^2 ? A. √n < n < n^2 b √n < n^2 < n c n <√n < n^2 d n < n^2 < &#87

Question 338587: If 0 < n < 1, which of the following gives the correct ordering of
√n, n, and, n^2 ?
A. √n < n < n^2
b √n < n^2 < n
c n <√n < n^2
d n < n^2 < √n
e n^2 < n < √n
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
If 0 < n < 1, which of the following gives the correct ordering of
√n, n, and, n^2 ?
A. √n < n < n^2
b √n < n^2 < n
c n <√n < n^2
d n < n^2 < √n
e n^2 < n < √n

√n = square root (sqrt) of n, n, and n^2, 0 < n < 1
n = 0 --> sqrt(n) = 0, n^2 = 0, the values are all 0 and all equal
n = 1 --> sqrt(n) = 1, n^2 = 1, the values are all 1 and all equal
n = 0 and n = 1 are the end points of the interval (0,1) and are not included

checking 2 values of n
n= 0.9 which is close to 1
and n=10^(-6) which is close to 0

n = 0.9 = 9/10
sqrt(n) = 3/(sqrt(10)) = (3sqrt(10))/10 = 0.9487 to 4 places
n^2 = 81/100 = 0.81
order: n^2, n, sqrt(n)
n^2 < n < sqrt(n)
n = 0.000001 = 1/1000000 = 1/(10^6) = 10^(-6)
sqrt(n) = 10^(-3) = 0.001
n^2 = 10^(-12) = 0.000000000001
order: n^2, n, sqrt(n)
n^2 < n < sqrt(n)

e n^2 < n < √n is the answer when 0 < n < 1

RELATED QUESTIONS
List all elements of the set A ∩ B, where A = {n ∈ N | n = 2^n − 1} and (answered by richard1234)

Dear math teacher, I am having difficulties with the following problem: 4 times nC2 (answered by Theo)

Which of the following pairs of expressions could represent consecutive odd numbers? n... (answered by 55305,jim_thompson5910)

1. Let X(k) denote the N-point DFT of the N-point sequence x(n) a) If x(n)... (answered by Fombitz)

If n is any negative number, which of the following must be positive? (A) n/2 (B) 2n (answered by stanbon)

if n is any negative number ,which of the following must be positive ? a n/2 b (answered by Alan3354,jim_thompson5910)

Which of the following expressions must be positive for all values of n? n^2 + n... (answered by josgarithmetic)

n!(n+2)=n!+(n+1)! (answered by LinnW)

find the sum of following series:... (answered by Edwin McCravy)

SOLUTION: If 0 < n < 1, which of the following gives the correct ordering of &#8730;n, n, and, n^2 ? A. &#8730;n < n < n^2 b &#8730;n < n^2 < n c n <&#8730;n < n^2 d n < n^2 < &#87

SOLUTION: If 0 < n < 1, which of the following gives the correct ordering of √n, n, and, n^2 ? A. √n < n < n^2 b √n < n^2 < n c n <√n < n^2 d n < n^2 < &#87