LakerFan24 wrote:
Bunuel wrote:
If n denotes a number to the left of 0 on the number line such that the square of n is less than 1/100, then the reciprocal of n must be
A. less than -10
B. between -1 and -1/10
C. between -1/10 and 0
D. between 0 and 1/10
E. greater than 10
We have\(n<0\) and\(n^2<\frac{1}{100}\)
\(n^2<\frac{1}{100}\) -->\(-\frac{1}{10}<n<\frac{1}{10}\) , but as\(n<0\) -->\(-\frac{1}{10}<n<0\) .
Multiply the inequality by\(-\frac{10}{n}\) , (note as\(n<0\) , then\(-\frac{10}{n}>0\) , and we don't have to switch signs) -->\((-\frac{1}{10})*(-\frac{10}{n})<n*(-\frac{10}{n})<0*(-\frac{10}{n})\)
A. less than -10
B. between -1 and -1/10
C. between -1/10 and 0
D. between 0 and 1/10
E. greater than 10
We have\(n<0\) and\(n^2<\frac{1}{100}\)
\(n^2<\frac{1}{100}\) -->\(-\frac{1}{10}<n<\frac{1}{10}\) , but as\(n<0\) -->\(-\frac{1}{10}<n<0\) .
Multiply the inequality by\(-\frac{10}{n}\) , (note as\(n<0\) , then\(-\frac{10}{n}>0\) , and we don't have to switch signs) -->\((-\frac{1}{10})*(-\frac{10}{n})<n*(-\frac{10}{n})<0*(-\frac{10}{n})\)
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