Difference between revisions of "2006 AMC 10B Problems/Problem 7"
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== Problem == | == Problem == | ||
| − | Which of the | + | Which of the following is equivalent to <math> \sqrt{\frac{x}{1-\frac{x-1}{x}}} </math> when <math> x < 0 </math>? |
<math> \mathrm{(A) \ } -x\qquad \mathrm{(B) \ } x\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } \sqrt{\frac{x}{2}}\qquad \mathrm{(E) \ } x\sqrt{-1} </math> | <math> \mathrm{(A) \ } -x\qquad \mathrm{(B) \ } x\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } \sqrt{\frac{x}{2}}\qquad \mathrm{(E) \ } x\sqrt{-1} </math> | ||
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<math> \sqrt{\frac{x}{1-\frac{x-1}{x}}} = \sqrt{\frac{x}{\frac{x}{x}-\frac{x-1}{x}}} = \sqrt{\frac{x}{\frac{x-(x-1)}{x}}} = \sqrt{\frac{x}{\frac{1}{x}}} = \sqrt{x^2} = |x|</math> | <math> \sqrt{\frac{x}{1-\frac{x-1}{x}}} = \sqrt{\frac{x}{\frac{x}{x}-\frac{x-1}{x}}} = \sqrt{\frac{x}{\frac{x-(x-1)}{x}}} = \sqrt{\frac{x}{\frac{1}{x}}} = \sqrt{x^2} = |x|</math> | ||
| − | Since <math>x<0</math> | + | Since <math>x<0,|x|= \boxed{\textbf{(A)}-x} </math> |
| − | + | == See Also == | |
| + | {{AMC10 box|year=2006|ab=B|num-b=6|num-a=8}} | ||
| − | + | [[Category:Introductory Algebra Problems]] | |
| − | + | {{MAA Notice}} | |
Latest revision as of 13:46, 26 January 2022
Problem
Which of the following is equivalent to 
when 
?
Solution
Since 
See Also
| 2006 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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