Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 4, 2011"
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== Solution == | == Solution == | ||
− | Since <math>0<x<1</math>, <math>\lfloor x\rfloor=0</math> and <math>\lceiling x\rceiling=1</math>, | + | Since <math>0<x<1</math>, <math>\lfloor x\rfloor=0</math> and <math>\lceiling x\rceiling=1</math>, the equation becomes <math>10x^2+\frac{0}{1}=\frac{9}{10}</math> <cmath>10x^2=\frac{9}{10}</cmath> <cmath>x^2=\frac{9}{100}</cmath> <cmath>x=\boxed{\frac{3}{10}}</cmath> |
Latest revision as of 10:40, 4 June 2011
Problem
AoPSWiki:Problem of the Day/June 4, 2011
Answer
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Solution
Since , and $\lceiling x\rceiling=1$ (Error compiling LaTeX. Unknown error_msg), the equation becomes