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 
![\[10x^2=\frac{9}{10}\]](http://latex.artofproblemsolving.com/0/4/3/043059f16c62146c978c8239464a5b3e2dc7a99f.png)
![\[x^2=\frac{9}{100}\]](http://latex.artofproblemsolving.com/1/6/4/164c696916e8bc0c2b3e89a69f4db48ad66cf87d.png)
![\[x=\boxed{\frac{3}{10}}\]](http://latex.artofproblemsolving.com/e/d/2/ed284e9ee95cf2267852db6ea69b7f5b9852c69a.png)
