Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 4, 2011"
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| − | {{:AoPSWiki:Problem of the Day/June 4, 2011}}{{toadd}} | + | ==Problem== |
| + | {{:AoPSWiki:Problem of the Day/June 4, 2011}} | ||
| + | ==Answer== | ||
| + | <math>\boxed{\frac{3}{10}}</math> {{toadd}} | ||
| + | |||
| + | == Solution == | ||
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| + | 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
To add a solution or an answer, click the + button on the right sidebar.
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)
