(Undo revision 8333 by JulieDunbar (talk))
(Undo revision 8334 by JulieDunbar (talk))
Line 3: Line 3:
 
$$x=1a \over 2b$$
 
$$x=1a \over 2b$$
  
$${ T }_{ ij }=\sum _{ ij }^{  }{ { e }^{ ij } } \left\lceil \ln { { X }_{ ij } }  \right\rceil$$
+
:<math>{ T }_{ ij }=\sum _{ ij }^{  }{ { e }^{ ij } } \left\lceil \ln { { X }_{ ij } }  \right\rceil <math>

Revision as of 22:44, 30 August 2017

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

$$x=1a \over 2b$$

<math>{ T }_{ ij }=\sum _{ ij }^{ }{ { e }^{ ij } } \left\lceil \ln { { X }_{ ij } } \right\rceil <math>