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-6c^2 + 3c^2

 Nov 18, 2014

Best Answer 

 #8
avatar+118723 
+5

$$-6c^2 + 3c^2=-3c^2$$

.
 Nov 19, 2014
 #1
avatar+2195 
+3

$${\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{{\mathtt{c}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{c}}}^{{\mathtt{2}}} = {\mathtt{3}}{\mathtt{\,\times\,}}{{\left(\underset{{\tiny{\text{Error: Unknown Identifier}}}}{{\mathtt{c}}}\right)}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{{\left(\underset{{\tiny{\text{Error: Unknown Identifier}}}}{{\mathtt{c}}}\right)}}^{{\mathtt{2}}}$$

.
 Nov 18, 2014
 #2
avatar+7188 
+3

$${\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{{\mathtt{c}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{c}}}^{{\mathtt{2}}} = {\mathtt{3}}{\mathtt{\,\times\,}}{{\left(\underset{{\tiny{\text{Error: Unknown Identifier}}}}{{\mathtt{c}}}\right)}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{{\left(\underset{{\tiny{\text{Error: Unknown Identifier}}}}{{\mathtt{c}}}\right)}}^{{\mathtt{2}}}$$

.
 Nov 18, 2014
 #3
avatar+2195 
0

 I did it before you this time. LOL

 Nov 18, 2014
 #4
avatar+7188 
0

I was trying to phrase it differently

 Nov 18, 2014
 #5
avatar+2195 
0

Yeah. I'm still counting it as I got it though. Hahaha!

 Nov 18, 2014
 #6
avatar+7188 
0

Whatever................

 Nov 18, 2014
 #7
avatar+2195 
0

LOL. Come on you got all but 1 of them first yesterday.

 Nov 18, 2014
 #8
avatar+118723 
+5
Best Answer

$$-6c^2 + 3c^2=-3c^2$$

Melody Nov 19, 2014

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