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Jan
31

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$Latex$

$TheSolutionof$

The solution to \[\sqrt{x} = 5\] is \[x=25.\]

Evaluate the sum $\displaystyle\sum_{i=0}^n i^3$.

2x^2 + 3(x-1)(x-2)&=&2x^2 + 3(x^2-3x+2)\\
&=& 2x^2 + 3x^2 - 9x + 6\\
&=& 5x^2 - 9x + 6

\documentclass{article}
\begin{document}
\begin{eqnarray*}
2x^2 + 3(x-1)(x-2)&=&2x^2 + 3(x^2-3x+2)\\
&=& 2x^2 + 3x^2 - 9x + 6\\
&=& 5x^2 - 9x + 6
\end{eqnarray*}
\end{document}

$\mathbf{\left(x-1\right)\left(x+3\right) }$

$$
\int_0^{2\pi}\cos(mx)\,dx = 0 \hspace{1cm}
\mbox{if and only if} \hspace{1cm} m\ne 0
$$
http://frodo.elon.edu/tutorial/tutorial/node25.html

\text {The solution to}\: \sqrt{x} = 5 \: \: \text {is} \: \: \, x = 25

$\text {The solution to}\: \sqrt{x} = 5 \: \: \text {is} \: \: \, x = 25$
http://www.mathhelpforum.com/math-help/latex-help/126399-how-make-sentence-spacing.html

$\[ \cos(\theta + \phi) = \cos \theta \cos \phi
- \sin \theta \sin \phi \]$
http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/StdFuncts.html



1 comments:
gravatar
nikk said...
January 31, 2010 at 4:38 AM  

$\[ \cos(\theta + \phi) = \cos \theta \cos \phi
- \sin \theta \sin \phi \]$

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