dsd
$
\begin{align*}
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{align*}
$
Saturday, February 27, 2010 | 1 Comments
Basic Latex 4
\begin{document}
\begin{align*}
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{align*}
\end{document}
$
\begin{align*}
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{align*}
$
\begin{align*}
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{align*}
$
Friday, February 26, 2010 | 0 Comments
Basic Latex 3
$\int_{0}^{1}\frac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}dx=\frac{22}{7}-\pi$
${ x + a = a + x = x}$
$$
\frac{x^n-1}{x-1} = \sum_{k=0}^{n-1}x^k
$$
$$
\frac{a/b-c/d}{e/f-g/h}
$$
$\frac{1}{2}$
$\frac{2}{x+2}$
\documentclass{article}
\begin{document}
%You won't see this in the final document.
You do see this.
\end{document}
Evaluate the sum $\displaystyle\sum\limits_{i=0}^n i^3$.
$
2x^2 + 3(x-1)(x-2) & = 2x^2 + 3(x^2-3x+2)\\
&= 2x^2 + 3x^2 - 9x + 6\\
&= 5x^2 - 9x + 6
$
$\frac {x^2+5}{3x +1}$
$$
\sum_{k=0}^\infty\frac{(-1)^k}{k+1} = \int_0^1\frac{dx}{1+x}
$$
$$
50 apples \times 100 apples = lots of apples
$$
\begin{eqnarray*}
1+2+\ldots+n&=&\frac{1}{2}((1+2+\ldots+n)+(n+\ldots+2+1))\\
&=&\frac{1}{2}\underbrace{(n+1)+(n+1)+\ldots+(n+1)}_{\mbox{$n$copies}}\\
&=&\frac{n(n+1)}{2}\\
\end{eqnarray*}
\begin{document}
\begin{align*}
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{align*}
\end{document}
$
\begin{align*}
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{align*}
$
----------------
$\begin{align*}
2x^2 + 3(x-1)(x-2) & = 2x^2 + 3(x^2-3x+2)\\
&= 2x^2 + 3x^2 - 9x + 6\\
&= 5x^2 - 9x + 6
\end{align*}$
Friday, February 26, 2010 | 0 Comments
Basic Latex 2
$a$
$f(x) = 3x + 7$
f(x) = 3x + 7
Let $f$ be the function defined by $f(x) = 3x + 7$, and
let $a$ be a positive real number.
http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/MathMode.html
$a/b$
$\sqrt[3]{x+y}$
\frac{num}{den}
$\frac{1}{2}$
$a + b= c$
$\frac{w+a}{d+3}=\frac{r}{t}$
$\cos^2 x +\sin^2 x = 1$
\[\LaTeX\]
\frac{1}{2}
Friday, February 26, 2010 | 0 Comments
Basic Latex 1
The solution to \[\sqrt{x} = 5\] is \[x=25.\]
The solution to
\[\sqrt{x} = 5\]
is
\[x=25.\]
$frac{x+y}{y-z)$
http://www.forkosh.dreamhost.com/source_mathtex.html
http://www.public.asu.edu/~rjansen/latexdoc/ltx-2.html
http://texblog.net/latex-archive/maths/eqnarray-align-environment/
http://andy-roberts.net/misc/latex/latextutorial10.html
http://andy-roberts.net/misc/latex/latextutorial9.html
Friday, February 26, 2010 | 0 Comments
test 1
nie jawapan saya
http://arawna.blogspot.com/2009/07/cara-buat-spoiler-di-blogspot.html
$5y$
$a/b$
becomes
$a/b$
Sunday, January 31, 2010 | 1 Comments
http://arawna.blogspot.com/2009/07/cara-buat-spoiler-di-blogspot.html
$5r+3$
Sunday, January 31, 2010 | 1 Comments