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Code manually copied from #45 as that was the easiest way to get it up to date. Co-Authored-By: zeozeozeo <[email protected]>
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When $a \ne 0$, there are two solutions to $(ax^2 + bx + c = 0)$ and they are | ||
$$ x = {-b \pm \sqrt{b^2-4ac} \over 2a} $$ | ||
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- Block equations: | ||
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$$ \frac{\partial \rho}{\partial t} + \nabla \cdot \vec{j} = 0 \,. \label{eq:continuity} $$ | ||
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- Stokes' theorem: | ||
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$$ \begin{equation} | ||
\int_{\partial\Omega} \omega = \int_{\Omega} \mathrm{d}\omega \,. | ||
\label{eq:stokes} | ||
\end{equation} $$ | ||
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- Maxwell's equations: | ||
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$$ | ||
\begin{align} | ||
\nabla \cdot \vec{E} & = \rho \nonumber \\ | ||
\nabla \cdot \vec{B} & = 0 \nonumber \\ | ||
\nabla \times \vec{E} & = - \frac{\partial \vec{B}}{\partial t} \label{eq:maxwell} \\ | ||
\nabla \times \vec{B} & = \vec{j} + \frac{\partial \vec{E}}{\partial t} \nonumber \,. | ||
\end{align} | ||
$$ | ||
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- L'Hôpital's rule | ||
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$$ | ||
\begin{align} | ||
\lim_{x\to 0}{\frac{e^x-1}{2x}} | ||
\overset{\left[\frac{0}{0}\right]}{\underset{\mathrm{H}}{=}} | ||
\lim_{x\to 0}{\frac{e^x}{2}}={\frac{1}{2}} | ||
\end{align} | ||
$$ | ||
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- More stuff | ||
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$$ | ||
\begin{align} | ||
z = \overbrace{ | ||
\underbrace{x}_\text{real} + i | ||
\underbrace{y}_\text{imaginary} | ||
}^\text{complex number} | ||
\end{align} | ||
$$ | ||
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- Multiline subscripts: $$ | ||
\begin{align} | ||
\prod_{\substack{ | ||
1\le i \le n \\ | ||
1\le j \le m}} | ||
M_{i,j} | ||
\end{align} | ||
$$ | ||
--- | ||
## Edge cases | ||
Cheese is $10.40 + $0.20 tax |
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