Test: Difference between revisions
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< | <math>E=mc^2</math> | ||
<math>2 + 2 = 4</math> | |||
</ | |||
<math>x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}</math> | |||
<math>a^2 + b^2 = c^2</math> | |||
<math>e^{i\pi} + 1 = 0</math> | |||
<math>\int_0^\infty e^{-x^2} dx = \frac{\sqrt{\pi}}{2}</math> | |||
<math>\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}</math> | |||
<math>\begin{pmatrix} | |||
1 & 2 \\ | |||
3 & 4 | |||
\end{pmatrix}</math> | |||
<math>\frac{d}{dx} \left( x^2 \right) = 2x</math> | |||
<math>(x + y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k</math> | |||
<math>\lim_{x \to 0} \frac{\sin x}{x} = 1</math> |
Revision as of 15:07, 19 November 2024
[math]\displaystyle{ E=mc^2 }[/math]
[math]\displaystyle{ 2 + 2 = 4 }[/math]
[math]\displaystyle{ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} }[/math]
[math]\displaystyle{ a^2 + b^2 = c^2 }[/math]
[math]\displaystyle{ e^{i\pi} + 1 = 0 }[/math]
[math]\displaystyle{ \int_0^\infty e^{-x^2} dx = \frac{\sqrt{\pi}}{2} }[/math]
[math]\displaystyle{ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} }[/math]
[math]\displaystyle{ \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} }[/math]
[math]\displaystyle{ \frac{d}{dx} \left( x^2 \right) = 2x }[/math]
[math]\displaystyle{ (x + y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k }[/math]
[math]\displaystyle{ \lim_{x \to 0} \frac{\sin x}{x} = 1 }[/math]