* how to export (latex) a image without using figure
@ 2021-04-04 10:31 Uwe Brauer
2021-04-04 11:17 ` Juan Manuel Macías
0 siblings, 1 reply; 3+ messages in thread
From: Uwe Brauer @ 2021-04-04 10:31 UTC (permalink / raw)
To: emacs-orgmode
Hi
Currently
#+CAPTION: La función La función $x^2 e^{-\alpha x} = \frac{1}{\alpha}$, $\alpha=-\ln(1-p)$ con $p=0.01$ con $p=0.3$
#+NAME: fig:plotcalor23
[[./images/dfp_03.png]]
Gets translated to
\begin{figure}[htbp]
\centering
\includegraphics[width=.9\linewidth]{./images/dfp_03.png}
\caption{\label{fig:plotcalor23}La función La función \(x^2 e^{-\alpha x} = \frac{1}{\alpha}\), \(\alpha=-\ln(1-p)\) con \(p=0.01\) con \(p=0.3\)}
\end{figure}
For reasons that needs a longer explanation I would need.
\includegraphics[width=.9\linewidth]{./images/dfp_03.png}
\captionof{figure}{\label{fig:plotcalor23}La función La función \(x^2 e^{-\alpha x} = \frac{1}{\alpha}\), \(\alpha=-\ln(1-p)\) con \(p=0.01\) con \(p=0.3\)}
How can this be achieved?
Thanks
Uwe Brauer
^ permalink raw reply [flat|nested] 3+ messages in thread
* Re: how to export (latex) a image without using figure
2021-04-04 10:31 how to export (latex) a image without using figure Uwe Brauer
@ 2021-04-04 11:17 ` Juan Manuel Macías
2021-04-04 11:33 ` Uwe Brauer
0 siblings, 1 reply; 3+ messages in thread
From: Juan Manuel Macías @ 2021-04-04 11:17 UTC (permalink / raw)
To: Uwe Brauer; +Cc: orgmode
Hi Uwe,
Try:
#+ATTR_LaTeX: :float nil
#+CAPTION: La función La función $x^2 e^{-\alpha x} = \frac{1}{\alpha}$, $\alpha=-\ln(1-p)$ con $p=0.01$ con $p=0.3$
[[./images/dfp_03.png]]
Best regards,
Juan Manuel
Uwe Brauer writes:
> Hi
>
> Currently
>
> #+CAPTION: La función La función $x^2 e^{-\alpha x} = \frac{1}{\alpha}$, $\alpha=-\ln(1-p)$ con $p=0.01$ con $p=0.3$
> #+NAME: fig:plotcalor23
>
> [[./images/dfp_03.png]]
>
> Gets translated to
>
> \begin{figure}[htbp]
> \centering
> \includegraphics[width=.9\linewidth]{./images/dfp_03.png}
> \caption{\label{fig:plotcalor23}La función La función \(x^2 e^{-\alpha x} = \frac{1}{\alpha}\), \(\alpha=-\ln(1-p)\) con \(p=0.01\) con \(p=0.3\)}
> \end{figure}
>
> For reasons that needs a longer explanation I would need.
>
>
>
> \includegraphics[width=.9\linewidth]{./images/dfp_03.png}
> \captionof{figure}{\label{fig:plotcalor23}La función La función \(x^2 e^{-\alpha x} = \frac{1}{\alpha}\), \(\alpha=-\ln(1-p)\) con \(p=0.01\) con \(p=0.3\)}
>
>
> How can this be achieved?
>
> Thanks
>
> Uwe Brauer
>
>
>
>
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2021-04-04 10:31 how to export (latex) a image without using figure Uwe Brauer
2021-04-04 11:17 ` Juan Manuel Macías
2021-04-04 11:33 ` Uwe Brauer
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