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
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
>
>
>
>
[-- Attachment #1: Type: text/plain, Size: 342 bytes --] >>> "JMM" == Juan Manuel Macías <maciaschain@posteo.net> writes: Hi Juan, > 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]] Muchas gracias, I mean thank you very much. Uwe [-- Attachment #2: smime.p7s --] [-- Type: application/pkcs7-signature, Size: 5673 bytes --]