My code are follows:  \documentclass[10pt]{book} \usepackage[T1]{fontenc} \usepackage{newtxmath,newtxtext} \usepackage{linguex}% \usepackage{tikz}% \usetikzlibrary{decorations.text,calc,arrows.meta}% \usepackage{tikz-qtree}%  \usepackage{adjustbox} %\usepackage{showframe} \begin{document}  \ex. \Tree [.CauseP [.DP \edge[roof]; {`the zog'} ] [ [.{\scshape cause} ] [ [.{\scshape past} ] [.vP \edge[roof]; {`dissolved the house'} ]]] ]  Throughout this monograph, I will focus on the reusable linguistic ingredients of the system and how they are efficiently deployed in the recursive combinatorics in building propositions. Reusability (and by extension polysemy) will be a seen as a design feature of the system rather than a bug. As a methodological principle, then, I will be guided by the fact of polysemous items and seek analytic unities to underlie them.  \ex. \textit{Achievements and accomplishments with result states}\\ %\begin{tikzpicture}[baseline=(current bounding box.north)] \begin{adjustbox}{max width=0.91\textwidth} \Tree [.InitP [.{\textsc{initiator}} ] [ [.Init ] [.ProcP [.{\textsc{undergoer}} ] [ [.Proc ] [.ResP [.{\textsc{resultee}} ] [ [.Res ] [.XP \edge[roof]; {\sc ground/final state} ] ] ] ] ] ] ] %\end{tikzpicture} \end{adjustbox}  \end{document}   and the output came as:    Is this possible to make the diagonal lines to be straight as marked in the below picture?    Diagonal line should be straight in all Trees. Please suggest...
I'm new to tikz-qtree, and can't find an automatic way to do something that qtree did very nicely.  With qtree, if you don't have a label for a single intermediate node, you get a straight line from the grandmother to the granddaughter.  With tikz-qtree, you get a "kink".  Compare the following:  With qtree:  \documentclass{article} \usepackage{qtree}  \begin{document}  \Tree [ Wiiiiiiiiiide [ nar- -row ] ]  \Tree [ nrw [ wiiiiiiiiiiiii- -iiiiiiiiiiiide ] ]  \end{document}   With tikz-qtree  \documentclass{article} \usepackage{tikz,tikz-qtree}  \begin{document}  \Tree [ Wiiiiiiiiiide [ nar- -row ] ]  \Tree [ nrw [ wiiiiiiiiiiiii- -iiiiiiiiiiiide ] ]  \end{document}   Is there any way to get the nice straight line in tikz-qtree, or do I have to adjust it manually?  Thanks in advance.
Is is possible to skip the confirmation step while adding a repository? When I add a repository by running  sudo add-apt-repository ppa:ubuntu-x-swat/x-updates   It outputs this dialog:  Updated versions of X.org drivers, libraries, etc. for Ubuntu.  This PPA is for stable upstream releases of X.org components.  If you're looking for something even more bleeding-edge, please see the xorg-edgers PPA.  While Ubuntu does not "officially/formally" support these packages, if you discover problems when installing these debs please feel free to report bugs to launchpad.  However, please make sure to clearly state that you are running packages from this PPA so we know the fixes need to come here.  If you are upgrading from one release to another with this PPA activated, please install the ppa-purge package and use it to downgrade everything in here beforehand. sudo ppa-purge ppa:ubuntu-x-swat/x-updates will do it.  [Directions for packaging drivers can be seen at https://wiki.ubuntu.com/X/DriverBuilding]  More info: https://launchpad.net/~ubuntu-x-swat/+archive/x-updates Press [ENTER] to continue or ctrl-c to cancel adding it   I want to skip the confirmation step  Press [ENTER] to continue or ctrl-c to cancel adding it   How can I automatically confirm the addition of a PPA? How do electrons know which path to take in a circuit? The current is maximum through those segments of a circuit that offer the least resistance. But how do electrons know beforehand that which path will resist their drift the least? $n$ Lines in the Plane How am I to "[u]se induction to show that $n$ straight lines in the plane divide the plane into $\frac{n^2+n+2}{2}$ regions"? It is assumed here that no two lines are parallel and that no three lines have a common point. Further, this is not a non-Euclidean problem, but I wouldn't mind a discussion on the non-Euclidean nature of the problem.  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Does $\langle\Omega|\mathrm{e}^{\mathrm{i}Px}=\langle\Omega|\mathrm{e}^{\mathrm{i}0x}$? $(\langle\Omega| =$ ground state of the interacting theory) Let $\langle\Omega|$ be the ground state of an interacting theory, just as Peskin &amp; Schroeder(PS) describes on page 82 and page 213. On page 213 PS do the following   $$\tag{1}\langle\Omega|\phi(x)|\lambda_{\vec{p}}\rangle = \langle\Omega|\exp(+\mathrm{i}Px)\phi(0)\exp(-\mathrm{i}Px)|\lambda_{\vec{p}}\rangle \\ =\langle\Omega|\phi(0)|\lambda_{0}\rangle\exp(-\mathrm{i}px)\bigg|_{p^0=E_{\vec{p}}} $$  In the above $|\lambda_0\rangle$ is an eigenstate of the full interacting hamiltonian $H$. The ket $|\lambda_{\vec{p}}\rangle$ is the boost of $|\lambda_0\rangle$ with momentum $\vec{p}.$ Furthermore $E_{\vec{p}} = \sqrt{p^2+m_\lambda^2}$ and $\vec{P}|\lambda_0\rangle = 0.$  I assume PS used $$\tag{2}\langle\Omega|\mathrm{e}^{\mathrm{i}Px}=\langle\Omega|\mathrm{e}^{\mathrm{i}0x}$$ in $(1)$.   Is $(2)$ true?   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