The tracks appear to be visible out in the open; Ron and Harry could easily follow them in COS.
Can non-magical persons see the Hogwarts Express during its journey to and from Hogwarts?
Why are addition and multiplication commutative, but not exponentiation? We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not.  My question is why.  As background there are plenty of mathematical schemes that can be used to define these operators. One of these is   where $H_0(a,b) = b+1$   (successor op) $H_1(a,b) = a+b$   (addition op) $H_2(a,b) = ab $   (multiplication op) $H_3(a,b) = a^b$   (exponentiation op) $H_4(a,b) = a\uparrow \uparrow b$   (tetration op: $a^{(a^{(...a)})}$ nested $b$ times ) etc.    Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$  Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?  Thanks. "1 in 10 are" or "1 in 10 is"? Take the examples:   "One in ten children are dyslexic."  "One in ten children is dyslexic." "One in ten children has dyslexia." "One in ten children have dyslexia."   The "one" is singular so 2 and 3 should be correct. But the "one in ten" is a fraction" so 1 and 4 should be correct. And yet I think I usually say 1 and 3. Which is/are correct?! Computing second partial derivative with polar coordinates Consider the polar coordinates with $x=r\cos\theta$ and $y=r\sin\theta$.  I can show using the chain rule that $$\frac{\partial}{\partial x} = \frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta}$$   What is the method to compute $\dfrac{\partial^2}{\partial x^2}$? I don't know how to do it. If two vectors $a$ and $b$ make angle $\alpha$ and $\beta$ with the $x$-axis, prove, using vectors, that:   $$\cos(\beta - \alpha) = \cos \alpha \cos\beta + \sin\alpha \sin\beta$$   I tried taking components of each vector along the plane, but that didn't get me anywhere. I tried taking component of vector $b$ along vector $a$, but I also got stuck there. Can anyone point me in the right direction? I was speaking to an English learner and said, “All you have to do is read a lot.” And they thought that sentence wasn’t grammatically correct because I dropped the word to between is and read.  They thought it should be “All you have to do is to read a lot.” That sounds weird to me.   How do I explain to them the reason you drop the to? Or am I incorrect and it really should have a to? Restore Missing "Settings" (gnome-control-center) Icon in Ubuntu 18.04 I have been running Ubuntu 18.04 for a couple of weeks. I upgraded from 16.04 and all went smoothly.   I am not sure what I changed, but this morning I noticed that the "Settings" Icon that launches gnome-control-center is missing from "Show Applications" and the applications menu as well. When I searched from "Show Applications" for gnome-control-center or just settings, nothing showed up.   So, I figured I must have accidentally removed it and figured a reinstall of the package would resolve the issue. I have searched and found several postings where this worked. I tried:  sudo apt install gnome-control-center   And then rebooted - which reinstalled the gnome-control-center and I could then launch it from the command line. However, there was still no Settings Icon in any of the menus.   I also tried many other suggestions including:  sudo apt-get install --reinstall gnome-control-center sudo apt-get install ubuntu-desktop sudo apt install ubuntu-desktop --reinstall sudo apt-get install unity-control-center   However, none of these has been successful in restoring the icon into "Show Applications" or any applications menu.   Anyone have any suggestions on something I haven't tried? Would be much appreciated.  Or, is there a way to just manually recreate the icon?   Output of locate gnome-control-center.desktop:  /usr/share/applications/gnome-control-center.desktop  /home/dccrens/.local/share/applications/gnome-control-center.desktop /home/dccrens/.gnome/apps/gnome-control-center.desktop   I am running Ubuntu 18.04 desktop on kernel 4.15.0-45-generic. In the  library, they are overloading the () operator.  Is this a convention in C++? For callbacks, etc.?  I have seen this in code of a co-worker (who happens to be a big Boost fan). Of all the Boost goodness out there, this has only led to confusion for me.  Any insight as to the reason for this overload? Get selected element's outer HTML How to automatically restart a linux background process if it fails? In my .bib file, I have two entries with exactly the same author names:  @article{Seshadrinathan2010A-Subjective-St,     Author = {K. Seshadrinathan and R. Soundararajan and A. C. Bovik and L. K. Cormack}, ....  @article{Seshadrinathan2009Study-of-Subjec,     Author = {K. Seshadrinathan and R. Soundararajan and A. C. Bovik and L. K. Cormack}, ....   When cited after each other, and typeset with the IEEEtran bibliography style, the .bbl file reads:  \bibitem{Seshadrinathan2010A-Subjective-St} K.~Seshadrinathan, R.~Soundararajan, A.~C. Bovik, and L.~K. Cormack, ``A   subjective study to evaluate video quality assessment algorithms,''   \emph{SPIE Proceedings Human Vision and Electronic Imaging}, 2010.  \bibitem{Seshadrinathan2009Study-of-Subjec} ------, ``Study of subjective and objective quality assessment of video,''   \emph{IEEE Transactions on Image Processing}, 2009.   As you can see, the author names have been replaced with ------. It looks like this in the final PDF:    Is this normal behavior? Windows 10: How to change a theme slideshow time interval? It's all int the title; I find myself unable to set a precise time interval defining how often the background pictures of the theme I am currently using change.  The only similar option available in Windows 10 settings applies only to single background images and not to multiple background images included in a theme.  Thanks in advance,  N. I have some files/directories on an external hard drive connected using USB to my Windows XP laptop. I am trying to delete those files but every time I try, I get an error "Cannot delete ....: Access Denied. Make sure disk is not full or write protected and that the file is not currently in use".  I checked that the directory is not is use. However, the permissions are set to read-only. Even if I remove the "read-only" checkbox, in Properties->Attributes, it automatically comes back again. How do I fix this? Is it a virus that is automatically changing it back to read-only?  regards,  JP I'm using Mathematicato produce some of my graphics (when i'm not able to just TikZ them easily). All these images are availabvle as .eps and .pnghaving the same file name - let's use img.eps / img.png in the following.  I would like to set a preference for graphicx to use only one type, i.e. using  \includegraphics[width=.6\textwitdh]{img}   I would like to specify, which img to use with one switch, (package) option or variable.  I tried  \DeclareGraphicsExtension{.eps}   to restrict the allowed image extension to .eps, but still the .png file is included. It seems, .pngis preferred over .eps.  Though this is of course restricted to XeTeX it might be interesting for others, if you use .png together with some lossy format like .jpg to have a file extension preference.   Is there a package option for that or an easy command (global document option) for that?  To provide an MWE is - I think - difficult, because it would be necessary to provide at least 3 files. Selecting attributes or rows of polygons that intersect two attributes of overlaying polygon using ArcGIS Pro Must I enter Europe with Schengen visa from the country where I applied? If I applied for a Schengen visa from "country A" and I got the visa but I had some commitments in "country B" and I have to finish it before going to Europe should I get back to "country A" to travel to Europe or can I use the visa from any where? why is the set of all binary sequences not countable? Proving $\mathbb{Q}$ isn't homeomorphic to $\mathbb{Z}$ Prove that $\mathbb{Q}$ isn't homeomorphic to $\mathbb{Z}$  I know that a homeomorphism is a bijection function from two sets where both the function and its inverse are continuos.   So I figure that if I show that there is no continuos invective mapping from $\mathbb{Q}\rightarrow\mathbb{Z}$ then I have done the proof. But I do not know how to do this (if his is the right way). How can I start to show this? I'm trying to install bang and olufsen drivers, or HP hd audio driver, as only 2 of the 4 speakers work and the 2 amps don't work in ubuntu  lspci -v | grep -i audio 00:03.0 Audio device: Intel Corporation Broadwell-U Audio Controller (rev 09)     Subsystem: Hewlett-Packard Company Broadwell-U Audio Controller 00:1b.0 Audio device: Intel Corporation Wildcat Point-LP High Definition Audio Controller (rev 03)     Subsystem: Hewlett-Packard Company Wildcat Point-LP High Definition Audio Controller   The volume is terrible. Is there any way to fix this?  The speakers work perfectly on the pre-installed Windows 8. Disable mouse speed check between monitors in windows 8.1 update 1 In windows 8.1 and below, the mouse would seamlessly move through all my monitors.  However, in windows 8.1 update 1 the mouse needs to be going at a certain speed before it goes to a different monitor. How do I disable this?  A google search has turned up nothing, so I'm not sure what exactly is going on Intersection of two parabolae Problem: Consider two parabolae such that their axes of symmetry form a right angle. Prove that all four points of intersection lie on a common circle (it is an assumption that there exist such four points). Could you help me obtain a non-algebraic solution? (The algebraic one I did is given below.)  Algebraic solution:  There is a very simple algebraic approach, just introduce a Cartesian coordinate system such that the parabolae are given by equations:  $$y = \alpha_1 x^2 + \beta_1 $$ $$x = \alpha_2 y^2 + \beta_2 $$  and by scaling, adding and simplifying we can get the following ordinary circle equation (the right-hand side is positive because parabolae intersect): $$ \left(x-\frac{1}{2\alpha_2}\right)^2 + \left(y - \frac{1}{2\alpha_1}\right)^2 = \frac{-\beta_1}{\alpha_1} + \frac{-\beta_2}{\alpha_2} + \left(\frac{1}{2\alpha_1}\right)^2 + \left(\frac{1}{2\alpha_2}\right)^2$$  Non-algebraic solution: ??