How can I stop objects like images or tables to move to unwanted position?
How to influence the position of float environments like figure and table in LaTeX?
Find $\lfloor {\alpha}^6 \rfloor$ If $\alpha$ is a real root of the equation   $$x^5-x^3+x-2=0$$   find the value of $\lfloor {\alpha}^6 \rfloor$.      This one totally stumped me. We are asked to calculate $\lfloor {\alpha}^6 \rfloor$ without actually calculating the root or using wolfram alpha or any other calculator. I found that the above equation has only one real root by sketching its graph using calculus. I was also able to use the intermediate value theorem to conclude that $1&lt;\alpha&lt;2$ , but this is of little use while calculating $\lfloor {\alpha}^6 \rfloor$. Please help! I have a quaternion, coming from a system with the following:(intertial sensor)  Right handed. Forward direction: Y axis Right direction: X axis Up direction: Z axis   I need to convert this into a coordinate system that is: (Unreal engine)  left-handed. Forward direction: X axis Right direction: Y axis Up direction: Z axis   I have tried negating the axis, and angle, I have tried switching values, i cannot get this to work. All help greatly appreciated! I am working in C#, with Microsoft.Xna.Quaternion. Incorrect indentation for tag listing questions loaded via AJAX I am seeing a weird indentation for questions loaded via AJAX (i.e., by clicking the n question(s) with new activity link). The following screenshot was taken before I clicked the link:  And this is after the new question was loaded:  A browser refresh resolves the problem (as can be expected). It was tested with Chrome 88 and Firefox 85 on  (Catalina). Showing that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set Problem 4.3.32 in Linear Algebra, Lay:  Let $V$ and $W$ be vector spaces, let $T:V\to W$ be a linear transformation, and let $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ be a subset of $V$. Suppose that $T$ is a one-to-one transformation [...]. Show that if the set of images $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set.  All I can think of doing: If $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then there exist weights $c_1, \dots, c_p$, not all zero, so that $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = \mathbf{0}$$ Since $T$ is linear: $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = T(c_1\mathbf{v}_1) + \cdots + T(c_p \mathbf{v}_p)=T(c_1 \mathbf{v}_1 +\cdots + c_p \mathbf{v}_p)=\mathbf{0}$$ So the set $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ must be linearly dependent. Am I missing something? And why must $T$ be on-to-one for this to be true? An example showing that it is not true if $T$ is not one-to-one would be great! Setting Zero values back to NULL I have a php file which I will be using as exclusively as an include. Therefore I would like to throw an error instead of executing it when it's accessed directly by typing in the URL instead of being included.  Basically I need to do a check as follows in the php file:  if ( $REQUEST_URL == $URL_OF_CURRENT_PAGE ) die ("Direct access not premitted");   Is there an easy way to do this? Quitting a job in the US on a work visa, going to Canada and re-entering on a tourist visa (ESTA waiver) I have been working in the USA on an E-3 visa and intend to return home to Australia, but I want to travel around the US before I do so. I am under the impression that after leaving my current company, I will have 10 days to leave the country.   Will I be able to go into Mexico or Canada and re-enter with an ESTA waiver as a tourist? How to modify existing, unpushed commit messages? I wrote the wrong thing in a commit message.  How can I change the message? The commit has not been pushed yet. How to escape quotes in shell? I'm having trouble with escaping characters in bash. I'd like to escape single and double quotes while running a command under a different user. For the purposes of this question let's say I want to echo the following on the screen:  'single quote phrase' "double quote phrase"   How can I escape all the special chars, if I also need to switch to a different user:  sudo su USER -c "echo \"'single quote phrase' \"double quote phrase\"\""   Of course, this doesn't produce the right result. HTTP GET request in JavaScript? I need to do an  request in JavaScript. What's the best way to do that?  I need to do this in a Mac OS X dashcode widget. Which media player has seek preview with thumbnails? I was looking for a good video player to replace VLC, something that is as good as vlc but has seek preview. I am tired of fast forwarding in vlc  EDIT: I meant seek preview with thumbnails I use TeX Live 2010 and my editor is Texmaker Bidi. It was working probably good but from 2 months ago I've encountered with the error:   !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ! ! Fatal fontspec error: "cannot-use-pdftex" !  ! The fontspec package requires either XeTeX or LuaTeX to function. !  ! You must change your typesetting engine to, e.g., "xelatex" or "lualatex" ! instead of plain "latex" or "pdflatex". !  ! See the fontspec documentation for further information. !  ! For immediate help type H &lt;return&gt;. !...............................................   when I'm trying to load xepersian LaTeX.  but I do not know how can I change pdfLaTeX to XeLaTeX.    This is the second edit of my question .   the code is :    ‎\documentclass{article}‎ ‎\usepackage{pstricks}‎ ‎\usepackage{‎pst‎-‎bezier}‎ ‎‎\usepackage{xepersian}‎  ‎\settextfont{XB Zar}‎  ‎\setlatintextfont{Junicode}‎  ‎\setdigitfont{XB Zar}‎  ‎\begin{document}‎  ‎!‎سلام دنیا‎‎ ‎\begin{pspicture}(‎3‎,3)‎ ‎\psbcurve(1,2)(5,7)(2,2)(12,3)‎ ‎\end{pspicture}‎ ‎ ‎\end{document}   but the output does not include the shapes . When constructing the matrices for the two CNOT based on the target and control qubit, I can use reasoning:  "If $q_0$==$|0\rangle$, everything simply passes through", resulting in an Identity matrix style $\begin{bmatrix}1&amp;0\\0&amp;1\end{bmatrix}$ in the top left. "If $q_0==|1\rangle$, we need to let $q_0$ pass and swap $q_1$, resulting in a Pauli X $\begin{bmatrix}0&amp;1\\1&amp;0\end{bmatrix}$ in the bottom right.  $CNOT \equiv \begin{bmatrix}1&amp;0&amp;0&amp;0\\0&amp;1&amp;0&amp;0\\0&amp;0&amp;0&amp;1\\0&amp;0&amp;1&amp;0\end{bmatrix}$  "If $q_1$==$|0\rangle$, everything simply passes through", results in leaving $|00\rangle$ and $|10\rangle$ unaffected. "If $q_1==|1\rangle$, we need to let $q_1$ pass and swap $q_0$, mapping $|01\rangle$ to $|11\rangle$ and  $|11\rangle$ to $|01\rangle$  $CNOT \equiv \begin{bmatrix}1&amp;0&amp;0&amp;0\\0&amp;0&amp;0&amp;1\\0&amp;0&amp;1&amp;0\\0&amp;1&amp;0&amp;0\end{bmatrix}$  This all seems to check out, but here comes my question: I would like to know if there is a more mathematical way to express this, just as there would be when combining for instance two hadamard gates: $H \otimes H \equiv \frac{1}{\sqrt{2}}\begin{bmatrix}1&amp;1\\1&amp;-1\\ \end{bmatrix} \otimes \frac{1}{\sqrt{2}}\begin{bmatrix}1&amp;1\\1&amp;-1\\ \end{bmatrix} = \frac{1}{2}\begin{bmatrix}1&amp;1&amp;1&amp;1\\1&amp;-1&amp;1&amp;-1\\1&amp;1&amp;-1&amp;-1\\1&amp;-1&amp;-1&amp;1 \end{bmatrix}$  And a bonus question:  How I can, using notation like "CNOT", show which qubit is the control bit and which qubit the target bit? primitive n-th roots of unity Show that the primitive n-th roots of unity have the form $e^{2ki\pi/n}$ for $k,n$ coprime for $0\leq k\leq n$.  Since all primitive n-th roots of unity are n-th roots of unity by definition they all have that form, the question is, how to show $k$ and $n$ are coprime. Device not detected in Eclipse when connected with USB cable Proof of $\gcd(a,b)=ax+by\ $ [Bezout's identity] I am trying to find out whether $A = \begin{bmatrix}     3 &amp; 1 \\     0 &amp; 2 \\ \end{bmatrix}$ and $B = \begin{bmatrix}     4 &amp; 1 \\     -2 &amp; 1 \\ \end{bmatrix}$ are similar. Both matrices have the same trace, same rank, same determinant and the same eigenvalues. So I think they are similar. The eigenvalues are $3, 2$. Now I'm not sure how to find if they are indeed similar.   I am trying to find if there is a basis for $\mathbb{R^2}$ that contains only eigenvectors but when trying to solve $(3 I - A)v = 0$ there is no solution... What should I do from here? Thunderbird - remove unused dictionaries from menu The boundary of union is the union of boundaries when the sets have disjoint closures How do you configure an OpenFileDialog to select folders? In VS .NET, when you are selecting a folder for a project, a dialog that looks like an OpenFileDialog or SaveFileDialog is displayed, but is set up to accept only folders.  Ever since I've seen this I've wanted to know how it's done.  I am aware of the FolderBrowserDialog, but I've never really liked that dialog.  It starts too small and doesn't let me take advantage of being able to type a path.  I'm almost certain by now there's not a way to do this from .NET, but I'm just as curious how you do it from unmanaged code as well.  Short of completely reimplementing the dialog from scratch, how do you modify the dialog to have this behavior?  I'd also like to restate that I am aware of the FolderBrowserDialog but sometimes I don't like to use it, in addition to being genuinely curious how to configure a dialog in this manner.  Telling me to just use the FolderBrowserDialog helps me maintain a consistent UI experience but doesn't satisfy my curiosity so it won't count as an answer.  It's not a Vista-specific thing either; I've been seeing this dialog since VS .NET 2003, so it is doable in Win2k and WinXP.  This is less of a "I want to know the proper way to do this" question and more of a "I have been curious about this since I first wanted to do it in VS 2003" question.  I understand that Vista's file dialog has an option to do this, but it's been working in XP so I know they did something to get it to work.  Vista-specific answers are not answers, because Vista doesn't exist in the question context.  Update: I'm accepting Scott Wisniewski's answer because it comes with a working sample, but I think Serge deserves credit for pointing to the dialog customization (which is admittedly nasty from .NET but it does work) and Mark Ransom for figuring out that MS probably rolled a custom dialog for this task.