after installing Ubuntu 16.04, when I choose in grub startup menu to boot in Windows 10 it restart PC but loop again in grub start up menu. I tried - boot repair (to correct it - "recomended change") - sudo update-grub also. So as I wrote after installing Ubuntu 16.04, when I select Win10 in grub, it restart computer, but again shows grub bootup menu... Basic information - older desktop (no UEFI), - 1. installed Win Xp, 2. Windows 10, both load without problem, - Win10 shut down, but fast boot was not disabled (in ubuntu I can see other drives, but ubuntu cannot mount them) - 3. installed Ubuntu 16.04  As first on HDD I had 2 NTFS partitions (50 and 70GB), on the first was Win XP, I split it ( 50GB to 20GB Win XP and new 30GB) and on new part I install win 10. After this I split the second partition (70GB in 30GB and 40GB) and on the 30GB I install Ubuntu 16.04 (the partition during installation was split into 20GB - system and rest /shows 8,6GB/ as "HOME"), after few restart I try to choose Win 10 during booting, but as I wrote earlier, I cannot access Windows.   When you can help me, thank you for your answer/help  PS.: When I correct installation (using Win 10 installation disk) - fix MBR, I can run Win10 (including Win XP) - but of course not Ubuntu. So then I again reinstall Ubuntu 16.04, but problem occurs again, and I cannot access my Window installations...!
I installed Windows on my computer, followed by an installation of Ubuntu. However, now I'm unable to boot into my Windows install.  What can I do to fix this?
If a superconductor has zero resistance, does it have infinite amperage? Last upgrade crashes network manager (no internet connection, no applet) While upgrading today, something caused my wireless network manager to disappear. Not even connecting via wired connection.  This error is being shown by files:  **(nm-applet:2716): WARNING **: Could not initialize NMClient /org/freedesktop/NetworkManager: The name org.freedesktop.NetworkManager was not provided by any .service files   (nm-applet:2716): nm-applet-WARNING **: Error connecting to ModemManager: Error calling StartServiceByName for org.freedesktop.ModemManager1: GDBus.Error:org.freedesktop.DBus.Error.Spawn.ExecFailed: Cannot launch daemon, file not found or permissions invalid   (nm-applet:2716): nm-applet-WARNING **: Could not find ShellVersion property on org.gnome.Shell after 5 tries   (nm-applet:2716): nm-applet-WARNING **: Failed to register as an agent: (2) The name org.freedesktop.NetworkManager was not provided by any .service files    Is there a patch I can download through windows and send through to my ubuntu, or is it a case of having to reload everything? I tried running two versions back but still no internet access. How to render Entity Reference fields in the current language I am trying to render fields from an entity reference in a twig template but they are displaying in the default language instead of the selected one. My twig code:  {% for location in node.field_locations %}   &lt;div class="home-location"&gt;     &lt;div class="wording"&gt;       {{ location.entity.title.value }}        &lt;span&gt;{{ location.entity.field_sub_title.value }}&lt;/span&gt;      &lt;/div&gt;    &lt;/div&gt;  {% endfor %} Database design for products with bundles of products I have the formula $x! + y! = z!$ and I'm looking for positive integers that make it true. Upon  inspection it seems that x = y = 1 and z = 2 is the only solution. The problem is how to show it.   From the definition of the factorial function we know $x! = x(x-1)(x-2)...(2)(1)$  So we can do something like this:  $$ [x(x-1)(x-2)...(2)(1)] + [y(y-1)(y-2)...(2)(1)] = [z(z-1)(z-2)...(2)(1)]$$   we can then factor all of the common terms out on the LHS.  $$ [...(2)(1)][x(x-1)(x-2)... + y(y-1)(y-2)...] = [z(z-1)(z-2)...(2)(1)]$$ and divide the common terms out of the right hand side  $$[x(x-1)(x-2)...] + [y(y-1)(y-2)...] = [z(z-1)(z-2)...]$$  I'm stuck on how to proceed and how to make a clearer argument that there is only the one solution (if indeed there is only the one solution).   If anybody can provide a hint as to how to proceed I would appreciate it. Group theory proof of existence of a solution to $x^2\equiv -1\pmod p$ iff $p\equiv 1 \pmod 4$ I've read through the elementary proof of why there exists a solution $x$ to $x^2\equiv -1\pmod p$ iff $p\equiv 1 \pmod 4$ for $p$ an odd prime. Is there a group theory generalization for this fact as well? It is quite easy to calculate the total number of functions from a set $X$ with $m$ elements to a set $Y$ with $n$ elements ($n^{m}$), and also the total number of injective functions ($n^{\underline{m}}$, denoting the falling factorial). But I am thinking about how to calculate the total number of surjective functions $f\colon X \twoheadrightarrow Y $.  The way I thought of doing this is as follows: firstly, since all $n$ elements of the codomain $Y$ need to be mapped to, you choose any $n$ elements from the $m$ elements of the set $X$ to be mapped one-to-one with the $n$ elements of $Y$. This results in $n!$ possible pairings. But the number of ways of choosing $n$ elements from $m$ elements is $\frac{m!}{(m-n)!\,n!}$, so the total number of ways of matching $n$ elements in $X$ to be one-to-one with the $n$ elements of $Y$ is $\frac{m!}{(m-n)!\,n!} \times n! = \frac{m!}{(m-n)!}$.  Now we have 'covered' the codomain $Y$ with $n$ elements from $X$, the remaining unpaired $m-n$ elements from $X$ can be mapped to any of the elements of $Y$, so there are $n^{m-n}$ ways of doing this. Therefore I think that the total number of surjective functions should be $\frac{m!}{(m-n)!} \, n^{m-n}$.  Is this anything like correct or have I made a major mistake here? Uninstall apps locked by phone vendor I'm using a Samsung I9000 and I have some annoying apps that came pre installed by the phone vendor, like a music store, a custom web browser and other memory leaking apps.  Those crappy apps can't be uninstalled neither by Manage applications tool, nor by editing the apps tiles in the front menu.  I guess I have to be a root user or something to get rid of those apps, but I don't know how. General Information about Eigenvalues for an 3x3 symmetric matrix How do we find general information about the eigenvalues of an arbitrary 3x3 symmetric matrix without resorting to explicitly computing the solutions to the cubic characteristic equation?:  \begin{bmatrix}     a   &amp; b &amp; c  \\     b    &amp; d &amp; e \\     c &amp; e &amp;f \end{bmatrix}  (In this case the sum and product of the eigenvalues, and the sum of the reciprocals of the eigenvalues.) Since diagonalization preserves the trace and determinant of matrices, would it be best to just find the corresponding diagonal matrix and read off the sum and product of the eigenvalues from there? The problem is I can't think of a way to diagonalize a matrix without explicitly computing the eigenvalues. For example:     The word rarely turns up outside of those contexts.      The word turns up rarely outside of those contexts.   Which one is correct and why? How would one go about proving that the rationals are not the countable intersection of open sets? I'm trying to prove that the rationals are not the countable intersection of open sets, but I still can't understand why   $$\bigcap_{n \in \mathbf{N}} \left\{\left(q - \frac 1n, q + \frac 1n\right) : q \in \mathbf{Q}\right\}$$   isn't a counter-example. Any ideas? Thanks! Prove by mathematical induction that: $\forall n \in \mathbb{N}: 3^{n} > n^{3}$ Prove by mathematical  induction that:   $$\forall n \in \mathbb{N}: 3^{n} &gt; n^{3}$$  Step 1: Show that the statement is true for $n = 1$:  $$3^{1} &gt; 1^{3} \Rightarrow 3 &gt; 1$$  Step 2: Show that if the statement is true for $n = p$, it is true for $n = p + 1$  The general idea I had was to start with $(p+1)^{3}$ and during the process substitute in $3^{p}$ for $p^{3}$ as an inequality.  $$(p+1)^{3} = p^{3} + 3p^{2} + 3p + 1 &lt; 3^{p} + 3p^{2} + 3p + 1$$  Now, if it can be shown that:  $$\forall p \in \mathbb{N}: 3p^{2} + 3p + 1 \leq 2 \cdot 3^{p}$$  ...the proof is complete. This is because $3^{p+1} = 3 \cdot 3^{p}$ and one of those three have already been used.   We do this by mathematical induction. First, the base case of $n = 1$:  $$3\cdot 1^{2} + 3 \cdot 1 + 1 \not \leq 2 \cdot 3^{1}$$  ..which turns out to be false.  What are some more productive approaches to this step? How to draw circle in html page? number of prime ideals in algebraic integers Why does change in speed of a wave make it refract? When a light wave enters a medium with a higher refractive index (e.g. from air to standard glass) and its speed decreases, why does that make it refract/bend?  I understand that wavelength decreases and frequency stays the same and therefore its speed decreases, but I can't find anywhere whatsoever why the speed decrease cause the wave to refract. So could someone please explain this? Show that every prime $p&gt;3$ is either of the form $6n+1$ or of the form $6n+5$, where $n=0,1,2, \dots$ It would be helpful if we could see the number of upvotes cast for our comments, or other people's comments in response to us directly in the user profile in either the activity tab or the response tab respectively. It should be quite easy to implement (e.g. sharing a single partial view for displaying comments).  It could be useful, since we can evaluate the legitimacy of the comment and address horrible mistakes sooner. Here is my proof, I would appreciate it if someone could critique it for me:  To prove this statement true, we must proof that the two conditional statements ("If $\mathcal{P}(A)⊆ \mathcal{P}(B)$, then $A⊆B$," and, If $A⊆B$, then $\mathcal{P}(A)⊆ \mathcal{P}(B)$) are true.    Contrapositive of the first statement: If $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$  If $A \nsubseteq B$, then there must be some element in $A$, call it $x$, that is not in $B$: $x \in A$, and $x \notin B$. Since $x \in A$, then $\{x\} \in \mathcal{P}(A)$; moreover, since $x \notin B$, then $\{x\} \notin \mathcal{P}(B)$, which proves that, if $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$. By proving the contrapositive true, the original proposition must be true.    To prove the second statement true, I would implement nearly the same argument, so that isn't necessary to write. So, does this proof seem correct? Also, was the contrapositive necessary? Or is there another way to prove the initial statement? Petersen graph is not a Cayley graph How can I show that the Petersen graph is not a Cayley graph?  I don't know very much about Cayley graphs, I know that they are vertex-transitive, but so is the Petersen graph. It probably has to do with the group structure of $\Gamma$ in $Cay(\Gamma,S)$ (which is a group of order $10$, i.e., it is either the cyclic group $C_{10}$ or the dihedral group $D_{10}$). Then I am guessing we go for some sort of contradiction(?). My group theory is a bit rusty, I would very much appreciate someone showing me how to do this one! Shortcut for "null if object is null, or object.member if object is not null"