I have recently become interested in trying out Ubuntu 15.10, just for the sake of fun. However, I am confused as to how to dual boot Windows 10 and Ubuntu 15.10 on my Toshiba Qosmio X75-A7295  The computer is a rather high-spec one, with one 256 SSD on which Windows 10 is installed (as in, any important files required to run Windows 10), and a 1 TB hard drive that is used to hold any extra files, like documents and what not.   The 256 GB SSD is almost full, leaving only the 1 TB hard drive open for installing Ubuntu.  However, the computer is EFI boot, so I believe that the boot partition is on the SSD  What advice would you give me when attempting to install Ubuntu in a dual-boot set-up?
I'm absolutely new to Linux. I would like to know how to install Ubuntu alongside the pre-installed Windows 8+ OS.  Should I do it with Wubi, or through the Live USB/DVD?  What steps do I need to take to correctly install Ubuntu?
Putting char into a java string for each N characters I have a java string, which has a variable length.   I need to put the piece "&lt;br&gt;" into the string, say each 10 characters.   For example this is my string:  `this is my string which I need to modify...I love stackoverlow:)`   How can I obtain this string?:  `this is my&lt;br&gt; string wh&lt;br&gt;ich I nee&lt;br&gt;d to modif&lt;br&gt;y...I love&lt;br&gt; stackover&lt;br&gt;flow:)`   Thanks N circles in the plane Currently, the number of "votes cast" excludes votes on deleted content. Once something gets deleted that you voted on, that vote vanishes from your stats. This of course happens frequently especially with down-votes - if you're a good voter, stuff you downvote often ends up getting deleted.  Seeing as the idea behind the public display of total votes you've cast is to encourage voting, is this really fair?   Also, doesn't it distort greatly all voting stats, especially when looking at candidates for moderator elections and such? Disabling submit-button until the user has completed the form How do you think about disabling the submit-button until the user has completed the form or the required fields?   I find similar questions here on ux.se and somebody answered that the user will think that the form is broken or not immediately realize whats going on.  I think hiding is not that good but a disabling (e.g. grayscale, opacity, ...) plus a short text in the button would improve usability because the user realise that he still has to fill out something.  Is there anybody with experience or examples besides the usual error-messages/validation after submitting? How can I decode HTML characters in C#? I have email addresses encoded with HTML character entities. Is there anything in .NET that can convert them to plain strings? Primary ideals in Noetherian rings I'm not sure how to solve this problem. Can you guys provide some input/hints?      Let $A=(x_1,y_1)$, $B=(x_2,y_2)$ and $C=(x_3,y_3)$ be three points in $\mathbb{R}^{2}$.      Show that the area of $\Delta ABC$ is given by   $\frac{1}{2}\left| \det(M) \right|$, where   $$M = \begin{pmatrix} 1 &amp; 1  &amp; 1\\  x_1 &amp; x_2 &amp; x_3\\  y_1 &amp; y_2 &amp; y_3 \end{pmatrix}$$ Proving any number of the form $2^n$ can't be written as the sum of k consecutive numbers. I think i have proved that any number of the form $2^n$ can't be written as the sum of k consecutive positive integers, but i would be grateful if i could have some clarification as to whether this is a correct and valid proof or not.  Let $g=x+(x+1)+(x+2)+(x+3)+...+(x+(k-1))$, hence g is the sum of k consecutive positive integers.  $g=kx+\sum_{d=1}^{k-1}d$  Hence, $g=kx+\frac{k(k-1)}{2}$  Rearranging for x we find, $x=\frac{g}{k}-\frac{(k-1)}{2}$  So, for certain values of $k$, we can be sure that a positive integer value of $x$ can be found, and hence $g$ can be written as the sum of $k$ consecutive positive integers.There are two cases for $k$ to considor:  Case 1)  $k$ is odd. If $x$ exists, it must be the case $k$ | $g$, as for odd values of $k$, $\frac{(k-1)}{2}$ will alwyas be an integer, and hence $x$ will.  Case 2)  $k$ is even. $k$ does not divide $g$, but $\frac{g}{k}$ must be a multiple of $\frac{1}{2}$, and hence equals $\frac{h}{2}$, for some odd $h$.  Letting $j=2^n$, assume that we can write $j$ as the sum of $k$ consecutive positive integers.  If $k$ is odd, then if a $k$ exists, it must divide $j$, by Case 1. But, $j$ has no odd factors as it is a power of two, hence $k$ could not be odd.  If $k$ is even, by Case 2 it mut be the case $k$ does not divide $j$. Also, $\frac{j}{k}=\frac{h}{2}$, for some odd $h$.  But, this implies $2j$ has an odd factor, $h$. but, as $2j$ is a power of $2$, it can't have odd factors. So, $k$ can't be odd.  Hence, $k$ is neither odd or even, and this is a contradiction to our assumption. Hence no $x$ exists, and hence and number of the form $2^n$ can't be written as the sum of $k$ consecutive positive integers.  Any feedback on this proof would be appreciated. Already I have lost all hope of looking at late answers etc. It's just a parade of meh. I can't make any kind of intelligent decision because I can't see the other answers to the question, for instance. There's no Looks Good, only Skip, and those feel different to me. So I've been more interested in .  But people keep approving crap while I'm looking at it. Two or three "the post was already approved; please visit to edit" and I'm just "screw this whole thing".  Can we do something like one reject bounces all existing votes: you need three approves in a row? Because a lot of people are stupidly approving and actually driving me away from the queue.    Note this is less about letting bad edits through and more about keeping reviewers motivated. Is "connected, simply connected" Redundant? Here are my definitions of &quot;connected&quot; and &quot;simply connected.&quot;  A topological space $X$ is connected if and only if it is not the union of two nonempty disjoint open sets. A topological space $X$ is simply connected if and only if it is path-connected and has trivial fundamental group (i.e. $\pi_1(X)\simeq\{\mathrm{e}\}$ and $|\pi_0(X)|=1$).  It is a classic and elementary exercise in topology to show that, if a space is path-connected, then it is connected. Thus, if a space is simply connected, then it is connected. Yet, despite this implication, I've read several cases where the words &quot;connected, simply connected&quot; appear together. For example, in Kobayashi and Nomizu's Foundations of Differential Geometry, Volume 1, page 252, the following is written: &quot;Let $M$ be a connected, simply connected analytic manifold with an analytic linear connection.&quot; In  and  (Links to Journal Storage, also known as JSTOR), they use &quot;connected, simply connected&quot; on their first page. The first of these takes it a step further, referring to &quot;closed, compact, connected, simply connected $4$-manifolds,&quot; where a closed manifold (unfortunate terminology) is defined as a compact manifold without boundary. My question is simple: Why use the words &quot;connected, simply connected&quot; when being simply connected implies connectedness? Thank you. Declare IDisposable for the class or interface? Cause of black artifacts in render? Method overloading in Objective-C? Is there a way to install a newer version than in the repository?  Ubuntu installs vagrant 1.4.3 at the moment, but I need at least 1.5 A well-known application of the MVT is to prove that any $f: \mathbb{R} \to \mathbb{R}$ with $f'= 0$ is constant. But of course, the MVT relies fundamentally on the properties of $\mathbb{R}$, and in particular does not hold in $\mathbb{Q}$. A standard counter-example is the function $f :\mathbb{Q} \to \mathbb{Q}$ defined by $ f(x)=0 $ if $ x &lt;\sqrt{2}$, and $f(x)=1$ if $ x &gt; \sqrt{2}$, for this function is locally constant. There are several ways of thinking about the difference between $\mathbb{R}$ and $\mathbb{Q}$ : $\mathbb{R}$ is connected while $\mathbb{Q}$ is totally disconnected, closed bounded intervals are compact in $\mathbb{R}$, but not in $\mathbb{Q}$, $\mathbb{R}$ satisfies the least upper bound axiom, $\mathbb{Q}$ does not.   Now, I wanted to see how far one could push this counterexample - is there a function $f :\mathbb{Q} \to \mathbb{Q}$ such that $f'=0$ and $f$ is strictly increasing? Clearly, any slight modification of the usual counterexample where $f$ is locally constant will not do. Also, one can show that if $f$ is uniformly differentiable with derivative zero, it must be constant, even in $\mathbb{Q}$, so I knew I was looking for non-uniformly differentiable but still differentiable $f$.   Eventually, I came up with a construction, albeit not a very explicit one: the idea is that, since $\mathbb{Q}$ is countable, one can enumerate it $q_{1}, q_{2}, q_{3}, \cdots$, and then define $f: \mathbb{Q} \to \mathbb{Q}$ inductively and recursively. If one could define $f$ in such a way that $ \forall \ q_{n} \in \mathbb{Q} \ \exists \ c_{n} \in \mathbb{Q} : \forall \ x \in \mathbb{Q} \ |f(x)-f(q_{n})| \leq c_{n}|x-q_{n}|^2$, and $f$ is strictly increasing, we would be done. The recursive definition essentially defines $f(q_{n+1})$, after having defined $f(q_{1}),f(q_{2}),f(q_{3}), \cdots , f(q_{n})$, in such a way that these inequalities hold and $f$ is strictly increasing, and at each step the new $c_{n}$ is chosen large enough so that these inequalities do not become incompatible at subsequent steps ( I am willing to provide more details of this construction if asked).  Note that, as $n \to \infty$, we must have $c_{n} \to \infty$, so that $f$ is not uniformly differentiable. Interestingly, the only key fact about $\mathbb{Q}$ used in the construction is that it is countable, so in a sense the construction provides a proof that $\mathbb{R}$ is uncountable, given the MVT.  While the construction works, it is not very simple nor explicit, which is perhaps to be expected, and the function must be evaluated point by point in the order given by the enumeration of $\mathbb{Q}$. So I was wondering if anyone else knows or can come up with a simpler construction of a function $ f : \mathbb{Q} \to \mathbb{Q}$ such that $f'=0$ and $f$ is strictly increasing. Any thoughts and ideas are welcome.  Edit: I've asked the same question on MO: How typeset a numeric table from a CSV file I have a CSV file (db.txt)like this:  30.0,   0.0,    0.0 60.0,   1.9098, 5.8779 90.0,   6.9098, 9.5106 120.0,  13.09,  9.5106 150.0,  18.09,  5.8779 180.0,  20.0,   0.0   I need to typeset it. My problem is related with the wrong alignment of decimal point and the number of decimals (must be a constant for each column):  \renewcommand*{\dtlrealalign}{r}  \DTLloaddb[     noheader,     keys={x,y,theta},     headers={         \shortstack{$\boldsymbol{\theta_{2,i}}$},         \shortstack{X},         \shortstack{Y}              }   ]{db}{db.txt} \begin{table}[t]   \caption{Database file}   \centering   \DTLdisplaydb{db} \end{table}     Any advice, please? Importing a Single Symbol From a Different Font Is there a general way to change the font of a particular symbol without switching packages?  For example, suppose I'm using Computer Modern for my entire document, but I want the "subset" operator to look as it does in mathabx.  How can I redefine it to appear in this way? I have modeled a simple pocket watch in Blender. Every time I import it into Unity3D, regardless of whether I export it from Blender as an FBX or 3ds file, it always comes out missing faces! Why is this happening? I have never had problems importing models into Unity3D like this before.  Blender Screenshot    Unity Screenshots     It seems to me that different faces appear when I rotate it differently. How can I prevent this? Thanks! Why is the [I'm Done] button unavailable after only flagging comments in the First Posts review queue? I just flagged two comments during a review of a question in the First Posts queue (on another SE site). Then I wanted to select the button I'm Done, but it's unavailable (I can only select No action needed). Even if I refresh my page (while still on that review page), it remains the same.  If however I also downvote the question (only temporary, as a test, since I don't want to downvote the question being reviewed), then it seems to work as I'd expect it to: the button I'm Done becomes available, and No action needed becomes unavailable.  Why is the I'm Done button unavailable after only flagging comments in the First Posts review queue?  Notes:   I'm not convinced that you're supposed to review the actual post, and not (only) the comments underneath it (as mentioned in 's comment). Because "if" I would post a new comment, or just upvote an existing comment, I can actually select I'm Done. While in this case I flagged existing comments which IMO had become obsolete. Flagging comments is "acting on comments". And if my flag gets accepted (in this case: "no longer needed"), and the comment is actually deleted (by the mod), then it makes the question a bit more readable (= don't waste your time digesting a comment which is obsolete). Therefor "IMO", in the end I also "acted on the post". Think about all those posts with a zillion of comments, of which many are obsolete (e.g because the post got updated already). I am not sure (yet) if a similar issue exist in any other review queue (i.e. the Late Answers queue). Can I change the color of the last selected face, so that it is the same as the other selected faces?  Here, I've selected the left face then the right face. Note the right face is some sort of checkered or dithered pattern, and just the outline has the nice orange 'selected' color.    Here I've selected the top face, note the right face has now the nice color.    And here I've deselected the top face. Now, both of the remaining selected faces have the nice color. This is the result I'd rather see instead of image 1.