I wanna cut a mesh by specifying the angle, but knife mode only cut 45° with C key.  I want to cut by the angle around specifying the vertex.
Instead of the constrained 45 degrees that happens when you press C, is there a specific angle I can get for example 35 degrees instead?   It's currently cut at 45 degrees in the attached image and I need it to be 35.
How to add home directory path to be discovered by Unix which command? How to combine two jQuery results How do you combine two jQuery search results? eg:  var $allFoos = $('.foo'),     $allBars = $('.bar')     $allFoosAndBars = $allFoos + $allBars;   Obviously, I just made up that last line, but I hope it makes it sorta clear what I mean. To be clear, the example is greatly simplified, and it could be any arbitrary sets i'm talking about, so $('.foo, .bar') is not what I'm after. Software recommendations Clear item with custom name? Floating point calculations in LaTeX? How to move windows that open up offscreen? I have a dual monitor setup, and I recently played around with the positioning settings, and some of my single window applications do the favour of preserving where they were last closed, and opening in the same position later. Unfortunately, that now places them out of the viewable area of my screens!  Is there some way to force a particular window into the viewable area? Please recommend undergraduate-level textbooks on electrodynamics which emphasize practical applications and real life examples. Please describe the book's level and contents and its intended audience in as much detail as possible.   Please provide both applications from everyday life for curious students as well as engineering applications from a physics perspective. Given the following problem     Let $U$ and $V$ be finite dimensional vector spaces, such that $\dim(U) = \dim(V)$. Prove that $U$ and $V$ are isomorphic.   I was just wondering if somebody could critique and validate my proof.  Proof  Two vector spaces $U$ and $V$ are isomorphic if and only if there exists a bijection $\phi: U \to V$. That is, an invertible linear map between the two. Let $\dim(U) = \dim(V) = n$ and let $\left\{u_1, \dots, u_n\right\}$ and $\left\{v_1, \dots, v_n\right\}$ be bases for $U$ and $V$, respectively. It follows that any vector $u \in U$ can be written as  $$ u = c_1u_1 + \cdots + c_nu_n $$  We define $\phi: U \to V$ as  $$ \phi(c_1u_1 + \cdots + c_nu_n) = c_1v_1 + \cdots + c_nv_n $$  Observe that $\phi$ is invertible. That is, given a vector  $$ v = d_1v_1 + \cdots + d_nv_n $$  in $V$, we define $\phi^{-1}: V \to U$ as  $$ \phi^{-1}(d_1v_1 + \cdots + d_nv_n) = d_1u_1 + \cdots + d_nu_n $$  Since there exists a bijection between $U$ and $V$, they are isomorphic. Let $p$ a cut point of a connected space $(X, \mathcal{T})$ and suppose $C$ and $D$ form a separation  of $X-\{p\}$, i.e $$X-\{p\}=C \cup D$$ Prove that $C \cup \{p\}$ is conected. I have this idea: Since $p$ is a cut point then $X-\{ p \}$ is not connected, and since $C$ and $D$ form a separation of $X- \{p \}$ then $X-\{p \} = C \cup D$, and also $C \cup D$ is not connected. And also $$C \cap \overline{D}= \overline{C} \cap D= \emptyset$$ I need an idea to continue please Why is Ubuntu 14.04 so slow on my laptop? I just installed Ubuntu 14.04 on a Dell laptop with 1.7Ghz Pentium M and 1Gbyte memory.  Graphics are Intel® 852GM/855GM x86/MMX/SSE2, OS type is 32bit.  The whole system is VERY slow. When I launch the applications manager, for example, I need to wait in order to see the letters I typed appear!  It is as if the CPU was overloaded and no memory available, but I only have Firefox running.  How can I make the system run smoother? It used to have Windows XP installed and it worked fine. How does Java's serialization work and when it should be used instead of some other persistence technique? I've been lately trying to learn more and generally test Java's serialization for both work and personal projects and I must say that the more I know about it, the less I like it. This may be caused by misinformation though so that's why I'm asking these two things from you all:  1: On byte level, how does serialization know how to match serialized values with some class?  One of my problems right here is that I made a small test with ArrayList containing values "one", "two", "three". After serialization the byte array took 78 bytes which seems awfully lot for such low amount of information(19+3+3+4 bytes). Granted there's bound to be some overhead but this leads to my second question:  2: Can serialization be considered a good method for persisting objects at all? Now obviously if I'd use some homemade XML format the persistence data would be something like this  &lt;object&gt;     &lt;class="java.util.ArrayList"&gt;     &lt;!-- Object array inside Arraylist is called elementData --&gt;     &lt;field name="elementData"&gt;         &lt;value&gt;One&lt;/value&gt;         &lt;value&gt;Two&lt;/value&gt;         &lt;value&gt;Three&lt;/value&gt;     &lt;/field&gt; &lt;/object&gt;   which, like XML in general, is a bit bloated and takes 138 bytes(without whitespaces, that is). The same in JSON could be   {     "java.util.ArrayList": {         "elementData": [             "one",             "two",             "three"         ]     } }   which is 75 bytes so already slightly  smaller than Java's serialization. With these text-based formats it's of course obvious that there has to be a way to represent your basic data as text, numbers or any combination of both.  So to recap, how does serialization work on byte/bit level, when it should be used and when it shouldn't be used and what are real benefits of serialization besides that it comes standard in Java? Proof of inequality using AM-GM inequality prove that $x^2 + y^2 + z^2 \ge xy + xz + yz$  I tried proving it the way I prove $x^2 + y^2 \ge 2xy$. But it just doesn't seem to work. I tried using AM-GM. Should I be using GM-HM or AM-HM? I really don't know what the problem is and I'd appreciate any help and a full proof explanation. Thanks. If you have a focus for your spell casting do you need to pay materials? In D&amp;D 5E, if you have a focus (like bards have instruments and clerics have holy symbols) for your spell casting do you need to pay for materials? In XFCE4 desktop environments, alt+scroll wheel zooms my entire desktop. I'm finding this very annoying because sometimes I flick the scroll wheel to go to the top of a document then press the alt key for a some unrelated keyboard shortcut and find myself zoomed in so far I can't see anything at all. Is there any way to disable this feature? What does the caret (‘^’) mean in C++/CLI? How to completely remove virtual packages? It turns out that removing a virtual package with apt-get only remove the package itself,  leaving behind the real packages that were installed by it  So how do you remove a virtual package, along with all the real packages that it installs? I am trying to solve a 2 part question, the first part is to expand $\sqrt{1-x}$ up to and including the term $x^2$ which I did.  This gives me $1-(0.5)x-(0.125)x^2 + ...$  However, the 2nd part of the question says that by taking $x=(1/64)$ in the expansion of part (i) -> which refers to the one above, deduce that $\sqrt 7 = 10837/4096$. How do i do this part?  EDIT (After Hint Given By SimpliFire):  $$ \sqrt{1-\frac1{64}}=\frac{\sqrt{63}}8=\frac38\sqrt7 $$ $$ =\frac83\sqrt{1-\frac1{64}}\;=\sqrt7 $$ $$ =\frac83\left(1-\frac12\left(\frac1{64}\right)-\frac18\left(\frac1{64}\right)^2\right)\; $$ $$ =\frac83\left(1-\frac1{128}-\frac1{32768}\right)\; $$ $$ =\frac83\left(\frac{32768}{32768}-\frac{256}{32768}-\frac1{32768}\right)\; $$ $$ =\frac83\left(\frac{32511}{32768}\right)\; $$ $$ =\frac{260088}{98304} $$ $$ \frac{10837}{4096}\;\approx\sqrt7 $$  I'm not sure if I can present it like this, if it is okay i will post it as my answer :X I'm currently working my way through Harvard's online abstract algebra lectures (if you're interested, you can find them ).  The lectures come complete with notes and homework problems.  Of course, since I don't actually go to Harvard, I can't hand in the homework assignments to find out if I'm doing them correctly.  So I've decided to try to crowd source the grading of my solutions.  I'll post individual questions as I finish them and wait for comments.  I would like to get critiques of not just my reasoning but the style of my write ups as well.  Also, any alternative approaches to the problem would be welcome.  I've looked for forums dedicated to these kinds of OCW courses but have not been able to find any.  This surprises me.  It seems like, with the advent of these free online educational resources, an online meeting place for those who take advantage of them would be a natural offshoot.  So this might be a bit of an experiment.  Unless I'm the 873rd person to post something like this here.  If that's the case, sorry for being so long winded.  Anyway, on with the question.  This one is assigned in the 4th lecture.     Let $V$ denote the Klein 4-group.  Show that $\operatorname{Aut}(V)$ is isomorphic to $S_3$.   (I've moved my solution below to keep this question from showing up in the unanswered list.)  Thanks... Got less output with print0 option on find? I have read the man page and other references, but I am still confused by the behavior of find with -print0 option.  -print0      This primary always evaluates to true.  It prints the pathname of              the current file to standard output, followed by an ASCII NUL              character (character code 0).   First command:  $ find /some/path/ -iname "*intro*jpeg" -o -iname "*intro*jpg" 2&gt; /dev/null /some/path//asdf199Intro-5kqlw.jpeg /some/path/199intro-2jjaj.JPEG /some/path/199intro-3.jpg /some/path/wuer199intro-4.JPG /some/path/xbzf199INTRO-1.jpg   Second command:  $ find /some/path/ -iname "*intro*jpeg" -o -iname "*intro*jpg" 2&gt; /dev/null -print0 /some/path/136intro-3.jpg/some/path/wuer136intro-4.JPG/some/path/xbzf136INTRO-1.jpg   I can see that the the filenames from the second command are null character separated, but why do I get 3 output as opposed to 5 above? Which arena do free and crown chests use when determining the cards you will get? If I am in arena X when I collect my tenth crown for the crown chest, or a free chest becomes unlocked, then proceed to climb to arena X+1 before opening the chests, which arena will be used to determine the cards I get? Will it use the arena I was in when I unlocked the chests in, or the arena I opened them in?