I accidently ran chown myname /usr -R and now the login screen won't show up! I have tried various things in the root terminal, but nothing works. What can I do?
I stupidly ran sudo chown -R carlos:carlos / and carlos is my user. This changed all files to be Carlos and I could not boot up again (because the root files were now owned by #1000) and I cant get sudo acess in live cd (because the owner of /usr was changed to my user). Can anyone help me? FYI I have mounted my hard drive.
How can I figure out why my washing machine is tripping the GFCI receptacle? I have a Kenmore 80 Series (Model No. 110.24872300) washing machine which up to this point has been working fine. Since it is near a sink, the outlet it is plugged into is a GFCI outlet. On the last load I came back after letting it run for some time to find that it had tripped the GFCI in the middle of the wash cycle - the washer was still full of soapy water. I reset the outlet and let it run again, only to come down the next morning and find it full of water still (it had tripped during the rinse cycle this time). I reset the outlet again and this time it finished the rest of the cycle.  My guess is that water is somehow getting onto the internal components of the washer and causing it to short - is there any way I can test this theory or fix the problem myself? How can PCB trace have 50 ohm impedance regardless of length and signal frequency? Hmm, this seems to be just another question on line impedances.  I understand that when we say "transmission line" effects we talk about things like cross talk, reflections and ringing (I guess that is just about it). These effects are not present at low frequencies where the PCB trace behaves like an "ideal" transmission medium, more like we expect a wire to behave in our early school days.  I also understand that the 50&nbsp;ohm value comes not from the line resistance which is going to be very small and less than 1&nbsp;ohm. This value comes from the ratio of L and C on the line. Changing C by changing the trace height above ground plane or changing L by changing the trace width shall change the impedance of the line.  We all know that the reactance of L and C is dependant on the signal frequency as well. Now my questions:   Why should we not call this as line reactance only rather than line impedance? How can it be just 50&nbsp;ohm? It has to be signal frequency dependent right?  E.g 50&nbsp;ohm at 1&nbsp;MHz  Will the world end if I chose to do a 100&nbsp;ohm or 25&nbsp;ohm trace instead?  I know that while we like to say 50&nbsp;ohm as a magic number, it will be within some range around 50&nbsp;ohm and not 50.0000 ohm exactly.  Is there any time when the actual resistance of a PCB trace may matter? We have a Barbarian on our team that very much likes to switch up his weapons; he goes from a Dwarven Hammer, to a pair of Hand-Axes, to occasionally just going bare-fisted. No issues there, he's fairly adept with all 3.  However, I have only just recently picked up on the way that he rolls his attacks for the two Hand-Axes: He rolls the first attack, adds his modifiers etc. Then he rolls the second attack, this time without modifiers, which is all well and good. But I'm not sure if this action uses his bonus action or not, or is considered a separate action (though, being a level 4 Barbarian, he only has one action per turn), or is merely part of the same action.  If we have a look at the Two-Weapon Fighting feature from the Fighter's fighting styles:     Two-Weapon Fighting   When you engage in two-weapon fighting, you can add your ability modifier to the damage of the second attack. (PHB5e p.72)   Similarly, the Dual Wielder feat also uses the Two-Weapon Fighting feat:     Dual Wielder   You master fighting with two weapons, gaining the following benefits:      ...         You can use two-weapon fighting even when the one-handed melee weapons you are wielding aren’t light. (PHB5e p.164, emphasis mine)      So; yes, this gives a bonus to the second weapon attack, but having asked a  to clarify which action this secondary weapon attack belongs to, it was determined that this uses your bonus action;     [...] the second attack requires a bonus action   So after that long winded explanation, here is the question: Without any additional feats, can you wield and use two weapons at once, and if so, how many actions does it require to make a full attack with both weapons in a turn? Why do archery arrows tilt downwards in their descent? Recently I paid a lot (almost one month of my salary) for repairing my car. I am wondering if there is a way to get deduction on federal or state taxes (USA, California). If $h : Y \to X$ is a covering map and $Y$ is connected, then the cardinality of the fiber $h^{-1}(x)$ is independent of $x \in X$. In "Knots and Primes: An Introduction to Arithmetic Topology", the author uses the following proposition   Let $h: Y \to X$ be a covering. For any path $\gamma : [0,1] \to X$ and any $y \in h^{-1}(x) (x = \gamma(0))$, there exists a unique lift $\hat{\gamma} : [0,1] \to Y$ of $\gamma$ with $\hat{\gamma}(0) = y$. Furthermore, for any homotopy $\gamma_t (t \in [0,1])$ of $\gamma$ with $\gamma_t = \gamma(0)$ and $\gamma_t(1) = \gamma(1)$, there exists a unique lift of $\hat{\gamma_t}$ such that $\hat{\gamma_t}$ is the homotopy of $\hat{\gamma}$ with $\hat{\gamma_t}(0) = \hat{\gamma}(0)$ and $\hat{\gamma_t}(1) = \hat{\gamma}(1)$.  The author then follows with; "In the following, we assume that any covering space is connected. By the preceding proposition, the cardinality of the fiber $h^{-1}(x)$ is independent of $x \in X$."   I am not sure why this results is true. This is my attempt at explaining it to myself. Take two different $x_1, x_2$ and their fibers $h^{-1}(x_1), h^{-1}(x_2)$ such that $y_1 \in h^{-1}(x_1)$ and $y_2 \in h^{-1}(x_2)$. Take two paths $\gamma_1, \gamma_2$ such that $\gamma_1(0) = x_1$ and $\gamma_2(0) = x_2$. Then we get two lifts $\hat{\gamma_1}, \hat{\gamma_2}$ with $\hat{\gamma_1}(0) = y_1$ and $\hat{\gamma_2}(0) = y_2$. Now, since our covering space is connected we can continuously deform $\hat{\gamma_1}$ into $\hat{\gamma_2}$ and conclude that every elements in $h^{-1}(x_1)$ is also in $h^{-1}(x_2)$ and vice versa. I feel that this is wrong but cannot figure out the right way to see this. Any help would be appreciated. Prove that $\overline {A^\complement} = (A°)^\complement $   To show that $\overline {A^\complement} \subseteq  (A°)^\complement$   We first show $A^\complement \subseteq  (A°)^\complement ...(*) $  We know that since $ A°\subseteq A \qquad (*)$ is automatically follows .  The part I didn't understand is the following:     Then we conclude that since $ (A°)^\complement$ is closed    $\therefore \overline {A^\complement} \subseteq  (A°)^\complement$ Suspending with root on USB I have a laptop running Ubuntu 14.04 from a root file system on USB storage. This is not working well, because after wake up from suspend, ext4 will frequently try to write to the root file system before USB is ready.  Here is what I see in the kernel log when this happens, notice how I get a bunch of I/O errors on sda1, and then one second later the USB storage drive is finally detected by the kernel.  [ 2826.517419] wlan0: associated [ 2826.517452] IPv6: ADDRCONF(NETDEV_CHANGE): wlan0: link becomes ready [ 2827.575371] EXT4-fs warning (device sda1): ext4_end_bio:317: I/O error -5 writing to inode 1733735 (offset 0 size 0 starting block 12629950) [ 2827.575380] Buffer I/O error on device sda1, logical block 12629694 [ 2827.575400] EXT4-fs warning (device sda1): ext4_end_bio:317: I/O error -5 writing to inode 3148603 (offset 0 size 8192 starting block 12844470) [ 2827.575404] Buffer I/O error on device sda1, logical block 12844212 [ 2827.575411] Buffer I/O error on device sda1, logical block 12844213 [ 2827.575448] EXT4-fs warning (device sda1): ext4_end_bio:317: I/O error -5 writing to inode 3015015 (offset 0 size 90112 starting block 6588832) [ 2827.575453] Buffer I/O error on device sda1, logical block 6588576 [ 2827.575461] Buffer I/O error on device sda1, logical block 6588577 [ 2827.575465] Buffer I/O error on device sda1, logical block 6588578 [ 2827.575469] Buffer I/O error on device sda1, logical block 6588579 [ 2827.575473] Buffer I/O error on device sda1, logical block 6588580 [ 2827.575477] Buffer I/O error on device sda1, logical block 6588581 [ 2827.575481] Buffer I/O error on device sda1, logical block 6588582 [ 2828.857284] sd 0:0:0:0: [sda] No Caching mode page found [ 2828.857293] sd 0:0:0:0: [sda] Assuming drive cache: write through   At first there is no visible indication outside of the kernel log, that the problem has triggered, but if I let Ubuntu stay running beyond this point, then the file system will get errors and eventually switch to read only mode. At that point I have to reboot into recovery mode and run fsck.ext4 manually from a root shell in order to repair the file system.  Is there some setting I can change such that access to the root device after waking from suspend can be delayed until the USB drive is ready? AuraClientService.postProcess: error in processing [RangeError: Maximum call stack size exceeded] - Client Side Action I have a lightning component that is using ui:inputSelect.   The component code:              &lt;div class="slds-form-element__control"&gt;                 &lt;ui:inputSelect class="slds-select" aura:id="quals" change="{!c.getPreCodes}"&gt;                     &lt;aura:iteration items="{!v.questions}" var="q"&gt;                          &lt;ui:inputSelectOption aura:id="question" text="{!q.Id}" label="{!q.Question_Text__c}"/&gt;                     &lt;/aura:iteration&gt;                 &lt;/ui:inputSelect&gt;             &lt;/div&gt;   The controller:      getPreCodes : function(cmp, event, helper){         var selected = cmp.find("quals").get("v.value");          var action = cmp.get("c.getPreCodes");         action.setParams({             "qualID" : selected         });         action.setCallback(this, function(response){             var state = response.getState();             if (state === "SUCCESS") {                  console.log('precodes: ', response.getReturnValue());              }         });          $A.enqueueAction(action);     }   The basic logic is that the uiSelect is dynamically populated on init (this works fine). When a user makes a selection the "change" parameter function is fired to retrieve a list from the database and pass it to another component.   However, the change function {!c.getPreCodes} seems to be called recursively causing a      "Something has gone wrong. AuraClientService.postProcess: error in   processing [RangeError: Maximum call stack size exceeded] . Please try   again."   error. If I comment out the enqueueAction the function only runs once (nothing is called recursively) so I suspect it has something to do with the controller call. What is the difference in meaning between "proposal" and "proposition"? I would like to create a batch file, that will add registry key, not manually, but full automaticly by .BAT file.  goes about adding everything but not the key.  shows how to add everything, and how to add key manually, but it requires user interference, which I am trying to avoid.  Is there possibility to add KEY (non REG_*) to "folder?" in registry by one simple batch file? Let $\displaystyle f$ be differentiable, $\displaystyle f(x)=0$ for $|x| \geq 10 $ and $g(x)=\sum_{k \in \mathbb Z}f(x+k).$ Recently I noticed that many people post .htaccess files here with:  &lt;IfModule mod_rewrite.c&gt;   Sometimes this even appears several times in the file! It obviously checks to see if mod_write is in fact enabled but the statements it protects are necesary to those sites anyway. So if the IfModule fails, the website is broken anyway.  Is there an advantage in having the &lt;IfModule mod_rewrite.c&gt; fail rather than the RewriteRule statements below it?  mod_write now is so essential and ubiquitous that I do not check for it anymore. Might this cause a vulnerability of some kind? Show that $a_n=\sin(n)$ does not converge     My idea:  Take two subsequences: $a_{n_k}=\sin(\frac {\pi k} 2)$ , $a_{n_l}=\sin(\frac {2\pi l} 3)$  So: $\forall n$ : $\lim_{n\to\infty} a_{n_k}=1$, $\lim_{n\to\infty} a_{n_l}=-1$  So two inifinite subsequence converge to different limits thus the sequence $a_n$ doesn't converge.   Is that correct ?  Edit:   The contrapositive of the defintion of a limit of a sequence:   $\exists\epsilon&gt;0 : \forall n\in N : \exists n&gt;N \Rightarrow |x_n-L|&gt;\epsilon$  Take $\epsilon =1$ and we know that $\sin(n)$ is bounded so lets take it's supermum: 1  $|1-1|&gt;\epsilon=1\Rightarrow 0&gt;1 \Rightarrow$ Contradiction. Should Dalvik Cache be wiped in Android OS that supports ART? When switching custom ROMs (in ART), is it necessary to wipe the Dalvik Cache?  Wiping the Cache is necessary, but should wiping Dalvik (which doesn't exist in Android Lollipop and afterward) be necessary?  What impact will it cause if we wipe Dalvik (something not present)?  In ART, what does the Dalvik cache contains? (is it an empty folder?) Let $F$ be a field. $(x-c)|(p(x)-p(c))$.   So if $x-c|p(x)=q(x)-\frac{r}{x-c}$ so for $x-c|p(x)-p(c)=s(x-c)-\frac{r}{x-c}$. Is that correct? Or is there a different way to go about proving this? I have proved that $x-c|p(x)$ has a remainder of $p(c)$ so It wuld make sense that $p(x)-p(c)$ would be $x-c$. The question I want to ask is related to the answer of this question:  .   My aim is to separate my bibliography into two parts : one part for a personnal bibliography, and a second one for the regular bibliography.  To do that, I use the technique of the post I cited :  \documentclass{article} \usepackage[T1]{fontenc} \usepackage[french]{babel} \usepackage[style=alphabetic,maxnames=6,natbib=true]{biblatex} \bibliography{biblatex-examples}  % Variants of each could be added \newcommand{\firstname}{Donald~E.} \newcommand{\lastname}{Knuth}  \DeclareBibliographyCategory{byname}  \DeclareIndexNameFormat{byname}{% Test could be refined   \ifboolexpr{ test {\ifdefstring{\lastname}{#1}}            and test {\ifdefstring{\firstname}{#3}}               } {\addtocategory{byname}{\thefield{entrykey}}} {}}  \AtDataInput{%   \indexnames[byname]{author}}  \begin{document}  \section{Knuth's books}  \begin{refsection}  \nocite{*}  \printbibliography[category=byname,heading=none]  \end{refsection}  \section{Not Knuth's books}  \printbibliography[notcategory=byname,heading=none] \end{document}   The problem is that only the second bibliography appears, and when I remove the refsection, then all the bibtex entries of my input files are printed, separated as they should.  EDIT: the aim is to have all publications (even the ones not cited) by author knuth in the first section and all the other cited publications in the second one. Counterexample for $A,B\triangleleft G,G/A\cong B \Rightarrow G/B\cong A$ How can one request reconsideration of a   after a certain period of time, let us say after one year? Logging in as the same user multiple times in Remote Desktop Windows Server 2008 Quick question: I have a situation where I need to let multiple people on different PCs log into one server 2008 machine as administrator simultaneously over remote desktop. I have the CALs for it, it's just not set up correctly. When one user tries to log in, it boots the other out. What I need is to present to them a different session, just each as logged in as admin. Sorry for the slightly rambling post, I'm new here. Thanks!