I have a Windows 7 computer that I put iTunes on.  I have an external hard drive.  I had iTunes organize my music in the external hard drive into a different directory.  I now find that all my mp3's are missing.  I assume they haven't all been deleted off my hard drive, so how can I recover them?
Is there a way to recover deleted data from a SSD / USB Flash drive?  All I could find were recovery for damaged drives, but I want to recover deleted data.  So how do I do this?
Lost my Downloads folder both from Dock and Finder I did something stupid apparently and lost my Downloads folder both from Dock and then when I opened Finder to drag it down to the Dock it somehow got deleted in the desktop.  Now although I can see the folder exists (e.g. by searching in Spotlight) I do not know how to restore it in the Dock and in Finder as well. Any help please? What is the name of this game involving racing to the other side of the board and putting down barriers? Does anyone recognize this game, where players take turns in a square board, each turn a player can move his pawn or place a wall and the objective is to reach the other side of the board before the other player.  Here is a picture of a near end game. Representative point of a cluster with L1 distance The representative point of a cluster or cluster center for the k-means algorithm is the component-wise mean of the points in its cluster. The mean is chosen because it helps to minimize the within cluster variances (which is to say that it is minimizing within cluster squared Euclidean distance, since its the same). Now, when I want to use a different distance metric, like L1 (Manhattan distance), I would be minimizing the within cluster absolute deviations (if I choose to). I was looking at what the centers calculation would be like, because the mean won't reduce the new objective function, and I found this on the Matlab documentation of their kmeans function for the distance parameter     cityblock - Sum of absolute differences, i.e., the L1 distance. Each centroid is the component-wise median of the points in that cluster.   Now, how did they come up with the component-wise median? How does a median minimize absolute deviations? Using a web service is often an excellent architectural approach. And, with the advent of WCF in .Net, it's getting even better.  But, in my experience, some people seem to think that web services should always be used in the data access layer for calls to the database. I don't think that web services are the universal solution.  I am thinking of smaller intranet applications with a few dozen users. The web app and its web service are deployed to one web server, not a web farm. There isn't going to be another web app in the future that can use this particular web service. It seems to me that the cost of calling the web service unnecessarily increases the burden on the web server. There is a performance hit to inter-process calls. Maintaining and debugging the code for the web app and the web service is more complicated. So is deployment. I just don't see the advantages of using a web service here.  One could test this by creating two versions of the web app, with and without the web service, and do stress testing, but I haven't done it.  Do you have an opinion on using web services for small-scale web app's? Any other occasions when web services are not a good architectural choice? What is the physical significance of dot & cross product of vectors? Why is division not defined for vectors? I get the physical significance of vector addition &amp; subtraction. But I don't understand what do dot &amp; cross products mean?   More specifically,    Why is it that dot product of vectors $\vec{A}$ and $\vec{B}$ is defined as $AB\cos\theta$? Why is it that cross product of vectors $\vec{A}$ and $\vec{B}$ is defined as $AB\sin\theta$, times a unit vector determined from the right-hand rule?   To me, both these formulae seem to be arbitrarily defined (although, I know that it definitely wouldn't be the case).   If the cross product could be defined arbitrarily, why can't we define division of vectors? What's wrong with that? Why can't vectors be divided? Hide deleted answers with certain characteristics Today I exceeded 10K reputation on Stack Overflow, and I see deleted answers.  I am now faced with pages like  where half the answers are "deleted."    Among these are many non-answers deleted by diamond moderators, and a self-deleted answer deleted on Nov 25 '09.  I find this very distracting, especially in my chosen sort order of "oldest" where these deleted answers are NOT displayed at the bottom.  I wish to be able to hide deleted answers based on combinations of characteristics such as:   age since deletion deleted-by-owner deleted by diamond moderators zero undelete votes   This would be a setting every user could enable for himself.    Here are HTML excerpts that may be useful for userscript filtering.  This shows the code that appears for a moderator deletion, as well as the date stamp.  &lt;div class="deleted-answer-info"&gt;deleted by &lt;a href="/users/10936/dori"&gt;Dori&lt;/a&gt;&lt;span class="mod-flair" title="moderator"&gt;♦&lt;/span&gt; &lt;span title="2011-06-26 09:13:34Z" class="relativetime"&gt;Jun 26 at 9:13&lt;/span&gt;&lt;/div&gt;   Here is the same class for deleted by owner:  &lt;div class="deleted-answer-info"&gt;deleted by owner &lt;span title="2009-11-25 19:12:11Z" class="relativetime"&gt;Nov 25 '09 at 19:12&lt;/span&gt;&lt;/div&gt; asp.net mvc put controllers into a separate project I'm just learning asp.net mvc and I'm trying to figure out how to move my controllers into a separate project.  Typically when I have designed asp.net web apps before, I created one project for my models, another for my logic, and then there was the web.    Now that I'm learning asp.net mvc I was hoping to follow a similar pattern and put the models and controllers each into their own separate projects, and just leave the views/scripts/css in the web.  The models part was easy, but what I don't understand is how to make my controllers in a separate project be "found".  Also, I would like to know if this is advisable.  Thanks! 80's/90's movie: man and woman driving through a red desert planet How to use \def to redefine \end{split}? When I use \def to redefine \end{split}, there is an error.  \documentclass{article} \usepackage{amsmath}  \begin{document} \def\be{\begin{equation}} \def\ee{\end{equation}} \def\bs{\begin{split}} \def\es{\end{split}}  \be     \bs         a\\         b     \es \ee  \end{document}   The error is:      Runaway argument?      a\ b \es \ee       ! Paragraph ended before \split was complete.   However, if I don't use \def\es{\end{split}} and just modify the code to be  \documentclass{article} \usepackage{amsmath}  \begin{document} \def\be{\begin{equation}} \def\ee{\end{equation}} \def\bs{\begin{split}} \def\es{\end{split}}  \be     \bs         a\\         b     \end{split} \ee  \end{document}   it's OK. What's the problem? Thank you! Is there a rule of thumb for objective questions asked out of curiosity? I have a strange issue with Windows 10  When I open .nef files they load fine. However after a second or two the contrast changes noticeably. I tried Windows 10 Photos and Picasa Image Viewer.   What am I trying to accomplish? I just want to open files quickly to check for focus and composition before importing into Lightroom  Lightroom does not have this issue   Here is the example opening with Windows 10 Photos    And here is after a second when it 'fully' loads    Picasa Photo Viewer after it loads a bit. It initially opens fine but then it goes all purple How do I format a String in an email so Outlook will print the line breaks? I'm trying to send an email in Java but when I read the body of the email in Outlook, it's gotten rid of all my linebreaks. I'm putting \n at the ends of the lines but is there something special I need to do other than that? The receivers are always going to be using Outlook.   I found a page on microsoft.com that says there's a 'Remove line breaks' "feature" in Outlook so does this mean there's no solution to get around that other than un-checking that setting?  Thanks What is the closure of an open ball $B_X(\mathbf{a},r)$ in $X=\mathbb{R}^n$? Suppose we have the open ball $B_{X}(\mathbf{a},r)$ and the closed ball $\bar{B}_{X}(\mathbf{a},r)$ of radius $r$ about $\mathbf{a}\in\mathbb{R}^n=X$ with the Euclidean metric $d_2$.  What is the closure of an open ball $B_{X}(\mathbf{a},r)$ in $(\mathbb{R}^n,d_2)$?  First, I have defined the sets like this:  $$B_X(\mathbf{a},r)=\{\mathbf{y}\in\mathbb{R}^n:d_2(\mathbf{a},\mathbf{y})&lt;r\}$$ $$\bar{B}_X(\mathbf{a},r)=\{\mathbf{y}\in\mathbb{R}^n:d_2(\mathbf{a},\mathbf{y})\le r\}$$  Then, let $U=B_X(\mathbf{a},r)$ and claim the closure is $$\bar{U}=\{\mathbf{x}\in\mathbb{R}^n:\exists (\mathbf{x}_n)\in U\text{s.t}~\mathbf{x}_n\to \mathbf{x}\}=\bar{B}_X(\mathbf{a},r).$$  Since closed balls are closed then we have $$\bar{U}\subset\bar{B}_X(\mathbf{a},r).$$  But, how can I prove the other direction $\bar{B}_X(\mathbf{a},r)\subset\bar{U}$?  Since it's a containment argument I know that I have to take any element from $\bar{B}_X(\mathbf{a},r)$ and show it is in $\bar{U}=\{\mathbf{x}\in\mathbb{R}^n:\exists (\mathbf{x}_n)\in U\text{s.t}~\mathbf{x}_n\to \mathbf{x}\}$.  But I do not know how to get an expression for $\mathbf{x}_n$.  I would appreciate any help. As, the above link shows, other countries have similar murder rates but they are not in the radar of international travelers.  For example, as seen in the list, Puerto Rico has similar statistics. But, personally, I never heard of Puerto Rico to be a dangerous place to travel.  Why is it considered so dangerous to travel in South Africa? Is that due to the incompetent law enforcement agencies? Google Drive files opening in new window vs new tab? This might be a duplicate of  but that question is 5 years old and unanswered.  When I'm in drive.google.com and I open a G Suite file (either double-click or select &amp; press enter) in Firefox it opens in a new window whereas in Chrome it opens in a new tab.   Why do they do that? Is this a browser-specific quirk or a Google setting (that I haven't found)? Is there a way to control the behavior? I'd like them both to open in tabs, but presumably the same answer would be relevant in the opposite direction too. Retrieve the position (X,Y) of an HTML element relative to the browser window I want to know how to get the X and Y position of HTML elements such as img and div in JavaScript relative to the browser window. How to know that your machine learning problem is hopeless? Imagine a standard machine-learning scenario:     You are confronted with a large multivariate dataset and you have a   pretty blurry understanding of it. What you need to do is to make   predictions about some variable based on what you have. As usual, you   clean the data, look at descriptive statistics, run some models,   cross-validate them etc., but after several attempts, going back and   forth and trying multiple models nothing seems to work and your   results are miserable. You can spend hours, days, or weeks on such a   problem...   The question is: when to stop? How do you know that your data actually is hopeless and all the fancy models wouldn't do you any more good than predicting the average outcome for all cases or some other trivial solution?  Of course, this is a forecastability issue, but as far as I know, it is hard to assess forecastability for multivariate data before trying something on it. Or am I wrong?   Disclaimer: this question was inspired by this one  that did not attract much attention. It would be nice to have detailed answer to such question for reference. Layers with same coordinate system should align/overlap in ArcMap but do not? ubuntu 18.04 LTS bluetooth [0cf3:3004] discovery not working Rolle's Theorem and the Mean Value Theorem Let $f\colon \mathbb{R} \to \mathbb{R}$ have derivatives of all orders. Suppose that $a &lt; b$ and that $f(a)=f(b)=f'(a)=f'(b) =0$. Prove that $f'''(c) = 0$ for some $c$ in $(a,b)$.