I use my USB flash (thumb) drive quite often to move data between several computers (Macs and PCs.) When I plug in this flash drive into Windows 7 it shows this message:     "Do you want to scan and fix ...?"     If I click "Scan and fix" the message goes away for a little while only to come back again.  Is there a setting in Windows to make this message go away because it is very annoying to click through this window every time I plug in my drive?  PS. I run this same flash drive on a Mac computer and it never tells me that there's an issue with the drive.  PS2. I have other flash drives that keep showing this same message on Windows 7. They all cannot have the same issue, can they :)
Every time I insert the SD card from my camera into the card reader, I get this dialog:    Of course there is no need to scan it. Is there a way to disable it?
dev/sda1: clean, 552599/6111232 files, 7119295/24414464 blocks   After I turn on my laptop this message appears. The system never boots, the message just stays there. Are commands complete sentences? seems to indirectly ask the question, but the upvoted answer says &quot;the actual sentence is...&quot;. So my question is this: Are commands considered to be grammatically sound, complete sentences in the English language? Some examples:  Go. Add a note. Do work. Received W-2 but was never paid I just now received a W-2 from a company that I worked for 3 days as a contractor. I had to leave country for 2 months due to a family emergency. Since the employment duration was very short, I was not paid. I was told that they don't want to charge the client because 3 days is not enough contribution.  Since I had to leave country the same day, I did not argue with them and told it's okay if they feel right.  Now I have a W-2 from them saying they have paid me about $1500. I contacted them but I have not received any reply yet.  I just want to be prepared. What can I do here? Is there even an automatic ban on meta? Somewhere I heard the automatic question ban on Meta is set lower than on the main site. But looking at the profile of  it looks like there is absolutely no automatic ban on meta.  Is this intended behaviour? The limit of a complex integral If $f(z)$ is continuous in the neighborhood of the point $z=a$, then $$\lim_{r\to 0^+} \int_{\vert z-a\vert =r} \frac{f(z) dz}{z-a}=2\pi i\,f(a)\tag{1}$$  In my desperate intents of proving the above exercise, I tried to prove a 'particular' case of it, which is when $z=0$.   So if $z=0$ and as $f$ is continuous on a neighborhood of zero, given $\epsilon&gt;0$, there exists $\delta&gt;0$ such that $|f(re^{i \theta})-f(0)|\le \epsilon$ for $r\in[0,\delta)$ and $\theta\in\mathbb R$. Hence,  $\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right| &lt;2\pi\epsilon \Rightarrow \limsup_{r\to0}\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right| \le2\pi\epsilon.$  If $\epsilon\to0$, then $\limsup_{r\to0}\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right|=0 \Rightarrow \lim_{r\to0}\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta=0.$  Is this proof correct?  How can I make the proof for $(1)$ following the 'particular proof'? Can someone help me? How do you disable browser autocomplete on web form field / input tags? How do you disable autocomplete in the major browsers for a specific input (or form field)? I was wondering if there is such a thing as an shapefile PRJ to PostGIS SRID lookup table?  Something that can translate the most standard shapefile PRJ definitions into the likely SRID.  When using PostGIS and pgAdminIII, if you use the postgisgui to import your shapefiles, the SRID is left as "-1".  It seems like the tool should be able to parse the Esri PRJ and determine the correct (or at least a couple of options) that are the likely SRID, rather than just leave the default.  Or does the importer have the capability to reproject on the fly if you choose another SRID?  It may seem lazy on my part, but to me it seems curious that this function hasn't already been put in place.  Does anyone know if this concept is in the works, or good reason's why it has been left out? Custom blocks created as a plugin in a module not showing for Block::load() I have a couple of blocks that have been created using the ... extends BlockBase syntax in my module. They work perfectly fine, and I can add them on the blocks layout page, can add them into DS as a field, etc. They definitely exist fine.  However, when I try and render them using Block::load(), they won't show up, unlike other blocks that work fine.  For example, if I have the following docblock:  /**  * Provides a 'Cute Cats' Block.  *  * @Block(  *   id = "cute_cats",  *   admin_label = @Translation("Cute cats yo"),  * )  */  class CuteCats extends BlockBase { ... }   ... and then try and call it like this:  $block = \Drupal\block\Entity\Block::load('cute_cats'); return render(\Drupal::entityTypeManager()   -&gt;getViewBuilder('block')   -&gt;view($block));   Any idea why? Thanks in advance. "Optimiser" vs "optimizer"? I have seen words which are spelled either with a "z" or with a "s"  like "optimizer" and " optimiser".  I thought that the ones with "z" are wrong but on searching the web I found that both are correct. Why so? Is there any difference in the use also? Can I run PC steam games on a Mac? I recently downloaded 'Blood Bowl' on Steam to my Mac, not realizing that it was not Mac-compatible. Is there any software or any way that will allow me to play it, as obviously I've already paid for it? Cheers. Best way to add KeyEvents and faster Quit Why (and how) did using cat on binary files mess up the terminal? If I understand the cat manual correctly:       concatenate files and print on the standard output   cat will take files as argument and print them on standard output. What I don't get is if I use the command:    cat img.png &gt; copy.png   I will obtain 2 png files identical while if I just    cat img.png     I have all chance that my terminal get messed up and misinterpret what I type.     How's that possible?   Binary values are still binary data. Why it does not simply shows a series of 0 and 1 or the interpretation of those binary data in ASCII or whatever the encoding in terminal is?   Is this behavior also possible by cating a text file containing strange characters?   Should a mechanism to prevent this behavior  like try{}catch{} statement should be implemented? Truncated 200GB table but disk space not released I recently moved to Linux, and I am in love with the experience. Opinions apart, I am having problems when running sudo dpkg --configure -a  My computer stops processing the command here:  You may know Adobe Flash Player is no longer supported, and I believe that could be why the command is unable to load pepperflashplugin-nonfree. I have tried fixing the problem by checking older posts here, but none of them was able to fix this problem. Printing an array to a file with each element of the array in a new line in bash Accessing laptop's hard drive when OS won't boot In a theoretical physics homework problem, I have to show the following: $$\partial_\nu T^{\mu\nu} = - j_\nu F^{\mu\nu}$$  Where $T$ is the Energy-Momentum-Tensor, $j$ the generalized current and $F$ the Field-Tensor. We use the $g$ for the metric tensor, I think in English the $\eta$ is more common.  I know the following relationships:   Current and magnetic potential with Lorenz gauge condition: $$\mathop\Box A^\mu = \mu_0 j^\mu$$ Energy-Momentum-Tensor: $$T^{\mu\nu} = \frac1{\mu_0} g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu} + \frac1{4\mu_0} g^{\mu\nu} F_{\kappa\lambda} F^{\kappa\lambda}$$ Field-Tensor: $$F^{\mu\nu} = 2 \partial^{[\mu} A^{\nu]} = \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}$$ d'Alembert operator: $$\mathop\Box = \partial_\mu \partial^\mu$$ Bianchi identity: $$\partial^{[\mu} F^{\nu\alpha]} = 0$$   So far I have set all the definitions into the formula I have to show, but I only end up a lot of terms from antisymmetrisation and product rule. I also drew all what I have in , but I still cannot see how to tackle this problem.  Could somebody please give me a hint into the right direction? "ImportError: No module named _struct" configuring python-minimal When I run sudo apt-get upgrade, the dpkg error appears and interupts the process.  Error details:  Setting up python-minimal (2.7.3-0ubuntu7.1) ... Traceback (most recent call last):   File "/usr/local/lib/python2.7/runpy.py", line 162, in _run_module_as_main     "__main__", fname, loader, pkg_name)   File "/usr/local/lib/python2.7/runpy.py", line 72, in _run_code     exec code in run_globals   File "/usr/local/lib/python2.7/compileall.py", line 16, in &lt;module&gt;     import struct   File "/usr/local/lib/python2.7/struct.py", line 1, in &lt;module&gt;     from _struct import * ImportError: No module named _struct dpkg: error processing python-minimal (--configure):  subprocess installed post-installation script returned error exit status 255 dpkg: dependency problems prevent configuration of python:  python depends on python-minimal (= 2.7.3-0ubuntu7.1); however:   Package python-minimal is not configured yet.   And this problem leads to a series of dependency problems when processing the following packages:    python  gwibber-service  libgwibber3  libgwibber-gtk3  gwibber  gwibber-service-facebook  gwibber-service-identica  gwibber-service-twitter  python-all  python-dev  python-all-dev  python-apt  python-problem-report  python-apport  python-libxml2  unity-lens-gwibber  unity-scope-video-remote   Finally, nothing I can do for this problem. And several days later, the root directory is full and I even could not uninstall any software because of this problem. Evaluating the limit of a sequence given by recurrence relation $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Is my solution correct? Problem  The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation:   $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$.   Evaluate the limit $\lim_{n\to\infty} a_n$.  Solution   Show that the sequence $(a_n)_{n=1}^\infty$ is monotonic. The statement $$V(n): a_n &lt; a_{n+1}$$ holds for $n = 1$, that is $\sqrt2 &lt; \sqrt{2+\sqrt2}$. Let us assume the statement holds for $n$ and show that $V(n) \implies V(n+1)$. We have that $$a_n &lt; a_{n+1}.$$ Adding 2 to both sides and taking square roots, we have that $$\sqrt{2+a_n} &lt; \sqrt{2+a_{n+1}},$$ that is $a_{n+1} &lt; a_{n+2}$ by definition. Find bounds for $a_n$. The statement $$W(n): 0 &lt; a_n &lt; 2$$ holds for $n=1$, that is $0 &lt; \sqrt2 &lt; 2$. Let us assume the statement holds for $n$ and show that $W(n) \implies W(n+1)$. We have that $$0 &lt; a_n &lt; 2.$$ Adding two and taking square roots, we have that $$0 &lt; \sqrt2 &lt; \sqrt{2+a_n} &lt; \sqrt4 = 2.$$ The limit $\lim_{n\to\infty} a_n$ exists, because $(a_n)_{n=1}^\infty$ is a bounded monotonic sequence. Let $A = \lim_{n\to\infty} a_n$. Therefore the limit $\lim_{n \to\infty} a_{n+1}$ exists as well and $\lim_{n \to\infty} a_{n+1} = A$. (For $(n_k)_{k=1}^\infty = (2,3,4, \dots)$, we have that $(a_{n_k})_{k=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$, from which the statement follows.) We have that $a_{n+1} = f(a_n)$. That means that $A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}$. Solving the equation $A = \sqrt{2 + A}$, we get $A = -1 \lor A = 2$. Putting it all together, we get that $A = 2$, because the terms of the sequence are increasing and $a_1 &gt; 0$.   Is my solution correct? What can I do when getting “This question body does not meet our quality standards”?