I have this weird issue with chrome. When I launch it in desktop mode, everything is blurry. However when I go to its settings and relaunch in windows 8 mode, everything is super clear and crisp.  I have a QHD+ screen and firefox is so far the only browser which works out of the box, but I want to use chrome. However I don't want to use it in windows 8 mode (metro crap). I've scoured the internet for answers but the registry hack doesn't seem to work anymore.   Why is it that I have to launch chrome in windows 8 mode to get hidpi support?  Note: the chrome://flags page says that high dpi is not available on my platform.. what?
Google Chrome doesn't seem to respect my DPI setting in Windows. I run at 150 DPI, but Chrome ignores that, meaning that images and text are teeeeeeny tiny.  I hit:   Ctrl + + Ctrl + +   to zoom in. Unfortunately every new tab, or new window, defaults to the 100% teeny tiny zoom level.  Update: It's also important to note that you cannot click buttons in flash (e.g. YouTube videos), because Chrome mis-handles mouse coordinates. (Fixed 6.0.472.55)  Is it possible for either:   Chrome to respect my Windows dpi setting? Chrome to remember my zoom preference?   Checked version: 4.1.249.1045    Half fixed    While Chome still doesn't honor the Windows DPI setting, it does now remember your zoom level.  Version: 5.0.375.55    Five-Eighths Fixed    While Chrome still doesn't honor the Windows DPI setting, it does now offer a default "page zoom" option:    Unfortunately there is no zoom level that matches my current DPI setting: 136% (131dpi).  Version: 10.0.648.133  Zoom flag  I tried scoducks' suggestion of the new hidden dpi flag. It's truly awful. It seems to cause Chrome to simply re-scale the final rendered Window; causing text to become blurry (rather than sharper):
CSS background opacity with rgba not working in IE 8 'was' versus 'had been' When I asked the driver/ to come with me to the temple/ he had been reluctant./ No error.   I think instead of 'had been', there should be 'was'. Even the answer key points that the error is in the 'he had been reluctant' part. But I am not sure why exactly is 'had been' wrong.  If we replace the first part with 'when I had asked the driver', would that make 'had been' in the third part correct? Where can I ask questions regarding Electronics? I am currently doing some electronics and microcontroller programming, and I have a couple of questions to ask the people here.    Most of the questions are regarding wiring on breadboards and components.    On which of the websites (Stack Overflow, Server Fault, or Super User) should I ask these particular questions? 3D software like GeoGebra Does it exist a free interactive geometry software, like GeoGebra, which works for 3D geometry? I would be able to draw spheres, great circles, and so on. Story where vampires are created by drop absorbed on neck? I have been trying to identify a book that I read maybe 25-30 years ago (1987-1992), in which there are vampire or vampire-like creatures, but instead of being created in the traditional way (e.g. bite on the neck), something drops a drop of a fluid on the back of their neck, which is then absorbed into the body and transforms them. I have a vague notion that some sort of tree is what creates these drops. The drops may have had a shiny metallic appearance.  I've tried to find this book before and always came up short. I'm starting to think the whole thing was a dream and I never read it at all. But I really feel like it was a book or at least a short story. Count number of increasing functions, nondecreasing functions $f: \{1, 2, 3, \ldots, n\} \to \{1, 2, 3, \ldots, m\}$, with $m \geq n$. I stumbled upon a question given like:  Let $m$ and $n$ be two integers such that $m \geq n \geq 1$.   Count the number of functions $$f: \{1, 2, · · · , n\} \to \{1, 2, · · · , m\}$$ of the following two types:  (a) strictly increasing; i.e., whenever $x &lt; y, f(x) &lt; f(y)$, and  (b) non-decreasing; i.e., whenever $x &lt; y, f(x) \leq f(y)$.  I tried in the following way.  For (a). We can say $f(n)&gt;f(n-1)$ AND $f(n)&gt;f(n-2) \ldots f(n)&gt;f(1)$ (total $n-1$ elements)  Similarly for $f(n-1)&gt;f(n-2), f(n-1)&gt;f(n-3) \ldots f(n-1)&gt;f(1)$ (total $n-2$ elements)...  and $f(2)&gt;1$  Like this $$(n-1) + (n-2) + \ldots + (1) = \frac{n(n-1)}{2}$$  Is this correct?  For (b). Since there's equality, it will be $$n + (n-1) + ... (1) = \frac{n(n+1)}{2}$$  Is my approach correct? Any help is appreciated. Thanks in advance. How to efficiently count the number of keys/properties of an object in JavaScript? What's the fastest way to count the number of keys/properties of an object?  It it possible to do this without iterating over the object?  i.e. without doing  var count = 0; for (k in myobj) if (myobj.hasOwnProperty(k)) count++;   (Firefox did provide a magic __count__ property, but this was removed somewhere around version 4.) Run command on another(new) terminal window LINQ to Entities does not recognize the method 'System.String Format(System.String, System.Object, System.Object)' How do I identify which Linux distro is running? We have some new hardware in our office which runs its own customized Linux OS.   How do I go about figuring which distro it's based on? CUDA Driver API vs. CUDA runtime When writing CUDA applications, you can either work at the driver level or at the runtime level as illustrated on this image (The libraries are CUFFT and CUBLAS for advanced math):   (source: )    I assume the tradeoff between the two are increased performance for the low-evel API but at the cost of increased complexity of code. What are the concrete differences and are there any significant things which you cannot do with the high-level API?  I am using CUDA.net for interop with C# and it is built as a copy of the driver API. This encourages writing a lot of rather complex code in C# while the C++ equivalent would be more simple using the runtime API. Is there anything to win by doing it this way? The one benefit I can see is that it is easier to integrate intelligent error handling with the rest of the C# code. Why did the first Linux developers choose to implement a non-preemptive kernel? Is it to save synchronization?  As far as I know, Linux was developed in the early '90s, when PCs had a single processor. What advantage does a non-preemptive kernel give in such PCs? Why, however, the advantage is reduced by multi-core processors? Best practice for Rails App to run a long task in the background? I am a super newbie for Ubuntu and basically need Ubuntu for a Python package.  I would like to run a Python file in say e:/username/folder1/folder2/python.py. I tried every method online and the only reply is bash: cd./e: No such file or directory. I am not sure whether it is because of the bash windows or virtualenv.  When I use bash on Ubuntu on Windows, ls, it says VIRTUALENV_DIR. cd, it says /home/username. The cd .. etc changes the directory a little, but not another drive, nor the folder I would like it to find. float to int unexpected behaviour Sql Server 'Saving changes is not permitted' error ► Prevent saving changes that require table re-creation How can I improve Ubuntu overall system performance? Order discrete x scale by frequency/value What is the maximum or minimum tire width I can fit on my bicycle We often get questions on bicycles.stackexchange about fitting wider or narrower tires to bicycles. We have a large number of these with answers suiting the specific bike or style of bicycle in question, but I have been unable to find one that provides a generic answer covering off all things that need to be considered. For most bicycles, the answer is the same regardless of the specific style and sizes of tires in question.   This is intended as a canonical question that we can point closed questions to. I want to show that for $S,T\in B(H)$ bounded operators on Hilbertspace with $S\geq 0,T\geq 0$ and $ST=TS$,  we have $S+T\geq 0$, and $ST\geq 0$.  $T\geq 0$ means $(Tx,x)\geq 0$.    To me it seems that $((S+T)x,x) = (Sx,x)+(Tx,x)\geq 0$. But for $ST$ i like some help.   There is a theorem in Rudin (12.32) that says that for every $T\in B(H)$ we have an equivalence:   $T$ is positive $\Leftrightarrow$ $T$ is self adjoint, i.e. $T=T^*$ and $\sigma(T)\subset [0,\infty)$.   Since $S, T$ are self adjoint we know that $ST=TS = S^*T^* = T^*S^*$ ($ST$ is selfadjoint since $S,T$ commute). Also I read somewhere that if $S,T $ commute then $\sigma(ST)\subset \sigma(S)\sigma(T)$, but I am not sure if this is true. Clearly I cant prove it, but if it were true by this ''Theorem'' it would follow that $ST \geq 0$.   Is there a more straightforward way?