I want to add this into my Drupal site in a Drupal way!  &lt;!--[if IE]&gt; &lt;link type="text/css" rel="stylesheet" href="IE.css" media="all" /&gt; &lt;![endif]--&gt;   I know it can be done easily by adding the above tag directly into html.tpl.php but I don't want hack the Drupal core and I want to follow a Drupal way for this.  What would be the best possible solution?
The following methods can be used to add CSS for specific versions of IE:   Paste the code into html.tpl.php:  &lt;!--[if lt IE 9]&gt;&lt;link type="text/css" rel="stylesheet" src="/sites/all/themes/mytheme/less/ie.css"&gt;&lt;![endif]--&gt; Make use of the plugin  Use drupal_add_css() function   I have always used option #1 above, and adding yet another plugin for something so simple seemed crazy. BUT working with LESS and calling in statically means the file does not get processed by the LESS pre-processor, so the question:  How do you include a stylesheet using drupal_add_css() for specific versions of IE?
In theory, is it possible that the loss on the train set will increase together with the train set accuracy? (in the normal case the train set loss should go down) And vice versa, can the train loss decrease along with the accuracy?   Note: This is obviously different than the classic underfit / overfit cases which are easier to grasp. Getting coordinates by clicking on QGIS Canvas with PyQGIS? I was trying to think but cannot figure it out. For instance, if the interaction is small, for instance limited to a windows, the observers in each universe will see that the other goes in reverse. But what if they can start a more meaningful interaction? (for instance, if one observer crosses the window and moves to the other universe, will anything change? why and how fast? (at least at the beginning his thinking will move in the same time arrow than in his universe, but the physical processes around him will still violate the second law? Global and local variables in R ionice without effect Why $∀x(P(x)→Q)$ is equivalent to $∃xP(x)→Q$ Given $x$ occurs free in $P$. If $x$ does not occur free in $Q$, then $∀x(P(x)→Q)$ is semantically equivalent with $∃xP(x)→Q$. How to understand this statement. And also, need an example to show that the formula will not necessarily be equivalent if $x$ does occur free in $Q$.  I guess what struggles me is I'm not quite understand what does "free" mean in this case. But it would be better to have answer for the above question. I used MiKTeX 2.9 on windows. I have downloaded a package I want to install from the internet (manually, not through MiKTeX). The archive (*.lmza) containing the package was saved to a directory on my C:\ drive. Through MikTeX -> Maintenance -> Settings -> Packages -> Change... I can select the directory in which I have saved the archive containing the package. However, if I choose Packages shall be installed from a directory and choose the location of the package, I get this error message: C:\my\package\is\here does not seem to be a local package repository. What does it take for a directory/folder to become a local package repository? Restrict certain mods to certain worlds xslt processing on select attribute on apply-templates Inequality ${n \choose k} \leq \left(\frac{en}{ k}\right)^k$ This is from page 3 of  ().  Copying the relevant segment:     Stirling’s approximation tells us $\sqrt{2\pi n} (n/e)^n \leq n! \leq e^{1/12n} \sqrt{2\pi n} (n/e)^n$. In particular we can use this to say that   $$ {n \choose k} \leq \left(\frac{en}{ k}\right)^k$$   I tried the tactic of combining bounds from $n!$, $k!$ and $(n-k)!$ and it didn't work. How does this bound follow from stirling's approximation? MicroSD card is set to Read-only state. How can I write data on it? My text introduces multi-quibt quantum states with the example of a state that can be &quot;factored&quot; into two (non-entangled) substates. It then goes on to suggest that it should be obvious1 that the joint state of two (non-entangled) substates should be the tensor product of the substates: that is, for example, that given a first qubit $$\left|a\right\rangle = \alpha_1\left|0\right\rangle+\alpha_2\left|1\right\rangle$$ and a second qubit $$\left|b\right\rangle = \beta_1\left|0\right\rangle+\beta_2\left|1\right\rangle$$ any non-entangled joint two-qubit state of $\left|a\right\rangle$ and $\left|b\right\rangle$ will be $$\left|a\right\rangle\otimes\left|b\right\rangle = \alpha_1 \beta_1\left|00\right\rangle+\alpha_1\beta_2\left|01\right\rangle+\alpha_2\beta_1\left|10\right\rangle+\alpha_2\beta_2\left|11\right\rangle$$ but it isn't clear to me why this should be the case. It seems to me there is some implicit understanding or interpretation of the coefficients $\alpha_i$ and $\beta_i$ that is used to arrive at this conclusion. It's clear enough why this should be true in a classical case, where the coefficients represent (where normalized, relative) abundance, so that the result follows from simple combinatorics. But what accounts for the assertion that this is true for a quantum system, in which (at least in my text, up to this point) coefficients only have this correspondence by analogy (and a perplexing analogy at that, since they can be complex and negative)? Should it be obvious that independent quantum states are composed by taking the tensor product, or is some additional observation or definition (e.g. of the nature of the coefficients of quantum states) required?  1:  (bottom of p. 18) &quot;so the state of the two qubits must be the product&quot; (emphasis added). Explicitly finding the sum of $\arctan(1/(n^2+n+1))$ This is my first time through playing Elder Scrolls and initially I grabbed every quest that has come available...  At first I thought I wanted to be part of the Imperial Legion but I feel like I picked the wrong side and am now regretting my choice.  So...is there any way to un-join the Legion and join up with the Stormcloaks, or will I have to wait and join them next time? There are several Q&amp;A threads that explain how to download youtube videos using the terminal.  However, I would also like to learn how to extract the video's soundtracks as MP3 files--also using only the terminal.  Answers briefly explaining how to use youtube-dl or other similar utilities before explaining how to extract the MP3 would be ideal for the sake of having all the information in one place--even though this aspect has been covered in other posts. Inexact measurement and wavefunction collapse As is usually said, measurement of an observable $q$ leads to collapse of wavefunction to an eigenstate of the corresponding operator $\hat q$. That is, now the wavefunction in $q$ representation is $\psi(q)=\delta(q-q_0)$ where $q_0$ is result of measurement.  Now, in reality measurements are never exact, so the wavefunction should not have that sharp peak. Instead it should be broadened depending on precision of measurement. In this case, can we still introduce an operator, an eigenstate of which would the new wavefunction appear? Is it useful in any way? Or does the new wavefunction depend too much on the way it was measured so that each instrument would have its own operator? How would such opeartor look e.g. for a single-slit experiment? So I was taught that:  Kinetic Energy (of electron) = Energy (of photon) - Ionization Energy  If E(photon) = IE, then KE=0 of the electron.   What does this physically/theoretically mean?   My current thoughts/interpretation is that enough energy/force is applied to ionize the electron so it is 'sufficiently far' from the atom, and then I guess it just moves with whatever speed it is moving at with natural kinetic laws, since no more energy/force is being applied....?  Any clarifications would be greatly appreciated, thanks in advance. My group of 8 level 3 players are about to run through the Thundertree Ruins and will probably have a hand at fighting the dragon. My question stems from the fact that a single poison breath attack from the dragon will outright kill all but 1 player assuming they fail their saving throws.  I understand this encounter is supposed to be ridiculously deadly, but I don't want to be punishing my players for testing the lengths at which they can fight stuff. A fight should be challenging and they should be able to try and run away, but instant obliteration seems a bit much.  All PCs are level 3 (HP in parenthesis):   Wood elf moon druid  (27) dwarven frenzy barbarian (44) fire genasi archer ranger (37) (pseudodragon companion (12)) wood elf assassin rogue (27) human wild magic sorcerer (22) dragonborn fighter (29) human monk (24) human fighter (32) Are real guns used in movies? I am trying cmd --> MMC --> Server Manager and then connecting to my Server 2012 no GUI  but i keep getting the following error The WinRM client cannot process the request  I am unable to upload images right now  This is not related to other post "How to Remote Manage Windows Server 2012 in Server Manager?" because title and description of that post is quite different from what i am asking