Often when I am modeling I would like to align the view straight on to a face, perpendicular to the normal.  Is there a way to do this?
I would like to align the view straight on to a face, perpendicular to the normal. How do I align the 3D viewport to the face of a mesh?
I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that $|a+b| \leq |a|+|b|$. Any help would be appreciated :) I'm developing a PHP application and I'm wondering about the best way to include multi-language support for users in other countries.  I'm proficient with PHP but have never developed anything with support for other languages.  I was thinking of putting the language into a PHP file with constants, example:  en.php could contain:  define('HZ_DB_CONN_ERR', 'There was an error connecting to the database.');   and fr.php could contain:  define('HZ_DB_CONN_ERR', 'whatever the french is for the above...');   I could then call a function and automatically have the correct language passed in.  hz_die('HZ_DB_CONN_ERR', $this);   Is this a good way of going about it?  -- morristhebear. How to concatenate text from multiple rows into a single text string in SQL server? give a variable to an operator For a workflow tool I made a panel in which it displays each separate library that is linked in my file. For each library I want to add some buttons that does stuff to the library in question.  So in this example I have my panel which shows for example 2 libraries and for each one it creates a "reload" button.  Now my question is.. How do I make it so the reload button only reloads the library I want to reload? I would probably have to give the button an id or something so it knows for which library it was created...  my code so far:  import bpy  libs = bpy.data.libraries  class Reloadlib(bpy.types.Operator):     """Reloads the current Library"""     bl_idname = "my.reload_lib"     bl_label = "Reload"      def execute(self, context):         libs[?????].reload()         self.report({'INFO'}, 'Reloading ' + libs[?????].name)         return {'FINISHED'}  class LibPanel(bpy.types.Panel):     """Creates a Panel in the scene context of the properties editor"""     bl_label = "Scene Libraries"     bl_idname = "SCENE_PT_layout"     bl_space_type = 'PROPERTIES'     bl_region_type = 'WINDOW'     bl_context = "scene"      def draw(self, context):         layout = self.layout         scene = context.scene         layout.label(text="Libraries:")          for i in libs:             layout.label(text= i.name.replace(".blend",""))             row = layout.row()             sub = row.row()             sub.operator("my.reload_xref", icon = "FILE_REFRESH")  def register():     bpy.utils.register_class(Reloadlib)     bpy.utils.register_class(LibPanel)  def unregister():     bpy.utils.unregister_class(Reloadlib)     bpy.utils.unregister_class(LibPanel)  if __name__ == "__main__":     register()   Where the question marks are is where I should be able to make the difference between the first [0] or second [1] library in the libs[] array.  In future I would also like to ad an 'open' and an 'unlink' button for each library so I can make changes to the other file. But this works by the same principle as the reload button. Conclusively stop wake timers from waking Windows 10 desktop How can I get free or trial ExactTarget instance I was recently told (  ):     You cannot delete a class in production directly. You will need delete   the class from your sandbox and then deploy the deletions to your   production org.   When you deploy from sandbox to production, the   missing classes will come up in red and you can choose to deploy these   deletions to Production.   However, I'd like some clarification because finding information on how to do this is like pulling teeth, unfortunately.  Am I supposed to:   Delete the Trigger in Sandbox Make a deployment package in Sandbox Add the Trigger which is going to be listed in RED now Deploy (and that deletes it from Production)?   I am not seeing deleted Triggers in Red in Sandbox.  Are they NOT listed under "Triggers" for deployment, but elsewhere?   Frustrated = true.   Does this ONLY apply to Classes??  UPDATE:  I just tried again, with a CLASS, and it still doesn't show up at all as a Deployable package (red or not) after I delete it in the Sandbox.  UPDATE 2:  I was just able to delete a class, that I know was written via IDE from a third party, in both Sandbox and Production.  So now I'm wondering if it has to do with where it was originally deployed from (if from IDE, IDE can delete it - it from Sandbox, you have to jump through hoops).  :-( Prove that $f'(0)$ exists and $f'(0) = b/(a - 1)$ Problem:   If $f(x)$ is continous at $x=0$, and $\lim\limits_{x\to 0} \dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|&gt;1$, prove that $f'(0)$ exists and $f'(0)=\dfrac{b}{a-1}$.   This approach is definitely wrong:    \begin{align} b&amp;=\lim_{x\to 0} \frac{f(ax)-f(x)}{x}\\ &amp;=\lim_{x\to 0} \frac{f(ax)-f(0)-(f(x)-f(0))}{x}\\ &amp;=af'(0)-f'(0)\\ &amp;=(a-1)f'(0) \end{align}  I will show you a case why this approach is wrong:     \[f(x)= \begin{cases} 1,&amp;x\neq0\\ 0,&amp;x=0 \end{cases}\]   $\lim_{x\to0}\dfrac{f(3x)-f(x)}{x}=\lim_{x\to0} \dfrac{1-1}{x}=0$   but $\lim_{x\to0}\dfrac{f(3x)}{x}=\infty$,$\lim_{x\to0}\dfrac{f(x)}{x}=\infty$   Does anyone know how to prove it? Thanks in advance! I don't want to do this through trial and error, and the best way I have found so far was to start dividing from 1.     $n! = \text {a really big number}$   Ex. $n! = 9999999$   Is there a way to approximate n or solve for n through a formula of some sort?  Update (Here is my attempt):  Stirling's Approximation: $$n! \approx \sqrt{2 \pi n} \left( \dfrac{n}{e} \right ) ^ n$$  So taking the log:     $(2\cdot \pi\cdot n)^{1/2} \cdot n^n \cdot e^{-n}$       $(2\cdot \pi)^{1/2} \cdot n^{1/2} \cdot n^n \cdot e^{-n}$       $.5\log (2\pi) + .5\log n + n\log n \cdot -n \log e$       $.5\log (2\pi) + \log n(.5+n) - n$   Now to solve for n:     $.5\log (2\pi) + \log n(.5+n) - n = r$       $\log n(.5+n) - n = r - .5 \log (2\pi)$   Now I am a little caught up here. Is the matrix diagonalizable for all values of t? Proving a binomial sum identity $\sum _{k=0}^n \binom nk \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}$ Is there a way I can avoid imperial ships? Whenever I am traveling on foot, imperial ships will fly by and drop down troops for me to fight. In a normal situation, this is no big deal and I will fight or flee to continue with my travels.  However this will also occur when I am doing a quest that requires me to locate and pick up items. This is an inconvenience as I cannot pick up items while in combat and I have to kill them before I can continue.  Is there a way I can "hide" and allow the ships to by pass my location? To briefly put forward my question, can anyone beautifully explain me in your own view, what was the main intuition behind inventing the ideal of a ring? I want a clarified explanations in these points:   Why is the name "ideal" coined?. In English 'ideal' means "One that is regarded as a standard or model of perfection or excellence." Why did people gave the name of ideal to such group? And why are the ideals not present in the case of groups? And give me a very fantastic intuition and motivation behind the ideals and what are the roles served by them in advanced mathematics.   Thanks a lot, to every one. Suppose $\sum a_n$ converges absolutely and $\sum b_n$ converges. Give an example where the Cauchy product does not converge absolutely. Problem:  Suppose $\sum_{n=0}^{\infty} a_n$ converges absolutely and $\sum_{n=0}^{\infty} b_n$ converges. Give an example where the Cauchy product, defined as $\sum_{n=0}^{\infty} c_n$ where $c_n=\sum_{k=0}^{n} a_k b_{n-k}$, does not converge absolutely.  Question:  I have been scratching my head over this for a while. What would an example be? Product and sum of positive operators is positive 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? Cauchy Problem for inviscid Burgers' equation Consider the Cauchy Problem of finding $u(x,t)$ such that    $$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=0,x\in\mathbb{R},t&gt;0$$ $$u(x,0) = u_0(x), x\in\mathbb{R}$$ Which choices of the following functions for   $u_{0}$ yield a $C^{1}$ solution $u(x,t)$ for all   $x\in\mathbb{R},t&gt;0$         $u_{0}(x)=\frac{1}{1+x^{2}}$   $u_{0}(x)=x$   $u_{0}(x)=1+x^{2}$   $u_{0}(x)=1+2x$.      If I use the existence and uniqueness theorem for Cauchy problem i get the corresponding determinant is non zero so all are true according to me. But in answer key only option 2nd and 4th is given. Please help me to solve the problem. Thanks a lot. Can screen layouts or workspaces be locked? Today I started having the following issues in Pokémon GO:   I can't spin pokéStops and get rewards When I catch pokémon, even the lowest CP ones, they escape every time. I've tried using razz berries, that doesn't work. Neither does using great balls.   I do have Internet. I'm not on wifi. I can send messages and browse Internet.  NOTE: I did try to modify my gps location and I was unsuccessful. I only got a pokémon or two and deleted them immediately. I do apologize for trying to hack it but I'm now fully legit. Placement of the asterisk in pointer declarations I've recently decided that I just have to finally learn C/C++, and there is one thing I do not really understand about pointers or more precisely, their definition.  How about these examples:   int* test; int *test; int * test; int* test,test2; int *test,test2; int * test,test2;   Now, to my understanding, the first three cases are all doing the same: Test is not an int, but a pointer to one.  The second set of examples is a bit more tricky. In case 4, both test and test2 will be pointers to an int, whereas in case 5, only test is a pointer, whereas test2 is a "real" int. What about case 6? Same as case 5? The square roots of different primes are linearly independent over the field of rationals