My old passport, in which ECR was stamped, expires in 2022. I have a Dubai work visa on this passport , but as I was told that it'd be a problem for me to travel to Dubai, I got a new passport with ECNR with a new passport number.  So will it be a problem for me to travel or do I have to transfer my UAE work visa to my new passport?
I have Dubai work permit mailed to me by my employer. It contains my old passport number which has ECR stamp. I then got a new passport issued without ECR stamping, but with different passport number. I have both passports.  Can I travel to Dubai now?
I was playing modern with the owner of a shop.  He played a  which he used to bring in 7 lands: 5 basic mountains and 2  cards.  He then claimed that each  triggered 5 times (killing me).  I am doubting this now.    Here is my reason: The two  cards entered at the same time as the other 5 basic lands.  That means when they were on the battlefield and their ability was in effect, the other 5 basic lands were already there too.  That seems too late to trigger a "Enters the Battlefield" effect.  So here is my question.  Am I right or wrong?  If I am wrong, please show me how the enter the battlefield effect can trigger for something that enters the battlefield at the same time as itself.  If I am right, please show me the rules that prove it.  I now have two shop owners that are convinced that this is a legal play.  If it is not legal, I am going to need clear rules citations to convince them.    Side note:  If I am wrong and it is a legal play, then I have a follow up scenario.  Say I have 2  cards and 5 other human cards in my graveyard.  If I play a spell that lets me return all of them to the battlefield at the same time (say ), would the  cards get 6 +1/+1 counters on them?    To me, if the trick that was played on me with  is legal then my  cards should also get the +1/+1 counters. A couple years ago I ran upon a  video demonstrating how researchers used x-rays given off by tearing tape off its spindle in hopes to miniaturized and cheapen future x-ray devices. As of late I have been wondering what interactions are causing this. I understood from the video that the electrons are being ejected due to the added energy to the system (peeling). This instinctively pointed me towards QED where interactions between electrons produce a photon for short periods. 1)Are those same photons from electron interaction the resulting x-rays? 2)Are the frequencies of the photons being limited by the $\Delta$energy$\Delta$time uncertainty, and/or $\Delta$position$\Delta$momentum uncertainty (at the vertex of the ripping tape, where high densities of electrons may accumulate) ? Note: I've also considered that the van der Waals force may come into consideration, not certain though. Looking for a QED based descriptions/answers. Any elementary references to this work is appreciated. For some reason, when I try to run any programs (I am trying to run a rootkit hunter: rkhunter) as root on the command line with sudo like this:  sudo rkhunter   Then I get this message:  You must be the root user to run this program   and then I get these lines:  sudo: error in /etc/sudo.conf, line 0 while loading plugin 'sudoers_policy' sudo: /usr/lib/sudo/sudoers.so must only be writable by owner sudo: fatal error, unable to load plugins   I try typing in just sudo but that comes up with the same three errors as above. Also in /etc/ sudo.conf does not exist.  I cannot run any other programs as root through sudo. It was also strange because I could run as root a day ago. I would like it to work.  Just so you know, I am the only user of this computer, and the administrator, running Lubuntu 14.04.1. Basic question about Merging faces I'm a beginner with blender so this is probably a very basic question.  I have this basic model below. I've ended up with a lot of unwanted geometry that I want to simplify. I wanted to start by merging the two selected faces below but no matter what I try to delete or merge it just destroys the geometry.  How can I achieve what I want to do (which is combining all of the unwanted geometry, either starting with this merge or some other way?) Proving that $a \mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb Z$ Given $m \in \mathbb{Z}$, let $m\mathbb{Z}$ denote the set of integer multiples of $m$, i.e. $m\mathbb{Z} := \{mk\mid k \in \mathbb{Z}\}$.   Now let $a,b \in \mathbb{Z}$ with $a,b$ not both $0$. Prove that $a\mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb{Z}$.   I am trying to write a proof for this, but I am unsure of what method to use. Also I am confused by $mk\mid k$, because wouldn't $m=1$ for this to be true. I'm new to QGIS and have no programming skills.  Is there an easy/uncomplicated way to create multiple buffers around a line feature? fake proof of $\forall a. \forall b. a = b \to 1 = 0$ I saw a less formal version of this fake proof that claimed to prove $2=1$ but because it assumed $a=b$ from the start I knew why it was wrong. It does seem however that the proof can be used to prove an certain implication I find very odd however. Clearly $\forall a. \forall b. a = b \to 0 = 1$ is false because I could pick $a$ and $b$ to be $1$ and then I would have derived 0 = 1 with no other conditions. So there must be something wrong with this proof that I am not seeing.  proof of $\forall a. \forall b. a = b \to 0 = 1$:  $a = b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (assume)\\ a+a=a+b  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x)=a+x)\\ 2a=a+b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  (a+a=2a)\\ 2a-2b=a+b-2b \,\,\, (x=y \to f(x) = f(y), f(x)=x-2b)\\ 2(a-b)=a+b-2b \,\, (distributive)\\ 2(a-b)=a-b \,\,\,\,\,\,\,\,\,\,\,\,\,\, (-b=b-2b)\\ 2=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x) =x/(a-b))\\ 1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x) =x-1) $  now that I have "proven" $\forall a. \forall b. a = b \to 0 = 1$ I can now pick a and b to be 1.  $ 1 = 1 \to 1=0 \,\,\,\,\,\, (a=1, b=1, above)\\ 1 = 1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\text{reflexivity})\\ 1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\text{modus ponens}) \\ $  So now it seems that I have managed to derive $1=0$. Clearly I messed up given that no one has been able to show $ZFC$ inconsistent and ostensibly this fits well within the most studied aspects of $ZFC$. Either that or a random guy that makes meme's online found the most profound result in mathematics ever.  So where have I missed the mistake? I am looking to replicate black &amp; white maps in QGis, using hatching patterns and greyscales to indicate different categories of a continuous variable (example below).  It is possible/straightforward to do so in QGis? At first sight I only see color gradients and symbol fills.   ). What the #$@&%*! is that called? Is there a name for the use of symbols in place of curse words, for example #$@&amp;%*!? How to put a variable in anchor tag Values of $\sum_{n=0}^\infty x^n$ and $\sum_{n=0}^N x^n$ Why does the following hold:  \begin{equation*} \displaystyle \sum\limits_{n=0}^{\infty} 0.7^n=\frac{1}{1-0.7} = 10/3\quad ? \end{equation*}  Can we generalize the above to      $\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ ?   Are there some values of $x$ for which the above formula is invalid?  What about if we take only a finite number of terms? Is there a simpler formula?     $\displaystyle \sum_{n=0}^{N} x^n$   Is there a name for such a sequence?    This is being repurposed in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: . Having line color vary with data index for line graph in matplotlib? GoogleApiClient.isConnected() return always false I have a map of an area and georeferenced photos. I need to be able to bring the pictures as points and then be able to click at a certain point and an image pop up.   I have revisited this issue, i managed to get this far but the pop up doesn't work.   See image: Javascript - Get Image height I need to display a bunch of images on a web page using AJAX. All of them have different dimensions, so I want to adjust their size before displaying them. Is there any way to do this in JavaScript?  Using PHP's getimagesize() for each image causes an unnecessary performance hit since there will be many images. How to quickly and easily create a grid with Adobe Photoshop? Suppose I need to draw a grid in Photoshop, each square being the same size, and a certain amount of pixels tall and wide, and the grid lines having a certain pixel width. Like this:     Currently I do this by drawing one square, copy+pasting+moving+merging it a few times to create one row, copy/pasting that a few times to create more rows above each other.  It works, but feels very clumsy, I'm sure there ought to be a more efficient way to do this?  What are some efficient methods for creating a grid of equally sized squares? How can I clip an image via a bezier-path? My options to link large image into AI file My .psd file has a resolution of 6m x 3m @ 150 DPI. This makes it 35433px x 5906px in size. Photoshop CS6 saves the file as .tiff or .psb only. When I first import into Illustrator I get an error that the file is not readable but I have tried many options. Last one is not supported by AI.  So I'm puzzled... is there a way to link large images into Illustrator? For $n \geq 2$, prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$ For $n \geq 2$ prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$  We need to use induction.   The Principle of Mathematical Induction, Theorem 4.2.1, states that $n_0 \in \mathbb{Z}$. For each integer $n \geq n_o$, let $P(n)$ be a statement about $n$. Suppose that the following two statements are true:   $P(n_o)$ $( \forall n \geq n_o)[P(n) \rightarrow P(n+1)]$   Then, for all integers $n \geq n_o$, that statement $P(n)$ is true.  Suppose,  $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$, the equation, is true for some natural number $k$.  $\frac{1}{2^2})*...*(1- \frac{1}{k^2}) = \frac{k+1}{2k}$  The induction hypothesis $P(k) \rightarrow P(k+1)$ must hold for any natural number $k, P(k),$ from the basis and inductive step.  $(\frac{1}{2^2})*...*(1- \frac{1}{k^2})*(1- \frac{1}{(k+1)^2}) = \frac{k+1}{2k}(1- \frac{1}{(k+1)^2})$  $\leftrightarrow$ $\frac{k+1}{2k}(\frac{(k+1)^2}{(k+1)^2}- \frac{1}{(k+1)^2})$  $\leftrightarrow$ $(\frac{(k+1)}{2k})( \frac{(k+1)^2-1}{(k+1)^2})$  $\leftrightarrow$ $( \frac{(k^2+2k+1-1}{(2k)(k+1)})$  $\leftrightarrow$ $( \frac{(k^2+2k}{(2k)(k+1)})$  $\leftrightarrow$ $( \frac{k+2}{(2)(k+1)})$  We need to demonstrate that   $(\frac{1}{2^2})*...*(1- \frac{1}{n^2}) = \frac{n+1}{2n}$  holds true for $n \geq 2$  The question is do I just substitute a number that is greater than two after I did the induction? I have a polygon layer in QGIS, and would like to represent the polygons like in the below image:    I have tried two different approaches, none of which are satisfactory.    Using a line pattern fill, the entire polygon is filled with the line pattern. I need a blank area in the middle which is not filled in.    Using a marker line fill, I am able to keep a blank area in the middle, but the markers are rotated to align with the edges of the polygon. I need the markers to preserve the same rotation along all edges, like in the example above.  Any ideas?