is fine, but there is a small problem: I can't place the table where I want to. For example  \documentclass{article} \usepackage{makecell} \setcellgapes{5pt}  \begin{document} This text should be above     \begin{table}[] \centering     \caption{Gruppo 6}     \label{my-label} \makegapedcells \begin{tabular}{ | >{\centering}p{2.75cm} | *{3}{>{\centering\arraybackslash}p{1.75cm} |}}     \hline\hline \textbf{Title / Title} & \textbf{X}    & \textbf{Y}    & \textbf{Z}    \\   \hline\hline \textbf{A}  & a\%           & b\%           & c\%           \\  \hline \textbf{B}  & d\%           & e\%           & f\%           \\  \hline \textbf{C}  & g\%           & h\%           & i\%           \\  \hline \textbf{D}  & j\%           & k\%           & l\%           \\  \hline \end{tabular}     \end{table} \end{document}

How to influence the position of float environments like figure and table in LaTeX?   This is a general question and should collect useful answers for all users. I hope we can use this as a reference


I am trying to get a small PDF output from pdftex.  From  I found the useful command  gs -sDEVICE=pdfwrite -dCompatibilityLevel=1.4 -dPDFSETTINGS=/ebook -dNOPAUSE -dQUIET -dBATCH -sOutputFile=small.pdf big.pdf   that allows me to set the image resolution for the PDF. But this command removes all the document hyperlinks generated by \cite and \ref in LaTeX.  So, how can I decrease the PDF size but keep the hypertext information?

I usually compress my tex PDFs with ghostscript:  pdflatex -shell-escape file.tex  gs -sDEVICE=pdfwrite        \    -dCompatibilityLevel=1.4   \    -dPDFSETTINGS=/prepress      \    -dNOPAUSE -dQUIET -dBATCH      \    -sOutputFile=file_compressed.pdf \     file.pdf   Normally that works well, but if I have relative links to local files, the links don't work afterwards.  Minimal example:  \documentclass{beamer} \usepackage{hyperref}  \begin{document}     \begin{frame}         \href{run:./audio.mp3}{test}     \end{frame} \end{document}   The link works in file.pdf but not in file_compressed.pdf. Is there any pdflatex or ghostscript setting that retains the link in the compressed version?  I use Acrobate Reader 10.1.14 for viewing, Ghostscript 9.10 and pdfTeX 3.14159265-2.6-1.40.15 (TeX Live 2014) on OS X 10.9.5.    Update: The breakage of hyperlinks had nothing to do with compressing the PDFs. The breakage happened for all re-processing of PDFs containing hyperlinks in its input PDF.


I am on CentOS 6.6.  I have a program that runs analysis and outputs its data in html files.  The file structure is  dir1/dir1.html dir2/dir2.html dir3/dir3.html   I want to open all three html files from a bash one liner, e.g.  for i in dir*; do firefox --new-tab $i/${i}.html; done   This opens the files individually, such that the first html file must be closed before the second one opens.  If I try to send it as a background process, e.g.   for i in dir*; do firefox --new-tab $i/${i}.html&; done   I get:   -bash: syntax error near unexpected token `;'   I also tried the same method as in  post, but it still opens them individually.  How can I open these all in the same tab using a little bash one-liner?

What's the difference between executing multiple commands with && and ;?  Examples:  echo "Hi\!" && echo "How are you?"   and  echo "Hi\!"; echo "How are you?"


I have a notebook with 2 graphic cards.  How can I install the HD4000-3000 drivers and GeForce 710M without GT-GTX?  (Yes I know the system settings > proprietary drivers but this option is OUTDATED!)  And when I'm launching csgo with this easy install I'm getting an OpenGL driver error telling to update drivers etc.

I just ordered the Nvidia GTX card. I have a dilemma, though. Should I keep using the driver which is available in "additional drivers" in Ubuntu, or should I install the driver from the Nvidia site?  So which driver is the best for me?


We did maintenance on our SharePoint server which included 40 updates (6 product updates and 34 security updates) to our SharePoint Enterprise Server 2013.  Now when the users go to any list on any browser (Chrome, FireFox, IE) instead of the list displaying, they get the following error:  TypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefined TypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property 'replace' of undefinedTypeError: Cannot read property '_events' of null   If they browse with a mobile device, it displays fine.

Is anyone here experience problems in SharePoint after installing the latest JAN 2016 on 13th Jan 2016 (Western Australian Time) updates for SharePoint 2013. I am guessing it could be KB3114503 update giving me problems.  I am getting the same problem on other machines after running the same updates too.  The problem I am receiving is, on every list view I get this error message below:  TypeError: Unable to get property 'replace' of undefined or null reference   TypeError: Unable to get property 'replace' of undefined or null referenceSys.ArgumentNullException: Sys.ArgumentNullException: Value cannot be null. Parameter name: element


This is a homework question. The scenario is as follows: A dishwasher can accommodate 20 plates in one loading. With every loading, there is a 20% chance that 1 plate will not be cleaned properly. With X representing the number of plates that are not cleaned properly, find P(2X > 4). n = 20, p = 0.2  I have no idea how to find P(2X > 4). Any hints or thoughts on this is greatly appreciated

This is a Binomial question. I am required to find P(2X>4), and given that X~B(10,0.2), how do I convert it to 2X~B(n,p)?  Have tried converting from P(2X>4) to P(X>2), but understand that it doesn't work that way.  Do i have to use integration or any formula that i should refer?


Let $U$ be bounded, with a $C^1$ boundary. Show that a ''typical'' function $u\in L^p(U) (1\leq p < \infty)$   does not have a trace on $\partial U$. More precisely there does not exist a bounded linear operator   $$T:L^p(U)\rightarrow L^p(\partial U) $$   such that $Tu=u|_{\partial U}$   whenever    $u\in C(\bar{U})\cap L^p(U)$

This question is from PDE by Evans, 1st edition, Chapter 5, Problem 14. It has been posted  previously, however, I cannot quite put all the information together from the responses there. Hopefully you can help me now. The problem is as follows:     Let $U$ be bounded with a $C^1$ boundary. Show that a ''typical''   function $u \in L^p(U) \ (1 \leq p < \infty)$ does not have a trace on   $\partial U$. More precisely, prove there does not exist a bounded   linear operator      \begin{equation} T:L^p(U) \to L^p(\partial U) \end{equation}      such that $Tu = \left. u \right|_{\partial U}$ whenever $u \in C(\overline{U}) \cap L^p(U)$.   So what we're trying to find is a bounded sequence of functions that go to infinity at the boundary, correct? Because then $Tu = \left. u \right|_{\partial U}$ is undefined.  I like the idea of using the $\mathrm{dist}(x, \partial U)$ function to do this. I was thinking about defining the function  \begin{equation} u(x) = \frac{1}{\mathrm{dist}(x,\partial U)}, \end{equation}  which must be in $L^p(U)$ since $U$ is bounded (hence integral of $\varepsilon$ over $U$ is finite, right?). But this function is not continuous at the boundary, and so $u \notin C(\overline{U})$. Can this be modified somehow?  Another option is to use that the boundary is $C^1$. So for each $x^0 \in \partial U$ there exists an $r > 0$ and a $C^1$ function $\gamma : \mathbb{R}^{n-1} \to \mathbb{R}$ such that $U \cap B(x^0,r) = \{ x \in B(x^0,r) \ | \ x_n > \gamma(x_1,\ldots, x_n) \}$. So around every point on the boundary we can find a ball where the n'th coordinate is greater than the $C^1$ function $\gamma$. How can we use this?


Quick question regarding editing PostGIS data in ArcMap 10.0.  Far as I can tell, this can only be accomplished through third-party plugins like GISSquirrel or ST-Links.  Does anyone know of an alternative way that omits third-party involvement?  As an alternative, is it possible to edit remotely-hosted data served up in ArcGIS via a Geoserver WFS?  Again, the research I've done indicates you can view the data in ArcGIS, but no edit capabilities.   I have my data stored remotely in PostGIS and served through Geoserver.  Of course, I'm trying to find a way that allows editing of the data within the ArcGIS environment.  Well aware that I can do this in QGIS, just wondering why I can't find a solution for ArcGIS.

I have been working with QGIS/PostGIS for a while now, without problems, but now I have to change QGIS for ArcGIS Desktop 10.1.  I am having trouble adding a layer from PostgreSQL and being able to edit that layer. I want to work directly with the database, and if I change something in the map it should be reflected on the database.  Can I do this without a Geodatabase?


PHP populates div using result set from SQL with the following code: <div id="products-div"> <?php while ($row = $result->fetch_assoc()) {    generate_category_product_div($row); } ?> </div>  After i update this div with the following jquery/ajax post it seems that the click event handler is missing from any div in data. It only works when i reload the page. $(function() {    $('.category').on('click', function() {        var category = this.id;        $.post('category.php', {          category: category,       })       .done(function (data) {          $('#products-div').html(data);       });    }); });  Any help will be much appreciated.

I have a bit of code where I am looping through all the select boxes on a page and binding a .hover event to them to do a bit of twiddling with their width on mouse on/off.  This happens on page ready and works just fine.  The problem I have is that any select boxes I add via Ajax or DOM after the initial loop won't have the event bound.  I have found this plugin (), but before I add another 5k to my pages with a plugin, I want to see if anyone knows a way to do this, either with jQuery directly or by another option.


I have Ubuntu 14.04 dual boot with Windows 8  So I decided I need to reinstall Ubuntu to increase its space.   I went to Windows 8 and deleted the Ubuntu partitions and now I can't turn on my computer or anything and I keep getting  every time.  My highest priority is not to touch other partitions.  I already have tried this:

I would like to see a full how-to guide on how to install Ubuntu.


I understood Proliferate to be an ability that allows you to, given any number of permanents and players of your choosing, add another counter of each existing counter type that is already on there. Like adding another poison counter to a player that already has one, or buffing a creature with a +1/+1 counter even more, or giving an opponent's creature with at least one -1/-1 counter another one. There are some edge cases I'm unsure of:  What if an attacker is exalted (+1/+1 until end of turn) and they ? Do they get an additional +1/+1? Does it affect combat in the current turn? If they get an additional +1/+1, does it wear off at the end of the turn like the exalted buff? Similar situation with : could you proliferate the +1/+1 buff it gives other creatures?

Does 's ability count as a +1/+0 counter? I want want to use it for other cards such as 's ability ("each creature you control with a +1/+1 counter gets first strike"). I cant find confirmation anywhere.


No matter how many ways I think of it - just how Socrates is a Man, any differentia is a Quality. So for eg, one much used example is 'rational' - man is a rational animal, where man is species, animal the genus and rational a differentia. Isn't rational a Quality? Doesn't it come under the category of Quality?

I will make it clear beforehand, that the little bit I know about Aristotle is from Aquinas (secondary material) and Deleuze (Difference and Repetition), so there may be some misunderstandings on my part. Anyway, the reason why Aristotle cannot say that Being is the uppermost genera (above all the categories) is because of the difference. Basically an animal can be rational or winged - but substance is only the animal, man or a bat. However, there's difference (rational or winged) which is still a Being but doesn't fall into the categories. However, isn't rational or winged just Quality?


On stackoverflow I have Emacs as a favorite tag, and many tags involved with HTML ignored. However, there could be the scenerio of an HTML user wanting advice about Emacs, I would still like to see those questions. Will I still see the questions tagged both Emacs and HTML?

Can you mark the same tag as both interesting and ignored? If you can, what is the resulting appearance of a question with that tag?


I want to find a bijective function from $(\frac{1}{2},1]$ into $[0,1]$. So, What is a bijective function $f:(\frac{1}{2},1]\to[0,1]$?

How to define a bijection between $(0,1)$ and $(0,1]$?   Or any other open and closed intervals?   If the intervals are both open like $(-1,2)\text{ and }(-5,4)$ I do a cheap trick (don't know if that's how you're supposed to do it): I make a function $f : (-1, 2)\rightarrow (-5, 4)$ of the form $f(x)=mx+b$ by \begin{align*} -5 = f(-1) &= m(-1)+b \\ 4 = f(2) &= m(2) + b \end{align*} Solving for $m$ and $b$ I find $m=3\text{ and }b=-2$ so then $f(x)=3x-2.$  Then I show that $f$ is a bijection by showing that it is injective and surjective.


I really want to make sushi (and sashimi). However, I'm concerned about the health risks of eating raw fish.  I made a research and I found this:     EU   regulations forbid the use of fresh raw fish. It must be frozen at   temperatures below −20 °C (−4 °F) in all parts of the product for no   less than 24 hours.[66] As such, a number of fishing boats, suppliers   and end users "super-freeze" fish for sushi to temperatures as low as   −60 °C.[67] As well as parasite destruction, super-freezing also   prevents oxidation of the blood in tuna flesh, thus preventing the   discoloration that happens at temperatures above −20 °C.[68]   Source: .  So I wonder if it's safe if I freeze the raw fish in my refrigerator's freezer for a day or two? Any advice on making safe sushi?

Is it safe to consume raw fish purchased at the fish counter of a typical suburban grocery store? If not, do you have any tips for finding sushi grade fish. Also, are there requirements for safe sushi/sashimi beyond the use of safe fish?


How to enable boot messages, kernel messages and other logs of various services to be printed on screen during boot up?

Is there a way to easily turn on/off showing the boot messages (loading the services) when Ubuntu starts? Is it something in Grub2?  I am running 10.04.


Because of mildew we waste breads. How can I save breads from mildew?

I always keep bread around, home-made or otherwise.  Is there a way to keep it from molding, or at least delay the process?  I've tried the fridge with some success (extra day or two), but I don't like my bread that cold to eat.  I've tried zip-top bags with no success (seems to mold faster).  Bread-box didn't seem to make any difference in time.  Thanks


On my visitor visa to the UK application form,the question “have they always had the same nationality?”which is under parent details  I put No instead of Yes..and I have already paid and submitted my application but I haven’t been to the embassy..what can I do please advice ..and I plan on going to travel alone and not with any parent of mine

I'm from the Philippines, and I plan on travelling to the UK with my sister to visit a relative.  I have just submitted my visa application and have already paid online, when I noticed that I've made a clerical error. In the 'visit or stay' portion of section "Do you have any friends or relatives in the UK?", I clicked the 'no' box, instead of 'yes'.  My sister's application is correct. She selected 'yes'. Our appointment is due next month. I am planning to email them about this, and also inform them on the day of application submission.  Is this the right path to take?


in matrix multiplication  if $AB=I$, do $A$ and $B$ need to be inverses of each other? Can $BA$ not equal I for the same two matrices?

If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.    I do not understand anything more than the following.   Elementary row operations. Linear dependence. Row reduced forms and their relations with the original matrix.   If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?  P.S.: Please avoid using the transpose and/or inverse of a matrix.


I want to convert float value i.e 3.14159267 in std::string I don't want to use sprintf  #include <iostream>   // std::cout #include <string>     // std::string, std::to_string  int main () {   std::string pi = "pi is " + std::to_string(3.14159267);   std::cout<<pi<<std::endl;   return 0; }   I got result as      3.141593

In my  I was printing a double using cout that got rounded when I wasn't expecting it.  How can I make cout print a double using full precision?


Suppose the set theory we are working with implies that $\forall x \forall y \exists(x \cup y)$.  Would that implication alone further imply the existence of an arbitrary union?:  $$\forall S\exists(\bigcup_{x\in S}S)$$  As such could be defined as:  $$\bigcup_{x\in S}S=x_1\cup x_2\cup x_3\cup \dots \forall x_i\in S$$  This is trivially true for finite sets. However, would infinite sets complicate the matter?  I should also point out that I understand that most set theories (such as ZF) first define the arbitrary union (perhaps even to avoid this question), which would retroactively account for binary unions. For the purpose of this question, assume that the arbitrary union is not directly implied within the axiomatic system we're working with, meanwhile the binary union is.

In ZFC, the axiom of union is the statement that the union of any set exists. Suppose we omit this axiom, and instead add the axiom that binary unions exist. Can we still derive the full union axiom?


I have a door that was installed and the contractor put in some spacer wood. I didn't paint them and these were exposed to some rain (I'm in Los Angeles where it rains a few days a year)  Here's a picture of the wood up close. Are those black spots mold? I tried to rub them off and they wouldn't come off. Do I need to treat it or can I just paint over that?

Back in the day I've heard that bleach is a good way to kill mold. The other day a home inspector suggested that Dawn dish soap actually works much better than bleach. I've heard from another home inspector to use the store-bought mold killer.  Does anyone know what really works and what doesn't for a DIYer?


How do you figure out which OpenVZ contain is under attack from a dDoS?  I know it is an attack because the b/w and incoming traffic shot WAY up.   Can this be done with netstat? Are some attacks not going to show up on netstat like UDP if they hit a port with no service running? Is there a monitoring service I could maybe install on the host node?

This is a  about DoS and DDoS mitigation.   I found a massive traffic spike on a website that I host today; I am getting thousands of connections a second and I see I'm using all 100Mbps of my available bandwidth.  Nobody can access my site because all the requests time out, and I can't even log into the server because SSH times out too!  This has happened a couple times before, and each time it's lasted a couple hours and gone away on its own.  Occasionally, my website has another distinct but related problem: my server's load average (which is usually around .25) rockets up to 20 or more and nobody can access my site just the same as the other case.  It also goes away after a few hours.  Restarting my server doesn't help; what can I do to make my site accessible again, and what is happening?  Relatedly, I found once that for a day or two, every time I started my service, it got a connection from a particular IP address and then crashed.  As soon as I started it up again, this happened again and it crashed again.  How is that similar, and what can I do about it?


I want to rotate my track to form different track elements in this dive coaster, but the curve won't allow me to add banking/rotation.I want to achieve this type of effect, where the track rotates around the curve.    Here is my current set up. I can't complete this immelmann without banking the track out.

Is there an easy way to control how the extruded width of a curve object is oriented, for example like this     or this     The curve in the images above has only two control points, and I got the effect by rotating one of the points. However, for a curve with lots of control points it would be somewhat difficult or at least tedious to achieve this. Is there a simpler way?


I am at UK on work permit for 88 days, i want to invite my wife and kid from India to spend 2 weeks in UK to see all historical places here, which VISA i have to opt for them.

I am resident of India and just got UK Business visa ( 6 months, multiple entry) .. I am going for a week on business trip. I want to take my wife along with me and at the end of trip.. I want to extend my trip for 3 days and we want to spend some time on leisure.   What visa I should apply for my wife.. visitor visa? should she mention in the visa form that she is going along with me and staying with me while I am on work and later we are spending time on leisure.   While filling my business visa form, there was a field asking if anyone else is going with me.. I selected no.. as there were no plans.  Also is it ok to spend time on leisure on UK Business visa


How do we prove that $\Big[\sqrt n \Big]+ \sum_{j=1}^n \bigg[ \dfrac nj\bigg]$ is an even integer for all $ n \in \mathbb N$ ? (where  $\Big[ \space \Big]$ denotes the "greatest integer" function)

Let $n$ is a natural number. Prove that $$\left\lfloor\frac{n}{1}\right\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+....+\left\lfloor\frac{n}{n}\right\rfloor+\left\lfloor{\sqrt{n}}\right\rfloor$$   is even.   I used induction and the inequality $$x-1<\lfloor{x}\rfloor\le{x}$$ to prove it.  Is there any other, nicer way to do it?


In the alternate timeline in season 9 of Smallville, which we saw via Lois' memories, how did Zod and his followers have powers under a red sun?  Does blue kryptonite cause one to have abilities under a red sun, as Zod and his men were laced with blue kryptonite?  Keep in mind that in this alternate timeline Zod did not receive a transfusion from Kal -El.  So how did Zod and his army have powers under the red sun?

We see that in the 9th season of Smallville, both in the first and the ninth episode, the Kandorians - whose genes were laced with blue kryptonite - have powers (example, Alia).  How can they have their powers under a yellow sun? In the first episode Alia had all the powers of a Kryptonian under a yellow sun even after her cells were laced with blue kryptonite, and in the ninth episode she was seen using super speed under a yellow sun.  Also in the alternate timeline(where Lois went) Major Zod and his men had kryptonian abilities under RED sun but Kal- El did not. Why? How can this be possible? How can they have their powers under a yellow sun and red sun?


Please show me how I can do that. And I want to use many times. Help me write a macro so that I can use every time I need.  For example,   \begin{question}      Texts in the question field.  \end{question}     Update: This question is already solved by other linked questions.

I'm looking for pretty formatting for warnings, and others for some info or tips.  For the moment, I use the following formatting but maybe someone has one better idea.    To obtain this, I use the following code.  \documentclass[10pt,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage[x11names,svgnames]{xcolor} \usepackage{graphicx} \usepackage{fourier-orns} \usepackage[framemethod=tikz]{mdframed} \usetikzlibrary{calc}  \makeatletter % Default settings for warnings     \newcommand{\warningSymbol}{\raisebox{0.8\depth}{\danger}}      \definecolor{warningColorText}{named}{Red3}     \definecolor{warningColorLine}{named}{Red3}     \definecolor{warningColorBack}{named}{LemonChiffon1}     \definecolor{warningColorBackSymbol}{named}{white}  % Inline mode     \newcommand{\warning}[1]{%         \textcolor{warningColorText}{\warningSymbol{}\,#1}%     }  % Block mode     \tikzset{         warningsymbol/.style={             rectangle,             draw  = warningColorText,             fill  = warningColorBackSymbol,             scale = 1,             overlay         }     }      \mdfdefinestyle{warning}{%         hidealllines      = true,         leftline          = true,         skipabove         = 12,         skipbelow         = 12pt,         innertopmargin    = 0.4em,%         innerbottommargin = 0.4em,%         innerrightmargin  = 0.7em,%         rightmargin       = 0.85em,%         innerleftmargin   = 1.1em,%         leftmargin        = 0.85em,%         middlelinewidth   = .2em,%         linecolor         = warningColorLine,%         backgroundcolor   = warningColorBack,%         fontcolor         = warningColorText,%         firstextra        = {             \path let \p1=(P), \p2=(O) in ($(\x2,\y1-4)$)             node[warningsymbol] {\warningSymbol};         },%         secondextra       = {             \path let \p1=(P), \p2=(O) in ($(\x2,\y1-4)$)              node[warningsymbol] {\warningSymbol};         },%         middleextra       = {             \path let \p1=(P), \p2=(O) in ($(\x2,\y1-4)$)              node[warningsymbol] {\warningSymbol};         },%         singleextra       = {             \path let \p1=(P), \p2=(O) in ($(\x2,\y1-4)$)              node[warningsymbol] {\warningSymbol};         },%     }      \newmdenv[style=warning]{@Warning}     \newenvironment{Warning}{\let\warning\relax\begin{@Warning}}{\end{@Warning}} \makeatother  \usepackage{lipsum}  \begin{document}  \warning{IMPORTANT !} \lipsum[1]  \begin{Warning}      \lipsum[1] \end{Warning}  \end{document}


I am almost 30 years old. I got my bachelor's many years ago, and I've been enjoying myself while in the industry. I am a practical person.  However, for some reasons, I felt my technical skills were regressing, so I decided to go back to school, to enhance and update my skills.  I chose a topic that is quite abstract, which is stressful to me. Also, I feel threatened by grades and such ... which sometimes make me regret having put myself in such a "delicate" situation.  Should I be patient (there will be better days) ? Or should I go back to the industry and discover things myself?

I really love research. I started a research project in graduate school for my M.A. in English. My research topic is a very viable and interesting topic for a dissertation. I'll be 60 on my next birthday and I know that many Ph.D programs are traditionally geared toward younger students.I was wondering are there any programs that I can continue this research in without facing the issues of ageism in academia?


Is there any way or any script to permit only Google maps in my firewall and not the other Google apps like GMail or accounts.google.com?  The problem is that when I inspect the links, maps.google.com is redirected to google.com/maps and there is a possibility to accidentally allow GMail or other services if I dynamically permit google.com. I am using IPtables as firewall.  Any solution or idea how to solve this problem?

Suppose I want to block Facebook permanently. To do this, I have followed following processes:     #host -t a www.facebook.com   Sample output:     www.facebook.com has address 69.171.224.40   Find CIDR     #whois 69.171.224.40 | grep CIDR   Sample output:     CIDR:           69.171.224.0/19   To prevent outgoing access to www.facebook.com  Approach 1:     #iptables -I FORWARD -p tcp -d 69.171.224.0/19 -j DROP   Approach 2:     #iptables -I FORWARD -p tcp -d www.facebook.com -j DROP   Both Approaches work well. In approach 1, IP address may be changed so it will not block Facebook permanently. I don't know about approach 2 whether it will block Facebook permanently or not. If above approaches are not right way to block a domain permanently, how can I do it?


In The Deathly Hallows, Hermione uses dittany to mend Ron's arm after it got Splinched and part of it was scooped out: (emphasis mine)     “Harry, quickly, in my bag, there’s a small bottle labeled ‘Essence of Dittany’–" [...] The wound now looked several   days old; new skin stretched over what had just been open flesh.   In The Goblet of Fire, we find out that Moody has a piece of his nose missing, similar to what Ron had with his arm: (emphasis mine)     The mouth looked like a diagonal gash,   and a large chunk of the nose was missing.   Since this is similar to what happened with Ron, it should make sense that someone as skilled as Moody would think to apply Dittany immediately after... but he didn't. Why is this?

While reading Sorcerer's Stone, I just finished the chapter where Harry walks in on Snape and Filch bandaging Snape's leg.  In a previous chapter Neville falls off of his broom and we see that magic can easily heal his broken wrist.  If magic can heal an injury like broken bones, why wouldn't Snape heal his leg, which was just torn up by a 3-headed dog?  My initial thought was that only certain people specialize in healing magic, but while poking around this site I came across this question:   This shows that Snape had at least some healing abilities.  So why didn't he heal his own leg?  Are there limitations that prevented Snape from healing his own wounds?


Can I safely remove a usb device from a win 7 computer that is turned off? Or do I need to turn on the computer and "unmount" the drive 1st?

I have a portable USB flash drive that I use with my netbook.  When I'm on the go, it is convenient to be able to simply close the lid and remove my flash drive without first having to use Windows's Safely Remove Hardware routine.  The netbook changes to sleep mode (ACPI S3 mode) when I close the lid.  Is it possible to do this safely with Windows 7 Home Premium?  What about for hibernate (S4) and off modes?


Let's say I have following:  .wrap { height:400px; } .header { height:100px; } .footer { height:100px; }

   Is there any CSS solution to have .middle_row be dynamically stretched to 200px?

I am working on a web application where I want the content to fill the height of the entire screen.  The page has a header, which contains a logo, and account information. This could be an arbitrary height. I want the content div to fill the rest of the page to the bottom.  I have a header div and a content div. At the moment I am using a table for the layout like so:  CSS and HTML    #page {     height: 100%; width: 100% }  #tdcontent {     height: 100%; }  #content {     overflow: auto; /* or overflow: hidden; */ }

...

...

     The entire height of the page is filled, and no scrolling is required.  For anything inside the content div, setting top: 0; will put it right underneath the header. Sometimes the content will be a real table, with its height set to 100%. Putting header inside content will not allow this to work.  Is there a way to achieve the same effect without using the table?  Update:  Elements inside the content div will have heights set to percentages as well. So something at 100% inside the div will fill it to the bottom. As will two elements at 50%.  Update 2:  For instance, if the header takes up 20% of the screen's height, a table specified at 50% inside #content would take up 40% of the screen space. So far, wrapping the entire thing in a table is the only thing that works.


While solving some linear equation I Stopped on the following: How to rearrange coefficients and constants to find integer solutions for: $2xy+3x+3y=106$. If we take $x$ as a common factor from the first two terms then we get: $x(2y+3)+3y=106$. Any suggestion how to go farther..

How many integer solutions does $xy+9(x+y)=2006$ have? Here $x$ and $y$ are both integers.  My trying:        I have tried to solve this problem. But I have no idea to solve this. Please help


I just installed VScode with anaconda on Ubuntu 18.04 and after installation it won't open.  I did the same on Ubuntu 16 LTS but it worked fine. I tried to uninstall it from anaconda navigator and install it with  but it's still not working. When I click on the icon just nothing happens.   code --verbose   Gtk-Message: 17:32:24.579: Failed to load module "overlay-scrollbar"  Gtk-Message: 17:32:24.621: Failed to load module "canberra-gtk-module" [main 4:32:25 PM] Startup error: ,, Error: EACCES: permission denied, mkdir '/home/mohsen74x/.config/Code/CachedData',Error: EACCES: permission denied, mkdir '/home/mohsen74x/.config/Code/logs/20180528T173225' [main 4:32:25 PM] Lifecycle#kill()

My Mendeley-desktop application suddenly started crashing on Ubuntu 16.04. It says:  Another application has your Mendeley database open. Closing the other program or logging out of your computer may fix the problem.  It flashes for a second with this written on the login screen. I have another account 'Administrator' which I think has it opened since when I log in as 'Administrator', the application runs fine. Here is what I have tried:   reinstall in my account. remove it from 'Administrator' account and install in mine.   I cannot get it to work on my account. How does it get it to stop capturing the said database?  Note: I can open it in my account using  sudo nautilus   But I want to be able to run it with my account.


Is there an example of a function $f:[0,1] \times [0,1] \to \mathbb{R}$ such that for all $x \in [0,1]$ the function $\phi(y) = f(x,y)$ is continuous in $y$ and for all $y \in [0,1]$ the function $\psi(x) = f(x,y)$ is continuous in $x$ but the function defined by $$J(x) = \int_0^1{f(x,y)}dy$$ is discontinuous at some $x \in [0,1]$?  I'm asking this because a theorem about the continuity of parametric integrals requires the function $f$ to be continuous in both variables at the same time (continuous as a function from $X \to \mathbb{R}$ with $X = [0,1]\times [0,1]$, but I can't find a counterexample for the case mentioned above.   EDIT: I know the difference between a function being continuous in two variables separately and jointly. What I want to know if there is a function that is continuous in two variables separately and the integral above is discontinuous. The function $$f(x,y) = \frac{2xy}{x^2+y^2}$$ is an example for a function that is not jointly continuous. But the parametric integral of this function IS still continuous.

Consider a function $f(x,y):[0,1] \times [0,1] \rightarrow R.$ What is the difference between $f$ continuous in each argument and jointly continuous?


I have trouble drawing this diagram in tikz, could anybody help me?

I am stuck on drawing a P-V graph through latex. Could anybody help me code on this?   I appreciate your help!


I wrote a simple program in JAVA having 4 classes:  Main - > CarFaxApp  GUI / Logic - > CarFaxFrame  Object -> Car  Validator -> CarFaxValidator  Once the program is running and I start to enter the necessary data in the textbox's for the car (String vin, String make, String model, int year). Everything works fine, but when I hit buttonADD(), I get a error.  private HashMap<String,Car>hmCar; //Hashmap to hold Cars, the key pair value consists of (Vin, Car Object)  The error happens on hmCar.put(car.getVin(), car); which I dont understand why, I tried debugging up to the point and car does hold the correct values.  Here a screen shot of the error:

What are Null Pointer Exceptions (java.lang.NullPointerException) and what causes them?  What methods/tools can be used to determine the cause so that you stop the exception from causing the program to terminate prematurely?


The action of a relativistic point particle is its negative rest energy along its worldline, the parameter being its own proper time. $$ S = - mc^2 \int d\tau $$    Action is energy multiplied with time. How can I imagine the "action" of an unaccelerated point particle which does not seem to deploy any kind of energy?   Further explanation: An oven may emit a certain number of Jsec in a given time, but not an unaccelerated point particle.

I got badly lost in derivation of relativistic formulas for energy and momentum.  I stumbled upon relativistic action as follows (which should explain relativistic motion of a classical particle):  $$ S = \int Cds=C\int_{t_i}^{t_f}\sqrt{c^2-(x')^2}dt $$  Where $C$ is some constant (depends on what kind of physics we put in equation) and $s$ is relativistic interval. Later on Lagrangian $$ L(x')\equiv C\sqrt{c^2-(x')^2} $$ is used in derving relativistic energy and momentum.  I am familiar with Lagrangians and symmetry rules which connect energy and momentum to Lagrangian formalism. What I do not understand is this action - weren't action sum over time? Why all of a sudden it is sum over relativistic interval?


I use the version of Iceweasel on Debian Stretch GNU/Linux on an x86_64 machine. Sadly, I need to use Adobe's abominable Flash technology.  I know that the way you install the Flash plugin on Debian is using the flashplugin-nonfree package, after making setting the appropriate APT source settings. Well, I have, and it's installed and up-to-date - as far as the package goes, but, woe is me, when I open sites with Flash in Firefox it tells me that my Flash player is out-of-date!  So, first question:   Is this my fault somehow (e.g. misconfiguration or what not), or is it just that the Stretch package maintainers haven't updated their version?   And then:   Should I manually install the up-to-date plugin from adobe.com, or is there some other Debian-ish alternative? If the answer to the previous question is "yes", then - should I just overwrite /usr/lib/flashplugin-nonfree/libflashplayer.so with the downloaded version, and overwrite the local settings, like Adobe's .tar.gz's README suggests, or should I do something else and Debian-specific?

I run Debian wheezy with Iceweasel as browser and have installed the Flash plugin similar to . Today I read that not only Flash on Windows and Mac also Adobe Flash Player 11.2.202.350 and earlier versions for Linux are vulnerable to . When checking my version in Iceweasel > Tools > Add-ons I was surprised to see that it is an older version, even though I did all upgrades with apt-get.   Does Debian provide updates for the Iceweasel plugin Flash at all? Are they not automatically provided? Do I always have to manually update the plugin?


Thanks Eric Lippert's comprehensive comments in the comment section! I learn a lot from his comments. To share my learning with future readers, I've used parenthesis to wrap the wrong assumptions in my original question below so you can compare. And I have put the corrected assumptions in bold.   I read through this post , and understand that we should not call overriddable methods in base class constructor, because the base class constructor (wrong: will invoke the virtual member in the base class when we want it to invoke the overridden member in the subclass) will invoke the overridden member in the subclass, but before subclass has fully initialized.  Now I have a base class with an abstract property, (wrong: not a virtual method), which is a still virtual member except that it must be overridden, and I'm using it in my constructor. Then I have my subclass overriding this property. My expectation is that when I create a new instance out of my subclass, the base constructor will be using the overridden property in the subclass. And this is what actually happened.  I'm wondering if this is a good practice, because I still get the same warning saying "Virtual member call in a constructor", (wrong: although technically I'm not using virtual members). I don't see any obvious anti-pattern here, the only potential one is that if I further subclass my existing subclass, things might get confusing, but I'm marking my first subclass sealed.  Below is an executable code sample, it will output: Name1, Name3  And I get the above warning in the base constructor line this.NeededNames.Contains(name)  So my code sample is simple enough to work because its virtual member NeededNames does not rely on anything in the TestNameList class. But as Eric explained in comments, if a subclass happens to have states that are depended by its abstract/virtual members, when the base constructor runs, the overridden member in the subclass will be called, but it will not be ready yet because the states have not been set up by the subclass constructor. In this case, the subclass instance would not be initialized properly.   This sample is not a good practice although it works, because in reality we don't know what subclass could others create from our base class, and if a new subclass has states that need to be set up during the subclass construction, then we would not be able to instantiate it correctly from the base class constructor. Outside of this question, to make this sample work in a better way, I created an Initialize method in the base class that will be invoked after every subclass is instantiated.  public class Program {     public static void Main()     {         NameListBase list = new TestNameList();         list.Test();     } }  public abstract class NameListBase {     protected IList<string> ActualNames { get; } = new List<string>();      IList<string> Names { get; } = new List<string>     {         "Name1", "Name2", "Name3", "Name4"     };      protected abstract IList<string> NeededNames { get; }      protected NameListBase()     {         foreach (string name in this.Names)         {             if (this.NeededNames.Contains(name))             {                 this.ActualNames.Add(name);             }         }     }      public abstract void Test(); }  public sealed class TestNameList : NameListBase {     protected override IList<string> NeededNames { get; } = new List<string>     {         "Name1", "Name3"     };      public override void Test()     {         Console.WriteLine(string.Join(", ", this.ActualNames));     } }

I'm getting a warning from ReSharper about a call to a virtual member from my objects constructor.   Why would this be something not to do?


What is a good simple easily installable app to measure download speed on Ubuntu 14.04?

I want the terminal to output the current upload and download speeds of my network connection. I do not want to monitor the speeds in real time. I just want a "static" value to be able to write it in a file.  How do I achieve it?


I want to be able to make my if statement restart the program if "Yes" is inputted, end the program if "No" is imputed, and say "Capitalization is important" if neither "Yes" or "No" is inputted.   Scanner sc = new Scanner(System.in);     final String vowels = "aeiouAEIOU";     System.out.println("Enter your word:");     String word = sc.nextLine();     while (!word.equalsIgnoreCase("done"))     {         String beforVowel = "";         int cut = 0;         while (cut < word.length() && !vowels.contains("" + word.charAt(cut)))         {             beforVowel += word.charAt(cut);             cut++;         }         if (cut == 0)         {             cut = 1;             word += word.charAt(0) + "w";         }         System.out.println(word.substring(cut) + beforVowel + "ay");        Scanner keyboard = new Scanner (System.in);       System.out.print("Do you wish to convert another setence ");       String name = keyboard.nextLine();       if(name = "Yes"){           new Prog508a().launch;       }       if(name = "No"){       break;       }       else{       System.out.print("Capitalzation is important.");       }

I've been using the == operator in my program to compare all my strings so far. However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.  Is == bad? When should it and should it not be used? What's the difference?


I have a headless computer configured to use a dhcp client which is directly connected to my laptop using an ethernet cable. I'm virtually sure I have seen a one liner before (using dnsmasq?) which would enable me to launch a dhcp server on that interface on my laptop, so that the headless computer would connect to it. This oneliner kept running in the foreground and displayed dhcp requests, so that I would see when the headless computer connected and would know its ip address. Could you tell me what that oneliner is? I tried searching for quick/temporary dns setups and skimmed dnsmasq's manpage, but couldn't find anything.

I have a device that I need to connect to over SSH. The device is connected to my workstation via a direct ethernet connection. I'm attempting to assign the connected device an IP address somehow that I can SSH to, however all of the guides I'm finding have the user configure the IP from whatever device they're working with (namely Raspberry Pi's and so on). This is not something I can do with this device as I've no physical interface to work with.  I've been recommended to setup a DHCP server on my workstation so it would take care of this for me, however I've no idea how to configure it and none of the guides I have followed have been helpful. With that approach, I seem to have been able to bind an IP to the interface the device is connected on, however SSHing to that IP just brings me back to my own machine.  So my main question is: How do I assign an IP address to a device directly connected to my computer via ethernet cable without having any kind of access to the device (no interface, keyboard, monitor, etc). If the answer is to set up a DHCP server on my machine, how do I properly configure this and get an IP I can SSH to?  For reference, I am using Ubuntu 16.04 with OpenSSH installed. The device is also running some flavor of Linux and has SSH software installed, but again there's no way to interact with it except through SSH. I also do not have access to my router, so plugging the device in there and letting the router do the work is not an option.


Somehow, most probably from one or another usb-born Linux, I got cute little directory named C: located on my C: NTFS partition, due to its peculiar name I can not operate on its directory tree as deleting C: is a no no operation after all...  rmdir, del, move, copy all have no effect, and even show some signs of undefined behavior (for example del asks to delete \*). Checkdisk failed to complete. (I have a bad feeling about this.) Environment is up-to-date Windows 8.1 in English localization with full admin privileges. I will be doing full format of this disk in few days anyway and all data are backed up so crazy ideas have green light.

I have a song that Windows can't play because there is a question mark in the name of the file.  "Where Have All the Cowboys Gone?.ogg" // as an example   So I try to rename it and Windows complains whether I try it in Explorer or from command prompt.  Error I get when trying to copy, rename, or move is:     The Filename, directory name, or volume label syntax is incorrect   Is there a Windows way to force a rename in this case?   Update  I'll keep an eye on this question, but after 13 answers and many attempts (aside form 3rd party solutions) it seems that Windows can't do this (or at least my windows can't, no short names). So I'm accepting the answer which was my original solution anyway of using Linux. It would be nice to see Windows handle this somehow, so don't stop just because I've accepted this answer, the question still stands!


I'm trying to do something similar to this is Blender:     Animation here:   I can get the shape and animation to look quite similar based on the linked answer, but how do I actually add the text uv texture to my object?

Do you know I can do something like this?  How to animate text over a twisted mesh, or Text flowing along an infinity loop      I get that there is a shape (cube / rectangle / cylinder being extruded, made into the shoe, but, how am I best doing a texture, and then make it loop etc.


Blackboard and Zoom on Firefox both require microphone permissions even if the microphone is never to be used. You can mute your microphone in those services and on your physical computer but I am still uncomfortable with giving those services the microphone permissions (not just because you can forget to mute your mic). Is it possible to create a dummy microphone that Firefox will detect and I can give permissions to so I can get rid of this paranoia?

I run a CentOS Server with VNC enabled and some services require a microphone to work properly, but it doesn't have to be used for recording, just a microphone device available. So I was wondering if I can create a virtual recording device that can be used as a microphone, but it doesn't have to input any sound. I run alsa tools but doing modprobe sound-aloop says that the operation is not permitted.


I've recently met with this new law and the way my teacher worded is so off for me.  Shouldn't the equal and opposite reaction cancel out the action because they are equal?  Apparently not?  Why?

Given Newton's third law, why is there motion at all? Should not all forces even themselves out, so nothing moves at all? When I push a table using my finger, the table applies the same force onto my finger like my finger does on the table just with an opposing direction, nothing happens except that I feel the opposing force. But why can I push a box on a table by applying force ($F=ma$) on one side, obviously outbalancing the force the box has on my finger and at the same time outbalancing the friction the box has on the table? I obviously have the greater mass and acceleration as for example the matchbox on the table and thusly I can move it, but shouldn't the third law prevent that from even happening? Shouldn't the matchbox just accommodate to said force and applying the same force to me in opposing direction?


I'm looking for a proof of this theorem by Dirac:  If a graph is $k$-connected for $k \ge 2$, then for every set of $k$ vertices in the graph there is a cycle that passes through all the vertices in the set.  I know a proof probably uses Menger's theorem but I haven't been able to find one or come up with one.

Let $G$ be a $k$-connected graph. Meaning, $G$ has no fewer than $k$ vertices, and for every set of $k-1$ or fewer vertices, if we remove them from $G$, the graph stays connected (Of course, $G$ itself is also connected).  I want to prove that for any $k>1$, if $G$ is $k$-connected, then every set of $k$ vertices is contained in a cycle.  I have tried some ways - mainly using induction by removing one of the vertices of the set from the graph, and/or using Menger's theorem to construct the cycle. But I always encounter problems with making sure that the cycle I'm building deosn't have repeating edges etc.  Help would be greatly appreciated :) Thanks!


Rudin's Theorem 3.3 states:  Suppose $\{s_n\}$, $\{t_n\}$ are complex sequences, and $\lim\limits_{n \to \infty} s_n = s$, $\lim\limits_{n \to \infty} t_n = t$. Then   (a) $\lim\limits_{n \to \infty} \left(s_n + t_n\right) = s + t$;   (b) $\lim\limits_{n \to \infty} cs_n = cs$,  $\lim\limits_{n \to \infty} \left(c + s_n\right) = c + s$, for any number $c$;   (c) $\lim\limits_{n \to \infty} s_n t_n = st$;   (d) $\lim\limits_{n \to \infty} \frac{1}{s_n} = \frac{1}{s}$, provided $s_n \neq 0$ ($n = 1, 2, 3, \ldots$), and $s \neq 0$.  I proved (a) and (b) without trouble. I'm trying to understand his proof of (c). He notes that for any $s$ we have: $$  s_n t_n - st = (s_n - s)(t_n - t) + s(t_n - t) + t(s_n - s). $$ My first question is: where does this come from? By brute-forcing the right-side, I can tell that this amounts to adding and subtracting $st$, $ts_n$, and $st_n$. But is there another reason why this holds? Rudin calls it an "identity,' but I've never seen it before. Then, given $\epsilon > 0$, he takes $N_1, N_2$, so that $n \geq N_1$ implies $|s_n - s| < \sqrt{\epsilon}$ and $n \geq N_2$ implies $|t_n - t| < \sqrt{\epsilon}$. So for $n \geq \max(N_1, N_2)$, we have \begin{align*} |(s_n - s)(t_n - t) - 0| = |s_n - s||t_n - t| < \sqrt{\epsilon} \cdot \sqrt{\epsilon} = \epsilon,  \end{align*} so $\lim\limits_{n \to \infty} (s_n - s)(t_n - t) = 0$. Rudin then claims that the remaining terms in the first identity converge to $0$ by using $(a)$ and $b$. The first term converges, as we just proved, and the second and third are scaled versions of $t_n$ and $s_n$, so they converge. (I'm not sure how to word this exactly, since I need to have convergence first to use this identity, but I'm using the identity to prove they converge, so this is a very circular argument. I'm trying to figure out how best to word it.) With that aside, we have: \begin{align*} \lim\limits_{n \to \infty} [(s_n t_n - st) & = \lim\limits_{n \to \infty}  (s_n - s)(t_n - t) + s(t_n - t) + t(s_n - s)] \\ & = \lim\limits_{n \to \infty} (s_n - s)(t_n - t) + \lim\limits_{n \to \infty} s (t_n - t) + \lim\limits_{n \to \infty} t(s_n - s) \\ & = 0 + \lim\limits_{n \to \infty} (st_n - st) + \lim\limits_{n \to \infty} (ts_n - ts) \\ & = s \lim\limits_{n \to \infty}t_n - st + t \lim\limits_{n \to \infty} s_n - ts \\ & = st - st + ts - ts \\ & = 0 \end{align*} So $\lim\limits_{n \to \infty} (s_n t_n - st) = 0$. I'm trying to figure out exactly how this implies the result, because I can't just invoke part (a)/b to break the limit apart, as that requires convergence of $s_n t_n$, which is what I need. I think, however, it can be used to prove the fact directly. Given $\epsilon > 0$, we use this limit to find $N$ so that for $n \geq N$, we have $|s_n t_n - st| < \epsilon$ for all $n \geq N$, so $(s_n t_n) \to st$. As a general principle, am I allowed to just "break it apart," or was that just a coincidence? I would appreciate any help and any feedback on what I've written above.

I intuitively understand proof with limits, but I'm not sure on how to write a formal proof for this example.  For each $n \in \mathbb{N}$, let $a_n$, $b_n$ be real numbers. Also, let $a_{\infty}$, $b_{\infty}$ be real numbers. Suppose that $\lim_{n\to \infty} a_n = a_{\infty}$ and $\lim_{n\to \infty} b_{n} = b_{\infty}$.  Prove that $\lim_{n\to  \infty} (a_n b_n) = a_\infty b_\infty$.


I failed in PhD candidacy and planning to apply in a new graduate school. Should I mention or hide the failure in my SOP as it is a negative in my profile. Please help

I failed my written PhD qualifying exam twice in Computer Science from a middle-ranked university. Instead of kicking me out, I can get a terminal master's degree. Please note that this will be my second master's degree.    Now I am planning to re-apply to PhD programs at other Universities. My questions are:   Since I am getting a master's degree, is it okay not to mention about failing in the PhD qualifying exam in statement of purpose (SOP)? Please note that the Graduate committee informed that it will not be reflected in the transcript. If the answer is yes for the previous question, then my advisor is ready to give a strong recommendation, so can I ask him not to mention about failing in the PhD qualifying exam in his recommendation letter?   Please note that the written qualifying exam is about getting good grades in 4 subject areas of Computer Science, it has nothing to do with research. This is one of the main reasons I do not want to mention in SOP, I wanted to highlight my research potential saying that I submitted a 10-page research paper in a top tier conference in one year. I have industry experience of more than 5 years.  Please help, since my goal is to get into academics, help students and do independent research. I want to pursue my dream irrespective of this setback.


What does the .blend1 file do? It automatically appears every time you create a .blend file. I only use the .blend file. When I double-click on the .blend file, it opens Blender. When I double-click on the .blend1 file, it doesn't have a software to run the file. What is the .blend1 file used for?

When I click save or hit Command + S what appears to be a new version of the document is created. So if I save 2 times, I get 2 documents, filename.blend1 and filename.blend2.  What exactly are these files and why is Blender creating them?


I'm trying to use @beforeSuite and @AfterSuite to run my test in one browser instance. however it run the first the first test class but fail with null pointer exception when it the second class.  Here is my code below:  LaunchBrowser.java class  public class LaunchBrower {      protected WebDriver driver;      public WebDriver getDriver() {         return driver;     }      @Parameters({ "browserType", "appURL" })     @BeforeSuite     public void setUp(@Optional String browsertype, @Optional String appURL) {         System.out.println("Launching google chrome with new profile..");         driver = getBrowserType(browsertype);         driver.manage().timeouts().implicitlyWait(60, TimeUnit.SECONDS);         driver.manage().window().maximize();         driver.navigate().to(appURL);     }      private WebDriver getBrowserType(String browserType) {         if (driver == null ) {             if (browserType.equalsIgnoreCase("chrome")) {                 return  new ChromeDriver();             }             else if (browserType.equalsIgnoreCase("InternetExplorer")) {                 return new InternetExplorerDriver();             }         }         return driver;     }       @AfterSuite     public void tearDown() {         if (driver != null)             driver.quit();     }     }   LoginPageTest class  public class LoginPageTest extends LaunchBrower {      protected WebDriver driver;     private LoginPage loginpage;     private MyProfile profilepage;      Logger log = Logger.getLogger("Results:");      @BeforeClass(alwaysRun = true)     public void setUp() {                loginpage = PageFactory.initElements(getDriver(), LoginPage.class);         //loginpage =  new LoginPage(driver);         //driver=getDriver();     }      @Test(groups = "LoginPage")     public void verifyLogin() throws InterruptedException {         //LoginPage login =  new LoginPage(driver);         System.out.println("Sign In functionality details...");         //System.out.println("Sign In functionality details seee...");         Thread.sleep(10000);         //login.enterUserName("11111111");         Assert.assertEquals("11111111",loginpage.enterUserName("11111111"));         log.debug("Correct Username Pass");         //System.out.println("Correct username...");         Assert.assertEquals("fe9245db",loginpage.enterPassword("fe9245db"));         log.debug("Correct Password Pass");         loginpage.clickOnLogin();         log.debug("Login Pass");     } }   MyProfileTest java class  public class MyProfileTest extends LaunchBrower {      protected WebDriver driver;     private MyProfile profilepage;      @BeforeClass(alwaysRun = true)     public void setUp() {         profilepage = PageFactory.initElements(getDriver(), MyProfile.class);         //driver=getDriver();     }      @Test(groups = "Myprofile")     public void verifyMyprofile() throws InterruptedException {         System.out.println("My profile test...");         //MyProfile profile = new MyProfile(driver);         profilepage.ClickToggleLink();         profilepage.ClickMyProfile();     } }

What are Null Pointer Exceptions (java.lang.NullPointerException) and what causes them?  What methods/tools can be used to determine the cause so that you stop the exception from causing the program to terminate prematurely?


Can someone help explain this question to me?  Let $f : X \to Y$ be a function and $E \subseteq X$ and $F \subseteq X$. Show that in general:  1) $f(E-F) \not\subseteq f(E)-f(F)$  2) $f(E \bigtriangleup F) \not\subseteq f(E)\bigtriangleup f(F)$

Let $f:X\to Y$ be a function and $E\subseteq X$ and $F\subseteq X$.   Show that in general $f(E − F)\nsubseteq f(E) − f(F)$.   I have no idea about how to prove this; and could anyone please explain the basic theory of functions(by relating to this question. Thank you.


$\emptyset$ is called both open and closed. Trying to understand this. It's complement is $R$ for instance. Does that mean R is both open and closed? R is unbounded for sure, but every cluster point of R is contained in R (which satisfies the definition of being closed). R is closed looked from this point of view. This would make $$\emptyset$$ open.  But R could also be considered as open, since every point $p_o$ has a neighbourhood which is interior to $R$. Is this why $\emptyset$ is considered both open and closed, therefore $R$ is considered both open and closed?  Another example :  Considered the set $$S={x^2+y^2≥0}$$ Can this be considered both open and closed then?. This condition is satisfied for all $x,y$ therefore this set would be the $R$ itself. Complement of this set is the empty set, and since empty set is considered both open and closed as proved above. This set is also considered both open and closed.

I think this is a basic question, but it's hard to wrap my head around.


I have this problem in my textbook. I spent many days to come up with the solution. While I'm quite sure about the first part of my proof, I'm unable to verify the second part (which is related to the REARRANGEMENT of the sequence and is the most important). Some users have left comments and I responded to every of these comments, but they seem to stop interact with me.     Any $n\in\mathbb N$ can be written as the sum of a strictly increasing sequence of Fibonacci numbers.     My attempt:  Let $(f_n\mid n\in\mathbb N^*)$ be the Fibonacci sequence such that $f_1=1,f_2=2,$ and $f_{n+2}=f_{n+1}+f_n$.  Assume the contrary, then set $T$ consisting of positive integers that can NOT be expressed as the sum of a strictly increasing sequence of Fibonacci numbers is non-empty.  Let $m_0=\min T$, then $m_0=m+2$ for some $m$. As a result, $m,m+1\notin T$ and therefore can be expressed as the sum of a strictly increasing sequence of Fibonacci numbers.  We define: $E_n$ is an F-expression of $n\iff\begin{cases} \ t\in E_n\implies t\text{ is a Fibonacci number} \\ \sum_{t\in E_n}t=n \\ \end{cases}$  It' clear that $k\notin T\implies$ there exists an F-expression of $k$.  Since $m\notin T$, then there exists an F-expression of $m$ and, for any $E_m$, $2\in E_m$ (If not, $\{2\}\cup E_m$ will be an F-expression of $m+2$, and consequently $m+2\notin F$, which is a contradiction). Similarly, $1\in E_{m+1}$, then $E_{m+1}\setminus\{1\}$ is an F-expression of $m$, then $2\in E_{m+1}\setminus\{1\}$, then $2\in E_{m+1}$. Hence $1,2\in E_{m+1}$ for any $E_{m+1}$.  Let $m_1=\min\{n\in\mathbb N\mid f_{n+1}\notin E_{m+1}\}$, then $(n\leq m_1\implies f_n\in E_{m+1})$ and $f_{m_1+1}\notin E_{m+1}$. We have two cases in total:     I'm not sure about the correctness of the below part. Please help me check it out.    $m_1=2k$   As a result, $m+1=f_1+f_2+...+f_{m_1}+...=(f_1+f_2)+(f_3+f_4)+...+(f_{2k-1}+f_{2k})+...=f_3+f_5+...+f_{2k+1}+...=f_3+f_5+...+f_{m_1+1}+...$  Thus $m+2=1+(f_3+f_5+...+f_{m_1+1}+...)=f_1+f_3+f_5+...+f_{m_1+1}+...$, then $m+2$ can be expressed as the sum of a strictly increasing sequence of Fibonacci numbers, and consequently $m+2\notin T$, which is a contradiction.   $m_1=2k+1$   As a result, $m+1=f_1+f_2+...+f_{m_1}+...=f_1+f_2+...+f_{2k}+f_{2k+1}+...=f_1+(f_2+f_3)+...+(f_{2k}+f_{2k+1})+...=f_1+f_4+...+f_{2k+2}+...\implies m=f_4+...+f_{2k+2}+...$ $\implies m+2=2+f_4+...+f_{2k+2}+...=f_2+f_4+...+f_{2k+2}+...$ Then $m+2$ can be expressed as the sum of a strictly increasing sequence of Fibonacci numbers, and consequently $m+2\notin T$, which is a contradiction.  Thus $T=\emptyset$. This completes the proof.

Proof that every natural number $n\in \mathbb N$, can be written as the sum of different Fibonacci numbers between $F_2,F_3,\ldots,F_k,\ldots$.  For example: $32 = 21 + 8+3 = F_8+F_6+F_4$  Research effort:  The base step it's simple:  Let $k=1$ it can be britten as $k=1=F_2$  For the inductive step I considered:  Let $k = k  F_2 =F_2+F_2+\cdots+F_2 = \sum_{i=2}^w a_iF_i, a_i =\{0,1\} $  Then if $k$ suffice I want to see if $k+1$ suffices too... But I'm not realy seeing how to use the inductive hypothesis, so I assume It's wrong.  Any thoughts on which can the inductive step be?


There's a kid in my High School D&D group with A.D.H.D and he has a lot of trouble sitting still and distracts the other players. He's having a lot of fun goofing off, and doesn't seem very interested in playing but he insist that he wants to play.   I've considered asking him to stop playing for a little while so that the other players can learn the game and help him out more, but I think that might drive him away. He's a good kid and doesn't have a lot of friends, but he's keeping other people from playing the game, I've seen he get out of his chair and literally start wrestling with one of the other players. What should I do?

I regularly run an Adventurers League table at a local game cafe. For the last few weeks I've been trying to incorporate a college age player into the group who hasn't worked out at other tables.  He's been displaying some disruptive behavior like drawn-out fights over rulings, fudging stats, and getting frustrated when I point out mistakes in his use of features. Rude and sudden outbursts--although enthusiastic ones--have already deterred several new players from returning to the store. He's gotten slightly better, but the same exact mistakes occur session after session.   When I approached him one-on-one before a session to figure out how we can work together better, he became slightly irritated and claimed that he's forgetting things as a result of his ADHD and that there's nothing he can do about it. He used a similar explanation later when I pointed out that this was the fourth time I had to repeat a specific setting description to him while he was on his phone.  I think he feels targeted due to his ADHD, after he has claimed I'm being unfair with rulings and challenges, especially when faced with scenarios that make it difficult for his monk to simply walk up and punch enemies.  I want to work with this player if possible. Since I have no perspective on ADHD I'm blind to how much of this is justified, but I'd be loath to let him push players around at my table. How can I effectively address this player's behavior in a way that won't close him off further?  What measures can I take to make this player feel welcome at my table without allow his distractions to disrupt the existing group, or letting him get away with being overly aggressive? Since this is Adventurers League, I don't really have a way to approach him in private, but I could try talking to him before the session again. But at what point can I safely say it's just not going to work and he's going to keep on fighting the other players/myself?


shows this:    Navigating to  reveals that the {{> yield}} code markup was removed.

The answer summaries shown in search results appear to strip out code block markdown from the text. For example, the second item in  doesn't show the code block markdown content of this phrase:     "As you remove characters, Cleaner becomes shorter, so you're likely to reach a point where x is too large."


I  this problem before  .I have satisfied explaination given by  sir and  sir . I also found  .  I have read the Wikipedia posts for  and  and I have concluded that a function should be defined on given range for satisfying conditions of continuity and bounded function .  I posted my answer at GateOverflow(a site old Q/A of GATE ) , but they not accepted my answer and they have given reason answer key is correct , but not my explanation .        Let $f(x)=x^{-(1/3)}$ and $A$ denote the area of region bounded by   $f(x)$ and the $X-$axis, when $x$ varies from $-1$ to $1$. Which of the   following statements is/are TRUE?         $f$ is continuous in $[-1, 1]$   $f$ is not bounded in $[-1, 1]$   $A$ is nonzero and finite        Given answer key of this question $:$   False True True     In my opinion all given statement should be false .    My doubts are $:$   If a function is not continuous then is it possible for bounded function in given range ? What about function in given problem . Is it possible bounded area zero or infinite ? What about statement $(3)$ of problem ?

Let $f(x)=x^{-(1/3)}$ and $A$ denote the area of region bounded by $f(x)$ and the X-axis, when $x$ varies from -1 to 1. Which of the following statements is/are TRUE?   f is continuous in [-1, 1] f is not bounded in [-1, 1] A is nonzero and finite     I try to explain :    graph is  :     False , since left limit is not equivalent to right limit. True , since  f(x) rises to $infinite$ at $x=0$. True , since we can calculate it by doing integrating the function


I'm having trouble assigning values to a specific bash index, but apparently only when the index variable is set using a while read loop. Taking this code as a test example: #!/bin/bash  read -d '' TESTINPUT << 'EOF' 1,100 2,200 8,300 EOF  declare -A ARRAY echo "$TESTINPUT"| while read _l; do     i=$(cut -d, -f1 <<< $_l)     j=$(expr $i + 0)     value=$(cut -d, -f2 <<< $_l)     ARRAY[$j]=$value done  for i in {4..6}; do     ARRAY[$i]=$i done  for i in {1..10}; do     echo "$i ${ARRAY[$i]}" done  The output clearly shows that in the case of the while loop, the array variables don't get set, while with the for loop, using a range of {4..6} doesn't seem to have any issue. $ ./test_array.sh 1  2  3  4 4 5 5 6 6 7  8  9  10  Notice I also tried to convert the index variable to an integer using j=$(expr $i + 0)  But that doesn't seem to work either. Any ideas?

Why do I get different values for $x from the snippets below?  #!/bin/bash  x=1 echo fred > junk ; while read var ; do x=55 ; done < junk echo x=$x  #    x=55 .. I'd expect this result  x=1 cat junk | while read var ; do x=55 ; done echo x=$x  #    x=1 .. but why?  x=1 echo fred | while read var ; do x=55 ; done echo x=$x  #    x=1  .. but why?


As I will be returning from Brazil, I passed thru Heathrow in London on April 13 4 pm and travel to Incheon by April 14, 12 nn. It is another less than 24 hours airside as I will not go sightseeing. Being a Philippine Passport holder, do I need a Visa for the UK then?

Is there a website or some other way to find out if I need a transit visa for a short stop in the UK on my way somewhere else? Ideally this should give me the answer for all nationalities, and take into account any other visas I may have.


I bought a new IPhone 3GS and can't download Messenger, Facebook, hike etc. Why?  How to download?

I reset my iPhone 3Gs but am unable to install Facebook Messenger back on my iPhone as it requires iOS 7. How can I install Facebook Messenger on my iPhone 3Gs running iOS 6.1.6?


i'm trying to see if two Song() objects are the same to eventually throw a duplicate song entry Exception but i've encountered some troubles. The if statement to check if they're equal just does not want to accept them as equal. I made the console print their values on each song entry as a test and they were printing as the same but the if statement was not initializing. Here is my code:  case 1: //ENTER A SONG TO DATABASE      try {         Song newSong = enterSong();         int logSize = database.getLogicalSize();         if (logSize>0){             System.out.println(newSong.getName() + " " + database.getSongName(0));             if (newSong == database.getSong(0)){                 throw new Exception("Exception: duplicate song entry!");             }         }         database.addSong(newSong);     }     catch (Exception e){         System.out.println(e.getMessage());     }  break;

I've been using the == operator in my program to compare all my strings so far. However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.  Is == bad? When should it and should it not be used? What's the difference?


I have a working OpenVPN configuration on Windows 7. I am able to connect to it via the GUI as well as from the command line (openvpn --config .\openvpn.ovpn).  Here is the config file:  ## acme.ovpn ## client proto tcp dev tun persist-key persist-tun ca ca.crt cert mycert.crt key mykey.key remote <REMOTE_IP> 443 route <SOME_OTHER_IP> 255.255.255.0 resolv-retry infinite status openvpn-status.log verb 3   I copied all my config files over to an EC2 instance of RHEL 7.3 and . However, when I try to run sudo openvpn --config ./openvpn.ovpn), the trace ends up showing   Wed Feb  8 14:52:27 2017 Restart pause, 5 second(s) Wed Feb  8 14:52:32 2017 WARNING: No server certificate verification method has been enabled.  See http://openvpn.net/howto.html#mitm for more info. Wed Feb  8 14:52:32 2017 Socket Buffers: R=[87380->87380] S=[16384->16384] Wed Feb  8 14:52:32 2017 Attempting to establish TCP connection with [AF_INET]<OPENVPN_IP>:443 [nonblock] Wed Feb  8 14:52:42 2017 TCP: connect to [AF_INET]<OPENVPN_IP>:443 failed, will try again in 5 seconds: Connection timed out Wed Feb  8 14:52:57 2017 TCP: connect to [AF_INET]<OPENVPN_IP>:443 failed, will try again in 5 seconds: Connection timed out Wed Feb  8 14:53:13 2017 TCP: connect to [AF_INET]<OPENVPN_IP>:443 failed, will try again in 5 seconds: Connection timed out Wed Feb  8 14:53:28 2017 TCP: connect to [AF_INET]<OPENVPN_IP>:443 failed, will try again in 5 seconds: Connection timed out Wed Feb  8 14:53:43 2017 TCP: connect to [AF_INET]<OPENVPN_IP>:443 failed, will try again in 5 seconds: Connection timed out Wed Feb  8 14:53:58 2017 TCP: connect to [AF_INET]<OPENVPN_IP>:443 failed, will try again in 5 seconds: Connection timed out   Please note that the EC2 instance is just an openvpn client, not a server, as has been misunderstood. I am trying to establish a connection from it to a server, which is what the above log placeholder  is the IP of.  Why does the same configuration fail on the cloud VM that succeeds locally on Windows 7 and what should I do to fix it?  This question has also been asked .  RELATED:

The server works fine via the Amazon assigned DNS entry, but I cannot reach it (using a browser) via the Elastic IP address Amazon assigned the box. Ping does not work either. I am trying to confirm it is reachable before I add the IP address to my own DNS entries.


Suppose you have a set $\textit{S}$ with a generalized inequality $\preceq_k$.  Is it possible for such a set to have a $\textit{unique}$ minimal element that  is not a minimum element? I feel like it should be possible, but am struggling  to come up with an example.

A minimal element (any number thereof) of a partially ordered set $S$ is an element that is not greater than any other element in $S$.  The minimum (at most one) of a partially ordered set $S$ is an element that is less than or equal to any other element of $S$.    Let's consider the power set $\mathcal P (\{x,y,z\})$ together with the binary relation $\subseteq$.  The Hasse diagram shows what element(s) we're looking for:    It's easy to see that:   $\emptyset$ is a minimal element $\emptyset$ is the minimum   Now if we remove $\emptyset$ and consider $\mathcal P (\{x,y,z\})\setminus \emptyset$ instead, we get the following:   $\{x\}$, $\{y\}$ and $\{z\}$ are minimal elements there is no minimum   (1) We know that a minimum is unique and it is always the only minimal element.  (2) And from the example above, it seems that, if a sole minimal element exists, it is always the minimum.  But I read that (2) is false. Why?


I have a new Macbook 16" running 10.15.3 OSx with an AMD Radeon Pro 5500M 4 GB for the graphics card and I'm unable to get the desktop to extend using just Display Port chaining.   The monitors I'm trying to chain together are a  and a .  The 27" is a new monitor that I received with the laptop however the 24" I've had for several years and was using in a chained model with my Lenovo laptop that I'm moving away from.  I have tried the following configurations without any success and the best that I've been able to achieve is that the 24" mirrors the 27" but will never extend to it's own desktop.  Chained  MBP USB-C -> 27" USB-C (MST Enabled) -> DP -> 24" DP (DP 1.2 Enabled) = "Mirror"  MBP USB-C -> 27" DP (MST Enabled) -> DP -> 24" DP (DP 1.2 Enabled) = "Mirror"  MBP USB-C -> 27" DP (MST Enabled) -> DP -> 24" DP (DP 1.2 Disabled) = "Mirror"  MBP USB-C -> 24" DP (DP 1.2 Enabled) -> DP -> 27" DP (MST Enabled)  = "Out of Sync Error on 27"  MBP USB-C -> 24" DP (DP 1.2 Enabled) -> DP -> 27" DP (MST Disabled)  = "Out of Sync Error"  UnChained (Using two cables to drive them at the same time)  MBP USB-C -> 27" USB-C (MST Enabled) = "Extended"  MBP USB-C -> 24" DP (DP 1.2 Enabled) = "Extended"  What is causing the chaining to only allow mirroring?  ** UPDATE ** (3/22 @ 13:27 PST)  I've attempted to go back to a know good configuration using the original two monitors that I used with my Lenovo setup to try chaining those two monitors together removing the 27" from the equation and even with that configuration the two external monitors will only mirror each other.  It seems like the USB-C ports on the new MacBooks are not true Display Port 1.2+ compatible and requires individual connections to each external monitor that you want to extend your desktop to.  What gives apple?

I'm thinking about buying the new MacBook Pro with 4 USB-C ports.  I want to attach two screens, mouse and keyboard and ethernet to the MacBook Pro while using the fewest number of USB-C ports possible.  So what I need is an adaptor with:   two DP or HDMI ports (or one DP port with daisy chaining support) two or more USB ports an ethernet port   Best case would be a hub to which all devices are connected with a single connection to the MacBook Pro, including power supply. Worst case if attaching the MacBook Pro to the devices would require me to use all four of USB-C ports.  Given the answers up to far, I can tell that:   Daisy chaining might or might not work with the MacBook Pro (since it requires DP1.2 which is supported by the graphics chipset, but it's uncertain if it's supported by the MacBook Pro itself) I can attach the monitors with two cables as a last resort, leaving me with two ports for the remaining peripherals.   What I still don't know:   Would daisy chaining work with a hub or does it require the first monitor to be directly attached to the MacBook Pro? Does some kind of hub exist supporting attachment of all listed devices with a single upstream connection to the MacBook Pro?


There's a trade paperback book that I purchased at my college bookstore back in 1990-1991 that really caught my imagination, but I can't remember the title or author to go find it again.  The story revolved around the Hindu crew of a commercial starship discovering an ancient escape pod drifting while illegally traveling in an exclusion zone, in the solar system of a human colony world. The exclusion zone was a former battlefield from thousands of years before, between two other alien races, and is filled with debris and wrecks from the combat. Naturally, things don't go as planned. The parasitic organism, a survivor from the previous war, attacks the crew member spacewalking to investigate, and then starts attacking the rest of the crew. Several crew members escape in a shuttle, trying to get to a nearby world colony.   The parasite cobbles together a missile to attempt to destroy the shuttle, but upon realizing that there's a planet of potential hosts nearby, it also builds it's own escape ship out of parts of the abandoned ship, and follows the survivors. The survivors land at the planets main city and spaceport, but their story about the parasite isn't believed. The pursuing parasite eventually lands and infiltrates the starport, eventually takes over a nearby prison and uses the inmates as biological feedstock for converting them into what was essentially lanky, hivemind humanoid infantry with wrinkly, hound dog features (yes this sounds really weird, but bear with me). Eventually, the humans are fighting a losing battle against a quickly growing insurgency in the colony's main city, pulling back into the mountains and hinterlands to escape. It ends with human ships arriving and glassing the planet with nukes to try to contain the infection.  The alien parasites appearance changes throughout the story; as I recall it resembled a broomstick with centipede legs initially on the Hindu starship, but it eventually infected and converted humans by inserting a converter into their bodies, either through the mouth or anus. This created a weird, thin  human/hound dog like hybrid by rewriting the hosts DNA.( hound dog like, as in long ears and wrinkled snouts and faces with rolls of skin covering their eyes) There is actually a scene in the prison, towards the end where a hive leader is being created by physically joining several humans who become connected at the stomach and morph into a single entity. A lot of body horror elements are involved.  There was a sequel that explored the parasite race's past, but I can't remember it's title either. These books have stuck in my mind for decades... any help in recalling their titles and the author would be great. Thanks

In which book did alien protagonists absorb and mutate humans, turning their flesh into biological weapons that they then used against other humans? I believe the book starts with a human encountering a lone member of the alien species, who promptly subsumes him and uses his flesh to make weapons that it can use. It then goes in search of more raw material.. I think the human discovered one of the aliens when he was atempting to salvage a small object adrift in space.  If I recall correctly, the human then crashed his ship on earth. I think the aliens were more insectile (lots of legs) and could control a human by inserting a proboscis into their spine. Once they didn't need the intact human, they injected them with something (nanotech?) that let them repurpose the flesh and bones into new forms. This could be done to dead humans too, as long as the body hadn't significantly decayed. The aliens had previously lost an interstellar war and were believed to be extinct. The species had different casts, with the one at the start of the story being an infiltrator/scout. I probably read the book between 1983 and 1999.


Why was I not rewarded with the reputation for the 3 upvotes?

I've found some questions and those two badges "Epic" and "Legendary" regarding the "daily reputation cap". What is it and how can I hit it?


I understand the governing force causing the earth's cycle around the sun is gravity. It can be described by Keplar's law. But what causes the earth's spin? what is its governing law?

The origin of spin is some what a puzzle to me, everything spin from galaxies to planets to weather to electrons.  Where has all the angular momentum come from? Why is it so natural?  I was also thinking do photons spin? we always think of the wave as a standard 2d sin wave but could this rotate in 3d? What implications would this have?   And what about spacetime how does all the spinning effect?  This has always been avoided in all lectures and classes I ever went to.


To whom it may concern,   I compiled the Legrand Orange Book template via TeXmaker successfully. I did not modify anything on the template. However, when I am running Biber (2.13) to build bibliography section by inputting  @book{book_key,     address = {City},     publisher = {Publisher},     author = {Smith, John},     title = {Book title},     year = {2012},     volume = {3},     series = {5},     edition = {1},     pages = {123--200},     month = {1} }  @book{book_key1,     address = {City},     publisher = {Publisher},     author = {Smith, John},     title = {Book title},     year = {2012},     volume = {3},     series = {5},     edition = {1},     pages = {123--200},     month = {1} }  @article{article_key,     address = {City},     publisher = {Publisher},     author = {Smith, James},     title = {Article title},     year = {2013},     volume = {15},     number = {6},     pages = {1--8},     month = {3} }  @article{article_key1,     address = {City},     publisher = {Publisher},     author = {Smith, James},     title = {Article title},     year = {2013},     volume = {14},     number = {6},     pages = {1--8},     month = {3} }   to bibliography.bib file. Unfortunately, only references with book_key and article_key appear on the pdf but not the rest. I attempts to find the solutions myself such as deleting main.bcf and main.bbl before recompile but still cannot solve it now. Could anyone suggest me on this?

Is there any way to have BibTeX or biblatex insert a bibliography without having citations in the body of the document itself? Maybe a way to hide the in-text citations, so that BibTeX still sees them and includes them in the bibliography, but does not print them in the body?


I annotate special points in my diagram and use a node/pin approach:  \documentclass[border=3mm,tikz]{standalone} \usepackage{pgfplots}  \begin{document} \begin{tikzpicture} \begin{axis}[     axis lines = middle,     xlabel = {$x$},     ylabel = {$y$},         ]     \addplot [domain=0:10]{x};     \node[coordinate, pin={[align=left,pin distance = 10mm]-30:{Test Text Test\\ Text Test}}]                at (axis cs:4,4) {}; \end{axis} \end{tikzpicture} \end{document}   I want to change the anchor for the pin line / pin egde -- here are two possibilities:       Is that possible?

I was looking for a possibility to adjust the anchor of the text of a pin. Tikz does not allow this out of the box.   This question was already asked on the mailing list , Sept. 2009 without any reply till today or feature addition to pgf/tikz:  \documentclass[tikz]{standalone}  \begin{document}      \begin{tikzpicture}[          every pin/.style={anchor=west},          every pin edge/.style={anchor=west}]          \node [pin=60:test] at (0,1) {};          \node [pin={[anchor=west]60:test}] at (0,0) {};      \end{tikzpicture}  \end{document}   There is also a question with a suggested workaround here on  by defining an aligned pin style consisting of another node positioned at the end of the pin using \tikzlastnode:  \tikzset{     aligned pin/.style args={[#1,#2]#3:#4}{         pin={[inner sep=0pt,%               pin distance=#2,% new option, default = 3ex               label={[append after command={%                     node[inner sep=0pt,%                         xshift=-0.1cm,% modified, default = 0                         at=(\tikzlastnode.#3),%                         anchor=#1,%                     ]{#4}%                 }%             ]center:{}}%         ]#3:{}}%     } }   In this code, I have added a second unnamed option for pin distance within the square brackets.  MWE using the workaround showing the desired result:  \documentclass[tikz]{standalone}  \tikzset{     aligned pin/.style args={[#1,#2]#3:#4}{         pin={[inner sep=0pt,%             pin distance=#2,% new option, default = 3ex             label={[append after command={%                     node[%                         inner sep=0pt,%                         xshift=-0.1cm,% modified, default = 0                         at=(\tikzlastnode.#3),%                         anchor=#1,%                     ]{#4}%                 }%             ]center:{}}%         ]#3:{}}%     } } \begin{document} \tikz[pin distance=15mm, every pin/.style={fill=blue,text=white}] \node [circle,draw,     aligned pin={[west,3ex]0:X-Value},%     aligned pin={[west,3ex]45:Y-Value},%     aligned pin={[west,3ex]90:Z-Value}] {my circle}; \end{document}     I would like to add further options such as pin edge to adjust the style of the edge. Another addition would be xshift to finetune the position of the text (and yshift).  Is it possible to change the workaround to allow named options with default values if not specified?  Usage could look like this (without using necessarily all options):  node[aligned pin={[pin anchor=west,pin distance=1cm,pin edge={thick}]90:{txt}}]{};   If someone is interested in the implementation of the pin feature in tikz see  and do a textsearch for pin.  Update after answer by Qrrbrbirlbel:  To achieve almost the same result as my MWE use following code  % use preamble from Qrrbrbirlbel answer \begin{document}  \begin{tikzpicture}[every pin/.append style=draw, every pin edge/.style={thick}] \node[c] [ pin={[pin anchor=south west,pin edge pin anchor=south west,pin distance=6ex]90:Z-Value}, pin={[pin anchor=west,pin edge pin anchor=west,pin distance=6ex]45:Y-Value}, pin={[pin anchor=west,pin edge pin anchor=west,pin distance=6ex]0:X-Value} ] at (0,0) {}; \end{tikzpicture}  \end{document}     However, I would like to finetune the text positioning using inner sep=0 and a slight shift:  \newlength{\xshiftlength} \setlength{\xshiftlength}{0.5\widthof{Z}}   Following works:  \tikzset{   pin inner xsep/.style={tikz@pin@post/.append style={inner xsep=#1}} }   Apparently xshift cannot be added in the same way. It shifts more than only the text. How could a pin text xshift be added?


The code is to shorten names.  When am executing in Hackerrank there is NullPointerException on temp.length() , temp.toUpperCase() , temp.substring().   The First Line contains an integer N, which denotes the number of testcases.  Sample Input  3  Mohandas Karamchand Gandhi  Mervin R Sundharam  Muthu  Sample Output  M K Gandhi  M R Sundharam  Muthu  import java.io.*; import java.util.*;  public class Solution {      public static void main(String[] args) throws IOException {         Scanner sc= new Scanner(System.in);         int size= sc.nextInt();         BufferedReader br = new BufferedReader(new InputStreamReader(System.in));          String ar[] =new String[size];         for (int i= 0; i < size; i++) {            ar[i]=br.readLine();        }          for(int i= 0;i<size;i++){             String temp=" ";             temp=ar[i];             int pos=0;             for (int j= 0; j < temp.length(); j++)                 if(temp.charAt(j)==' '){                     System.out.print(temp.toUpperCase().charAt(pos)+ " ");                     pos=j+1;                 }             }             System.out.println(temp.toUpperCase().charAt(pos)+temp.substring(pos+1));         }     } }

What are Null Pointer Exceptions (java.lang.NullPointerException) and what causes them?  What methods/tools can be used to determine the cause so that you stop the exception from causing the program to terminate prematurely?


In the hierarchy I have "GPencil and GP_Layer". I would like to delete this from the hierarchy. But mainly, every time I use the grease pencil, Blender puts a lime-green line in the timeline. I would like to get rid of those too, or at least hide them in the Timeline.

Is it possible to use the grease pencil without creating keyframes when you draw: I just want a single drawing so I don't need keyframes, but I need to be able to scrub through the timeline when drawing.  I know I can move the keyframes or delete them, but it would be easier not to have any.  I have already tried turning the "automatic keyframe" button from the timeline off.


I need to solve $18^{19^{20}} \mod21$ and problems like it without a calculator. I believe there is a way to decompose this into a problem solvable with Fermat's little theorem. I just can't figure out how.

How do I efficiently compute $a^b\bmod c$:  When $b$ is huge, for instance $5^{844325}\bmod 21$? When $b$ is less than $c$ but it would still be a lot of work to multiply $a$ by itself $b$ times, for instance $5^{69}\bmod 101$? When $(a,c)\ne1$, for instance $6^{103}\bmod 14$?  Are there any other tricks for evaluating exponents in modular arithmetic?  This is being asked in an effort to cut down on duplicates, see  and .


How can we prove this $$(a, b)= \cup_{n=1}^\infty [a + 1/n, b - 1/n],$$ ?  First: it is clear that $\left [ a + \frac{1}{n}, b - \frac{1}{n}\right ] \subseteq (a, b)$ for every n, then $\cup_{n=1}^\infty [a + 1/n, b - 1/n] \subseteq (a, b) $. what about the other inclusion?

So, I have proven already that every open set in $\mathbb{R}$ can be written as a countable union of disjoint open intervals; i.e., that $\mathcal{O} = \bigcup_{i=1}^\infty (a_{x_i}, b_{x_i})$.  Now, consider the closed interval $\left [ a + \frac{1}{n}, b - \frac{1}{n}\right ] \subset (a, b)$.  Since the midpoint of $(a,b)$ is $\frac{b-a}{2}$, we need $\displaystyle n > \frac{2}{b-a}$.  Otherwise, $\displaystyle a + \frac{b-a}{2} = a + \frac{1}{ \frac{2}{b-a}} = b - \frac{1}{ \frac{2}{b-a}} = b - \frac{b-a}{2}$, and our interval will be degenerate.  Note that $\forall (a,b) \in \mathcal{O}$,   $\displaystyle \bigcup_{n = \frac{b-a}{2}+1}^{\infty} \left[ a + \frac{1}{n}, b - \frac{1}{n} \right] = (a,b)$.  So, we have $\displaystyle \mathcal{O} = \bigcup_{i=1}^\infty \left ( \bigcup_{n = \frac{b-a}{2} + 1}^\infty \left[ a_{x_i}+\frac{1}{n}, b_{x_i}-\frac{1}{n} \right] \right)$.  But, I don't like this. It looks like I'm still just taking a union of open sets.  Please help me get this to look like the union of closed sets.  (For unbounded intervals, I know what to do - it should follow from this relatively easily).


Hello I am building a version of Craps where it randomly rolls two dice and stores the information in two different rolls, where how many times it rolled a 7 and how many times it rolled an 11 in 100 rolls. I have managed to get it where I can output how many rolls I got for each and how many wins altogether, but now when I prompt the user if they want to play again I cant see to get it to execute the while method I created for the rolling loop. Is there a way I can get it to re-execute that in the last if statement and if so what would I put in that if statement to achieve this or is there some other kind of method I should be doing here instead? It was recently closed and pointed me to a string comparison page on stack that didn't answer my question. I understand how to compare strings and what == does. However that doesn't tell me why  or what is needed to re-execute a previously performed while loop after prompting the user for action.  import java.util.Random; import java.util.Scanner;  public class Craps_ws7 {     public static void main(String args[])     {         Scanner keyboard1 = new Scanner(System.in);         System.out.println("Welcome to my Craps Game!");         boolean stop = false;         int sevenRoll = 0;         int elevenRoll = 0;         int totalWins = 0;         int gameCount = 0;         while(gameCount <= 100)         {             gameCount = gameCount + 1;             Random generator = new Random();             int die1 = generator.nextInt(6) + 1;             int die2 = generator.nextInt(6) + 1;             int dieTotal = die1 + die2;               if (dieTotal == 7)              {                 System.out.println("You rolled: " + dieTotal);                 System.out.println("You Win!");                 sevenRoll++;                 totalWins++;              }              else if(dieTotal == 11)              {                 System.out.println("You rolled: " + dieTotal);                 System.out.println("You Win!");                 elevenRoll++;                 totalWins++;              }              else              {                 System.out.println("You rolled: " + dieTotal);                 System.out.println("You lose!");              }         }     System.out.println("You rolled a total of " + sevenRoll + "  sevens and " + elevenRoll + " elevens for a total of " + totalWins + " wins out of 100!" );     System,out.println("Would you like to play again? [y/n]");     string ans = keyboard1.nextLine();     if (ans = "y" || ans = "Y")     {      }     else      {         System.out.println("Thank you for playing, Good Bye!");     }     } }

I've been using the == operator in my program to compare all my strings so far. However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.  Is == bad? When should it and should it not be used? What's the difference?