My Colombia visa is on my previous passport. Since I have run out of pages, I have obtained a new passport. Can I still travel with both passports and enter the country?
If your visa is on an expired passport but still valid, can you still use it with your new passport, or does the old passport have to have a recent expiration date? If so, how do you transfer your visa to your new passport?
BAV99 forward voltage Finding how multiplication and addition behave on $\mathbb{F}_4$ without any result I'm currently tutor in an undergraduate course and the students are asked to find how addition and multiplication behave on the field with 4 elements: $\{0,1,x,y\}$. In the solution the teacher gave me, he uses the fact that $(\mathbb{F}_4^*, \cdot )$ is cyclic with order 3 to find that $x$ and $y$ are both generators and that $(\mathbb{F}_4,+)$ is a group of order 4 to deduce that $0=1+1+1+1=(1+1)(1+1)$ and hence, $(1+1)=0$ since a field is integral.   However, I'm not satisfied with this reasoning because this question is asked to student in first year and they juste have been introduced to fields (and this was part of an analysis course so they won't see any result soon). So I tried to find how it behaves by myself, using as few results as possible.  Multiplication is the easy part, $0\cdot a$ and $1 \cdot a$ is trivial for all $a\in \mathbb{F}_4$. Then, $x\cdot y\not = 0$ because it is invertible, and it can't be equal to $x$ or $y$ because that would imply the other element to be equal to $1$. Hence, $x\cdot y=1$. We are left to find what is $x^2$ (and similarly, $y^2$). $x^2$ cannot be $0$, nor $x$. If $x^2=1$, then $y=(x^2)y=x(xy)=x\cdot 1=x$. Contradiction. Hence $x^2=y$ and by the same reasoning, we have $y^2=x$.  It was easy with the multiplication, but I can't figure out any non-trivial result concerning addition. My intuition tells me that distributivity should be used here but I'm feeling like it's just converting a problem into another one because we don't know anything about addition.  Does anyone here have an idea of how addition can be "found" without using any advanced result (or as few results as possible)?  Thanks in advance. The international Stack Overflow sites in Spanish, Portuguese, Russian, and Japanese seem to have a bug in their quick site descriptions (under their titles), for they all start with the word  For...  in English, instead of their respective languages As an example, the Spanish Stack Overflow site description is:  For programadores y profesionales de la informática  Obviously, this isn't something too big of a deal, but since they offer those sites in their respective languages, it would be nice if this issue were fixed. You can see this under &quot;More communities&quot; in the top bar.    You can also see this on the  page, where the descriptions begin with  Q&amp;A for... The  contains a link to the new code of conduct,  . As of 2018-08-07T05:27:38+00, . Also, I cannot find any other link to the conduct yet, neither in the help center, nor in the site navigation, which is a bummer, given that we're invited to read the new CoC:     Really, please, take a moment to read the new Code of Conduct now. It’s not long, it’s not wordy-and-legalese-y and in our long held tradition, we feel that it does a good job of assuming that the vast majority of folks that use our sites always have the best intentions at heart.    Is the code of conduct not yet published?  Edit: As mentioned in the comments,  announces the conduct and links to , so maybe it's not public yet? What it says on the tin, a foreign friend of mine has asked and I can't tell him; apart from it sounding horrible.  For example:  "Why is it raining today?"   Instead of:  "Why it is raining today?" How can I deploy a connected app from my sandbox environment to my production environment? I built a changeset and I was able to include all my classes, but it was a surprise that salesforce has not provided any option to deploy a connected app.  So how do I go about deploying a connected app? Do I need to create it in every other environment? Sitemaps - one per subdomain or one for the base domain? I have more than 5 subdomains on a project, say developer, www, staff and some more, what is the best practice in creating sitemaps?   All subdomains that share a same sitemap at base domain (example.com/sitemap.xml) ? or different sitemaps for different subdomains?(developer.example.com/sitemap.xml and www.example.com/sitemap.xml)   Since the subdomains are several parts of the main application, what are the pros and cons of each, and what should I consider while choosing the one over the other? (one sitemap per subdomain or one for the whole site) Escape velocity from a rotating black hole JavaScript open in a new window, not tab I have a select box that calls window.open(url) when an item is selected.  Firefox will open the page in a new tab by default.  However, I would like the page to open in a new window, not a new tab.    How can I accomplish this? Rstudio is duplicating commands in the command line Postfix SMTP banner show multiple hostnames We have a Postfix SMTP server that's bound to two IP addresses and serves mail for two different domains. We can't change the domains to both use the same, single IP. Both IP addresses have reverse DNS   1.1.1.1 reverses to mail.domain1.com 2.2.2.2 reverses to mail.domain2.com   In our Postfix config I have  myhostname = mail.domain1.com myhostname = mail.domain2.com  inet_interfaces = 1.1.1.1, 2.2.2.2  smtpd_banner = $myhostname Mail Server   (And some other stuff, which I think is irrelevant)  Using , running the smtp test command on mail.domain2.com returns everything as OK, however running it on mail.domain1.com returns an error because Postfix is identifying itself as 'mail.domain2.com' in the SMTP banner, how do I get it to return the correct banner based on the external IP address used? Proving the given quadratic form is chi-squared $k$ Suppose $\underline{X}$ is an $m$-dimensional vector following multivariate Normal distribution i.e. $\underline{X}$~$N_m(\underline{\mu},\Sigma)$ where $\Sigma$ is positive definite. Let $B$ be a symmetric $m$ by $m$ matrix, having the following properties:  $B\Sigma B=B$  $\underline{\mu}'B\underline{\mu}=0$  $trace(B\Sigma)=k$, where $k&lt;m$   Prove that $\underline{X}'B\underline{X}$ follows chi-squared ($k$).  I proceeded a lot but got stuck in one last step where I think linear algebra is troubling me. Kindly have  a look at my &quot;solution&quot;.  The second condition can be written as $\underline{\mu}'B\Sigma B\underline{\mu}=0\implies B\underline{\mu}=\underline{0}$ as $\Sigma$ is Positive definite. Then after a little algebra and considering condition 2, we can write $\underline{X}'B\underline{X}=(\underline{X}-\underline{\mu})'B(\underline{X}-\underline{\mu})=(\underline{X}-\underline{\mu})'B\Sigma B(\underline{X}-\underline{\mu})$ But $\Sigma$ is positive definite hence we can have a symmetric matrix $\sqrt{\Sigma}$ defined such that $\sqrt{\Sigma}\sqrt{\Sigma}=\Sigma$. Using this, we can write $(\underline{X}-\underline{\mu})'B\Sigma B(\underline{X}-\underline{\mu})=(\sqrt{\Sigma}B(\underline{X}-\underline{\mu}))'(\sqrt{\Sigma}B(\underline{X}-\underline{\mu}))$ We want to show that this follows chi-squared ($k$). Now, it is obvious that as $(\underline{X}-\underline{\mu})$~$N(\underline{0},\Sigma)$, $\sqrt{\Sigma}B(\underline{X}-\underline{\mu})$~$N(\underline{0},\sqrt{\Sigma}B\Sigma B\sqrt{\Sigma})=N(\underline{0},\sqrt{\Sigma}B\sqrt{\Sigma})$ Now, $trace(\sqrt{\Sigma}B\sqrt{\Sigma})=trace(B\Sigma)=k$ and $\sqrt{\Sigma}B\sqrt{\Sigma}$ is idempotent (verified). So its eigenvalues are only $0$ or $1$. Since its trace is $k$, there are $k$ eigenvalues with value $1$ and other $m-k$ eigenvalues are $0$. Now here is the thing I want to prove: I want to show that$\sqrt{\Sigma}B\sqrt{\Sigma}$, which is the covariance matrix, is diagonal and has only $k$ $1$'s and remaining $m-k$ diagonal entries $0$. The reason is, then I will be able to say that $k$ components of $\sqrt{\Sigma}B(\underline{X}-\underline{\mu})$ are independent standard normals whose sum of squares is our quadratic form and thus it follows chi-squared ($k$). But I really could not show that $\sqrt{\Sigma}B\sqrt{\Sigma}$ has the desired property i.e. it is diagonal with rank $k$. But this is crucial, it is the finale. If this does not happen, then some components will be correlated with each other and we cannot say that our quadratic form has chi-squared ($k$). I would really appreciate if someone helped me to prove this final part. Find all solutions to $x^2\equiv 1\pmod {91},\ 91 = 7\cdot 13$ I split this into $x^2\equiv 1\pmod {7}$ and $x^2\equiv 1\pmod {13}$.  For $x^2\equiv 1\pmod {7}$, i did: $$ (\pm1 )^2\equiv 1\pmod{7}$$ $$(\pm2 )^2\equiv 4\pmod{7}$$ $$(\pm3 )^2\equiv 2\pmod{7}$$ Which shows that the solutions to $x^2\equiv 1\pmod {7}$ are $\pm1$.  For $x^2\equiv 1\pmod {13}$, i did: $$ (\pm1 )^2\equiv 1\pmod{13}$$ $$(\pm2 )^2\equiv 4\pmod{13}$$ $$(\pm3 )^2\equiv 9\pmod{13}$$ $$ (\pm4 )^2\equiv 3\pmod{13}$$ $$(\pm5 )^2\equiv {-1}\pmod{13}$$ $$(\pm6 )^2\equiv 10\pmod{13}$$Which shows that the solutions to $x^2\equiv 1\pmod {13}$ are $\pm1$.  Thus, I concluded that the solutions to $x^2\equiv 1\pmod {91}$ must be $\pm1$. I thought that $\pm1$ were the only solutions, but apparently I am incorrect! How do I go about finding the other solutions to this congruence? Reading over the Monk I saw that for Martial Arts:     When you use the Attack action with an unarmed strike or a monk weapon on your turn, you can make one unarmed strike as a bonus action.   This implies that non-monk weapons do not count towards being able to make this single unarmed strike. However, when looking over Flurry of Blows:     Immediately after you take the Attack action on your turn, you can spend 1 ki point to make two unarmed strikes as a bonus action.   The requirement doesn't seem to be there anymore, and I didn't see where this discrepancy was covered in . This  says that you can use flurry of blows as a bonus action on any form of attack.   So does the limitation really only exist on a single unarmed strike, but you can use whatever attack action you want to trigger Flurry of Blows? How to work the BibTeX? I have searched the tutorials but have not been successful in understand them. The tutorials always mensionam concepts that I do not know. I would like to understand the basics of BibTeX.  By basics I mean:   How do I install BiBTeX? How to build a file in bib? For this question assume that I already have the bibtex format each reference provided by databases like MathSciNet. See observation 1 below. What I write commands in the file format. Tex to call call the format file bib? How to exploit the format of the bibliography that comes out in PDF? The commands are made in the preamble of the file. Arquifo bib or format. Tex? What must be the BiBTeX with biblatex? Is there any tutorial (in pdf or video) that teaches step by step how to operate BiBTeX?   Observation 1 I already have the bibtex format each reference provided by databases like MathSciNet:   @article {MR2373353,  AUTHOR = {Avila, Artur and Lyubich, Mikhail},  TITLE = {Hausdorff dimension and conformal measures of {F}eigenbaum {J}ulia sets},  JOURNAL = {J. Amer. Math. Soc.},  FJOURNAL = {Journal of the American Mathematical Society},  VOLUME = {21},  YEAR = {2008}, NUMBER = {2},  PAGES = {305--363},  ISSN = {0894-0347},  MRCLASS = {37F35 (37F10 37F25)},  MRNUMBER = {2373353 (2009i:37112)},  MRREVIEWER = {Feliks Przytycki},  DOI = {10.1090/S0894-0347-07-00583-8},  URL = {http://dx.doi.org/10.1090/S0894-0347-07-00583-8}, } I am a grad student developing an interest for statistics. I like the material over-all, but I sometimes have a hard time thinking about applications to real life. Specifically, my question is about commonly used statistical distributions (normal - beta- gamma etc.). I guess for some cases I get the particular properties that make the distribution quite nice - memoryless property of exponential for example. But for many other cases, I don't have an intuition about both the importance and application areas of the common distributions that we see in textbooks.   There are probably a lot of good sources addressing my concerns, I would be glad if you could share those. I would be a lot more motivated into the material if I could associate it with real-life examples. Qualcomm Atheros AR9462 with Ubuntu 18.10 I can't enable the Wifi (nothing happens when press enable button) and in Wifi settings it says:     "No Wi-Fi Adapter Found"   I have tried several things from other questions/answers here but nothing worked. Here some information if this is helpful for the answer:     Device: Lenovo Miix 2 11   sudo lspci:  Network controller: Qualcomm Atheros AR9462 Wireless Network Adapter (rev 01)   sudo lshw -C network:  *-network DISABLED            description: Wireless interface    product: AR9462 Wireless Network Adapter    vendor: Qualcomm Atheros    physical id: 0    bus info: pci@0000:02:00.0    logical name: wlp2s0    version: 01    serial: 48:d2:24:8a:ea:e6    width: 64 bits    clock: 33MHz    capabilities: pm msi pciexpress bus_master cap_list rom ethernet physical wireless    configuration: broadcast=yes driver=ath9k driverversion=4.18.0-16-generic firmware=N/A latency=0 link=no multicast=yes wireless=IEEE 802.11    resources: irq:18 memory:f0400000-f047ffff memory:f0480000-f048ffff   sudo lsmod | grep ath:  ath3k                  20480  0 bluetooth             548864  12 btrtl,btintel,btbcm,bnep,ath3k,btusb ath9k                 151552  0 ath9k_common           36864  1 ath9k ath9k_hw              475136  2 ath9k_common,ath9k ath                    36864  3 ath9k_common,ath9k,ath9k_hw mac80211              794624  1 ath9k cfg80211              663552  5 wl,ath9k_common,ath9k,ath,mac80211   In system settings it shows the Wifi-Adapter Name:  Qualcomm Atheros AR9462   Edit:  lspci -knn | grep Net -A3; rfkill list:  02:00.0 Network controller [0280]: Qualcomm Atheros AR9462 Wireless Network Adapter [168c:0034] (rev 01) Subsystem: Lenovo AR9462 Wireless Network Adapter [17aa:3214] Kernel driver in use: ath9k Kernel modules: ath9k, wl 0: phy0: Wireless LAN     Soft blocked: no     Hard blocked: no 1: ideapad_wlan: Wireless LAN     Soft blocked: no     Hard blocked: yes 2: ideapad_bluetooth: Bluetooth     Soft blocked: yes     Hard blocked: yes I wrote a regular expression which works well in a certain program (grep, sed, awk, perl, python, ruby, ksh, bash, zsh, find, emacs, vi, vim, gedit, …). But when I use it in a different program (or on a different unix variant), it stops matching. Why? Simulate INSERT key in Mac OS X Long story short, I am using Citrix to remote into my PC at work, and several applications require the use of the INSERT key, which, as everyone knows, Apple chose to eliminate from their keyboards. I've tried everything I can think of, and never been able to find an easy way to send this keycode via Citrix.  I'm sure there are other uses for this, and I've read quite a few people's questions on the subject, but no one seems to have offered a viable solution.  I have the Apple USB keyboard (full-size, with numeric keypad), and in the position where the INSERT key would normally exist, Apple placed the "fn" key. Now, I still need this key from time to time, but what I'd love to see is perhaps Control-fn or similar to send the keycode for INSERT.  If anyone knows how to achieve this, preferably without installing 3rd-party software (adware?), I'm all ears. Power Supply Capacity Formula