Who's the kid at the funeral
Who's the random kid standing in the gathering at the end?
Why I cannot ping 192.168.0.108 (my computer IP?) Ipconfig says that my IP is 192.168.0.108 C:\Users\vbcbu&gt;Reply from 192.168.0.108: Destination host unreachable.ip 'Reply' is not recognized as an internal or external command, operable program or batch file.  C:\Users\vbcbu&gt;ipconfig  Windows IP Configuration   Ethernet adapter Ethernet:     Media State . . . . . . . . . . . : Media disconnected    Connection-specific DNS Suffix  . :  Wireless LAN adapter Local Area Connection* 12:     Media State . . . . . . . . . . . : Media disconnected    Connection-specific DNS Suffix  . :  Wireless LAN adapter Local Area Connection* 13:     Media State . . . . . . . . . . . : Media disconnected    Connection-specific DNS Suffix  . :  Wireless LAN adapter Wi-Fi 2:     Connection-specific DNS Suffix  . : www.tendawifi.com    Link-local IPv6 Address . . . . . : fe80::3804:2d04:a591:7142%6    IPv4 Address. . . . . . . . . . . : 192.168.0.108    Subnet Mask . . . . . . . . . . . : 255.255.255.0    Default Gateway . . . . . . . . . : 192.168.0.1  Ethernet adapter Bluetooth Network Connection:     Media State . . . . . . . . . . . : Media disconnected    Connection-specific DNS Suffix  . :  C:\Users\vbcbu&gt;  I can ping router at 192.168.0.1 I can't ping 192.168.0.108 C:\Users\vbcbu&gt;ping 192.168.0.1  Pinging 192.168.0.1 with 32 bytes of data: Reply from 192.168.0.1: bytes=32 time=2ms TTL=64 Reply from 192.168.0.1: bytes=32 time=24ms TTL=64  Ping statistics for 192.168.0.1:     Packets: Sent = 2, Received = 2, Lost = 0 (0% loss), Approximate round trip times in milli-seconds:     Minimum = 2ms, Maximum = 24ms, Average = 13ms Control-C ^C C:\Users\vbcbu&gt;ping 192.168.0.8  Pinging 192.168.0.8 with 32 bytes of data: Reply from 192.168.0.108: Destination host unreachable. Reply from 192.168.0.108: Destination host unreachable. Reply from 192.168.0.108: Destination host unreachable. Reply from 192.168.0.108: Destination host unreachable.  Ping statistics for 192.168.0.8:     Packets: Sent = 4, Received = 4, Lost = 0 (0% loss),  C:\Users\vbcbu&gt;ping 192.168.0.8  Pinging 192.168.0.8 with 32 bytes of data: Reply from 192.168.0.108: Destination host unreachable. Reply from 192.168.0.108: Destination host unreachable. Continuous bijections and their inverses Are there any theorems characterizing the continuity of the inverses of bijections between certain classes of particular spaces. E.g. it seems to me the inverse of any continuous bijection between connected subsets of $\mathbb{R}$ is continouos. It seems that most counterexamples require highly specific spaces. I'm thinking there might be something along the lines of any continuous bijection between homeomorphic spaces has a continuous inverse. Any of this make sense? How to get this only in small letters (lowercase)? I used the following code to give out the category name (without link) in a single post.  How would it be possible for the category name that is given out to only appears in small letters ?  I just think this has to be made with substring, but I dont get it :) Could somebody please help me :)  &lt;?php foreach((get_the_category()) as $category) { echo $category-&gt;cat_name . ' '; } ?&gt; Audits bug in the filtered review queue Just recently the . I've got  this morning which seems to be out of scope for my filtered list. I have filtered my list to only review VBA and Excel-VBA tags while the audit is tagged with perl http pipes which have nothing in common with VBA. I see a problem with the way the system currently works  Since I have a filtered list of tags, once I see a differently tagged question I already know it's an audit. Come on, that's too easy!  Can this please be fixed? update I have ran into  which I decided to edit - the question was tagged VBA and I thought it had nothing to do with VBA so I selected Edit which was the correct action. Now, once the first audit was clearly one to close - nothing to do with VBA and really easy to know that it actually was an audit - the second one was a very good audit. I am just thinking now if this is a proper bug or just a matter of malfunction in the system. I think the first one may have been made an audit for the wrong reason. Let's turn this into a discussion as I can't clearly work the system out. Define the sequence $aₙ=20+n²$ and $dₙ=\gcd(aₙ , aₙ₊₁)$ Find with proof the values dₙ takes. Why is the wild card character * so different between the zip and rm commands? I put together a script to do some file operations for me. I am using the wild card operator * to apply functions to all files of a type, but there is one thing I don't get. I can unzip all files in a folder like this  unzip "*".zip   However, to remove all zip files afterward, I need to do   rm *.zip   That is, it does not want the quotation marks. The unzip, on the other hand, does not work if I just give it the * (gives me a warning that "files were not matched").   Why is this different? To me, this seems like the exact same operation. Or am I using the wild card incorrectly?   in Unix do not really go into this, and I could not locate anything in the rm or zip docs.  I am using the terminal on a Mac (Yosemite). Give $a, b. c$ be the lengths of a triangle and $a+ b+ c= 3$. Prove that: $$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a}\geq  \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}$$ My try  We have England MO inequality: $$\left ( \frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \right )^{2}\geq  \left ( a+ b+ c \right )\left ( \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c} \right )$$  and $$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \geq  3=  a+ b+ c$$  But it cannot be used to prove the above inequality and how to use triangle inequality in this inequality. I need the help. Thanks! In many practical applications, one can consider the Earth-Centered Inertial () reference frame approximately as an inertial reference system, though strictly speaking, it is non-inertial.   Is there any quotable reference, where this claim is supported by a detailed estimation how small the effects are that one neglects if one considers the Earth's frame as inertial? How should an educational institute woo good faculty? What steps must an educational institution wanting to recruit good faculty take? There is a cycle in the academia:    Good doctorates from top universities invariably join other top universities as faculty members.  Good students aspire to study in top universities under excellent faculty members.   Suppose a new educational institution or department springs up and desires to set up a top-quality group, how does it break into this cycle? I suppose this could take some years, but is money-power the only thing that could be wielded by the institute to achieve its end? pkgloader is not finding l3regex I would like to try a package which seems interesting, but it some dependencies are not satisfied.   Here is my minimal example:  \documentclass{article}  \usepackage{pkgloader} \LoadPackagesNow  \begin{document}     pckloader trial \end{document}   And I get the following error: ! LaTeX Error: File 'l3regex.sty' not found.. Therefore, I tried to update my distribution with the linux commands  tlmgr update --self tlmgr update --all   But the package l3regex is still not foundable. Plus, my computer is not lying since l3regex.sty is not in my folder /texmf-dist/tex/latex/*.  So, is it something normal ? How can I workaround this problem ? How to replace ${} placeholders in a text file? Viruses: Adaptation to a new host through repeated host jumps Why would a JavaScript variable start with a dollar sign? Is the term "errored out" a grammatically correct phrase, or just a colloquial one? Let's say I have written an equation in my paper and numbered it:  y = a + bx    (1)   Then, later in the paper, I want to remind the reader by showing the same equation with the same equation number.  Sure, I can repeat the equation, but LaTeX gives it a new equationn number:  y = a + bx     (2)   How do I force LaTeX to use the original equation numbers? Find the last two digits of $3^{45}$ I was wondering if there is a simpler way to find the last to digits of a power such as $3^{45}$. I reduced it modulo 100 to get the answer, which is 43. But I was curious if there was a simpler, or more eloquent way to get this result.  Here's my method:  Using the fact that $3^4=81\equiv -19 \pmod{100}$ and $19^5=2,476,099\equiv -1 \pmod{100}$  $$ 3^{45}\equiv 3(3^4)^{11}\equiv 3(-19)^{11}\equiv 3(-19)(19^5)^2\equiv 3(-19)(-1)^2 \equiv -57 \equiv 43 \pmod {100} $$  I just felt like this was sort of a messy way to find out that the last two digits are 43. I am trying to figure out this:  c = 1 def f(n):     print c + n  def g(n):     c = c + n  f(1) # =&gt; 2 g(1) # =&gt; UnboundLocalError: local variable 'c' referenced before assignment   Thanks! I'm trying to connect 32 I²C "root of trust"  devices.    I'm thinking about a good way that does that, I mean what kind of communication should I use? A digital multiplexer that selects a device? but based on what? switching on the VDD source? but that needs reinitialization of the device and that takes time. What is the difference between ++i and i++? What is the difference between therapy and treatment?