I have a question with a query in PostGIS. I try to calculate the area from a polygon. It actually works but the area-result is not correct. My Query is:  SELECT name, ST_Area(way)/10000 AS hectares FROM osm_data ORDER BY hectares DESC LIMIT 1;   The problem is, that the amount of ha is completely wrong. So i tryed to change the EPSG in the standard 4326 EPSG. But when I use than the query it shows me an error:  SELECT name,ST_Transform ((way, 4326)ST_Area(way))/10000 AS hectares FROM planet_osm_polygon ORDER BY hectares DESC LIMIT 1;      SQL state: 42601   Where is the mistake?!
I am trying to get from a particular point more informations about the surroundings (which are also points). So I wrote this:  SELECT *    FROM osm_point    WHERE ST_DWithin(way,(ST_SetSRID(ST_MakePoint(12.2385578688105, 51.8435615931295), 4326, 1000);   But there is an error     ERROR: FEHLER:  LINE 3: ...T_MakePoint(12.2385578688105,   51.8435615931295), 4326, 1000) ^ SQL state: 42601 Character: 133   What is wrong?
ArcPy Utility Network Trace Not Returning Any Result Centering in tabularx and X columns How can I center the column when using X in tabularx environment like in this example?  \begin{tabularx}{\textwidth}{@{}lXXXXX@{}}  1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\    \end{tabularx} How to hide "internal" polygon boundaries? How can I determine if someone's SSH key contains an empty passphrase? Some of my Linux &amp; FreeBSD systems have dozens of users. Staff will use these "ssh gateway" nodes to SSH into other internal servers.  We're concerned that some of these people use an unencrypted private SSH key (A key without a . , because if a cracker ever gained access to their account on this machine, they could steal the private key and now have access to any machine which uses this same key. For security reasons, we require all users to encrypt their private SSH keys with a passphrase.  How can I tell if a private key is not-encrypted (e.g. Does not contain a passphrase)? Is there a different method to do this on an ASCII-armored key vs. a non-ASCII-armored key?  Update:  To clarify, assume I have superuser access on the machine and I can read everybody's private keys. If Chrome is already open, it opens new sites (e.g. from Windows shortcuts, application shortcuts, or from the Run dialog) in new tabs of the existing instance.   I'd prefer to open new sites in a separate window (a new instance of Chrome).   I don't see that in the options - is it possible? $\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite Let $a\ge 2$ be an integer. Show that for positive integers $m,n$, we have $a^n - 1$ divides $a^m - 1$ if and only if $n$ divides $m$.   I am having trouble showing this. I've seen a similar problem on here but it only shows one direction for the proof. I would like to see both directions.  This is what I have so far: Let $m = nq + r$, then $a^m - 1 = a^{nq + r}- 1 = a^{nq}a^r - 1 = a^{nq}a^r - a^r + a^r - 1 = a^r(a^{nq} - 1) + a^r - 1$. Since $0\le r\lt n$, $r = 0$ then $a^m - 1$ = $a^{nq} - 1$ so $a^m$ = $a^{nq}$ . Thus $m = nq$ hence $n\mid m$. I cant use lhopital's rule here is my attempt:  Limit of $(\frac{n^{2}-2n+1}{n^{2}-4n+2})^{n}$ =$e^{\lim_{n\to \infty} n\times \ln(\frac{1 -2-n +\frac{1}{n^{2}}}{1 -\frac{4}{n} +\frac{2}{n^{2}}})}$ I got stuck here counting triangles in a graph or its complement Given a simple graph $G$ [no loops or parallel edges] on six vertices,  let $G^c$ denote its complement. It is known that either $G$ or $G^c$ must contain a triangle  $T$ in it. [An example of a Ramsey number I think.]   My question came up when I tried to find such a graph on six vertices  which had only one $T$ in it or its complement.  I believe if I did things right there are none of these. This I tried to check by drawing a copy of each graph $G$  on six vertices which has a $T$ in it,  and then looking at the complements to see if any had  no $T$'s in them. I found none.  One way to pose my question for a general number $n$ of vertices  is then as follows: Consider the set of all graphs $G$ on $n$ vertices,  and for each one construct $G^c.$  Count the number of $T$ in $G$ and add that to the number of $T$ in $G^c.$  Finally find the minimal such sum for the given $n$  and call that say $M(n).$ So my above six vertex conjecture  is that $M(6)=2.$   By considering a cycle graph on five vertices we find $M(5)=0.$   So going further one could ask if anything is known  about this minimum function $M(n)$ for larger $n$ [such as bounds, etc.] And does it have a name?  I would also like to see a simpler way to handle the six vertex case, one not involving drawing all the graphs. I have a string containing the , and I need to convert it to a Java Date object.  String date = "1081157732"; DateFormat df = new SimpleDateFormat(""); // This line try {   Date expiry = df.parse(date);  } catch (ParseException ex) {   ex.getStackTrace(); }   The marked line is where I'm having trouble. I can't work out what the argument to SimpleDateFormat() should be, or even if I should be using SimpleDateFormat(). “Fire” a weapon before firearms existed? Did the verb “fire a weapon” exist before the actual introduction of firearms on battlefields?  More specifically, does it make sense for a creative work to have archers (or whatever ranged weaponry) be told to “fire!”, when the world they live in has not yet seen firearms?  It seems some kind of an anachronism to me, since before firearms, “fire” would never propel any projectile...  I've seen several movies do it; I can't remember them all, but for instance I verified it in The Lord of the Rings trilogy (left), Kingdom of Heaven (top right), and more recently Frozen (bottom right):    The first two do it several times (with Kingdom of Heaven also using it on ballistae, and The Return of the King even having Aragorn ask Legolas to ), whereas firearms never appear on the battlefields they feature (Kingdom of Heaven is set in the 12th century; as for The Lord of the Rings, apart from one occurrence of a witchcraft-ish bomb, it's archery and medieval siege engines all the way).  Frozen may be debatable, happening probably somewhere in the 18th or 19th century, but since we only see swords, spears and the like, it made me flinch to hear crossbowmen use “fire!”.  I've also noticed other movies avoiding this, using “loose!” instead, such as Troy (left) and Gladiator (right):    This makes me think it would be the right thing to say instead... although I'm still wondering why not use “shoot”, which seems simpler to me. I'm trying to solve the following problem:  Given two CRHF $H_1:2^{4n}\to2^{2n}$, $H_2:2^{2n}\to 2^n$, construct the following hash function $H^{*}=H_2(H_1(x))$ compressing from $2^{4n}\to 2^n$. We want to demonstrate that if $H_1$ and $H_2$ are collision resistant then $H^*$ must be collision resistant too.  I made the following reduction  However I'm still unsure on how to calculate ”BAD event” in which $A^{H^∗}$ outputs a collision for $H_{s_1}$, in this case $x^∗=x′$ and the second part of the reduction doesn’t work. Why would I want to use Interfaces? I understand that they force you to implement methods and such but what I cant understand is why you would want to use them. Can anybody give me a good example or explanation on why I would want to implement this. In a modern SPFx Web Part, how do I check if a URL exists? I'm buying a bus ticket to get to Boston 23 hours before my flight to ensure I reach my departure airport on time.  If I didn't arrange bus transport well in advance, what legal remedies would I have if a delayed bus caused me to miss my flight? Suppose that sequences of real numbers satisfy:  \begin{align*}   a_n &amp;= \sum\limits_{r=0}^n \binom{n}{r} b_r \\ \end{align*}  Prove that:  \begin{align*}   (-1)^n b_n &amp;= \sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s \\ \end{align*}  My work:  The $n=0$ case:  \begin{align*}   a_0 &amp;= b_0 \\   b_0 &amp;= a_0 \\ \end{align*}  Then $n=1$ case:  \begin{align*}   a_1 &amp;= \binom{1}{0} b_0 + \binom{1}{1} b_1 = b_0 + b_1 \\   -b_1 &amp;= \binom{1}{0} a_0 - \binom{1}{1} a_1 = a_0 - a_1 \\   b_1 &amp;= a_1 - a_0 = b_0 + b_1 - a_0 = b_1 \\ \end{align*}  The inductive step:  \begin{align*}   (-1)^{n+1} b_{n+1} &amp;= \sum\limits_{s=0}^{n+1} \binom{n+1}{s} (-1)^s a_s \\ \end{align*}  I'm not sure where to go from here. I tried substituting $a_s$ into that last equation and it didn't help. Any ideas? How to obtain and use the .bbl file in my tex document for ArXiv submission? I am submitting my first article to ArXiv and in the instructions they specify I have to include the .bbl file. Their instructions are:  "We do not run BibTeX in the auto-TeXing procedure. If you use it, include in your submission the .bbl file that BibTeX produces on your own machine; otherwise your references will not come out correctly. We do not run BibTeX because the .bib database files can be quite large, and the only thing necessary to resolve the references for a given paper is the .bbl file. The name of the .bbl file must match the name of the main .tex file for the system to process the references correctly."  My knowledge of LaTeX is basic, and I have not heard of .bbl file before. I am using TeXmaker. I have my tex file and my bib file. So what do I exactly do to obtain the .bbf file correctly and submit the article?  Thanks a lot! Why does the Stern–Gerlach quantum spin experiment conflict with classical mechanics? My understanding of the Stern–Gerlach experiment is that neutral (0 total charge) particles are sent through a non-homogeneous magnetic field, with the expectation that the field will push that particle's path higher or lower on a detector because of the collective spin of that particle. While the detector can detect particles in a two-dimensional surface, the results of the experiment are that particles appear in only two localized areas directly above the path of the particle stream, or directly below the path of the particle stream - spin up or spin down. The conclusions from these measurements are that the particle, when measured, will always have about the same magnitude.  Why does this not follow from classical mechanical theory related to magnetism? If you shot a magnet through a similar apparatus, I would expect the magnet to be rotated to align with the magnetic field in some way which, at high enough field strengths relative to the mass of the magnet, would cause us to measure basically the same magnitude as if the magnet entered the apparatus pre-aligned with the field.  How is my explanation incorrect? Cannot see files in my external HDD Someone told me I need a ticket out of Thailand in order to enter.  I am traveling to Thailand, on March 18th. I have a Ticket from Bangkok to Sweden booked for June 4th.   I will be staying in Thailand for 2 weeks when I arrive there, then traveling around Cambodia, Laos and Vietnam until June 4th when I leave.   With that ticket out of Thailand, will I be able to get in and get a (30 day) Visa-On-Arrival (or not because the ticket out is not within the first 30 days). Or do I need to show proof that I am leaving within 2 weeks of being there at first?