I was asked this question by a friend today, and the question really beats me. I know that gracias and merci are like practical workarounds, but is there no synonym for Thanks in English?
The context for my question might be a bit strange.   I have stuttering and therefore I have some difficulties pronouncing some words starting with particular phonetics. And I've found that if a sentence starts with a /th/ sound, it makes me feel uncomfortable.  Especially in the word is "thanks" or "thank you".  So I want to know if there's a popular alternative for "thank you" that may not seem strange and that unnatural. I've searched the website and I know there are alternatives like "I appreciate your help." but that sounds a bit overly formal for me. (Or is it? correct me if I'm mistaken.)
Assurance that ST_Envelope works with geography as expected Using PostGIS, I have a table of data with geographic points defined as last_known_coordinates GEOGRAPHY(POINT,4326). I want to find all of these within a geographic bounding box with given top left and bottom right geographic coordinates, and be sure that the calculations are geographic, not geometric, e.g. that the results account for the curvature of the earth. The query I'm using contains the fragment: last_known_coordinates &amp;&amp; ST_MakeEnvelope(?, ?, ?, ?, 4326)::geography As you can tell I'm a bit new to this, and I know what I want, but not necessarily how to do it correctly. Other answers hint that this is correct, but I'd like to know for sure: Will this point data and SQL fragment take into account the curvature of the Earth in the bounding box? What is the procedure for taking down Titans? System log full of session and slice messages Pretty much every application I use in full-screen (including Firefox, MacVim and the Terminal) have tabs on the top of the screen.  The problem is, when I move my mouse to reach the tabs I'll often accidentally touch the top of the screen, bringing in the OS menu OVER the tabs:    Is there any way to change this behavior? I'd prefer it if the menu bar would stay hidden when I'm using a full-screen app, even if my mouse does touch the top of the screen. Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials Suppose that $$Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} $$and $$P(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}.$$ How do I find $$\lim_{x\rightarrow\infty}\frac{Q(x)}{P(x)}$$ and what does the sequence $$\frac{Q(k)}{P(k)}$$ converge to?  For example, how would I find what the sequence $$\frac{8k^2+2k-100}{3k^2+2k+1}$$ converges to?  Or what is $$\lim_{x\rightarrow\infty}\frac{3x+5}{-2x+9}?$$   This is being asked in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: . How can I get a list of all packages in a repository section from command line? Usually a probability distribution over discrete variables is described using a probability mass function (PMF):  When working with continuous random variables, we describe probability distributions using a probability density function (PDF) rather than a probability mass function. -- Deep Learning by Goodfellow, Bengio,  and Courville  However,  is using PDF to describe the probability distribution over discrete variables:    Is this a mistake? or it does not much matter? Is there any work-around if you are somewhere where images are blocked? I'm currently working in a bank, and while about 99.7% of the internet is blocked, Stack Exchange thankfully isn't (yet) however imgur is.  It means I can't view images posted alongside questions. The only workaround I have found is to search Google images for the URL, but you get a  which is a bit useless.  Does anybody know any clever tricks for viewing the images accompanying questions when you are somewhere where imgur is blocked?  Are there any plans to route the images stored on imgur through a Stack Exchange URL? "IllegalFormatConversionException: d != java.lang.String" when padding number with 0s? In a comment on this , someone said that they weren't sure what functools.wraps was doing. So, I'm asking this question so that there will be a record of it on StackOverflow for future reference: what does functools.wraps do, exactly? Use of would in sentences Should I always disconnect event handlers in the Dispose method? What is the function represented by the power series $$ \sum_{k=2}^\infty\frac{x^k}{k(k-1)}\quad? $$  It looks like $\dfrac1{1-x}$ but I don't know. I am trying to create a beamer animation using tikz, but I can't even get basic rotating of a tikz image working as desired.  I have a rectangle specified by  \draw[rounded corners,dotted] (-3.5,0) rectangle (-0.5,2);   I would like to rotate this rectangle about the point (-1,.5) (which is near the bottom right corner of the rectangle).  I've tried:  {\rotatebox[origin=c,x=-1,y=.5]{130}{             \draw[rounded corners,dotted] (-3.5,0) rectangle (-0.5,2);}}   But this produces an error.  If I just use the origin=c option, it compiles, but it rotates the rectangle about the point (0,0).  I've also tried it with just the x=-1,y=.5 (without the origin option), and it fails to compile.  I have searched the documentation, which is where I learned about the options x=xdim and y=ydim.  However, I can't find a single example demonstrating how these options should be used.  Any suggestions would be greatly appreciated.    Also, if there is a better command for rotating things that will be more suitable for use in a beamer animation, please let me know.  Thanks! Flying EU to UK to EU. Can I get duty free alcohol? On the 20 of March this year I will be flying from Ireland to the UK and then on the another EU country.  Can I buy duty free alcohol (as in alcohol with no tax) leaving the UK ? Strikeout in math mode How can I strikeout words in math mode (similar to \sout with \usepackage{ulem} in text mode)? I've tried \usepackage{cancel} and \(\cancel{1+1=2}\) but that draws a diagonal line. Determinant of block matrix with commuting blocks I know that given a $2N \times 2N$ block matrix with $N \times N$ blocks like   $$\mathbf{S} = \begin{pmatrix} A &amp; B\\  C &amp; D \end{pmatrix}$$  we can calculate $$\det(\mathbf{S})=\det(AD-BD^{-1}CD)$$ and so clearly if $D$ and $C$ commute this reduces to $\det(AD-BC)$, which is a very nice property.   My question is, for a general $nN\times nN$ matrix with $N\times N$ blocks where all of the blocks commute with each other, can we find the determinant in a similar way? That is, by first finding the determinant treating the blocks like scalars and then taking the determinant of the resulting $N\times N$ matrix.  Thanks in advance! I think the answer is yes but I know a lot of people disagree. So, I would like to ask these people when exactly does it stop being me.  Let's say I want to upload my brain into a computer using the following procedure:  I have each of my neurons replaced by electronic neurons. By electronic neurons I mean chips with the necessary hardware and software to perform the exact same function that the neuron being replaced was performing. This means that given the same inputs that the biological neuron was given it would produce the same outputs. Then I connect that electronic neuron to all the neurons that former biological neuron was connected to.   At the end of this process is it still me? If not when did it stop being me? When one neuron was replaced? When 1,000 were replaced? When 1,000,000 were replaced?  (assuming it is still me)  Then, I divide these electronic neurons into groups and for each group I perform the following:   Write software to simulate the group of electrical neurons in my computer such that given the same inputs that go into the group of electronic neurons in my brain, it will produce the same outputs. Replace each group of neurons with another chip that communicates with my computer. Each of the chips will communicate to the corresponding software module in order to determine how to transform its inputs into the necessary outputs. Connect all of the inputs and outputs of the group of electronic neurons to that new chip.   At the end of this process is it still me? If not when did it stop being me? When one group was replaced? When 1,000 were replaced? When 1,000,000 were replaced?  (assuming it is still me) Now, what I have is a bunch of these new chips in my head, connected to each other and for each of these chips I have a small neural network software module in my computer. So, now for each pair of chip/software module I do this:   Transfer all connections going into and out of the chip into the computer version (i.e. connect the software module to the other software modules that represent the other chips that this chip is connected to) Disconnect the chip from the other chips and remove it   At the end of this process is it still me? If not when did it stop being me? When one of the new chips was replaced? When 1,000 were replaced? When 1,000,000 were replaced?  If it is still me then I have successfully moved my brain into a computer without ceasing being myself.  So, again the question is if at the end it is not me anymore when exactly does it stop being me and why does it stop being me?  EDIT: This question assumes that our universe and everything in it (including ourselves) follows the (testable) laws of physics. As it stands it also assumes that neuroscience is mostly right, although the question could also be considered without that requirement if instead of neural simulations we were to implement physics simulation of the brain. Of course this is a lot less practical. However, if your view is that our minds have a supernatural component then the question does not make any sense.  It was also pointed out that this is related to the Ship of Theseus, i.e. is something still the same if all its parts have been changed. With regards to that I think it is useful to consider that most of the cells in our body (not including brain cells) are regularly replaced anyway and some people even get entire organs replaced. Yet, we don't say that they stopped being themselves. Note, however, that if you are assuming a supernatural component of the mind that this explanation is not really adequate or relevant. This is a follow up from this .  I'm using this slightly modified script to enumerate all installed MSI packages:  strComputer = "."  Set objWMIService = GetObject("winmgmts:" &amp; _     "{impersonationLevel=impersonate}!\\" &amp; _     strComputer &amp; _     "\root\cimv2")  Set colSoftware = objWMIService.ExecQuery _     ("SELECT * FROM Win32_Product")     If colSoftware.Count &gt; 0 Then     For Each objSoftware in colSoftware         WScript.Echo objSoftware.Caption &amp; vbtab &amp; _         objSoftware.Version     Next Else     WScript.Echo "Cannot retrieve software from this computer." End If   What is surprising however, is its abysmal performance. Enumerating the 34 installed MSI packages on my XP box takes between 3 and 5 minutes !  By comparison, the Linux box that sits besides is taking 7s to enumerate 1400+ RPMs... sigh  Any clues on this ? How to prove $a^3 + b^3 +c^3 - 3abc =1 $, If a,b,c are infinite series as given in the description box below? How do I prove that $$ a^3+b^3+c^3−3abc = 1 $$ if   $ a = 1+\frac{x^3}{3!}+\frac{x^6}{6!}......$ , $b=x+\frac{x^4}{4!}+\frac{x^7}{7!}....$ , $c=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}......$ ?  I have tried the following :-  I discovered that $a+b+c=e^x$ and that $a=\frac{d(b)}{dx}$, $b=\frac{d(c)}{dx}$, $c=\frac{d(a)}{dx}$ . But I am unable to proceed further with these information.