Here's a screenshot showcasing my problem:     I found  on this issue, but for some reason, it's not working for me. I'm using the Inset tool, and setting the depth to a negative number.  Do I need edge loops first for that to work? There's no way for blender to create vertices at the right spot automatically?
As the title says, how can I make this shape without resorting to a Boolean operation?    This is obviously a very simple example, but I find myself needing to do something like this relatively often when modelling.  Despite using Blender for a couple of years, and considering myself an intermediate to possibly advanced user, I still know of no easy way to make this kind of indentation in an object.  Obviously, if you are looking to make an indentation in the middle of the surface, you can simply select a face and extrude down. As you will know, this does not work at the edge of the surface.  I can create this shape using loop cuts, deleting the faces and then using snap to edge to extrude the edges down to the same level (and then filling in the faces), but it feels cumbersome and a lot of work for what feels like it should be very simple.
Let $E$ be an empty set. Then, $\sup(E) = -\infty$ and $\inf(E)=+\infty$. I thought it is only meaningful to talk about $\inf(E)$ and $\sup(E)$ if $E$ is non-empty and bounded?   Thank you. System settings is missing System settings is missing. I tried reinstall it as it mentioned , but it seems to be installed and deactivated. This problem occurred when I installed Elementary desktop environment and uninstalled it afterwards. What is meant by a "random variable"? Lasso regression feature selection I have been reading many articles on LASSO regression, and everyone claims that LASSO address multicollinearity showing contour plots of cost function touching the corner of the diamond(x1+x2). In the first place, I didn't understand how does this phenomenon translate to addressing multicollinearity when there are many groups of correlated variables in the data.  Verbatim from one of the research paper: "• If there are grouped variables (highly correlated between each other) LASSO tends to select one variable from each group ignoring the others". I'm a little confused as to when to use  for my physics class. For example, I'm asked to find the average speed of a race car that travels around a circular track with a radius of $500~\mathrm{m}$ in $50~\mathrm{s}$.  Would I need to apply the rules of significant figures to this step of the problem? $$ C = 2\pi (1000~\mathrm{m}) = 6283.19 $$  Or do I just need to apply significant figures at this step? $$ \text{Average speed} = \frac{6283.19~\mathrm{m}}{50~\mathrm{s}} = 125.664~\mathrm{m}/\mathrm{s} $$  Should I round $125.664~\mathrm{m}/\mathrm{s}$  to  $130~\mathrm{m}/\mathrm{s}$ since the number with the least amount of significant figures is two? How can I obfuscate (protect) JavaScript? I'm a big fan of the votes tab in the profile page, as it allows me to keep track of my usage of the moderation tools. Or at least it would, if votes cast on deleted posts didn't unceremoniously disappear.  Although I don't like it, I can live with that for (up|down|close|reopen)-votes. In fact, I didn't even notice it before gaining 10k rep on SU. But it renders the deletion and undelete subtabs pretty much useless, since it hides – by design – all successful deletion and (yet) unsuccessful undeletion votes.  @slhck suggested in an answer to  the following rationale for this behavior:     While this sounds strange, it's just a result of the … let's call it "Stack Exchange policy" of not showing lists of deleted posts to non-moderators. In my opinion this policy is a little incoherent, because your flagging record shows deleted posts [...]   Actually, the flagging record shows deleted posts to pretty much all users. So does the reputation tab of the user profile (as long as you earned/lost reputation on that post).  If that is in fact the reason, why can't a 10k user see a list of posts he voted on (and thus, already saw)? Using recordtypes of the same name across multiple SObject types I have one recordtype named 'India' in both Account and Contact Objects.My requirement is whenever i create a new account record of type 'India' a new contact record of type 'India' should be created automatically.  i tried with the following code:  trigger AccRecTypeTrigger on Account (before insert) {     List&lt;RecordType&gt; recType = [select id,name from RecordType where sobjectType = 'Account' and isActive = true];    Map&lt;String,Id&gt; m = new Map&lt;String,Id&gt;();    for(RecordType r : recType){      m.put(r.name,r.id);    }     List&lt;Contact&gt; conlst = new List&lt;Contact&gt;();    for(Account a : trigger.new){        a.recordTypeId = m.get('India');        Contact con = new Contact();        con.accountId = a.id;        con.recordTypeId = a.recordTypeId;        con.recordType.name = a.recordType.name;         conlst.add(con);       }    insert conlst; }   But i am getting following error when i try to save     Apex trigger AccRecTypeTrigger caused an unexpected exception, contact   your administrator: AccRecTypeTrigger: execution of BeforeInsert   caused by: System.NullPointerException: Attempt to de-reference a null   object: Trigger.AccRecTypeTrigger: line 14, column 1   Please help me how to achieve this...... Shooting many bullets toward the mouse position I try to shoot a bullet toward the mouse position by this code.  float deltaX = mousePosition.x - (aircraft.getPosition().x + aircraft.getLocalBounds().width / 2); float deltaY = mousePosition.y - (aircraft.getPosition().y + aircraft.getLocalBounds().height / 2); float angleRadian = (float)(std::atan2(deltaY, deltaX)); float angleDegree = (float)(angleRadian * 180 / PI);   It works fine with one bullet but when I add another one bullet to shoot together I use this code to specify start shooting point of each bullets.  sf::Vector2f offset1(aircraft.getPosition()); offset1.x += -10;//start shooting point first bullet.  sf::Vector2f offset2(aircraft.getPosition()); offset2.x += 10;//start shooting point second bullet.   this is a picture of this problem.   When I rotate aircraft to picture 2 the bullets are not the same as the first picture.  How can I solve this problem? What is the best way to draw figures with canvas and set their angle? Batch file to delete files older than N days Where do I get a SIS mirage 672 vga driver? I'm planning to fly from Ulaanbaatar to Melbourne via Seoul (and Kuala Lumpur).  However, can I actually enter South Korea using this visa waiver? I'm hoping to meet a friend there for a few hours before flying. Also, would traveling via Malaysia affect this policy?  Transit tourists bound for another country Eligible countries  Applicable to all countries (save for Macedonia, Cuba, Syria, Sudan and Iran) that are not granted visa-free entry into Korea.  Applicable to…  Those with a visa (excluding Japanses Group Visa, excluding e-visa) or a re-entry permit for U.S., Japan, Canada, Australia, or New Zealand (5 countries) and traveling to any one of those countries through Korea.  Those traveling from any one of the five countries mentioned above, whose final destination is not Korea.  However, please note that the first bullet points are not applicable to those travelling to/from Guam and Saipan, which are unincorporated territories of the United States.   Requirements  Must have a confirmed onward flight ticket for departure within 30 days after entering Korea, in addition to having no record of criminal offence in the five countries mentioned above. UMVUE of $e^{-\lambda}$ from poisson distribution Let $X_1,\ldots,X_n \sim$ Poisson$(\lambda)$. I wish to find UMVUE of $e^{-\lambda}$. Here $\overline{X}_n$ is complete and sufficient for $\lambda$ (hence for $e^{-\lambda}$??). Define $Y_i=1$ if $X_i=0,$ $Y_i=0$ otherwise. Clearly (I checked!) each $Y_i$ and hence $\overline{Y}_n$ is unbiased for $e^{-\lambda}$. Then I should be able to use the Lehmann Scheffe theorem to deduce the UMVUE. The estimator I get looks like $$\frac{1}{n} E\left\{\#X_i:X_i=0 \mid \sum_{i=1}^n X_i\right\}$$ whereas the correct answer is $$\left(\frac{n-1}{n}\right)^{\sum_{i=1}^n X_i}$$ I'd appreciate if anyone lets me know where I'm going wrong and guides me into the right path. Thank you. An exciting new feature is described in the :        There is a new NPC housing interface that can be used to assign living areas to NPC’s. This interface will also tell you if a room is missing any requirements.      How can this interface be accessed? How to make clickable links and cross-references in pdfs produced by LaTeX? How can I make things link url references, my index, table of contents, and cross references clickable links in documents produced by LaTeX? I generally use pdflatex to produce my documents. Zooming in to inspect texture details in 3D view clips some closer faces like shown in the image. It's not the camera view so changing camera settings doesn't help.   Where can the distance to the clipping plane be adjusted? NSArray becomes NSCFArray when passed My application got crashed because of not initializing the view Disproving the statement "If $f(z)$ is not an entire function, then $g(z) = f^2(z)$ is not entire.” If $f(z)$ is not an entire function, then $g(z) = (f(z))^2$ cannot be an entire function.”  From this statement why would saying &quot;Let $f(z) = \sqrt z$, assuming that it is a branch that takes $−1$ to $i$. Then it is not analytic on the branch cut, but $g(z) = (f(z))^2 = z$ is obviously an entire function.&quot; Not be a proper example to disprove this statement/ be a incorrect counterexample? I understand that for $\sqrt z$ to be a incorrect counterexample means it is not analytic on the branch -1 to I, therefore not entire. However I thought $\sqrt z$ would be analytic throughout the whole branch cut from -1 to i. Since on the complex plane, $\lim_{z \to -1} \sqrt z$ exists as it would approach $i$ in that case, and in the other case $\lim_{z \to i} \sqrt z$ it approaches $\sqrt i$ meaning that at least on that branch cut in the complex plane it is continuous everywhere, therefore the partial derivatives exist everywhere on the branch cut meaning it is analytic, but this assumption seems to be incorrect. What am I missing or not understanding here. Precisely, I don't understand why $\sqrt z$ would not be a proper counterexample to the statement.