Submit button needs to be clicked twice on IE
why does IE need for me to click twice
Consider the blow-up of the plane in one point. Let $E$ the exceptional divisor. We know that $(E,E)=-1$. What is the geometrical reason for which the auto-intersection of $E$ is $-1$? In general, what does it mean, geometrically, that a divisor has a negative self-intersection or that the "right" number of divisors gives rise to a negative intersection?  Thanks. I have a Windows 10 Home PC that is often unattended but doing important work. The work follows no particular schedule, and may take place at any time of day or night.  As things stand, Windows 10 (anniversary update) is configured to automatically restart and install updates during inactive times. The user can configure the inactive times, but the OS forces the user to have no more than 12 active hours a day. This means that the machine may well choose to restart at a time that is highly disruptive to our work and when there is no user around to prevent the restart.  For this reason, I would like to ensure that Windows never automatically restarts. How can I achieve this? I want to fill a part of a picture with more or less evenly and randomly distributed circles which should not overlap. It would be okay if circles are only partly inside the shape.  The density should be like what you see on this page  with values 80-100. (But I do not need so many circles, 30 should be enough, I could repeat the pattern).   I cannot use a grid and "wobble" the circles a bit: The circles are too large and it simply does not look randomly.   Currently I am thinking about a random attempt: Get some random point, measure its distance to existing circles, either paint or throw the point away. Try 100 times and hope that you get the right number of circles (and that it does not take too long). But perhaps someone has a better idea.   The example draws the circles simply manually:  \documentclass{article} \usepackage{tikz} \usetikzlibrary{backgrounds} \begin{document}   \begin{tikzpicture}[framed,gridded,radius=0.5cm]  \draw[red](0,0) rectangle (5,5);   \draw (1,1) circle;   \draw (2.3,1.1) circle ;   \draw (4.5,0.8) circle ;   \draw (5.1,1.8) circle ;   \draw (0.4,3.3) circle ;   \draw (2.1,2.8) circle ;   \draw (3.8,3.5) circle ;   \draw (4.8,4.2) circle ;   \draw (0.8,4.9) circle ;   \draw (2.1,4.1) circle ;   \draw (3.8,2.0) circle ;   \draw (3.5,0.6) circle ;   \draw (3.0,5.0) circle ;      \draw (4.1,5.1) circle ;             \draw (0.9,2.1) circle ;  \end{tikzpicture}  \end{document} Can words like "what" be the subject of a sentence? In a question like "Who hears a noise?",  is the subject of the sentence who?  I can think of a few tests for subjects like: "the subject is the phrase that inverts with the auxiliary to form a question". But this is a question and there is no auxiliary. We could say that's because it's a subject, but that would be just to presuppose that it's a subject in the first place.  I can think of tests like: "the form of the verb will agree with the number of the subject", but on the other hand who does not seem to have  fixed grammatical number, and it sometimes seems like who might actually be agreeing with noun phrases later in the sentence:   Who are the most prolific writers of our age? Who is the most prolific writer of our age?   There are also tests like "the subject is usually the first noun phrase in the sentence". But of course if we have a question like:   Who have you bitten?   ... you, not who seems to be the subject. The first noun phrase test is not very good.  I also know that some theories of grammar say that there is a gap in wh- questions that the wh- word is extracted from. So can this gap for instance be the real object of a question? If so, is [See community wiki post in the linked to question]  How can we show whether who is really the subject of my original example question? And how about other wh- words like what? Can they (also) function as subjects? I have sucessfully installed AniMove by activating experimental plugins (yay! thanks gis.stackexchange). But I can't find it in Qgis! Yes I have activated it. Where is it? Is this a common bug? I searched for it the toolbox, I looked up the names of the several single tools based on this  and only found "Kernel Density", which is filed under SAGA. But I don't have SAGA installed (and don't know if I need to?) Why can't a single ReLU learn a ReLU? As a follow-up to  I simplified even more and tried to train a single ReLU (with random weight) to a single ReLU.  This is the simplest network there is, and yet half the time it fails to converge.  If the initial guess is in the same orientation as the target, it learns quickly and converges to the correct weight of 1:      If the initial guess is "backwards", it gets stuck at a weight of zero and never goes through it to the region of lower loss:        I don't understand why.  Shouldn't gradient descent easily follow the loss curve to the global minima?  Example code:  from tensorflow.keras.models import Sequential from tensorflow.keras.layers import Dense, ReLU from tensorflow import keras import numpy as np import matplotlib.pyplot as plt  batch = 1000   def tests():     while True:         test = np.random.randn(batch)          # Generate ReLU test case         X = test         Y = test.copy()         Y[Y &lt; 0] = 0          yield X, Y   model = Sequential([Dense(1, input_dim=1, activation=None, use_bias=False)]) model.add(ReLU()) model.set_weights([[[-10]]])  model.compile(loss='mean_squared_error', optimizer='sgd')   class LossHistory(keras.callbacks.Callback):     def on_train_begin(self, logs={}):         self.losses = []         self.weights = []         self.n = 0         self.n += 1      def on_epoch_end(self, batch, logs={}):         self.losses.append(logs.get('loss'))         w = model.get_weights()         self.weights.append([x.flatten()[0] for x in w])         self.n += 1   history = LossHistory()  model.fit_generator(tests(), steps_per_epoch=100, epochs=20,                     callbacks=[history])  fig, (ax1, ax2) = plt.subplots(2, 1, True, num='Learning')  ax1.set_title('ReLU learning ReLU') ax1.semilogy(history.losses) ax1.set_ylabel('Loss') ax1.grid(True, which="both") ax1.margins(0, 0.05)  ax2.plot(history.weights) ax2.set_ylabel('Weight') ax2.set_xlabel('Epoch') ax2.grid(True, which="both") ax2.margins(0, 0.05)  plt.tight_layout() plt.show()     Similar things happen if I add bias: 2D loss function is smooth and simple, but if the relu starts upside down, it circles around and gets stuck (red starting points), and doesn't follow the gradient down to the minimum (like it does for blue starting points):    Similar things happen if I add output weight and bias, too.  (It will flip left-to-right, or down-to-up, but not both.) Java - Could not find the main class Adding the consensus to review's history I'm an Egyptian citizen, who holds Schengen visa and stay permit in Italy. Can I go to UK for some days with my stay permit and Schengen visa? REST API composite query: How do I use the result of a query subrequest in the referenceId of another subrequest? I am using Salesforce's REST API to (1) retrieve a list of object IDs according to a query, then (2) retrieve the records that correlate to those IDs.  I am familiar with the Salesforce composite documentation .    My request looks like this:  POST: /services/data/v48.0/composite  {     "allOrNone":false,     "compositeRequest": [ {         "method":"GET",         "url": "/services/data/v48.0/query?q=SELECT+Id+FROM+My_Entity+WHERE+(CreatedById+!=+'asdfghjkl')",         "referenceId": "myEntity"         },{         "method":"GET",         "url": "/services/data/v48.0/composite/sobjects/My_Entity?ids=@{myEntity.records[0].Id}",         "referenceId":"myObject"     }] }   My response:  {     "compositeResponse" : [ {         "body" : {             "totalSize" : 21,             "done" : true,             "records" : [ {                 "attributes" : {                     "type" : "My_Entity",                     "url" : "/services/data/v48.0/sobjects/My_Entity/a0D6g000xxxxxxx"                 },                 "Id" : "a0D6g000xxxxxxx"                 }, {*record2*}, {*etc*}             ]},             "httpHeaders" : { },             "httpStatusCode" : 200,             "referenceId" : "myEntity"         }, {         "body" : [ {         "errorCode" : "PROCESSING_HALTED",         "message" : "'myEntity.records' references an invalid Datatype. Only Strings and primitive data types are allowed to be referenced from previous operations."         } ],         "httpHeaders" : { },         "httpStatusCode" : 400,         "referenceId" : "myObject"     } ] }   The first subrequest (the query) executes without errors.  My problem is the second subrequest.  I understand from  that I can refer to the ID of a single item (e.g. myEntity.records[0].Id).  Question: How do I use all items from myEntity.records in the second subrequest? Windows 7 SP1 Windows Update stuck checking for updates I installed Windows 7 fresh and installed SP1. Now, when I try to check manually for Windows Updates it just hangs on the Checking for updates screen.    I tried running the tools in , but this did not fix the issue either:  No matter what I do it just hangs on the "Checking for updates..." screen and goes no further. VBA array sort function? I'm looking for a decent sort implementation for arrays in VBA. A Quicksort would be preferred. Or any other  other than bubble or merge would suffice.  Please note that this is to work with MS Project 2003, so should avoid any of the Excel native functions and anything .net related. Each number in a subset $S\subseteq \{1,\ldots,2n\}$ does not divide another one. Then $\max |S|$? This problem comes from a seemingly innocuous question from a professor during a lesson for a Math Olympiad course. [A part of this question is really a classic of number theory/combinatorics]  Problem: Let $S$ be a a non-empty subset of $\{1,2,\ldots,2n\}$. It is clear that if $|S|$ is "enough large" then, independently from the choice of $S$, there are for sure some distinct $a,b \in S$ such that $a$ divides $b$.  Then, what is the maximal $k$ such that, for some $S$ with cardinality $k$, if $a,b$ are distinct members of $S$ then $a$ does not divide $b$?  [Ps. I know the solution] Usage in sentence:  Person A: Ooh - look at that crow!  Person B: Well, actually, as someone who is a scientist who studies crows, I am telling you, specifically, in science, no one calls jackdaws crows. They're not the same thing. If you're saying "crow family" you're referring to the taxonomic grouping of Corvidae, which includes things from nutcrackers to blue jays to ravens. It's not one or the other, that's not how taxonomy works.  Person A: Are you always such an XXX?   "Pedant" is close, but not quite.  It would be a mix of "know-it-all", "condescending" and "arrogant". How can I stop certain apps from appearing in the Dash? Limit of $n$th root of $n$! Multivariable Delta Epsilon Proof $\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4}$ --- looking for a hint I have the limit $$\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4},$$ and would like to show with an $\epsilon-\delta$ proof that it is zero. I know with a situation like $$\left|\frac{x^4y}{x^4+y^4}\right|\leq y$$  or something similar, but I can't find a way to do the same thing here, as no single term in the numerator is of sufficient degree, although I think I could get this with a small hint. Is "laic" an alternative word to the word "layperson"? How did this question get a 'linked' question that isn't mentioned in comments or answers? Looking at  which has been closed as a duplicate (quite right), we see three 'linked' questions in the sidebar; however, only two of those questions are given as duplicate-close-targets.  which to my mind is the actual 'exact duplicate', isn't mentioned anywhere else except in the 'linked questions' sidebar itself! The  says:  The Linked sidebar... gathers up any links to this question provided by the community via comments, answers, or questions  So how did  get into the linked questions sidebar of ? (There's no explicit link in the other direction, in case you were wondering.) Is the trace of inverse matrix convex?