How to join two separate shapes with an intermediate one so my probleme is quite simple. i have 2 circles away from each other and not intersecting, I want to connect those 2 cirlces with a rectangle in the middle.  heres what I have    and heres what i want     however as u can see its not perfect. I would like to know how. (Ai cc 2018)
Illustrator: Is there an efficient and accurate way to join two circles by their tangents (and keep it editable)? It's easy to join two circles if they are the same size and horizontal/vertical, into a pill shape:    If the circles are at an add angle then it's still possible but you can no longer rely on smart guides.     You can add lines roughly where the tangents should be and then use the scissors tool on the circles, cut out the inner area and then join the two crescents into one shape.    Is there a better way to do this? Is there a way to do this that is still editable afterwards too?  There are some other questions that cover the same area as this but don't cover editability or joining circles that are not in line with each other (where smart guides can't help).
A virtual file containing the concatenation of other files Why does no physical energy-momentum tensor exist for the gravitational field? Starting  with the Einstein-Hilbert Lagrangian  $$ L_{EH} = -\frac{1}{2}(R + 2\Lambda)$$  one can formally calculate a gravitational energy-momentum tensor   $$ T_{EH}^{\mu\nu} = -2 \frac{\delta L_{EH}}{\delta g_{\mu\nu}}$$  leading to  $$ T_{EH}^{\mu\nu} = -G_{\mu\nu} + \Lambda g_{\mu\nu} = -(R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R) + \Lambda g_{\mu\nu}. $$  But then, in the paragraph below Eq. (228) on page 62 of  paper, it is said that this quantity is not a physical quantity and that it is well known that for the gravitational field no (physical) energy-momentum tensor exists.  To me personally, this fact is rather surprising than well known. So can somebody explain to me (mathematically and/or "intuitively") why there is no energy-momentum tensor for the gravitational field? Solving linear congruences by hand: modular fractions and inverses When I am faced with a simple linear congruence such as  $$9x \equiv 7 \pmod{13}$$ and I am working without any calculating aid handy, I tend to do something like the following:  "Notice" that adding $13$ on the right and subtracting $13x$ on the left gives: $$-4x \equiv 20 \pmod{13}$$  so that $$x \equiv -5 \equiv 8 \pmod{13}.$$  Clearly this process works and is easy to justify (apart from not having an algorithm for "noticing"), but my question is this: I have a vague recollection of reading somewhere this sort of process was the preferred method of C. F. Gauss, but I cannot find any evidence for this now, so does anyone know anything about this, or could provide a reference? (Or have I just imagined it all?)  I would also be interested to hear if anyone else does anything similar. If I've decided to update my SE avatar (in any of the various ways well described in the answer to this ), am I able to return to either the original stock image or previous avatar images I have used? What control type should I use - Image, MediaElement, etc.? I just notice it now as I am new to Illustrator.  When I am in RGB mode, blending transparency seems to be easier as mode such as Screen, Multiply or Overlay looks good naturally.  If I switch the document mode to CMYK then the image is getting messed up.  I'd like to know, if CMYK is not ideal for blending options? In a ring $(A,+, \cdot)$ if $aba = a$ then $bab = b$ and all non zero elements in $A$ are invertible. $r$ primitive root of prime $p$, where $p \equiv 1 \mod 4$: prove $-r$ is also a primitive root question is as follows:  Let $p$ be a prime with $p \equiv  1 \mod 4$, and $r$ be a primitive root of $p$.  Prove that $-r$ is also a primitive root of $p$.  I have shown that $-r^{\phi(p)} \equiv 1 \mod p$.  What I am having trouble showing, however, is that the order of $-r$ modulo $p$ is not some number (dividing $\phi(p)$ that is LESS than ($\phi(p)$).  By way of contradiction, I've shown that the order of $-r$ cannot be an EVEN number less than $\phi(p)$.  But my methodology does not work for the hypothetical possibility of an ODD order that is less than $\phi(p)$.  Any and all help appreciated.  Happy to show methodology for any of the parts I have managed to do, if requested. In my object record type i have created formula field. Now i want my formula field to work as below.  If Basement__c = NA and Property__r.Basement__c = NA OR  Basement__c = Crawl and Property__r.Basement__c = Crawl OR  Basement__c = Slab and Property__r.Basement__c = Slab Then it will return Type__c(formula field) = "MAtch"  And (Another rule) Basement__c DOES NOT contain any of the following "N/A,Crawl,Slab,None and Property__r.Basement__c Does not contain any of this then formula field will return "HAs basement"  How can i achieve this? Please help. What's the easiest way to run GUI apps on Windows Subsystem for Linux as of 2018? I searched around, and currently there are two methods suggested; installing an enhancement for Windows Subsystem for Linux and installing an XServer.  I want to know which method is the most hassle-free (easy to install AND to use), and which one is less memory-heavy.  I just want Synaptic and CMake. Why couldn't that be a builtin feature? This is rather difficult to describe with fancy pictures with red circles but it's easily reproducible.  Whenever I click Leave Closed in the re-open queue on MSO the "nominate this question for re-opening" pop-up appears. I normally press escape out of shock and a leave closed vote is recorded.  This is the latest review it's happened to me on:   You can also test it on , I clicked reopen this time and a leave closed vote was recorded but I also got a red popup stating that an error had occurred.  This is also occurring when the Edit &amp; Reopen button is clicked, the edit box opens for a second before being overridden by the popup. Prove $x \equiv a \pmod{p}$ and $x \equiv a \pmod{q}$ then $x \equiv a\pmod{pq}$ Why does Calculate button remain gray in Group Stats Plugin for QGIS? In my case I failed to use  plugin (2.0.30) with QGIS 2.14.11 and 2.18.3 on windows64.  Tested on several datasets: points, polygons, and csv.- But I am not sure what I have missed.    I want to click on this Calculate button. How t show $f=g$ almost everywhere A limit ordinal $\gamma&gt;0$ is said to be indecomposable iff $\nexists\alpha,\beta&lt;\gamma$ such that $\alpha+\beta=\gamma$.  In view of this definition, I’m trying to prove the equivalence of the following statements for limit ordinals $\gamma$:  (1) $\gamma$ is indecomposable.  (2) $\alpha+\gamma=\gamma$ $\forall\alpha&lt;\gamma$.  (3) $\gamma=\omega^\alpha$ for some $\alpha\in\mathrm{ON}$.  I’ve been able to get $(1)\implies(2)$ and $(3)\implies(1)$, but I have nothing more than scratch work for $(2)\implies(3)$.  Any suggestions would be appreciated. Does the order of declaration matter in Java/C#? In C++ I can't use a method if I declare it after the calling method.    Does this order matter in other languages like Java or C#? I am not sure which conjunction to use in the following context:        I know about programming but I have a little bit of confusion in Java programming.   I know about programming although I have a little bit of confusion in programming.    I know about programming however I have a little bit of confusion in Java programming. I am looking for a multivariable calculus book that is really physics oriented.  Anyone know of any?  EDIT: My wife is looking to brush up on multivariable at the same time she needs to brush up on the related physics. I am unable to find the bias of the sample variance estimator in this problem. Unfortunately I keep coming up with   Question: Suppose that the true linear model is   $y = X_1\beta_1 + X_2\beta_2 + e$,   where $e \sim MVN(0,\sigma^2 I)$, $X_1: N \times p_1, \beta_1: p_1 \times 1, X_2: N \times p_2, \beta_2: p_2 \times 1$. But we wrongly fit the model  $y = X_1\beta_1+e$ and calculate the estimators of $\beta_1$ and $\sigma^2$ as:  $\hat{\beta}_1 = (X_1'X_1)^{-1}X_1'y$,  $s^2 = \large \frac{y'[I-X_1(X_1'X_1)^{-1}X_1']y}{N-p_1}$.  (a) Show that, in general, $\hat{\beta}_1$ is a biased estimator with  Bias$(\hat{\beta}_1) = (X_1'X_1)^{-1}X_1'X_2\hat{\beta}_2$  Under what condition on $X_1$ and $X_2$ is $\hat{\beta}_1$ unbiased?  (b) Show that, in general, $s^2$ is a biased estimator with   Bias$(s^2) = \frac{1}{N-p_1}\beta_2'X_2'[I-X_1(X_1'X_1)^{-1}X_1']X_2\beta_2$  Is the bias positive or negative? Under what condition on $\beta_2$ is $s^2$ unbiased?  Attempt:  For part (a):  Bias$(\hat{\beta}_1) = E((X_1'X_1)^{-1}X_1'y - \beta_1)$  $\hspace{15mm}= (X_1'X_1)^{-1}X_1'E(y) - \beta_1$  $\hspace{15mm}= (X_1'X_1)^{-1}X_1'(X_1\beta_1 + X_2\beta_2) - \beta_1$  $\hspace{15mm}= (X_1'X_1)^{-1}X_1'X_1\beta_1 + (X_1'X_1)^{-1}X_1'X_2\beta_2) - \beta_1$  $\hspace{15mm}= I\beta_1 + (X_1'X_1)^{-1}X_1'X_2\beta_2 - \beta_1$  $\hspace{15mm}= \beta_1 + (X_1'X_1)^{-1}X_1'X_2\beta_2 - \beta_1$  $\hspace{15mm}= (X_1'X_1)^{-1}X_1'X_2\beta_2$  If $X_1$ and $X_2$ are orthogonal, then $\hat{\beta}_1$ is unbiased.  For part (b), I am unable to figure out how to derive:  Bias$(s^2) = \frac{1}{N-p_1}\beta_2'X_2'[I-X_1(X_1'X_1)^{-1}X_1']X_2\beta_2$  I can get   $\frac{1}{N-p_1}\beta_1'X_1'+\beta_2'X_2'[I-X_1(X_1'X_1)^{-1}X_1']X_1\beta_1+X_2\beta_2$  after taking the expectation of y on both sides.   Any help would be much appreciated. My two main questions:  Are all URLs that include your User Id tracked?  Can the visitors who use the tracked URLs have Stack Exchange accounts?   Specifically: Recently the ability to earn badges for sharing links to posts on Stack Exchange sites . I'd like to know when and how this tracking works. It's clear that if you go to a question, and make use of the link button, you will get a URL with your User ID that will be tracked if a visitor uses it to get to SO. For example, let's use  as an example. It's regular URL is:  https://stackoverflow.com/questions/309300/defend-php-convince-me-it-isnt-horrible  If you go to that page and use &quot;save link as&quot; in the link section of the question you will be given a tracked url with your User ID:  https://stackoverflow.com/q/309300/[USER ID]  My question is, are there other URLs that would be tracked? All of the following URLs take you to the Question / Answer, but are they tracked? For example what if I just tack my User ID onto the long url?  https://stackoverflow.com/questions/309300/defend-php-convince-me-it-isnt-horrible/[USER ID]  What if I add my User ID to an answer URL?  https://stackoverflow.com/questions/309300/defend-php-convince-me-it-isnt-horrible/309622#309622/[USER ID]  or even  https://stackoverflow.com/q/309622/[USER ID]   Is there a way to know that a URL is being tracked? Is there some sort of format to follow? Are the only tracked URLs the ones provided in the Question links? Finally, does this sentence from the blog post mean that visitors cannot have an account?  Also, the tracked IPs must originate from outside our existing network.