How to Delete multiple records in Linq to Entity?
How do I delete multiple rows in Entity Framework (without foreach)
Extending from two classes Initial note: Having read , the answers there do not help in this specific situation, as they rely on having lots space to spare - I have very little.  When I started my city, I designed a simple exit from the default highway that I thought was sustainable: I left enough space around it to upgrade the road as I needed to, which has worked up until now. However, I am now stuck with a six-lane one-way exit from the highway that isn't getting used: since cars seem to pick the most efficient lane to get where they want to go, regardless of traffic, the rightmost lane fills up, then spills over onto the highway. This blocks the highway, my bus routes, and more importantly my planned emergency routes.  My setup looks like this:    The red arrows show road direction, and the orange line is the traffic backed up. Note that in this image, I have replaced the six-lane one-way exit with a highway exit, as this is just about as efficient...  This situation is getting to be pretty bad, and is bringing half my city to a standstill. Its effects compound themselves further down the highway:    In this image, the orange line is the traffic directly caused by this exit not working. This traffic jam then blocks two exits/entrances further down, on the red crosses; the yellow line shows the traffic jam caused by these being blocked. In other words, my inefficient exit design is now blocking itself from moving.  So this is a two part problem.   Considering that I'm constructing a residential district in the space above left of the problem exit, and would like to keep the solution inside the power line, how can I redesign this exit to be more efficient? How do I solve the ring of traffic jams? Is it a simple case of just deleting and replacing a bit of road? n tasks assigned to n computers, what is the EX value of a computer getting 5 or more tasks? Say a central server assigns each of n tasks uniformly and independently at random to n computers connected to it on a network. Say a computer is ‘overloaded’ if it receives 5 or more tasks.  Q: Calculate the expected number of overloaded computers.  I thought of doing [1 - Pr(a computer is not overloaded)] but that leads me to a complicated expression of: $$1 - PR(NotOver) = 1 - \sum_{i=0}^4  \left( \frac{1}{n} \right)^{i} { \left( \frac{n-1}{n} \right)}^{n-i}$$ multiplying this by n would(hopefully) give the Expected value. But the answer seems not very elegant atall, is there something I'm missing or an easier way to tackle this? Thanks! I have what seems to me a very simple proof that $A_{n}$ is the only subgroup of $S_{n}$ of index 2.  Since I've seen other people prove it with what feel like really complicated methods (), I'm wondering if I've overlooked something.  Proof: Let $H$ be any subgroup of index 2 in $S_{n}$.  Then $H\cap A_{n}$ is a normal subgroup of $A_{n}$ and since $A_{n}$ is simple, $H$ is either the trivial or the improper subgroup of $A_{n}$.  If it's the trivial subgroup it doesn't have index 2 in $S_{n}$, and otherwise the theorem is proved. Which predictions of The Last Jedi: Darths & Droids Bingo were correct? How to move from screen 1 to screen 3 by moving cursor to the left and vice versa? If I have to move the mouse from the left screen all the way to the right screen, the distance is quite big.   Is there a way to virtually connect the sides of the screen as if they were arranged in a circle? I could then move from the left screen to the right screen, by simply moving the cursor to the left. Resizing Mac OS partition - macOS High Sierra Bijection between an open and a closed interval Recently, I answered to this problem:     Given $a&lt;b\in \mathbb{R}$, find explicitly a bijection $f(x)$ from   $]a,b[$ to $[a,b]$.   using an "iterative construction" (see below the rule).  My question is: is it possible to solve the problem finding a less exotic function?  I mean: I know such a bijection cannot be monotone, nor globally continuous; but my $f(x)$ has a lot of jumps... Hence, can one do without so many discontinuities?    W.l.o.g. assume $a=-1$ and $b=1$ (the general case can be handled by translation and rescaling). Let:  (1) $X_0:=]-1,-\frac{1}{2}] \cup [\frac{1}{2} ,1[$, and  (2) $f_0(x):=\begin{cases}     -x-\frac{3}{2} &amp;\text{, if } -1&lt;x\leq -\frac{1}{2} \\ -x+\frac{3}{2} &amp;\text{,     if } \frac{1}{2}\leq     x&lt;1\\ 0 &amp;\text{, otherwise} \end{cases}$,  so that the graph of $f_0(x)$ is made of two segments (parallel to the line $y=x$) and one segment laying on the $x$ axis; then define by induction:  (3) $X_{n+1}:=\frac{1}{2} X_n$, and  (4) $f_{n+1}(x):= \frac{1}{2} f_n(2 x)$  for $n\in \mathbb{N}$ (hence $X_n=\frac{1}{2^n} X_0$ and $f_n=\frac{1}{2^n} f_0(2^n x)$).  Then the function $f:]-1,1[\to \mathbb{R}$:  (5) $f(x):=\sum_{n=0}^{+\infty} f_n(x)$  is a bijection from $]-1,1[$ to $[-1,1]$.  Proof: i. First of all, note that $\{ X_n\}_{n\in \mathbb{N}}$ is a pairwise disjoint covering of $]-1,1[\setminus \{ 0\}$. Moreover the range of each $f_n(x)$ is $f_n(]-1,1[)=[-\frac{1}{2^n}, -\frac{1}{2^{n+1}}[\cup \{ 0\} \cup ]\frac{1}{2^{n+1}}, \frac{1}{2^n}]$.  ii. Let $x\in ]-1,1[$. If $x=0$, then $f(x)=0$ by (5). If $x\neq 0$, then there exists only one $\nu\in \mathbb{N}$ s.t. $x\in X_\nu$, hence $f(x)=f_\nu (x)$. Therefore $f(x)$ is well defined.  iii. By i and ii, $f(x)\lesseqgtr 0$ for $x\lesseqgtr 0$ and the range of $f(x)$ is:  $f(]-1,1[)=\bigcup_{n\in \mathbb{N}} f(]-1,1[) =[-1,1]$,  therefore $f(x)$ is surjective.  iv. On the other hand, if $x\neq y \in ]-1,1[$, then: if there exists $\nu \in \mathbb{N}$ s.t. $x,y\in X_\nu$, then $f(x)=f_\nu (x)\neq f_\nu (y)=f(y)$ (for $f_\nu (x)$ restrited to $X_\nu$ is injective); if $x\in X_\nu$ and $y\in X_\mu$, then $f(x)=f_\nu (x)\neq f_\mu(y)=f(y)$ (for the restriction of $f_\nu (x)$ to $X_\nu$ and of $f_\mu(x)$ to $X_\mu$ have disjoint ranges); finally if $x=0\neq y$, then $f(x)=0\neq f(y)$ (because of ii). Therefore $f(x)$ is injective, hence a bijection between $]-1,1[$ and $[-1,1]$. $\square$ How do you handle "that that"? The double "that" problem Allowing users to compile / run PLSQL and SQL As a DBA - assuming my users are knowledgeable with PLSQL and SQL - why would I not grant my users privileges to compile and run their own PLSQL code. We have an MI database environment where the data has undergone ETL. The tables contain a large numbers of rows but not necessarily large rows in size. The analysis - will be - I assume of calculus in nature. The users are wanting to run their own code on the data for the purposes of ad-hoc analysis - within their own schemas. They come from a SAS background which from my understanding has allowed them to write and run their own analysis of data. Albeit, SAS runs on its own dataset - extracted from Oracle or elsewhere.  In my opinion this is a dangerous set-up - but without answering my own question - what are the pitfalls? Remove border of top left table cell Do I have to do any configuration to start brace expansion? When I run mkdir {1..10}, it just creates a dir naming {1..10}. I'm passing through Canada, on my way to somewhere else. I'm staying only a short time.   Is there a website, or some other way of find out out if I need a visa? Solve for $x$ in $6x^2-25x+12+\frac{25}{x}+\frac{6}{x^2}=0$ I simplified to get $6x^4-25x^3+12x^2+25x+6=0$  Or $6 (x^2+1)^2+25x(1-x^2)=0$  Then I substituted $z=x^2+1$, to get $6z^2+25\sqrt{(z-1)}(2-z)=0$  I can't find a way to proceed further. Git - how to force merge conflict and manual merge on selected file Given an infinite number of monkeys and an infinite amount of time, would one of them write Hamlet? Of course, we've all heard the colloquialism "If a bunch of monkeys pound on a typewriter, eventually one of them will write Hamlet."  I have a (not very mathematically intelligent) friend who presented it as if it were a mathematical fact, which got me thinking... Is this really true?  Of course, I've learned that dealing with infinity can be tricky, but my intuition says that time is countably infinite while the number of works the monkeys could produce is uncountably infinite.  Therefore, it isn't necessarily given that the monkeys would write Hamlet.  Could someone who's better at this kind of math than me tell me if this is correct?  Or is there more to it than I'm thinking? What is the angle between those two matrices over $\mathbb{C}$? Let $\left&lt;\;,\;\right&gt;$ be an inner product defined by $\left&lt;A,B\right&gt;=$ tr$\left(A\overline{B}^{T}\right)$ over $M_n\left(\mathbb{C}\right).$  I am looking for the angle between: $$A=\begin{pmatrix}2+i &amp; 1\\-1&amp;0\end{pmatrix}\quad B=\begin{pmatrix}3-2i &amp; 1+i\\0 &amp; -i\end{pmatrix}$$  First we can see that:$$\overline{A}=\begin{pmatrix}2-i &amp; 1\\-1&amp;0\end{pmatrix}\quad \overline{B}=\begin{pmatrix}3+2i &amp; 1-i\\0 &amp; i\end{pmatrix}$$  $$\Downarrow$$  $$\overline{A}^T=\begin{pmatrix}2-i &amp; -1\\1&amp;0\end{pmatrix}\quad \overline{B}^T=\begin{pmatrix}3+2i &amp; 0\\1-i &amp; i\end{pmatrix}$$  And therefore:   $\left&lt;A,B\right&gt;=$ tr$\left(A\overline{B}^{T}\right)=\left(2+i\right)\left(3+2i\right)+1-i=5+6i$ $\left&lt;A,A\right&gt;=\left(2+i\right)\left(2-i\right)+1+1=7$ $\left&lt;B,B\right&gt;=\left(3-2i\right)\left(3+2i\right)+\left(1+i\right)\left(1-i\right)+1=16$   Now as the angle between any two vectors is defined we get that: $$\cos{\alpha}=\frac{\left&lt;A,B\right&gt;}{\Vert{A}\Vert\Vert{B}\Vert}=\frac{5+6i}{\sqrt{112}}$$  $$\Downarrow$$  $$\alpha = \;?$$  I didn't even know a complex number could be an image for the cosine function.  Thanks for any help. Is it true that the order of $ab$ is always equal to the order of $ba$? How do I prove that if $a$, $b$ are elements of group, then $o(ab) = o(ba)$?   For some reason I end up doing the proof for abelian(ness?), i.e., I assume that the order of $ab$ is $2$ and do the steps that lead me to conclude that $ab=ba$, so the orders must be the same. Is that the right way to do it? How do you do load testing and capacity planning for web sites? How can I install .ipa file to my iPhone simulator? I need to install .ipa file to my iPhone simulator.   I tried to change the extension .ipa to .zip and extracted it.  And I got the Payload folder and moved the application inside to my iPhone simulators applications folder found at: /Library/Application\ Support/iPhone Simulator/6.1/Applications/.   Application appears on iPhone Simulator, But the application did't running. How to solve the problem then how to run the app.