I have a class which has public Assembly LoadedAssembly { set; get; }  var assemblyName = AssemblyName.GetAssemblyName(fullPathToDLLFile); myClass.LoadedAssembly = Assembly.Load(assemblyName);   This class I keep under Application.Current.Properties["MyClass"].  I try to null it like Application.Current.Properties["MyClass"] = null; but it doesn't help.  I see that the loaded assembly works (It has a timer inside and it is working).  How do I can clean it?  (I don't use something like   AppDomain d = AppDomain.CreateDomain("NewDomain", null, domaininfo); to load assembly.)
I would like to know how to unload an assembly that is loaded into the main AppDomain.  I have the following code:  var assembly = Assembly.LoadFrom( FilePathHere );   I need/want to be able to unload this assembly when I am done.  Thanks for your help.
Transformation of Random Variable $Y = X^2$ I'm learning probability, specifically transformations of random variables, and need help to understand the solution to the following exercise:      Consider the continuous random variable $X$ with probability density function $$f(x) = \begin{cases} \frac{1}{3}x^2 \quad  -1 \leq x \leq 2, \\ 0 \quad \quad \text{elsewhere}. \end{cases}$$ Find the cumulative distribution function of the random variable $Y = X^2$.   The author gives the following solution:   For $0 \leq y \leq 1: F_Y(y) = P(Y \leq y) = P(X^2 \leq y) \stackrel{?}{=} P(-\sqrt y \leq X \leq \sqrt y) = \int_{-\sqrt y}^{\sqrt y}\frac{1}{3}x^2\, dx = \frac{2}{9}y\sqrt y.$  For $1 \leq y \leq 4: F_Y(y) = P(Y \leq y) = P(X^2 \leq y) \stackrel{?}{=} P(-1 \leq X \leq \sqrt y) = \int_{-1}^{\sqrt y}\frac{1}{3}x^2\, dx = \frac{1}{9} + \frac{1}{9}y\sqrt y.$  For $y &gt; 4: F_{Y}(y) = 1.$    Previous to this exercise, I've managed to follow the solutions of two similar (obviously simpler) problems for a strictly increasing and strictly decreasing function of $X$, respectively. However in this problem, I don't understand the computations being done, specifically:    How does the three intervals $0 \leq y \leq 1$, $1 \leq y \leq 4$ and $y &gt; 4$ are determined? In the two previous problems I've encountered, we only considered one interval which was identical to the interval where $f(x)$ was non-zero. In the case where $0 \leq y \leq 1$, why does $P(X^2 \leq y) = P(-\sqrt y \leq X \leq \sqrt y)$ and not $P(X \leq \sqrt y)$? I have put question marks above the equalities that I don't understand.   I think I have not understand the theory well enough. I'm looking for an answer that will make me understand the solution to this problem and possibly make the theory clearer. Debian 8 apt-get update showing 404 Not Found I got the following output every time I run apt-get update on my Debian 8.   Is there a way to fix this?   Err http://ppa.launchpad.net jessie/main amd64 Packages                         404  Not Found Ign http://ppa.launchpad.net jessie/main Translation-en                 Err http://ppa.launchpad.net jessie/main amd64 Packages                   404  Not Found Ign http://ppa.launchpad.net jessie/main Translation-en                 Fetched 164 kB in 3s (47.0 kB/s)                                        W: Failed to fetch http://ppa.launchpad.net/certbot/certbot/ubuntu/dists/jessie/main/binary-amd64/Packages  404  Not Found  W: Failed to fetch http://ppa.launchpad.net/ondrej/php/ubuntu/dists/jessie/main/binary-amd64/Packages  404  Not Found  E: Some index files failed to download. They have been ignored, or old ones used instead.   And this is what I have in /etc/apt/sources.list  deb http://ftp.debian.org/debian/ jessie main deb-src http://ftp.debian.org/debian/ jessie main  deb http://security.debian.org/ jessie/updates main deb-src http://security.debian.org/ jessie/updates main  # jessie-updates, previously known as 'volatile' deb http://ftp.debian.org/debian/ jessie-updates main deb-src http://ftp.debian.org/debian/ jessie-updates main  deb http://ftp.debian.org/debian jessie-backports main   And this is the output of cat /etc/apt/sources.list.d/*  deb http://ppa.launchpad.net/certbot/certbot/ubuntu jessie main # deb-src http://ppa.launchpad.net/certbot/certbot/ubuntu jessie main deb http://ppa.launchpad.net/ondrej/php/ubuntu jessie main # deb-src http://ppa.launchpad.net/ondrej/php/ubuntu jessie main deb http://ppa.launchpad.net/ondrej/php/ubuntu jessie main # deb-src http://ppa.launchpad.net/ondrej/php/ubuntu jessie main deb https://packages.sury.org/php/ jessie main deb https://packages.sury.org/php/ jessie main As things stand, will House Stark go extinct? According to the family tree in A World of Ice and Fire Ned Stark's children are in fact the only surviving Starks. And as of S06E09;      Bran is the only legitimate, male Stark.   In the Season Six Finale it is revealed that;      Jon is in fact the Son of Lyanna Stark and most likely, Rhaegar Targaryen.   This means that if Jon was legitimised and there was record of his birth in the Citadel;      He would become a Targaryen.   So I guess my big question is: Can Bran father children? I find it to be quite an important issue considering House Stark's long term survival depends on it.  In season 4 (I can't remember which episode) Locke Bolton gives Bran a cut on the leg to confirm his identity. Bran reacts only with fear and not pain, indicating that he has no feeling below his waist.  Does this mean that he wouldn't be able to...get it up? And if so, why hasn't the show addressed House Stark's impending doom? It looks like if I load dynamic content using $.get(), the result is cached in browser.  Adding some random string in QueryString seems to solve this issue (I use new Date().toString()), but this feels like a hack.   Is there any other way to achieve this? Or, if unique string is the only way to achieve this, any suggestions other than new Date()? Let $A$ be the set of all functions $f:\mathbb{R}\to\mathbb{R}$ that satisfy the following two properties. Let $A$ be the set of all functions $f:\mathbb{R}\to\mathbb{R}$ that satisfy the following two properties:  (1) $f$ has derivative of all orders, and  (2) for all $x,y \in \mathbb{R}$, $f(x+y)-f(y-x)=2xf'(y)$.  Which of the following sentences is true?  (a) Any $f\in A$ is a polynomial of degree less than or equal to 1  (b) Any $f \in A$ is a polynomial of degree less than or equal to 2  (c) $\exists f \in A$ which is not polynomial  (d) $\exists f \in A$ which is a polynomial of degree 4  It is a problem from TIFR GS-2018. I have found polynomials upto degree 2 hold this properties but have to exclude the possibility of (c) and (d). I also found out if $f'(x)$ has at most one real root and clearly graph of $f$ is symetric about that root. How can I proceed further? Is the usage of text embedded in an image a bad practice, for very controlled uses? What are your thoughts on writing your text in images?  I'm not saying throughout the whole website... But in certain key locations like image carousels. And I'm not talking about the special cases where you have font types that browsers can't render. I'm talking about text that uses standard fonts and is not too big (3/4 bullets).  Should I embed the text in the images that compose the carousel? Or should I apply the text to the images as "real text" HTML?  What are the advantages/ disadvantages of using text in an image? Isomorphism of quotient modules implies isomorphism of submodules? Let $A$ be a commutative ring, $M$ an $A$-module and $N_1, N_2$ two submodules of $M$.  If we have $M/N_1 \cong M/N_2$, does this imply $N_1 \cong N_2$?  This seems so trivial, but I just don't see a proof... Thanks! Is my processor 64 or 32 bit? I am having a problem running the CDC change capture job against some of my test databases.  The error message I'm getting is the following:     Executed as user: NT AUTHORITY\NETWORK SERVICE. Cannot execute as the database principal because the principal "dbo" does not exist, this type of principal cannot be impersonated, or you do not have permission. [SQLSTATE 42000] (Error 15517).   How can I fix this? Best way to save data in a Java application? Let $f(z)$ be entire function. Show that if $f(z)$ is real when $|z| = 1$, then $f(z)$ must be a constant function using Maximum Modulus theorem Let $f(z)$ be entire function. Consider the functions $e^{if(z)}$ and $e^{−if(z)}$ and applying the Maximum Modulus Theorem, show that if $f(z)$ is real when $|z| = 1$, then $f(z)$ must be a constant function.  (We take $f(z)=u(z)+iv(z)$)  I am confused as so far I have $|g(z)|=|e^{if(z)}|=|e^{-v(z)}|$ and then since $f(z)$ is real, $f(z)=u(z)$ and $v(z)=0$ so I assumed it would follow that $|g(z)|=|e^{v(z)}|=1$.  Similarly, $|g(z)|=|e^{-if(z)}|=|e^{v(z)}|=1$.  Using Liouville I assumed one could say that both $g(z)$ and $h(z)$ are bounded entire functions, they are constant and so it follows that $v(z)$ is constant, meaning that both its partial derivatives are equal to 0 and, due to Cauchy Riemann, both of the partial derivatives of $u(z)$ are equal to zero.  It would then follow that $f(z)$ is constant.  I don't know how to go about the question using the Maximum Modulus Theorem, also I feel I am overlooking the importance of $|z|=1$ perhaps?  Any help would be much appreciated!! I killed my first wife once I found out there is a mercenary you can marry who gives you up to 300 gold a day (in Whiterun). The only requirement, of course, being that I had to hire her first. So I hired her, and then I popped the question, to which she replied the usual: tell the priest to arrange the wedding.   When we reached Riften I told her to wait, just outside the bee and barb, and arranged the wedding. I went through the process, and the second it finished she left, I followed but as I left she was nowhere to be seen. I have tried going back to my house, the pub where I hired her and that same spot in Riften about 6 times each now, some during the day, some during the night and cannot find her at all.   Now this isn't to much of a problem because I can re-marry...oh wait, to remarry, my spouse has to die, so I can't claim gold off her and can't remarry, and the autosaves have been used up fast travelling to the places I thought she might've been. My last save is something like 6 levels ago, and I don't want to have to load it.   What should I do about this situation? Uncountable subset with uncountable complement, without the Axiom of Choice Let $X$ be a set and consider the collection $\mathcal{A}(X)$ of countable or cocountable subsets of $X$, that is, $E \in \mathcal{A}(X)$ if $E$ is countable or $X-E$ is countable. If $X$ is countable, then $\mathcal{A}(X)$ coincides with the power set $\mathcal{P}(X)$ of $X$. Now suppose that $X$ is uncountable. Assuming the axiom of choice, we can conclude that $\mathcal{A}(X) \ne \mathcal{P}(X)$, since $|X| = |X| + |X|$. So the question is:     Can we prove in ZF that $\mathcal{A}(X) \ne \mathcal{P}(X)$ for every uncountable set $X$?   I'm assuming that a set $X$ is uncountable if there is no injective function $f : X \rightarrow \mathbb{N}$. Can the determinant of a matrix can be made $0$? An entry of an $n \times n$ matrix with nonzero determinant is defined as interesting if by changing this entry (and only this entry) the determinant of the matrix can be made $0$.   Is it true that each entry of every matrix with nonzero determinant is interesting? Is it true that there is an interesting entry in each row of a matrix with nonzero determinant? "Contact us"... and then what? How to know if my servers should use jumbo frames (MTU) What is the difference between 'tell' and 'say' How to format numbers as currency strings How to find out client ID of component for ajax update/render? Cannot find component with expression "foo" referenced from "bar" How do you get the rendered height of an element?  Let's say you have a &lt;div&gt; element with some content inside. This content inside is going to stretch the height of the &lt;div&gt;. How do you get the "rendered" height when you haven't explicitly set the height. Obviously, I tried:    var h = document.getElementById('someDiv').style.height;   Is there a trick for doing this? I am using jQuery if that helps.