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The spell Armor of Agathys states:      You gain 5 temporary hit points for the duration. If a creature hits you with a melee attack while you have these hit points, the creature takes 5 cold damage.   Now, assume my buddy uses Inspiring Leader to give me 10 temporary hit points — does the effect end? I would like a brief explanation of the following command line:  grep -i 'abc' content 2&gt;/dev/null Chrome 100% CPU usage, again - software_reporter_tool.exe software_reporter_tool.exe belonging to Chrome running at 100% cpu, fan full speed.  Deleting it doesn't fix anything, because it gets redownloaded again! Disabling "Protect me from bla bla" doesn't make it stop running.  Please anyone have any solutions? When I view most glass from the side it's green which I've found out is due to impurities in the glass specifically from iron oxide. Why is it when I view the larger face from an oblique angle, it isn't nearly as green? I cannot personally notice any different on the piece I have next to me even when I hold it at an angle that would be almost looking at the edge of the glass. It is pretty small (about 2.5" x 5" x .0625" or about 61mm x 127mm x 2mm, l x w x h) but I feel like it's big enough that I'd be looking through enough glass to get the green. What is the benefit of breaking up a continuous predictor variable? I'm wondering what the value is in taking a continuous predictor variable and breaking it up (e.g., into quintiles), before using it in a model.   It seems to me that by binning the variable we lose information.   Is this just so we can model non-linear effects?  If we kept the variable continuous and it wasn't really a straight linear relationship would we need to come up with some kind of curve to best fit the data? Suppose $n$ divides $3^n + 4^n$. Show that $7$ divides $n$. Suppose $n \geq 2$ and $n$ is a divisor of $3^n + 4^n$. Prove that $7$ is a divisor of $n$.  My work so far:  I had a hypothesis that if $n| 3^n + 4^n$, then $n = 7^k$ for some $k\in\mathbb{N}$. But this is not necessarily so. Take $n = 7⋅379$, where $3^7 + 4^7 = 7^2⋅379$. Then, $3^7+4^7$ divides $3^n + 4^n$, and since $n$ divides $3^7+4^7$, we must have $n|3^n+4^n$. Does there exist a continuous bijection from $(0,1)$ to $[0,1]$? Of course the map should not be a proper map. The axiom of choice asserts that for any set $X$ there exists a function $f : (2^X − \{\emptyset\}) \rightarrow X$ such that for any nonempty $A \subseteq X$, $f(A) \in A$. Show that this is equivalent to the assertion that every relation includes a function with the same domain. I am given this problem in my assignment. My approach is following. I have a set $X$. Then it has a choice function $g : \mathcal{P}(X) \setminus \left\{\emptyset\right\} \rightarrow X$. Suppose $X$ has a relation $R \subseteq X \times X$. $dom(R) \subseteq X$. Then the required function $f : dom(R) \rightarrow X$ is $$f(x) = g(\{y\ |\ (x, y) \in R\})$$ Conversely, suppose $X \supseteq Y$ is a non-empty subset. Let $$R_Y = \{(x, y)\ |\ y \in Y, x \in Y, (p, q) \in R_Y \Rightarrow p = x\}$$ Then there is a function $f_Y : dom(R_Y) \rightarrow X$. But $dom(R_Y)$ is singleton from the definition of $R_Y$. Then $im(f_Y)$ is also singleton. We define our choice function as $f = \{(Y, y)\ |\ y \in im(f_Y)\}$ Are my arguments correct?  Edit.1: what I tried to do for converse is, for any subset $Y$, I construct a relation $Y_1 \times Y$ where $Y \supseteq Y_1 = \{y\}$. Then there is a function $f_Y : Y_1 \rightarrow Y$. Then if $f$ is my choice function I define $f(Y) = f_Y(y)$. May be I should define $R_Y$ as $$R_Y = \{(x, y)\ |\ y \in Y; x \in Y; (p, q), (p', q) \in R_Y \Rightarrow p = p'\}$$ I understand there may be multiple $R_y$'s. But can't I define a singleton set as $\{s\in S\ |\ a, b \in S \Rightarrow a = b\}$? This can be $\{s\}$ $\forall s \in S$. But isn't it a valid specification?  Edit.2: This is the final solution with the help from @Asaf.  If we assume axiom of choice then for any $X$ there is a choice function $f : \mathcal{P}(X) \setminus \left\{\emptyset\right\} \rightarrow X$ such that $f(A) \in A$. Suppose $R$ is a relation between $P$ and $Q$. Then we have a choice function $f_Q$ for $Q$. Let $f_R(p) = f_Q(\{q\ |\ pRq\})$. The function $f_R : P \rightarrow Q$ and $dom(f_R) = dom(R)$. So this is our required function.  Conversely, if every relation includes a function with same domain. Let $R$ be a relation between $\mathcal{P}(X) \setminus \left\{\emptyset\right\}$ and $X$ such that $aRb \Rightarrow b \in a$. Now $R$ includes a function $f : \mathcal{P}(X) \setminus \left\{\emptyset\right\} \rightarrow X$. Then $xRf(x) \Rightarrow f(x) \in x$. So $f$ is the choice function for $X$. How to connect Android with MySQL using Mysql JDBC driver Denoting the end of example/remark What is the etiquette to denote the end of an example or remark? For proofs, it's the QED symbol (white box). I know it's not necessary for everyone, but I would like to use it to break up the text a little. What should I use? A circle or filled box or what? Let $\mu(\Omega)$ be a probability measure (i.e. $\mu(\Omega) = 1$), and let $f,g$ be non-negative measurable functions on $\Omega$ such that $fg \geq 1$. Show that $1 \leq (\int f^p )(\int g^p)$ for all $0 &lt; p &lt; \infty$   I'm a bit at a loss of what to do. I know that since $fg \geq 1$, we may obtain:  $$\int_{\Omega} d\mu = \mu(\Omega) = 1 \leq \int_{\Omega} fg = \int_{\Omega} |fg|$$ The right-most integral is true since $fg$ is positive and it is equal to $||fg||_1$.  By Hölder's inequality, we obtain:  $$1 \leq \int |fg| \leq ||f||_p ||g||_q = (\int f^p)^\frac{1}{p}(\int g^q)^\frac{1}{q}$$  where $p$ and $q$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$. Raising the left-most and right-most sides of this inequality to the power $p$, I may obtain:  $$1 \leq (\int f^p)(\int g^q)^\frac{p}{q}$$  I now wish to show that $(\int g^q)^\frac{p}{q} \leq \int g^p$, but I've been at a loss of how to do it. Any tips or hints would be greatly appreciated. Chat gives me really, really specific time gap notifications Does $\int_a^bf(x) = 0$ imply $f(x) = 0$ for a Riemann Integral function? Let $a &lt; b$ and $f:[a, b] \rightarrow \mathbb R$ be a Riemann integral function. Suppose that $f(x) \geq 0$ for all $x \in [a, b]$. I want to show whether $\int_a^bf(x) = 0$ implies $f(x) = 0$ or not, but I'm not sure where to start. I don't know much about Riemann integral functions in general. I only know that $$\underline \int_a^b f= \overline \int_a^b f.$$ Also, I know the fact that if $f$ is continuous, then $f(x) = 0$ for all $x\in [a, b]$. Can anyone provide any hints for me, because I'm really confused about this whole thing. I've setup a HTTPS and FTP site in IIS but how can I add users to use the FTP site?  Does they necessarily need an account with password in the computer to be able to use it? How to add a Skype indicator? Define a linear operator $T:C[0,1] \to C[0,1]$ as follows: $$Tf(x) = f(x) + \int_0^x f(u)du$$ It is easy to show that $T$ is a bounded linear operator. The statement also (1) claims that $T$ is one-to-one and onto, and (2) asks to compute $T^{-1}$. (Hence $T$ is an isomorphism due to a corollary of Open Mapping Theorem)  I need help to prove $T$ is onto: given any $g \in C[0,1]$, how to decompose it into a sum between a continuous $f$ and its antiderivative. I have trouble decomposing $g(x)=|x|$, not to mention bizarre functions like Weierstrass function. It seems impossible.  Thanks in advance for help! Why do variable names often start with the letter 'm'? The FAQ mentions bounty, but sometimes a question just doesn't really have a right answer:   No solution exists The question as written is unanswerable and unsalvageable There may be a technically correct answer, but the asker is unable to utilize it for some reason.   In these cases, a bounty, and more answers, won't really help. What is the appropriate etiquette in these cases? Should the question be closed? Left hanging? Test Class Apex Page Parameters How to do research in two very different fields?