If $f:[a,b]\to R$ is a continuous function, prove that there exists $c \in [a,b]$ such that $\int_{c-a}^{b-c}f(x)dx=0$.
If $f,g : [a,b] \to (0,\infty)$ are continuous functions, prove that there exists $c \in (a,b)$ such that: $$\int_{a}^{c} f(x) \ \mathrm dx = \int_{c}^{b} g(x) \ \mathrm dx$$
Omission of 'for' with various quantified time intervals: influence of verb Can a Dungeons and Dragons 4e character take D&D Essentials feats? In Heroes of the Fallen Lands many of the core feats have been streamlined, some paragon feats have become available to all levels.  Before that feats were tiered by level: heroic, paragon, epic.  Can a character that was made before the Heroes of the Fallen Lands publication use Heroes of the Fallen Lands feats? With ubuntu 19.10 I have been attempting to pair my Airpods with my laptop. There seems to be a problem with pairing and system is unable to add them as Bluetooth device. This happened after an upgrade from 19.04. Changing ControllerMode in /etc/bluetooth/main.conf seems to have no affect. Any help appreciated. I want to show that $$f: \mathbb{R}_{\geq0} \to \mathbb{R}_{\geq0}$$ $$x \mapsto x^s$$ is Holder continuous with Holder exponent $s \in \mathbb{R}$, where $0&lt;s \leq 1$. So what I want to show is that $\exists \hspace{2 mm} C \in \mathbb{R}_{\geq0}$ sucht that for all $ x,y \in \mathbb{R}_{\geq0},  $ $$|x^s -y^s| \leq C|x-y|^s$$ and therefore, assuming, wlog $\hspace{1mm} x&gt;y$ $$(x^s -y^s) \leq C(x-y)^s.$$  I thought about Bernoulli's inequality but couldn't make that work.I thought about the binomial theorem, but didn't know how to handle the fact that $s \in \mathbb{R}$. We have set up GA on several websites to gather information about traffic to the site. We know about the organic keywords that are displayed under the Acquisition section of the traffic report. But we have been asked to find a way to display "not provided" keywords typed in by the end user in our Google Analytics report.  Is it possible to accomplish this? How do I show that $X = \left\{ (x,y) \in \mathbb{R}^2 \mid x \in \mathbb{Q}\text{ or }y \in \mathbb{Q}\right\}$ is path connected?    Note that $X$ is a topological space with subspace topology $\tau = \left\{ U \cap X \mid U\text{ is open in }\mathbb{R}^2\right\}$.  I only know two things (with respect to path connectedness):   A topological space $X$ is path connected iff for every $p,q \in X$, there is a continuous function $f: [a,b] \to X$ with $f(a)=p$ and $f(b) = q$.    Continuity $f$ is defined as for every open set in $X$, its preimage is open in $[a,b]$ (with subspace topology induced by $\mathbb{R}$ (with standard topology) )   Suppose $X$ and $Y$ are two topological spaces and $f: X \to Y$ is a continuous surjection. If $X$ is path connected, then $Y$ is path connected. What is the name of the default font in math mode? I draw figures in Inkscape. When I label elements within the figures with variable names that I have used in the underlying TeX document, I would like them to look exactly the same as in the document. (e.g. l does not look the same as $l$)  What is the name of the math mode font so I can select it correctly in the Inkscape font list?  If the exact font should not be available, what is a similar looking font that is present on most systems? How can I get sql string, generated by Linq2SQL ORM? I noticed in this thread the OP successfully managed to connect a 30" cinema display to the new Macbook Pro USB-C, with full resolution:    I have the official Apple dual DVI adapter, going to a displayport > USB-C adapter like yours, however the screen does not fire up.  I then tried a multi adapter instead of the single displayport > USB-C, which worked but only at 1280 resolution:    Would the official Apple Dual DVI adapter not work? Why with two classes, LDA gives only one dimension? i would like to know how to prove that if $a^2+b^2=c^2$ then   a) $3|a$ or $3|b$  b) $5|a$ or $5|b$ or $5|c$  c) $4|a$ or $4|b$   Also that $a^2+b^2+c^2+1$ can't be divisible by 8  Thank you very much! Why do some Unicode text messages crash the iPhone OS? What does $\sum_{k=0}^\infty \frac{k}{2^k}$ converge to? This problem comes from another equation on another question (. I tried to split it in half but I found out that $$\sum_{k=0}^\infty \frac{k}{2^k}$$ can't be divided. Knowing that $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$ I wrote that $$\sum_{k=0}^\infty \frac{k}{2^k}=\sum_{k=0}^\infty \left(\frac{\sqrt[k] k}{2}\right)^k=\frac{1}{1-\frac{\sqrt k}{2}}=\frac{2}{2-\sqrt[k] k}$$ But that's not what I wanted. Could anyone help me? I know this sounds silly but I would like some objects to follow physics while others not to collide each other.   I tried to achieve them by setting their position exclusively. But then it will ignore all physics.   Is what I am trying to do even possible? $ \int_{0}^{ \infty} \int_{0}^{ \infty} \frac { e^{-(x+y)}}{x+y} dx dy $ About multi-root search in Mathematica for transcendental equations Sort a Map by values Calculate: $$\int_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)\mathrm{d}t$$ My try: $\displaystyle\int_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)dt\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\cos(\mathfrak{I\textrm{z}}+3(-i\log z)dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}}^{(\mathfrak{I\textrm{z}}+3(-i\log z))i}dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}^{\mathfrak{I\textrm{z}}}}z^{3}dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}^{\mathfrak{I\textrm{z}}}R}z^{3}dz$ and there is nothing here that is not holomorphic, therefore according to Cauchy theorem it must be exactly $0$. Equivalence between least squares and MLE in Gaussian model I am new to Machine Learning, and am trying to learn it on my own. Recently I was reading through  and had a basic question.   Slide 13 says that "Least Square Estimate is same as Maximum Likelihood Estimate under a Gaussian model". It seems like it is something simple, but I am unable to see this. Can someone please explain what is going on here? I am interested in seeing the Math.   I will later try to see the probabilistic viewpoint of Ridge and Lasso regression also, so if there are any suggestions that will help me, that will be much appreciated also. Is it legal to develop a game using D&D rules? For a while now I've been thinking about trying my hand at creating a game similar in spirit and execution to Baldur's Gate, Icewind Dale and offshoots. I'd rather not face the full bulk of work in implementing my own RPG system - I'd like to use D&amp;D rules.  Now, reading about the subject it seems there is something called "The License" which allows a company to brand a game as D&amp;D. This license seems to be exclusive, and let's just say I don't have the money to buy it :p.  Is it still legal for me to implement and release such a game? Commercially or open-source? I'm not sure exactly which edition would fit the best, but since Baldur's Gate is based of 2nd edition, could I go ahead an implement that?  in short: what are the issues concerning licensing and publishing when it comes to D&amp;D?  Also: Didn't see any similar question...