Squealing of rim brakes on a bike My bike for 2 days started to squeal whenever I use my back brakes it has been really annoying and I am trying to fix it. Please give me any suggestion that can help this problem
How do I stop my rim brakes making a squeaking/squealing sound? I bought a Schwinn Frontier Sport few months ago and started riding happily. After a few weeks my front rim brakes started creaking. It's so annoying in public places. My rear brake doesn't produce any sound but isn't as powerful as the front brake.  The retailer says cleaning would solve the front brake's issues and I got it cleaned, but nothing changed. Is there something I can do something myself to fix this?
Splitting a semicolon-separated string to a dictionary, in Python I have a string that looks like this:  "Name1=Value1;Name2=Value2;Name3=Value3"   Is there a built-in class/function in Python that will take that string and construct a dictionary, as though I had done this:  dict = {     "Name1": "Value1",     "Name2": "Value2",     "Name3": "Value3" }   I have looked through the modules available but can't seem to find anything that matches.    Thanks, I do know how to make the relevant code myself, but since such smallish solutions are usually mine-fields waiting to happen (ie. someone writes: Name1='Value1=2';) etc. then I usually prefer some pre-tested function.  I'll do it myself then. Show that a matrix $A$ is singular if and only if $0$ is an eigenvalue. I can't find the missing link between singularity and zero eigenvalues as is stated in the following proposition:  A matrix $A$ is singular if and only if $0$ is an eigenvalue.  Could anyone shed some light? Apple Macbook Dual Monitor MST Cannot select vertices from different object while in edit mode Whenever I enter edit mode, with 2 separate meshes (cubes), I can only select vertices/edges/faces of 1 mesh.  Example (the screen shot did not include the mouse cursor):    Also, when I enter edit mode with both meshes selected, one of the meshes has a red border and cannot be selected (it also has this border in Object mode): Let $n \geq 1$ be an odd integer. Show that $D_{2n}\cong \mathbb{Z}_2 \times D_n$. I define a map $$\phi:D_{2n} \rightarrow \mathbb{Z}_2 \times D_n$$ by $R \mapsto (0,r^{\frac{n+1}{2}})$ and $M \mapsto(1,m)$. Then I am stuck at showing the map is bijective. By the way, do we need to show that $(0,r^{\frac{n+1}{2}})(1,m)(0,r^{\frac{n+1}{2}})^{-1}=(1,m)$ ? If we can prove this, then what can we conclude? In older versions of iOS, if I searched for an app with Spotlight, it would tell me which folder the app was in. (E.g., this feature still works on my iPad, which is running iOS 9, but not on my iPod Touch, which is running iOS 11.)  This feature seems to have been removed at some point. When I search for an app under iOS 11 using Spotlight, it will find the app and let me run it, but it doesn't tell me which folder it is in.  How can I find out which folder an app is in now? Help with proving that $\pi$ is irrational I was trying to prove that $\pi$ is irrational, just to see if I could do it. So far, I've tried to do this by using the fact that the sum   $$S=\sum\limits_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$ and arguing that $S$ is irrational instead, and thus implying that $\pi$ is also irrational.  To do this, I thought I could use partial sums $S_n$ of $S$:  $$S_n=\sum\limits_{k=1}^n \frac{1}{k^2}=\frac{A_n}{B_n}$$ where the fraction $A_n/B_n$ is written in lowest terms. We can note that the sequence $\lbrace S_n \rbrace$ is a strictly increasing sequence, with $\pi^2/6$ as its lowest upper bound, so I thought that maybe the sequences $\lbrace A_n \rbrace$ and $\lbrace B_n \rbrace$ both have no upper bound, and the sequences would tend to infinity, and if I were able to prove this, I thought I could use it to argue that $S$ is an irrational number, using the fact that every rational number can be written as the quotient between two finite integers.  We note that $A_n\geq B_n$ for all natural $n$, as $S_n\geq 1$, so all we would really need to prove in that case is that $\lbrace B_n \rbrace \rightarrow \infty$(if this assertion is even correct).  At first I thought that both $\lbrace A_n \rbrace$ and $\lbrace B_n \rbrace$ would be increasing sequences, but after checking with Maple I noticed that they weren't, sadly enough($S_9=\frac{9778141}{6350400}$ and $S_{10}=\frac{1968329}{1270080}$). However, they do indeed seem to get very large very quickly, so I'm thinking that my hypothesis about $\lbrace A_n \rbrace$ and $\lbrace B_n \rbrace$ is correct.  But I have trouble proving my hypothesis, and I'm kind of stuck, no knowing what to do. Is there a way to prove that $\lbrace B_n \rbrace \rightarrow \infty$? And of course, is this approach to prove that $\pi$ is irrational logically sound, or is it fundamentally flawed in some important aspect? In the latter case, what idea should I try next? Evaluate the limit      $$ \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} $$   My Attempt:  $$ \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} = \lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^{n^2}\cdot \frac{1}{e^n} $$  How can I solve the problem from this point? Is there a proof for this Fibonacci relationship? I was looking at the decomposition of Fibonacci numbers using the definition of $F_n = F_{n-1} + F_{n-2}$, and noted the pattern in the coefficients of the terms were Fibonacci numbers. It appears to hold, and I believe it's true, but I haven't seen a proof for it. Does one exist?  $F_n = 1F_{n-1} + 1F_{n-2}$ $F_n = 2F_{n-2} + 1F_{n-3}$ $F_n = 3F_{n-3} + 2F_{n-4}$ $F_n = 5F_{n-4} + 3F_{n-5}$ etc  This can be generalized to $F_n = F_xF_{n-(x-1)}+F_{x-1}F_{n-x}$ Two-finger swipe 'breaks' heads-up notifications on Lollipop? I have a Nexus 7 (2013) and a Moto X (2014) running Lollipop. On both of these, if I get a heads-up notification (i.e., those that show up on top of the screen when the device is unlocked) and swipe down on it with 2 fingers, it 'breaks' further heads-up notifications:   I can't tap on them to open the responsible app For, e.g., an incoming phone call with 'answer' and 'dismiss' buttons, tapping the buttons does nothing. I can't answer phone calls anymore!   I can still swipe to the side to dismiss the notifications, and they still disappear on their own after a few seconds. I haven't been able to find a way to get the correct behavior back aside from a restart.  Yeah, the obvious workaround is to not two-finger swipe on them, but it's second nature as that same gesture expands notifications in other contexts.  I'm guessing the problem is generic, but I've triggered it on Hangouts and Phone notifications.  Do others experience the same thing? Any known 'fixes' aside from restarting? How to change Function (Fn) Key behavior on wireless Logitech keyboard I do have a wireless Logitech K750 keyboard (silver Mac version) connected to a Lenovo ThinkPad (via the unifying USB receiver, together with a wireless Logitech mouse M705) and currently running Ubuntu 12.04.  By default the F-keys are mapped as multimedia/function keys, e.g. Play  /Pause. As I don't use them very often, I want the F-keys to behave like the "real" F-keys (F1-F12). How can I do that?  I know there is an option in the BIOS settings, but this only toggles the built-in keyboard of the notebook, not on my wireless keyboard.  I know there is a way to change this for the Apple wireless keyboard (), but there is no "parameters" directory in /sys/module/hid_logitech_dj/.  Any ideas? Or is there maybe a way to do this in a later Ubuntu version (13.04)?  UPDATE:  lsusb -d 046d: shows Bus 002 Device 005: ID 046d:c52b Logitech, Inc. Unifying Receiver  UPDATE 2: Thanks a lot for your help Lekensteyn. Finally  works just fine. In the "Known Issues" of Solaar it says that it does not work with the Nano Receiver. But I have one and it works just fine. Just don't forget to read the installation instructions and run the rules.d/install.sh script. It also recognized my M705 Mouse. How to read contacts on Android 2.0 I want to determine whether a certain n-dimensional, simply-connected vector field is conservative. I can very easily calculate its Jacobian in numerical form.  In 3 dimensions, the problem is easy: the Jacobian is a 3x3 matrix, and the curl can be obtained from its elements:  [jacobian[2,1] - jacobian[1,2],  jacobian[0,2] - jacobian[2,0],  jacobian[1,0] - jacobian[0,1]]  A curl of zero would mean that the vector field is conservative.  But what if my Jacobian has 100 elements? The curl is only defined for 3 dimensions, but is there a formula for some high-dimensional generalization? I'm starting to do designs meant for printing, but I'm fairly new to this medium so I don't really understand how bleed works and how much bleed should I specify for:   Business cards Large format prints (42 inches and bigger) Magazines   Should I extend the design into the bleed areas? Lightsabers are not made of light, nor have they anything to do with lasers (even if there are people who mistakenly refer to them as "laser swords").   Blaster bolts are also not lasers. They travel significantly slower than light, they travel even slower than ordinary bullets fired by gunpowder. (called slug throwers on the Star Wars universe)  Was there any use of lasers ever presented in the Star Wars universe, either as we use them today, or as weapons?  By laser I mean something which behaves like laser does, even if it's not called by that name in a movie or book. If something is called a laser (in the script or by fans) but clearly doesn't work like a laser (for example, it travels slowly, and in visible blocks) it doesn't count. Pixelated wallpaper/background I installed Ubuntu 14.04 on my Lenovo IdeaPad Y510P today. Everything worked fine, but after installing my graphics card driver the background became pixelated (Nvidia GeForce GT755m). I downloaded the driver through Additional Drivers and selected Version 331.113 (Proprietary, tested). It was using Nouveau before.  When I apply Nouveau it becomes clean again, but it is much weaker when I open games, it gives me very low FPS even in Counter-Strike. With Nvidia driver FPS and everything else seems fine, except things look pixelated and when I minimize windows and maximize again, the window goes black.    I attempted to solve the problem myself by removing the old Nvidia driver and manually installing the latest version, but it did not help.  Please let me know if you need more details. Is God either immoral or not omnipotent? The usual (Christian) justification for suffering/evil in the world created by a benevolent God is freedom of the will. However, the more interesting question is not about the source of evil (which the free will may very well be) but why God chose to create the world such as it is. Why not create a world with freedom of the will, but without suffering? Even assuming that God is bound by logic such combination doesn't seem to be logically impossible: creatures certainly can do good of their own free will. There are two alternatives.  1) God is powerful enough for all practical intents and purposes, enough to be called God, but not omnipotent. He created only the best of all possible worlds (Leibniz). This position is perhaps the most defensible philosophically, it also avoids logical problems with , etc., but it does diminish deity's status and invites unpleasant questions like what limits God (logic only, inert matter, outer chaos, other beings), is there someone/thing more powerful, etc.    Interestingly enough, the original Christian theodicy wasn't free will. St.Augustine says poetically "the canvas of creation requires both the black and the white paint". "Requires" here can be interpreted differently,  is to explain the black by the "just punishment" for the original sin, which brings us to Adam and Eve, and back to the free will. But if God is still omnipotent and hence omniscient creating creatures he knew would sin, and then punishing them for it is hardly an improvement.   2) However, Augustine can be read more straightforwardly. Evil is "required" for the perfection of the world just like black paint is required for the perfection of a painting (evil can be interpreted as lack of good rather than something onto itself but that is beside the point). God is immoral, and he is beyond our conception of morality, his notion of perfection is his own. This is much bolder, and similar to Plotinus's position on the existence of the One, the One can not exist because it is "prior to all existence". But it does evoke the unpleasant taste of "ends justify the means". It is one thing to use "God is beyond our logic and comprehension" to get rid of unliftable stones, it is quite another to convince people to worship an immoral perfectionist.  Formally, there is another logical possibility of asserting both morality and omnipotence, but defending the morality of creating a world full of suffering, for a higher purpose perhaps. For example, the suffering only appears bad to us, not to an absolute being in its wisdom. But this simply redefines morality and is only different in phraseology from the Augustine's position. The question is asked in human terms and has to be answered in kind. One is free to argue that God is beyond human logic and/or morality, but it still leaves him immoral as commonly understood.   There does not appear to be a way to square this circle, one has to accept Leibniz's or Augustine's alternative (or both) and deal with the downsides. Option 2) looks more consistent, but it is rarely embraced, at least not openly. Perhaps, for "public relations" reasons it is more attractive to dismiss the question with something like "this is beyond us" or change the subject to the free will.      What are the modern theological positions on the morality/omnipotence in catholisism, protestantism, etc.? Is the question relevant to other religions?   EDIT: I am not sure that it came through but by immoral I did not mean morally deficient or evil. Rather at the God's plane of being morality becomes meaningless, like tastes or smells are meaningless when applied to elementary particles. I'm taking Calculus II next semester so I don't have all the tools from it to back me up but something came up in a discussion this evening that asked me to prove that the following sequence converges. As I'm new to formal math I'm hoping someone can verify my proof, especially step $6$. I must prove that $$\lim_{n\to\infty}P_n=b$$ where $$P_{n+1} = P_n + \frac{b-P_n}{2}$$ $$P_0 = a$$ and $a,b \in \mathbb{R}$ such that $a&gt;b$ My proof goes as follows:   We must prove that $$\lim_{n\to\infty}P_n=b$$  That is, $\forall\epsilon&gt;0$ there exists a $\delta$ such that there exists an $n$ satisfying $$n&gt;\delta \iff b-P_n&lt;\epsilon$$  First, let us prove that $P_n$ is bounded by $b$ by contradiction. Let us choose an $i$ such that $P_i &lt; b$ and $P_{i+1}&gt;b$. Then we have $P_{i+1} = P_i + \Delta P$ where $\Delta P &gt; b-P_i$. but this is impossible because by the definition of the recurrence $\Delta P = \frac{b-P_i}{2} &lt; b-P_i$. Therefore, $P_i$ is bounded from above by $b$.  Additionally, it is trivial to show that $P_{i+1} &gt; P_i$.  With this in mind, choose an arbitrary $\epsilon$. We need to show that there exists an $n$ such that $P_n &gt; b-\epsilon$. Assume the contrary, that $P_n$ is bounded by $\epsilon$.  Using a similar argument to what which was used before, assume $P_n$ is bounded by $\epsilon$ and then there exists a $P_i$ such that $P_{i+1}\not &gt; P_i$. But this results in a contradiction with (4)  Therefore, $P_n$ is not bounded by $\epsilon$ and we have proved what we want to prove. Proper display of fractions Cannot install WPS Office I'm using Ubuntu GNOME 16.10 and was trying to install WPS Office. I tried to install using software centre But it wasn't installing without giving any error logs. Then I opened it with Gdebi installer. Here is what I saw.    I have already trying enabling multiverse repositories.  Edit: I have downloaded the amd64 package from the WPS official site, but even now installing through terminal I still get the missing dependencies error.