Word to describe a 'homemade' way to fix something.
single word that means which can be constructed at home
What hinge is this? Bit of a newbie to DIY here.  We've moved in to a house that came with a summer house but the double doors were a bit wonky/wouldn't open easily without breaking in the door. I need to get 4 new hinges for the doors but I'm not sure where I can find these ones. I've looked online on UK sites such as Screwfix but as I'm not sure of the particular name I'm not having much luck.  I've attached a drawing (sorry it's the best I can do as my camera isn't working). My Paint skills aren't the best, it's actually a cylinder shape and half of it spins/moves (this is what is broken/doesn't move anymore so it breaks in to the door).    Cheers, Jack Blown out blue/red light making photos look out of focus Proving a Composition of Functions is Bijective If I have $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f$ is onto and $f\circ f\circ f = f$, how can I prove that $f$ is bijective? I know that I only have to prove that it is 1-to-1 because I'm given the fact that it's onto, but how can I use the fact that $f\circ f\circ f=f$ to prove that it is 1-to-1?  Thanks! "I feel bad for you" versus "I feel badly for you" Duplicate explanation should encourage improvement of original The new closed-as-duplicate banner says:     This question has been asked before and already has an answer. If those answers do not fully address your question, please .   I think it should instead say     This question has been asked before and already has an answer. If answers there do not fully address your question, please  or  to the pre-existing one.   &mdash; to discourage re-re-asking the same question. (I also changed "those answers", which has no plural , to "answers there", which doesn't need one.)    As of September 19, 2013, the original text quoted above has returned here. Can all circles of radius $1/n$ be packed in a unit disk, excluding the circle of radius $1/1$? This problem occurred to me when I came across a similar problem where the radii were taken over only the primes. That question was unanswered, but it seems to me infinitely many circles of radius $1/2, 1/3, 1/4...$ can fit into a unit disk. The area of all those circles would be $\pi \sum_2^\infty 1/n^2 = \pi^3/6 -\pi$, which is less than the area of the unit disk $\pi$. But can the circles actually be packed with no overlaps? How do people understand astromech droids? First of all, am I being crazy in thinking that if $\lambda$ is an eigenvalue of $AB$, where $A$ and $B$ are both $N \times N$ matrices (not necessarily invertible), then $\lambda$ is also an eigenvalue of $BA$?  If it's not true, then under what conditions is it true or not true?  If it is true, can anyone point me to a citation?  I couldn't find it in a quick perusal of Horn &amp; Johnson.  I have seen a couple proofs that the characteristic polynomial of $AB$ is equal to the characteristic polynomial of $BA$, but none with any citations.  A trivial proof would be OK, but a citation is better. Book identification: technologically-enhanced man (1970s/80s YA) I'm very new to classical propositional logic, and my lecturer is using a system with the following axioms:  A1. X→(Y→X)  A2. (X→(Y→Z))→((X→Y)→(X→Z))  A3. (¬Y→¬X)→(X→Y)  Use uniform substitution and Modes Ponens (MP) as rule of inference. Deduction Theorem also allowed.  I'd like to deduce the following:  (a) X, Y ⊢ ¬(X→(¬Y)) (conjunction, but expressed in terms of NOT and IMPLIES)  (b) X→((X→Y)→Y)  (c) (X→Y)→(((¬Y)→X)→Y)  (d) (((¬Y)→X)→Y)→(X→Y)  Remark: The last two are to show that the axiom (A3) can be replaced with a different axiom:  B3. (¬Y→¬X)→(((¬Y)→X)→Y).  Solution: Bram28 proves (b) and (c) below. I demonstrate (a) and (d).  (d) is alright; first, I claim that X, ((¬Y→X)→Y) ⊢ Y.  Axiom (A1) gives ⊢X→(¬Y→X) so that ((¬Y→X)→Y)⊢X→Y by Hypothetical Syllogism (HS), which Bram28 shows to be valid.  Now that we have ((¬Y→X)→Y)⊢X→Y just apply the deduction theorem to get what we want: ⊢((¬Y→X)→Y)→(X→Y).  (a) is slightly trickier. Axiom (A1) gives ⊢ Y→(X→Y) so that X, Y ⊢ (X→Y).  Next, by part (b), we have as a theorem that ⊢X→((X→¬Y)→¬Y).  Therefore X, Y ⊢(X→¬Y)→¬Y.  But ⊢(¬¬(X→¬Y))→(X→¬Y) by DN Elim, which Bram28 shows to be valid.  By HS, X, Y ⊢(¬¬(X→¬Y))→¬Y.  By Axiom (A3) and MP, X, Y ⊢(Y→¬(X→¬Y)).  Also, rather trivially X, Y ⊢Y.  Finally by MP we conclude X, Y ⊢¬(X→¬Y) as required. Let $P(X)$ denote the power set of $X$ (the set of all subsets of $X$) with a partial order given by inclusion.  If $F: P(X) \to  P(X)$ is monotone (order preserving), then $F$ has a fixed point.  How are we going to prove that without using the term (complete lattice)?! Let $G=\langle a\rangle$ be a cyclic group of order $n$. Show that, for every divisor $d$ of $n$, there exists a subgroup of $G$ whose order is $d$. Proving the geometric sum formula by induction $$\sum_{k=0}^nq^k = \frac{1-q^{n+1}}{1-q}$$  I want to prove this by induction. Here's what I have.  $$\frac{1-q^{n+1}}{1-q} + q^{n+1} = \frac{1-q^{n+1}+q^{n+1}(1-q)}{1-q}$$  I wanted to factor a $q^{n+1}$ out of the second expression but that 1- is screwing it up... I've been having an argument with a colleague about this sentence, could you please let me know which one of us is correct:     There are no shortage of applications for our product in this   space.   She is convinced that are should be replaced by is, and I think it should stand as it is. Thanks for your help! database disk image is malformed with dnf on Fedora 29 I am trying to copy multiple files named "F3.bam" two level of directories to up and then rename these files with the name of the sub-directory after copy.  For example:  /samples/mydata1/RUN1/ID_date/PCR2/TIME1/F3.bam /samples/mydata2/RUN1/ID2_date4/PCR2/TIME7/F3.bam /samples/mydataxxx/RUN1/IDxxx_datexxx/PCR2/TIMExxx/F3.bam   Expected results:  1. The files are first copied two directories level up:  /samples/mydata1/RUN1/ID_date/F3.bam /samples/mydata2/RUN1/ID2_date4/F3.bam /samples/mydataxxx/RUN1/IDxxx_datexxx/F3.bam   2. The files are renamed according to the name of the current sub-directory:  /samples/mydata1/RUN1/ID_date/ID_date_F3.bam /samples/mydata2/RUN1/ID2_date4/ID2_date4_F3.bam /samples/mydataxxx/RUN1/IDxxx_datexxx/IDxxx_datexxx_F3.bam   Ideally a bash loop would be great (working on a Mac). How can Finder empty Trash more quickly than "rm -rf /Volumes/Foo/.Trashes/501/*"? The "Trash" of a 16-GB usb2 flash drive had 3.4-GB of data. I emptied it as such:     $ cd /Volumes/FooDrive/.Trashes/501   $ rm -rf *   This took over 5-seconds. That seemed pretty long. So, as a test I just decided to empty the Trash of all of my drives via Finder. Other drives: 2-TB usb3 500-GB usb2 16-GB usb2 250-GB ssd  I measured the size of the Trash as such:       $ du -ch /Volumes/WhateverDrive/.Trashes   There was 15-GB in the Trash of the 500-GB drive, and 5-GB in the Trash of my other 16-GB flash drive. I didn't measure the other drives.  Finder appeared to take much less than 5-seconds to empty the Trash for all those drives.   Does Finder really empty the Trash so much more quickly than /bin/rm? How?   Does Finder just make the "clicking sound" (signaling the trash is empty) before the Trash on all the drives is really empty? Error: ! Font OT1/mathkerncmss/m/sl/10.95=mathkerncmssi10 at 10.95pt not loadable: Met ric (TFM) file not found jQuery/JavaScript to replace broken images Terence Tao, Analysis I, Ex. 5.5.2: Entry point needed