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What is an ultra-wide lens? What is the point at which a lens is considered "ultra-wide angle" vs just wide angle? If I have an object implementing the Map interface in Java and I wish to iterate over every pair contained within it,  what is the most efficient way of going through the map?  Will the ordering of elements depend on the specific map implementation that I have for the interface? Showing that a group of order $pq$ is cyclic if it has normal subgroups of order $p$ and $q$ This is question 2.9 #9 from Topics in Algebra by Herstein:     If $o(G)$ is $pq$ where $p$ and $q$ are distinct prime numbers and if $G$ has a normal subgroup of order $p$ and a normal subgroup of order $q$, prove that $G$ is cyclic.   Let $N$ denote the normal subgroup of order $p$ and $M$ denote the normal subgroup of order $q$. Here are a couple things I noted while exploring the problem:   $N$ and $M$ are cyclic, since they have prime order. $G = NM$ [1].   Any hints where to look next?    [1] Since $N,M$ are normal, $NM \le G$. Then $o(NM) \mid o(G)$, so $o(NM)$ is either $1$, $p$, $q$, or $pq$. The first three don't work because they force one of $N$ or $M$ to be $(e)$, so we have $o(NM) = pq$. Therefore, $NM = G$. Softmax vs Sigmoid function in Logistic classifier? What decides the choice of function ( Softmax vs Sigmoid ) in a Logistic classifier ?  Suppose there are 4 output classes . Each of the above function gives the probabilities of each class being the correct output . So which one to take for a classifier ? Problem using wildcard with apt When I run this command :  apt install libkf5*   I get the following error :  Unable to locate package libkf5*    I'm sure that the packages whose names start with libkf5 exist (tab-completion says so) . So the problem is not about the absence of those packages.(Note the asterisk at the end of that error message , the wildcard's not working at all) I use apt v.1.9. Thanks in advance. Selecting through one-to-many relationship I'm trying to make a species distribution project where I can select a species from a non-spatial list of species and the selection would highlight (select) all the counties to which the species has been attributed.  Simplified example data: species, a non-spatial table of species (CSV with header): "species" "Mus borealis" "Mus australis"  species_distributions, a non-spatial relation table (csv with header): "species","county" "Mus borealis","North" "Mus borealis","Central" "Mus australis","South" "Mus australis","Central"  counties, a spatial feature class representing county boundaries (first row = field name): "county" "North" "South" "Central" "West"  So if I select "Mus borealis" from the species table, the "North" and "Central" counties would be highlighted in the map. Seems like a simple scenario, but I have been unable to uncover a solution on my own. The closest I've found is a plugin "Select by relationship" but it doesn't appear to be available for QGIS 3 (I'm using QGIS 3.4.3). I have a PostGIS server available. Why not use development provisioning instead of ad hoc? Are there any cheap, fast lenses capable of highschool sports photography? I recently took on my first paying client, and she is also into photgraphy. She usually does sports for her highschool yearbook. They just have an entry level Canon DSLR with a kit zoom on it that goes up to 200mm, but it's not very fast (f/4 or 5.6 or so). She told me that she has a tough time - especially at basketball games - getting shots with any kind of reasonable lighting, and has to do a lot of work in post just to get them reasonable.   mentions the Cannon EF 300mm f/2.8L IS USM II for professional sports photographers. At $6k USD I suspect the yearbook won't be springing for that lens anytime soon.  So I was curious, are there any reasonable quality, fast lenses for Canon cameras that are lower priced (certainly no more than say, $1k)? I'd love to be able to point them to a couple of options. I assume the largest print they'd be making would be a full spread in the yearbook - maybe two 17"x11" (two 8.5" side by side) $gh = hg, \ \gcd(|g|, |h|) = 1\Rightarrow|gh| = |g||h|\ \ (|a|$ = order of element $a)$ Let $G$ be a group and $g,h \in G$. I need to prove that if $g$ and $h$ commute and their orders are coprime, then $|gh| = |g||h|$, that is, the order of their product is the multiple of their orders.  Since $gh = hg$, then $(gh)^{ lcm(|g|, |h|)} = e$, so, $|gh|$ divides $lcm(|g|, |h|)$. Since $\gcd(|g|, |h|) = 1$, then $lcm(|g|, |h|) = |g||h|$.  So, $|gh|$ divides $|g||h|$. A cardinal $\kappa$ such that $2^{\lambda}<\kappa$ for all $\lambda<\kappa$ is regular? A cardinal $\kappa$ such that $2^{\lambda}&lt;\kappa$ for all $\lambda&lt;\kappa$ is regular?  I would appreciate very much an answer What happens in electron-electron collisions? $n$ people attend the same meeting, what is the chance that two people share the same birthday? Given the first $b$ birthdays, the probability the next person doesn't share a birthday with any that went before is $(365-b)/365$. The probability that none share the same birthday is the following: $\Pi_{0}^{n-1}\frac{365-b}{365}$. How many people would have to attend a meeting so that there is at least a $50$% chance that two people share a birthday?   So I set $\Pi_{0}^{n-1}\frac{365-b}{365}=.5$ and from there I manipulated some algebra to get $\frac{364!}{(364-n)!365^{n}}=.5\iff (364-n)!365^{n}=364!/.5=.....$   There has to be an easier way of simplifying this. Deriving Chinese Remainder Theorem from gcd Bezout identity We want to find the solution $x$ to the congruence system  $$\begin{align} x &amp;\equiv r_1 \!\!\!\pmod{\!m_1}\\ x &amp;\equiv r_2 \!\!\!\pmod{\!m_2}\end{align},\ \ {\rm where}\ \  \gcd(m_1, m_2) = 1$$  These can be rewritten as  $$\begin{align} x &amp;= r_1 + m_1 j\\ x &amp;= r_2 - m_2k\end{align}$$  for unknown integers $j, k$. So we set those two equations as equal and rearrange them to  $$m_1j + m_2k = r_2 - r_1$$  Now assume we perform $\text{egcd}(m_1, m_2)$, which gives us the solution to  $$m_1u + m_2v = \gcd(m_1, m_2)$$  Let the integer $\,h\,$ make $\,h \gcd(m_1, m_2) = r_2 - r_1$.  Then $\qquad hm_1u + hm_2v = h\gcd(m_1, m_2) =  r_2 - r_1 = m_1j + m_2k$  At this point I am lost because I don't think we can just assume  $hm_1u = m_1j$ and extract $j = hu$.   How do I get the values of $j$ or $k$ so I can get the value of $x$? Why do some Kree have white skin? This seems bizarre to me.  All the Kree we've seen thus far (MCU based) were blue skinned... Ronan the Accuser being the most famous  What do Kree people look like?  Is there anything concrete with their appearance?  It seems odd that Danvers (Captain Marvel) and Jude Law's character (Yon-Rog?) and the "Supreme Intelligence" are all white skinned. Restoring backups failing Ubuntu 16.04 Invalid variance: The type parameter 'T' must be contravariantly valid on 'UserQuery.IItem.ItemList'. 'T' is covariant Options for crowdsourcing GIS data? I am looking at collecting spatial information from a range of stakeholders to identify opportunities within a landscape.  However, the data that we are collecting are not presently mapped and so we wish to use an interactive method to get this information onto a map.  Which online websites/methods would people recommend for circulating to a non-technical audience to try and get them to map what they know?    Where the emphasis is on ease of use by the stakeholder. Consider the sentence      E-books are on the rise, but they haven't suppressed paper books though.    This usage seems to be quite common, but when I learned English  I was taught to use "however" where I now read "though". My questions on this use of "though" are:     Is this a new trend? Is it restricted to American English? Is "though"  considered more colloquial than "however" ?    And on a syntactic level: Do you put a comma before "though"? If $X_1, ..., X_n$ are independent identically-distributed random variables, what can be said about the distribution of $\min(X_1, ..., X_n)$ in general? I was wondering what the cardinality of the set of all real sequences is. A random search through this site says that it is equal to the cardinality of the real numbers. This is very surprising to me, since the cardinality of all rational sequences is the same as the cardinality of reals, and it seemed fairly intuitive to me that if cardinality of a set $A$ is strictly greater than the cardinality of the set $B$, then cardinality of $A^{\mathbb{N}}$ should be strictly greater than cardinality of $B^{\mathbb{N}}$. It turns out to be false.  Some technical answers have appeared in this forum elsewhere but I do not understand them. As I am not an expert in this topic, could some one explain me in simple terms why this is happening?  Also is the cardinality of all functions from reals to reals also the same as the cardinality of reals?