When I attempt to add XY Data to my map the lat and lon coordinates are projected in meters for example (-82, 34) becomes -82 meters and 34 meters. The GCS and Projection for both the new layer and the data frame are WGS 1984 Web Mercator (auxiliary sphere). Why is the new layer's XY Data displaying incorrectly?
I am trying to import an Excel table with around 10,000 XY points into ArcMap 10.2. I am using State Plane NAD 1983 Alabama West (US Feet) as the Coordinate System in my map data frame. I then add the excel table to my Table of Contents in the map then select Display XY Data. Next, I choose the same coordinate system as my data frame then select OK. Once it creates the Events layer for my points they are not in the correct location.   Should I switch everything in my map to WGS84 or is there a solution for my points to be added onto my current map with the state plane coordinate system?
Getting a domain name off of the blacklist I have a friend who purchased a domain name recently that he found out (after buying it) is apparently blocked by many corporate firewalls or proxies b/c it's been tagged as porn by the blacklist db they're using.    Are there any common agencies he can appeal to in order to get the domain name off the black lists?  It's not just one corporation, either, he's looking at marketing a subscription-based service to a certain industry, so any contacting a particular firewall/proxy admin and have them manually update the blacklist won't scale well.  Thanks in advance. Paul How do I join several objects into one? What are the details of "Objective-C Literals" mentioned in the Xcode 4.4 release notes? Pdf of the square of a standard normal random variable Formula based on picklist I am trying to create a formula that will put a 1 in a hidden field if the value of a picklist is equal to a specific value.  I keep getting stuck here and I can't figure out what I am missing?  IF(ISPICKVAL( Examiner_Type__c ="Pending Examiner Status","1","",)) How many numbers of 6 digits, that can be formed with digits 2,3,9. And also divided by 3? How many numbers of 6 digits, that can be formed with digits 2,3,9. And also divided by 3?  $$$$ I was trying to to add 2^6 (when there is no 2)+ C(6,2)2^2 (when 2 can be in two places)+C(6,3)2^3 (when 2 can be in three places)... Calculate $E(Y)$ and $Var(Y)$ given the random variable which is exponentially distributed The random variable $X$ is exponentially distributed with parameter $\lambda = \dfrac{1}{2}$. Also, define $Y = -2X + 1$. Calculate $E(Y)$ and $Var(Y)$.  So we have   $$f_X(x) = \dfrac{1}{2} e^{-\dfrac{1}{2}x} $$  if $x\ge0$ and $0$ otherwise.  But how to calculate $E(Y)$ and $Var(Y)$ ? How to change password using usermod? I am superadmin of a server and like to change password of an existing user . How can I do that ?  I tried   usermod -p 'new-password' john   but it didn't worked ? How to move a file? How is the default user avatar generated? Example of $f,g: [0,1]\to[0,1]$ and Riemann-integrable, but $g\circ f$ is not? "In" and "on": How can I decide which one to use for vehicles? Understanding English (Listening skills) What's wrong with this Wheatstone bridge? I was trying to solve this unbalanced Wheatstone bridge and found that the current $i$ in the mid wire is 9A .This result was achieved by using node voltage method.    Then I rearranged the same circuit to the form shown in the second image.  and at this point I flipped the right half of the circuit so as to obtain the equal resistors on the same side.    now I again rearranged the circuit as shown.    Here, the final circuit  is a balanced Wheatstone bridge and thus the current i must be 0.  So clearly something went wrong in there.  I saw a post  before this one just to make sure that there is no mistake in rearranging the circuit in the way I have done.   It was answered by the user PhillipWood as being correct and was unfortunately put on hold.  So can someone please point out what exactly went wrong with the above method. Do web browsers use different outgoing ports for different tabs? Many older movies seem to have vivid reds and particularly realistic or noticeable skin tones (they certainly seem to stand out from typical skin tones on TV or the majority of movies). This applies even in DVD or online versions and is not, given the range of material I've noticed the effect in, an artifact of special post processing or deliberate colour manipulation.  All the movies I've noticed the effect in are from the film era, especially from when colour became common to the late 1960's. I used to think it was just Technicolor but I've seen things recently that don't claim to be Technicolor that also show the effect.   claims the effect was due to the final print being constructed late in the process for separate black and white film strips for each primary colour allowing the final dyes to be very vivid and clear:     The rich colors that the Technicolor process gave came in part from the fact that the color was not added to the process until the final stages. The color information was recorded and processed as separate black and white images which were relatively easy to control and preserve.   Is this a good explanation? If the magnetic field doesn't polarize does it follow the electric field path of propagation? or does it vanish? $G$ solvable $\implies$ composition factors of $G$ are of prime order. I'm trying to prove the equivalency of the following definitions for a finite group, $G$:  (i) $G$ is solvable, i.e. there exists a chain of subgroups $1 = G_0  \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_s = G $ such that $G_{i+1}/ \ G_i$ is abelian for all $0 \leq i \leq s-1$  (ii) $G$ has a chain of subgroups $1 = H_0  \trianglelefteq H_1 \trianglelefteq \cdots \trianglelefteq H_s = G $ such that $H_{i+1} /\  H_i$ is cyclic for all $0 \leq i \leq s-1$.  (iii) All composition factors of $G$ are of prime order.  (iv) G has a chain of subgroups $1 = N_0  \trianglelefteq N_1 \trianglelefteq \cdots \trianglelefteq N_t = G $ such that each $N_i$ is a normal subgroup of $G$, and $N_{i+1}/ \ N_i$ is abelian for all $0 \leq i \leq t-1$.  It was simple to show that (iii) $\implies$ (ii) and (ii) $\implies$ (i). Now, I'm trying to prove that (i) $\implies$ (iii), and I'm getting stuck.   I know that If $G$ is an abelian simple group, then $|G|$ is prime, so all I need to show is that each $G_{i+1}/ \ G_i$ in the definition of solvable is simple.  If I knew that $1 = G_0  \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_s = G $ were a composition series, then each $G_{i+1}/ \ G_i$ would be simple by definition. But the definition of solvable doesn't assume a composition series. I know every group has a unique composition series, but I'm not sure how to connect that to the subgroups in the definition of solvable.   More generally, I guess I just don't understand how being abelian connects to being of simple or of prime order.  Any pushes in the right direction would be appreciated! Placement of images in row using subfigure I want to put four figures in a single row. The code I am trying to use is taken from .  This puts three figures in a row. When I add a fourth one, I get three images in row and the last one appears in the second row. But I want all of them in one row. Here is my code.  \documentclass[12pt,twoside]{report}% Use this line for the print version of the thesis   \usepackage{graphicx} \usepackage{caption} \usepackage{subcaption} \begin{document}  \begin{figure}         \centering         \begin{subfigure}[b]{0.3\textwidth}                 \centering                 \includegraphics[width=\textwidth]{gull}                 \caption{A gull}                 \label{fig:gull}         \end{subfigure}%           \begin{subfigure}[b]{0.3\textwidth}                 \centering                 \includegraphics[width=\textwidth]{gull}                 \caption{A gull2}                 \label{fig:gull2}         \end{subfigure}%          ~ %add desired spacing between images, e. g. ~, \quad, \qquad etc.           %(or a blank line to force the subfigure onto a new line)         \begin{subfigure}[b]{0.3\textwidth}                 \centering                 \includegraphics[width=\textwidth]{tiger}                 \caption{A tiger}                 \label{fig:tiger}         \end{subfigure}         ~ %add desired spacing between images, e. g. ~, \quad, \qquad etc.           %(or a blank line to force the subfigure onto a new line)         \begin{subfigure}[b]{0.3\textwidth}                 \centering                 \includegraphics[width=\textwidth]{mouse}                 \caption{A mouse}                 \label{fig:mouse}         \end{subfigure}         \caption{Pictures of animals}\label{fig:animals} \end{figure} \end{document}   As I have mentioned above, it works for three figures.  But I can not use it for four figures. What should I change?   What does  width=\textwidth mean?   I changed    \begin{subfigure}[b]{0.3\textwidth}   to  \begin{subfigure}[b]{0.2\textwidth}   but it didn't help.  Then, I changed     \includegraphics[width=\textwidth]{mouse}   to  \includegraphics[width=0.5\textwidth]{mouse}   Still no solution. Strictly Upper Triangular $n\times n$ Matrices form a Nilpotent Lie Algebra over $\mathbb{C}$ for $n\geq 2$ I have been attempting this problem for a while, it is an assignment problem so I don't want somebody to just post the answer, I'm just looking for hints.  Let $\mathfrak{u}(n,\mathbb{C})$ be the Lie algebra of strictly upper triangular $n\times n$ matrices over $\mathbb{C}$.  I am trying to show that $\mathfrak{u}(n,\mathbb{C})$ is a nilpotent Lie algebra for $n\geq 2$.  This is what I have looked at so far.  I have done some explicit calculations (I shall omit the bulk of them and just summarise below) for $n=2,3,4$ and have come up with some ideas.  Let $L=\mathfrak{u}(n, \mathbb{C})$.  If $n=2$,  $L^1 = [L, L] = &lt;0&gt;_{\mathbb{C}}$  So $L$ is nilpotent.  If $n=3$,  $L^1 = [L, L] = &lt;e_{13}&gt;_{\mathbb{C}}$  $L^2 = [L, L^1] = &lt;0&gt;_{\mathbb{C}} = \left\{ 0 \right\}$  So, $L$ is nilpotent.  If $n=3$  $L^1 = [L,L] = &lt;e_{13},e_{14},e_{24}&gt;_\mathbb{C}$  $L^2 = [L,L^1] = &lt;e_{14}&gt;_{\mathbb{C}}$  $L^3 = [L, L^2] = &lt;0&gt;_{\mathbb{C}} = \left\{ 0 \right\}$  So, $L$ is nilpotent.  So my hypothesis so far is that for an arbitrary $m\in\mathbb{N}$, $L^{m-1}=\{0\}$  My instinct is to do this by induction. My base case, $n=2$ has already been shown. Now I assume that $\mathfrak{u}(n,\mathbb{C})$ is nilpotent for $n=k$ ($k\in\mathbb{N}$) and consider the case for n=k+1.  However, I have no idea how to link these two cases together. I have read in Introduction to Lie Algebras and Representation Theory - J.E. Humphreys that $L^k$ should be spanned by basis vectors $e_{ij}$ where $j-i=k+1$ - however in my calculations for $L^1$, I obtain an $e_{14}$ which does not satisfy 4-1=2.  Am I missing something completely obvious? Or have I made a mistake somewhere? (I have checked my calculations numerous times, but cannot find any errors!)  Thanks,  Andy.