So one thing which is still giving me a real hard time is incoming traffic from outside the city limits. I just can't seem to figure out a good way to deal with it and I always end up with a ton of gridlock. Here I've got a 6 lane one way road going around, and I'd think that would alleviate congestion some but it sure doesn't, besides which everyone seems to use only one lane anyways.  I've tried actual roundabouts but seem to have trouble getting them to work right.
I noticed that I got some initial traffic problems at the point where the initial highway is connected to my city. At the beginning I simply elongated both roads that came from that highway, but this doesn't seem to scale.  What I wanted to do now is to create a single six-lane highway from these two four-lane roads. I'd be open to continue to use two roads in parallel here, but I didn't manage to add any way of crossing between the lanes when I tried that idea.  So, how do I merge two one-way roads into a single two-way road? Or, if my idea is really bad, what is a good way to connect the initial highway access to my fledgling city?  Here's a screenshot of the roads I want to connect, if my description wasn't entirely clear:
"I who has" vs "I who have." Possible Duplicate:          Am aware it is correct to say "I have written..." and "I am a doctor who has writen..."  I have, however, severally come across the sentence "I, Tertius, who have written...". Shouldn't it be "I, Tertius, who has written..."? How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? "I think him to be about 50" or "I think he is about 50"? I have two options. Which one is correct?      a) I think him to be about 50.   b) I think he is about 50.   If both are correct, should I avoid one or the other? Whenever I try to cut out a circular hole in either a cylinder or sphere the polygons around the circular hole always have distorted normals after smoothing. Is there anyway to fix this. I've tried using the remesh modifier but wind up with some pretty ugly results. Convert /etc/network/interfaces to netplan for floating IP I am new to the netplan on Ubuntu 18.04, just discovered it exists today. I have an interface that I am trying to add to create a floating IP on google cloud. It is based on this example:   I tested the example, and it works, but now I try to do the same on ubuntu, and I don't know how to convert this:  cat &lt;&lt; EOF &gt;&gt; /etc/network/interfaces auto eth0:0 iface eth0:0 inet static     address 10.190.1.1     netmask 255.255.255.255 EOF   to netplan.  The output for ls /etc/netplan is '50-cloud-init.yaml'. and the output for cat /etc/netplan/*.yaml is:  network:     version: 2     ethernets:         ens4:             dhcp4: true             match:                 macaddress: 42:01:0a:8e:00:3e             set-name: ens4   my ifconfig:  # ifconfig  ens4 ens4: flags=4163&lt;UP,BROADCAST,RUNNING,MULTICAST&gt;  mtu 1460     inet 10.142.0.62  netmask 255.255.255.255  broadcast 0.0.0.0     inet6 fe80::4001:aff:fe8e:3e  prefixlen 64  scopeid 0x20&lt;link&gt;     ether 42:01:0a:8e:00:3e  txqueuelen 1000  (Ethernet)     RX packets 9430  bytes 1635180 (1.6 MB)     RX errors 0  dropped 0  overruns 0  frame 0     TX packets 13383  bytes 1513428 (1.5 MB)     TX errors 0  dropped 0 overruns 0  carrier 0  collisions 0   What is the equivalent of what he said to put in the file, for netplan? How do I check the integrity of a storage medium (hard disk or flash drive)? I am trying to define an alternative \item command that would allow me to use a different color for some of the items. My code has the following form:  \documentclass{article} \usepackage{color}  \definecolor{gray}{rgb}{0.7,0.7,0.7} \newcommand{\grayitem}[1]{{\color{gray} \item #1}}  \begin{document}  \begin{enumerate} \item First item. \grayitem Second item. \item Third item. \end{enumerate}  \end{document}   I was hoping that  \grayitem Second item.   would produce the same result as  {\color{gray} \item Second item.}   but for some reason, these are not equivalent. They produce the following results, respectively:  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;   Why are these two lines of code not equivalent? What am I doing wrong and how can I change the definition of \grayitem to produce the second result?  I have also tried removing a pair of braces, like this  \newcommand{\grayitem}[1]{\color{gray} \item #1}   but then the third item becomes gray as well.  The solution should also work for multi-paragraph items. In Chamber of Secrets, in the first Duelling Club class, Snape suggests Malfoy use the Serpensortia spell against Harry. This of course reveals that Harry is a Parselmouth, but how could Snape anticipate this? Was there any reason for Snape to suspect Harry is a Parselmouth? Or was it just a way to laugh at how Harry would get frightened? Is the Formula Logically Valid? I just purchased an unlocked iPhone 5 from the Apple store today, and I can't activate it to use it because it says "please" insert a valid SIM card to use it. I'm not using this phone as a "phone" to make calls. I'm a developer and am just looking to use it for development purposes. How can I go about using the iOS 5? Does the sequence $$\displaystyle \frac{n}{\sum\limits_{k=1}^{n}\Big(\frac{k}{k+1}\Big)^k}$$   converge?   Attempt. Since $\Big(\frac{k}{k+1}\Big)^k \rightarrow 1/e\neq 0$ and the terms are positive, the series $\sum\limits_{k=1}^{\infty}\Big(\frac{k}{k+1}\Big)^k$ diverges to $+\infty$. I find hard to determine if $n$ or the sum goes faster to $+\infty.$   Thanks in advance. Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ How to reinstall PulseAudio (Ubuntu 12.04) Distribution of range of Uniform $(0,1)$ distribution Let $X_1,...,X_n$ be iid as $U \sim [0,1]$ with pdf $f(x) = 1(0 &lt;x&lt; 1)$. Determine the range distribution.   I have managed to determine the distribution of the minimum and the maximum, however I cannot/do not know how to determine the distribution of range.   Minimum:  $f_N = n(1-x)^{n-1}$ and maximum $f_M= nx^{n-1}$   I am aware that the calculation of range distribution is based on convolution: $f_R=\int f_{N,M}(x,x-r)dx$, However I cannot get the correct result, which according to my lecture slides is : $f_R= n(n-1)r^{n-2}(1-r)$  Could someone break this down for me ? My math background is not great A Sequence converges to a limit $L \in \mathbb R$ if, and only if, an $N$-th tail of the sequences converges to $L$ for some $N \in \mathbb N$ I have been assigned this problem for homework:     Show that, if $a &lt; b + \epsilon$ for every  $\epsilon \gt 0$, then $a\le b$.   I have tried to go about this using Induction, but I don't know what the base case would be. It is obvious to me in my mind, but I don't know how to put it into mathematical terms on paper. any hints? Layers with same coordinate system should align/overlap in ArcMap but do not? I put 4 different layers into my ArcMap document, but only one of them will show up on the map! I should be seeing the other 3 layers as well. When I do a Zoom to layer, I can see one layer, but I cannot figure out how to make the other 3 show up at once on top of each other.  Before I opened ArcMap, I used ArcCatalog where I right-clicked 3/4 of the shapefiles and  clicked Properties &rarr; Coordinate system &rarr; Import and then clicked on my Trees layer, which has the coordinate system I want the other 3 to have. But when I opened ArcMap the 4 layers did not overlap, nor could I move them above or below each other. K-means is a widely used method in cluster analysis. In my understanding, this method does NOT require ANY assumptions, i.e., give me a dataset and a pre-specified number of clusters, k, and I just apply this algorithm which minimizes the sum of squared errors (SSE), the within cluster squared error.  So k-means is essentially an optimization problem.  I read some material about the drawbacks of k-means.  Most of them say that:   k-means assumes the variance of the distribution of each attribute (variable) is spherical;  all variables have the same variance; the prior probability for all k clusters is the same, i.e., each cluster has roughly equal number of observations;   If any one of these 3 assumptions are violated, then k-means will fail.  I could not understand the logic behind this statement. I think the k-means method makes essentially no assumptions, it just minimizes the SSE, so I cannot see the link between minimizing the SSE and those 3 "assumptions". I often get an error saying      unable to update ICE authority.    Is there a way out of this? Are fractions with prime numerator and denominator dense?