I'm using Excel (2013). I have a workbook with numerous different worksheets, and would like to have them take up two rows of worksheet tabs rather than just one. Can I achieve this somehow? I can't find an appropriate setting in the Excel Options dialog.
Some days I'm working with extremely large Excel files.  They have too many sheets that can't be shown in one sheet tab. (see image)      So, I'm wondering whether or not we can show Excel sheet on more than 1 row. Many tries with Google don't help much.    Many tries but I still can't make Excel show the sheet bar the way I want (≥ 2 rows), so I'm working with some VBA Scripts kopischke suggested.    FYI, right click on |&lt; &lt;&lt; &gt;&gt; &gt;| button on the bottom-right corner show you a list of up to 15 sheets.  For more, click on View more in the list appeared.
What is an adjective referring to “someone who puts people down”? I was just thinking about someone I went to school with who had a penchant for putting people down, and I was trying to think of a word that describes this, but couldn’t find it. It would be a synonym of caustic, abrasive, discouraging, and undermining, but these all have slightly different connotations (to wear away", cause discouragement or to make fall). I want an adjective that has the connotation of a hammer hitting you on the head. The word does not necessarily need to describe someone who does this on purpose, out of evil intent. Is there such a word?    The word should give you a mental picture of someone barraging you with negative comments, but it not necessarily affecting you. Think of someone who so often has something negative to “contribute” that many can just either ignore it or roll their eyes at it, and that is the extent of its effect. I keep getting the mental picture of the word buffet as in, how waves buffet the shore. A constant hitting, without the hitting necessarily being devastating in its destruction. Remove category & tag base from WordPress url - without a plugin How do you allocate more RAM to Minecraft? There is a lot of lag when I play Minecraft. I think this is due to the low amount of RAM I've allocated to Minecraft. (495MB)  How do you allocate more RAM to Minecraft?  I checked my other computer (Intel i5 Core) and it had the same amount of RAM allocated to Minecraft. However, Minecraft did not lag on that computer.  After I installed Optifine on this computer, the lag stopped (I don't know why cause I read that Optifine splits the load onto 2 cores, but I have only one core (At least I think so, a sticker on the laptop says Intel CORE i7)). Problem with camera orientation in motion tracking sorry for my English, I write from Italy. I'll try to explain my problem. I'm tracking a footage with a solver error 0.2973. When I try to set the floor, the orientation of the camera is wrong (see screenshot).  So I tried to manually orient the camera, but even if well oriented respect to the xyz axes, geometry is incorrect as you can see from the screenshots below (the selected marker appears to be on the same plane but it is lower).  The video was recorded by a roof of a house so much higher than the road. It seems that Blender is not able to recognize this height and set the floor in a wrong point of view. Can someone help me solve this problem? Thank you Here is my proof, I would appreciate it if someone could critique it for me:  To prove this statement true, we must proof that the two conditional statements ("If $\mathcal{P}(A)⊆ \mathcal{P}(B)$, then $A⊆B$," and, If $A⊆B$, then $\mathcal{P}(A)⊆ \mathcal{P}(B)$) are true.    Contrapositive of the first statement: If $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$  If $A \nsubseteq B$, then there must be some element in $A$, call it $x$, that is not in $B$: $x \in A$, and $x \notin B$. Since $x \in A$, then $\{x\} \in \mathcal{P}(A)$; moreover, since $x \notin B$, then $\{x\} \notin \mathcal{P}(B)$, which proves that, if $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$. By proving the contrapositive true, the original proposition must be true.    To prove the second statement true, I would implement nearly the same argument, so that isn't necessary to write. So, does this proof seem correct? Also, was the contrapositive necessary? Or is there another way to prove the initial statement? SEVERE: A child container failed during start java.util.concurrent.ExecutionException: org.apache.catalina.LifecycleException: Array with labeling columns Is there an intelligent way to label (or name) columns in a matrix? I would like to have something like  \[M=\left[\begin{array}{c|cc} 1 &amp; 2 &amp; 3\\ 4 &amp; 5 &amp; 6 \end{array}\right],N=\left[\begin{array}{c|cc} 7 &amp; 8 &amp; 9\\ 10 &amp; 11 &amp; 12 \end{array}\right]\] \[A \ B \ C \ D \ E \ F\]   where A, B,..., F are the labels (or rather names) of the first, second, ..., sixth columns, respectively. What I want is to label A be exactly below the first column, label B be below the second column, etc., of course, nicely aligned. So far I have forced some extra spaces before the labels manually, and also experimented with the \phantom command. Raspberry Pi + IRFZ44N to control a 5050 RGB LED strip: current seems to be limited If $f:[a,b]\to \mathbb{R}$ is continuous and nonnegative and $\int_a^b{f}=0$, then $f(x)=0$ for all $x\in [a,b]$ I've got a capture the flag arena. One issue we keep running into during play, is that of clutter.  We have /gamerule keepInventory false, so the pitch gets rather cluttered with armor, weapons and building materials, such that it becomes difficult to pick the flag up, due to full inventories.  I've discovered the following command to remove all drops from the arena:  /kill @e[type=Item]   However, this removes all the floating items from the ground. I'd like, if possible, to preserve the flag from being destroyed.  I have this command for detecting the flag:  /testfor @e[type=Item,x=40,y=1,z=40,dx=40,dy=27,dz=80] {Item:{id:"minecraft:banner",Damage:4s}}   But I'm not sure how to combine the two in a way that it will destroy everything except for the flag.  Any ideas? An infinite Penney's game. Context: I need to solve one variation of the . Problem: Two players ($A$ and $B$) throw a coin until one of the victory sequences appears. For player $A$ the victory sequence is $HTT$, for player $B$ the victory sequence is $TTH$. What is the probability that $A$ wins and what is the expected number of throws given that $A$ won? My attempt: Intuitively, it seems that player $B$ is more likely to win. Basically, I reliase that we can use the Markov-Chains, but my idea is only draw a binary tree and consider the sequences on the n-th step. My another idea was to calculate the expected number of throws to get, say, $HH$ that would be equal to $r_{HH}$: $$r_{HH} =  \frac{1}{4}E(number|HH) + \frac{1}{4}E(number|HH)+ \frac{1}{4}E(number|TT)+ \frac{1}{4}E(number|TH) = \frac{1}{4}(9 + \frac{5}{2} r_{HH})$$  And get $r_{HH}$. My issues: I don't really understand how to join these ideas to get a proper solution. XeLaTeX update [20201004] and item ball in Beamer Let's say it again, we all make mistakes, and I have just made one.  A brief history: I was doing some stuff on a VPS (Debian) I'm renting, when I noticed some strange behaviour. Using the netstat command I saw an non-authorized connection through SSH. I didn't know what to do, so I decided to close his connection using iptables:  iptables -A INPUT -p tcp --dport ssh -s IP -j DROP   But I am tired, and I wrote  iptables -A INPUT -p tcp --dport ssh -j DROP   and I kicked myself (and everyone else) out...  How do I fix this? Let $A$ be a complex matrix such that $A^n = I$, show that $A$ is diagonalisable.  How do I do this?    I would hazard a guess that since $A^n = I$ that means that $A^n$ is obvious diagonal, and I suppose that a diagonal matrix can only be formed by the square of another diagonal matrix, but I am not sure if that is the case. Also I think this neglects the complex cases.  Is there a Jordan block argument that I should use? I need to monitor the network activities of a process and the bandwidth of the connection. So I decided to use nethog and netperf combine with grep and awk to write the values to a file.   nethog command: nethogs -t  eth0 | grep firefox | awk '{ print $3 }' (I want to take the upload/download speed). The 1st problem is that command doesn't print anything.  More detail: command: nethogs -t  eth0 | grep firefox output:  usr/lib/firefox/firefox-bi�)/4956/1000  0.338867    0.239063 /usr/lib/firefox/firefox-bi�)/4956/1000 0.543555    0.274219 /usr/lib/firefox/firefox-bi�)/4956/1000 0.794531    0.489844 /usr/lib/firefox/firefox-bi�)/4956/1000 0.794531    0.489844 /usr/lib/firefox/firefox-bi�)/4956/1000 0.749023    0.589844 /usr/lib/firefox/firefox-bi�)/4956/1000 1.30098 3.47617 /usr/lib/firefox/firefox-bi�)/4956/1000 1.90449 8.0127 /usr/lib/firefox/firefox-bi�)/4956/1000 5.31641 25.0033 /usr/lib/firefox/firefox-bi�)/4956/1000 8.60762 42.0176   I want the two values at the bottom right. But as you can see the output ain't consistent (the space), add | awk '{print $3}' most of the time return nothing. And I can't fix this.  netperf command: netperf -H 192.168.1.10 and the output is:  Recv   Send    Send                           Socket Socket  Message  Elapsed               Size   Size    Size     Time     Throughput   bytes  bytes   bytes    secs.    10^6bits/sec    524288 524288 262144    10.00    **718.62**   I want to print the value on the bottom right but have no idea how to. That's the 2nd problem => solved by  fedorqui!  Can you guys help me out? Any ideas will be appreciated! Can a 20V 4.51A power supply be used with a computer that requires a 19V 3.42A power supply? How to prove that $\lim_{n\rightarrow \infty}\frac {n^{2}}{2^{n}}=0$ & $\lim_{n\rightarrow \infty}\frac {n^{2}}{n!}=0$? Suggested improvements to badges index page ()  Currently,  shows the possible badges, number of total recipients, and brief descriptions. However, I have a few suggestions (mostly minor changes) that could make it a bit nicer.   There is no way to see (on that page) which badges I've earned multiple times — they all appear with a green checkmark. It would be nice to show something like "x 4" like in the badges section of the user profile page. For example:      There is no way to see which badges can be awarded multiple times — you have to click on a specific badge to see yea or nay. It would be nice to have a small, simple icon next to badges that can be earned multiple times, similar to the checkmark showing which badges I've earned. The layout would be nicer if the number of recipients for each badge were in a separate column — since the badge images are different sizes, the overlap and ragged right alignment makes the numbers harder to read. An alternative view that groups badges by "achievement level" (gold/silver/bronze) would be nice, as would a view that groups progressive degrees of the same accomplishment. (e.g. Nice Answer &rarr; Good Answer &rarr; Great Answer OR Popular Question &rarr; Notable Question &rarr; Famous Question) I have noticed that the confidence interval for predicted values in an linear regression tends to be narrow around the mean of the predictor and fat around the minimum and maximum values of the predictor. This can be seen in plots of these 4 linear regressions:    I initially thought this was because most values of the predictors were concentrated around the mean of the predictor. However, I then noticed that the narrow middle of the confidence interval would occur even if many values of were concentrated around the extremes of the predictor, as in the bottom left linear regression, which lots of values of the predictor are concentrated around the minimum of the predictor.  is anyone able to explain why confidence intervals for the predicted values in an linear regression tend to be narrow in the middle and fat at the extremes? Redirect 301 with hash part (anchor) #