Here's a quick backstory as to why I'm making this post. If you don't want to read it, skip to the bold. I've just recently picked up Minecraft after a couple years and started playing around with it. After I got comfortable messing around with it, I noticed that they implemented command blocks into the game and I wanted to see if I remembered how to use them. I messed around with command blocks for a while before I wanted to give myself a weapon. I did a little bit of research on command blocks, tried several different commands, but nothing can get me what I want. So I thought to ask you guys, after all, some of the commands I tried were on here. This is where you start reading I want a netherite sword with the title &gt;The Legendary Sword of Power&lt; (How original) and the lore being &gt;The Ancient Ones used this weapon to kill the powerful Wither.&lt; On top of that, I wanted to give it a couple overpowered enchantments. If anyone could help me with this (or just give me a command), that would be highly appreciated. Thank you!
I'm used to creating Java Edition Minecraft maps, where NBT can be used to set many different block, item, and entity properties, like an item's CanPlaceOn/CanDestroy list, a chest's items, and more! To open my maps up to more players, I've started creating maps on Minecraft Bedrock Edition as well. But NBT is inaccessible from commands in BE, which I knew before, but didn't know how much of a problem I would have creating these complicated command chains. One of the things that I've always been doing is giving players items with custom item names, lores, enchantments, and other properties. But there is a problem. All this would have been done in NBT in Java Edition. In Bedrock, NBT is inaccessible from commands, meaning the commands I used in JE don't work. Are my dreams crushed?
Using Model to equally distribute points, spatially within known geometry using ArcGIS ModelBuilder? Is it possible to create a model in ArcGIS ModelBuilder where you simply state number of points and an area in the form of a polygon to get the model to try and equally space out the points based on the geometry of the polygon and number of points? Change the title size in scrartcl I'd like to use a smaller title. I'm using the scrartcl documentclass from the . I can change it's color font but not it's size.  \setkomafont{title}{\normalfont\bfseries\small}   Everything works exept the size. Display error message if same record exist with combination of some fields? I have taken three fields from case object called AccountName, IntegrationTo__c and SfType__c (checkbox).  I have created one case record with a specific combination of three fields.  For the second time I create a record with that combination, then  display error message only if SfType__c is true in first record otherwise allow  for record creation.  trigger AvoidDuplicate on Case(before insert)  {          set&lt;string&gt; newNameSet = new set&lt;string&gt;();     newNameSet = new set&lt;string&gt;();      set&lt;string&gt; newEmailSet = new set&lt;string&gt;();      set&lt;string&gt; dbNameSet = new set&lt;string&gt;();      set&lt;string&gt; dbEmailSet = new set&lt;string&gt;();      for(Case newLead : trigger.new){          newNameSet.add(newLead.Account.Name);          newEmailSet.add(newLead.IntegrationTo__c);      }      for(Case dbLead : [select id, Case.Account.Name, IntegrationTo__c from Case where IntegrationTo__c IN: newEmailSet OR Case.Account.Name IN: newNameSet]){          dbNameSet.add(dbLead.Account.Name);          dbEmailSet.add(dbLead.IntegrationTo__c);      }      for(Case newLead : trigger.new){          if(dbNameSet.contains(newLead.Account.Name) &amp;&amp; dbEmailSet.Contains(newLead.IntegrationTo__c))              newLead.addError('You are inserting Duplicate Case');      }  }   i wrote this this not checking the account field name and integration__c values only with combination of account and integration__c shows error Limit of derivative is zero. Does it imply a limit for f(x)? Do there exist an infinite number of 'rational' points in the equilateral triangle $ABC$? Let's call a point $P$ which satisfies the following condition 'a rational point'.  Condition: Each distance $PA, PB, PC$ from a point $P$ to three vertices $A, B, C$ of an equilateral triangle $ABC$ which has edge-length $1$ is rational number.  I'm interested in this point because I found the following three:  1. There exist an infinite number of rational points on an edge of the equilateral triangle $ABC$.  2. There exist an infinite number of rational points on the circumference of the circumscribed circle of the equilateral triangle $ABC$.  3. There exists a rational point in the equilateral triangle $ABC$.  Supposing that $A(0,\frac{\sqrt3}{2}), B(-\frac12,0), C(\frac12,0)$, one example for the above 3 is the following: $$P\left(\frac{61}{2058},\frac{220}{1029}\sqrt3\right), \left(PA, PB, PC\right)=\left(\frac{73}{147}, \frac{95}{147}, \frac{88}{147}\right).$$  I found this example by using computer. I found the other examples, but I don't know there exist an infinite number of rational points in the equilateral triangle $ABC$. Then, here are my questions.  Question 1: Do there exist an infinite number of rational points in the equilateral triangle $ABC$ ?  Question 2: Can we find all rational points in the equilateral triangle $ABC$ ? $\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) &gt; 1$.  Why should analytically continuing to $\zeta(-1)$ give the right answer? How to use both binary and continuous variables together in clustering? I need to use binary variables (values 0 &amp; 1) in k-means. But k-means only works with continuous variables. I know some people still use these binary variables in k-means ignoring the fact that k-means is only designed for continuous variables. This is unacceptable to me.  Questions:     So what is the statistically / mathematically correct way of using binary variables in k-means / hierarchical clustering? How to implement the solution in SAS / R? Why are Perl 5's function prototypes bad? If apostrophe + s is the acceptable way of denoting a genitive in English, is it possible that possessive pronouns, such as hers, ours and yours, started life as possessive adjectives with apostrophe + s?  E.g. her's, our's, your's, their's? Perhaps, even his' ? Its and my obviously stick out from this tempting pattern.  Is there any evidence that this might be the case, or is the 's completely unrelated? Is there an official Marvel Cinematic Universe viewing order? Make Launchpad Appear on screen with mouse copy and paste doesn't work in terminal What is the best way to remove duplicate rows from a fairly large SQL Server table (i.e. 300,000+ rows)?  The rows, of course, will not be perfect duplicates because of the existence of the RowID identity field.  MyTable  RowID int not null identity(1,1) primary key, Col1 varchar(20) not null, Col2 varchar(2048) not null, Col3 tinyint not null Non-root login on EC2 instance GRUB does not detect Windows I have finally installed Ubuntu on my second drive. When I start my computer GRUB only offers me to boot Ubuntu, not Windows 7. What needs to be done so that I can choose between Ubuntu and Windows in GRUB?  When I press F12 for boot menu at startup and I choose Windows Boot Manager it boots into Windows 7.    I ran command sudo fdisk -l and here is log ():  WARNING: GPT (GUID Partition Table) detected on '/dev/sda'! The util fdisk doesn't support GPT. Use GNU Parted.   Disk /dev/sda: 1000.2 GB, 1000204886016 bytes 255 heads, 63 sectors/track, 121601 cylinders, total 1953525168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 4096 bytes I/O size (minimum/optimal): 4096 bytes / 4096 bytes Disk identifier: 0xc3ffc3ff     Device Boot      Start         End      Blocks   Id  System /dev/sda1               1  1953525167   976762583+  ee  GPT Partition 1 does not start on physical sector boundary. This is a variation of the  questioned some time ago.  For a complex Hilbert space $H$ let $T: H \rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum  $$\sum_{i}{\langle |T|e_{i}, e_{i} \rangle} &lt; \infty$$  converges, where $|T| = (T T^{*})^{\frac{1}{2}}$ is the absolute value of the operator.  Assume that  $$\sum_{i}{\langle Te_{i}, e_{i} \rangle}$$  converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?  The progress on the problem is the following: Given an arbitrary bounded operator $T: H \rightarrow H$, one can use the following decomposition $$T = \big( \frac{T + T^{*}}{2} \big) ^{*} + i \big( \frac{T^{*} - T}{2i} \big) ^{*}$$  The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.  For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to  $$(UT U^{-1})(f(x)) = g(x) f(x)$$  where  $$U: H \rightarrow L^{2}(X, \mu)$$ $$g \in L^{\infty}(X, \mu)$$  Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement? Chocolate and oil mixed When I melt 16 oz of chocolate I add 4 oz of oil how long can I keep this and how should it be stored. Someone said something about botulism ?? Let $U$ be connected. A function $u \in H^1(U)$ is a weak solution   of Neumann's problem   \begin{equation} (*)\qquad\left\{ \begin{array}{rl} -\Delta = f &amp; \text{in } U \\ \frac{\partial u}{\partial \nu} = 0 &amp; \text{on } \partial U \end{array} \right. \end{equation}   if   $$ \int_U Du \cdot Dv \; dx = \int_Ufv \; dx $$   for all $v \in H^1(U)$. Let $f\in L^2(U)$. Prove $(*)$ has a weak solution   if and only if   $$ \int_U f \; dx =0. $$   For the only if part I set $v=1$. However I do not see where to start the if part. I was thinking of the Lax-Milgram theorem. That's where I'm now. The keyboard I am trying to create has a lot of Unicode and looks like this:   Being unsuccessful with the following tutorial:    (There are a few errors in the tutorial, and it uses imprecise language such as "Cutting" things instead of "Deleting" them, and it is formatted strangely)   I was wondering if anybody would be so kind to provide a better tutorial with step by step instructions? In a LaTeX article, the math characters are more "curvy" than the text characters:  \documentclass[12pt]{article} \begin{document}  $x^2 + y^2 = z^2$ \end{document}     In a beamer presentation, the math characters do not have the same curves:  \documentclass{beamer} \begin{document} \begin{frame}  $x^2 + y^2 = z^2$ \end{frame} \end{document}     Question  How can I get the beamer math to look just like the article math?  I would prefer if the solution was a single line (that I could place at the top of my LaTeX file) as opposed to something that needs to appear next to every piece of math.