Minecraft commands CanPlaceOn - Granite
How do I give myself an item that can only be placed on a block with a certain damage value?
What are valid values for "data defined" labeling settings? I'm using QGIS 1.8. I'd like to customize label features on an existing shapefile using the new label engine and it's "data defined settings" option. I'd like to know what the attribute table field types should be for each of the settings: All 7 font options, all 2 buffer options, and all 6 position options. Should they all be numeric fields, or some text, or both? I want to create these fields in the attribute table correctly so that I can map them to the label field settings. I can't find information on the proper field types anywhere. Thanks for your thoughts. Optimal step size in gradient descent Why I'm seeing a lock besides the partition I'm trying to modify with gparted? I wanted to create more space for Ubuntu on my hard disk in favor of my Windows partition. I booted the livecd and resized the NTFS partition to 100 GB. Then I wanted to resize my Ubuntu (ext4) partition to fill up the created unallocated space.  A screenshot of my current disk. (With the livecd there's no 'key' icon after sda6)    My first thought was just right click on sda6 → move/resize → done. Unfortunately I cannot resize or move the partition. However I can resize the NTFS partition.  I guess it is because the extended sda4 partition is locked. I couldn't see an unlock possibility though…  So how do I resize the ext4 partition anyway, probably by unlocking the extended partition, but how? I was doing a package removal with apt-get remove but then realized I should have done a --purge along with it to remove the configuration files.  Can I remove the packages configuration files easily or do I need to reinstall the package and then remove with a --purge? sum of the residue at all poles of the rational function p/q is 0. We are given two polynomials p, q with deg(q)>1+deg(p). Show that the sum of the residue at all poles of the rational function p/q is 0.   How do I go about solving this? Making mathematical animations with Mathematica Let $A$ be a square matrix of order $n$ such that $A^2 = I$.  Prove that if $1$ is the only eigenvalue of $A$, then $A = I$. Prove that $A$ is diagonalizable.   For  (1), I know that there are two eigenvalues which are $1$ and $-1$, how do I go about proving what the question asks me? Word whose form is contrary to its meaning? Is there a word for a word whose form is contrary to its meaning?  For example, "quotidian" is anything but quotidian. I'm currently writing an import script/addon.  The file format is one vertex per line: x, y, z, u, v With 3 lines for each triangle (no quads).  So far I've got vertex's, edges and faces all importing correctly.  Now I just need to import the UV coordinates for each vertex.  However I'm having trouble trying to find a way to do this.  The best result so far was copying code from the UV operator template, which gave me a UV Map, but it puts the whole model at one point. using System; using System.Security.Permissions; using Microsoft.SharePoint; using Microsoft.SharePoint.Security; using Microsoft.SharePoint.Utilities; using Microsoft.SharePoint.Workflow;  namespace Change_Title_Event.Change_Title {     /// &lt;summary&gt;     /// List Item Events     /// &lt;/summary&gt;     public class Change_Title : SPItemEventReceiver     {        /// &lt;summary&gt;        /// An item is being added.        /// &lt;/summary&gt;        public override void ItemAdding(SPItemEventProperties properties)        {            properties.AfterProperties["Title"] = "Kevin: Changed by event receiver at" + DateTime.Now.TimeOfDay.ToString();            base.ItemAdding(properties);        }       } }   Where simple example, which works perfect on my local Sharepoint 2010. It is sandboxed.  But when I package this solution, it do not work on Sharepoint Online. I does not give any Errors either.  I think I am doing something wrong, something simple, but I can't simply find anything on Google about it.  Here is another who asks on the MSDN forum, but gets a very wierd reply, that I do not understand. Function $f(z)$ such that $|f(z)| > |z|$ for every $z\in\mathbb{C}, 1/f(z)$ I know how to use the subfig package to fit many subfigures (each with its own caption) into a larger figure (with a general caption).  I know how to use the sidecap package, the put a figure's caption on the side of the figure.  What I do not know is how to display a caption inside the figure, as if it were a subfigure.  More precisely, I have three subfigures that I would like to fit on a 2x2 layout, with the remaining quadrant used for the caption.    Is that easy to achieve? Optimization when Cost Function Slow to Evaluate Speed $v$ of a wave on a tightrope Are there any official specifications regarding the torque for camera mounts? I've been wondering why I've never seen any official specifications of the maximum allowed torque that can be exerted upon lens mounts. It seems to me like a very useful specification, but I can't seem to find it from any manufacturer.  indicates that there might be no specifications for Nikon mounts. Are there really no manufacturer that publish these specification of their mounts? better Curve/Text to Mesh generation? Proving that $\lim_{x \to 0} \frac{f(x)}{|x|}=1$ implies $f$ is not differentiable at $0$. $$\lim_{x \to 0} \frac{f(x)}{|x|}=1$$ Then prove that $f$ is not differentiable at $0$. I was given this assignment and I'm not sure about my proof.. it feels like I'm assuming too much and not explaining with enough detail. Is there anything I need to add to make it sufficient? Assume toward contradiction that $f$ is differentiable at $0$. Then by definition we know that $f$ is continuous at $0$. Therefore $\lim _{x\to 0}f\left(x\right)=f\left(0\right)$ We know that $\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}=1$ Therefore we can conclude $\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \left|x\right|=1\cdot 0=0$ But we can also rewrite this as $\lim _{x\to 0}\left(f\left(x\right)\right)\cdot \frac{\left|x\right|}{\left|x\right|}=f\left(0\right)\cdot 1=0$ Therefore $f(0) = 0$. Since $f$ is differentiable at $0$ we have $\lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0}=\lim _{x\to 0}\frac{\left(f\left(x\right)-0\right)}{x}=\lim \:_{x\to \:0}\frac{\left(f\left(x\right)\right)}{x}\cdot \frac{\left|x\right|}{\left|x\right|}=\lim \:\:_{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\frac{\left|x\right|}{x}$ Since the one sided limits of $\frac{f\left(x\right)}{\left|x\right|}$ exist, and are both equal to 1, we have $\lim \:_{x\to 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^+}\frac{\left(f\left(x\right)\right)}{x}\cdot \:\:\:\frac{x}{x}=1\cdot 1=1$ $\lim \:_{x\to 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^-}\frac{\left(f\left(x\right)\right)}{-x}\cdot \:\:\:\frac{-x}{x}=1\cdot -1=-1$ Since the limit doesn't exist, we have a contradiction to the fact $f$ is differentiable at $0$. Therefore $f$ is not differentiable at 0. Multiple forms in delphi How can I find the points at which two circles intersect? Given the radius and $x,y$ coordinates of the center point of two circles how can I calculate their points of intersection if they have any? i keep getting the script error problem every time i try using polygon to raster tool in ArcGIS 10.1. Please help me how I can solve this?  screenshot embedded