I'm making a shallow hemisphere for a 3d printing project. I have to add some "depth" to the hemisphere for the printer to be happy so I made a copy of the current object and shrunk it down a bit. In order to close the shape up I need to add faces to hold these two objects together.  I selected the edges of both hemispheres but when I type "F" it fills each hemisphere independently to itself. I've bool > union these two hemispheres together thinking that would help but it hasn't.  How can I get these two circular shapes to close together making one shape rather than having them close independently making two overlapping shapes?  This is my first 3d project from blender so if there is a better way to be doing this please let me know. Cheers.
Hey I want to connect the bottom loop with the top loop so that it forms a solid.  I have tried selecting multiple vertices and pressing F but that just creates a large face which I do not want. I have tried using grid fill bu that doesn't work either. The solution that seems to work is by selecting four vertices and pressing F and then selecting the next 4 and pressing F and on and on. That just takes too long.
How to convert a string of bytes into an int? How can I convert a string of bytes into an int in python?   Say like this: 'y\xcc\xa6\xbb'  I came up with a clever/stupid way of doing it:  sum(ord(c) &lt;&lt; (i * 8) for i, c in enumerate('y\xcc\xa6\xbb'[::-1]))   I know there has to be something builtin or in the standard library that does this more simply...  This is different from  for which you can use int(xxx, 16), but instead I want to convert a string of actual byte values.  UPDATE:  I kind of like James' answer a little better because it doesn't require importing another module, but Greg's method is faster:  &gt;&gt;&gt; from timeit import Timer &gt;&gt;&gt; Timer('struct.unpack("&lt;L", "y\xcc\xa6\xbb")[0]', 'import struct').timeit() 0.36242198944091797 &gt;&gt;&gt; Timer("int('y\xcc\xa6\xbb'.encode('hex'), 16)").timeit() 1.1432669162750244   My hacky method:  &gt;&gt;&gt; Timer("sum(ord(c) &lt;&lt; (i * 8) for i, c in enumerate('y\xcc\xa6\xbb'[::-1]))").timeit() 2.8819329738616943   FURTHER UPDATE:  Someone asked in comments what's the problem with importing another module. Well, importing a module isn't necessarily cheap, take a look:  &gt;&gt;&gt; Timer("""import struct\nstruct.unpack("&gt;L", "y\xcc\xa6\xbb")[0]""").timeit() 0.98822188377380371   Including the cost of importing the module negates almost all of the advantage that this method has. I believe that this will only include the expense of importing it once for the entire benchmark run; look what happens when I force it to reload every time:  &gt;&gt;&gt; Timer("""reload(struct)\nstruct.unpack("&gt;L", "y\xcc\xa6\xbb")[0]""", 'import struct').timeit() 68.474128007888794   Needless to say, if you're doing a lot of executions of this method per one import than this becomes proportionally less of an issue. It's also probably i/o cost rather than cpu so it may depend on the capacity and load characteristics of the particular machine. How to return multiple objects from a Java method? I have no root access on this machine.  I would like to know if there is a way I can download Ubuntu packages and install them as non-root?  Probably in my ~/bin or ~/usr/share or something like that? Would that work? Is $\langle1-i\rangle$ a maximal ideal in $\mathbb{Z}[i]$? I'm trying to show that $\langle1-i\rangle$ is a maximal ideal of $\mathbb{Z}[i]$.  I started by assuming there is some ideal $A$ that properly contains $\langle1-i\rangle$, and then I want to show that $1 \in A$, and then I will have that $A= \mathbb{Z}[i] $, so $\langle1-i\rangle$ is maximal.  Is this the right approach?  Thanks  Second Attempt: Because $\forall a+bi \in \mathbb{Z}[i]$, we have $a+bi = (a+b) -b(1-i)$, then we have $a+bi - (a+b) = -b(1-i) \in \langle1-i\rangle$, which implies $a+bi +\langle1-i\rangle = a+b +\langle1-i\rangle$.  So every coset is of the form $c +\langle1-i\rangle$, where $c\in \mathbb{R}$.  Furthermore, $-i(1-i)^2=2 \in \langle1-i\rangle$. Write $c = 2q + r$, where $q,r\in \mathbb{R}$ and $r=0,1$.  Then because $2 \in \langle1-i\rangle$, we have $2q\in \langle1-i\rangle$ and therefore $2q= 2q+r-r = c-r\in \langle1-i\rangle$, so $c+\langle1-i\rangle = r+\langle1-i\rangle$, and every coset is of the form $0 + \langle1-i\rangle$ or $1 +\langle1-i\rangle$. Thus, $1 +\langle1-i\rangle$ is the only nonzero element, and is invertible with itself as its inverse.  Thus, $\mathbb{Z}[i]/\langle1-i\rangle$ is a field with two elements, and by a theorem, $\langle1-i\rangle$ must be maximal.  Is this correct? Is there an idiom that means "revealing a secret unintentionally"? I am thinking there could be an idiom that states that you or someone revealed a secret unintentionally? Can you think of an idiom like that? I am looking for an idiom I can use to write a blog post about the recent incident where Secretary Pompeo unintentionally revealed the dirty secrets behind the CIA. The idea behind the sum of powers of 2 I know that the sum of power of $2$ is $2^{n+1}-1$, and I know the mathematical induction proof. But does anyone know how $2^{n+1}-1$ comes up in the first place.  For example, sum of n numbers is $\frac{n(n+1)}{2}$. The idea is that we replicate the set and put it in a rectangle, hence we can do the trick. What is the logic behind the sum of power of $2$ formula? Image uploads from the editor strip preceding line breaks Recently, using the image uploader in the post editor has starting stripping preceding line breaks; placing the image directly after any previous text instead of at the current cursor position. This means that the inserted image is placed inline as part of the previous line of text rather than in a new paragraph as it did previously...  Before inserting the image (note the cursor position):        After inserting the image:        This also happens after the Markdown for a horizontal rule, which destroys the rule and gives us a bunch of hyphens followed by an image instead:        Seems the behaviour is a bit inconsistent... Sometimes I get the behaviour shown above and sometimes it leaves a single newline, I'm not sure when or why but single line breaks are ignored in Markdown so the rendered result is the same regardless.  Obviously it's useful to be able to have inline images, but I'm pretty certain most images aren't intended to (and shouldn't) be. how do I get my code to use the value of the $HOME variable? I'm trying to put together a generic script that will check the existence of several key directories for various users on different servers. Additionally, I want to leverage each user's $HOME variable.  For eg, let's say this were true:   on server 1: jdoe's home is /home/jdoe  on server 2: jdoe's home is /opt/jdoe2 server 3 hasn't been built yet; we won't know where they build his $HOME until the server is built. on server 4: mysql's home is /opt/home/mysql   This is what I have for my important directories (ordered from most to least impt):  $ cat mylist.txt $HOME/most_impt_dir1 $HOME/most_impt_dir2 $HOME/most_impt_dir3 $HOME/misc $HOME/junk   ...I want to find the most impt dir owned by this user.   Here's what I'm trying:  for i in `cat mylist.txt` do    if [[ -O $i ]] &amp;&amp; [[ -d $i ]]; then     echo "found it: $i"     break   else     echo "$i is not it."   fi  done   The above code does not work for anything in my list because it is literally checking for dirs beginning with $HOME. How do I get my code to use the value of the user's $HOME variable? Three linked question on non-negative definite matrices. 1.a symmetric matrix in $\mathbb{M}_n(\mathbb{R})$ is said to be non-negative definite if $x^Tax≥0$ for all (column) vectors $x\in \mathbb{R}^n$. Which of the following statements are true? (a) If a real symmetric $n\times n$ matrix is non-negative definite, then all of its eigenvalues are non-negative. (b) If a real symmetric $n\times n$ matrix has  all of its eigenvalues are non-negative , then it is non-negative definite. (c) If $ A\in \mathbb{M}_n(\mathbb{R})$, then $AA^T$ is non-negetive definite. 2. only one of the following matrices is non-negetive definite. Find it. (a) $\begin{pmatrix} 5 &amp; -3 \\ -3 &amp; 5 \end{pmatrix}$. (b) $\begin{pmatrix} 1 &amp; -3 \\ -3 &amp; 5 \end{pmatrix}$. (c) $\begin{pmatrix} 1 &amp; 3 \\ 3 &amp; 5 \end{pmatrix}$.      3.let $B$ be the real symmetric non-negative definite $2×2$ matrix such that $B^2=A$ where $A$ where is the non-negetive definite matrix in question $2$.write down the characteristic polynomial of $B$.      My thoughts..  For1.  (a) &amp; (b) are true but not sure about (c). For 2. (a) is the correct option since it has all positive eigen values.they are $2,8$. For 3. Eigen values of $B$ will be $\sqrt{2}$ and $\sqrt{8}$. So the answer is $x^2-3√2x+4=0$.      Can anybody help me to verify the solutions of the above problems. Most instructions for equipment and software make a habit of shortening sentences by dropping words of certain parts of speech - usually definite and indefinite articles.  For example:     PULL (THE) LEVER TO OPERATE      CARS MAY BE TOWED WHEN (THEY ARE) PARKED ILLEGALLY      PLEASE USE (THE) OTHER DOOR   Is there a name for this custom? And what is the best way to describe it? If $a^3 =a$ for all $a$ in a ring $R$, then $R$ is commutative. Let $R$ be a ring, where $a^{3} = a$ for all $a\in R$. Prove that $R$ must be a commutative ring. I want to use symbols from wasysym and marvosym packages:  \usepackage{wasysym} \usepackage{marvosym}   but it gives an errors as both packages have the command of \CheckedBox. I do not want to use this command, but since both packages have it, it makes confusion for latex. How can I ignore one command to have both packages? Cross-referencing between different files How can continue the equation of the rectangle red in next line?   \documentclass[11pt,a4paper,twocolumn]{article}  ...  \begin{equation*}     \left| \dot{r} \right| =     \sqrt{{- \, {e}^{-t} \left( \cos t + \sin t \right)}^{2} \, + \, {{e}^{-t} \left( \cos t + \sin t \right)}^{2} \, + \, {(- \, {e}^{-t})}^{2}} \end{equation*} The equivalent electric field of a magnetic field $f^{-1}([-\infty,r))$ measurable for all rational numbers $r$ If $X$ is measurable space, $f : X\to [-\infty,+\infty]$, function such that $f^{-1}([-\infty,r))$ measurable for all rational numbers $r$,prove than $f$ is measurable  I know that $f$ is measurable iff set $\{x\in X : f(x)&lt;a\}$ is measurable for all $a\in \mathbb{R}$ but for $a\in\mathbb{Q}$ I don't know how to proceed. Do I cast the result of malloc? How to Backup and Restore Android OS without Rooting Is there really any difference between the jackknife and leave one out cross validation? The procedure seems identical am I missing something? "Forgot" vs "Forget" Is the following correct, or is there more to it?  "I forgot his name" — I knew his name, but I forgot it.  "I forget his name" — I keep forgetting his name. Where using "forget" basically means that you tried to remember this information before as well, but you couldn't then either.