I Wanted To Use Ubuntu From USB So I Burned The Iso File By Using "Rufus.exe",,  It Worked Fantastic! But There Is a Problem Which Is, If I Installed Google Chrome Browser For Example, I Can't Find It The Next Time I Boot From The Usb!   So How To Keep Everything I Install For The Next Time I Boot From That USB? Is That Possible Or I Have To Install Ubuntu Side By Side With Windows To Do It ??
I want to make a persistent live Ubuntu USB drive with more than 4GB of storage.   I know there is an option of installing directly on the drive but I don't want to do this as it will restrict the drive from working on different computers.   One option is to make an additional partition with extra space but if I do this how can I get newly installed software to be stored on this partition?
Automatically convert quotations in the form of "abc" to become ``abc" I did not know that latex had different symbol for opening and closing quotation marks. I've used it like "abc" every where and just realised that I really should have used ``abc". Is there anyway I can convert all the quotations to become like ``abc"? Doing it manually will take forever as I have a lot of quotations. Why are there humans in the Star Wars Universe? Can't boot without Flash Drive plugged in I had Ubuntu 12.04 Beta installed on my computer. When 12.04 was finally released, I made a bootable USB Flash Drive using Startup Disk Creator.  Then I decided to check if this drive works properly and to reinstall the system on my desktop. I must add, my desktop behaves a bit strange when it comes to bootable USB's, it recognizes them as HDD.  In BIOS I changed priority of boot so USB Flash Drive (recognized as HDD) was first. Successfully booted, I installed Ubuntu. Everything worked fine, but...  Now I cannot boot from my real HDD. Every time I want to boot, I put the USB Flash Drive into my computer, boot, safely remove it and everything works.  What do I need to do to repair boot? Which graphics formats can be included in documents processed by latex or pdflatex? I'm looking for a high current DC switch (able to switch something like 5V@30A) and I've encountered several DPDT that might do what I need, however they are only rated for AC applications (3A@250VAC or 6A@125VAC). Are these rating convertible to DC? or is that something that only the facturarer would know? Is it possible to query the content of the Time-based Workflow Queue from an Apex unit test to assess that a specific Workflow Rule has been executed and added a new item to the queue? I didn't see anything alike in the documentation, but wish it could be possible to improve testing and stability overtime.  Thanks Which is better: ... or ... Many of the people have asked almost the same question.  I'm trying to render a big mesh + fur/grass file with a few characters on the screen.  when I hit render, after a number of processing, copying data to gpu etc... i get following error.  Mem: 2960.07M, Peak:2960.07M | CUDA error: Out of memory in cuMemAlloc(&amp;device_pointer, size).  I'm using Geforce GTX 690 which contains 4GB memory.   Can somebody tell me how can i get GPU rendering, while 2960M is far less then 4196M of memory.   I know somebody would suggest me to use CPU.. but i'm curios why not GPU?   I've already compressed texture, etc   Edited  I don't think any other application is using any of my 1GB remaining memory. everything is closed but blender is there, also the GPU utilities are showing 2000/2000 MB are free on one card and 1900/2000MB is free on 2nd card.    i'm using GTX 690. with two GPUs of 2GB each. If somebody thinks that one GPU should have memory matching to the cuda memory/or Peak memory then I'm doubtful. Because in preferences Blender does know that I have 2GPUs where I can select either first GPU or second GPU or even both.   When I render some of smaller objects/scenes both of my GPU renders simultaneously with two blocks.  I'm running following specs  Xeon E5-1660 0 24GB - 1600mhz GTX690 divides, not divides and cardinalities In group theory it happens very often that you want to say: "The order of $H$ divides / does not divide the order of $G$." But the length of | is then always the same:  $ |H| | |G| $ produces |H|||G|  which is not ideal at all. (The negation \nmid has just this length too.) The length of "divides" should be longer than the "cardinality length". I could use   $ |H| \big| |G|$  but then there is no good way to say "does not divide" with the same length. Also, \big| is no solution when the expression has to be used in subscripts.   I thought it was just me, but I asked other group theorists, and nobody had a solution. I hope there is one.  I am seeking a division sign which is a little longer than the usual | , allows negation (with the same length) and proportionally correct use in subscripts (also under \sum etc.). It does not look professional if the "order sign" has the same length as the "divides" sign. (And what about combinations with absolute values, norms - although personally I don't use them as often as group orders.) Suppose I have some theorem in the paper  Section 1 Theorem 1.1. Let ...  And then later in the paper I want to recall the theorem by reprinting it  Section 4 We recall Theorem 1.1: Theorem 1.1. Let ...  What's the proper way to do this? Will something accelerate forever if a constant force is applied to it on a frictionless surface? If so how? I don't understand the fact that let's say hypothetically I was pushing a block on a frictionless surface, and I kept applying the same force on the object how it would keep accelerating! Wouldn't there have to be a certain velocity where it would stop getting faster? If not then what is actually acting on the object to cause it to accelerate infinitely? I read this sentence on an educational website:     Now times have changed and you are ready for situations involving forces in two dimensions.   Shouldn't there be a comma before and, since the subject of the sentence has changed (things vs. you)?      Now times have changed, and you are ready for situations involving forces in two dimensions.   What is the exact comma rule for joining two clauses with and? Subgroup of $\mathbb{Q}$ with finite index Consider the group $\mathbb{Q}$ under addition of rational numbers. If $H$ is a subgroup of $\mathbb{Q}$ with finite index, then $H = \mathbb{Q}$.  I just saw this on our exam earlier and was stumped on how to show this. Any ideas? Algorithm to match one input file with given numbers of file Failed to load map. Error contacting Google servers. This is probably an authentication issue Prove if $\operatorname{ord}(a)=m$, $\operatorname{ord}(b)=n$, and $\operatorname{gcd}(m,n)=1$, then $\operatorname{ord}(ab)=mn$.   I was reading this and was thinking how this proof would look like. I tried to do it and am not sure if this is correct. Here is what I did:  If $a$ and $b$ commute then $(ab)^{mn} = a^{mn} * b^{mn} = (a^m)^n * (b^n)^m = 1 * 1 = 1.$ So $ord(ab) | mn$.  Now, take $k = \operatorname{ord}(ab) = m^\prime * n^\prime$ where m' is relatively prime with $n$ and $n'$ is relatively prime with $m$. By the result above $m' | m$ and $n' | n$. Now we have $((ab)^k)^{m/m'} = 1$ since $(ab)^k=1$. But on the other hand, by the commutativity:  $$((ab)^k)^(m/m') =$$  $$((ab)^(m' * n'))^(m/m') =$$  $$ a^{n'm} * b^{n'm} = $$  $$(a^m)^{n'} * b^{n'm'}=$$  $$b^{n'm'} = 1$$  This implies that $n' * m'$ is divisible by $n$. But $m'$ is relatively prime with $n$, so we must have $n' = n.$ By symmetry, $m' = m$. So $ord(ab) = mn$.  Just to say this again, I want to prove if $a$ and $b$ commute and $m$ and $n$ are relatively prime then   $$ord(ab)=mn.$$   The comments are vague. I guess this must be the answer then and there probably is not another way to do this. In swimming, is there a competitive advantage to being in one lane or another? Upon observing swimming competitions in the Olympics, it is not clear to me whether one of the eight lanes is more advantageous versus another. I've seen vague claims but nothing backed up with science.  In competitive swimming, is there an advantage to being in one lane or another? Furthermore, does the stroke (ie: breast, fly, freestyle, etc.) affect this effect? I'm having some trouble proving the following statement using mathematical induction:   $$\frac{1}{2}n^{\frac{3}{2}} \leq \sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4} + ... + \sqrt{n}      ,\text{ (for all sufficiently large n)}$$   I'm sort of confused because $n^{\frac{3}{2}}$ = $n  n^{\frac{1}{2}}$ which means that there are n of the largest possible root values on the left hand side while the right hand side has values like $\sqrt{1}$, $\sqrt{2}$, etc. As n grows sufficiently large wouldn't the left hand side outgrow the right hand side no matter what the constant multiple is? If I'm thinking about this correctly just let me know, if not, any help would be appreciated. Exporting feature class into multiple feature classes based on field values using ArcGIS Desktop? I have a feature class with over 2,000 features, and I need to make them all individual feature classes based on a field.   Is there a way to do this? Conjugate Variables, Noether's Theorem and QM