I'm new with using Linux, so I don't know, how to install downloaded programms (downloaded with Firefox). Could anybody please give me a discription. Best would be in german, but english is ok too.
How can I install applications like Google Chrome on Ubuntu?  Are there any commands to install an application?    Index of answers:
My Android device won't connect. I've never got MTP to work, I've only ever got it to connect by USB mass storage, which was how it was working most recently. But now, since 13.04, it seems that option has been taken away entirely.  I need to be able to connect by USB to transfer files.  When I plug in the device I get an error that says Unable to open MTP device '[usb:002,003]':    After some web searching, I found that I might need a program called go-mtpfs, so I tried that, and it failed:  $ go-mtpfs /media/MyAndroid 2013/07/16 16:41:43 compiled against libmtp 1.1.5 Device 0 (VID=04e8 and PID=685e) is a Samsung Galaxy Y. 2013/07/16 16:41:43 found device Samsung: Galaxy Y (04e8:685e) @ bus 2, dev 3 :  LIBMTP PANIC: Unable to find interface &amp; endpoints of device 2013/07/16 16:41:43 rdev.open failed: open: open returned nil   How do I get my USB device to just connect, automatically, like it used to? Scientific visualization - How to replicate instances around a sphere? I want to replicate a protein structure around an icosahedron. Basically a vesicle of protein structure is what we want:  When I am instancing the protein structure using each vertex of the icosahedron it looks like:  However, the tails are supposed to be directed towards the center. How can I make that work? Matrix algebra: If $A^2=0$, Proof rank(A) $\le \frac{n}{2}$ Why is Lightroom not rendering what I see in every other JPG viewer? I want to color sketches in Photoshop, but when I prepared the sketch (using Levels, Hue, and other modifications) to turn it into workable line art, some details were missing. I think it depends on how you draw your sketch.  What are important steps for drawing a sketch (with pen or pencil) that is to be colored in Photoshop? A step-by-step instruction would be appreciated.  For instance, here it is the sketch    This is after my preparations in Photoshop Use of the comment package IMAP: how to move a message from one folder to another (using the IMAP commands, not with the assistance of any other mail package) The change in energy of an object can be determined by the work equation, where work is the change in energy:  $$ W = F \cdot d $$  I conceptualize the transfer of energy as simply a series of small "packets" of energy being transferred at every Planck length. These small "packets" of energy add up to the total energy transferred (i.e. work). I'm not sure if this conceptualization is correct, so correct me if I am wrong.  However, it makes me wonder why the amount of energy transferred is dependent on distance and not time.  $$ m_1 = 10~kg \\ m_2 = 20~kg \\ W_1 = (10~N)\cdot(5~m) = 50~J \\ W_2 = (10~N)\cdot(5~m) = 50~J \\ W_1 = W_2 \\ t_1 \neq t_2 $$  If I apply a constant force on an object, why isn't the energy transferred at a constant rate with respect to time? The energy transfer rate varies dependent on how long it takes to cover the set distance.  In other words: why is the energy transferred consistent per unit of distance, and not per unit of time? I have    iPhone 5 (with iOS 8.3) MacBook Pro Early 2013 (with OS X Yosemite 10.10.3) iPad Air (with iOS 8.3)   My iPhone is paired with the MacBook, so e.g when my phone rings, I can see it in my computer. But when I want to AirDrop photos from my iPhone the MacBook, then I can't see the MacBook in my iPhone AirDrop targets.  I can see the iPad, I can see my wife's iPhone, but not the MacBook.  On MacBook, I can see the iPad, my wife's phone and even my iPhone and I can drop to there, but not the other way around.  Before I started troubleshooting, the iPhone and MacBook were mutually invisible (I only saw iPad from both). After a while of trying things and sending files back and forth, my MacBook started to see the iPhone, but I did not manage to make it work from the other way. Still, the iPhone seems to be much "less" visible than the iPad - which I see almost instantly and always, the iPhone only seems to appear after a while.  Just to make it clear:   both iPhone and MBP have BT turned on both have AirDrop visibility set to Everyone   I would be glad for any suggestions to fix this - what settings to check, logs, whatever. iOS: Compare two dates "If $1/a + 1/b = 1 /c$ where $a, b, c$ are positive integers with no common factor, $(a + b)$ is the square of an integer" If $1/a  +  1/b  =  1 /c$   where $a, b, c$ are positive integers with no common factor, $(a + b)$ is the square of an integer.  I found this question in RMO 1992 paper !  Can anyone help me to prove this ?  May be it's too easy and simple . But , I'm just novice in number theory ! Is the "the ... the ..." construction governed by some grammatical rule? a) The more I study, the less I learn. b) More I study, less I learn.  Could we rewrite the sentence shown under the letter a) in the form shown under the letter b) without breaking any grammatical rule? If not, why not? Let $\Omega$ be an open subset in $\mathbb{C}$. Let $\{f_n\}$ be a sequence of holomorphic functions on $\Omega$ such that $f_n\to f$ pointwise and converges uniformly on any compact subset $K\subseteq \Omega$. Then by Cauchy Theorem and Morera Theorem, $f$ is holomorphic. Let $f_n$ and $f'$ be the derivatives of $f_n$ and $f$ respectively. Prove that $f_n'\to f'$ uniformly on any compact subset $K\subseteq \Omega$.   How to prove? Two figures side by side I am a Physics undergrad, and just started studying Topology. How do you define neighborhood and open set in Topology.Wikipedia gives a circular definition.   An  is defined as follows.      In topology, a set  is called an open set if it is a neighborhood of every point    While a  is defined as follows:     If $X$ is a topological space and $p$ is a point in $X$, a neighbourhood of $p$ is a subset $V$ of $X$, which includes an open set $U$ containing $p$   which itself contains the term open set.  How do you define it exactly? How can adding a 2nd IV make the 1st IV significant? I have what is probably a simple question, but it is baffling me right now, so I am hoping you can help me out.  I have a least squares regression model, with one independent variable and one dependent variable. The relationship is not significant. Now I add a second independent variable. Now the relationship between the first independent variable and the dependent variable becomes significant.  How does this work? This is probably demonstrating some issue with my understanding, but to me, but I do not see how adding this second independent variable can make the first significant. Map of with object references as keys? The close dialog says there are two off-topic votes, but none of the off-topic reasons have votes next to them I've found strange behavior when closing this  - I definitely see that there are 3 closing votes and 2 of them are at "Off-topic" section:   -but there are no explanations on section itself:   -while, normally, it's expected that it will appear. In custom comments section there are also no closing-votes displayed.   Perhaps it's just my browser (which is FF 26 x64 under Win) - so marking this as a "possible bug". Thank you for your attention. Consider this equation :   $$\sqrt{\left( \frac{dy\cdot u\,dt}{L}\right)^2+(dy)^2}=v\,dt,$$  where $t$ varies from $0$ to $T$ , and $y$ varies from $0$ to $L$.  Now how to proceed ?   This equation arises out of following problem :   A cat sitting in a field suddenly sees a standing dog. To save its life, the cat runs away in a straight line with speed $u$. Without any delay, the dog starts with running with constant speed $v&gt;u$ to catch the cat. Initially, $v$ is perpendicular to $u$ and $L$ is the initial separation between the two. If the dog always changes its direction so that it is always heading directly at the cat, find the time the dog takes to catch the cat in terms of $v, u$ and $L$.  See my solution below :   Let initially dog be at $D$ and cat at $C$ and after time $dt$ they are at $D'$ and $C'$ respectively. Dog velocity is always pointing towards cat.  Let $DA = dy, \;AD' = dx$  Let $CC'=udt,\;DD' = vdt$ as interval is very small so $DD'$ can be taken straight line.  Also we have $\frac{DA}{DC}= \frac{AD'}{ CC'}$ using triangle property.  $\frac{dy}{L}= \frac{dx}{udt}\\ dx = \frac{dy.udt}{L}$  $\sqrt{(dx)^2 + (dy)^2} = DD' = vdt \\ \sqrt{(\frac{dy.udt}{L})^2 + (dy)^2} = vdt $  Here $t$ varies from $0-T$, and $y$ varies from $0-L$. Now how to proceed? I am a bit lost how one has deduced the formula for electric field with electric dipole because of some inconsistency between different sources. The  contains a delta function in the formula, but  gives only the formula without the delta function,  $$\bar{E} = -\nabla \phi = \frac{-p}{4\pi \epsilon_{0} r^{3}} ( \hat{e}_{z}-3\hat{u}_{r}(\hat{u}_{r} \cdot \hat{u}_{z}) ) $$  Is it intentionally left out?  My notes mention that the dipole is with a circular thing $P=Q d$ where $d$ is distance and $Q$ is charge.