Is UNIX password the root password . i changed UNIX password ? I did not find the right answer before posting this
Change password on root user and user account
How do different difficulties work on story missions? When selecting a story mission, I can choose between "Easy" and "Hard".  Next to the difficulty, the game displays some kind of level.  However, when you start the mission, you still end up in the common map area with other players.  What exactly changes when you select a higher difficulty?  Is it possible to see other players playing on "Easy" when you select "Hard" before entering the mission zone where respawning is restricted? A formula for Perspective measurement If I have a photograph of a rectangular object whose image recedes with perspective, is it possible to work out the scale of measurements along that object?  For example: I have a photograph of a (metric) ruler.  The ruler is graduated in centimetres as shown, 30 cms.  Is it possible to work out the distance between each centimetre?  The screenshot shows the photograph.  The representation on the Picture Plane shows the nearest edge of the ruler to be 50mm and the farthest edge to be 15mm. By measuring the photo, I see that the distance between 0 and 1 is 7.5mm and the distance between 27 and 28 is 0.9mm.  Is there a way to get these results from a formula?  It doesn't need to be 100% accurate: it's for photometric software I'm writing, so it's only going to be used as a rough guide. The equation $X^{n} + Y^{n} = Z^{n}$ , where $ n \geq 3$ is a natural number, has no solutions at all where $X,Y,Z$ are intergers. How can I schedule timer to power on and off device? Views indicator inaccurate, shows one views on question with two close votes Two close votes and only one person has viewed the question?    How did this happen? Show  $$\int_0^\infty \frac{\ln(x)}{e^x} = -\gamma$$ (gamma is Euler-Mascheroni constant).  Can anyone please prove this result?  Also  $$ \int_0^\infty \frac{\left( \ln(x) \right)^2}{e^x}\mathrm dx. $$ Prove that: $(f(x))^{2} + (f'(x))^{2} \leqslant \max(a,b)$ where $(a,b) \in \mathbb{R}^2$ How to fully backup non-rooted devices? I know I'm asking for the close-to-impossible -- as a real full-backup certainly requires root permissions. But still there are many folks around not wanting to root their devices -- afraid of voiding their warranties or bricking their devices, or just not having their devices supported by any rooting method.  I also know there were a lot of questions asked already on this topic, and I checked all of them having the backup tag -- none covers the entire topic, all just have pieces, and especially people new to Android (but other non-power-users as well) would have a hard time figuring out the best way.  So I want a detailed canonical answer to the question:  Not involving root-solutions, how to achieve a backup as close as possible to a full backup?  Please do not give simple one-liner answers like "Use app xyz [full-stop]." If mentioning an app (and I doubt a single app would be the answer here), include what parts are covered by it -- as well as ideas on how to fill the gaps it leaves. (And to make it absolutely clear: I know the first thing coming to mind is Titanium Backup -- I use it myself, but this does not apply here: This question is strictly about methods for non-rooted devices!) Moreover, the answer should be as generic as possible in terms of applying to a wide range of devices (i.e. it should not be restricted to one device or manufacturer).  I already mentioned having investigated all questions here tagged backup. These will certainly be helpful providing details for answers here, so I will list up the topmost ones concerning my question:            All these provide some details for my question (so feel free to get inspired by them) -- but I feel there are still details missing. Also it would be helpful to have a summary -- say, an answer including the full description of "the maximum possible".  A last criterium: While I'd consider cloud-based solutions legit, I'd favour local solutions. If you can provide both: all the better. But some people value privacy quite high (I'm one of them), and thus would not trust their data to some cloud service.    Thanks to Ryan, we now have a solution for a really complete backup of all apps and their data -- which I didn't dare to hope for! The only disadvantage: His solution only works for less than 10% of all Android users; namely those whose devices are running at least with Android 4.x.  So again, I ask the close-to-impossible: Any solutions applicable at least on Gingerbread (2.3.x) or, better, even Froyo (2.2.x) -- so they cover the majority of current devices? These solutions may involve multiple tools (try to keep it as simple as possible though; think of your mother to follow your instructions ;)  And keep in mind: NO ROOT!    To further motivate high-quality answers (and hopefully avoid low-quality ones), I will be a harsh one here: As I rarely do, I will down-vote answers not fitting the mentioned criteria on one end -- but on the other end, I will setup a bounty for the best answer (which cannot be done immediately, but earliest after 2 days). Two parties are trying to resolve some dispute that involves interest of both. Sometimes it's hard for them to reach an agreement on a fair basis. This is typically when another party who has no interest involved and no bias towards either party comes into play in order to settle the dispute between the two conflicting parties. The term "third party" seems being used quite often. I would also call it a "middle-standing" party or we could say this party is neutral.  These terms seem NOT formal enough to me though. I'd like to see if anybody can come up with a better one. During my calculations I ended up with the following product: $$P=\prod_{n=1}^\infty e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}$$ I tried to express it in terms of series by taking the logarithm  $$S=\ln P=\sum_{n=1}^\infty \ln\left(e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}\right)$$ but I also got stuck. Numerical calculation suggests that it is equal to $$P\stackrel{?}=\frac{\sqrt{2\pi}}{e}$$ but I am not able to prove the conjecture. Any idea about how to evaluate the product? Any help would be appreciated. Thanks in advance. I know that nobody likes duplicate questions, but I haven't found my answer on any of the other questions with this topic.  I built a computer with a  (made by Gigabyte).  It supposedly has 512 CUDA cores.  I followed  tutorial, (which enabled the Nvidia proprietary drivers) and was able to enable dual monitors and all the desktop graphics that are needed.  Unfortunetly, the GPU processing option still doesn't show up in Blender.  I've tried installing nvidia-cuda-toolkit, and that didn't make a difference.  I also tried to update my drivers from nvidia-346 to nvidia-349, and this had no apparent effect either.  What can I do to enable GPU rendering?  (I've tried both blender installed and just downloaded from the site). Hello I am Indian Citizen currently studying in Australia and I am on a student visa. In the month of November 2019 I am travelling to India for vacations and I booked a ticket from Melbourne to Bangkok and than form Bangkok to India with a layover time of 7 hours in Bangkok and I am not going to leave the airport but my flights are different, so do I need to get and Airport Transit Visa in Bangkok. Finding local maximum and minimum YA dystopia novel of battle school with unusual military control scheme Given this (very) tricky determinant, how can we calculate it easily?  $$\begin{pmatrix} \alpha + \beta &amp; \alpha \beta &amp; 0 &amp; ... &amp; ... &amp; 0 \\ 1 &amp; \alpha + \beta &amp;  \alpha \beta &amp; 0 &amp; ... &amp; 0 \\ 0 &amp; 1 &amp; \alpha + \beta &amp; \alpha \beta &amp; ... &amp; ... \\ ... &amp; ... &amp; ... &amp; ... &amp; ... &amp; 0 \\ ... &amp; ... &amp; .... &amp; ... &amp; ... &amp; \alpha \beta \\ 0 &amp; 0 &amp; 0 &amp; ... &amp; 1 &amp; \alpha + \beta   \\ \end{pmatrix} \in M_{n\times n}$$  EDIT:     I have to prove it is equal to $\frac{{\alpha}^{n+1} - {\beta}^{n+1}}{\alpha - \beta}$   Any help is appreciated, I just could not find a trick to ease it up! I'm on a new computer and I haven't downloaded Data Loader in years but now, I'm having all kinds if issues.  So apparently, I have to download Zulu JDK 11 (msi) first, Then download and unzip Data Loaded and install. Which opens a command prompt... After I follow all of the command prompt steps, then I get an error.  Am I supposed to install Java too? And if so, do I download Java or Zulu first?  Apparently a lot of people are having this same problem with no resolve.  Any help would be much appreciated. Definitionally, we have that  $$ \int_\gamma f\ \overline{dz} = \overline{\int_\gamma \overline{f}\ dz} $$  Now let $\int_\gamma f\ dz = w = x +yi$.  Question 1: Is it not the case that $\int_\gamma \overline{f}\ dz = \bar{w} = x - iy$?  Question 2:  Assuming that $\int_\gamma \overline{f}\ dz = \bar{w} = x - iy$, then is it not the case that  $$ \overline{\int_\gamma \overline{f}\ dz} = \overline{x - iy} = x+iy = \int_\gamma f\ dz? $$  Basically, I'm failing to see how $\overline{\int_\gamma \overline{f}\ dz}$ isn't just $\int_\gamma f\ dz$. How can you prove that a function has no closed form integral? I'd like to output a list of all installed packages into a text file so that I can review it and bulk-install on another system.  How would I do this? Cannot find hard drives after installing Ubuntu 13.10