I deleted the Xubuntu partition, hoping it would just delete Xubuntu and everything would be fine. But now it still boots to grub, and will not boot to windows. I have no idea how boot to Windows, it's not under any of the boot options or devices.
I have absolutely no experience with Linux, and I desperately need to get my computer back up and running again with Windows.  How do I remove Ubuntu and reinstall Windows?    Editor's note: many of the answers are about removing Ubuntu from dual-boot but keeping Windows (which is a bit complicated), while other answers are about removing Ubuntu from single-boot (which is easy: basically just format the disk while installing Windows). The question as written is ambiguous between dual-boot or single-boot.
Changing the size of the mouse pointer in Ubuntu MATE 18.04 does not work Proving that $\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ How would we go about proving that $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots +\frac{1}{n(n+1)} = \frac{n}{n+1}$$ Native Youtube app suddenly stopped working on iPad 1 It's been some weeks now, whenever I try to used my Youtube app, it tells me it's impossible to connect to Youtube. Since it's an iPad 1, I have had trouble finding information online.  Did Google or Apple stop supporting the app? Is it finally obsolete? Is there another app I could use that works on iOS 5?  Just to be clear, I did not tinker with my device. I did not change its parameters, I did not jailbreak it, it connects to the Internet just fine and I've tried the app on different Internet connections. Also, this is not a login problem, the app just refuses to let me browse videos, whether I'm logged in or not.  Thanks for your help. The  (ESC), states that for every natural number $n \geq 2$, there exists a set of natural numbers $a, b, c$ , such that the following equation is satisfied:  $$\frac{4}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\tag{1}$$  The basic approach to solving this problem outlined by Mordell [Ref1] is described below  By defining $t$ and $m$ as positive integers greater than zero and $q$ a positive integer greater than one we can observe that  a) There is always a solution for even $n$, since if $n=2^qt$ we have the trivial solution $$\frac{4}{4t}=\frac{1}{t}$$  In the remaining case $n=2(2t+1)$, a solution in the form of two Egyptian fractions can always be found  e.g. $$\frac{4}{2(2t+1)}=\frac{2}{2t+1}=\frac{1}{t+1}+\frac{1}{(t+1)(2t+1)}$$  b) If $(1)$ is a solution for some particular prime $n$ then all composite numbers $mn$ divisible by $n$ are also solutions, thus  $$\frac{4}{mn}=\frac{1}{ma}+\frac{1}{mb}+\frac{1}{mc}$$  will also be a solution. This means that we can simplify the analysis to the cases where $n$ is a prime greater than 2.  Using Mordell's approach we have just shown that we only need to consider the cases where $n$ is prime and where $n \equiv 1 \pmod{2} \;\;[meaning \;\;n=2t+1]$  The argument continues...  Mordell goes on to show in turn that the search can be reduced further to the cases when $$n \equiv 1 \pmod{4} \;\;[meaning \;numbers \;\;n=4t+1]$$ $$n \equiv 1 \pmod{8} \;\;[meaning \;numbers \;\;n=8t+1]$$ $$n \equiv 1 \pmod{3} \;\;[meaning \;numbers \;\;n=3t+1]$$ $$n \equiv 1,2,4 \pmod{7} \;\;[meaning \;numbers \;\;n=7t+1,n=7t+2 \;or\;n=7t+4 ]$$ $$n \equiv 1,4 \pmod{5}  \;\;[meaning \;numbers \;\; n=5t+1 \;or\;n=5t+4]$$  Assembling these results together, Mordell showed that the conjecture can be proved in this context except for the cases when $$n \equiv 1,11^2,13^2,17^2,19^2,23^2 \pmod{840}$$  Mordell stated that since the first prime meeting this condition is 1009, this is proof that the conjecture holds for $n&lt;1009$.  This basic approach can be pursued further. Other workers have shown that the conjecture holds for much higher values of $n$ using similar methods as can be seen on the above Wikipedia page.   Note that other intermediate results can be constructed from the above congruence's, e.g. $n \equiv 1 \pmod{24}$.  The question is:  Are there any other elementary approaches to solving this problem than the one outlined by Mordell (and described above)?  [Ref1] Louis J. Mordell (1969) Diophantine Equations, Academic Press, London, pp. 287-290. In the top left corner of the post-race menu, there is a prompt to press + to add (or remove) one star out of six. What do these stars do? i would like to know how to prove that if $a^2+b^2=c^2$ then   a) $3|a$ or $3|b$  b) $5|a$ or $5|b$ or $5|c$  c) $4|a$ or $4|b$   Also that $a^2+b^2+c^2+1$ can't be divisible by 8  Thank you very much! I've seen a lot of games in the LCS and All-Star, and they sometimes changed bot and top lanes.   When is this a good practice? What do you gain from doing this? Should I do this to avoid counters or if my solo laner has more sustain than theirs? Prove there's a simple path of length $k$ in a simple graph $G$ where all the vertices have degree of at least $k$.   Relevant definitions:  $G$ is a simple graph that consists of a vertex set $V(G) = \{v_1, v_2, ..., v_n\}$ and an edge set $E(G) = \{e_1, e_2, ..., e_m\}$ where each edge is an ordered pair of vertices. The edge $\{u,v\}$ is denoted $uv$.  A walk of length $k$ is a sequence $v_0,e_1,v_1,e_2,...,e_k,v_k$ of vertices and edges such that $e_i=v_{i-1}v_i$ for all $i$. A path is a walk with no repeated vertex.   My attempt:  Induction, for $k=1$ it's obvious.  Suppose for $k-1$ and we'll prove for $k$.  Let $G$ be a simple graph such that $\forall v \in V : d(v)\ge k$.  We can assume that there are at least $k+1$ vertices since there are $k$ neighbours to every vertex.  Let $v_0$ be some vertex, it has $k$ neighbours, we'll move to one of its neighbours say $v_1$ and remove $v_0$. So now there are $k$ vertices with a degree of at least $k-1$ and from the induction hypothesis we'll have a path of length $k$.  Is using only $k$ vertices from $n$ right? Does it keep the generality when using only all the neighbours of $v_0$? Inversion in "only [adverb] have they" I have seen this construction quite often:      Online ads have been around since the dawn of the Web, but only in   recent years have they become the rapturous life dream of Silicon   Valley.   What is the rule there?. When your sentence doesn't start with pronoun + verb, invert them as verb + pronoun?. I know it sounds awkward but is it possible (grammatically correct) to use something similar to:      Online ads have been around since the dawn of the Web, but only in   recent years they have become...   And in any case, does this only work with have (or has)? Maybe it works fine with 'had' but I can't think of an example right now. What makes a Feynman diagram real or virtual? Simple question: as the title says, what makes a real Feynman diagram real, and what makes a virtual diagram virtual? Or in other words, how do I tell whether any given diagram is real or virtual? I've never gotten a really satisfying explanation of this. I would imagine it has something to do with virtual particles, but all internal propagators are virtual particles and I know for a fact that having internal lines doesn't make a diagram virtual. Normally Star Trek seems to based on some sort of logic, but the time travel in "Star Trek: Enterprise" seems to have no logic at all, or  am I just missing it?  Example: If they go back in time to change something it takes time until it has any effect in the future? I can't see any reason for this. Word meaning the reverse of 'Xenophobia' Xenophobia refers to the irrational fear or hatred of foreigners. But recently I've seen people displaying (in newspaper articles and other places) an irrational fear or hatred of their own countrymen. Is there a word for that? Can I kill the Golem pre-hardmode? Show $a^p \equiv a \mod 2p$ I've stucked at the following problem:  Let $p$ be an odd prime number. Show that for all $a\in\mathbb{Z}$  $$a^p \equiv a\ (\mathrm{mod}\ 2p)$$  How can I prove it? I know within Fermat's theorem that $a^p \equiv a\ \mathrm{mod}\ p$. Ubuntu 18.04 opening system settings logs me out What is the meaning of the density of a distribution at a point? I know how to use PDFs to calculate probabilities, but I don't think I understand them. For example, at $X=0$ the PDF of the standard normal distribution is $\approx 0.4$. Does this have any useful meaning? Strip whitespace from jsp output I have seen people showing one slide progressively by pressing Enter. Could anyone tell me how to realize that in Beamer? Is JavaScript a pass-by-reference or pass-by-value language? How to disable Windows 10 memory compression? my former wrong assumption: it drains my notebook battery and I've no shortage of RAM. In Task manager you see ntoskrnl.exe is eating more CPU time than any other process. this feature does not appear to be optional and came 2 days ago with the 1st Win10 Update (Win10Pro in my case).  Edit:  I want to disable it, at least to exclude it as a culprit for my (unproven) assumptions. That process never came to my attention in the last 2 months and in task manager's long name it's called "system memory and compressed memory".