I've messed things up a bit.  I was running Windows 7 ultimate 64-bit in my HP laptop. I was using live USB of Ubuntu 11.10 and then I tried to install Ubuntu on hard disk using the same live USB.   To install Ubuntu, I chose 50 GB free partition and 325 GB swap space or something.   While installing Ubuntu, my pen-drive came out and everything stopped. Then I booted again using the pen-drive and finished installation of Ubuntu.   Now the problem is I can't access my Windows partition as they were in extended partition plus I am not able to boot Windows 7 from the booting screen and I can't even repair or recover it.   Whenever I try to boot Windows it shows the message HARDWARE INACCESSIBLE, TRY TO REPAIR YOUR WINDOWS USING DISC. When I tried using disc, there was no instance of OS available to repair.   I can't even install fresh Windows 7 as the setup asks me to load drivers of Hard disk.  I have important data in the Windows root drive as well as in other partitions. How can I recover that data?
I installed Windows on my computer, followed by an installation of Ubuntu. However, now I'm unable to boot into my Windows install.  What can I do to fix this?
Best way to save data in a Java application? "Pending" vs "Impending". Are they synonyms? They appear to me to mean almost if not exactly the same thing, but I am not sure. Are there differences in meaning between them? How can I wrap text around a bottom-right div? Every time I try to do something seemingly-simple in CSS, it doesn't work.  I have a content div that contains a 460x160 image. All I want to do is position the image at the bottom-right corner and wrap my text around it.  &lt;div id="contents"&gt;     &lt;img src="..." /&gt;     text text text text text text ... &lt;/div&gt;   So I want it to look like:  ------------------ | text text text | | text text text | | text text -----| | text text |    | ------------------   Doing it with a top-left or top-right image is cake:  #contents img { float:right; }  ------------------ | text text |    | | text text -----| | text text text | | text text text | ------------------   Now how do I push that to the bottom? The best I've come up with so far are:  #contents img { float:right; margin-top: 90%} // really needs to be 100%-160px  ------------------ | text text      | | text text      | | text text -----| | text text |    | ------------------   In this case the text does not print in the margin, so there is white space above the image.  #contents { position:relative; } #contents img { position:absolute; right:0; bottom:0; }  -or-  // move the img tag under the text in the html and: #contents img { float:right; margin-top:-160; }  ------------------ | text text text | | text text text | | text text -----| | text text | te | ------------------   In this case the text prints over or under the image.  So... how can I accomplish this? Troubleshooting "Windows cannot be installed to this disk. The selected disk has an MBR partition table" Search for emails from contacts with multiple email addresses Install App Image such as beaker browser and pin it to dock (favorites) in Ubuntu 20.04? What communication can be intercepted by network I'm connected over VPN to? I have to connect to client's network over VPN, and I was wondering if the client can intercept (and read) my Google Talk messages (or any other IM communication) while I'm VPN-ing.  My main concern is them having access to our internal conversations.  Suggest workarounds if you have any. Just doing this for revision, seems much harder than it should be, should I use $x=\tan u$ ? Any help appreciated. How to get straight quotation marks? How do I get straight quotation marks, i.e., " and ' in TeX?  (No, \verb changes the font and this is unacceptable.)  [Edit]  I forgot to mention that I’m using XeLaTeX, and fontspec is a necessity.  The first two proposed answers did work, without fontspec. PHP validation/regex for URL I've been looking for a simple regex for URLs, does anybody have one handy that works well?  I didn't find one with the zend framework validation classes and have seen several implementations. Consistency of an order statistic in exponential distribution I have two questions.   1) If $X_1,X_2,X_3,...,X_n$ constitute a random sample of size $n$ from an exponential distribution, show that $\bar X$ is a consistent estimator of the parameter $\lambda$.    This is my attempt: Mean of exponential distribution is $\lambda^{-1}$. Thus by central Limit theorem $E(\bar X)=\lambda^{-1}$. Using mean square consistency, if  $\lim_{n\to \infty} E[(\bar X -\lambda )^2]=0$ then $\bar X $ is consistent for $\lambda$. Since $$ E[(\bar X -\lambda )^2]= E[(\bar X)^2]+E(\lambda)^2-2\lambda E(\bar X) , $$  I get $$ \lim_{n\to \infty} E[(\bar X -\lambda )^2] =    \lim_{n\to \infty} {2 \over \lambda ^2 }+\lambda^2-2 . $$ Clearly this is not equal to $0$. How can I show this is consistent? I tried using Chebechev inequality as well. But even using that I couldn't.    2) Is $X_{(n)}$ a consistent estimator for $\lambda$? If $P[|X_{(n)}-\lambda|&lt;\epsilon]$ approaches $1$ as $n$ goes to infinity for all $\epsilon &gt; 0$ then $X_n$ is a consistent estimator for $\lambda$.    The probability density function of $X_{(n)}$ is $f_n(x) = n \lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}$,  for $X&gt;0$.  Thus \begin{align*} P[|X_{(n)}-\lambda|&lt;\epsilon]  &amp;= P[\lambda -\epsilon&lt;X_{n}&lt;\lambda+\epsilon] \\  &amp;=P[0&lt;X_{n}&lt;\lambda+\epsilon] \\  &amp;=\int_0^{\lambda+\epsilon} n \lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}\\  &amp;= (1-e^{- \lambda X_{(n)} }), \end{align*} where $X{(n)}$ takes values in $[0 , \lambda+ \epsilon]$.  Thus, can I say $\lim_{n\to \infty}[1-e^{-\lambda(\lambda+\epsilon)}]^n=1$? Order of elements is lcm-closed in abelian groups How can I prove that if $G$ is an Abelian group with elements $a$ and $b$ with orders $m$ and $n$, respectively, then $G$ contains an element whose order is the least common multiple of $m$ and $n$?  It's an exercise from Hungerford's book, but it's not homework. I could not solve it when I take the course on groups and I think it should be easy. There is a hint that says to first prove when $m$ and $n$ are coprimes. I did this part. But I have no idea how to solve the general case.  Thanks. How can I attach a file to a post? I want to attach some files to a post on one of Stack Exchange sites. What is the best way to do this? Label equations arbitrarily Show that G is a group, if G is finite, the operation is associative, and cancellation law holds I am trying to work through Hartshorne's book and while working through one of the exercises, I need to show the following:  Let $k$ be an algebraically closed field. Let $Y \subseteq A^3$ be the set $Y=\{(t,t^2,t^3)\;|\;t \in k\}$. Show that $Y$ is an affine variety.  The rest of the problem I have completed. I'm just not sure how to show that this set is closed and irreducible. I assume that they intend for $A^3$ to have the Zariski topology. When deriving the quadratic equation as shown in the  () the main proof contains the step: $$ \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}} $$ the square root is taken from both sides, so why is  $$\sqrt{4a^2} = 2a$$  in the denominator and not $$ \sqrt{4a^2} = 2\left |a  \right | $$ Could somebody explain this to me? Thank you very much Significant figures vs. absolute error On , $N_A = 6.022\;141\;29 \times 10^{23} \text{ mol}^{-1}$ has 9 significant figures and a standard uncertainty of $0.000\;000\;27 \times 10^{23} \text{ mol}^{-1}$.  First, can it be written as the following?  $$ N_A = 6.022\;141\;29 \times 10^{23} \pm 0.000\;000\;27 \times 10^{23} \text{ mol}^{-1} $$ If yes, consider the following quote from  (emphasis added):     Whenever you make a measurement, the number of meaningful digits that you write down implies the error in the measurement. For example if you say that the length of an object is 0.428 m, you imply an uncertainty of about 0.001 m. To record this measurement as either 0.4 or 0.42819667 would imply that you only know it to 0.1 m in the first case or to 0.00000001 m in the second. You should only report as many significant figures as are consistent with the estimated error. The quantity 0.428 m is said to have three significant figures, that is, three digits that make sense in terms of the measurement. Notice that this has nothing to do with the "number of decimal places". The same measurement in centimeters would be 42.8 cm and still be a three significant figure number. The accepted convention is that only one uncertain digit is to be reported for a measurement. In the example if the estimated error is 0.02 m you would report a result of 0.43 ± 0.02 m, not 0.428 ± 0.02 m.   and the following quote from  (emphasis added):     The absolute error places bounds telling you that the actual value is expected to fall between x - D x and x + D x. The absolute error is an easy way to report the error and makes it easy to check that you are not reporting too many significant figures. For example, suppose you had a measurement that was reported to be 10.02 ± 0.3 m. Because the reported value is uncertain in the tenths place, it is wrong to report it to the hundredths place. It should be reported to the same decimal place as the absolute error: 10.0 ± 0.3 m. Another point is that the absolute error and the reported value must have the same units.   If there is such a convention (cf. Rice's lab guide) and it is wrong to report more significant figures than is warranted by the absolute error (cf. the joint project's guidelines), and NIST should have adhered to such convention and striven to be correct, is it true that NIST adheres and is correct because the standard uncertainty has two least significant digits (i.e., $27$)?  Had the standard uncertainty been $0.000\;000\;2\text{ mol}^{-1}$, would NIST have reported Avogadro's Number as $N_A = 6.022\;141\;3\times 10^{23}$ with 8 significant figures? Can't cast to to unspecific nested type with generics shortcut for commands I use following commands very often, and they're too long.  sudo /etc/init.d/kerio-kvc start sudo /etc/init.d/kerio-kvc stop   how can I create a shortcut command for them? something like sudo kerio start or sudo kerio stop