Can i install ubuntu 18.04 on this specs Dell 3521 core i5 ram 8 hd 1 terabyte amd 2 gb ?!
Will my device work with Ubuntu?
past simple or present perfect Do superheroes/villains like Quicksilver or Flash with super speed get tired due to running? I mean a average human can run at his full speed for 15 to 20 mins before getting tired. Trained athletes can go on running for much more time without tiring out.  Do Superheroes like Quicksilver or Flash ever get tired due to excessive running? I want to advance in the ranks of Senator Lavigny-Duval. How can I get merit points? I know I can participate in expansion goals, but is there a better way than that? How do I enable error reporting in PHP? Difference between "per" and "a" Can I smoothly upgrade from one LTS to next LTS release? What is the coefficient of the term $a^i b^j c^k$ in the expansion of $(a+b+c)^n$? What is the coefficient of the term $a^i b^j c^k$  in the expansion of   $(a+b+c)^n$ ?   Given solution:     Before collecting like terms, each term in the expansion is a product   of $n$ factors each  of which is $a, b,$ or $c$.  Those terms which   have exactly i a’s, j b’s and k c’s can be combined into   the single term $a^i b^j c^k$ and the number of such terms will be   equal to the number of n letter strings containing i a’s, j b’s and k   c’s, which by the Mississippi rule is $\frac{n!}{(i!\times j! \times  k!)}$.   I understand that this problem is actually about the derivation of the coefficient of the multinomial theorem,but I am not sure about the combinatoric reasoning,what I mean is that I understand the analogy problem but what I couldn't is how the analogy holds here,Could anybody explain this to me? What is the command to truncate a SQL Server log file? I need to empty an LDF file before sending to a colleague.  How do I force SQL Server to truncate the log? Why can some holodeck matter leave the holodeck, while other holodeck matter (notably people) cannot? How to prove that $\frac{x^2}{yz+2}+\frac{y^2}{zx+2}+\frac{z^2}{xy+2}\geq \frac{x+y+z}{3}$ holds for any $(x,y,z)\in[1,2]^3$ Prove that for $x,y,z\in [1,2]$ the following inequality holds: $$\frac{x^2}{yz+2}+\frac{y^2}{zx+2}+\frac{z^2}{xy+2}\geq \frac{x+y+z}{3}.$$ I tried to apply the Cauchy-Schwarz inequality or the power mean inequality, but my attempts were unsuccessfull, so I am asking for help. Master database is corrupt, instance won't start - what are my options? Help! My master database is corrupt, I can't even bring the SQL instance online! What are my options to get my server back up?   I do have a backup of master, but the  asks me to start the instance in single-user mode, which I can't do!  (Note: I'm leaving this question unspecified as to SQL version, so as to be a more widely-applicable reference. There are some similar questions on DBA.SE, but none that involve the server unable to start.) Does the series $\sum \limits _{n=0}^{\infty} \cos(n\pi)$ converge or diverge? Does the series $\sum \limits _{n=0}^{\infty} \cos(n\pi)$ converge or diverge?   On substituting values I get alternate $1$ and $-1$.    So taking sum of infinite GP, I get $\dfrac 1 2$. So it looks like it converges. The question is $$\text{If }\, p, \ q\in \mathbb{N}, \;1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}.\qquad \text{Prove that } 1979\mid p.$$ So my solution went like this:  $$S=1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}\Rightarrow S=\sum_{k=1}^{1319}\frac{1}{k}-2\sum_{k=1}^{659}\frac{1}{2k}=\sum_{k=660}^{1319}\frac{1}{k}$$ Now this is equivalent to finding the sum of the roots of the equation $$\left(x-\frac{1}{660}\right)\left(x-\frac{1}{661}\right)\dots \left(x-\frac{1}{1319}\right)=0$$. Rewriting this equation, we get $(660x-1)(661x-1)\dotsc(1319x-1)=0$ (we can discard the $\displaystyle \frac{1319!}{659!}$ since $1979$ is a prime). Calculating the sum of the roots, we get the required value to be equal to $(1320+659)\times 330$ which is obviously divisible by $1979$ (again discarding the coeffecient of $x^{660}$).  Now my question is: Using the same method of converting the problem into finding the sum of roots of a polynomial, we get from the original equation something like this: $$(x-1)\left(x+\frac{1}{2}\right)\dots \left(x-\frac{1}{1319}\right)=0 \Rightarrow (x-1)(2x+1)\dotsc(1319x-1)=0$$ So, the sum of roots would be $-1+2-3+4-\dotsc+1318-1319=-660$ But this is not divisible by $1979$. Why does the former solution work while the latter doesn't? Is there a flaw in my reasoning? Note: I know that the question can be solved by some other methods. Please only explain my given doubts and don't instead give another solution  I realised my mistake after some hours of pondering over it. (In the 1st equation) To get the required sum I would need to take calculate $\displaystyle -\frac{\text{coeffecient of }x^{659}\,}{\text{coeffecient of }x^{660}}$. However what I've done is calculate this:  $\displaystyle -\frac{\text{ coeffecient of }x}{\text{coeffecient of }x^{660}}$ which is actually equal to $\text{sum of roots taken }659 \text{ at a time} $. It was just a coincidence that the value I calculated was divisible by $1979$. The correct way, indeed, would be through the link supplied Why are the orbits of planets in the Solar System nearly circular? Except for Mercury, the planets in the Solar System .  Is this property special to the Solar System?  :     Most exoplanets with orbital periods of 20 days or less have near-circular orbits of very low eccentricity. That is believed to be due to tidal circularization, an effect in which the gravitational interaction between two bodies gradually reduces their orbital eccentricity. By contrast, most known exoplanets with longer orbital periods have quite eccentric orbits. (As of July 2010, 55% of such exoplanets have eccentricities greater than 0.2 while 17% have eccentricities greater than 0.5.) This is not an observational selection effect, since a planet can be detected about equally well regardless of the eccentricity of its orbit. The prevalence of elliptical orbits is a major puzzle, since current theories of planetary formation strongly suggest planets should form with circular (that is, non-eccentric) orbits.   What is special about the Solar System that orbits of planets here are nearly circular, but elsewhere they are moderately or highly eccentric? I have a table story_category in my database with corrupt entries. The next query returns the corrupt entries:  SELECT *  FROM  story_category  WHERE category_id NOT IN (     SELECT DISTINCT category.id      FROM category INNER JOIN         story_category ON category_id=category.id);   I tried to delete them executing:  DELETE FROM story_category  WHERE category_id NOT IN (     SELECT DISTINCT category.id      FROM category        INNER JOIN story_category ON category_id=category.id);   But I get the next error:      #1093 - You can't specify target table 'story_category' for update in FROM clause    How can I overcome this? I have seen some hypotheses that the recent series corporate actions and statements are a sign of SO/SE being in the process of being, or about to be, sold. Arguments cited include the corp statements being made in paywalled areas and addressed to an audience that seems very distinct from the community, the choice of CEO, the push for a 'service sector' flavour of tensely smiling hospitality, and surely some more I can't recall in a coherent form.   Just how plausible and implausible, and how founded or unfounded, is such a hypothesis? Is it, as far as people more knowledgeable either about business in general or about the specific case of SE/SO, strategically consistent or even outright connected with the corp's behaviour (and a likely cause thereof)? NEWID() In Joined Virtual Table Causes Unintended Cross Apply Behavior My actual work query was an inner join, but this simple example with cross join seems to nearly always reproduce the problem.  SELECT * FROM (     SELECT 1 UNION ALL     SELECT 2 ) AA ( A ) CROSS JOIN (     SELECT NEWID() TEST_ID ) BB ( B )   With my inner join I had many rows for which I added to each a GUID using the NEWID() function, and for about 9 out of 10 such rows the multiplication with the 2-row virtual table produced the expected results, just 2 copies of the same GUID, while 1 out of 10 would produce different results. This was unexpected to say the least and gave me a really hard time trying to find this bug in my test data generation script.  If you take a look at the following queries using as well non-deterministic getdate and sysdatetime functions, you won't see this, I don't anyway-I always see the same datetime value in both final result rows.  SELECT * FROM (     SELECT 1 UNION ALL     SELECT 2 ) AA ( A ) CROSS JOIN (     SELECT GETDATE() TEST_ID ) BB ( B )  SELECT * FROM (     SELECT 1 UNION ALL     SELECT 2 ) AA ( A ) CROSS JOIN (     SELECT SYSDATETIME() TEST_ID ) BB ( B )   I'm currently using SQL Server 2008 and my work around for now is to load my rows with GUIDs into a table variable before finishing out my random data generation script. Once I have them as values in a table as opposed to virtual table, the problem goes away.  I have a workaround, but I'm looking for the ways to workaround without actual tables or table variables.  While writing this I tried without success these possibilities: 1) placing the newid() into a nested virtual table:  SELECT * FROM (     SELECT 1 UNION ALL     SELECT 2 ) AA ( A ) CROSS JOIN (     SELECT TEST_ID     FROM (         SELECT NEWID() TEST_ID     ) TT ) BB ( B )   2) wrapping the newid() within a cast expression such as:  SELECT CAST(NEWID() AS VARCHAR(100)) TEST_ID   3) reversing the order of appearance of the virtual tables within the join expression  SELECT * FROM (     SELECT NEWID() TEST_ID ) BB ( B ) CROSS JOIN (     SELECT 1 UNION ALL     SELECT 2 ) AA ( A )   4) using uncorrelated cross apply  SELECT * FROM (     SELECT NEWID() TEST_ID ) BB ( B ) CROSS APPLY (     SELECT 1 UNION ALL     SELECT 2 ) AA ( A )   Just before finally posting this question, now I tried this with success it seems, a correlated cross apply:  SELECT * FROM (     SELECT NEWID() TEST_ID ) BB ( B ) CROSS APPLY (     SELECT A     FROM (         SELECT 1 UNION ALL         SELECT 2     ) TT ( A )     WHERE BB.B IS NOT NULL ) AA ( A )   Anyone have any other more elegant, simpler workaround? I really don't want to use cross apply or correlation for a simple row multiplication if I don't have to. How to save picture to iPhone photo library? Google App Engine: Is it possible to do a Gql LIKE query? Evaluate the sum: $\sum\limits_{n=0}^{\infty} \frac1{F_{(2^n)}}$ Evaluate the sum:  $$\sum_{n=0}^{\infty} \frac{1}{F_{(2^n)}}$$  where $F_{m}$ is the $m$-th term of the Fibonacci sequence. I need some support here. Thanks.