Why the 20.04.1 is still not available in the update manager?
Why isn't an upgrade to 20.04 from 18.04 available yet?
It seems that a List object cannot be stored in a List variable in C#, and can't even be explicitly cast that way.  List&lt;string&gt; sl = new List&lt;string&gt;(); List&lt;object&gt; ol; ol = sl;   results in Cannot implicitly convert type System.Collections.Generic.List&lt;string&gt; to System.Collections.Generic.List&lt;object&gt;  And then...  List&lt;string&gt; sl = new List&lt;string&gt;(); List&lt;object&gt; ol; ol = (List&lt;object&gt;)sl;   results in Cannot convert type System.Collections.Generic.List&lt;string&gt; to System.Collections.Generic.List&lt;object&gt;  Of course, you can do it by pulling everything out of the string list and putting it back in one at a time, but it is a rather convoluted solution. Why does the jquery change event not trigger when I set the value of a select using val()? "Easy" vs. "simple" Are those two words equivalent or is there a difference? When would I say something is easy rather than simple, or vice versa? Lost email settings and no icon for Mail I've lost my Settings → Mail, Contacts, and Calendars on my iPhone 5.  No icon for Mail and no Mail settings.  How do I get it back? What is difference between int.class and Integer.TYPE in java? Duplicate user in reputation league So I understand that this has happened in the past: , but all of the &quot;duplicate&quot; posts are marked as  so here is a new bug report: On the , I am listed twice:  While that's fun and all, it isn't exactly right. Placing point over pixel with maximum value falling in grid overlay using QGIS Call a javascript function after 5 sec of last key press I'm looking for a high current DC switch (able to switch something like 5V@30A) and I've encountered several DPDT that might do what I need, however they are only rated for AC applications (3A@250VAC or 6A@125VAC). Are these rating convertible to DC? or is that something that only the facturarer would know? Drawing Roman laurel leaves (SPQR) in TikZ I was trying to get the symbol of the old .  I'm actually interested in the laurel leaves, and not so much in the banner per se.  What I would like to have is something like this (in Tikz, as oppose to simply \includegraphics).  With the possibility to change the stylish A to any other simple roman letter.    I'm a complete newbie with TikZ, and even though I know this question sounds like do this for me, I would really appreciate if someone could at least throw in an sketch where to start. Which prepositions are followed by accusative pronouns instead of reflexive pronouns? I have read answers to questions like  but I couldn't find a general rule for using "you" or "yourself", "him" or "himself" that could explain things I see in some examples.  For example, I think that one normally say (according to examples I see in ):   You bring something with you. You have something in you. You have something on you. You pressed her to you.   But on the other hand one would say:   You draw attention to yourself. He killed himself. You can try it out for yourself.   As a general rule, it seems that the reflexive pronoun is used when the object is the same as the subject of the verb. But for prepositional phrases, it seems more complicated. Do some prepositions (like "with", "in", "on") always come with the accusative form of the pronoun, and some other (like "for") don't? What about "to", that seems to be followed by both in different circumstances? How to recover a deleted NTFS partition with data Difference between least squares and minimum norm solution Consider a linear system of equations $Ax = b$.   If the system is overdetermined, the least squares (approximate) solution minimizes $||b - Ax||^2$. Some source sources also mention $||b - Ax||$. If the system is underdetermined one can calculate the minimum norm solution. But it does also minimize  $||b - Ax||$, or am I wrong?   But if least squares is also a minimum norm, what is the difference, or the rationale of the different naming? Suppose that $f(x)$ is continuous on $(0, \infty)$ such that for all $x > 0$,$f(x^2) = f(x)$. Prove that $f$ is a constant function. Suppose that $f(x)$ is continuous on $(0, +\infty)$ such that for all $x &gt; 0$,$f(x^2) = f(x)$. Prove that $f$ is a constant function. My attempt is to show that for any point $a \neq b$ , we have $f(a)=f(b)$. But I have no idea on how to get this. Anyone can help? How to install Canon LBP2900b drivers? Possible Duplicate:          I am a little bit confused about asking a question:     Can you please tell me my next work?   or     Could you please tell me my next work?   Which is more correct? Defaults to true.  If set to true replaces all instances of known swear words with replacements (comical or * based). If false renders the content as originally presented.  Benefits:    Reduces moderation requirements necessary to ensure that people behind corporate firewalls are not blocked by an errent swear word.   Stops idiots like me complaining about it. This is the code that's giving me trouble.   f = Frame(root, width=1000, bg="blue") f.pack(fill=X, expand=True)  l = Label(f, text="hi", width=10, bg="red", fg="white") l.pack()   If I comment out the lines with the Label, the Frame displays with the right width.  However, adding the Label seems to shrink the Frame down to the Label's size.  Is there a way to prevent that from happening? Movie about hostile robots disguised as footballs I saw this on TV in mid 90s I think. One or more people travel to another celestial body (might or might not be the/a moon or mars), they find an American football. The football transforms into a robot that is hostile. They later find LOTS of these footballs/disguised robots.  Since there are lots of robotic projects with "football robots" or " robots", it is very hard to filter search results using Google etc. I want to express $$\sum_{n=-\infty}^\infty \dfrac{1}{(z+n)^2+a^2}$$ in closed form. What comes to mind is the formula $$\pi\cot\pi z = \dfrac{1}{z}+\sum_{n\ne 0}\left(\dfrac{1}{z-n}+\dfrac1n\right)=\dfrac{1}{z}+\sum_{n=1}^\infty\dfrac{2z}{z^2-n^2}$$  and also $$\dfrac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty \dfrac{1}{(z-n)^2}.$$ But neither of these gives the term $(z+n)^2+a^2$ that we want. Perhaps we can adjust somehow?