I saw this around 7-8 years ago on Toonami/Cartoon Network and I'm still not quite sure it was an anime or an animated TV show. I watched this series completely in Spanish since that's my main language but sadly I was too young to remember anything else about the animation style or quotes from it. The opening is the thing I'm less sure of but it basically recapped the story, and it was something like...     a king having a belt (a belt? could be something else) that had elemental gems on it,   the antagonist tried to get the gems so the king hid the gems all around the world, now the villain's trying to get it back and the heroes are trying to get them so the villain doesn't get them.   I remember 3 episodes, to an extent,     The first was the heroes being trapped on quicksand, and they end up meeting with an old man and an old woman that save them, they spoke in gibberish and gave them food and a place to stay. Some of the protagonists were suspicious about this but were obviously told they were exaggerating. Eventually they found out the old people were speaking backwards and were trying to eat them, the old couple transform into monsters and they defeat them.   Another episode is the heroes looking for the water gem and finding themselves on a beach with a huge castle floating right in the middle of the water. They tried their best to reach it (not sure how they did?) only to have one of them reach the castle and find a huge blue dragon with the gem on it's forehead.   The next episode is just the rest of the characters getting to the castle and defeating the dragon.
Okay so I remember almost nothing of this show. The only thing I know is that I used to watch this show on Cartoon Network and it had a hero who used a sword to fight monsters and a girl who used bow and arrows. Also, the monsters kind of looked like computer generated and there was a book and stones which were somewhat related to the story?
How can I make an object show up on different layers I'm running into trouble trying to mount a large iso:  dev@dev-OptiPlex-745:~$ sudo mount -o loop /home/dev/Hämtningar/matlab2011a_64.iso /cdrom mount: warning: /cdrom seems to be mounted read-only. dev@dev-OptiPlex-745:~$    Can you tell me how I should do it? How variance is defined? Pipelined Sed does not work on found filename inside Bash command substitution when invoked from Find "-exec" It looks like 'find', 'bash' and 'sed' in some cases does not work as one expects.  The following example should first create file 'sample.txt', then find the file and finally process it by '-exec' command. The executed command prints found filename, test specimens, and modified filename. The 'sed' command itself is used to replace 'txt' to 'TXT'.  touch sample.txt find ./ -maxdepth 1 -name "*.txt" -exec echo {} $(echo Specimen_before.txt {} Specimen_after.txt |sed -e "s/txt/TXT/g") \;   The expected output is:     ./sample.txt Specimen_before.TXT ./sample.TXT Specimen_after.TXT   Instead it produces:     ./sample.txt Specimen_before.TXT ./sample.txt Specimen_after.TXT   (the example has been tested also with old-school command substitution through backquotes '`' with the same result)  What am I doing wrong? What is this sand covered insect? I have found in a wooden house, in the mountains of central Spain this very curious insect:    I was wondering if that was its skin or whether it's decomposition or sand or saw-dust.  That particular region of Spain does have stink-bugs ... but this one doesn't look like anything I've seen before.  A google image search returns mostly arachnids ... but it seems to have only 6 legs. Avoid synchronized(this) in Java? Can the sum of the first $p$ factorials ever be a perfect power for $\ p>3\ $? Has $$\sum_{j=1}^p j!=q^r$$ , where q,p,r are positive integers, and r > 1 , a solution ?   I solved partially, if r is even, then RHS is a perfect square, and there is no doubt in that. Therefore, the left side must be a perfect square as well. But for p>3, the last digit of LHS is 3 which is not the last digit of any square. Hence, p&lt;4, and hence manually checking, only solutions are p =1,3. But i cannot generalize when r is odd.  Any solution to the next part would be helpful! When we measure the speed of a moving element we do it with the help of a reference frame. Now if we need to measure the speed of time, is it possible? Does time actually have a speed? I am seeking advice on which memory to use with my motherboard and CPU. Below are my components and some relevant information extracted from their official websites.   NOTE: Although this is a very specific case, a general answer is equally good for me — and also alot more useful for everyone else!  CPU: AMD Phenom II X4 810   ASUS M4A785TD-V EVO listed as one of many  motherboards.   Motherboard: ASUS M4A785TD-V EVO   4 x DIMM, Max. 16 GB, DDR3 1800(O.C.)*/1600(O.C.)/1333/1066/800 ECC,Non-ECC,Un-buffered Memory Dual Channel memory architecture (Complete list of specs: ).   NOTE II: I wouldn't want to do any overclocking as I have no experience with that. So I think the 1333 MHz is the most appropriate. Since the higher frequencies are marked with "O.C."  Questions:   How can I be certain that a specific RAM will work well (or not work) together with my CPU and motherboard? (The Qualified Vendor List is a good starting point, but it's a bit limited and sometimes a bit outdated). Is it true that choice of memory is not independent of the processor? How?    I'm worried I'll be missing something essential when buying my next RAM chips :) Can $n!$ be a perfect square when $n$ is an integer greater than $1$?  Clearly, when $n$ is prime, $n!$ is not a perfect square because the exponent of $n$ in $n!$ is $1$.  The same goes when $n-1$ is prime, by considering the exponent of $n-1$. What is the answer for a general value of $n$?  (And is it possible, to prove without Bertrand's postulate. Because Bertrands postulate is quite a strong result.) I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ which are continuous and satisfy $f(x+y)=f(x)+f(y)$ I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $f(x+y)=f(x)+f(y)$. I know that there are other questions that are asking the same thing, but I'm trying to figure this out by myself as best as possible. Here is how I started out:  Try out some cases:  $x=0:$ $$f(0+y)=f(0)+f(y) \iff f(y)=f(0)+f(y) \iff 0=f(0) $$ The same result is for when $y=0$  $x=-y:$ $$f(-y+y)=f(-y)+f(y) \iff f(0)=f(-y)+f(y) \iff 0=f(-y)+f(y)\iff \quad f(-y)=-f(y)$$ I want to extend the result of setting $x=-y$ to numbers other that $-1$, perhaps all real numbers or all rational numbers. I got a little help from reading other solutions on the next part:  Let $q=1+1+1+...+1$. Then  $$f(qx)=f((1+1+...+1)x)=f(x+x+...+x)=f(x)+f(x)+...+f(x)=qf(x)$$ I understood this part, but I don't understand why this helps me find all the functions that satisfy the requirement that $f(x+y)=f(x)+f(y)$, but here is how I went on:  Thus  $$f(qx)=qf(x)$$ and it should follow that $$f \bigg (\frac {1}{q} x\bigg)= \frac{1}{q}f(x)$$ where $q\not =0$, then it further follows that  $$f \bigg (\frac {p}{q} x\bigg)= \frac{p}{q}f(x)$$ where $\frac{p}{q}$ is rational, and lastly it further follows that  $$f (ax)= af(x)$$ where $a$ is real. Thus functions of the form $f(ax)$ where $a$ is real satisfies the requirement of  $f(x+y)=f(x)+f(y)$.  I don't know how much of what I did is correct\incorrect, and any help would be greatly appreciated. Also is there any way that I can say that functions of the form $f(ax)$ where $a$ is real are the only functions that satisfy the requirement of  $f(x+y)=f(x)+f(y)$? Or do other solutions exist?  Again, thanks a lot for any help! (Hints would be appreciated, I'll really try to understand the hints!) What does ':' (colon) do in JavaScript? I'd like to find a place to store stuff while I'm out adventuring, without spending 5000 on a house (at least, until I can afford that...). Where can I safely store items without worrying about them disappearing?   There are two possible places that come to mind,  (to which I have the key and was told I could stay), and the small bedroom given to me at the  (which I was told was mine, despite seeing J'zaro sitting in it later). Can I store items safely in either of these places? Where has all the RAM gone? Can I store all the content of my previous iPhone in a new iPhone using iTunes? How do I take the limit as $n$ goes to $\infty$ of $\frac{\sqrt{n}}{\log(n)}$? How do take this limit:  $$ \lim_{n\to\infty} \frac{\sqrt{n}}{\log(n)}$$  I have a feeling that it is infinity, but I'm not sure how to prove it. Should I use L'Hopitals Rule? Prove that if $\gcd(a,b)=1$ then $\gcd(a^m, b)=1$ Is there a string math evaluator in .NET? If I have a string with a valid math expression such as:  String s = "1 + 2 * 7";   Is there a built in library/function in .NET that will parse and evaluate that expression for me and return the result? In this case 15. Storable Action - System.LimitException: Too many DML statements: 1 @AuraEnabled(cacheable=true) public static String  deleteContacts(List&lt;String&gt; contactIds) {     List&lt;Contact&gt; returnList = new List&lt;Contact&gt; ();     String query = ' SELECT Id,Name, LastName,Department,MobilePhone, Email FROM Contact WHERE id IN : contactIds ';     for (Contact thisContact:  Database.Query(query)) {          returnList.add(thisContact);     }     try {       delete returnList;       return'deleted successfully';     }     catch(Exception ex){        return 'Problem occoured';     }  }  JS Method deleteSelected(){     deleteContacts({         contactIds :this.selectedIds     })      .then(result =&gt; {         this.dispatchEvent(             new ShowToastEvent({                 title: 'Success',                 message: result,                 variant: 'success',             }),         );         return refreshApex(this.getContactList, { searchKeyWord: '$searchKeyWord' });     })     .catch(error =&gt; {         this.error = error;     }); Detecting TCP Client Disconnect