Visualforce email templates- date time, and timezone rendering for a date field?
Visualforce Date formatting
Fiction novel or movie about sentient dinosaurs Clip tool gives ERROR 000732? Is $\int \frac{1}{2x} \, dx $ equal to $\frac{\ln|2x|}{2}+ C$ or $\frac{\ln|x|}{2}+ C$ Reasoning for $\frac{\ln|x|}{2}+ C$ :  $$\int  \frac{1}{(2x)}  \, dx = \frac{\ln|2x|}{2}+ C$$   Check: By taking derivative we get $\frac{1}{x}$  Reasoning for $\frac{\ln|x|}{2}+ C$ :  $$\int  \frac{1}{2x}  \, dx = \frac{1}{2}\int  \frac{1}{x}  \, dx =\frac{\ln|x|}{2}+ C$$  Check: By taking derivative we get $\frac{1}{x}$  Question: Is  $\int  \frac{1}{2x}  \, dx $ equal to $\frac{\ln|2x|}{2}+ C$ or $\frac{\ln|x|}{2}+ C$? When in many rooms, user's info goes off screen So, I wanted to check what low-number rooms are still active so I joined a bunch of rooms in the chat.  This is the result:    Bug also exists when expending rooms on the side:    Since the position is fixed I can't scroll down.  Proposed fix:  Only show the first 10 rooms, and show ... pointing to the user's profile for other rooms. For example, if I revisit a memory of mine where I'm inside a castle and talking to someone, can the present-me explore different parts of the castle meanwhile the past-me continues her conversation? Is that possible? Split comma-separated values Reading a huge .csv file So, if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ means $1=ax+by$, and want to show $a^2x+b^2y=1$. By squaring $1=ax+by$ both sides, I get, $1=(ax)^2+b(2axby+by^2)$. It doesn't help my proof. Please help me with this proof. Create alphabetical pager "Member for: today" Very new members will have this text displayed in their profile:    Should this not be changed to say "1 day" or "3 hours", etc? Especially as the "last seen" also works in this way.  Note that members who joined 24 hours ago have been members for "2 days", etc.  You'd say "x has been a member for 1 day" over "x has been a member for today"; saying "member for today" just doesn't feel right to me. Let $(a_n)$ be a strictly increasing sequence of positive integers such that: $a_2 = 2$, $a_{mn} = a_m a_n$ for $m, n$ relatively prime. Let $(a_n)$ be a strictly increasing sequence of positive integers such   that: $a_2 = 2$ and $a_{mn} = a_m a_n$ for $m, n$ relatively prime. Show that $a_n = n$, for every positive integer $n$.   This is a result apparently due to Paul Erdős, and  has a proof by induction.  I tried like this, $a_{10}=a_2a_5$. After this what we can do?    [Editor's comment] Possibly due to the apparent simplicity of the conditions it may be difficult to appreciate the subtleties of this question. If we try and construct a counterexample like $a_3=4$, $a_4=5$, $a_5=6$, then the requirements dictate $a_6=8$, $a_{10}=12$, $a_{15}=24$. At this point we realize that we have been speeding. For monotonicity forces $a_9\le 11$, and therefore $a_{18}\le22&lt;a_{15}$, violating the requirements. It is not obvious why something similar ruins all the modifications to the sequence $a_n=n$. [/comment, JL] Show that $E(XE[X|G])=E[(E[X|G)^2]$ Would a very small creature inside a sealed container carried by a character take damage from AOE spells? Assume character A is in the AOE of a damaging spell.  Character A is carrying a closed container inside his backpack, alongside several sheets of parchment.  Damaging spell description specifically states that items carried by a creature are not damaged even on a failed save. Unattended objects are damaged, though.  Inside said container there is a small creature. Creature has no problem living inside container for a long amount of time.  Does this creature take damage from the spell? If it does not, what mechanic prevents damage? Downgrading MacBook Pro 16” to Mojave Did anyone succeed in installing Mojave on the MacBook Pro 16”? It ships with Catalina but my corporate requires Mojave to connect to the corporate network.  I  but once I select booting from the Mojave external drive I get a “forbidden” icon and the system shuts down. $A_n:Y\rightarrow Z$ are operators that strongly converge to $A$. Also, $\|A_n\|_\text {op}\le c$ for $c&gt;0$. Given a compact operator $T:X\rightarrow Y$, I need to show that $A_nT$ converges to $AT$ in operator norm (all spaces in this question are Banach spaces).     I was unable to prove this and I also do not understand why we need to assume that $T$ is compact. Any ideas? I have a .txt file with numbers ordered like this (on the same row):  106849_01373    106849_01967    106850_00082    23025.7_01059      I would like to convert them like that:  106849_01373 106849_01967 106850_00082 23025.7_01059   I have no idea which command to use. Can somebody help me with this? Why does chmod succeed on a file when the user does not have write permission on parent directory? I am trying to understand file/dir permissions in Linux.  A user can list the files in a directory using   cd test ls -l    Even if the user issuing above commands does not have read, write or execute permission on any of the files inside the test directory, still he can list them because he/she has read permissions on the test directory.   Then why in the following scenario user B can change permissions of a file he owns but does not have write permissions of the parent directory?  User A, makes a test directory and gives other users ability to write in it:  mkdir test chmod o+w test   User B, creates a file in test folder.  cd test touch b.txt   User A removes write permission of others from the directory  chmod o-w test   User B, can successfully change permissions, even though permissions are part of directory and this user does not have write permission on the parent directory of the file he owns  chmod g-r b.txt   why does chmod not fail since the user cannot modify the directory which has the file information - permissions etc? I'm using Ubuntu 16.04 and there's no fglrx! Badass and BAMF are both modern words with approximately the same meaning: "Someone who is awesome to an extreme level, thereby leveraging unquestionable authority." Is there another noun or title that would convey the same level of awe, admiration and toughness that calling someone a badass does, without the vulgarity?  Thanks for any help. :) In the simple linear regression case $y=\beta_0+\beta_1x$, you can derive the least square estimator $\hat\beta_1=\frac{\sum(x_i-\bar x)(y_i-\bar y)}{\sum(x_i-\bar x)^2}$ such that you don't have to know $\hat\beta_0$ to estimate $\hat\beta_1$  Suppose I have $y=\beta_1x_1+\beta_2x_2$, how do I derive $\hat\beta_1$ without estimating $\hat\beta_2$? or is this not possible?