I browse from a shared IP address (my university gives no other option) and I, for the first time ever, decided to answer a question on Stack Exchange.  I have apparently been answer banned but I have no deleted answers or even deleted posts.  I have a reputation of 10, not very good but surely not answer ban worthy?  I read on the FAQ question  that this might have something to do with my IP address.   Are there any workarounds other than paying for a VPN?
If you are reading this, you may have been affected by one of our post quality bans. If so, it is important that you  to understand what has happened and how to regain the permissions that you lost. While trying to ask a question, one could see:  We are no longer accepting questions from this account. See  to learn more.  Likewise, for answers:  We are no longer accepting answers from this account. See  to learn more.   Why am I getting this message? Are deleted posts taken into account too? Is a question/answer ban the same as a suspension? How do I avoid getting a question ban? How long do I have to wait before I can post again? What can I do to release the ban? How can I reactivate my account? Can I simply create a new account? I'll just ask somewhere else on the SE network, and they'll migrate my question to the correct site! Will a ban on one Stack Exchange site affect my standing on other sites in the network? Does this apply to meta sites too? My account is in good standing. Why am I still blocked?
jquery stop child triggering parent event Setup for a wine bottle with multiple labels I have modeled a wine bottle and created various wine label designs for a client. The labels were created in Illustrator and so I have PNGs. I want to present a render of the same bottle with each label the client. The labels are not simple rectangles.   Should I:    "model" each label and wrap it on the bottle, or  try to map each label image to a portion of the bottle's surface?    I've seen lots of tutorials showing option #2, but something is telling me that the glass of the bottle should be separate from the "paper" label.  Also, with #1, couldn't I use the same bottle blend file and use a different layer for the each label, then turn on a different layer for each new render?  Maybe there's an even better strategy?  thanks, How can I show notify-send messages triggered by crontab? Similar Matrices and Linear Transformations Why does adding a curve modifier move my object to the center of the curve? I'm using a curve modifier and an array modifier. My object is at one end of the curve but as soon as I choose the curve for the curve modifier it moves to the center. How can I fix the problem so that I can have the array going from one end of the curve to the other? extended stars-and-bars problem(where the upper limit of the variable is bounded) In my passport, my name is Firstname middlename1 middlename2 lastname, and in the ticket is LASTNAME/FIRSTNAME MR.  Will be a problem? I'll travel from Brazil to Bali, Indonesia and according to the agency I bought the ticket from, there's no problem and it's common.  The airline is Qatar Airlines and I have looked , which says that the passenger name can't be changed after the booking has finished.  But I don't know if it is a problem. Why does:   $$ \sum_{k=0}^{n} k \binom nk p^k (1-p)^{n-k} = np $$ ? Taking the derivative of:   $$ \sum_{k=0}^{n}  \binom nk p^k (1-p)^{n-k} = (1 + [1-P])^n = 1 $$   does not seem useful, since you would get zero. And induction hasn't yet worked for me, since – during the inductive step – I am unable to prove that:   $$ \sum_{k=0}^{n+1} k \binom {n+1}{k} p^k (1-p)^{n+1-k} = (n+1)p  $$   assuming that:   $$ \sum_{k=0}^{n} k \binom {n}{k} p^k (1-p)^{n-k} = np  $$ Proof of Dobinski's formula for Bell numbers Verification that $\lim_{x \to a}\frac{f(x)-f(a)}{x-a}$ and $\lim_{h\to 0}\frac{f(a+h) -f(a)}{h}$ are equivalent definitions of the derivative. I wanted to verify that for definition of the derivative it is true that:  $$ \lim_{x \to a}\frac{f(x)-f(a)}{x-a}= \lim_{h\to 0}\frac{f(a+h) -f(a)}{h}$$  If I denote $h=x-a$,  we can  let $x\to a$, this means that $h \to 0$. Also notice that $f(x)=f(a+h)$ $$ \lim_{x \to a}\frac{f(x)-f(a)}{x-a}=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$$ As desired. Is the main idea here really just to use a variable substitution? I suppose there is some formal theorem I also use here that when $f(x) \to y$  we can let $g(f(x)) \to g(y)$ if the function $g$ is continuous. What is a decoupling capacitor (or smoothing capacitor as referred to in the link below)?  How do I know if I need one and if so, what size and where it needs to go?    mentions many chips needing one between VCC and GND; how do I know if a specific chip is one?   Would an  4-bit parallel access shift register used with an Arduino need one? (To use my current project as an example) Why or why not?  I feel like I'm starting to understand the basics of resistors and some places they're used, what values should be used in said places, etc, and I'd like to understand capacitors at the basic level as well. Very specific paragraph shape Why does the multi-paragraph quotation rule exist? When do you use “learnt” and when “learned”? Is learnt UK English and learned US? Is it that simple?  I’m used to using learnt, but my US spellchecker says it is wrong. Restoring a backup to an older version of SQL Server Cropping Shapefiles in QGIS? It used to be in settings under the friends tab. The friends tab no longer exists! Where do I find that infuriatingly missing setting now? Additive function $T: \mathbb{R} \rightarrow \mathbb{R}$ that is not linear. A function $T:V \rightarrow W$ is additive if $T(x+y) = T(x) + T(y)$ for every $x, y \in V$. Prove that there exists an additive function $T: \mathbb{R} \rightarrow \mathbb{R}$ that is not linear.  My attempt: Let $T$ be the function $T: \mathbb{R}$ (over the field $\mathbb{Q}$) $\rightarrow \mathbb{R}$ (over the field $\mathbb{R}$). The set $\{1, \sqrt{2}\} \subseteq \mathbb{R}$ is linearly independent for the vector space $\mathbb{R}$ over $\mathbb{Q}$. Then, there must exist a linearly independent set $W \subseteq \mathbb{R}$ (over $\mathbb{Q}$) such that $\{1, \sqrt{2}\} \subseteq W \subseteq \text{span}(W) = \mathbb{R}$ (over $\mathbb{Q}$).  I have been told that the function defined as $T(1) = 1$ and $T(w) = 0$ for all $w \in \text{span}(W) \setminus \{1\}$ is additive but not linear, but I cannot see why this is? I can see why it is not linear, clearly $T(\sqrt{2} \cdot 1) = 0$ but $\sqrt{2} T(1) = \sqrt{2}$. But, why is $T$ additive? For example, $T(1+1) = T(2) =0$ but $T(1) + T(1) = 1+1 = 2$? Is there a mistake somewhere? What is this socket/port/connector on my laptop? I've have this socket on my laptop (Lenovo Z570).  What is this port and what is it used for? And what does that symbol mean (I expected a LOCK symbol)? A geometric sequence has its first term equal to $12$ and its fourth term equal to $-96$.   How do I find the common ratio? And find the sum of the first $14$ terms