Which of the following two sentences is correct:  Jon (and Michael) is smart. Jon (and Michael) are smart.   This example is clearly contrived but my question is whether the sentence should be grammatically correct when you erase the text in parentheses (Jon is smart) or should it be grammatically correct when read with the text in parentheses (Jon and Michael are smart).
If a set of parentheses lies between a subject and its verb, and the parentheses contain an substitutive subject whose singularity/plurality disagrees with the original subject, whose singularity/plurality should be chosen for the verb?  In other words, in the following example, should "questions" (and its verb "are") be singular, or should they remain plural as shown?     Many (if not every) questions on this StackExchange are answered.   My intuition tells me that the two words in question should remain in plural forms, since the text in parentheses only interrupts the sentence (and the sentence would be grammatically incorrect if everything in parentheses were removed and the words were in singular form).  On the other hand, when read aloud (assuming one reads the text in parentheses), this has an uncomfortable sound to it, and I've seen others write in what would be the above example's singular-form case, so I'm curious to find out which is correct.  And, thinking about it, I suppose the same question would apply when commas are used in place of parentheses.
What happens if you use Polyjuice potion after saving hair or nails? Suppose you cut a five year old's hair, save it for 10 years, and then use it for a Polyjuice transformation. What age do you turn into: The toddler or the teenager? Limit of derivative is zero. Does it imply a limit for f(x)? How to select the most recent set of dated records from a mysql table Does something like  exist for topological (eg. braided) circuits?   Alternatively, any ideas on how @ is getting  (or something similar)? Why can't I destroy or place blocks? I'm playing on a Survival world with cheats on, "easy" difficulty, and at night in-game. I didn't have a sword or anything to make a sword with, and I was starving, so I switched to Creative to give myself a stack of food and a bed to set my spawn point. When I switched back to Survival, I couldn't destroy or place blocks. I'm on single player, and my Internet is fine, and I can destroy blocks still if I switch to creative. I've tried resetting the button one, and reloading the map, and it still doesn't work. I've even tried deleting everything I gave myself in Creative. This happened before on a different world and I ended up deleting it.  This only happens on this world. How do I fix it?  If I know I'm in the correct game mode, what else can cause this?  Are there different causes in multiplayer? How do I convert all files in a folder to a different line ending? (on Windows) I don't want to watch the prisoner cart ride to the execution: is it possible to skip this?   If so, how? Can I replace the intro movie file by some empty video? If $0<a<b$, prove that $a<\sqrt{ab}<\frac{a+b}{2}<b$ I get this error message   NMI watchdog: BUG: soft lockup - CPU#2 stuck for 23s! [plymouthd:305]    on the startup of my system (even with live-cd).  My PC is a dell inspiron 7559 with intel graphic and nvidia.  UPDATE - this issue also happens when attempting to install ubuntu 18.10 using LiveUSB Pass a PHP string to a JavaScript variable (and escape newlines) How do I sort custom objects alphabetically in an array in Java? Proving the inequality $\frac{n^k}{k^k}\leq\binom{n}{k}\leq\frac{n^k}{k!}$ involving binomial coefficient. What is the best way of proving inequalities including binomial coefficients in general? Lets look at this example:     $$\frac{n^k}{k^k}\leq\binom{n}{k}\leq\frac{n^k}{k!}$$   What I started with is:  $$\binom{n}{k}\frac{1}{n^k}\leq\frac{1}{k!}$$ $$\frac{n!}{k!(n-k)!}\frac{1}{n^k}\leq\frac{1}{k!}$$ $$\frac{(n-1)!}{(n-k)!n^{k-1}}\leq1$$ This is where I got stuck. How can I prove that is is lower than or equal to 1? Could you give me a hint? If $z_{1},z_{2},z_{3}$ are three complex number Such that $|z_{1}| = 2\;\;,|z_{2}| = 3\;\;, |z_{3}| = 4\;\;$ Then $\max$ of $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{1}|^2$  $\bf{My\; Try::}$ Let $z_{1}=2\left(\cos \alpha+i\sin \alpha\right)$ and $z_{2}=3\left(\cos \beta+i\sin \beta\right)$ and $z_{3}=4\left(\cos \gamma+i\sin \gamma\right)$ So $$f(\alpha,\beta,\gamma) = 58-\left[12\cos(\alpha-\beta)+24\cos(\beta-\gamma)+16\cos(\gamma-\alpha)\right]$$ Now How can I calculate $\max$ of $f(\alpha,\beta,\gamma)$ Help me Thanks Not getting email alerts on answers on Stack Overflow Not getting email alerts on answers on my questions on stackoverflow.com   Used to get it earlier (>1 month back), but not getting alerts in recent times Email field in profile is filled  In prefs, "I'd like email notifications of activity on questions I've asked" is ticked.   What is missing?  Is it possible that signing up on another Stack Exchange community (say, Server Fault), on which by default, "I'd like email notifications of activity on questions I've asked" is unchecked affects same preference on stackoverflow.com? If so, it is a bug. TypeError: 'str' does not support the buffer interface If I set multiple Alarms, can I tell them apart? Proving Pascal's Rule : ${{n} \choose {r}}={{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$ I am an Indian citizen studying in Italy on an Italian visa that expires 19 November, 2015. I have permesso di soggiorno. I want to visit the United Kingdom for my vacations for a week during March. I know the procedures already, however I am a little bit confused about the amount that I must show as a proof of support with my bank statements. I will stay for a week with my friends who are also students in Italy and will be in London for 5 days. What would be the amount in Euros that I need to show for my visit to UK? equality of two operators... Please help me with the following problem( give some hints or references):  Let $X$ be a Banach space and $B(X)$ be the algebra of bounded linear operators on $X$. Suppose that $A$ and $B$ are two operators in $B(X)$ such that for every $T \in B(X)$ we have $\sigma(AT)=\sigma(BT)$. Show that $A=B$.  Here $\sigma(A)$ denotes the spectrum of $A$, i.e., the the set of all complex numbers $\lambda$ such that $\lambda I − A$ has no inverse in $B(X)$.  Thanks in advance. How to use both binary and continuous variables together in clustering? I need to use binary variables (values 0 &amp; 1) in k-means. But k-means only works with continuous variables. I know some people still use these binary variables in k-means ignoring the fact that k-means is only designed for continuous variables. This is unacceptable to me.  Questions:     So what is the statistically / mathematically correct way of using binary variables in k-means / hierarchical clustering? How to implement the solution in SAS / R?