If my understanding is correct, we create antibodies after vaccines and after getting infections. So, after 10 vaccines and 2 contracted diseases, I can have in my body antibodies for 12 types of diseases.  Is there a limit to the number of different antibody types that we can have in the body at one point?   Also, do we spend resources keeping those numbers up when the antibodies are not used?  Thanks.
Before the seasonal flu vaccine was recommended for all healthy adults, people in developed countries might get vaccinated against on the order of 10 different pathogens and develop immunity for some more via natural exposure.  Now the same person is expected to take a seasonal flu vaccine for a new strain every year - won't this put them in an uncharted territory as far as effects on the immune systems are concerned within a couple of decades?
Asking users to accept answer when they comment that it answers their question terminal won't launch after install python 3.5 After I installed Python 3.5 and ran the commands below, my terminal and many other programs won't launch any more.  cd /usr/bin sudo rm python3 sudo ln -s python3.5 python3    How can I restore the default Python version again to fix these problems? Would someone give a thorough explanation of 1.14 Minecraft's /data modify? Data defined label position for more than one scale in one layer I'm looking for a better labeling-solution for my projects. It seems to be difficult - but maybe I'm just blind?   My projects: I build  printed bike-hike-books. The main scale is 1:75k and there are a city scale of 10k and a overview scale of 200k. Because I like to have a "cartographic" quality of labels placements, I touch every single label and move it to it's best position. I have to do this for every single scale so my labeling columns for 3 scales look like this ('scale 1' stands for 10k 'scale 2' for 75k  and 'scale 3' for 200k):    But before I try to explain in detail, I build this simple Example-Project where you can see and test my (old) labeling solution for 3 scales:   Well, it is working but:    You need labeling columns for every scale  You need a layer for every scale   I'm looking for a much better solution without duplicate layers and tons of columns. I count on your creative minds!  EDIT: I added this two very dense map as examples:    There is only one single possible position for "St. Paulus Dom" and it's not any quadrant:   Every label has to be moved manually. Some of the labels are not in a quadrant. The bar at the very bottom of the screen has gone away? Prove $n\mid \phi(2^n-1)$ If $2^p-1$ is a prime, (thus $p$ is a prime, too) then $p\mid 2^p-2=\phi(2^p-1).$  But I find $n\mid \phi(2^n-1)$ is always hold, no matter what $n$ is. Such as $4\mid \phi(2^4-1)=8.$  If we denote $a_n=\dfrac{\phi(2^n-1)}{n}$, then $a_n$ is , but how to prove it is always integer?  Thanks in advance! EDIT: Summary  I think I've found the best definition of what I want thanks to the discussion with @M. Tibbits:  When I open or refresh the SO, I get a notification of new activities that are of interest to me. Today these are (hard-coded): answers to my questions and comments addressed to me. I want a bit more control over it, that's all.     Yes, I've read  and , but I think I add new (good :) ) reasons for re-asking it:  The first question was answered before this so handy "inbox" notification feature was implemented. So now using favorites for "watching" a question is wrong, IMHO, for 2 reasons:   it splits essentially the same functionality (of getting notified when there is an activity you are interested in) between 2 very different user interfaces it's cumbersome - I need to: 1) add question to favorites; 2) click on my profile dropdown; 3) click on the "favorites" line; 4) see all my favorites with recent activity; 5) select the one I watch; 6) repeat this process occasionally to get updates; and 7) possibly remove the question from favorites at the end. All this as opposed to this little red button when there are news.   Favorites are different in nature that watching. I add a question to Favorites when I find the answers useful to me in my work, so I may want to have it handy in the future. By watching an interesting question I will be able to decide whether I want it in faviorites, after I see the answers.  EDIT: I see that I was not clear enough.  I want to add a question to Favorites when:   it has some good answers the question and answers are relevant in my work I want to keep them handy as my long-term (persistent) knowledge base   I want to watch a question for short while (matter of days/weeks) when:   it is interesting / relevant I want to follow it to see what are the answers. Then I can decide whether I want it in my favorites   As you see, the main differentiators are good answers and long/short term. For short-term, automatic notification is better.  Bottom line, all I ask is "watch this question" checkbox (which may expire after say a week if there is a need to free resources) How to get UTF-8 working in Java webapps? What is the use of a persistence layer in any application? Prove that $g$ is tangent to the graph of $f$ Hahn-Banach Theorem for separable spaces without Zorn's Lemma Let us consider that $u\in C^2(\Omega)\cap C^0(\Omega)$ and satisfying the following equation .  $\Delta u=u^3-u , x\in\Omega$ and   $u=0 $   in  $\partial \Omega$  $\Omega \subset \mathbb R^n$ and bounded .   I need hints to find out what possible value $u$ can take ?  Thank you for ur hints in advance . Let $S$ be the set of $3 \times 3$ matrices $\rm A$ with integer entries such that $$\rm AA^{\top} = I_3$$ What is $|S|$ (cardinality of $S$)?   The answer is supposed to be 48. Here is my proof and I wish to know if it is correct.  So, I am going to exploit the fact that the matrix A in a set will be orthognal, so if the matrix is of the form \begin{bmatrix}         a_{11} &amp; a_{12} &amp; a_{13} \\         a_{21} &amp; a_{22} &amp; a_{23} \\         a_{31} &amp; a_{32} &amp; a_{33} \\         \end{bmatrix}  Then each column and row will have exactly one non-zero element which will be +1 or -1. Thus, I have split possibilities for the first column into three cases and counted the possibilities in each case as follows :- $$a_{11} \neq 0$$ or $$ a_{21} \neq 0$$ or $$ a_{31} \neq 0$$  In case 1), we obviously have two possibilities(+1 or -1) so we consider the one where the entry is +1. Now, notice that the moment we choose the next non-zero entry, all the places for non-zero entries will be decided because of the rule 'each column and row will have exactly one non-zero element'. Meaning, if b and c are remaining two non-zero entries, we only have two possibilities left  \begin{bmatrix}         1 &amp; 0 &amp; 0 \\         0 &amp; b &amp; 0 \\         0 &amp; 0 &amp; c \\         \end{bmatrix}  or  \begin{bmatrix}         1 &amp; 0 &amp; 0 \\         0 &amp; 0 &amp; c \\         0 &amp; b &amp; 0 \\         \end{bmatrix}  Using the fact that b and c are simply $$\pm1$$  In each of the above matrices, we get 4 possibilities for each of the matricies. Thus, 8 possibilities in totality. Basically, we are getting 8 possibilities on the assumption that  $$a_{11} = 1$$  Thus, we get 16 possibilities on the case that $$a_{11} \neq 0$$  Following, the second and third cases analogously, we get a total of 16 possibilities in each of them and 48 possibilities in total.    Source :- Tata Institute of Fundamental Research Graduate School Admissions 2016 How do I permanently allow a website to access my location? In iOS, I only need to be asked once whether I want to allow a website to share my location data. Is it possible to permanently do this in OS X version of Safari as well? On proving that a finitely generated group has a finite number of subgroups with index $n$. Theorem: If $G$ is a finitely generated group, then it has a finite (maybe zero) number of subgroups of index $n$ for any $n\in \mathbb{N}$.   is a sketch of a proof. It consists of assuming such a subgroup $H$ exists, and allowing $G$ to act on $G/H$ by left multiplication. Such action defines a homomorphism $$\phi :G \rightarrow \text{Sym}(G/H)\cong \mathbb{S}_n$$  If $G=\langle g_1, \ldots ,g_r\rangle$, then the homomorphism is uniquely determined by where it sends the generators $g_i$. In particular, there are at most $(n!)^r$ such homomorphisms.  The last step of the proof requires one to notice that a different subgroup $H'$ of index $n$ in $G$ would produce a different homomorphism, and here is where I'm struggling. I've been able to show that $\text{ker}(\phi )=\text{core}(H)$, yet haven't been able to rule out the possibility of $\text{core}(H)=\text{core}(H')$. Thor: Ragnarok ends with the remaining people of Asgard aboard a refugee ship that meets Thanos' ship. Avengers: Infinity War opens with      Thanos and his minions having commandeered the ship and several bodies can be seen on the floor.   This begs the question, what happened to these characters:   Korg Miek Asguardian extra #3 Valkyrie Problem:  $F = \frac{(G m_1 m_2)}{r^2}$  An object of unknown mass is placed directly between the Earth and the Moon such that the forces of gravity acting on the object are equal from both the Earth and the Moon. The distance from the Moon to the object is exactly one-tenth of the distance from the Moon to Earth. Using the above formula, determine the distance from the Moon to the Earth.  Given is:  The factor $G≈6.67∙10−11\mathbf{N(\frac{m^2}{kg^2})}$ is a universal constant.  Mass of Earth $≈5.972∙10^{24}\mathbf{kg}$   Mass of the Moon as $≈7.348∙10^{22}\mathbf{kg}$  Mass of unknown object = $m$  My solution:  $F = \frac{(6.67 *10^-11) m (7.348*10^{24})}{R^2}$  $F = \frac{(6.67 *10^-11) (5.92* 10^22) m}{10R^2}$  As gravitational force acting on the object is equal   $\frac{(5.92\cdot 10^22)}{100r^2} = \frac{(7.348*10^{24})}{r^2}$  $R = 1.24 \cdot 10^4 KM$  Is this a viable solution to the problem? Please provide feedback. Thank you! Why does range(start, end) not include end? Show/Hide CSS Ignored When Inside an SLDS Grid Column Can I boil milk unattended (and still keep the pan clean)?