Can warlock spell slots be used to cast above warlock casting level when multiclassing? One of my players is considering multiclassing warlock/cleric. I have not played warlocks or had other players use them so I am looking into the casting mechanics. After looking at other threads I think I have the idea down. Something like a warlock 1/cleric 5 would have the cantrips and 1 warlock slot that can cast at first level and the cleric would provide up to 3rd level spell slots. my confusion comes from  Pact Magic: If you have both the spellcasting class feature and the Pact Magic class feature from the warlock class, you can use the spell slots you gain from the Pact Magic feature to cast spells you know or have prepared from classes with the spellcasting class feature, and you can use the spell slots you gain from the spellcasting class feature to cast warlock spells you know.  so they would be able to cast warlock spells up to 3rd level with the cleric 3rd level slot, as evident from answers to , but would they also be able to cast fireball (in the case of a light domain cleric) using their warlock slot but doing so at 1st level? If so how would this translate to an end game build? would the cleric then have the equivalent to 2 9th level slots? I am assuming this does not work this way and the initial build listed essentially just has a slightly expanded spell list and an extra 1st level slot that comes back on a short rest, but I just want to make sure I am not missing something so I can explain this fully in the future.
Can I cast a Warlock spell with another (higher) spell slot from a different multiclass? Reviewing an answer to a recent question I posted, I had a thought about Armor of Agathys. In the event of a Sorcerer 18 / Warlock 2 build, would the character be able to cast Armor of Agathys with a 9th level spell slot, thus getting 45 Temporary Hit Points and dealing 45 damage when hit?   Is the spell in any way limited to being cast using Warlock spell slots or to a maximum of a 5th level spell slot?
What functionality do 'marks' offer in the El Capitan Terminal? Since upgrading to OS X El Capitan, I've noticed a change in the Terminal: Executed lines show an opening bracket on the left, and a closing bracket on the right, as shown in this screenshot:    Similar questions have been asked here:          Some of the questions are referring to these brackets as marks, and the Terminal's Edit menu has a couple of entries related to marks, for example:   Edit > Marks > Mark as Prompt and Send Return Edit > Clear to Previous Mark Edit > Navigate > Jump to Previous Mark     Is there a summary of what these marks can be used for? I have not seen any reference to this new feature in any of the El Capitan feature overviews. What should I opt for?   Option A) very well paid PhD, around €3000 per month at a respectable university or;  Option B) a non-paid PhD, where I have to pay €2000 per year at one of the top 10 universities of the world?  EDIT: Actually they are both UK universities, Option B  is ranked between the top 10 in the world in engineering and tech, and I do get a scholarship that pays 18000 pounds which covers the difference between international student tuition and EU tuition. In Option A, I will be doing research in a good university and will be working alongside giants in the industry. I think my son was trying to guess the passcode! What do I do?  I can't wait 42 years! This question is from PDE by Evans, 1st edition, Chapter 5, Problem 14. It has been posted  previously, however, I cannot quite put all the information together from the responses there. Hopefully you can help me now. The problem is as follows:     Let $U$ be bounded with a $C^1$ boundary. Show that a ''typical''   function $u \in L^p(U) \ (1 \leq p &lt; \infty)$ does not have a trace on   $\partial U$. More precisely, prove there does not exist a bounded   linear operator      \begin{equation} T:L^p(U) \to L^p(\partial U) \end{equation}      such that $Tu = \left. u \right|_{\partial U}$ whenever $u \in C(\overline{U}) \cap L^p(U)$.   So what we're trying to find is a bounded sequence of functions that go to infinity at the boundary, correct? Because then $Tu = \left. u \right|_{\partial U}$ is undefined.  I like the idea of using the $\mathrm{dist}(x, \partial U)$ function to do this. I was thinking about defining the function  \begin{equation} u(x) = \frac{1}{\mathrm{dist}(x,\partial U)}, \end{equation}  which must be in $L^p(U)$ since $U$ is bounded (hence integral of $\varepsilon$ over $U$ is finite, right?). But this function is not continuous at the boundary, and so $u \notin C(\overline{U})$. Can this be modified somehow?  Another option is to use that the boundary is $C^1$. So for each $x^0 \in \partial U$ there exists an $r &gt; 0$ and a $C^1$ function $\gamma : \mathbb{R}^{n-1} \to \mathbb{R}$ such that $U \cap B(x^0,r) = \{ x \in B(x^0,r) \ | \ x_n &gt; \gamma(x_1,\ldots, x_n) \}$. So around every point on the boundary we can find a ball where the n'th coordinate is greater than the $C^1$ function $\gamma$. How can we use this? pgfplots legend style for plots with error bands Does a card have a keyword if it has the same effect as said keyword? I was looking at  and found the effect surprisingly similar to the keyword Surveil. To compare, first the wording of Contingency Plan and then the wording of the Surveil keyword:       Look at the top five cards of your library. Put any number of them into your graveyard and the rest back on top of your library in any order.     And       701.41a To "surveil N" means to look at the top N cards of your library, then put any number of them into your graveyard and the rest on top of your library in any order.     They quite clearly are the same, so can I assume the card states "Surveil 5" instead? And if so, does it trigger abilities which state "Whenever you surveil ..."? Why do Windows 10 foreground apps mysteriously launch as background processes? Let $A$ be an infinite set. For each $a \in A$, let $x_a$ be a non-negative number. Let the value of the series $\sum_{a \in A} x_a$ be the value of the supremum of its finite partial sums, i.e.:      $$\sum_{a \in A} x_a = \sup_{n \geq 0} \sup_{(a_1, \ldots , a_n) \subset A} \left(x_{a_1} + x_{a_1} + \ldots + x_{a_n}\right)$$   (Supremum is well defined even when its value is infinite)       If $\sum_{a \in A} {x_a}$ is finite, prove that the set $A' = \left\{{a \in A | a &gt; 0}\right\}$   is countable.   My guess is that we have to write $A$ as a union of finitely many sets. Is it appropriate to remove excessive commenting from code? I am referring to this question on SO , where there is a massive amount of commenting within the code (the question has since been edited to correct this).  Is it allowed to remove these for the sake for readability (the comments just specify some lesson details, method comments etc)? How does substituting Butter for Margarine/Shortening affect the Recipe I know that Margarine / Shortening are usually softer than butter, but what other differences are there? Can I just melt the butter and call it good or will this effect my final Product? I have seen Margarine / Shortening a lot in cookies, but I am looking for a general answer. I'm trying to run an interactive command through paramiko.  The cmd execution tries to prompt for a password but I do not know how to supply the password through paramiko's exec_command and the execution hangs.  Is there a way to send values to the terminal if a cmd execution expects input interactively?  ssh = paramiko.SSHClient() ssh.connect(server, username=username, password=password) ssh_stdin, ssh_stdout, ssh_stderr = ssh.exec_command("psql -U factory -d factory -f /tmp/data.sql")   Does anyone know how this can addressed? Thank you. Limits, find a limit that exist in absolute value but not outside the absolute value. Showing that $f_0 (x_1, \ldots, x_m) \mathrm tr A =\displaystyle{ \sum_{i=1}^n} f_0(x_1, \ldots, Ax_i,\ldots, x_m)$ Question:  Consider $f: (-\epsilon, \epsilon) \to \mathbb R^{m^2}$ a differentiable path of matrices $m \times m$ such that $f(0) = I_m$ and the function $g: I \to \mathbb R$ is defined by $$g(t) = \det f(t)$$  Show that $g'(0) = \mathrm{tr} \,a $ (trace of the matrix $a$), where $a = f'(0)$.   Attempt:  First we use that $\det f(t) = \det (f_1(t), \ldots, f_m(t)) $, where $f_i$ is the $i$-th row of the matrix $f(t)$ $m \times m$. Considering the operator $': \mathbb R^m \to \mathbb R^m$, $f(t) \mapsto f'(t)$  With this in mind $$\begin{align}g'(t) = \det'f(t) &amp;= \sum_{i=1}^{m} \det (f_1(t),\ldots,f'_i(t),\ldots, f_m(t)) \\&amp;\underbrace{=}_{(*)}\det(f_1(t), \ldots, f_m(t))\, \mathrm tr f'(t) \end{align}$$  For $t = 0$ we have   $$\begin{align}g'(0) &amp;=\det(e_1, \ldots, e_m) \,\mathrm tr f'(0) \\&amp;= 1 \dot\, \mathrm tr \,a \\&amp; = \mathrm tr \, a\end{align}$$  I'm trying to come up with a proof using the fact that $\det$ is a $m$-linear form.   In order to prove $(*)$ I need the following equality, which according to this  holds.   If $A: V \to V$ is a linear operator and $f_0$ is a alternated $m$-linear form, such that $f_0 (e_1, \ldots, e_m) = 1$ then   $$f_0 (x_1, \ldots, x_m) \mathrm tr A = \sum_{i=1}^m f_0(x_1, \ldots, Ax_i,\ldots, x_m) $$  Could someone give a proof to this? Dynamic date range query I'm trying to create two cells that will update my query: A >= date"A1" and A &lt;= date "B1".  My goal is to create a search page for the data, where the end user could insert a date or the =today() function and it would update the query. I am currently getting a formula parse error from:  =QUERY(Data!6:1397,"SELECT A,B,C,D,E,F,G,H,I,J,K,L WHERE A &gt;= date"A1"and  A &lt;= date"B1"") I have to prove that if $E$ is a connected space, then so is $\overline{E}$ (the closure of $E$) a connected space. I tried to prove the contrapositive. So suppose that $\overline{E}$ is not connected in a metric space $X$. This implies there are $A\subset X$ and $B\subset X$ such that $A\cap B=\emptyset$, $A\cap \overline{E}\neq \emptyset$, $B\cap \overline{E}\neq \emptyset$ and $\overline{E}\subset A\cup B$. I will try to prove that this $A$ and $B$ will also work for $E$. Suppose that $A\cap E=\emptyset$ and $B\cap E=\emptyset$, this implies that $A\cup B$ is a subset of the limit points of $E$, because $A\cap \overline{E}\neq \emptyset$ and $B\cap \overline{E}\neq \emptyset$. This contradicts the fact that $\overline{E}\subset A\cup B$, so we have to conclude that $A\cap E\neq \emptyset\neq B\cap E$. Since $E\subset\overline{E}\subset A\cup B$, we have that $E$ is also not connected.  My question is whether my proof is correct, because I have a little bit doubt. If it's not correct, what is the best I can do? Thanks in advance!   Edit1: The definition we use for a connected space $E$ in a metric space $X$ is that if $E$ is connected, then there are no open $A,B\subset X$ such that $A\cap B=\emptyset$, $A\cap E\neq \emptyset\neq B\cap E$ and $E\subset A\cup B$.  Edit2: $A$ and $B$ must be open. De Rham Cohomology of the tangent bundle of a manifold I would like to compute the de Rham cohomology of the tangent bundle $TM$ of a manifold $M$. It seems to me that we can just homotopy each fibre to a point, and that this would give a homotopy equivalence between $TM$ and $M$. Since de Rham cohomology is a homotopy invariant, this would imply that $H^*_{\text{dR}}(TM) \cong H^*_{\text{dR}}(M)$.     Is it always true that the tangent bundle of a manifold is homotopic to the manifold itself by the homotopy of each fibre with a point? Is a visa to Poland required for a Russian spouse of a Polish citizen? I am a holder of a Russian passport and my husband is a Polish citizen. We both live and work in the UK. I have received my family member of EEA national card, good for 5 years. We would like to travel to Poland together. On the website of the Polish embassy it clearly states that if we are traveling together, I do not require a visa, however in the embassy they were sure that I do need a visa. Could anyone please clarify? Can not find the tag library descriptor for “http://java.sun.com/jsp/jstl/core” How can I transform the elliptic curve $E/\mathbb{C}$ of the form $$y^2=4(x-e_1)(x-e_2)(x-e_3)$$ with $e_1&gt;e_2&gt;e_3\in\mathbb{R}$ roots of $E$ into a Weierstrass-Form $y^2=x^3+ax+b$? How many spell slots does a Level 3 Warlock have? The Warlock Table in the PHB (p106) states that a Warlock has the following:   1 First-level spell slot at level 1. 2 First-level spell slots at level 2. 2 Second-level spell slots at level 3.   Does this mean that a third level Warlock has a cumulative 5 Spell Slots? (3 First-Level slots and 2 Second-Level slots)  Or do the new slots just replace old slots between levels such that a 3rd-level Warlock ultimately has only 2 second-level spell slots?