I am new to multiclassing and new to spells with 5e. Currently I am a Level 3 Rogue (Arcane Trickster) and just leveled up to level 4. For my fourth level, I was going to put a level into the wizard class as it makes sense for my backstory. However, I'm confused by the calculation for determining what spell slots I have. According to the  rules in the PHB (p. 164-165), to determine what spell slots I have, I would take my rogue level divided by 3 and then rounded down (1), and add it to my number of levels in the wizard class (1), for a total of 2. Then looking at the 2nd row of the  table on page 165, it would seem that I'd have three 1st-level spell slots. But that doesn't make sense to me, as I originally had two 1st-level spell slots as an Arcane Trickster rogue, and a 1st-level wizard has two 1st-level spell slots. Would a multiclassed Arcane Trickster rogue 3 / wizard 1 have only three 1st-level spell slots? Or would they have four?
How do I determine how many spell slots I have if I multiclass into more than one spellcasting class? For example, I have a character that is a level 3 Wizard, a level 5 Eldritch Knight Fighter, a level 4 Arcane Trickster Rogue, a level 4 Paladin, and a 4 Warlock. How many spell slots do I have and how do I determine this?
Using [Authorize] without SimpleMembershipProvider Can you take different archetype options at different levels? Once you've picked a subclass option at level 3, could you later take a different subclass option, or do you lock into a specialty at level 3 and then run with it for the rest of that class' progression?  For example, as a rogue, if you took the "Thief" archetype at level 3, giving you access to Fast Hands and Second Story Work, could you then take the "Assassin" archetype to give you the specific benefits that archetype gives instead of the "Thief" ones?  The  seems to imply that you couldn't:     At 3rd level, you choose an archetype that you emulate in the exercise of your rogue abilities: Thief, detailed at the end of the class description, or one from another source. Your archetype choice grants you features at 3rd level and then again at 9th, 13th, and 17th level.   but I'd like a more certain answer, especially one that refers to archetypes in general. Also I'm not ashamed to admit I'd prefer a source with some authority that tells me you can mix and match, but I am looking for the truth of the rules when all is said and done...   Different to  because the no on that demonstrates that you can't re-take a level of the same class. Taking different features for different levels is not the same thing, and would work differently. Similar to  but that one was specific to monks - which is why it didn't show up on any searches I did, and why it wouldn't on the searches of others. This question is specific enough to be clear and precise, but broad enough to be useful to others who aren't in the exact same situation/using the same class. This plant was found growing in Germany. I have never seen it before, reverse Google image search did not help either. Can you help me identify what type of plant this is?      Thanks, looks like it is something like this:    Related question: Specific system of differential equations I have the following system of equations:  \begin{eqnarray}\frac{dx}{dt} = x(1 - x^2 - y^2) \\ \frac{dy}{dt} = y(4 - x^2 - y^2) \end{eqnarray}  I want to prove that if a solutions starts (at time $t = 0$) in the region   $$ G = \{(x, y) \in \mathbb{R}^2 : x &gt; 0, y &gt; 0 \text{ and } 1 &lt; x^2 + y^2  &lt; 4\} $$  that it remains there for all $t \geq 0$. I have already proven that, from $G$, a solution cannot cross the $x$-axis where $0 &lt; x &lt; 2$ or the $y$-axis where $0 &lt; y &lt; 2$ (since those line segments are orbits), but the pieces of the circles with radius 1 and 2 (where $x &gt; 0, y &gt; 0$) are the lines I haven't done yet. In this question we can focus on just the circle with radius 1 since the proof for the other circle is completely analogous to it.  In my book, there is a similar problem where they show that if a solution crosses a certain line, that solution attains its maximum or minimum value for $x$ and $y$, and then they derive a contradiction. However, in this case, for $n \geq 1$ the $n$-th derivative of $x$ equals 0 on the circle with radius 1. (If the second derivative were negative, $x$ would attain its maximum on the circle, a contradiction since $\frac{dx}{dt} &lt; 0$ in $G$, so $x$ is decreasing.)  I don't see why a solution can't cross the circle at all. On the circle, $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} &gt; 0$, so after crossing the circle, the solution would go back up to $G$.  Is the circle itself an orbit?  EDIT I think I have found the answer. Using polar coordinates, I found $$ \frac{dr}{dt} = r - r^3 + 3r \sin^2(\theta) $$ Now, $$ \frac{dr}{dt} &gt; 0 \Leftrightarrow 0 &lt; r - r^3 + 3r \sin^2(\theta) \leq r - r^3 + 3r = 4r - r^3 $$ since $0 \leq \sin^2(\theta) \leq 1$. 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