dnd 5e – Is it possible to cast two level 9 spells without having a long rest in 5e?

Spell Scrollsone

As long as you get enough of them and you have enough time to cast them all, the use of 9th level spell scrolls will allow you to cast many 9th level spells without a long rest.

Epic Boons

The DMG has rules for awarding epic bonuses (starting on page 231) that include:

Blessing of high magic

You gain a level 9 spell slot, as long as you already have one.

Breath of spell

You can cast any spell you know or have prepared without spending a spell slot. Once you do, you will not be able to use this blessing again until a long break is over.

Any of these blessings would allow him to cast Foresight (or another level 9 spell) twice before taking a long rest.


Monsters do not have the same restrictions as player characters and may also have more than one 9th level spell slot. An example of this in published adventures is (spoilers):

Acererak from the tomb of annihilation

1. Suggested by NautArch

Web development: How can I calculate how many developers I need and how long it takes to complete the project?

I am trying to create an LMS system using Angular and NativeScript. It is a LMS web application such as Adobe Connect (although a simplified version) to be written in Angular 8 for example, and a mobile version that has a different user interface and some different features that will be written in NativeScript-Angular, but is Connected to website and both use a back-end. Therefore, the project has 2 front-end interfaces written in Angular and NativeScript-Angular and a back-end that will be written in NodeJS.

As this is my first project (start), and I have a low budget, and web / mobile applications are not my experience (my experience is in the field of AI, but I have some student experiences in mobile / web applications), Need some advice on:

  1. How many front-end developers are needed?
  2. How many backend developers?
  3. Should I do the work in parallel or should I finish in the front at the beginning, then go and do the background part?
  4. How long does it take to do the job? (I need some estimates)

And many other questions that I will try to make in the comments.

It has been a long time

Wow, I just had a bit of nostalgia when reading some old messages. It's been more than 10 years since I posted on Digital Point.

What have all been doing? A lot has changed since I was here for the last time. Not many of you will probably remember me, but hey everyone goes ahead.

Just a trip down the path of memory. Those were the old days of internet marketing. Is there someone still in the game?



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Test verification of sec. 24 "Topology" of Munkres, the long line can not be embedded in real

I'm doing this exercise in the topology of Munkres (~ paraphrasing). I have two motivations for reading this book: 1, the topology is really pretty, but the main one, 2, is that I'm not yet an undergraduate student; therefore, I have never taken any kind of "evidence-based" course and would like to learn how to write legible tests. Appreciate:

  1. A verification of my test.
  2. Any advice on how to make my test more readable.

Remember that $ S _ { Omega} $ Denotes the minimum set ordered, uncountable. Let L denote the ordered set $ S _ { Omega} times[01)$[01)$[01)$[01)$ in the order of the dictionary, with its smallest element removed. The set L is a classic example in topology called Long line.

Theorem: the long line is connected to the route and locally homeomorphic to $ mathbb {R} $, but it can not be embedded in $ mathbb {R} $.

The structure of the test is given by Munkres.

(a) Let X be an ordered set; leave $ a <b <c $ Be X points. Show that $[ac)$[ac)$[ac)$[ac)$ has the order type of $[01)$[01)$[01)$[01)$ Yes, both $[ab)$[ab)$[ab)$[ab)$ Y $[bc)$[bc)$[bc)$[bc)$ have the order type of $[01)$[01)$[01)$[01)$.

Assume $[ab)$[ab)$[ab)$[ab)$ Y $[bc)$[bc)$[bc)$[bc)$ have the order type of $[01)$[01)$[01)$[01)$. Then there are order isomorphisms. $ f:[ab)rightarrow[0frac{1}{2})$[ab)rightarrow[0frac{1}{2})$[ab)rightarrow[0frac{1}{2})$[ab)rightarrow[0frac{1}{2})$ Y $ g:[bc)rightarrow[0frac{1}{2})$[bc)rightarrow[0frac{1}{2})$[bc)rightarrow[0frac{1}{2})$[bc)rightarrow[0frac{1}{2})$, why $[01)$[01)$[01)$[01)$ has the same kind of order that $[01/2)$[01/2)$[01/2)$[01/2)$. Define $ h:[ac)rightarrow[01)$[ac)rightarrow[01)$[ac)rightarrow[01)$[ac)rightarrow[01)$ as follows:

$$ h (x) = begin {cases}
f (x) text {if} x in[a, b) \
g (x) text {if} x in[BC)
end {cases} $$

$ h $ it's an order-isomorphism, like $ p <q $ it implies $ h (p) <h (q) $.

Conversely, it assumes $[ac)$[ac)$[ac)$[ac)$ has the order type of $[01)$[01)$[01)$[01)$. Then there is an order-isomorphism. $ f:[ac)rightarrow[01)$[ac)rightarrow[01)$[ac)rightarrow[01)$[ac)rightarrow[01)$. Restricting the domain to $[ab)$[ab)$[ab)$[ab)$ gives a new order-isomorphism $ g:[ab)rightarrow[0d)$[ab)rightarrow[0d)$[ab)rightarrow[0d)$[ab)rightarrow[0d)$, where $ d in (0.1) $. $[0d)$[0d)$[0d)$[0d)$ is of the order type of $[01)$[01)$[01)$[01)$; multiplying by $ 1 / d $ It is an order of isomorphism. Similarly, restricting the domain of $ f $ to $[bc)$[bc)$[bc)$[bc)$ gives order-isomorphism $ h:[bc)rightarrow[d1)$[bc)rightarrow[d1)$[bc)rightarrow[d1)$[bc)rightarrow[d1)$, where $ d in (0.1) $.

(b) Let X be an ordered set. Leave $ x_0 <x_1 <… $ Be a growing sequence of points of X; suppose $ b = sup {x_i } $. Show that $[x_0b)$[x_0b)$[x_0b)$[x_0b)$ has the order type of $[01)$[01)$[01)$[01)$ if each interval $[x_ix_i+1)$[x_ix_i+1)$[x_ix_i+1)$[x_ix_i+1)$ has the order type of $[01)$[01)$[01)$[01)$.

This is simple, having already tried. $ (a) $. Begin with $ i = 1 $: $[x_0x_1)[x_1b)$[x_0x_1)[x_1b)$[x_0x_1)[x_1b)$[x_0x_1)[x_1b)$ It is a partition of a set in the same way that we saw in $ (a) $; therefore, each must have the same type of order that $[01)$[01)$[01)$[01)$. We proceed by induction, considering now the whole. $[x_1b)$[x_1b)$[x_1b)$[x_1b)$ and the sequence $ x_1 <x_2 <… $ The test is the same.

(c) Leave $ a_0 $ denotes the smallest element of $ S _ { Omega} $. For each element $ a $ of $ S _ { Omega} $ different from $ a_0 $, shows that the interval $[a_0times0atimes0)$[a_0times0atimes0)$[a_0times0atimes0)$[a_0times0atimes0)$ of $ L $ has the order type of $[01)$[01)$[01)$[01)$.

The[[Innuendo: Proceed by transfinite induction. Either has an immediate predecessor in $ S _ { Omega} $, or there is a growing sequence $ a_i $ in $ S _ { Omega} $ with $ a = sup {a_i } $]

How can I order this ?: $[a_0a)$[a_0a)$[a_0a)$[a_0a)$ is accountable by the definition of $ S _ { Omega} $. Following the trail: yes $ a $ it has an immediate predecessor, $ b $, so $[btimes0atimes0)$[btimes0atimes0)$[btimes0atimes0)$[btimes0atimes0)$ has the order type of $[01)$[01)$[01)$[01)$ (All points in this set are of the form $ b times x $, where $ x en[01)$[01)$[01)$[01)$).

By induction backwards. $ a_0 $ any finite section of $ S _ { Omega} $of which $ a $ is the largest element, is the isomorphic order for $[01)$[01)$[01)$[01)$.

Yes $ a $ is not the largest element of a finite section of $ S _ { Omega} $, then there is a growing sequence $ a_i $ in $ S _ { Omega} $ with $ a = sup {a_i } $, this sequence is exactly every element of $ S _ { Omega} $ strictly less than $ a $, in order (which must be countable). Every $ a_ {i + 1} $ In this sequence has an immediate predecessor, namely: $ a_1 $, then each $[a_{i}times0a_{i+1}times0)$[a_{i}times0a_{i+1}times0)$[a_{i}times0a_{i+1}times0)$[a_{i}times0a_{i+1}times0)$ must be isomorphic order to $[01)$[01)$[01)$[01)$ for the previous argument. Then for $ (b) $, then you must $[a_0times0atimes0)$[a_0times0atimes0)$[a_0times0atimes0)$[a_0times0atimes0)$.

(d) Show that $ L $ The road is connected.

Extensible $ (c) $ to include sets of the form $[ab)abenL$[ab)abinL$[ab)abenL$[ab)abinL$ It's trivial – just add / subtract the sets of the ends – I'll assume this. This implies that $[a,b]$ has the same kind of order that $[0,1]$. So between two points $ a, b in L $, there is an order-isomorphism $ f:[0,1] correct arrow [a,b]$; which is necessarily a homeomorphism and therefore a continuous map.

Is that last step valid? How can I make that clearer?

(e) Show that each point of $ L $ It has a homeomorphic neighborhood with an open interval in. $ mathbb {R} $.

Leave $ x $ be our point, assume $ x neq a_0 $. Then there are some points, $ a, b in L $, such that $ a <x <b $. By $ (d) $, $ L $ The path is connected, so there is some homeomorphism. $ f:[c,d] correct arrow [a,b]$ such that $ f (c) = a $, $ f (d) = (b) $. Then, restricting the range of $ f $ to $ (c, d) $ we get a homeomorphism $ f & # 39; 🙁 c, d) rightarrow (a, b) $.

And finally…

(f) Show that L can not be embedded in $ mathbb {R} $or indeed in $ mathbb {R} ^ n $ For any $ n $.

The[[Innuendo: any subspace of $ mathbb {R} ^ n $ It has an accounting basis for its topology.

Suppose there was an imbedding $ f: L rightarrow mathbb {R} ^ n $, therefore a homeomorphism $ f & # 39 ;: L rightarrow Y $, $ Y subset mathbb {R} ^ n $ obtained by restricting the range of $ f $. $ Y $ It must have an accounting basis because $ mathbb {R} ^ n $ is metrizable, so $ L $ It must also have an accounting base for its topology, being homeomorphic with $ Y $.

Assume $ L $ I had an accounting basis, call this base $ U_ {i} $ where $ i in mathbb {Z} ^ {+} $. Each open set of $ L $ It must contain, in its entirety, at least one. $ U_i $. Leave $ x _ { alpha} $ Be the collection of all the points of the form. $ y times 0 $ where $ y in S $ {Omega} $. $ x _ { alpha} $ It is uncountable, being indexed by $ S _ { Omega} $. Then for (e), there is a countless collection $ V _ { alpha} $ of neighborhoods around $ x _ { alpha} $. Keep in mind that we can choose, with more force, $ V _ { alpha} $ in such a way that it is paired-disunion.

That last step feels neglected; Any suggestions?

Then each contains some $ U_ {i} $; So, there is some injective function. $ f: S _ { Omega} rightarrow mathbb {Z} ^ {+} $ defined by $ f ( alpha) = i $ as long as $ U_i subset V _ { alpha} $. This contradicts that $ S _ { Omega} $ it is countless

$ therefore $

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How long do you have to pay the subscription fee of $ 125 per month for Unity pro?

I'm a little confused about how the Unity Pro subscription works. If I had to get Unity Pro, do I still have to pay $ 125 per month while using Unity Pro? Or do I have to continue paying $ 125 per month until a certain period, after which I no longer have to pay anything?

Also, if I developed a game with the personal version of Unity and the game over $ 100K, would that lead to an immediate penalty or will Unity give me a deadline before which I should get Unity Plus / Pro?

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design: what should happen with a prominent application bar with a long title and two lines when scrolling

In the design of materials, if the title is long, we should place it in a prominent application bar like this one:
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The title should not be truncated or shrunk. So when the content moves up, should it transform the prominent application bar to normal? If so, what would happen to the long title, since it should not be truncated or reduced? Or should we let it be this way all the time?

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