Mote of Potential and Magic Inspiration

I wanted to know whether Magical inspiration used for damage triggers Mote of Potential’s Attack:

Magical Inspiration:

If a creature has a Bardic Inspiration die from you and casts a spell that restores hit points or deals damage, the creature can roll that die and choose a target affected by the spell. Add the number rolled as a bonus to the hit points regained or the damage dealt. The Bardic Inspiration die is then lost.

Mote of Potential:

Attack Roll. Immediately after the creature rolls the Bardic Inspiration die to add it to an attack roll against a target, the mote thunderously shatters. The target and each creature of your choice that you can see within 5 feet of it must succeed on a Constitution saving throw against your spell save DC or take thunder damage equal to the number rolled on the Bardic Inspiration die.

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View attachment 261529 View attachment 261530
Domain in Porkbun…

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How do I clearly foreshadow a potential out-of-combat death?

Let me first set the scene

The group I DM for is in a position where they are at the mercy of a group of people who are contemplating what to do with the party.

I have decided on a game of chance to decide their fate (It makes narrative sense), and at a certain point one of the PC’s may do so badly that their fate is to be killed (Again it makes narrative sense).

The particular method of death is that they are going to be pushed off a very high ledge.

While I do plan to give the player chance to do something to avoid this fate (Running, bribing, fighting, etc.) the gist is that if the player doesn’t do anything they will be pushed to their death.

The player may well trust the ‘enemy’ because there is a likelihood that other members of the party have had amusing fates, and be curious to find out their own.

Further to the above:

Some of my players read this site, so I was trying to be vague, but the players are already ‘captive’ in this scene, and due to language barriers the level of communication is minimal. As such I can’t forewarn the party, and I think having the enemy communicate clearly for the first time cheapens the scene.

The PC’s are literally sat at a table playing cards and losing may mean being thrown off a cliff. They don’t know the rules, can’t read the cards, can’t understand much of what is said around them and are 100% out of their element.


  1. Player draws the king > character doesn’t know what it means > NPC’s lead character somewhere > something potentially ominous happens such as being surrounded by NPC’s > character is given his stuff back and set free
  2. Player draws the queen > character doesn’t know what it means > NPC’s lead character to a cliff edge (The ominous happening) > character is pushed over


What techniques are there to ensure that this player knows they face potential death, without just outright telling them?

Basically when they get to the cliff edge I want them to have a good idea that they are going to be pushed off.

Note on answers: I am playing D&D 5e but I am very influenced by ideas from other systems so I am happy to hear about how to successfully pull this off as a GM, regardless of system. I don’t stick to RAW either. – Potential Health Niche Website with Keyword On Domain

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Does this site come with any social media accounts?


How much time does this site take to run?

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with more offpage SEO, you will get more traffics

Scalar potential function from 3D vectorial field

I am new at this forum and a beginner with Mathematica. Today I was studying multivariable calculus and I came with this problem. For example, in 2D case I have this code:

DSolve[{D[f[x, y], x] == y*E^(x*y), D[f[x, y], y] == x*E^(x*y)}, f[x, y], {x, y}] 

where $ \vec{F}=(ye^{xy},xe^{xy})$ and if the vector field were not conservative, DSolve would return unevaluated. But, if my vector field consists of 3 variables, how can I modify my previous code?

For instance, with this vectorial field $ \vec{F}(x,y,z) = (y^2z + 2xz^2, 2xyz, xy^2 + 2x^2z)$ , I tried:

DSolve[{D[f[x, y, z], x] == y^2 x + 2 x z^2, D[f[x, y, z], y] == 2 x y z, D[f[x, y, z], z] == x y^2 + 2 x^2 z}, f[x, y, z], {x, y, z}] 

but it doesn’t work. Thanks for your help.

Using Solve[] to find Eigenstates of a 1D Double Dirac Potential

I’d like to Solve

$ $ k^2 \equiv – \frac{2mE}{\hbar^2} = (- \frac{mA}{\hbar^2} (1+ e^{-2ka}))^2 $ $

for E, in terms of m, $ \hbar$ , A, a.

I tried using the following command:

Solve[-((2 m ene)/h^2) == (m^2 A^2)/h^4 (1 + E^(-2 a*Sqrt[-((2 m ene)/h^2)])), ene] 

Isn’t working well for this task. What do you recommend? At first glance it seems it could not be simple to solve "by hand".

Background: This problem comes from Solving a 1D Quantum well with 2 Symmetric Dirac’ Deltas $ \delta_a$ and $ \delta_{-a}$ , where $ A$ is the amplitude. – potential for a multi million dollar auto pilot website

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Potential vulnerability in JSON response returning base 64 encoded image data, with the response being vulnerable to MIME sniffing

A JSON response in the API of a webapp is returning the base64 of a user-uploaded image, and there’s no X-Content-Type-Options Header to prevent MIME sniffing.

Could this be a potential vulnerability such as an XSS for the webapp by using steganography to edit the image with a payload, uploading it, and then MIME sniffing the JSON response? (or by any other means?)

Potential Example of an algorithm failure

I was looking an algorithm to solve a problem of finding whether and array contains a quadruple with sum = k,(k is input) mentioned at GeeksforGeeks. In one solution the approach mentioned is below,which i think i not correct:

  1. Store all possible two elements pairs of array in some another auxiliary array and also store i,j indices of the elements.There will be $ C(n,2)$ pairs.
  2. Sort the auxiliary array.
  3. Take two pointers,left=0 and right=n-1 to auxiliary array.

3.1. If auxiliary[left]+auxiliary[right]==k and there is no common index in elements represented by left and right, we return true.

3.2 else If auxiliary[left]+auxiliary[right]<k, we do left++

3.3 else we do right– 3.4 And this goes in continues loop until left<right or we get element at 3.1

Now my doubt is that in case sum of left and right elements=k and there is some common element between left and right,then this algo will do right–. I want to understand that why we do right– here and no left++ and the other way.I think there must be some example where this scenario will make algorithm fail but i am not able to produce that.Can some tell that whether it will indeed fail or prove that it is always correct?