## dnd 5e – Would it be balanced to remove the level cap on my homebrew Water Jet elemental discipline for the Wot4E monk?

I have homebrewed various additional elemental disciplines for the Way of the Four Elements monk subclass. Although many of those are just adding more spells (erupting earth, tidal wave, etc), and others are just reskins of existing disciplines (such as fist of unbroken air, water whip, etc), one unique discipline that I came up with was this:

Water Jet (11th level required). You can spend 2 ki points as an action to unleash a jet of water in a line that is 30 feet long and 5 feet wide. Each creature in the line must make a Strength saving throw, taking bludgeoning damage equal to your Martial Arts die + your Wisdom modifier on a failed save, or half as much on a successful one. In addition, each target that fails its saving throw is pushed up to 20 feet away from you.

However, I wrote this a while ago, and I can’t remember why I decided that it should cost 2 ki (which is typically what the disciplines you have access to at 3rd level cost) but also be capped at 11th level or above.

Intuitively, this feels like it would be fine as-is for a 3rd level Wot4E monk to have. Either that, or if it is to have a level cap, the ki cost should match that of the other disciplines capped at that level (e.g. 4 ki points for an 11th level discipline, not 2 ki points).

Compared with the RAW elemental disciplines from the Player’s Handbook, would this discipline be balanced if I removed the level requirement?

## Outdoor fireplace with leather chairs, with snow cap mountains behind — which golf course in Jackson Hole, Wyoming? [closed]

1. I stumbled this on View of the mountains and a fireplace. : ViewPorn. I don’t know who under is right. Is this Sun River OR or Jackson Hole WY?

2. Which golf course?

3. Can I sit down these chairs and enjoy view? Or is golf course private?

Looks similar to a lodge in Sun River, OR, but I think the mountains are a bit too high.

Falcitone 1 year ago

This is at a golf course right outside of Jackson Hole, Wyoming. The view is of the southern portion of the Grand Tetons. I spent a year working there and did a double take when I saw this photo.

## Cong ty chuyen cung cap gia thi cong nha tot nhat hien nay

Tong hop bang gia thiet ke thi cong nha moi va tot nhat hien nay

Gia thi cong nha moi nhat

## keyboard – keycode cap 193

I have the issue that whenever I link a keypress to a keycode above 193, there is no event created in the X-server when that key is pressed (xev and xinput don’t recognize the keypress). Result is that I’m unable to map the key via the keyboard-shortcut menu. Only thing I could find about this is that it should be capped at 255, so why is it capped at 193?

I tested this on 2 systems both running ubunutu 18.04.
I also tried upgrading one of them to 20.04. Nothing helped.

Odd thing is that evtest and showkey do recognize the key and keycode.

Can someone explain why this is and maybe give me a potential fix so I can use the keycodes between 193 and 255?

## azure – How to support RDP on to VMs in a Hub and Spoke toplogy using AAD authentication (and CAP)

I have a hub and spoke topologogy made up of a central vnet and then peered VNETs. Inside each peered VNET is a VM.

This, it turns out, may be tricky.

Option 1:
I had hoped to place an Azure load balancer inside the hub and assign a single IP address to it. In my AAD I would the add a CAP for this IP. This would then allow me to RDP (via a jump box in the hub) onto each spoke VM.

Alas Azure load balancer doesnt support peered VNETs.

Option 2:
I considered using Bastion to access the VNets… but bastion doesnt support AAD or peered networks. So thats useless.

Option 3:
I could stick a public IP on each VM (and allow only outbound traffic via an NSG) with all public IPs using a prefix to simplify CAP. Although this rather confusing article https://docs.microsoft.com/en-us/azure/azure-resource-manager/management/azure-subscription-service-limits?toc=/azure/virtual-network/toc.json#networking-limits seems to imply that the number of public ips is restricted per subscription (to 10 for basic!?!)
Here I’d use IP prefixes to simplify CAP to a known range.

Option 4:
I could introduce a virtual appliance.
This sounds horrid and requires management etc

Are there other options Ive missed (I almost certainly have)?

## camera basics – Do I need to remove the flash hotshoe cap when taking pictures?

I just bought an Olympus Trip 35 and this is my first ever film camera. I went hiking yesterday and I took the camera to take shots. All of them were outdoors.

I was just wondering, the little cap that’s on top of the camera, covering the silver bit between the flash hotshoe rails, should that be taken out when taking pictures?

I’m very new to this, and it might sound stupid but I need to know.

## What is Dogecoin (DOGE)? PRICE AND MARKET CAP! | Proxies123.com

“Dogecoin” is a decentralized peer to peer open source cryptocurrency It is operated on a blockchain.

Dogecoin official release date: December 6, 2013.

Dogecoin official authors: Billy Markus, Jackson Palmer.

Do you invest or use Dogecoin?

## Is this Correct, the existence of cryptography requires \$UP cap Co-UP notsubseteq BPP\$

Is this Correct, the existence of cryptography requires $$UP cap Co-UP notsubseteq BPP$$?
Or does it require $$UP notsubseteq BPP$$?

## general topology – If \$f: X to Y\$ and continuous in \$D in mathcal{B}_{X}\$, then \$V\$ open implies \$f^{-1}(V) = U cap D\$ with \$U\$ open

I am reading through Yeh’s Real Analysis 3rd edition, pp. 21-22. Yeh proves the following theorem:

Given two measurable spaces $$(X, mathcal{B}_X)$$ and $$(Y, mathcal{B}_Y)$$ where $$X$$ and $$Y$$ are topological spaces and $$mathcal{B}_X$$ and $$mathcal{B}_Y$$ are the Borel $$sigma$$-algebras of subsets of $$X$$ and $$Y$$ respectively, if $$f$$ is a continuous mapping defined on a set $$D in mathcal{B}_X$$, then $$f$$ is a $$mathcal{B}_X$$/$$mathcal{B}_Y$$-measurable mapping of $$D$$ into $$Y$$.

In the above, it is understood that the domain and range of $$f$$ satisfy $$mathcal{D}(f) subset X$$ and $$mathcal{R}(f) subset Y$$ respectively.

Yeh starts off the proof as follows:

Let $$V$$ be an open set in $$Y$$. The continuity of $$f$$ on $$D$$ implies that $$f^{-1}(V) = U cap D$$ where $$U$$ is an open set in $$X$$ so that $$f^{-1}(V) in mathcal{B}_X$$.

I don’t follow this part of the proof, and I suspect it’s because my background in topology is lacking.

Every topology text I have that I’ve opened up has something like the following:

Let $$X$$ and $$Y$$ be topological spaces. $$f: X to Y$$ is continuous if $$f^{-1}(V)$$ is open in $$X$$ for every open set $$V$$ in $$Y$$.

Great, so that means that $$f^{-1}(V)$$ should be open, but why does it equal $$U cap D$$? Why does this matter anyway, since $$mathcal{B}_X$$ is the $$sigma$$-algebra generated by the collection of all open sets of $$X$$ anyway, so $$f^{-1}(V) in mathcal{B}_X$$ should be obvious?

## complexity theory – how to prove that \$NP cap co – NP\$ = { S | S such that there exist a Strong Deciding Algorithm for S}?

i need to prove that and i find it struggle:
given:

for deciding problem S:

a non deterministic algorithm $$A(x)$$ is strong deciding algorithm if:

1. $$x in S =>$$ fo every run of $$A(x)$$ returns “Yes” or “Maybe”, and there exist run which $$A(x)$$ returns “Yes”

2. $$x notin S =>$$ for every run of $$A(x)$$ returns “No” or “Maybe”, and there exist run which $$A(x)$$ returns “No”

now we need to prove i assume with equality between two sets that

$$NP cap co – NP$$ = { S | S such that there exist a Strong Deciding Algorithm for S}

thank you,