As if that is actually soled for that price :o.
As if that is actually soled for that price :o.
Have a look at your roof :*
I’m just sitting in the polling place beein’ bored cause none of you are showing up to vote.
Thats what the 2fa for most transfers is for.
highly scalled person
You might be on to something, it might have been the lizzard people!
Even before clicking on the link I was like, oh yeah, like Bernd Höcke.
Weigh some things you know the exact weight of like 1L of water or your phone (you can google most phones weights, without the case and only if you dont have a screen protector of course). I had the same issue at home but realized my scale was jus off most of the time.
A while ago a collective in germany used a software to merge the faces of two people into one photo and used it on their passport. On one hand the idea was to evade facial-recognition software which uses the passports photo. On the other to use the passport to smuggle refugees into europe by plane.
I am not sure the software was based on Machine learning but would strongly suspect it.
(Source in German: https://taz.de/Peng-Kollektiv-faelscht-Passbilder/!5534868/ )
Nah, TCP is still just kicking the box over, but just kicking it over again, if the reciever doesn’t kick back a box saying they got it.
It’s not like Hamas isn’t also doing this (both by directly killing Israelis and using their own population as a human shield)
omg I haven’t seen this in an eternity
Tweet saying:
If you encounter a guy dressed as a girl “as a joke” this Halloween offer them transition info and be welcoming because if it’s an exploring trans girl, it’ll be helpeful and if it’s a shitty cis dude, he will be furious. It’s a win/win.
As many other comments here (really liked the way /u/opperharlie put it) I think it’s a shitty take. First of all, there is drag which is totaly valid. But I think crossdressing in genereal -if it’s not done derogatory (which I never experienced)- is a totally fair costume. I am certain I am cis and have dresses feminin for costumes a few times and don’t see what the problem with hat is.
Try (100,100,100,100,100,101) or 50 ones and a two, should result in 102 and 4 as a max respectively. I tried using less numbers, but the less numbers you use, the higher the values (to be exact less off a deviation(%-difference) between the values, resulting in higher numbers) have to be and wolframAlpha does not like 10^100 values so I stopped trying.
thanks for looking it up:).
I do think the upper bound on that page is wrong thought. Incedentally in the article itself only the lower bound is prooven, but in its sources this paper prooves what I did in my comment before as well:
for the upper bound it has max**+log(n)**. (Section 2, eq 4) This lets us construct an example (see reply to your other comment) to disproove the notion about beeing able to calculate the max for many integers.
to be fair it does seem to work for any two numbers where one is >1. As lim x,y–> inf ln(ex+ey) <= lim x,y --> inf ln(2 e^(max(x,y))) = max(x,y) + ln(2).
I think is cool because works for any number of variables
using the same proof as before we can see that: lim,x_i -->inf ln(sum_i/in I} e^(x_i)) <= ln(.
So it would only work for at most [base of your log, so e<3 for ln] variables.
Wth, I always use them to pay when I have one…