The cardinals have it way better. The can siphon off millions to their personal account with no comeuppance.
Who would want to be the pope? All that gravity. Wearing white like a wannabe bride while everyone else has to wear black like bridesmaids at a funeral. How ridiculous. It's like all the downsides of being a rock star with none of the upsides. And your colleagues are aging power hungry pedophiles. No thanks.
Scientology, that brave and wonderful and obviously correct sci-fi religion involving hydrogen bombs and galactic warlords and the spirits of pissed off hydrogen-bomb blowees haunting people, who viciously attacks anyone who criticizes them, brings in billions as well.
Things that piss me off! Theres's a side bar add showing this 'genius' puzzle:
I spend a bit of time bec it's obv that beaker 3 gets filled first so there must be some kind of trick. That made me go to the tubes to see if it's there and sure enough
Mofo's at 'Brilliant' have beaker 4 higher up than beaker 3 just to make Howard think he's dumb and click the link for the answer - which it doesn't do - and then look it up where the set-up is different.
This is some kind of deceptive advertising and someone should be made to pay!
Exclusion method works. How could 1 be filled first?, etc. There may be a cork somewhere
Beaker 1 could be filled first if the water is under high enough pressure so that the plumbing is overwhelmed but we make the assumption that it isn't.
Quote:
Originally Posted by Zeno
Do you own a large hand gun? Or rifle?
Sometimes I think that I'm the only resident of Arizona w/o an assault rifle.
I want to share another 'Brilliant' puzzle and see if I got the answer right:
Spoiler:
My answer is 'Number 1.'
Spoiler:
I called the rectangle ABCD with AB=b AD=a and drew a line AF intersecting BC at F making angle BAF=f and another AE intersecting DC at E making angle DAE=w.
Then ABF area is 1/2*b^2*Tan(f)=4 ADE area is 1/2*a^2*Tan(w)=2, EFC area is 1/2*(a-b*Tan(f))*(b-a*Tan(w))=3 and the unknown area S of AEF is S=1/2*Sin(90-(f+w))*a/Cos(w)*b/Cos(f) which leads to S=1/2*a*b*(1-Tan(f)*Tan(w)). Then also S+4+3+2=a*b for the whole rectangle.
If you then see that Tan(f)=8/b^2 , Tan(w)=4/a^2 and eliminate a*b and the tangents from the above last 2 equations you get a single equation in S that has root S=7. Then from that you get the rest if interested to solve the system completely and see that F is the middle of BC for example etc.