Erotophobia is used as a psycological term, to describe a phobia directly related to sex, and as a political term, used for example by feminists and in literary analysis
by dragonsniper22 March 25, 2011
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Erato
• Eratocflurries
• eratosthenes
• Eratosthenia
• The Seethe of Eratosthenes
• eragon
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• Erat
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• Erotomania
Erratas is a myth on the internet, that describes a system used by large corporations to aim for specific victims, to copyright strike. According to the myth, the system will delete any reference of itself from the internet. Not much is known about the myth, but most suspect that it is just a story, and nothing serious. Sometimes, people use Eratas to go under the system.
by Krotokoic October 20, 2019
Get the Erratas/Eratas mug.Elegant, beautiful, down to heart girl,
she is beautiful inside out, she might be shy or quiet but crazy atimes, she gives a sweet grin on her cheek and full of kindness with a soft heart eventually
she is beautiful inside out, she might be shy or quiet but crazy atimes, she gives a sweet grin on her cheek and full of kindness with a soft heart eventually
by Obadara October 24, 2020
Get the ewatomi mug.A formal, professional way to talk about the act of grubbing, almost said in a British tone. A word that is only used by really cool people who have the title of Best Friend in the Whole Wide World. If the word is used by those with a lesser title, they will be beheaded.
by Lobster 2754 February 8, 2023
Get the Eaton mug.A pretty badass hero who has a flaming sword, an elf to hang out with, and a huge blue dragon. Plus the task of saving the world from an evil guy who wants to enslave everybody and kill all the good people. Yeah.
by thegirlwhogotLoki'dagain December 31, 2012
Get the Eragon mug.Originating from the Greek analogous hóper édei deîxai (ὅπερ ἔδει δεῖξαι), meaning "which had to be demonstrated". The phrase is traditionally placed in its abbreviated form (Q.E.D.) at the end of a mathematical proof or philosophical argument. Phrase synonymous with "Quite Easily Done."
∫|Ψ(x, t)|² dx (from -infinity to infinity)= e^(2Γt/ħ) ∫|ψ|²dx(from -infinity to infinity)
The second term is independant of t, therefore Γ=0 & ∫|Ψ(x, t)|² dx (from -infinity to infinity)=∫|ψ|²dx(from -infinity to infinity)=1 {Normalized}
Q.E.D. "quod erat demonstrandum"
The second term is independant of t, therefore Γ=0 & ∫|Ψ(x, t)|² dx (from -infinity to infinity)=∫|ψ|²dx(from -infinity to infinity)=1 {Normalized}
Q.E.D. "quod erat demonstrandum"
by justinbonito October 31, 2013
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