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ITT: Post Mathematics (2U) Questions/Problems (1 Viewer)

iBibah

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Noticed similar threads for other subs, so why not 2u :)
 

iBibah

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I might as well start with a question for people to attempt.
 

rsagar

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you have to somehow get 1/2 x base = (r + square root of r^2 - x^2)

hmmm... i'm working on it...
 

rsagar

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I might as well start with a question for people to attempt.
1/2B = x (so that's done)
to find the vertical height:
rotate r to make it horizontal the ad r + y which is the formula of the circle
there for h = (r + sq. root r^2 - x^2)
if you get what i mean
 

Leffife

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Uh yeah? It's maximum/minimum (2U course)
Wow I really love that question, very unique in someway. Is there any chance that I could save her for you? No terms and conditions applied :)
 

RealiseNothing

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1/2B = x (so that's done)
to find the vertical height:
rotate r to make it horizontal the ad r + y which is the formula of the circle
there for h = (r + sq. root r^2 - x^2)
if you get what i mean
r is just the radius. So the height of the triangle is just r+y, and use pythagoras to find y, and then the solution falls out straight away.
 

iBibah

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1/2B = x (so that's done)
to find the vertical height:
rotate r to make it horizontal the ad r + y which is the formula of the circle
there for h = (r + sq. root r^2 - x^2)
if you get what i mean
no need to explain to me, i wrote the question lol

just for people to have a go at

but yes your right so far (and note what realisenothing said)
 

RealiseNothing

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no need to explain to me, i wrote the question lol

just for people to have a go at

but yes your right so far (and note what realisenothing said)
You made this question? It's very nice imo. I might come up with one of my own soon.
 

Sy123

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SOLUTION


Very well written question, while people to Spiral's one Ill think of my own
 

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