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agdgdgdgwngoBest ResponseYou've already chosen the best response.1
You mean out of a shape?
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.1
curved hexagon O.o
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.1
do you mean a circle with an inscribed hexagon?
 2 years ago

athitayaBest ResponseYou've already chosen the best response.0
no not really, its hard to explain , sorry
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.1
that's how you normally approximate it, right
 2 years ago

amistre64Best ResponseYou've already chosen the best response.2
dw:1326317247010:dw
 2 years ago

amistre64Best ResponseYou've already chosen the best response.2
consider a given segment and approximate it with a straight line; distance = \(\sqrt{\Delta x^2 + \Delta y^2}\)
 2 years ago

amistre64Best ResponseYou've already chosen the best response.2
as you make the change in x smaller and smaller you get a better approximation
 2 years ago

amistre64Best ResponseYou've already chosen the best response.2
when you get to the infinitesimal level; the change in x and y are dx and dy
 2 years ago

amistre64Best ResponseYou've already chosen the best response.2
\[\int_{a}^{b}\sqrt{dx^2+dy^2}\ ds\]
 2 years ago

amistre64Best ResponseYou've already chosen the best response.2
...curved hexagon? hmm
 2 years ago

athitayaBest ResponseYou've already chosen the best response.0
thank you for your help
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.1
right; use amistre's formula to approximate the length of curves and arcs in graphs.
 2 years ago
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