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agdgdgdgwngo
 3 years ago
Best ResponseYou've already chosen the best response.1You mean out of a shape?

agdgdgdgwngo
 3 years ago
Best ResponseYou've already chosen the best response.1curved hexagon O.o

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

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

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1326317247010:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2consider a given segment and approximate it with a straight line; distance = \(\sqrt{\Delta x^2 + \Delta y^2}\)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2as you make the change in x smaller and smaller you get a better approximation

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2when you get to the infinitesimal level; the change in x and y are dx and dy

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2\[\int_{a}^{b}\sqrt{dx^2+dy^2}\ ds\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2...curved hexagon? hmm

athitaya
 3 years ago
Best ResponseYou've already chosen the best response.0thank you for your help

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