Little fun problem

I'd say about 7. Maybe a bit more

You'd have to calculate the integral, multiply that by 1/2 * bh (area of a triangle) and multiply again by the volume of a hexagonal figure.

>>7960923
yeah but what hexagonal figure

>>7960931
It's a double-sided dodecahedron with Mahler edges.

A lot. But not a ton.

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>>7960943
oh ok

I'm quite new, is this some meme? Cause the answer isn't hard

>>7960952
What's the answer then, asshole?

>>7960952
not a meme, just asking. I did this in photoshop in class the other day and was just wondering

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>>7960917
> 0

>>7960988
Holy shit, rekt

>>7960917

>count starting points
>calculate total intersections
>add all points together
>calculate all possible combinations of 3 points

17C3 - 16 = 664

{we subtract as there are 16 lines (8 edges and 8 diagonals)}

Is it 16c3 = 560?

>>7961168
>>7961216
>>7962797

Wrong.

>>7962859
Yes please do give the correct answer
its (17+intersection points) C3 and we subtract all collinear points of 3

bout 350

>>7963059

Well first of all it's 16, not 17.
Second, you haven't actually given a way to calculate anything, which is why you can't provide an actual number.
Third, you can plainly see that most sets of three intersection points do not form triangles.

https://youtu.be/K8P8uFahAgc
This might help

0 it's an octagon are you retarded