...
This
Other urls found in this thread:
desustorage.org
twitter.com
sage
Take your pedophile cartoons back to .
That's not too difficult
Haven't taken math since high school, so I hope I didn't fuck anything up
Anyway
When we consider opposing tetragons, we see that they consist of two triangles. A right angled triangle that's the same for each segment, and two traingles with base b and height h1 and h2.
The surface area of a triangle is equal to 1/2 base*height. Their combined surface is as follows:
A = 1/2 b * h1 + 1/2 b * h2
A = 1/2 b (h1 + h2)
h1 + h2 is constant, so no matter where we move the centre point, the combined surface area of the triangles remains the same. The combined area of opposing tetragons is therefor also constant.
Opposing tetragons have a surface area of 16 + 32 = 48 cm2 , so A = 48 - 20
A = 28 cm2
Did I do a good job? I used to love geometry
lel I made that image a few years back
prove it
Anime is the foundation of Veeky Forums. Veeky Forums has always been about anime if you don't like it don't come here.
The way I see it:
20+32+16+A=X
(B) (C) (D)
Split the difference of C and D and multiply that by 4. Then, subtract the summation of the parts and from the result.
4(D+(C-D))=Y
Y-(B+C+D)=A
A=24 cm^2
Isn't that assuming the overall shape a square and not a rectangle? I see hashmarks that seem to indicate variable length, but I could be wrong
The square comment was meant for
Hello summerfriend
At least lurk for a few fucking months before you start shitting up the place
yup
I'm sorry, are you trying to learn me something? Because I couldn't see pass the asshat. Try again?
25cm^2
What is your major / minor?
This
van der waals
thats not a number :^)
85
29 because thats what my theory says lol
This is clearly 0.
no
and not even close in any way.
>not getting my joke
...
What do you think the :^) was for?
[math]\det(\mathbf{A})=(-1)^n n[/math]
I got it by trying to construct A from the identity using elementary matrices and finding the determinant of them (which is just the product of the diagonal terms). This works because
[math]\det(\mathbf{EF})=\det(\mathbf{E})\det(\mathbf{F})[/math]
And only elementary matrices with a non-one value on the diagonal will have any contribution to the overall determinant and each time you increase n, there will be an additional elementary matrix with determinant [math]\frac{-n}{n-1}[/math]
Sorry, messed up the signs
[math]\det(\mathbf{A})=(-1)^{n+1} n[/math]
>it's another faggot math problems thread to show off our epenis
>inb4 no one even tries this
>not getting that getting the joke requires silence
Guys, I need enlightment on this, I need to study
[eqn] \lim_{n \to +\_infty} \sum_{k=1}^n (\frac{k}{n})^n[\eqn]
23
the angles in the original don't appear to be symmetrical, I don't think it's this simple
is it n! ?
Yes!
No
it doesn't matter what they appear to be, the sketch tells you they're equal
The angles aren't marked as right angles in either pic. Neither is it stated. I've literally spent the day trying to avoid saying this, dammit.
what angles are you talking about? if the original box's sides are y, then the second box's sides are y/sqrt2.
The corners of the box in OP's image aren't right angles
However, this is an autistic interpretation
you mean the curved parts? yeah, you're right, the presentation is a bit retarded. the solution still holds if we assume the outer lines are perpendicular
This
32-16 = 16
32-20 =12
12+16 = 28 cm^2
...
Use the GNS representation of the albegra and use locality to force the Stone integral representations of the translations to have support on the forward light cone.
subtract row 2 from row 1
subtract row 3 from row 2
subtract row 4 from row 3...
Get a lower triangle matrix with -1 all along the diagonal.
det = (-1)^n
not quite.
You dont subtract anything from the bottom rightmost diagonal element.
[eqn](-1)^{n+1}n[/eqn]
Finding a clever way to solve a problem is fun, writing a proof isn't. Especially when it's your homework.
>a well known fact stated in multiple books is homework
Nice projection idiot lmao
Loads of well known theorems are given as homework problems.
Take your pedophile cartoons back to .
Fucking degenerate.
desustorage.org
(The fan in blue) = X
(The fan in yellow) = A
(The triangle in red) = B
[eqn]S = 8X = 8(A - B)[/eqn]
[eqn]A = \frac{1}{2} r^2 \theta = \frac{1}{2} r^2 \frac{\pi}{12} = \frac{1}{24} \pi r^2[/eqn] because [math]\theta = \frac{\pi}{12}[/math]
[eqn]B = \frac{1}{2} (\frac{\sqrt{3}}{2} r - \frac{1}{2} r) \frac{1}{2} r = \frac{\sqrt{3}-1}{8} r^2[/eqn]
Therefore, [eqn]S = 8(\frac{1}{24} \pi r^2 - \frac{\sqrt{3}-1}{8} r^2) = (\frac{\pi}{3} - \sqrt{3} + 1) r^2[/eqn]
Especially, [eqn]r=10[/eqn]
[eqn]S = 100(\frac{\pi}{3} - \sqrt{3} + 1) [{cm}^2][/eqn]
But... there's nothing to solve.
[math]|A| = -n^2 / 144[/math] for [math]n \to \infty[/math]
nice meme
They really aren't. Any respectable text will not give you problems that you can just copy the answer straight from the texts
Mayuri best girl, why did she have to die 30 times ?