IS THIS FOR REAL? v.2. Are Asians Superior now?

Previous thread hit bump limit.

>Be Chink
>Be in middle school
>pic related

Solve for shaded area

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youtube.com/watch?v=B_yNI3gAHNk
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8?

bumping

come on Sup Forums i believe in you

42

Well the area is somewhere above 32 and below what looks like 36

post more pls

48?

>Are Asians Superior now?
Not empirically or historically, that would be whites and jews

we could be doing the same shit, but our education system panders to the lowest common denominator. They're more concerned with making sure everyone feels included rather than really challenging students.

How do you solve this? I suck at geometry

Your average chinese middle schooler isn't solving complicated integrals you fucking faggot.

You are all retards

The answer is Viewtiful Joe

Pi.radius squared.
Go nuts

[email protected]

32

So why don't you go back to China?

This doesn't look particularly difficult, is there a catch? You can figure out the radiuses of most of the circles involved and add/subtract.

Pic related from previous thread
>shaded aread

Sup Forums solved in ~200 posts
29.3

>middle school
have you seen what our middle school kids do?`

dont kek me

No idea, not relevant to my field and I was never educated on this.

t. Electrical engineering masters

Yeah I've been in that thread, it's just that OP is being stingy with his geometry problems.

It's not that hard, it's just quarter circles.
>Area of a circle = pi r^2
>start with biggest quarter circle and work down

Not going to waste 10 minutes on a middle school problem.

Shit like this pisses me off.

I mean I know you're just supposed to assume that the curves are circular. But you don't actually KNOW that unless equations are provided.

ESTIMATE for the shaded area would be a better way to word it.

wtf I hate doggos now

>complicated integrals
>solving 4*4 and eyeballing an area

Nike?

you dont need integrals to solve that, retard

I never went to middle school. So I'm guessing that's around age 14? I could do addition, subtraction, and multiplication by then. I also knew the formula for a fair number of geometric shapes including circles and quadrilaterals.

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this. though they aren't that complicated, just fucking tedious.

This is fucking easy. All those arcs are quarter-circles; the radius for each is known. Why not make it challenging with some function generated arcs?

Oh, right. Didn't realize it's perfect circle segments. Thought it were some arbitrary curves.

yea you do if you want any sort of accuracy

> I can totally do a backflip
> they're just tedious
> nah, I don't want to do one right now

...

start with area radius 12 quarter circle

Show the solution step by step, senpai.

It's deceiving at first . Just think of it as a lot of quarter circles and work your way from there. You don't need calculus.

>i passed all my calculus classes, im not doing that shit for fun. fuck you.

the absolute state of burger education, at least tell me that you are of shitskin variety.
anyway wait a bit and some bored user will solve it with nothing more complicated than pi r^2

It doesn't say those are circles though so it's impossible to solve.

> I can t-totally do it.

I went to one of the top 100 school districts in the country and never handled anything akin to these problems.

Can you explain it?

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Solving maths problems /=/ You will have a better life

t. guy who can't solve math problems

ausie here did here we dont do such problems in school either but cmon, its not hard enough for you to be unable to see solution without doing this previously

waiting warmly for solution

mathematician here. We're going to need double integrals and a third variable "t".

INT {INT { SQRT[pi*t^2]^3}dx}.

Good luck guys.

fine i'll do it.
>144 pi/4 - 2.5*16 - 64 + 16pi - 4 pi +8 -16pi + 32

Whatever that equals. Don't have a calc on me.

Some can't others can. If thats your thin than go for it but that doesn't mean you are superior.

You're assuming the lines are tangent to all those 1/4 circles, and that all the boxes are 4 x 4 units.

Jews own the west. THEY are the superior castes of our time!

ougi = best girl

step by step:
>biggest quarter circle has a radius of 12
>subtract pieces from that... which is what I did
when will you brainlets ever learn?

The absolute STATE of western education

There's probably missing something. I think you need some mathematical definitions of the curves / functions.

Nah based on the coordinates you can figure out the curves. Then you can make an equation that adds up the difference

If it was multiple choice that would probably work but if you need an exact answer you would have to do a little work.

t. inferior guy

Huh, interesting. Thanks.

I was always aabominably terrible at mathematics, it's a wonder I graduated in the end.

This is pretty basic-bitch shit, just create functions to represent the lines and integrate. Also

>Are Asians Superior now?

Lolno

i see the halves of 4 ellipses that need to be added/subtracted. i don't remember if there are pythagorean tricks or geometric proofs that can solve for the shorter radii or if they can be eliminated by simplifying the equation.

I was the poster in the last thread that was using algebra to solve the equation.

I worked out the previous answer and it was 100-(45π/2) = 29.31625

youtube.com/watch?v=B_yNI3gAHNk
ezest math problem of my lyf

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did I do it right user ?

You're forgetting this is Sup Forums and you need to spell out every single detail of a problem people will make up assumptions in order to come up with the most incorrect answer possible.

You can post a picture of two scales and ask "Which is heavier, a kilogram of steel or a kilogram of feathers?" and people will say
>GUYS WE CAN'T ASSUME THE LOCAL GRAVITY IS EXACTLY THE SAME ON BOTH SCALES

I could probably do this in middle school when I did math all the time. No way in hell I'll do it now though lol

My post in the last thread

same principle as previous

Can't be bovered to use a calculator

me too

the answer is 32pi - 64

don't forget to subtract 15 for tyrone.

Thanks user. Getting the segment by subtracting the right triangle was the part I was missing.

Easy to do geometrically once you realize that.

i'm an asian
((pi*8^2)/4 - (8^2)/2) - (4*pi - (4^2)/2) + ((pi*12^2)/4 - (12^2)/2) - ((pi*8^2)/4 - (8^2)/2)

shit and you wonder why you can't get laid.

I just used the application for it nerds.

And how do you know it's exactly quarters of circles and not some random curves?

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>not factoring out the constans
Omuich can't compete with REAL asians.

can confirm, did something similar
36pi - 64

too easy, op

Just find the semicircles area from the curves, reverse it to the total 12*12, and you will find it, brainlet

I thought this thread was for politics and not for fucking math problems. If you're gonna shill for the asians, at least bring up a point better than "they do harder math in school."

youtube.com/watch?v=B_yNI3gAHNk
It says in the video where the problem was shown, which faggot OP didn't give us because he wanted to mess with us

ok

> mad cuz he can't do math
> lashes out instead of learning

discuss

>be illiterate communist leftyfaggot
>do my math homework Sup Forums

fuck off to /leftycuck/

No you don't just divide into quarters of circles add and subtract these.

ITT angry snow niggers that don't like being told they're stupid

Is that a bad dragon dildo?

quick approximation:

113568 total pixels in the mainsquare.
25846 red pixels

Total area of the main square: 12*12 = 144
Percentage red: 25846/113568 = 0.22758

22.7% of 144 is : 0.22758*144 = 32.77152

Therefore the area of the red zone is approximately 32.77

So this one is pretty easy when you recognize the quarter segments...unless I'm missing it, the other problem was much harder and there was no quarter segment trick. Only brute force answers were found, no clean expression involving pi (29.27625)

I thought we invented computers so that we wouldn't have to deal with this kinda shit anymore.

No; it's ~36.53. You fucking killed all the astronauts.

The computer can't do it without an Asian telling it what to do.

They are the niggers of electronics if you will.