Sup Forums how much is reasonable to pay for an i5 2500k?

Sup Forums how much is reasonable to pay for an i5 2500k?

Solve what?

There's nothing to solve there, it's just a piece of art in a whiteboard.

You can get them for like $60-70 pretty easily, so I'd say that's the realistic max you should pay.

ok im buying one for $75

comes with 8gb of rams too so id say its an ok deeleyo

t. brainlet

0. You might pay some if it comes with a mobo.

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>he cannot identify a wonderful piece of art when he sees one
>call others brainlet
I bet you don't even have enough brain cells to distinguish Picasso from Dali.

Remove yourself.

pi times half of r squared will give you teh area of the loft left circle. from that you need to subtract one quarter of the area of the larger circle, subtracted from the area of the square.

Fuck I can't even describe it and that's still wrong.

It's the moon. Hello moon.

underrated post

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sHuT THE

FuCK UP

YOU DICK

Saw a few for 50$ or so
Problem is the mobo, u gotta buy them together or get it first bc sometimes they run for insane prices such as 75-100$ which is just a waste of moneis.

I'm eyeing cpu+cooler+mobo+ram combos which u can find for around 150$, i5 3570k too

is 2500k, mobo, and 8gb of rams for $95 good?

yea what kinda ram though? sandy bridges esp. i5 benefit alot from higher frequencies
1600vs 2333 can gain u 10-15 fps

Also yea u could get that and just find a cooler separately, bc stock is garbage

its frrequencey is like 2600 ok

f
u
c
k

y
u

gay

>t. engineer

You solved for the area of the little circle, not the yellow area.

(Area of yellow sliver) = (Area of smaller circle) - ¼(Area of larger circle) + 3(Area of red)

= πs2 - ¼π(2s)2 + 3(s2 - ¼πs2)
= (πs2 - πs2) + 3s2 - ¾πs2
= (3 - ¾π)s2

(Area of square) = (4s)2 = 16s2

(Area of yellow sliver) = [(3-¾π)s2] / 16s2 * (Area of square)
= [9/4π]s2 / 16s2 * (Area of Square)

(Area of yellow sliver) = 9π/64 * (Area of Square)

---

I stayed up for 10 minutes doing this. Thanks for the problem, I kinda like these

Where do you even get a good mobo for a sandy bridge these days?

I got an ivy-bridge and my backup mobo is dying as we speak and replacements are impossible to get without paying outrages prices or getting a very very shitty chipset.

she's dead jim

>+ 3(Area of red)
Your forgot the purple areas.

Fuck this attractor field.

You're missing the little areas that aren't covered by either of the circle though.

Fuck me I'm absolutely stumped. How would you even define that little area?

just ask /sci/ if it's a troll image or not

What do you mean by "troll image", you can solve it for sure somehow.
Find out the coordinates for the intersection points of the two circles, and then do partial integrations to find the area of the upper-purple.

Assuming that the big square's area is 1:
1/4*arccos(sqrt(2)/4)-arccos(5*sqrt(2)/8)+sqrt(7)/8


What do I win?

(pi * r)^2 = circle
(pi * (r/2))^2 = top left circle
(r * 2)^2 = square
r^2 = top left square

Square A - circle A = Area of corners.
Divide it by 4 to get area of one corner.
small square A - small circle A = Area of top left square corners.


(Now you divide large circle A to 4 and you'll get portion of top left lcA, then you plus four times small square corners to it) and remove it from whole small square area.

Did I miss anything except those barely visible small portions? I have my last entrance exam in an hour, hold me Sup Forums.

>How would you even define that little area?
You do not need to.

>integrations
Ugly.

It's not, I have seen solutions. I also have once implemented a MC method to test one solution and it seemed correct within the statistical uncertainties (6 digits or so).

See I did it your way.

You already know area of these portions, just sum them and remove from small square area.

Idiot.

350

>using the area of the yellow circle to find the area of the yellow circle
nice.

How hostile.
Sorry made mistake there.

I am trying to find the angle and am stuck. You win... everything.

>How hostile.
I am sorry. Should have said: "You have overlooked a detail."

Banana

So this might not be the best way to do this. I feel like a retard right now. I need to stop being retarded.

But first of all, you see that the only important thing is the inner circle of radius 1/2 centered at the origin and the outer circle of radius 1/4, centered at some radius from the origin. You compute that radius by R = sqrt(1/4^2 + 1/4^2) = sqrt(1/8) = sqrt(2)/4. Then you recognize that the whole thing is symmetric, so you rotate the whole thing until both circles have centers on the y axis.

then our equations are
x^2 + y^2 = 1/4
x^2 + (y - sqrt(2)/4)^2 = 1/16

you see that the spots where the two circles intersect have x values -sqrt(7/2)/8 and sqrt(7/2)/8

you then notice that for the region we're trying to find the area of, the y values are positive, so we solve each above equation for y in terms of it's positive square root.
y^2 = 1/4 - x^2
y = sqrt(1/4 - x^2)

(y - sqrt(2)/4)^2 = 1/16 - x^2
y - sqrt(2)/4 = sqrt(1/16 - x^2)
y = sqrt(1/16 - x^2) + sqrt(2)/4.

Since the latter equation is the top of the region, and the former equation is the bottom of the region, and the intersection x values earlier are the bounds of the region in terms of x, we can set up an integral, splitting in half using symmetry

2 * integral of sqrt(1/16 - x^2) + sqrt(2)/4 - sqrt(1/4 - x^2) from 0 to sqrt(7/2)/8

which yields 1/128( 9sqrt(7) + 8arctan(sqrt(7) ) according to wolfram alpha.

please tell me where I fucked up. Also I apologize in advance for not knowing latex or using code tags. I'm really tired and don't post on Sup Forums ever.

(pi * (0.5 ^ 2)) - ((pi * (2 ^ 2)) / 4)

Firstly, if the outer circle has radius .5, the big one has radius 1. Secondly, your calculation incorrectly includes three regions outside of the desired region.

craigslist and dare i say it r/hardwareswap
I got this for $80 a few months ago when my P8P67 died a terrible death

That was kind of fun.

Now I can sleep.

Because I'm lazy.

interesting. Somehow I copied the result to my integral wrong, but when I recomputed it I got

1/32( sqrt(7) - 8arcsin(sqrt(7/2)/4) + 2arctan(sqrt(7))

which is approx. 0.0365953148825869561

I did it! Thank you for verifying.

>Rule №1 - Never start a thread with image more interesting than question itself

>make thread
>picture more interesting than your question
>thread derailed

>solve for shaded area in terms of r
>draw axis as shown
>circles C1 and C2 are easily written in equation form
>integrate and subtract the area under C1 and C2 over the x-bound of their intersection, highlighted by the orange dots

too lazy to do out the math but thats the simplest way to solve it

I wanna die fuck the illumniati!
fuck em in the bungholio

you'd have to split it up into parts. If you look closely, C1 isn't a proper function of x

the way I did it, I just rotated the whole thing to avoid that.

yea did you see that my axis (drawn in red) is rotated 45 deg to account for that

Just buy a new one, goy

for some reason I thought you just drew a big weird 'X'. Genius.

Interested to know how this'll turn out.

I can solve it with basic geometry but integration and such seems more efficient.

cosine theorem for a triangle with sides 1,sqrt(2)/2 and 1/2

Draw the intesection points A and B between the two circles. Cutting off a circle with a line gives you a half-lens, which is easy to get the area for.

The yellow area is then just the area of the half lens between the line and the smaller circle, minus the area of the half lens between the line and the larger circle.

if mobo is OC compatible, and ram is decent brand and Hz then yeah it's a great price.
get an aftermarket cooler, the 212 evo is like 20 bucks.
the 2500k is still a great chip

Didn't calculate, but it looks good.

My final expression ended up being: (with r being the radius of the small circle):

r^2 * (Sqrt(7) + arccos(1/(2*sqrt(2))) - 4*arccos(5/(4*sqrt(2)))

Oops, there should be a 1/2 in front of the Sqrt(7)

Is steins;gate 0 any good?