/sci/ - Science & Math

[derp No.2487449]

Okay /sci/ here to test you.

Solve this

_________________
(5-2x)+(2x-5)=

[derp No.2487458]

Its actually very simple

[Anonymous No.2487463]

0

[Anonymous No.2487466]

0?

Am I missing something?

[derp No.2487471]

bump, are you stupid or what?

[Anonymous No.2487472]

Zero for all values of x.

[derp No.2487489]

It can't be that hard for you, after all you are Science&Math,

Or should I just spit out the answer?

[derp No.2487505]

Okay heres the answer faggots

(5-2x)+(2x-5)=5-2x+2x-5=4x

/sci/ is now the dumbest board on 4chan

[Anonymous No.2487510]

-4x^2 - 25

[Anonymous No.2487522]

>>2487505

[Anonymous No.2487527]

>>2487505
are you trolling? Fucking elaborate it if you think you've just made a fool of us here.

[Ek !/RHPIgb8po No.2487556]

>>2487505

I'm Derping so hard

[Rainbow Dash !!pGgXu8jfKhv No.2487567]

>>2487449
-infinity<x<infinity

[Rainbow Dash !!pGgXu8jfKhv No.2487581]

>>2487567
y=5-2x+2x -5
when y = 0

[Anonymous No.2487582]

isn't it just 0...

[Anonymous No.2487621]

>>2487510

It's not multiplication.

[Anonymous No.2487642]

itt: /sci/ passed 3rd grade.

[Anonymous No.2487679]

a=5
b=2x

(a-b)+(b-a)

No matter what values a or b may be, the answer will always be 0.
Why? Because (b-a) = -(a-b). And (a-b) = -(b-a). So we can also
write this as:

(a-b)+(-(a-b))

or (a-b)-(a-b)

And we all know anything subtracted from itself is 0.

[Anonymous No.2487780]

Define f(x) as (5-2x)+(2x-5).
f is clearly continuous and <div class="math">f(0)=0=f(2\pi)</div>.
So you can use Furierrepresantation to get something like
<div class="math">\sum_{k \in \mathbb{Z}} \frac{1}{2\pi}(\int_0^{2\pi}(5-2x)e^{-ixk}+\int_0^{2\pi}(2x-5)e^{-ixk})\int_0^{2\pi}e^{-ixk}</div>
Using partial integration you could easily solve it, but that's left as an exercise.

[Anonymous No.2487795]

>>2487780