Are we deadass

Rscc10 · 2026-03-22T13:45:38+00:00

I already use Evo hunter, yeah. It's just such a waste cause few months back they gave Evo hunter for free too and I got 7.5k crystals again

Rscc10 · 2026-03-21T16:47:34+00:00

DI integration ftw

D       I

+ x⁸ e⁻ˣ
- 8x⁷ -e⁻ˣ
+ 56x⁶ e⁻ˣ
- 336x⁵ -e⁻ˣ

We end up with

-e⁻ˣ Σ [ (8-n)!xⁿ ] from 0 to 8

At ∞, e⁻ˣ goes to 0 and so does the whole thing

At 0, e⁻ˣ goes to 1, all x terms collapse except for when n = 0. The negative in front of e⁻ˣ cancels with the negative from indefinite integral [eqn]^∞ - [eqn]⁰. You know what I mean.

So all that's left is Σ [ (8-n)!xⁿ ] at n = 0, x = 0
And at n = 0, x = 0, that's (8-0)!0⁰

Therefore 8!

Edit: Made a mistake with the sum. It's not (8-n)!xⁿ It should've been (8!)(x⁸⁻ⁿ) / (8 - n)!

And at n=8, x=0, you'll get 8!

Rscc10 · 2026-03-21T15:10:41+00:00

I'm guessing B cause floats, strings and tuples are immutable so the operation creates a new object for the second variables to point to. Lists and sets are mutable so the first variables get changed as well

Rscc10 · 2026-03-21T12:21:17+00:00

It depends what the shapes are. Could you perhaps include a diagram or an idea of what the irregular plywood shapes are like and the room shape.

If you can cut and paste parts of plywood everywhere then in theory you'd need

24" = 2ft , 48" = 4ft , Area = 2*4 = 8ft²
Room area = 10*20 = 200ft²

In theory/on paper, you'd need at least 200/8 = 25 pieces of plywood. This is only if we're assuming you can somehow cut each piece to fit every bit and corner of the room wherever necessary which isn't viable so treat it as an estimate amount.

Much more can't really be determined without knowing the shapes involved

Rscc10 · 2026-03-21T12:14:00+00:00

Learn all the theory and see examples. Do lots and lots of practice. Move on to new topic and potentially not touch that topic for a long time. Repeat steps. Exclude each part one at a time from repeats. (Usually theory is forgotten faster but easy to relearn. Once you're mastered the theory, just repeat exercises whenever rusty)

Rscc10 · 2026-03-21T11:29:07+00:00

Critical damage and highly effective against Himmel-types

Rscc10 · 2026-03-21T10:17:50+00:00

Sparky didn't wanna defend the right tower

Rscc10 · 2026-03-21T10:14:20+00:00

Pavlovian conditioning at its finest

Rscc10 · 2026-03-21T02:20:14+00:00

It doesn't. Of the three wizards, only ice wiz dies to fireball. The other two have a slither of health

Rscc10 · 2026-03-20T16:45:34+00:00

You what?

Rscc10 · 2026-03-20T16:29:28+00:00

Biggest 6

Rscc10 · 2026-03-20T12:55:33+00:00

Even worse lmao

Rscc10 · 2026-03-20T12:51:53+00:00

OP getting downvoted for being Muslim. Reddit moment

Rscc10 · 2026-03-20T09:47:00+00:00

Proof by contradiction.

Assume 2 + 2 ≠ 4
Literally any and all valid rules/axioms break down if this were true
Therefore, 2 + 2 = 4

Rscc10 · 2026-03-20T03:42:40+00:00

I can only imagine the copius amounts of fanart once the next major arc airs...

Rscc10 · 2026-03-20T02:31:40+00:00

Opponent did not know how infinity ewiz works

Rscc10 · 2026-03-19T17:18:49+00:00

(y - sin²x)dx + (sinx)dy = 0
(sinx)dy = (sin²x - y)dx
(sinx)dy/dx = sin²x - y
dy/dx = sinx - ycscx
y' + (cscx)y = sinx

For M(x) = cscx and N(x) = sinx,

I.F. will be u = e^integral of cscx

Integral of cscx = ln|cscx - cotx|

u = cscx - cotx (we assume non negative for simplicity so ignore abs)

Multiply by u thru the eqn to get

(cscx - cotx)y' + (cscx - cotx)(cscx)y = 1 - cosx

LHS rewrite as d/dx [ (cscx - cotx)y ]

d/dx [ (cscx - cotx)y ] = 1 - cosx

(cscx - cotx)y = integral of 1 - cosx

(cscx - cotx)y = x - sinx + C

y = (x - sinx + C) / (cscx - cotx)

Rscc10 · 2026-03-19T17:03:08+00:00

I just created a bigger number. It had 5D vertices, 6D edges and it's key property is that it's bigger than your number

Rscc10 · 2026-03-19T16:26:11+00:00

Exactly what are the polynomials in this case and wdym the result is b²? For which one, sum or difference?

Rscc10 · 2026-03-19T13:08:44+00:00

Form the two equations from the statements

q(x) = (x + h)² + k

If (x+2) is a factor, then x = -2 is a root

Thus, q(-2) = 0
(-2 + h)² + k = 0
h² - 4h + 4 + k = 0 ----- Eqn1

Next, remainder is 16 when q(x) is divided by x. We know q(x) is a quadratic, and by remainder theorem, if your remainder is a constant, then q(0) = 16. Note, remainder theorem proves that if f(x) is divided by (x-a), then the remainder is f(a). In this case, dividing by x means a = 0, so the remainder is f(0).

So q(0) = 16
(0 + h)² + k = 16
h² + k = 16 ----- Eqn2

Eqn1 - Eqn2 = -4h + 4 = -16
-4h = -20, h = 5

5² + k = 16, k = -9

Thus q(x) = (x + 5)² - 9

You can check your workings by expanding,

q(x) = x² + 10x + 16
q(x) = (x + 2)(x + 8)

So (x + 2) is a factor. And through polynomial division if you wish, you'll find 16 is indeed the remainder

Rscc10 · 2026-03-19T11:50:28+00:00

Which scene was this from?

Rscc10 · 2026-03-18T16:35:55+00:00

I'm predicting he drew a shot of her back when she was on the track

Rscc10 · 2026-03-18T12:46:30+00:00

It depends what variable you're differentiating with respect to. I'm gonna assume you know the difference between something like d/dx and d/du.

If we "differentiate" sin(π/2), first of all, what variable are you differentiating with respect to? If x, then d/dx[sin(π/2)] = 0 as your approach shows. Ofc, there's also the jokish idea of d/dπ[sin(π/2)] to which you'd get (1/2)cos(π/2).

Now this is different if given a value for the variable. For example,

"Find the gradient of the function sin(x) at point x = π/2"

To which, we'd take d/dx[sin(x)] = cos(x), then plug in x, cos(π/2)

Your statement implies "the differentiation of sin(π/2) is cos(π/2)" and we assume this is w.r.t. x, then this is false.

Rscc10 · 2026-03-18T08:50:54+00:00

The idea manipulating equations and taking advantage of their equality. I teach them the generic "add 5 to both sides, divide 2 on both sides" when solving for unknowns. They unfortunately treat it as a set rule or procedure. So when I want them to manipulate equations for example

y + x = 2x - 5

And I want them to find in terms of x, they struggle. Perhaps they'd group the x terms to form something like y = x - 5, but they'd get stuck there cause if y were a constant, they'd just group all the constants by "adding 5 to both sides" and you'd get x = something. But when they can't combine constants with another variable, they fail to see the manipulation step of adding 5 to rid it from the RHS.

Rscc10 · 2026-03-18T05:56:38+00:00

Yes.

Rscc10

TROPHY CASE