help-10

[static cedar]

Okay

[woeful edge]

what about the values mutliplied by b

[static cedar]

Uhh would it be -x=(1/5+3/5)

[woeful edge]

lovely

now u have a system of equations

formed of the two equations u created

now

do you know how to solve a system of equations...?

[static cedar]

Looking at it uhm no

[woeful edge]

oh god

[static cedar]

Idk if I’m meant to substitute 😭

[static cedar]

I PROBABLY DO

Is it simultaneously

[woeful edge]

u can do that yess

find y on an ewuation in terms of x and substitute it in the second, u would have an equation with one unknown

[static cedar]

Uh

[woeful edge]

oh god

[static cedar]

Is it (4/5+3/5x)=y

[static cedar]

I can do it trust 😔

[woeful edge]

theres a y missing here btw next to 3/5

[static cedar]

Ohhh whoops

[static cedar]

Wait no

Isn’t that for the -x?

This is what I have

[woeful edge]

u missed a y somewhere in ur work

[static cedar]

Where

Yeah I think ur right

[woeful edge]

see at the third line (from bottom going up)

+3/5by

when u factorized by b

u forgot to put the y

[static cedar]

Oh yeah

My bad

So then

I can substitute the (4/5+3/5x)

Into (1/5+3/5y)

[woeful edge]

yessssssss

[static cedar]

As this equals y

[static cedar]

This is so long

[woeful edge]

yeah its the two pathway find whatever its called idk

the way used in ur book

[static cedar]

Do I just multiply it

And that equals -x

[woeful edge]

mhm

it becomes [1/5 + 3/5(4/5+3/5x)] = -x

[static cedar]

Uh I got 26/25x=-x

[woeful edge]

and find x

then smt is wrong

theres smt wrong

[static cedar]

I still get 26/25

[woeful edge]

u would get x=0 which is not right

[static cedar]

Ah

[woeful edge]

wait what

ur wrong

[blissful finch]

What's the original question?

[woeful edge]

1/5 + 3/5(4/5 + 3/5x) = -x

[woeful edge]

in pinned messages

[woeful edge]

this means 1/5 + 12/25 + 9/25x = -x

[static cedar]

does this not add up to 26??

[woeful edge]

this means 17/25 = -34/25x

[blissful finch]

You have to show that the position vector for midpoint of MN and ST are equal

[woeful edge]

how

[static cedar]

Oh wait nvm

[woeful edge]

u cant add up terms with terms multiplied by the x

[static cedar]

Yeah I realised after

[woeful edge]

so in short

what is x equal to?

[static cedar]

-1/2

[woeful edge]

nice

y remains now

[static cedar]

But it’s meant to positive no

The 1/2

[woeful edge]

why

the correction said so?

[static cedar]

Markscheme

Yeah

[static cedar]

I mean it was -x so x on its own would be a 1/2 right

[woeful edge]

when u wrote OD = OT + TD

u then write in the spot of TD:
x(b - 3/5a)

[static cedar]

Yeah

[woeful edge]

yeah and in the mark scheme its reversed

its multiplied by -1

[static cedar]

ohhh

[woeful edge]

in the markscheme they treated as x like a constant like u can multiply hy -1 cause -1 times constant x remainds a constant

[static cedar]

Y is = 1/2

[woeful edge]

yep

[static cedar]

Is the question done once u find the values are the same

[woeful edge]

did u find that or u read the markscheme lol

[woeful edge]

no u have to say why when x = 1/2 and y = 1/2 D represents the midpoint of each MN and ST

[woeful edge]

ah ok

[static cedar]

I HAVE TO REASON TOO?

[woeful edge]

well

yes

if u wanna fully understand yes lol

[blissful finch]

I don't like the mark scheme

[woeful edge]

i know right

[static cedar]

The mark schemes are evill

I’ve lost hella marks cus of it previous tests

[blissful finch]

This is a much better way:

First, define the position vectors of the vertices of the parallelogram $OACB$ relative to origin $O$:
$$\vec{OA} = \mathbf{a}$$
$$\vec{OB} = \mathbf{b}$$
$$\vec{OC} = \mathbf{a} + \mathbf{b}$$

Using the given ratio $1:4$ for points $M, S, N, T$ on $OB, BC, CA, AO$:
$$\vec{OM} = \frac{1}{5}\vec{OB} = \mathbf{\frac{1}{5}b}$$
$$\vec{OS} = \vec{OB} + \frac{1}{5}\vec{BC} = \mathbf{b} + \frac{1}{5}\mathbf{a}$$
$$\vec{ON} = \vec{OC} + \frac{1}{5}\vec{CA} = (\mathbf{a} + \mathbf{b}) + \frac{1}{5}(-\mathbf{b}) = \mathbf{a} + \frac{4}{5}\mathbf{b}$$
$$\vec{OT} = \vec{OA} + \frac{1}{5}\vec{AO} = \mathbf{a} - \frac{1}{5}\mathbf{a} = \mathbf{\frac{4}{5}a}$$

Now we find the midpoint of $MN$. Let $P$ be the midpoint of $MN$. The position vector $\vec{OP}$ is given by:
$$\vec{OP} = \frac{1}{2}(\vec{OM} + \vec{ON})$$
$$\vec{OP} = \frac{1}{2}\left(\frac{1}{5}\mathbf{b} + \mathbf{a} + \frac{4}{5}\mathbf{b}\right)$$
$$\vec{OP} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$$

Next we find the midpoint of $ST$. Let $Q$ be the midpoint of $ST$. The position vector $\vec{OQ}$ is given by:

$$\vec{OQ} = \frac{1}{2}(\vec{OS} + \vec{OT})$$
$$\vec{OQ} = \frac{1}{2}\left(\mathbf{b} + \frac{1}{5}\mathbf{a} + \frac{4}{5}\mathbf{a}\right)$$
$$\vec{OQ} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$$

Since they are equal, we have proven (c).

[warm shale]

Ajay

[blissful finch]

So much easier and more elegant

[woeful edge]

ive stated a better way at the start

lol

[blissful finch]

Is this A levels?

[static cedar]

yepp

[blissful finch]

A levels like geometry

[static cedar]

Don’t get me started

Year 2 is much worse I’m not looking forward to it

[blissful finch]

Ever think about doing the IB?

[static cedar]

What’s that

[blissful finch]

International Baccalaureate

The A level equivalent

[static cedar]

I heard for my course that I wanna do alevel maths is quite favoured as a filler subject

so I chose it

Never heard of the IB

weird

[blissful finch]

Are you in the UK?

[static cedar]

yess

[blissful finch]

ohhh no wonder

[static cedar]

I’m finally done with the homework 🙏

thanks guys

[obtuse pebble]

@static cedar Has your question been resolved?

[obtuse pebble]

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[sinful locust]

is the square root of x 1...

[hybrid cosmos]

um, what's the context behind this question? could you please show the original question?

[hybrid cosmos]

the square root of x is just the square root of x, and is definitely not 1.

[sinful locust]

is 1/2squareroot x correct

[hybrid cosmos]

in what sense? what are you trying to do?

and what is 1/2 square root x?

[sinful locust]

is the square root of x 1 or 1/2squareroot of x

[mortal blade]

,tex .diff power

[warm shale]

Xavier 🌺

[pine wharf]

yes this

so is the square root of x 1 or 1/2 square root of x

[mortal blade]

Derivative of the square root

[mortal blade]

The square root of x depends on the value of x

[wild glade]

1/2 sqrt x

[hybrid cosmos]

exactly as stated it is neither, but I also have no idea what "1/2 square root of x" means.

[wild glade]

but do you have a value for x

[pine wharf]

💔

[mortal blade]

What you're asking for is possibly the derivative of the square root

[pine wharf]

how do i show

[hybrid cosmos]

three helpers, so I'm stepping back. all the best, OP.

[pine wharf]

1/2squareroot of x

ok hold on

ill show u guys the original equation

[wild glade]

okay

[pine wharf]

this

so

what i did

2(5x^3-square root of x)

(20x^3 - ??????)

[mortal blade]

Firstly that's now what you're supposed to do here

,tex .diff rules

[warm shale]

Xavier 🌺

[pine wharf]

?

[thin pawn]

He‘s right tho. Differentiating the outside, then the inside. He‘s just unsure of the derivative of sqrtx

Actually nope, my bad

[wild glade]

do you mean 5x^4 instead of 5x^3?

[thin pawn]

i forgor my calc 1 sobbingcrying

[pine wharf]

yes my bad

[mortal blade]

I was gonna comment what you were talking about but you realized lol

[pine wharf]

5x^4 i meanr

[wild glade]

yes

[mortal blade]

Anyway, I'm gonna step away, all yours

[thin pawn]

It‘s early in the morning

[pine wharf]

hlelp me dont leave

BRO NO

[wild glade]

I didnt say that

[rocky pine]

Use chain rule

[wild glade]

rewrite square root of x as x^ 1/2

[pine wharf]

yse

i used chai ruel

[wild glade]

he got the chain rule
he kust need the sqrt bit

[wild glade]

then differentiate this

[balmy mortar]

show what you've done

[pine wharf]

ok hold on

[hybrid cosmos]

the power rule applies for the square root too, just for your information.

[balmy mortar]

wait im so confused

why has the op switched accounts???

[pine wharf]

I COULDNT WAIT FOR 10 MINS

[wild glade]

what

[pine wharf]

SIRRY

[balmy mortar]

ok i was confused who was the op

[rocky pine]

$$\frac{d}{dx} \sqrt{x}=\frac{1}{2\sqrt{x}}$$

[warm shale]

crypt

[wild glade]

yes
rewrite the squaure root as x^(1/2)

[rocky pine]

Im really dumb

[wild glade]

dont say that

[rocky pine]

anyway just plug this into the formula you'll get teh answer

[wild glade]

at least you havent got 2 Es

unlike me

[balmy mortar]

$$\frac{d}{dx} x^{\frac12} = \frac12x^{\frac{-1}2}$$

[balmy mortar]

using the power rule.

[pine wharf]

so its

1/2x^-1/2

[rocky pine]

yeah, You can write it in radical form

[pine wharf]

thank you guys i got it

❤️

[wild glade]

❤️

[rocky pine]

yayy

[hybrid cosmos]

any other remaining question?

[pine wharf]

nuh uh

[sinful locust]

.close

[obtuse pebble]

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[rapid osprey]

$f:\mathbb{R} \to \mathbb{R} : f(x+y) + f(xy) = f(x) + f(y) + f(1+xy) \forall x,y \in \mathbb{R}$. f is continuous

[warm shale]

CherryMan

[rapid osprey]

we found that f(x) = a(x-1) + b(x-1)^2 works

but cant prove it

[fossil crag]

Did you manage to prove f(1) = 0

[rapid osprey]

yea

thats the motivation to test x-1 and (x-1)^2

and this fe "is linear"

f1,f2 being solutions means af1 + bf2 is a solution

[fossil crag]

I'm tempted to check x = y = t and x = y = -t

and then look at the odd part of f, g(x) = (f(x)-f(-x))/2

[obtuse pebble]

@rapid osprey Has your question been resolved?

[fossil crag]

have you tried this @rapid osprey

[rapid osprey]

i dont get much

[fossil crag]

$x = y = t$: $f(2t) + f(t^2) = 2f(t) + f(1+t^2)$

$x = y = -t$: $f(-2t) + f(t^2) = 2f(-t) + f(1+t^2)$

[warm shale]

Rafilouyear2026

[rapid osprey]

oh

[fossil crag]

equation1 - equation2

[rapid osprey]

g(2t)=2g(t)

[fossil crag]

what are the solutions to that

[rapid osprey]

g is continuous

so g

wait

2^x

no

ax

[fossil crag]

yes

g(x) = ax

It can be more obvious if instead of taking x = y you just replaced x by -x and y by -y

$$f(x+y) + f(xy) = f(x)+f(y) + f(1+xy)$$
Then replacing "x" by "-x" and "y" by "-y":
$$f(-x-y) + f(xy) = f(-x)+f(-y) + f(1+xy)$$

[rapid osprey]

g(x+y)=g(x)+g(y)

[warm shale]

Rafilouyear2026

[fossil crag]

exactly

ok, now we've looked at the odd part of f

we only need to figure out its even part

[fossil crag]

So suppose f is even in here

[rapid osprey]

how can we suppose f is even

[fossil crag]

as you said it, the equation is linear

mmh wait

nvm

[rapid osprey]

i mean we can

[fossil crag]

look at (f(x) + f(-x))/2

[rapid osprey]

ok

$g(x+y)-g(x)-g(y) = f(1+xy)-f(xy)$

[warm shale]

CherryMan

[rapid osprey]

$f(2t) + ct + f(t^2) = 2f(t) + 2ct + f(1+t^2)$

[warm shale]

CherryMan

[rapid osprey]

wait that means c=0??

did i make a mistake

cus f(x)=f(-x) isnt true

[fossil crag]

let's write it another way

[fossil crag]

g(x) = ax is the odd part of f

so let's look at h(x) = f(x) - ax, even

[rapid osprey]

wait

[rapid osprey]

is this wrong

i get c=0 which isnt true

[fossil crag]

yeah it looks wrong

[fossil crag]

since f(x) = h(x) + ax

[fossil crag]

we have h(x+y) + a(x+y) + h(xy) + axy = ...

simplifies to $h(x+y) + h(xy) = h(x) + h(y) + h(1+xy) + a$

[warm shale]

Rafilouyear2026

[rapid osprey]

right

[fossil crag]

h(1) = -a

let's see what else we can figure out about this

most likely swapping "y" with "-y" for example

or y = -1 directly

mmh no it's not interesting

[rapid osprey]

i still think this is worth looking at

x=y=-t

and wherever we get f(-2t) or f(-t) we replace with f(2t) +2ct or f(t) + ct

everything cancels out so we get 2ct=ct

[fossil crag]

I'm not sure with how you simplify stuff

[rapid osprey]

alright

f(2t) + f(t^2) = 2f(t) + f(1+t^2)

and f(-2t) + f(t^2) = 2f(-t) + f(1+t^2)

oh

nvm

i understood where i went wrong

[fossil crag]

yeah, also it wouldn't lead to anything new

since we already got something out of "x=y=t" + "x=y=-t"

it led us to determine the odd part of f

[fossil crag]

I guess technically

you can rewrite this as

$h(x+y) + h(1) + h(xy) = h(x) + h(y) + h(1+xy)$

[warm shale]

Rafilouyear2026

[fossil crag]

Looks cool, but I haven't thought on how to get something out of this yet

[rapid osprey]

hm

[fossil crag]

replacing y by -y maybe

h(x-y) + h(1) + h(xy) = h(x) + h(y) + h(1-xy)

[rapid osprey]

yes

we havent tried opposite signs

maybe thats where we use the odd part

h(-xy) and h(-y) ?

[fossil crag]

h is even

[rapid osprey]

oh

right

[fossil crag]

so h(x+y) - h(x-y) = h(1+xy) - h(1-xy)

y = 1: h(1+x) = h(1-x)

wait

[rapid osprey]

y=1 gives h(x+1) = h(1+x)

[fossil crag]

yeah mb

welp I guess x = y or x = -y

[rapid osprey]

h(2x)-h(0) = h(x^2+1) - h(x^2-1)

[rapid osprey]

.close

[obtuse pebble]

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[limber sandal]

How do I evaluate sin(5pi/12)

[limber sandal]

Without using a calculator

[vestal meadow]

use the identities

[limber sandal]

Is there a way to rewrite this to a standard angle?

[vestal meadow]

yes

[latent quiver]

try to separate it into the sum of two angles that you do know

[vestal meadow]

thats another way

[unborn knoll]

or just, double angle twice

[limber sandal]

I'll begin with this

then I try the other methods

pi/4 + pi/6

What's the next step from here?

I expand it using the sin(x+y)?

And then evaluate the angles from there?

[latent quiver]

apply the sum identity [
\6\sin{x+y} = \sin(x)\cos(y)+\sin(y)\cos(x)
]

[limber sandal]

Got it

[latent quiver]

yes

[limber sandal]

Makes sense

[limber sandal]

What is this method?

[unborn knoll]

eh nvm that it requires more algebra steps

[limber sandal]

I don't mind

[warm shale]

firestepper

[wide zinc]

,tex .double angle

[warm shale]

Heyy, Hilda's here

[limber sandal]

I see

sin(2x) = 2sinxcosx

[unborn knoll]

,, \sin(x)=\sqrt{\sin(x)^2}

[warm shale]

cupcake

[unborn knoll]

not always

but it's right in this case

[vestal meadow]

its not

[unborn knoll]

[vestal meadow]

xD

[unborn knoll]

oh

wait

wait yah

[vestal meadow]

$|\sin(x)|=\sqrt{\sin(x)^2}$

why TEX

[unborn knoll]

yess

It's right

In this case

since 5pi/12<pi

[vestal meadow]

ok yea in this case

[unborn knoll]

no space between dollar sign and the thing inside

[warm shale]

Roy

[unborn knoll]

just algebra bashing to get cos(2x)

[vestal meadow]

bruh

[unborn knoll]

then cos(4x)

[unborn knoll]

using the first formula here

5pi/12*4=5pi/3

which is a standard angle

[limber sandal]

hmm

Got it

Thanks for the help

[unborn knoll]

oh wait 5pi/6 is already a standard angle

so no need for cos(4x)

[limber sandal]

Ya

.close

[obtuse pebble]

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[vapid bobcat]

i understand why q is 4 but im not sure how to find p

[old lily]

what did u get for A

[vapid bobcat]

part a is in the question, its a "show that" question

unless u meant the point A?

[old lily]

yeah

obvi

[vapid bobcat]

(4/cosh(theta), 0)

[old lily]

ok

gtg sorry

[vapid bobcat]

nws

[obtuse pebble]

@vapid bobcat Has your question been resolved?

[vapid bobcat]

.close

[obtuse pebble]

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[rich igloo]

I have been analyzing multiplayer gamesets (all possible games with a certain rule set, such as nim, hackandbush, ect) and have encountered an issue. The specific type of multiplayer game I am using is one where the first player to be unable to move loses, then the next loses 'less' (think of this as the player to lose first gains 0 'points' second gains '1' and so on. The players try and maximize how many points they get), until the last player to be unable to move, who wins, this way, players will try to live as long as possible even when winning is not. The reason this is implemented so there is never a position when it is equally valued for a player to do more than one different move.

[rich igloo]

First, notation: the outcome class of a p player game is written as a p! length string where each letter represents who wins in a specific game order, where we go through each permutation to get the full string. For example, in a three player game, the first letter is who wins in the order left then middle then right, second is left then right then middle, third is middle then left then right, and so on. From this, we can find that the empty three player game is RMRLML. These strings are outcome classes.

In the p player gameset of nim, there are p outcome classes, each one being for when the nth player wins. However, I am not looking at normal nim, instead I am looking at a variation of nim where along with each game having a certain board, it also has what I am calling a player key. The player key is a tuple written (a,b,c...) with one value for each player, these value means that the player must make that many moves within their turn. Clarification: if a player is unable to make all moves necessary within their turn, they simply do not make a move at all and lose, instead of making moves until they lose.

What I have found: the maximum amount of outcome classes for a p player gameset is $p^{p!}$, for any two player gameset with player key m=(a,b), where a and b are different, the amount of outcome classes is $O(S,m)=4$ when n is the gameset of nim (notation is still iffy). I have also found that $O(Nim,(1,1,2))=19$, although I have not proven this yet, and my brute forcing method is not nearly good enough to generalize. I also know that O(Nim,(n…n)=|m|

[warm shale]

pinkishnova cgt queen

[rich igloo]

What I want to know: is there some way of taking any player key for the gameset of p player nim and get $O(Nim,m)=x$

[warm shale]

pinkishnova cgt queen

[obtuse pebble]

@rich igloo Has your question been resolved?

[rich igloo]

<@&286206848099549185>

[obtuse pebble]

@rich igloo Has your question been resolved?

[obtuse pebble]

@rich igloo Has your question been resolved?

[rich igloo]

slightly altering how the O function works, it takes in a set, like usual, these are the values that players can have, then the player key is a tuple as long as the set which gives how many players have that amount, so {1,2},(3,1) would give 3 players with 1 move and 1 player with 2

[obtuse pebble]

@rich igloo Has your question been resolved?

[obtuse pebble]

@rich igloo Has your question been resolved?

[obtuse pebble]

@rich igloo Has your question been resolved?

[winter marsh]

this is breaking my head a little

why has no one tried to help even a little

[rich igloo]

Do you want some clarification

[winter marsh]

no no i just need to read again

have you tried

reducing Nim to the 1D total-token game

then

[rich igloo]

What is that?

[winter marsh]

(G_m(T))

[rich igloo]

I’m sorry I don’t know what you are talking about

[winter marsh]

im thinking wait

so in your version of multiplayer Nim

the only thing that matters about a position is the total number of tokens left

you are looking at multiplayer gamesets

your win condition is

the first player to be unable to move loses (gets 0 points)

the next player to be unable to move loses less (gets 1 point)

…

the last player to be unable to move wins

So players are not trying to win or whatever

they are trying to survive as long as possible

right?

@rich igloo

[rich igloo]

Wait

[winter marsh]

okay waiting

[rich igloo]

Yes

[winter marsh]

ok

you then look at all permutations of turn order

and for each permutation you record who wins

[rich igloo]

Mhm

[winter marsh]

That gives a string of length (p!) which you call the outcome class or whatever

so far all good

[rich igloo]

Yep

[winter marsh]

now your special Nim variant

you are not using normal Nim

each player has a player key

m = (a,b,c,...)

[rich igloo]

No

There is one player key

That applies how many moves each player has

[winter marsh]

ok

so it applies to the whole game, not per player

[rich igloo]

Yep

[winter marsh]

so the player key tells us how many moves each player gets on their turn

[rich igloo]

Yes

Basically

[winter marsh]

and players may have different numbers of moves

but that information is encoded inside the single key

[rich igloo]

Mhm

[winter marsh]

so same Nim board, same rules

one global key that assigns move-counts to players

ok

[rich igloo]

Alright

[winter marsh]

mb i'm trying hard to understand

i have to lay it down

[rich igloo]

It’s alright

[winter marsh]

so claim is

if a player is unable to make all moves necessary within their turn, they make no move and lose

[rich igloo]

Yes

That’s part of the rules

[winter marsh]

i think there is a general way

[rich igloo]

To do what?

[winter marsh]

this is your question

[rich igloo]

Yes

[winter marsh]

am i wrong

[rich igloo]

Is there a direct formula for O(X,m)

[winter marsh]

no

[rich igloo]

Yes

That is what my question is asking

[winter marsh]

you can compute this by analyzing how the outcome class changes as the total number of tokens increases and counting the distinct outcome strings that appear

there's like no formula

[rich igloo]

I’ve tried that

[winter marsh]

what happened

[rich igloo]

Doesn’t seem to be a pattern

Plus it’s extremely computationally expensive

[rich igloo]

How do you know?

[winter marsh]

why is there no closed formula

a closed formula would mean

uhhh

you plug in m

you do some algebra

out comes x

no simulation

no case-by-case behavior

i don't think it can work here

because in your game

outcomes depend on long chains of eliminations

and then those chains depend on exact inequalities between move counts

[rich igloo]

well there is a formula for 2 player games

[winter marsh]

and then small changes can cause completely different elimination orders

and these changes happen at irregular token totals

[rich igloo]

and O(n,(1,1,1,1...)) games

[winter marsh]

if ther was a closed formula, you'd see symmetry, monotonicity etc

[rich igloo]

so there is not reason i see why there cant be a closed form solution

there is some symetry

kinda

here is some values

[winter marsh]

well having closed formulas in special cases does not imply a closed formula exists in general

[rich igloo]

i know

but why wouldnt there be anything for 3 player?

or in other words: can you prove that no formula exists

which i am having trouble with

[winter marsh]

i can't

that kind of statement is almost never provable lowkey

[rich igloo]

i know upper bounds, i know lower bounds, i know specific cases, i know some general cases

[winter marsh]

what counts as a “formula”

[rich igloo]

basicaly any statment that gives the exact right answer for all posible values plugged into O(X,m)

[winter marsh]

there is no known non-algorithmic statement that gives the exact value in general

[rich igloo]

how do you know?

[winter marsh]

https://klipy.com/gifs/golden-retriever-walking-away

[rich igloo]

there would have to be something intrinsicly diferent from 3 player games and up from 2 player games that makes it imposible to get a formula

[winter marsh]

i don't think anyone (including me) has ever exhibited one

[rich igloo]

which is something i have not found

[rich igloo]

yes, this is curently an unsolved problem

[winter marsh]

so no one has an exact, general statement for all O(X,m)

[rich igloo]

not yet at least

[winter marsh]

we checked bounds, cases, reductions and whatever

and the remaining gap is genuinely open

[rich igloo]

no

im the only person to work on this problem so far

i have checked not much

because my computer cant handle much

[winter marsh]

keyword only

and i feel like you'll be alone on this lowkey

i dont have a pc for that either

https://tenor.com/view/lorax-gif-23202671

[rich igloo]

and i think aprotching this from a math perspective would be more easy

alright, bye

[winter marsh]

i think

you're right to think that

that's the only relevant input i can give

[rich igloo]

which is why im doing it mathematicaly

[winter marsh]

should i try help freeze the problem and phrase it

or should i zip it lowkey

[rich igloo]

idk

up to you

[rich igloo]

I have an idea

[winter marsh]

I am all ears

[rich igloo]

Ok so the idea is in the works

I’m not entirely sure how to do it

I’m trying to see how the 0s work in the 2,2 game set, but the problem is there is a number that is identical to 0 but not equal to 0 and I’m very confused because this means addition doesn’t work

[kind snow]

guys what are you talking about

this seems intresting

do you mean with O(X, m) like "we dont have a clean formula or theorem that works for all values of X and m" and what is this big O
it isn't standard notation? anyways i just wanted to check out what you're talking about

[rich igloo]

big O gives the amount of outcome classes for a specific game set of nim

[rich igloo]

.close

[obtuse pebble]

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[sand roost]

questions like these might be better answered in #game-theory or perhaps #combinatorial-structures

your fellow cgt queens are there 😌

[rich igloo]

turns out its not cgt at all

well kinda

so i have to invent a new thing to answer this problem

[obtuse pebble]

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[errant aspen]

if i got 3 w's how can i solve the same equation ? will it be, w.x+w.x+w.x+b ? / i learned the equation from the other photo but i only know how to solve it if its 1 w. help please

[polar fossil]

it would be the dot product of w and (hours studied, previous scores, sleep hours, sample question papers practiced)

plus b

[errant aspen]

so is the equation going to be w.x+w.x+w.x+b ? sry im a bit dumb with math bare with me

instead of w.x+b

[polar fossil]

ummm it would be w.x but with a dot product

so like for i = 1 it would be
y = 1 * 7 + 0.7 * 99 - 0.5 * 9 + 0.2 * 1 + 0.1

see how i match up the coefficients from w with the values from the table

[errant aspen]

😭 so it is w.x+w.x+w.x+b

tysm

[polar fossil]

well there are four components.... not three

but like. yes i guess?

[errant aspen]

yeah i didn't notice

tyysmmm

do i write .close rn or what

.close

[obtuse pebble]

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[obtuse pebble]

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[proper thistle]

can someone explain the step 6

why do we go to positive r

for -1,3pi/2

[royal basin]

you face in the direction of angle 3pi/2 and then you moonwalk Michael Jackson style by 1 unit

[proper thistle]

i dont get it

🙁

[royal basin]

https://tenor.com/view/michae-jackson-moon-walk-gif-14999009

[proper thistle]

so i always face in the direction of the angle

and walk in the opposite direction

[royal basin]

you know how to think of coordinates as instructions/"directions" for how to get to a point, right?

[proper thistle]

so if it were (2,3pi/2) i would walk down

[royal basin]

down by 2 units yes

[proper thistle]

what

[royal basin]

the "opposite" direction is simply what negative r looks like

[proper thistle]

so i would land in negative y axis

[royal basin]

you would

[white widget]

guys can any1 help me i cant understand yk cosx/3=-1 its like 3p but how can i get -9p

[proper thistle]

yeah so lowkey i just face in the direction of the angle and walk whatever units but like michael jackson

[proper thistle]

this channel is taken

go to another channel bro

[white widget]

srry

[proper thistle]

@royal basin is that right

[winter cargo]

if value is negative at that angle

[proper thistle]

if its pos

then?

[winter cargo]

towards that angle

negative means away

[proper thistle]

so in negative we walk away from the the point

and in positive walk towards

[winter cargo]

yeah

[proper thistle]

can you explain this too

how do i graph it ugh

r = 1 + 2sin(theta/2). The question was to graph this polar curve and choose a parameter interval that produces the entire curve. I got values of r and theta from 0 to 2pi. i got (1,0), (1+root2, pi/2), (3,pi), (1+root2, 3pi/2), (1,2pi). i am now stuck

I am confused here again where to plot 3,pi

[obtuse pebble]

@proper thistle Has your question been resolved?

[obtuse pebble]

@proper thistle Has your question been resolved?

[icy notch]

r=3, theta=pi means that you cast a ray from origin so that the angle is pi between positive x semiaxis and the ray, and you take point on the ray that's in distance 3 from origin

The ray will be negative x semiaxis here

So the point in cartesian coordinates is (-3, 0)

[obtuse pebble]

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[obtuse pebble]

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[sweet fern]

Yo hello buddy

[sweet fern]

Bruh wrong server

Omg

Sorry

[split sluice]

.close please

[royal basin]

.close

[obtuse pebble]

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[royal basin]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[wraith mauve]

Can someone explain whether it's choice 1 or choice 2, and why?

[polar fossil]

easiest way is to draw the given relation and reflect it across y = x

[wraith mauve]

Can you elaborate?

[split sluice]

Graph the original function

The. Draw the line y=x on the same graph

[royal basin]

nop

actually the easiest way is to write down the equation describing r^-1

which is simply the same equation as r but with all x's replaced with y and vice-versa

[wraith mauve]

But the original function is V-shaped?
I don't see any inverted V, or did I do something wrong?

I don't understand

Is it choice 2?

Because |Y-1| can only be >= 0

From the inverted function X = |Y-1| + 2

So X cannot be less than 2?

Right?

[obtuse pebble]

@wraith mauve Has your question been resolved?

[hardy widget]

yes

[obtuse pebble]

@wraith mauve Has your question been resolved?

[obtuse pebble]

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[mental narwhal]

Hello, i need help with this <@&286206848099549185>

[inner sierra]

!15min

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[mental narwhal]

Oh, sry abt that

[inner sierra]

dw

so like $\left{a\right} = a - \lfloor a \rfloor$?

[mental narwhal]

Yes

[warm shale]

1 divided by 0 equals Infinity

[inner sierra]

hold up

so this is the definition that they gave you

try to expand the definition to $\frac{a}{b}$?

[warm shale]

1 divided by 0 equals Infinity

[mental narwhal]

I think i got it

Tx

[inner sierra]

alright

if you're done then

[inner sierra]

!done

[obtuse pebble]

If you are done with this channel, please mark your problem as solved by typing .close

[mental narwhal]

.close

[obtuse pebble]

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[obtuse pebble]

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[round stream]

I need help setting up the FBD for this one

[round stream]

I know in the two mass system you have that dividing line for the angles

But the three mass it's hard to discern where that 3rd mass would even be

<@&286206848099549185>

.close

[obtuse pebble]

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[obtuse pebble]

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[woeful nacelle]

can anyone explain how to do this

[woeful nacelle]

This is what I’m tryna do

But I don’t know how to cancel the sqr root

[daring ravine]

Sub y out?

[woeful nacelle]

?

[agile bane]

Well do you really need to cancel the sqrt?

[daring ravine]

you have y = x^2 / 2

use it

[woeful nacelle]

yeah ik

[agile bane]

Ye whats your new integral

[daring ravine]

Then?

[woeful nacelle]

but how do i integrate wiht teh sqaure root

[daring ravine]

u sub

[woeful nacelle]

ill sned the simplified version wait

[agile bane]

Isn't it1/2 x^2sqrt(x^2+1)?

Like the integral of this

[woeful nacelle]

yep

im just stuggling with the integrating part

but yeha i think u subs gonna work

[agile bane]

What you gonna sub

[woeful nacelle]

im trying with u=1+x^2

[agile bane]

Wait this is your integral right?

[woeful nacelle]

yeah

[agile bane]

I sont think u=x^2+1 is gonna work

[daring ravine]

x^2 = u - 1

[agile bane]

2xdx=du

And then u still left with an x

This is a hyperbolic sub

[woeful nacelle]

im usingx=sqrt(u-1)

[agile bane]

You covered hyperbolics?

[woeful nacelle]

nah

[agile bane]

You covered trig sub?

[woeful nacelle]

ye

[agile bane]

Alr I think a trig sub is kost suitable here

[daring ravine]

is there a typo in this problem actually

i dont think it should be this complicated but ok

[agile bane]

Maybe cause I think this is a lil diffucult for him considering they didnt cover huperbolics

[agile bane]

Alr @woeful nacelle what trig sub do you think is most suitable

[woeful nacelle]

im not sure

none of the inverse trig functions have sqrt(1+x^2)

[agile bane]

Ye but focus on the x^2+1

[daring ravine]

you dont need to exactly match the whole thing

[agile bane]

You wanna get a pure trig integral with no radicals

[woeful nacelle]

thats arc tan

[daring ravine]

x^2 + 1 = a^2 for some x

what could x and a be

think trig identities

[agile bane]

Nice so what sub should you use

[woeful nacelle]

u=1/arctanx

?

idk

[agile bane]

Nah you just said it before

[woeful nacelle]

but using arc tan would give u 1/1+x^2

right

as du/dx

[agile bane]

Well if u=arctanx doesnt that imply x=tanu

[woeful nacelle]

yeah

[agile bane]

But yeah I agree I think there might be a typo cause this problem gonna become nastier than its supposed to be

[agile bane]

Alr then if you differentiate now what do ya get

[woeful nacelle]

1/1+u^2

[agile bane]

But its 10 points so maybe its intended

[woeful nacelle]

fishy ngl

[woeful nacelle]

blud idts

[agile bane]

Nah think about it x=tanu

[woeful nacelle]

easier qs than this have been like 20

[woeful nacelle]

oh tan mb

sec^2 u

[agile bane]

Ph yeah idk cause for us most problems are like 3 marks a piece

[agile bane]

Alr so dx=sec^2du right?

[woeful nacelle]

yeh

[agile bane]

Alr now we can sub it back into the integral

[woeful nacelle]

ok

[agile bane]

But I mean look at this even the limits aren't nice numbers

[woeful nacelle]

but nothings rlly cacnelling out

lol yeha

is there anyway to do it not using trig

[agile bane]

Hyperbolics

[woeful nacelle]

cuz in class most of the probs we did were where the differential of the fucntion ended up being a complete sqaure

so teh sqrt cancelled out

[woeful nacelle]

havent done those

[agile bane]

Ye I guess trig sub it is

[agile bane]

Ye like this is a common problem like usually they won't assign nasty integrals like this, especially if you haven't covered hyperbolics

[woeful nacelle]

mmm

[agile bane]

But yeah just leave the limit as arctan(3)

[woeful nacelle]

imma confirm this

[agile bane]

Confirm what

[woeful nacelle]

if its not a mistake ill ask again

but thanks for the help

[agile bane]

Np

[obtuse pebble]

@woeful nacelle Has your question been resolved?

[obtuse pebble]

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[warped holly]

Could someone explain to me the strong form of the pigeonhole principle?

[warped holly]

I mean i got the simple form, but i have no idea what any of this means

[kind hawk]

do an example

take your favourite three numbers

plug them in for q_1, q_2, q_3

read everything again with those values substituted

[warped holly]

i did and it still doesnt make any sense

not even sure what these positive integers are meant to represent

[obtuse pebble]

@warped holly Has your question been resolved?

[warped holly]

no

[obtuse pebble]

@warped holly Has your question been resolved?

[ocean ingot]

Its like, you've assigned the n boxes the integers q_i

So the ith box has corresponding integer q_i

Since we have q_1 + q_2 + ... + q_n - n + 1 objects, at least one of the boxes must have as many objects in it as the number we gave it

[warped holly]

is it like their capacity?

i dont really get it

[ocean ingot]

If we make all the q_i = 2 the we get the weak form, where having n+1 objects and n boxes means at least 1 box has 2 objects in it

Not so much capacity as... a lower threshold

Like if box i has q_i or more objects in it a light above it will turn on

If we have q_1 + ... + q_n - n + 1 objects then at least one light will turn on

[warped holly]

maybe im just dumb because i swear i still do not get it

[ocean ingot]

Hmm

[warped holly]

i cant see it graphically

[ocean ingot]

How about...

If we had m objects, n boxes, and the numbers q_1,...,q_n where q_1 + ... + q_n = m

Then for each box i we could put exactly q_i objects in it and have no leftover, right?

[warped holly]

so q is kind of like how many items we want a box to have lets say

[ocean ingot]

Yeah!

[warped holly]

its intended capacity

[ocean ingot]

Capacity might not be the right word, since that's sorta like the max

[warped holly]

fair enough

but we do have a leftover dont we?

[ocean ingot]

Yeah

So this is basically saying that while not every box will have its number of objects in it

At least one box will

[warped holly]

wait i think i get it

so we're removing n so every box potentially is missing one item

and then we're adding one to potentially complete at least one box

[ocean ingot]

Yeah, then adding 1 so at least one must match its number

Yup!

[warped holly]

ohhhh i can see it now

i should probably draw this

thanks a lot

[ocean ingot]

That sounds like a great idea

Glad I could help!

[warped holly]

tyy i was going crazy over this haha

[ocean ingot]

Lmao no problem

[warped holly]

.close

[obtuse pebble]

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[obtuse pebble]

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[frigid burrow]

Draw the circuit and send

[dense yoke]

dropping in to say long time no see

ok out

[tawdry mesa]

wait a min

Idk if i draw it correctly

[frigid burrow]

That red part should not be there

[tawdry mesa]

how do i assign mesh currents?

[frigid burrow]

I assume that just say i1 current passes through cell1 and i2 from cell2

So i1 + i2 passes through the 2 ohm resistor

[tawdry mesa]

wait a min]

does this work?

[obtuse pebble]

@tawdry mesa Has your question been resolved?

[obtuse pebble]

@tawdry mesa Has your question been resolved?

[fossil pendant]

what is a mesh current

are you doing kirchoff's?

[fossil pendant]

might want to clean this up a bit fefore starting tbh

you can reduce some of the resistances.

[tawdry mesa]

yes

[fossil pendant]

idk what a mesh current is tho.

[tawdry mesa]

its like a looped circuit

[fossil pendant]

ok sorry I cant help you. Iam extremely dizzy

greatest of luck.

[obtuse pebble]

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[obtuse pebble]

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[neon ice]

@winter marsh

[neon ice]

this is my other ID

[winter marsh]

is this hehehehehe

[agile bane]

Why is bro laughing mid sentence

[winter marsh]

that was his username, “bro”

@proper thistle

[short crescent]

close this channel if you dont need math help, if you guys want to talk theres #discussion

[obtuse pebble]

@neon ice Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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[plucky mountain]

can i get a clue for 6b

[plucky mountain]

brain just doesnt want to brain today

[long current]

AB is parallel to DC, so BAD + ADC = 180

[plucky mountain]

ohhh

u right

wait

[uneven dagger]

You can also apply the alternate angle property

[plucky mountain]

w

[plucky mountain]

this doesnt make sense

wait

is my diagram inaccurate

theres that random angle in the middle

see

waiut

thats the angle i hjavbe to solve for

lmao

yeah i got it

ty

.close

[obtuse pebble]

Channel closed

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[plucky mountain]

.reopen

[obtuse pebble]

✅ Original question: #help-10 message

[plucky mountain]

.solved

[obtuse pebble]

Channel closed

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[swift vapor]

any1 can help me in linear eq in 2 varible

?

[plucky mountain]

#❓how-to-get-help

read it twin

[swift vapor]

ok

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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[kind prairie]

can anyone guide me with jee maths

[kind prairie]

yoo

[frozen wing]

Yes it

[solar elbow]

hi OP, you may find it faster to get helpers to look at your question by showing some example questions you need help with!

[frozen wing]

Ig

[frozen wing]

11th 12th or dropper?

[kind prairie]

10th

but i wanna study jee maths

[frozen wing]

Ok

[kind prairie]

cuz im too cooked in 10th boards

[hollow grail]

Bruh

[frozen wing]

Not a nice idea

[kind prairie]

;-;

i havent studied even a bit

[frozen wing]

Still tell the question

[kind prairie]

and its

4days

later

[kind prairie]

actually i wanted to learn a subject

[frozen wing]

Abe question bath

[kind prairie]

inequality

[frozen wing]

Matlab?

[kind prairie]

inequality bc

sets

topic hai na

sets relation ka

how to solve inequality

[winter cargo]

jee can wait

[kind prairie]

gng how do i study bout 10th boards, i am way too cooked

havent studied even once

the year

and now

in 4days

[frozen wing]

Bohot sari hoti hai example de ki tu kaise question kar rha hai

Kyuki olympiads walk bhi punch let’s hai jee mai, at least mere coaching ke mocks mei

[kind prairie]

u also giving jee

?

[frozen wing]

Yeah

[kind prairie]

cool

can u also give tips

?

[frozen wing]

Padhai kar

[kind prairie]

asking other stuff like tips and other shi is allowed here right

last time i got banned on a diff srv

[frozen wing]

Ig okay hai

Padhai kar itna mushkil nahi hai agar tu consistent hai

Vaise ab mains hard Hota Jaa rha hai

[kind prairie]

I see

[obtuse pebble]

@kind prairie Has your question been resolved?

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[gritty lintel]

gimme some important questions on cbse triginometry class 10th

[obtuse pebble]

@gritty lintel Has your question been resolved?

[high lily]

Don't you have books/worksheets/handouts in class?

[obtuse pebble]

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[obtuse pebble]

Available help channel!

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Remember:
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• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[winged jungle]

Can $x > \frac{4}{3}$ also be written as $\frac{4}{3} < x$?

[warm shale]

Vortac

[glad vapor]

yes

[winged jungle]

Thank you!

.close

[obtuse pebble]

Channel closed

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[obtuse pebble]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tacit timber]

not sure what b is asking for

[tacit timber]

I've done a and I think it should be correct?

[hardy widget]

yeah that matrix is good

part b is saying that the two populations always add to 3000

why does this make sense

also to verify that the poupulations actually do add to 3000 for t=1,2,3

[tacit timber]

oh ok I got a bit ocnfused with what formula its referring to and couldn't figure out how to use the matrix mb
think I got it now
thanks

.close

[obtuse pebble]

Channel closed

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[hardy widget]

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[coarse birch]

I have this question and I have no idea how to find the point A