help-13

[crimson sedge]

yeah i agree

[fallen heath]

okay so

we took any random arrangement, picked one such file containing a rook, and labelled the file containing it, as say F_1, then you mark the corresponding rank as R_1

[crimson sedge]

yes

so we can choose any other 4 rooks not in that row or column

right?

[fallen heath]

yes, but there's one better idea

oops that was a wrong image

since the rooks attacking each other are only dependent on the respective positions of row, column, this move should be invariant right?

[crimson sedge]

which ones did you move?

the rank wzs never filled

[fallen heath]

Yeah. I filled it cuz, we're considering a situation where "okay, we for sure have at least 5 files to pick from... and then, 5 ranks to pick from, but what if the 4 remaining rooks lie within the same file?"

[crimson sedge]

?

we have 7 files to pick from

where is the five coming from?????

[fallen heath]

7 yes, but we showed that there's at least 5

so in any possible arrangement, there's at least 5 to pick from

consider this arrangement.

is there really 7 files to pick from?

[crimson sedge]

yes

wait no

but its not 5

[fallen heath]

Yeah it's not 5 but

there you go

the one case where PHP is evident

even if you fill the 32 cells in the half-board, there's always 1 rook remaining

in this case you've only got 5 files

but also, you can very easily see you 5 rook pair

[crimson sedge]

you have 3 files to choose from

[fallen heath]

wdym 3?

[crimson sedge]

oh we are choosing from those

[fallen heath]

hmmm?

[crimson sedge]

so how can i write up the answer from the begining

[fallen heath]

hmm, even though I can tell you how this thing works and the process, I've never once been rewarded well for my combinatorics proof writings

[crimson sedge]

what did we start by saying?

i just need to go in order

[fallen heath]

we started by showing that the there exists at least 5 different rows and columns with at least 1 rook in them.

[crimson sedge]

ok i have that

next we choose one rook, right?

[fallen heath]

next you consider any random arrangement, and pick one rook, yes. [don't start writing your answer yet btw, coz if even once you get lost, you might end up scratching everything]

[crimson sedge]

ok

we then eliminate the file and rank it is in

what did we do after that @fallen heath

[fallen heath]

https://math.stackexchange.com/questions/478422/prove-that-if-33-rooks-are-placed-on-a-chessboard-at-least-five-dont-attack-on

here you go for a more formal proof writing

[crimson sedge]

but we just need to prove at least a pair out of those five rooks dont attack eachother

not that all the five rooks dont attack

[fallen heath]

what. no

[crimson sedge]

Show that you can find 5 among them such that any pair among them does not attack each other.

[fallen heath]

such that ANY PAIR among them does not attack EACH OTHER

[crimson sedge]

oh

[fallen heath]

so any pair you pick, should not be attacking each other

lol

idk why one would give an invariance question for h.w.

[crimson sedge]

invariance?

[crimson sedge]

i dont want to copy their proof

[fallen heath]

yep no worries

if you really plan on doing this yourself, here's a more accurate hint:

Color the board diagonally(symmetrically) in different colors. Your task is to prove that at least 5 rooks lie in one of these colors

@crimson sedge

notice how, as soon as you prove that there exists 5 rooks lying in one of these colors, you're done(since none of these colors attack each other)

[crimson sedge]

5 rooks in one color?

how does this relate to the question?

[crimson sedge]

how can i prove that?

[fallen heath]

that's what we've been trying to prove so far no?

[crimson sedge]

yeah

it si

[fallen heath]

that you've a pair of 5 such rooks

that, when arranged like this diagonally, do not attack each other

[crimson sedge]

^^

[fallen heath]

think by yourself

[crimson sedge]

it doesnt have to be possible

do i use php

i dont know what to do

oo

[fallen heath]

you have 33 rooks

you used the 33 = 8 x 4 + 1 argument column wise, row wise to conclude some info about rows, columns

think if you can conclude something using the same?

[crimson sedge]

right, the argument that there will be at least 1 rook in 5 files

[fallen heath]

about at least 5 rooks being in at least 1 color

[crimson sedge]

idk

its different

there is no fixed capacity for each diagonal @fallen heath

[fallen heath]

you're concerned about each COLORS

not diagonals

[crimson sedge]

oh

its the same thing

4*8+1

[fallen heath]

lmfao

now as you can see

whichever 5 rook pair lies in 1 of these colors, the rook in them don't attack each other

[crimson sedge]

ok

so it seems like we did something different

how should i write this up now??

[fallen heath]

not really. we followed up on the same PHP argument, just from a slightly more visible perspective

I was getting at this previously

but the colors seemed to have worked better

[crimson sedge]

how do i go right away to talking about colors lol

in my solution

[fallen heath]

XD

hmm.. start with defining colors for each diagonals :o

Like, "we colour each diagonal"

[crimson sedge]

so scratch everything we had before?

[fallen heath]

yeah, this approach is "more visible" so you won't have to worry about anything else

[fallen heath]

can you not see why any pair of rooks lying in any one of these colors won't attack each other?

[crimson sedge]

yes

[fallen heath]

then it won't be necessary to add more write up

so okay, you start with coloring the chessboard diagonally.

then you argue that, for any pair lying in one of these colors, the rooks won't attack each other

Got it?

[crimson sedge]

We can color each diagonal:
-- the image here
If we had at least 5 rooks in one color, they would not attack each other, and would satisfy the conditions. Each color spans across a capacity of 8 squares, and 32 is 4*8+1, which means that at least 5 colors must have at least 1 rook.

hows this?

[fallen heath]

It's fine

[crimson sedge]

it is?

[fallen heath]

yep

wait wait

heyyyyyyyy

[fallen heath]

you mean.. at least 1 color must have 5 rooks?

Your claim : "If we had at least 5 rooks in one color, they would not attack each other"
and your conclusion: "which means that at least 5 colors must have at least 1 rook."

.
.
don't match each other (@_@;)

[crimson sedge]

right

but that is literally what 4*8+1 means tho

@fallen heath

[jaunty mural]

5 colors must have at least one rook is true

However there is an alternate conclusion you can draw from php

[fallen heath]

shuri hereeeee

da heck @crimson sedge

what do you want to show ?

[crimson sedge]

?

[fallen heath]

what's the motive of doing all the coloring?

why did you color to start with?

what was your claim?

[crimson sedge]

to show that one color contains at leasy 5 rooks

[fallen heath]

correct

[crimson sedge]

but does 4*8+1 show that?

[jaunty mural]

can u see how php allows you to draw either conclusion?

[fallen heath]

how much capacity does each color have?

[jaunty mural]

think about it

[crimson sedge]

8

4*8 is 4 colors

[jaunty mural]

u mean how many colors are there

[fallen heath]

yes.

right

[jaunty mural]

thats the alternative we need

[fallen heath]

how many colors are there.

[crimson sedge]

8

[jaunty mural]

you can think about capacity or the number of the sets

so if there are 8 sets

start filling them up with 33 rooks

what must happen

[fallen heath]

so there are 8 colors

and at max only 8 rooks can be colored with same color

urrrrrrrr confusion

[crimson sedge]

?

[jaunty mural]

ansh is confuse

@crimson sedge hi

[crimson sedge]

crabbo too

[fallen heath]

massive confusion

[jaunty mural]

i have 8 baskets

i will now put 33 eggs in them

we want to conclude using php that at least 1 basket has 5 eggs

convince yourself this must always happen

[crimson sedge]

you put 4 in each

but you have 1 left over

[fallen heath]

yeah that's the approach

[jaunty mural]

exactly

[crimson sedge]

so one must have 5

[fallen heath]

🤦‍♂️

[jaunty mural]

that is php in a different way

so you can php on capacity or number of sets

we need num sets here

[fallen heath]

smh... the capacity/sets thing is confusing T_T

Just fill them up with equal numbers and show there's enough remaining for the PHP

[crimson sedge]

okj im done

thankkkks

[fallen heath]

:O

[crimson sedge]

biiiii

[jaunty mural]

@crimson sedge to follow up
What I initially suggested yesterday for the holes didn't work because of the counter examples. Wrong sets were suggested.

What would have been a much better approach for when u were stuck was to test the smaller examples.

3 rooks -> 2x2 chess board
9 rooks -> 4x4 chess board

This is how I thought about the diagonal argument. As always this is a good technique.

[crimson sedge]

yeah

this was a good technique

im gonna close this

[jaunty mural]

i mean the smaller cases technique

👋

[crimson sedge]

both

were good

.close

[cedar kiln]

Channel closed

Closed by @little wave

Use .reopen if this was a mistake.

[fallen heath]

:O

me curious about what the Z(mod 8) was about

[jaunty mural]

you are taking that board

that is integer lattice

Z^2

[fallen heath]

mhmm

(1,1) to (8,8)

[jaunty mural]

and mod 8 on each coord

00 to 77

[cedar kiln]

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[fallen heath]

claim!

[jaunty mural]

then the equivalence relation is what i said earlier

[fallen heath]

There are 33 rooks on a regular (8 × 8) chessboard. Show that you can find 5 among them such that any pair among them does not attack each other.

equivalence relation? (@_@;)

[jaunty mural]

(a, b) ~ (a + k, b + k) forall a b k

[fallen heath]

also, 00 to 77?

[jaunty mural]

consider this relation

[fallen heath]

that'd also include 19 and stuff..

[jaunty mural]

mod 8

(Z/8Z)^2

u have seen this or not

[fallen heath]

that's just (1,1) to (8,8) no?

[jaunty mural]

thats what i am describing fornally

00 to 77

including 07 and 70

64 squares

[fallen heath]

oh okie

[jaunty mural]

(a, b) ~ (a + k, b + k) forall a b k

then verify this is an equivalence r

and what the classes are

[fallen heath]

so equivalence relation as in...

what are relations anyways 0,0 mappings? but different from functions?

ohh then you show that the equivalence classes are 00, 01, 02, 03, 04, 05, 06, 07

[jaunty mural]

sorry

So firstly a relation is a ...

set of ordered pairs from a given set

So technically functions are types of relations (but thats not relevant here)

[crimson sedge]

we are still talking about the question lol

is this a different technique?

[jaunty mural]

I am talking about something more general

but using your Q as an example

In particular about relations and equivalence relations

They are a very very important concept in math, especially pure (but they work behind the scenes everywhere).

[fallen heath]

mhmm

understandable and commendable how you could relate

[jaunty mural]

well actually, if u want to put some time I thought of a good intuition for the whole idea of quotient

Without explaining the specifics

[fallen heath]

:o I'd like to know as much specifics as I can cover though... tbh

[jaunty mural]

Sure.

But I think I wouldve appreciated this intuition like 5 yrs ago =...=

😂

[fallen heath]

👀

[jaunty mural]

only recently it appeared because of quotient keeping coming up

[fallen heath]

you'd already read on Group Theory 5 yrs ago?

[jaunty mural]

We did very basics in high school but I did

start on it

[fallen heath]

wow 😱 nice

[jaunty mural]

The first time I encounctered quotient groups was maybe 5 yrs ago

And it doesnt make sense if u dont know wtf its meant to be

Anyways.... so back to relation --- do you get the concept

Any set of ordered pairs from the given set is a relation

$$\sim, :={(a, b) : a, b\in S}$$

[wraith dagger]

Shuri2060

[jaunty mural]

So ~ represents a set of ordered pairs, this is the usual initial definition given

[fallen heath]

Yep

Got it

and then there's symmetric, transitive and stuff

and the three combined is equivalent

[jaunty mural]

That is a definition on top of that

[fallen heath]

and the equivalent relations have partitions, called equivalence classes

[jaunty mural]

right

[fallen heath]

mhmm

[jaunty mural]

DId you get all of this?

[fallen heath]

(@_@;) none

[jaunty mural]

like why it makes sense

[fallen heath]

🚶‍♂️

[jaunty mural]

ok ok so first we have this

An alternative way of defining a relation

[fallen heath]

no waittttttt

I wanna read on sets for a bit before starting with relations

[jaunty mural]

$$a\sim b \iff (some condition)$$

[wraith dagger]

Shuri2060

[fallen heath]

and, I have an Algebra test tmrw

;-;

[jaunty mural]

you dont need to know that much about sets

[violet rapids]

Algebra 1

[fallen heath]

yeah almost everytime I was introduced to relations, it was only after a brief meeting with sets

but I really wanna know both better[ coz like, I feel like it'd be so nice knowing those ]

[jaunty mural]

What you did earlier to this board is form an equivalence relation

[fallen heath]

noticed after you shared your soln.

[jaunty mural]

All the squares of the same color are related to each other.

Its an equivalence relation <-> the sets partition the chess board

[fallen heath]

all the colors are (a, b) ~ (a + k, b + k)

[jaunty mural]

Right exactly

[fallen heath]

hence, the equivalence classes

[jaunty mural]

Now understanding what equivalence relations are, is v. important to the idea of quotients

For me the best way to describe a relation is that it glues stuff together

All the black squares are glued together, and are considered '1 unit', etc..

[fallen heath]

yeah we were given a brief lecture on equivalence and partitions, during that time

[jaunty mural]

Right so if we quotient the chess board with the given equivalence relation

Can you see what we end up with?

[fallen heath]

"quotient" ?

[jaunty mural]

or is that a bit of a leap 👀

[fallen heath]

how?

[jaunty mural]

haha i will explain first then

https://i.imgur.com/EllZp73.png

To divide 12/4

You can draw a 3x4 grid of dots

Form an equivalence relation on groups of 4 (thats the circles wrapping around them)

Then quotienting 12 by 4 means you treat each of the equivlance classes as a single point

[fallen heath]

lost me here

[jaunty mural]

1. Glue elements together
2. Treat the glued elements as 'one' element

[fallen heath]

hmmm

[jaunty mural]

We have the 3 classes right?

I define my equivalence relation to be what the loops are doing

Everything within (1) is related to each other, and same for (2), (3)

[fallen heath]

oh okay okay

got it

[jaunty mural]

So to 'quotient' 12 by 4, we do this process and end with 3 --- what you would show to a kid (without the complicated words).

[fallen heath]

1,2,3,4 | 5,6,7,8 | 9,10,11,12

[jaunty mural]

I just wanted to use this example to get across all mathematical quotients are the same concept

even normal division

On that grid, you form an equivalence relation saying a ~ b if they are in the same column

===
So the exact same idea is used for quotient groups, quotient rings, quotient topologies..... probably quotient anything

Except what makes these quotients special is the resulting thing still needs to be a group/ring/topology (you want to preserve some structure from your original thing)

That's why we have to form all those extra rules and stuff about normal subgroups, ideals, etc.

[fallen heath]

:c I think I'll have to go through these topics once before I can understand what you mean

it's getting too overwhelming

[jaunty mural]

👌

[fallen heath]

btw, I'll read a bit on rings and stuff today plus constructible numbers n all, are you free?

[jaunty mural]

for a few hrs from now I am

[fallen heath]

ah, hmm. okie

ty

[jaunty mural]

Did they teach you Cayley Graph for groups?

[jaunty mural]

If yes, that's a nice way to visualise this stuff when u look at some examples

[fallen heath]

no idea :c

[jaunty mural]

dw. I don't think I was ever taught it (probably just briefly mentioned)

[fallen heath]

mhmm

Imma go read now

.close

[cedar kiln]

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[cedar kiln]

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[sweet plaza]

Hello there, I'm trying to do a 3D curve interpolation, so far I figured out how to do Bezier-like curve (blue curve on the image), but what I actually want is a curve that goes through center point (green curve on the image). Then I need to expand it into a chain, so what I actually want to create is a function/equation that can get 3D coords of any point on curve for specific time/progress T anywhere between any set of points (let's say 15), which are known X,Y,Z (red dots on image). Any help is welcome.

[sweet plaza]

I'm aware, that I haven't described it really well, so feel free to ask.

[chilly warren]

A question, do you need the third curve to be the extreme point?

[sweet plaza]

I'm not really sure what that means to be honest. But I'll try to explain further.

This is how it would look like in 2D (although I need it in 3D), red dots are known X,Y,Z coordinates; green curve (or rather a path) is something the orange dot will travel on, so for any specific time I can get coordinates on the path.. orange dot coords are unknown, that is the output

[chilly warren]

Well, let's look at how bezier curve is defined.

$B[t] =L[t;L[t;P_0;P_1];L[t;P_1;P_2]]

[sweet plaza]

With some crazily placed 3D points it would look like this:

which is not a Bezier curve, Bezier is something I was able to do, but I was wrong, since it is not what I wanted.. therefore I'm trying to explain it with images

Bezier wouldn't go through P1 and P2

[sweet plaza]

Unfortunately I don't know how this is called.. probably a 3D spline curve?

[chilly warren]

cubic bezier curve

[sweet plaza]

Is it still bezier, even if it goes through all of the points?

[chilly warren]

Well, because you wanted it to be bezier, right?

It's just that the point it's crossing through won't be the control point

[sweet plaza]

because if I understand Bezier correctly, it has start point, end point and control points and the curve doesn't go through control points.. what I want is a path that curves and goes through all the specified points.. I'm not sure if I can use Bezier for that (probably not)

what I need is essentially a graph interpolation, but in 3D, something like this, where the path goes through all of the points:

[chilly warren]

It's like asking to find a polynomial interpolation given the point to cross.

Because you asked for polynomial function, it will be polynomial, even you don't define it from $ax^n + b x^{n-1}+ ...$

[wraith dagger]

Xwtek

[sweet plaza]

This might be good representation of what I want to accomplish.. I'm sorry for not being able to put it into right words and just pasting images..

[chilly warren]

Anyway, I found this link that you might find it useful

https://towardsdatascience.com/bézier-interpolation-8033e9a262c2

[sweet plaza]

I'll take a look, thanks

[chilly warren]

Basically, you make a cubic bezier curve for every two points and then you solve for the control points so the curve will look smooth

[sweet plaza]

anyway what I need in the end is something like this, where I can use T as input (with set of grey points) and get red point X,Y,Z (which are unknown)

I don't really mind which interpolation or curve is used (doesn't have to be Bezier), as long as the path is smooth and I can get the unknown point coordinates on the path between known points

I thing this is more like what I need https://en.wikipedia.org/wiki/Centripetal_Catmull–Rom_spline

okay, I figured it out using the Catmull-Rom Spline, thanks anyways

.close

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[bronze cave]

i need help on this question

[sand ether]

How to find slope

[cedar kiln]

@bronze cave Has your question been resolved?

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[cedar kiln]

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[quiet summit]

Having trouble understanding how to negate (c)

I have tried letting a,b be reals such that a < b and r be a rational number such that a < r < b.

[cedar kiln]

@quiet summit Has your question been resolved?

[quiet summit]

<@&286206848099549185>

[quiet summit]

.close

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[meager shuttle]

Hey, I fail to see how the equation was simplified between these two blocks. Should be fairly easy I hope

[chilly warren]

What is $\delta$?

[wraith dagger]

Xwtek

[meager shuttle]

Dirac delta function

[dire geode]

are you just stuck on algebra?

set $\alpha = \omega_1 - \omega_2$ and $\beta = \omega_1 + \omega_2$, re-write your initial $F_l$ and see if you can simplify

[wraith dagger]

riemann

[meager shuttle]

Possibly, I just don't understand what step was done between the gap. The script just said "summarize"

[dire geode]

there are terms in the first set of brackets that cancel the second set of brackets

[dire geode]

do this and if you're still stuck, write $g_{\pm\alpha}= \delta(\omega \pm \alpha)$ and $h_{\pm \beta} = \delta(\omega \pm \beta)$

[wraith dagger]

riemann

[meager shuttle]

No way, I feel insanely stupid.

I appreciate the help! Thanks

For some reason I limited myself to inside the brackets

.close

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[crisp trail]

how do i find the p in y=1/5x^2?

[dire geode]

what's p?

[crisp trail]

x^2=4py

the p

i’m trying to find the directrix

[dire geode]

$y = \frac{1}{4p} x^2$

[wraith dagger]

riemann

[dire geode]

do you see how i got ^?

[crisp trail]

umm

no

[dire geode]

divide both sides by $4p$

[wraith dagger]

riemann

[crisp trail]

noo omg

i’m trying to find

the p

in

y=1/5x^2

[dire geode]

right

[crisp trail]

and directrix formula is y=4pm

*py

x=4py**

my bad

i know that if x wasn’t squared it would be

[crisp trail]

1/5=4p

[dire geode]

nope

https://mathbitsnotebook.com/Geometry/Equations/EQParabola.html

[crisp trail]

that’s if the square isn’t there

but it is there

[dire geode]

this is your equation right?
$y = \frac{1}{5} x^2$

[wraith dagger]

riemann

[crisp trail]

yes

[dire geode]

this is very close, but still wrong

[crisp trail]

how would i make it correct

what r the steps

[dire geode]

do you understand this equation?

[crisp trail]

sure

[dire geode]

set it equal to this one

[crisp trail]

that’s a variation of x=4py correct?

[dire geode]

read this

[dire geode]

this is irrelevant to your problem

[crisp trail]

i read it

it’s almost the same as my current notes

[dire geode]

do this

[crisp trail]

so 1/5x^2=1/4p x^2?

[dire geode]

yup

rest is algebra, but do it slowly

[crisp trail]

gotchu

lemme try to solve

is it 5/4

[dire geode]

looks right

👏🏻

[crisp trail]

thank you

.close

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[hallow solar]

Hello https://media.discordapp.net/attachments/763586648209883176/943298470486364240/unknown.png

[hallow solar]

wait wrong picture

I found the answer

[obsidian coral]

What answer did you get?

[hallow solar]

10

10.625

[obsidian coral]

Which one?

[hallow solar]

wdym

[obsidian coral]

10 or 10.625?

[hallow solar]

10 boxes but 10.625 is the exact

[hallow solar]

10.625

[obsidian coral]

It's neither

[cold shard]

it’s 11. round it up

[obsidian coral]

Don't give the answer

[hallow solar]

but 11 or 10 isnt the answer

[cold shard]

mb

[obsidian coral]

What answer does your teacher want?

[hallow solar]

I have to make a equality of it I don't know how

[hallow solar]

Bruh she gave a paper then to solve it.

Why would she tell us the answer lol

[obsidian coral]

Do you know what inequality sign, at least, means?

[hallow solar]

yes.

[obsidian coral]

What sign is it?

[hallow solar]

=

[obsidian coral]

So your teacher just wants an equation with that problem

[hallow solar]

I think so

so uhm

So I think B should be Boxes

then 10 would be B > 10

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[gentle spire]

I was wondering how you could solve this with algebra, if it’s possible. I spent about an hour working on it before I broke and used a graphing calculator to see if there was a solution.

[toxic owl]

$\log{ab} = \log{a} + \log{b} \newline$
$\log{a/b} = \log{a} - \log{b} \newline$
$\log{a^b} = b\log{a}$

[wraith dagger]

[TEB] darthlothins

[gentle spire]

I’ve been using those, I just don’t know how to find the solutions for a and b. I isolated b on one side and substituted that to the other equation and that didn’t work

[toxic owl]

You can try to simplify in both equations in terms of $\log_2{a}$ and $\log_2{b}$. Then, let $\log_2{a} = x$ and $\log_2{b} = y$. Try solving for x and y after that

[wraith dagger]

[TEB] darthlothins

[gentle spire]

Alright, thanks! I’ll try that out!

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[placid ginkgo]

hello

[placid ginkgo]

I just need help with this

Not sure if my answers are correct

[dire geode]

[placid ginkgo]

<@&286206848099549185>

[frozen forge]

in the second excercise, how did you get the 2nd line?

[placid ginkgo]

grouping

[frozen forge]

if we open your final answer, we get 28x^4 + 236x^3 + 280x^2

unless i'm missing something, that doesn't seem to be correct?

[placid ginkgo]

i think u may have done something wrong

[frozen forge]

ah sorry

misread 2 as 7

[placid ginkgo]

yea

[frozen forge]

seems to be right, although dont take my word for it

[placid ginkgo]

alr ty

just waiting for any helper that can help me out

[cedar kiln]

@placid ginkgo Has your question been resolved?

[placid ginkgo]

<@&286206848099549185>

[cedar kiln]

@placid ginkgo Has your question been resolved?

[placid ginkgo]

Can someone help please

[digital raptor]

For #1, either the 10^2 or the 1^2 is wrong, i dont know which off the top of my head. #2 and #3 are good, and for #4, the answer would be the second line down, since it asks for the equation in point-slope form, not slope intercept form.

[placid ginkgo]

wdym second line down

[digital raptor]

$y-2=\frac{1}{3}(x-3)$ is the final answer

[wraith dagger]

opfromthestart

[silent bobcat]

For the first, since you have a "major vertical axis of length 10", it should concern the y axis, which is for y

You've put it for the x

Just write the general version of the cartesian equation of any ellipse

By precising the definition of each letters

From there, you can try to replace the information you've got, so that you're sure that your equation is the right one

Also when they talk about the "major vertical axis" and "minor axis"

I suppose they say "vertical axis" for one so that we know the other one is the horizontal one, since these are the only parameters for your ellipse

[placid ginkgo]

so I would switch 10^2 and 1^2

[placid ginkgo]

ok ty

[silent bobcat]

Not exactly

Maybe I should tell you that on vc, it would be more practical

@placid ginkgo Like in 384kbps

I'll stream a whiteboard so I draw stuff

[placid ginkgo]

ok

@silent bobcat im lagging or something

lemme restart discord

[placid ginkgo]

wouldnt it be y+2

since the y point is -2

[digital raptor]

it is y--2, which is the same as y+2

$y-y_1=m(x-x_1)$

[wraith dagger]

opfromthestart

[placid ginkgo]

@silent bobcat i think x comes first

so (x-2)^2/25 + (y+1)^2 = 1

can u repeat what u said again please

yea

wait so for 1 why isnt it (x-2)^2/25 + (y+1)^2 = 1

a is vertical

[placid ginkgo]

i am pretty sure this is correct

yea I was given this equation

yea can u please just look it over

isnt a > b

yea I can see

can u go back to the other screen

to see the answer again

i just wanna look over it again

so I was given this

so u are correct

yea

so a would be 5 right

then sqaure that

and get 25

so this looks better right

ok thank you so much

can you maybe check the other questions

[silent bobcat]

The other questions are good

🤌

[placid ginkgo]

ok ty

.close

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[forest dust]

How many 14-bit strings (that is, bit strings of length 14) are there which:
Start with the sub-string 011?
2048
Have weight 5 (i.e., contain exactly 5 1’s) and start with the sub-string 011?
165
Either start with 011 or end with 01 (or both)?
5632
Have weight 5 and either start with 011 or end with 01 (or both)?
?
``` I am now having a problem with part D. My first attempt was  C(12,2) + C(11,2)-C(9,2) which ends up being 66+55-26=85 which is still wrong but i dont know why

[fallen heath]

what's the reasoning behind C(12,2) + ....?

[forest dust]

C(12,2) is bit strings of weight 5 that start with 011

[fallen heath]

... hear yourself..

starts with 011, and 11 spaces left

where does the 12 come from?

and weight 5 means there's still 3 "1"s to adjust

[forest dust]

oh my gosh that was my mistake its only choose 2 for when its both

[fallen heath]

._.

not just that, I'm surprised why it's 12 choose as well

[forest dust]

12 choose is for ending with 01

[fallen heath]

at least for the starting with 011 scenario

[forest dust]

not start

i just put the order backwords

[fallen heath]

hmm. okie, correct your answer and lemme know

[forest dust]

yes that was my mistake, thank you. I really need to start writing this out so i can visualize it better, at least while im still learning

[fallen heath]

(@_@;) I'm still curious if you got the right one

Can you share what you got for the final ans?

[forest dust]

yeah my answer was correct, final answer was C(12,4)+C(11,3)-C(9,2)

[forest dust]

does that make sense or do you want me to elaborate on my thought process

[fallen heath]

,calc (11)(5)(3) + (11)(5)(9) - 36

[wraith dagger]

Result:

624

[fallen heath]

okie (ヘ･_･)ヘ┳━┳

[forest dust]

why did you format it like that with the 5 in the middle

[fallen heath]

just a little mental math lol, ignore

[forest dust]

ok haha

well thank you for the help

This discrete math is kicking my ass, the computations arent even hard its just remember what to do and visualizing it is my problem

.close

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[rocky rivet]

I'm having some trouble with this problem. I can't find a way to get delta x out of the denominator.

[cedar kiln]

@rocky rivet Has your question been resolved?

[cedar kiln]

@rocky rivet Has your question been resolved?

[violet rapids]

For this don't you multiply by the conjugate

[rocky rivet]

.close

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[fleet relic]

1.) Give two points of the line y = 2x – 4 where x coordinates are from -4 to +4 only

[fleet relic]

i need now!.

[glad kestrel]

what

[fleet relic]

i need answer.

[tropic oxide]

we don't give out answers here.

[fleet relic]

i mean

solution.

[calm sierra]

@fleet relic we dont give full solutions here

pls show ur own efforts to solve, then we can help u work toward a solution

[fleet relic]

ok its help me😑

[cedar kiln]

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[cedar kiln]

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[carmine shuttle]

Hello! I am researching yoghurt, and i need to standardize the protein content of the final product. Would anyone be able to help me? The initial protein concentration of the milk is 3,41%, and the protein concentration of skim milk powder is 35,11%. What i need to reach is a protein concentration of 4,1%. I have concluded that i need 650g of skim milk power, but i fear i may be wrong

[cedar kiln]

@carmine shuttle Has your question been resolved?

[carmine shuttle]

I am able to show an excel sheet of my calculations if it is of help

[cedar kiln]

@carmine shuttle Has your question been resolved?

[cedar kiln]

@carmine shuttle Has your question been resolved?

[upper abyss]

650g of skim milk powder in how much milk?

[cedar kiln]

@carmine shuttle Has your question been resolved?

[cedar kiln]

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[hidden merlin]

Hi

[cedar kiln]

Please don't occupy multiple help channels.

[hidden merlin]

Can someone help me solve 102x103

Using like an identity

I tried like (103-3) (100+2)

(100+3) (100+2) what should I do now?

I’m puzzled

[ripe hamlet]

100^2 + 500 + 6

[green jungle]

which ones do u need help with

[hidden merlin]

16th one

102x103

[hidden merlin]

What?

500+6?

[green jungle]

use a+b times a-b

hint^^

[hidden merlin]

Yes

(A-b)^2 (a+b)^2

Right?

[green jungle]

no

[hidden merlin]

Then?

Ohhh

Wait

(A+b) (a-b) a^2-b^2

This ind

One

?

[green jungle]

yes

[hidden merlin]

But

I know we should use that

I’m puzzled in where like what numbers should I take

[hidden merlin]

For this identity the a and b needs to be same?

And over here b is differenct

[green jungle]

no?

[hidden merlin]

Different

Yes

(105-2) (105-3)

Idk bro I’m puzzled

🤔

[green jungle]

yeah this

[hidden merlin]

Yes

But

Shouldn’t there be a

• and -?

[green jungle]

oh wait

[hidden merlin]

Yeee

Das whaaa

That’s where I’m confused

[hidden merlin]

I think we need to use another identity

[obsidian coral]

You can try what you said first

Apply distribution

Wait

[green jungle]

they've asked him to do an algaebric identity

[ripe hamlet]

It is algaebric identity

[hidden merlin]

Yes

Can u say me

Pls

All I know is 4 identity

[obsidian coral]

They told you which one

[hidden merlin]

No

They didn’t

[ripe hamlet]

105^2 + 105(-3 + -2) + (-2 * -3)

[hidden merlin]

I need to find out and do

Which property

Udnetitb

[obsidian coral]

Don't solve for the person

[hidden merlin]

Identity

[green jungle]

what identity does this fall under?

[hidden merlin]

Yeah which one?

[obsidian coral]

Distribution identity?

[hidden merlin]

No

That’s not an

[ripe hamlet]

(x+a)(x+b)= x^2 + x(a+b) + ab

[hidden merlin]

Algebraic identity

[hidden merlin]

Ohhh

But isn’t it

X+a and x+b

And not

X+a and x-b?

Ohhhh nvmmm

I’ll make it

100+3 and 100+2

Understood thx

[green jungle]

i mean that works but like i thought he was supposed to use like the basic four ones lol

[hidden merlin]

Yes

This is the 4th one only

1st is a+b whole square 2nd is a-b whole square 3rd (a+b) (a-b)

I didn’t think of this one

Oh wait

Tanjiha

(Xyz-4) (xyz-2)

What identity

Pls say

[obsidian coral]

Don't beg

[green jungle]

u can use the same one

[hidden merlin]

But

That one says

Not -

[green jungle]

yeah ik just flip the signs

[hidden merlin]

Brooo I’m not just being humble

[green jungle]

he wasn't?

[obsidian coral]

"pls say" makes it seem like you are begging

[green jungle]

he was just being polite

[obsidian coral]

That's not being polite

[green jungle]

alright

[hidden merlin]

.

[hidden merlin]

R u sure?

[green jungle]

yeah yeah

[obsidian coral]

There's a difference between properly asking "what identity did you use" vs "pls say"

[green jungle]

take a-b whole square for instance

a^2-2ab+b^2

[hidden merlin]

Ok thx

So

[green jungle]

look a the term in the middle

-2ab

they just flipped the signs

[hidden merlin]

(X-a) (x-b)

[green jungle]

a+b whole square

[hidden merlin]

X^2 -(a-b) x ab?

[hidden merlin]

Majority wins LOL

[hidden merlin]

.

[obsidian coral]

Okay, and?

Just saying

[green jungle]

dude drop it man we get it he was begging you win

[hidden merlin]

Yes yes

I was fully begging him bro

Ligit fulllllllyyy

Goodguyneo what does the property become?

[hidden merlin]

.

This?

[green jungle]

you mean the final answer to the one you were asking or just this

xyz whole square -6x + 8

that should be it i think

@hidden merlin

[cedar kiln]

@hidden merlin Has your question been resolved?

[hidden merlin]

Bro wait

One min

Hello?

U der

No tell me what identity

Can u solve the identity

(X-@) (x-b)?

[cedar kiln]

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[hidden merlin]

Omleave

Ok leave

.close

Nvm

It already closed

[cedar kiln]

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[oblique turret]

Suppose that you borrow $36000 from a bank, to be repaid in 12 monthly instalments, and the APR is 51%. Then find the monthly flat rate,

[violet rapids]

What equation

Would you use for this problem

[oblique turret]

That is the problem, I don't know

What I know is

If I were using the monthly flat rate, I would have to be paying back 36000/12 + 36000x Flat Rate % per month

Let Flat Rate be x

So the total payment would be (3000+36000x)*12

Which should be equal to the rate using APR

[cedar kiln]

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[reef hearth]

Is my solution correct , pls ping thanks

[rough pebble]

Yes it is correct

[reef hearth]

Thanks

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[safe sparrow]

is it possible to solve this any further?

[turbid lynx]

Well yes

Depends

What is the question

[safe sparrow]

i need to find if x is higher or lower then 0

[turbid lynx]

Well how do you know if you can solve it further

[safe sparrow]

well i dont know if i can solve further

[turbid lynx]

Ever heard of discriminants

[safe sparrow]

no

[turbid lynx]

Wait

[obsidian coral]

Try factoring

You mean discriminate?

[turbid lynx]

Yes that

[safe sparrow]

this above was the original problem i just cleaned it up a bit

[obsidian coral]

Is that 5x + 4?

Or a 9?

[safe sparrow]

5x+4

[obsidian coral]

Then that next line, 13x?

[safe sparrow]

yes

[obsidian coral]

You mental math that wrong, I think

[safe sparrow]

hmm, i guess

maybe the anwser is just x=R

yea i think thats it, i dont think you can find if just x is higher then 0,

.close

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[inner quest]

Can someone help me with graph problems. When it comes to Chain rule with graph I loss my mind

[steady imp]

wow yeah I dont even want to think about this problem

[inner quest]

Haha I'm not good with graphs

[violet rapids]

Is $(q \circ p)(x) = q(p(x))$

[wraith dagger]

sills

[violet rapids]

Yes it is

Chain rule time

[inner quest]

Yep

[ocean wadi]

so it's r'(3)=q'(p(3))*p'(3)

[violet rapids]

Good job on giving them the answer

👍

[inner quest]

I was doing it all wrong then 🤣

I don't like math

[cedar kiln]

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[crimson sedge]

i need a hint

Peter stands at the vertex A of a square ABCD. He can make one step at a time either clockwise or counter-clockwise (the length of the step equals the side-length of the square). In how many ways he can make 8 steps and land at the same vertex A?

[digital raptor]

Count how many go around no time, one time, two times.

That should make it a bit easier to count

[crimson sedge]

@digital raptor did you just become a helper?

[digital raptor]

Yep

[crimson sedge]

👍

[crimson sedge]

ok

0: inf
1: 0
2: 4
3: 0
4: 4
5: 0
6: idk
7: 0
8: idk

[digital raptor]

Not quite, by "go around", i mean that if it went clockwise 6 times and counterclockwise 2 times, it would go around the square once

[crimson sedge]

ok right

but, how can i count all the ways

this doesnt really make it easier

[digital raptor]

So, you have 8 different chances to turn. If to go around one time, 6 need to be clockwise, how many ways are there to choose which 6 they are?

Have you learned combinations and permutations?

[crimson sedge]

yes

[crimson sedge]

sorry but i dont really understand what you are saying

[digital raptor]

So you need to start and end on the same corner of the square. If you go 4 times clockwise, you end up on the same corner, same with 4 counterclockwise. Also, if you go clockwise and then counterclockwise, you also stay on the same square.

[crimson sedge]

hm yes

what does this mean

If to go around one time, 6 need to be clockwise,

[digital raptor]

If you have 6 clockwise and 2 counterclockwise, you go around the square 1.5 times, then go back .5 times, which add up to 1 full time around

I draw pictue

So, if you go 6 steps clockwise, and then 2 steps counterclockwise, you end up at A.

But the 6 clockwise dont need to come first, they could be any of the 8 moves needed. You should use combinations to figure out how many ways there are

[crimson sedge]

well

i need to know two things

what is the total number of objects in the set

what is the set we are working with??

[digital raptor]

The set is similar to ${0,1}^8$, where 0 represents a clockwise turn and 1 represents a counterclockwise turn

[wraith dagger]

opfromthestart

[crimson sedge]

what does {0, 1}^8 mean?

[digital raptor]

Do you know what the cartesian product of two sets is?

[crimson sedge]

no

i never learned that

[digital raptor]

If you are supposed to do this using sets, you probably need to know.

Basically, if you have a set A with some elements and a set B with some elements, AxB is a set of all pairs of elements.

So {a,b}x{c,d}={(a,c),(a,d),(b,c),(b,d)}

And a power is just the cartesian product with itself that many times

but you dont need to know what set its using to get the number of solutions, necessarily

[crimson sedge]

im not supposed to be using sets

im just trying to use combination

[digital raptor]

So how many different times can you make a decision?

[crimson sedge]

8

[digital raptor]

And how many choices do you have for each decision?

[crimson sedge]

2

[digital raptor]

So how do you get the number of choices for all of the decisions?

[crimson sedge]

$\frac{8!}{2!(8-2)!}$

[wraith dagger]

kaniii

[digital raptor]

That would be the number of them that went around once. The total number of choices that there are would be $2^8$

[wraith dagger]

opfromthestart

[crimson sedge]

ah yes

i answered the wrong question

[digital raptor]

Because that is $8\choose 2$, which is the same as saying how many times can i have two clockwise turns.

[wraith dagger]

opfromthestart

[digital raptor]

So what other combinations do you need to have?

[crimson sedge]

why does combination work here?

[crimson sedge]

wdym?

[digital raptor]

Because you have 8 options, and for this subsection exactly two of them must be clockwise, but we dont care where, and we dont care what order they are in. So we use combination.

Since if you go 6 clockwise and then 2 counterclockwise, you go a total of 4 clockwise, which is one cycle around the square

[crimson sedge]

this isnt the asnwer tho

[crimson sedge]

2 of them must be clockwise?

i dont really understand?

[digital raptor]

Well, this is only counting when you go one time around clockwise. you need to count for all the numbers of times you can go around in either direction, including 0 times

[crimson sedge]

??

isnt this counting for 8 times ?

@digital raptor

[digital raptor]

No, it is only counting for the case where it goes around clockwise once. Do you know how many ways there are to go around the square twice?

So if $8\choose 2$ is the number of combinations where it goes counterclockwise twice, what other combinations do you need to add to have all possible combinations?

[wraith dagger]

opfromthestart

[crimson sedge]

8?

im really confused about everything now

[digital raptor]

So, to go around the square no times, how many of each turn do you need to have?

[crimson sedge]

0

right?

[digital raptor]

Well, you need to have 8 in total, but the difference between the two should be zero.

[crimson sedge]

??

[digital raptor]

It says that you need to make 8 steps, and you want to go around in total 0 times, so the number of times you go clockwise needs to be the same number of times you go around counterclockwise

[crimson sedge]

i got to continue this tommorow since im int a class

sorry

.close

[cedar kiln]

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[lilac plaza]

im confused what conditions it's asking for

[modern compass]

All of them?

Oh. Wait. It just wants you to prove wether or not this is a parallelogram and justify your decision with properties/conditions

[lilac plaza]

can u give me an example ? like on this one

our teacher only showed us like

• both pairs of opposite sided are parallel
  -both pairs of opposite sides are congruent

etc

are those the conditions it's asking for?

[gaunt hamlet]

Those are the conditions given. For the first question, you are told those two angles are the same. You're also told about a few other angles. You have to determine if the shape is a parallelogram given that information

[lilac plaza]

alr alr thanks!!

.close

[cedar kiln]

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[sullen cobalt]

is it possible to concatenate a dfa with a dfa with no accept states

[sullen cobalt]

is it possible to concatenate these two dfa's

[cedar kiln]

@sullen cobalt Has your question been resolved?

[sullen cobalt]

<@&286206848099549185>

[dire geode]

What's a dfa

[sullen cobalt]

Deterministic Finite Automaton

[cedar kiln]

@sullen cobalt Has your question been resolved?

[cedar kiln]

@sullen cobalt Has your question been resolved?

[cedar kiln]

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[flint plinth]

Presumably you know that derivatives are involved here. Any thoughts on how to proceed?

[crimson sedge]

we are expected to use the formula (f(x+h)-f(x))/h

however

i finished that and do not know how to convert it to y=mx+b

[flint plinth]

That's only an approximation to the derivative, not the derivative itself.

[crimson sedge]

?

[glad kestrel]

i'm assuming you had to take the limit of that..?

[crimson sedge]

this is what i have so far

[glad kestrel]

well you haven't completed the limit

[crimson sedge]

but how would turn this into y= mx+b

?

you mean putting 0 in

[glad kestrel]

yeah

[crimson sedge]

that would cause it to become undefined

?

[glad kestrel]

so keep manipulating until you can evaluate the limit

[crimson sedge]

ok i will try

i cannot manipulate further

.CLOSE

.close

[cedar kiln]

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[light fern]

Did I do 12 correctly?

[hollow osprey]

no

oh wait u said that hold on

yeah it would be C