help-39

[light helm]

and i'm going to stop responding here, to minimise clutter

[cyan lotus]

thx, but I wanna talk on dm

[light helm]

just going to add one more thing.
people are wary of people insisting on talking only on dms as that correlates with suspicious behaviour

[cyan lotus]

Iam sorry, but it is my persoal issues that's why...

[flint basalt]

if you need personal 1-on-1 help it might be better to work with your teacher or a classmate in person; you won't find people here willing to talk in dms

[pearl pond]

@vocal heron Has your question been resolved?

[pearl pond]

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[hardy rose]

to fid the intersetction points for x^2 = 10(y+20) and x^2/16 - y^2/64 = 1

[hardy rose]

10(y+20)/16 - y^2/64 = 1

multiply everything by 64

40y + 480 - y^2 = 64

?

-y^2 + 40y + 480 = 64

[versed mica]

🤔

[wicked edge]

Rearrange first equation.

Then let y1 = y2.

[hardy rose]

i did

[versed mica]

what is 40 times 20?

[hardy rose]

(10y + 120)/16

[hardy rose]

why 20

[versed mica]

$64 \cdot \frac{10(y + 20)}{16} = 4 \cdot 10(y + 20) = 40(y + 20) = 40y + 40 \cdot 20$

[jolly parrot]

knief

[versed mica]

no clue where 480 came from

[hardy rose]

ill do it 1 by 1

(10y+200)/16 - y^2/64 = 1

10y and 200 because there was 10 multiplying it

[hardy rose]

here

[versed mica]

sure

200 times 4 still isn't 480

[hardy rose]

oh i did a mistake

wait ima erase it

[versed mica]

you did

[hardy rose]

multiply everything by 64 and for the fraction with 16 as the denominator divide it by 16

[versed mica]

uhh

[pastel umbra]

?

[versed mica]

what

[pastel umbra]

If you're multiplying trhough by 64

[versed mica]

is "it" supposed to be 64?

because as phrased it refers to the fraction

😭

[pastel umbra]

Then the fraction with denominator 16 becomes 4 times the numerator

[hardy rose]

yeah thats why u divide it by 16

[pastel umbra]

That is-

what?

[hardy rose]

thats what i did

[pastel umbra]

No it's already divided by 16

[hardy rose]

what

[pastel umbra]

oh

Are you calling 64 "it"?

[hardy rose]

you see the 10y + 200

that has 16 as the denominator

[versed mica]

this translates to $\frac{64 \cdot \frac{10(y + 20)}{16}}{16}$

[jolly parrot]

knief

[hardy rose]

we're multiplying this by 64

and then dividing it by 16

thats what i mean

[pastel umbra]

yh that's not "it", when you say this

"it" here refers to the fraction, not the 64 you're multiplying it by

[hardy rose]

forgot to specify just the numrtor

so when u do that it becomes -y^2 + 40y + 800 = 64

then u just do 800 -64

then quadratic equation

to find the intersection

alr thanks

🙌

.close

[pearl pond]

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[pearl pond]

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[summer imp]

<@&268886789983436800>

[pearl pond]

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[hoary relic]

[hoary relic]

That was fast

[pearl pond]

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[trim meadow]

i need help solving some parametric equations, we did this at the start of the year and i have no idea how to solve if their paths meet/if they collide/if theyre parallel

[trim meadow]

x=2t+8 y=3t-4 z=9t+4, x=9t-6 y=t+4 z=3t-8

[vital estuary]

first off, to check if theyre parallel, look at the coefficients of t

[trim meadow]

well the coeffs are different for like the xs ys and zs across both so i assume not

[vital estuary]

nice

ok, to check if they intersect, they have to have the same x, y, and z value for some t for each

you have to separate the two t values, so pick a different variable for the second one

[trim meadow]

i'll pick u ig

9u-6 u+4 3u-8

[vital estuary]

you now have 3 equations in 2 variables, if they are consistent, then they intersect

consistent meaning that after solving for t and u, the third equation should also work

[trim meadow]

oh hm

and if t = u then they collide?

[vital estuary]

yes

[trim meadow]

okay, thanks!

.close

[pearl pond]

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[pearl pond]

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[modest tartan]

i got
a>25 U a<-25

[modest tartan]

however answer is given a>25

my reasoning for a<-25 was that the 2nd term is always continous
so 1st term should be continous

for first term to be continous the gif part has to remain constant whole interval

the numerator lies between 1 and 25(both included)

if a>25 the gif becomes 0
if a<-25 then gif becomes -1

gif is greatest integer function or floor function

plz ping if you answering

[amber wind]

Is that GIF become discontinuous if it have integer value?

[modest tartan]

yeah

gif becomes discontinous if at integers

lhs and rhs become different

lhl and rhl

left hand lmt and rt hand lmt

[amber wind]

So i think if f to be continuous in [4,8] then i think either that GIF must not have inside value reach integers

or if it does, then the term multiplying it also a 0

[modest tartan]

indeed

the term multiplying cant be 0 for whole interval

so only was is for gif part to be const

[amber wind]

but if check sin(x-3)=0, no x satisfies it to be 0

[modest tartan]

3*

[amber wind]

in that given domain 4 8

[modest tartan]

oh in domain no

[amber wind]

yeah so we cross this scenario then

because it is not sufficient for any x in that domain

[modest tartan]

i alr understood how to do it...

just wanted to see why a<-25 isnt answer

[amber wind]

yeah so what did you do to calculate in the other scenario

[modest tartan]

find range of the numerator

[amber wind]

first you would need to substitute in x=4 and x=8 of the GIF to find the value of a that spans the domain

[modest tartan]

x=>[4,8]
x-3=>[1,5]
(x-3)^2=>[1,25]

[modest tartan]

nah u dont need to do that part

[amber wind]

really?

[modest tartan]

the thing i did faster and easier

[amber wind]

I thought intuitively i would start it first

[modest tartan]

nah

denominator isnt fixed

so the whole thing inside gif isnt fixed

[amber wind]

so you consider two cases for GIF?

[modest tartan]

so you cant say it would be max at start

[amber wind]

either is GIF is constant 0 or >=1

[modest tartan]

gif being constant is only way

it cant have multiple values

or else the function becomes discontinous

[amber wind]

Ok so i just calculated out that [1/a, 25/a] spans the interval 4 8

But for GIF to output an only constant integer value across this span, there cannot be any integers sitting between 1/a and 25/a, right

[modest tartan]

indeed

[amber wind]

so i think the constant value n either have 2 cases

0 or >= 1

whereas n here is an integer

[modest tartan]

where u getting the > sign

[amber wind]

that GIF can get the output greater than 1, no?

But we are not sure, so we say, suppose GIF produce a constant integer n, then n <= 1/a <= 25/a <= n+1

[modest tartan]

dude i mean no offence

but i jus came to check if my ans correct

[amber wind]

so it is false there

[modest tartan]

i do not get what u saying anymore

[amber wind]

lmao

but wait can you tell me again what did you do to solve so i can check where did you go of trail

[modest tartan]

find range of numerator

gif has to be const

a has to lie outside range of denominator (make it never become integer)

so a>25 or a<-25

[amber wind]

ahh i see why

the question stated that f must be continuous on 4,8

then it must be left-continuous at the right endpoint right

this means if you approach x = 8 from left hand side then the limit is exactly f(8)

[modest tartan]

i still dont get what u saying

[amber wind]

uhh then you can test by plugging in a = -25

then when x = 8, GIF = -1, but once you change x = 8.0001 (past 8), then GIF jumpt down suddenly

[modest tartan]

i got a<(-25) however ans was different

so i came here to see if i made mistake

[amber wind]

no you were right but you are just confusing on how that GIF handles limits at the boundaries

[modest tartan]

i put a>(-25) so boundry case becomes useless

[dry skiff]

I can only verify that if a negative value for a that satisfies the question exists, it must be < -25.

[modest tartan]

thanks

[dry skiff]

If it exists

[modest tartan]

ima go do more questions

[modest tartan]

it was a subjective question

so yeah..

[dry skiff]

I may be missing a nuance

I'm not the brightest

[modest tartan]

it should exists

doesnt have anything stopping it

well thanks

.close

[pearl pond]

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[pearl pond]

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[dense storm]

How do i do q3

[dense storm]

Im thinking something like p(x,y)?

[pearl pond]

@dense storm Has your question been resolved?

[dense storm]

<@&286206848099549185>

[dense storm]

<@&286206848099549185>

Do i use like s = f(0)?

[rocky stone]

which question

?

3?

[dense storm]

Ye

[odd kelp]

yo wsp

which one

[dense storm]

3

[odd kelp]

damn dude that's

crazy

@delicate temple

try kr

[amber wind]

!nopdf

[pearl pond]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[amber wind]

!nosols

[pearl pond]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dense storm]

Why does f(1+yd) = f(1+y)

Am i not understanding an important part of functions

O wait

I read the wrong thing

Oop

[sharp delta]

Say cos(x+ycos(x)+ y = xy +cos(x+y) You'll understand then

[amber wind]

!nosols

[pearl pond]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[amber wind]

dude keep sending solutions

[sharp delta]

It isn't a solution

Its an example

[amber wind]

that is the solution

the word solution is right there

[sharp delta]

I said ""Say""

[amber wind]

what?

[sharp delta]

I gave an example

[hollow cobalt]

They are not talking about what you said

They were replying to the image sent by Krish

[amber wind]

Bro mod ping that guy

Krish ignore my factoids

[dense storm]

What does f cannot constantly equal d mean

[sharp delta]

<@&268886789983436800> This guy is giving solutions

[dense storm]

Yo what is this solution 😭🙏
Or am i too dumb for this

[sharp delta]

You need to apply some critical thinking

[amber wind]

yea

[sharp delta]

What that means

I have misunderstood it

[amber wind]

The one like !nosols

[dense storm]

Oh i didnt see the fsoc

Ok im starting to get it🥲

[pearl pond]

@dense storm Has your question been resolved?

[merry carbon]

We don't generally allow full solutions to be given to problems, and we are doubly unhappy when they're taken from AI

You have a day to be timed out, which is [nicer than we need to be](#changelog message)

[pearl pond]

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[wild fable]

hi so i guess the easiest way to start is by seeing what happens to L(b1) and L(b2) but what do i do after?

[wild fable]

ohhhh

oh ya

.close

[pearl pond]

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[eager jewel]

mb

[sharp delta]

Whats your doubt

[copper echo]

Hello, @sharp delta!

[eager jewel]

who pinged me

[sharp smelt]

me

I asked is this a post jee re-lock in

lock in for uni now 🔥

[eager jewel]

then why did u delete 😭

[eager jewel]

no i am trying my best to lock out rn, started netflix+discord+insta+roblox+minecraft+doomscrolling

[iron basin]

🙏

[pearl pond]

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[crimson haven]

I got it wrong

[crimson haven]

Pls help

[dense jasper]

(In the future, please make the first message you send the actual question since that's what the bot pins. This saves us the effort of having to change the pin manually.)

[crimson haven]

Oh

Sorry I thought it send my pic first

[dense jasper]

vertex, not axis of symmetry. The axis of symmetry is simply the vertical line that "splits the graph in half", namely x=3

so you should have also drawn the line x=3 dotted in or something along those lines

[crimson haven]

Oh ok

[dense jasper]

that being said, this is the only real issue I see right now

is there something else that's giving you pause

[crimson haven]

I checked my answer according to my book and I got it wrong

Shield I show the answer

Wait nvm

I looked at the wrong page again

[dense jasper]

rip

[crimson haven]

I’m so sorry

[dense jasper]

In the future, this is context that you should include when asking for help

[dense jasper]

all good

[pearl pond]

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

[crimson haven]

.close

[pearl pond]

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[crimson haven]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[crimson haven]

Can u help me w this too.

[crimson haven]

Do I use my original question for y or my new one

[vivid grove]

Can you provide the exact context?

[crimson haven]

This one

Question 4

No I mean 18

[vivid grove]

The working out is correct for the turning point if that is what you are asking.

[crimson haven]

Oh ok ty

Bye

.close

[pearl pond]

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[pearl pond]

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[pliant stone]

[hoary relic]

Whar

<@&268886789983436800> troll?

[humble stone]

https://cdn.discordapp.com/attachments/807809192537882647/1383537414043205803/togif-24.gif

[pliant stone]

😂

It's a mistake

I thought it's already claimed channel

Larp

[pliant stone]

It's kindda funny

[hoary relic]

Next time dont do trolling like that ig

[pliant stone]

Ngl

[humble stone]

@hoary relic whats ur major

[pliant stone]

Alr

[humble stone]

any minors

or just math

[pliant stone]

Advanced maths

[humble stone]

https://tenor.com/view/son-im-crine-son-im-crine-gif-5176045023480767429

[hoary relic]

(You know undergrad is self selectable right)

[pliant stone]

I solve olympiad question in eating time

[hoary relic]

(I chose that role so i can access advanced channels)

[pliant stone]

Hey can you help me?

I have olympiad exam in 3 months

Ioqm

Give me tips

It's my first ime

[humble stone]

i forgot about this discord server till i got to modern physics

[pliant stone]

What should I focus on more

[humble stone]

but ima just ai my way through it

[pliant stone]

Number theory

Algebra

Comvinatrics

Or geometry

In which should I focus more to qualify

[livid raven]

Algebra maybe

Cuz algebra will be in all those

Do flashcards for number theory

[humble stone]

gemini guided learning isnt that bad too

[pliant stone]

Any more tips

[humble stone]

and their notebook in gemini

[livid raven]

Hmm I'm still in 9th grade so no unfortunately

[pliant stone]

You're indian

[pliant stone]

So

You eman

Mean

I

Solve my problems from gemini

Right?

[humble stone]

lowkey

[pliant stone]

Why

[humble stone]

u should have it for free using ur university email

wdym why

[pliant stone]

Yup you're right too

But I want to be

A

Pro

Solver

Problem solver

[humble stone]

u dont have to just copy down the answer looking at the steps can help

[pliant stone]

In js 3 months 💀

[pliant stone]

Any books?

Is book required ?

[humble stone]

u can just find the textbook on some pdf

[pliant stone]

Alr alr

Tahnks

Thanks

[humble stone]

on gemini u can upload the textbook and like use that

[pliant stone]

Wooow

[humble stone]

or walk into a random lecture to ur nearest uni

[pliant stone]

I'm a school student

I'm in high school

[humble stone]

take calc its in bascially everything

[pliant stone]

Alr alr

[humble stone]

the ap exam gets easier every year

and taking it in college is way harder

[pliant stone]

Whats ap?

[humble stone]

are you american

[pliant stone]

You know ioqm

Indian

[humble stone]

nah

[pliant stone]

Ioqm is first stage of ino

Imo

[humble stone]

ngl ur tweets are kinda crazy

[pliant stone]

Wdym 😂

[pearl pond]

@pliant stone Has your question been resolved?

[pliant stone]

No

Yes

Yesyes

[pearl pond]

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[inner ruin]

The figure shows the graph of the function
y = A sin kx + B

Determine the constants A, B, and k.

[late moon]

So you're missing the value of k?

[inner ruin]

Yuh

[main oriole]

try tracing this perfectly straight blue line here

[late moon]

Did you try getting it?

[main oriole]

notice anything?

[inner ruin]

I am taught that in order to get k, you take the distance between 2 tops

[late moon]

Right

[inner ruin]

The problem there is that the tops are a bit off of a clear value

[late moon]

Then I suggest getting the distance from the axis of propagation (the line that B corresponds to)

[late moon]

Like what MarcoMa210 was indicating to

[inner ruin]

I don't understand 🙁

[main oriole]

sin(x) is periodic right?

it repeats?

[inner ruin]

Yuh

[main oriole]

so we know the y value is at -1 when x is 0

and so we grab the next x value where y = -1

it's this green line here

[inner ruin]

Is the answer 45 * 2.5 ?

[main oriole]

and why would it be that?

no if you're thinking like that then it's 30° * 1.5

because you add half a grid space

[late moon]

Hmm

[inner ruin]

This shit is so hard 🙁

[late moon]

Do you know the period of regular sin(x)

[inner ruin]

180 ?

[late moon]

Yeah

[inner ruin]

Hold on

Isn't it 360/180?

Nvm.

.closw

.close

[pearl pond]

Channel closed

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[pearl pond]

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[proven horizon]

is there a smart way to do this using determinant properties or do i need to expand every one and show?

[vivid grove]

Which question?

[proven horizon]

any

[vivid grove]

For Q2) you can subtract rows, and that will keep the determinant the same.

[proven horizon]

yeah it worked

oh i think i can do the same in 04

so gonna be x, y and z in the diagonal

and 0 in the rest

[cursive wraith]

So all you have left is question 3 right

[proven horizon]

yes

[cursive wraith]

ok

recall that multiplying a row or a column by x makes the determinant multiply by x

[proven horizon]

yea

[cursive wraith]

in the first determinant, you can spot some a and a^2, while in the second, you can spot some a^2 and a^3

can you spot how we could manipulate the first to look more like the other?

[proven horizon]

hmmm i think i can take the tranpose of the first and then multiply the first row by a, second by b, and so on and then the thid column gonna be abc for every one

so i can take off

but idk

[cursive wraith]

we can do that yeah

just remember that everytime you multiply a column/row by a number, you gotta compensate back in the determinant

so $\begin{vmatrix}a&b\c&d\end{vmatrix} = \frac 1x\begin{vmatrix}ax&b\cx&d\end{vmatrix}$

[jolly parrot]

Rafibirthday2003

[cursive wraith]

(provided x isn't 0)

[proven horizon]

yeah this works

and then we can exchange columns 2 times

and the sign gonna be the same

[cursive wraith]

exactly

[proven horizon]

thank u

.close

[pearl pond]

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[cursive wraith]

now there is a slight hiccup just in case a,b or c is 0 (because in that case what we did isn't valid)

you just have to check that it still works in those cases

[pearl pond]

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[lusty gull]

guys can anyone help me with group theory?

[brisk steeple]

!da2a

[pearl pond]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[brisk steeple]

why NOT

[lusty gull]

aight thank you

could we go dm?

[hoary relic]

Not recommended, since they might be wrong and no one else would be there if it were to be in dms. Its best that you post the question here

Also like why would you not want to post the question here

[lusty gull]

because its not really a question is more like idk whats going on in the topic

its about orbits and stablisers

[hoary relic]

In that case you could go to #groups-rings-fields or #advanced-algebra

If you cant chat in there , read #get-advanced-access

[pearl pond]

@lusty gull Has your question been resolved?

[summer imp]

You can always just point out stuff you’re having trouble understanding

[pearl pond]

@lusty gull Has your question been resolved?

[pearl pond]

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[north talon]

let $P$ be a point inside triangle $ABC$ such that $\angle BPC = 180 - \angle A$. Let the reflections of $P$ across $BC,AC,AB$ be $A'$, $B'$,$C'$ respectively. Show that $A',B',C',A$ are cyclic

[jolly parrot]

Copter

[north talon]

well clearly A' lies on (ABC), and by equal sides we know BC'A' is isosceles and similarly for other triangles

is that sufficient to just angle chase?

guys i suck at geometry pls help

[bitter herald]

Ooo so tbis is one of the easier ones that I can help you with

[bitter herald]

Yes it should be an isosceles

[north talon]

yay🥺

[lament dawn]

yes it should be

[bitter herald]

I think the angle at A should be half of angle C'AB'

[north talon]

it is,yea

cause <DAC'= <DAP

<@&286206848099549185>

[bitter herald]

Oh but also

Look at P's pedal triangle into D, E and F

Its the same as that of A'B'C' up to a scaling factor

[pearl pond]

@north talon Has your question been resolved?

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[stoic imp]

d in D and c not in D so d < c?

[honest oyster]

Is it possible to give a translation?

[stoic imp]

D is a dedekind cut so D is a decreasing set

D doesn't have a maximum

ah I see what prof means now

suppose d in D, with D being decreasing set

now suppose c < d

if c < d then since D is decreasing set then c in D which is absurd

[rare scaffold]

comments such as these are unnecessary and unhelpful

[velvet falcon]

sry

i mean im new to this

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

.close

[pearl pond]

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[pearl pond]

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[loud wraith]

On an 8×4 chessboard, a king is placed at the bottom-left corner 𝐴. The king can move one square at a time in any of the possible directions.

How many distinct paths are there for the king to reach the bottom-right corner B in exactly 7 moves?

[umbral shadow]

to get from column 1 to column 8, the king has to move exactly 7 squares to the right.
since the king only has 7 moves total to reach the destination, what does that tell us about the direction of every single move he makes horizontally?

[fleet ridge]

Think it might be 8 tall

[loud wraith]

see the image

the king can move in any direction

[fleet ridge]

Diagonal too right?

[loud wraith]

ofcourse it can't move straight up

[fleet ridge]

Basically the king

[loud wraith]

yup

[fleet ridge]

Always has to move horizontally

At any time

Whether it is diagonally or straight horizontal

[loud wraith]

yes

[fleet ridge]

So how many options does the king have at any point in time?

Say the first move. To which squares can the king move?

[loud wraith]

either the diagonal one or the adjacent horizontal one

[fleet ridge]

Thats 2

Right?

[loud wraith]

yup

[fleet ridge]

So for every point in time, the king has 2

Except

At a certain time

I min

Ill show the diagr

[umbral shadow]

let's get this step by step the king starts at the bottomleft corner which is row 1, every single move must take him one column to the right
so his very first move, what are the only rows he can land on?

[fleet ridge]

If the king is here

Then he can't move straight foreward

[loud wraith]

BRO

Bruh

the king is at the start

[fleet ridge]

If the kign is there...

[loud wraith]

oh

sry

yeah

[fleet ridge]

Np

[loud wraith]

that point is the topmost point it could get

[fleet ridge]

Also

Ill mark the places

Where the king has only one legal move

I think this is self explanatory

[loud wraith]

crct

[fleet ridge]

So

Umm

I woudl say coutn every possibel route

But thats too tedious

[loud wraith]

are you crazy

[wraith jacinth]

i have an idea-

[loud wraith]

we cant count every possible path

practically impossible

[loud wraith]

hm

[wraith jacinth]

in some squares the king has 3 choices
-# 3 possible paths

and this is true for all of this squares in row 2

now for the bottom row each square only has 2 choices continue forward or go up-

and for the top row u can only touch it- once or twice per path

[loud wraith]

crct

[loud wraith]

at 2 squares

now

?

[wraith jacinth]

well we will first calculaute how many posibble path for the first row

then second row then thid row and so on-

this is my idea atleast there is probably a more optimail way to do it

basically the idea here is that u count once (the bottom path) and then u multiply how much options u have at each point

[loud wraith]

but in the last 2 rows, we can also make zigzag motion and it can move diagonally. so practically it will be very difficult to count all the possible paths

[wraith jacinth]

we are bassicly calculating each postion what possibe otipons exsist for it-

so that is accounted for because we will also be counting those paths in te 2nd row that lead back to the first row

[wooden flare]

if we repeat a "winning" method for each layer of squares we produce an image like this

[fleet ridge]

I have an idea

[wooden flare]

each path is legal, since we always advance 1 square

[fleet ridge]

Since if you go up you must come down the same amount

[wooden flare]

so we make an "arc" of solutions

thus, how many unique solutions arise from this set?

[wraith jacinth]

-# thank you btw (that graph is nice)

[wooden flare]

-# no problem, ill step away for now to let you continue. Figured i'd contribute my understanding of it, perhaps it finds itself helpful

[wraith jacinth]

-# no no feel free to take over i am really lost 😭 or if ur busy atleat stay and watch over incase i say smthin wrong

[wraith jacinth]

this can be done algeberically through a fucntion maybe?

[wooden flare]

-# ill be going for dinner in a lil bit, but ill watch over for now

[fleet ridge]

You could think of a move as moving horizontal up or down

[wooden flare]

I'd argue: see how many paths each "layer" has (since each layer is a new set of solutions, and there should be a function which can describe this [im thinking combinatorics rn])

[fleet ridge]

And find combinations for it

[wraith jacinth]

true true that is another path for a solution

[fleet ridge]

But I have no idea how to ensure that an up move always comes before a down move

[wraith jacinth]

-# well its onyl true for row 1

[fleet ridge]

-# Wdm?

[wraith jacinth]

on other rows u can go up or down it doesnt matter (atleast except for the top row)

[fleet ridge]

Well no im treating the problem as a whole. If you move up 2 times, you HAVE to move down 2 times

[wooden flare]

-# this is true (i.e. it follows my above sketch)

[fleet ridge]

But after that im stuck. I know how to get a combination between moving horizontal, up and down, but i dont knwo how to ensure up happens before down

[wooden flare]

imo, easiest way would just be brute force

probs less time consuming than finding an actual formula for it

but you can find a "rule" instead of a function

at step N, I have X moves I can make, thus for some path K on step N, I have X solutions

where each path K, is 1 layer higher than the previous

from there its number crunching, so I see brute force being easiest approach

[fleet ridge]

After googling its related to a Dyck (yes thats the name) or a Ballot sequence

@loud wraith do you know what they are?

[loud wraith]

what

[wooden flare]

at step N, for K>1, where K=1 is the starting case (ground layer if that makes sense). We have 3 moves for every subsequential step N+1, however at some point we must make 3 moves (down, right, down-right) and then 2 moves (right, down-right)

that should be the # of paths for each layer K

[loud wraith]

total permutations are too much

for the first and second row, total number of choices will be 7!/2

no wait

[fleet ridge]

U cld try this. But I have no idea what it is unfortunately

[wooden flare]

-# ^ ^ with some added help from knowing how many possible moves can be made at each step

[loud wraith]

leave it guys

y r just telling your views on the question instead of solving it

didnt even analyzed the problem properly. you cant just take any move straight up.

.close

[pearl pond]

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[humble root]

I think when they said "up" they really meant diagonally up

You did establish earlier that you can't move straight up

[wooden flare]

"up" would fail the problem in every case, since min(moves) will always be 8

[pearl pond]

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[humble root]

The asker has left, but this is actually an interesting combinatorics question.
My idea is to break apart this problem into four cases: 0 diagonal movements, 2 diagonal movements ... 6 diagonal movements. The first case is trivial, just 1 possible scenario. The second case is also straight forward:
We are essentially arranging RRRRRDD, where R means to go right, and D means to move diagonally. The first diagonal has to go up, the second diagonal has to go down. So that is 7!/(5! * 2!) different cases.

The 4 and 6 diagonal cases are a little more interesting. It is interesting because you can have both up diagonals first, or you can have one up diagonal and one diagonal. In that case, you need to consider further casework and find the number of ways to arrange DDDD properly. You find that there is only two possible cases; UUdd, UdUd (where U = up and d = down). The constraint is that you cannot move down more times than you have moved up, which makes sense based on the scenario. That means you have a few constraints:

1. The very first diagonal must be up always
2. The next two diagonals must not be down
   That means you multiply 7!/(3! * 4!) by 2.

For the 6 diagonal case, you can apply a similar set of logic;

1. The very first diagonal must be up always
2. The next three diagonals must not be down
3. The next two diagonals must not be down
   (On inspection, I think this diagonal case is also a little tricky than just the above cases, but can still be managed if you brute force it)

Then from here, you just add all the cases up. I think my reasoning covers everything.

[humble root]

.close

[pearl pond]

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[wraith jacinth]

-# pretty neat solution nice ;3

[humble root]

It is pretty bashy but I mean you sort of expect as such with a scenario that has a lot of free movements and hidden constraints

But yeah thanks 👍 There might be a way to refine it and/or look at it in a different perspective

[errant solstice]

-# if I may add, if one is interested in problems of this nature, one may look up what Motzkin paths are!

[pearl pond]

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[sharp sequoia]

i need help with this

[sharp sequoia]

my teacher says that the miles are the radius but i dont know how to measure that out on a graph

[humble root]

Wdym measure

Oh like

You're not sure how to draw a circle of radius 7?

[sharp sequoia]

likee if the miles was 7, then the radius would be 7 units but i dont know how to do the 7 units

OH WAIT

[sharp sequoia]

i think so

[humble root]

Wait

[sharp sequoia]

i could just add 7 to 2.5?

[humble root]

So you do know how to draw a circle of radius 7

[humble root]

Yep

That'd give you a point on the circle 👍

[sharp sequoia]

9.5 would be a point on the circumference 🙂‍↕️

i will try

[humble root]

9.5 is a number, what might be the actual point?

[humble root]

Excellente 👍

[sharp sequoia]

i meant (2.5, 9.5)?

[humble root]

Yep. I had an idea you knew, but just to double check

[sharp sequoia]

i think maybe i would say (4.5, 3) is the epicenter?

[humble root]

Yeahhh so the funny thing about this question is that all three graphs seemingly don't intersect, which is very interesting.

[humble root]

I'd say you are roughly right 👍

[sharp sequoia]

i dont know how specific my teacher would want it though

[humble root]

You could find the centroid of these three points and call that the epicenter

[sharp sequoia]

like if i should just go (5, 3)

OH YES

[humble root]

wait I just realised

[humble root]

THere should be a negative symbol on the 4.5 😭

Did you mean positive 4.5 or -4.5

[sharp sequoia]

(-4.5, 2.9) is more accurate 🙂‍↕️

[humble root]

I don't know what this is but I'm going to trust your graph is fine 👍

[sharp sequoia]

wait and then for the last part about the earthquake… am i supposed to make a circle with a radius of 10 around the epicenter we just found?

i said yes for it

[humble root]

yep

[sharp sequoia]

i dont know what to explain 😔

[humble root]

Oh well

I mean you said it yourself;

You should start by drawing the circle with radius 10

Then consider if you were standing at the point (4, -6)

[sharp sequoia]

i dont know how to explain the answer

[humble root]

Welll

I guess

What if you answer the question straight up?

As in,

Do you think you'd feel something from the earthquake at that point?

[sharp sequoia]

the epicenter is the origin of the earthquake which extends 10 miles out… THAT MEANS to draw a circle with a radius of 10 around the epicenter (approximately -4.5, 2.9) by doing so, you’ll find that a person standing at (-4, -6) does feel the earthquake 🙂‍↕️

[humble root]

Good.... why?

Why does the person feel the earthquake?

[sharp sequoia]

becauseeeeee theyre standing in the earthquake 10 miles big circle

[humble root]

Perfect

[sharp sequoia]

😭 i think i said that already OH OKAY

[humble root]

The person is standing inside of the circle, hence they feel the earthquake

[humble root]

I was being picky because you didn't explicitly say so :)

Teachers are usually not this picky... hopefully

[sharp sequoia]

mine isnt but this is on my test tomorrow

[humble root]

Ahhh

Good luck for your test!

[sharp sequoia]

can i have more help?

i got this question originally wrong for having a poor explanation

i need lots of help with explaining things

[humble root]

All goods! Well, explain it to me how you normally would first :)

[sharp sequoia]

i said “the point wouldn’t be in the circle because just by looking, 8 goes beyond 5” i can see why i got it wrong

[humble root]

Hmmmm okay, well why did you get it wrong?

[sharp sequoia]

cited evidence from my math teacher

😔

[humble root]

wait well

Sorry what I meant was

[humble root]

Do you know why this specific reason is not quite right

I.e do you know the actual answer to this question?

[sharp sequoia]

hmmmmmmmmmHMMMMMMM

because i didnt actually do the problem?

[humble root]

Oof okay

I was sort of expecting a specific reason 😭

But no worries, I'll go through it with you

[humble root]

The specific reason why your reason is wrong is because it doesn't actually mention the properties of a circle

[sharp sequoia]

oh yes

[humble root]

To determine whether a point is on or off a circle, you really need to actually mention the circle in some way

So, maybe start with that :)

As in, what's the circle here?

[sharp sequoia]

and i was also supposed to like simplify the square root of 39 and 17 as in like turn them into whole numbers rounded to the nearest tenth?

[humble root]

Wait wait wait

Let's not jump ahead just yet

LEt's take things slowly :)

[sharp sequoia]

😊

[humble root]

A circle has two key components; what are they?

[sharp sequoia]

its round and has infinite amount of sides

???

[humble root]

Okay yes, that is true

Is there another important two properties of circles?

Maybe something a little simpler

Theres two things that every circle has that is very very important

[sharp sequoia]

origin…

slash center

[humble root]

Good! Okay the circle has a center, perfect.

[sharp sequoia]

circumference

[humble root]

What else?

[sharp sequoia]

points?

[humble root]

Close

Not circumference technically

I mean circumference is important

[sharp sequoia]

radius diameter!!!

[humble root]

But this thing is more important

Perfect!

So a circle has two things:

A radius, and a center

[humble root]

Now, based on this question

Can you figure out what the radius and center of this circle is?

[sharp sequoia]

i dont think it specifically says how big the circle is

cause 8 can be in the circle and so can 5?

[spice edge]

have you done the equation of a circle

[sharp sequoia]

well the center is given: at the origin

[sharp sequoia]

whats that???

[spice edge]

its a way to graph a circle x^2 +y^2=r^2

[sharp sequoia]

OHHHHHHHHHH

[spice edge]

i think thats learned later for you

[sharp sequoia]

nuh uh

[spice edge]

ill let the other dude explain i think hes better than me

[sharp sequoia]

that was on the test

[spice edge]

ah

[sharp sequoia]

🙂‍↕️ so i should plug in the points

[spice edge]

@humble root i know how to prove if a line is a tangent to a circle, but how do u prove if a point is on a circle?

[spice edge]

yes

i think

and work out the radius based on the first given point

[humble root]

Hint: ||As you say, x^2 + y^2 = r^2
If the point isn't on this circle, I wonder what happens to this equation?||

[spice edge]

its false

ok thank you

[humble root]

I gotta go soon anyway, did you want to explain?

[spice edge]

@sharp sequoia do you understand?

im alright we just had a test on this lol

[humble root]

Oh lmao

[sharp sequoia]

😭

[sharp sequoia]

me too!

[humble root]

Anyways, as David said:

You might remember that

The radius is the distance from the center to any point on the circle.

How then might you find the radius of the circle?

[sharp sequoia]

i need to try the plugging in method first

i truly believe it will work

[humble root]

Alright, try it :)

[spice edge]

u have to find the radius too

[humble root]

Tell me what you get

[sharp sequoia]

okay how do i find the radius…?

probably distance formula from the center

[humble root]

Okay okay, we're getting somewhere

How would you set that up?

[sharp sequoia]

i dont know actually im just thinking

is the radius 8

[humble root]

What is the distance formula?

do you remember?

[sharp sequoia]

square root of… (x2-x1)^2 + (y2-y1)^2

[humble root]

Perfect!

WHat's (x1, y1)? What's (x2, y2)?

Wait the radius is 8? I don't think so

wait

nvm I can't add things

Yeah

The radius is 8

[humble root]

You can figure this out by using the distance formula, and substituting the points here

[sharp sequoia]

i just removed 0 because it wasnt needed so 5 squared is 25 and then square root of 39 becomes 39 when squared

[humble root]

Yep.

[sharp sequoia]

a little tongue twisty

square

okay and then i think i get the get

[humble root]

https://media.discordapp.net/attachments/908078029778083872/1506950941901852803/IMG_1294.jpg?ex=6a1020ee&is=6a0ecf6e&hm=a47da395d71f5f7412be5120a6f50675eb889bc63a8cd67540843b5f4ebad78b&=&format=webp&width=2485&height=788

Nice.

So is (8, sqrt(17)) on the circle?

[sharp sequoia]

because then ill find the radius of 8 sqrt 17

[humble root]

(8, sqrt(17)) yes

[sharp sequoia]

ITS NINE

[humble root]

Uhuh and what does that imply?

[sharp sequoia]

its not on the circle

and i can explain the process in a bunch of words

of finding the answer i mean

[humble root]

Why?

[sharp sequoia]

because its radius is bigger than 8 meaning its beyond the circle’s circumference

ummmmmmmm

[humble root]

👍

Good!

I'd describe it like this:
The radius of the circle is 8, but the distance of the point (8, sqrt(17)) is 9. That means the point lies outside of the circle.

Alternatively, the rightmost endpoint of the circle is (8, 0). That means (8, sqrt(17)) is definitely further than the circle, since you are just then going up from the endpoint.

However, the most important point is - when you have a question referencing circles, you should probably bring up the actual circle!

The best strategy to answer any question is to actually deal with the specified thing in the question, whether it be a circle, or any other graph.

Anyways, I gtg now. Hopefully that answered your question 👍

[sharp sequoia]

i get the get!

i appreciate your help

have a good day GOODBYE 👋

.close

[pearl pond]

Channel closed

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[pearl pond]

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[desert cliff]

how do we solve this?

[vivid grove]

!steps

!show

[pearl pond]

Show your work, and if possible, explain where you are stuck.

[desert cliff]

I havnt attempted because i dont know what the question is asking

[vivid grove]

So what part of the question do you not understand?

The key idea of the question is calculate the area under the curve

i.e. area under g(x)

[desert cliff]

basically this

actually only this

and calculating the upper sum

[pearl pond]

@desert cliff Has your question been resolved?

[desert cliff]

<@&286206848099549185>

What’s k

[cursive wraith]

you're making a subdivision of the interval [7,8]

So, as suggested, you're going to split [7,8] over intervals [7, 7 + 1/n], [7 + 1/n, 7 + 2/n], ....

the k is just to quickly indicate how the points of the subdivision are defined

[cursive wraith]

k will take any value between 0 and n, exactly as written to the right of the brackets

so we can have k = 0

that gives the subdivision point 7 + 0/n = 7

we can have k = n

that gives the subdivision point 7 + n/n = 8

and any value of k in between

so 7+ 1/n for k = 1

7 + 2/n for k = 2

etc...

[cursive wraith]

We very often use "k = 0 to n" instead of the writing on the right so we can compress neatly what we want to say, instead of having to write down every single point with the "dot dot dot" writing

Like with summation

instead of writing $1^2 + 2^2 + 3^2 + ... + n^2$ to write that we're summing every square of integers between $1$ and $n$, we'll write
$$\sum_{k=1}^n k^2$$

[jolly parrot]

Rafibirthday2003

[desert cliff]

ah that makes sense

that k=1 just means were starting from 1

but k takes on any values

and n is the last value

[cursive wraith]

exactly

and k will take on every (integer) value between the starting and the ending value

[desert cliff]

yepp got it

so how will i know how many subintervals

[cursive wraith]

ok, k takes on any value between 0 and n

meaning the subdivision is made of how many points (not intervals)?

[desert cliff]

I also guessed the width to be 1/n so I might need clarification on that too

[cursive wraith]

Well I guess we can start with the width, either is fine

your choice

[desert cliff]

oh we can just start with this

im just tryingg to think

[cursive wraith]

So, there are as many points as there are possible values of k, right?

If I ask you to count, let's say, from 0 to 5

How many integers would you count

[desert cliff]

n=6

[desert cliff]

yeah points and values of k is the same thing

[cursive wraith]

number of integers you would count is 6

[cursive wraith]

here I chose the "n", the end value, equal to 5

So in general, when I ask you to count from 0 to n

how many integers do you count

[desert cliff]

0, 1, 2...n?

[cursive wraith]

yeah

how many integers did you list in total here

(accounting for the dot dot dot)

[desert cliff]

n+1 right

[cursive wraith]

yes

so, our subdivision is made of n+1 points

How many subintervals would that make?

[desert cliff]

actually woudnt this be n

and the sub intervals be n+1?

[cursive wraith]

nah, it's the opposite

here we are at the classic fencepost problem

you said it that 0, 1, 2, 3, 4, 5 makes it 6 points in total

[desert cliff]

because i was counting the interval outside the end point

[cursive wraith]

the interval [7,8]?

Doesn't count in the subinterval count

in any case, think of the subintervals as fences

[cursive wraith]

every subinterval [a,b] must be delimited by their endpoints, the "fenceposts"

[desert cliff]

the points = n

and intervals = n+1

[cursive wraith]

what are those two intervals on the edges doing?

The one on the left doesn't have a beginning point

and the one on the right doesn't have an endpoint

[cursive wraith]

Back to this

[desert cliff]

thats true

[cursive wraith]

an interval [a,b] always has a beginning point, a, and an endpoint, b

so it's the other way around

https://en.wikipedia.org/wiki/Off-by-one_error more onto this problem if you're interested

It's a common mistake

[desert cliff]

ok that makes sense

[cursive wraith]

So n+1 points, n subintervals

[desert cliff]

so points = n+1
and intervals eequal n

yep

[cursive wraith]

ok

now we have n subintervals

their width should sum up to what?

(remember what the subintervals are a "sub" of)

[desert cliff]

sum up to the original interval

[cursive wraith]

They should sum up to the width of [7,8], yes

Which is = ?

[desert cliff]

I assume partitions are another word for sub intervals

[cursive wraith]

In this case yes, almost

Partition is another word for "subdivision"

"We partition [7,8] into intervals of the form ..."
Is the same as
"We divide [7,8] into intervals of the form..."

[desert cliff]

so partitions = subdivision = the black points?

[cursive wraith]

Ehhh, the partition is more like the set of subintervals

A partition is a way of splitting up a set into different subsets such that :

• Everything that was in the original set is in one of the subsets
• No two subsets intersect each other

We often conflate subintervals and the points because one is exactly linked to the other

And it's easier to write {point1, point2, point3, ...} than {interval1, interval2, ...} because intervals are longer to write

I hope that was clear enough

[desert cliff]

fair enough

yes that was clear

thank you

.close

[pearl pond]

Channel closed

Closed by @desert cliff

Use .reopen if this was a mistake.

[desert cliff]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[desert cliff]

wait one more thing

whats Pn?

I know what S is

[bitter herald]

nth partition?

[desert cliff]

upper sum and lower sum

[bitter herald]

I didnt read the earlier context but from that image that seems to be the natural conclusion

[desert cliff]

I assume its just the points right?

[bitter herald]

Yes

Riemann sum u make partitions to be ur rectangles widths

Its what that is

[desert cliff]

so how do wwe calculate it?

[cursive wraith]

Well you go back to the definition of your lower/upper riemann sums

[pearl pond]

@desert cliff Has your question been resolved?

[desert cliff]

.close

[pearl pond]

Channel closed

Closed by @desert cliff

Use .reopen if this was a mistake.

[pearl pond]

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[pearl pond]

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[ashen ivy]

<@&268886789983436800>

[pearl pond]

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[main oriole]

his username is "devil" in italian

no way he does mrbeast scams

[ivory basin]

Would be in character

[pearl pond]

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[acoustic tangle]

Solve x + |x| = 0

[acoustic tangle]

Im a bit confused how to consider the cases

If x<0, x=0 and x>0?

[shrewd rune]

you should divide this ecuation into inecuations

[acoustic tangle]

Idk what that word means

[shrewd rune]

exactly that

so what is |x|

[acoustic tangle]

the mod of x

Like absolute value

[shrewd rune]

yes, so the module of x is always positive

[acoustic tangle]

Yeah

[shrewd rune]

so lets base ourselves on x>0 and x<0

try with x being a positive number

any positive number you want

[acoustic tangle]

You get x=x

[shrewd rune]

not really

just substitute "3" for example

x + |x| = 0

[acoustic tangle]

3 + 3 = 0

[shrewd rune]

which is untrue right?

[acoustic tangle]

Yh

[shrewd rune]

now try a negative three

[acoustic tangle]

-3 + 3 =0