help-17

[dense perch]

try

[sudden peak]

omg let me try that

\left(5y+2\right)^{2}+\left(5x+11\right)^{2}=1

ugh wait pls hold

omg thank you

absolute legend

so what can i do for her hand now 😭

[dense perch]

hehe maths

[dense perch]

well you know better lol

1. use youtube
2. experiment

[sudden peak]

okk thanks

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[true silo]

how do you get the remainder in the box method how did they get that as the remainder. Im fine with everything but that

[true silo]

nvm guys got it!

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[snow juniper]

could someone explain how to calculate c)

i think i start it with 12c6, but im not sure what to do now

[vocal sleet]

@snow juniper Has your question been resolved?

[snow juniper]

<@&286206848099549185> soz for ping but could anyone help

[foggy rune]

for a) is it not 12 choose 3? ie there are 12 vertices, choose 3 to form a distinct triangle from these vertices

[snow juniper]

yes thats the answer to a

[foggy rune]

for b) this was my thought process:
fix a triangle of 3 random vertices. then there are 9 choose 3 other triangles with distinct vertices that you can pair this triangle with. so there are 9 choose 3 pairs between this fixed triangle and another triangle with distinct vertices from it

so count (12 choose 9) (9 choose 3) pairs.

however, every 1 pair counted is counted twice so divide by 2:

(12 choose 9) (9 choose 3)/2

i am not too confident in this so maybe someone more informed in helpers can confirm or deny this approach

[snow juniper]

ty the answers correct

could you help with c)

[foggy rune]

i am a bit tired, but i can look at this in the morning

[vocal sleet]

@snow juniper Has your question been resolved?

[snow juniper]

<@&286206848099549185> i still need help with c), i would appreciate it if anyone can help

[edgy sapphire]

once you choose the 6 points the picture is essentially always the same. try counting how many ways you can make 2 non-overlapping triangles out of 6 points, there's only a few

[calm sand]

If the triangles have only 1 edge in common,
then do they overlap each other?

[edgy sapphire]

good question, answer is depends and the problem excludes those cases

[calm sand]

I wanna confirm if the question allows triangles with only 1 edge in common to be counted as they overlap each other or not 🤔

[calm sand]

IF we assume that triangles with only 1 edge in common don't overlap each other
THEN
there're only 10 triangles that can coexist without overlapping each other.

It's quite intuitive when you draw all possible line segments limited by those points and start from the same point,
i don't wanna spend too much brain power on it at the moment so i don't know how to rigorously prove this yet, i'll leave this for you guys.

The total amount of overlapped triangles is just (12C3) - 10

[snow juniper]

Okay so just to confirm, listing it out aeems to be the easiest/most intuitive way?

[calm sand]

The "pairs" thing is just devided by half or something like that.

And the questions are NOT rigorous.

It should be the MAXIMUM amount of triangles or pairs can be drawn,
not just "can be drawn" alone,
you can fewer than the maximum amount.

[calm sand]

I'm not sure if there's a more intuitive way to solve it. 🤔

I didn't just list them out,
i observed the way those points form triangles,
so i can tell the maximum amount even if there're many more points.

I assume that you're in highshool? @snow juniper

[snow juniper]

Yes

[calm sand]

I usually encountered this kind of questions when i was in highshool.

Question about statistics was and still is one of my favorite kind of question 😄

[snow juniper]

actually wait i think i figured everything out

thank you for your help

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[dawn crystal]

https://artofproblemsolving.com/wiki/index.php/Newton's_Sums

using this when i am solving this question i am getting different answer then my teacher (he used other method)

[dawn crystal]

am i doing something wrong like is not not applicable or what

ok

well

sir did it wrong

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[fickle tusk]

how do i slove this

[ebon sand]

Alright, can you split 80 as a product of something and 5?

[fickle tusk]

16 * 5

[ebon sand]

Yes

Now root of (16x5) = (root 16 )*( root 5 )yes?

[fickle tusk]

yes

[ebon sand]

What’s root 16?

[fickle tusk]

4

[ebon sand]

yupp

[paper depot]

how is this meant to be interpreted at all

[fickle tusk]

ohh, so the answer would be 4root5 - 5root5 which gives us -1root5

[ebon sand]

so root80 = 4 root5

Yuppp

that’s correct

[fickle tusk]

thank youu

[ebon sand]

Wlcmmm

[fickle tusk]

so radicals can be in negative as well?

[ebon sand]

They can

I mean they can be multiplied with negative numbers

But

If the radical is even like a square root or a fourth root

The number inside the radical should always be positive or zero

[fickle tusk]

ohk

thankss

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[bleak prawn]

Part (b)

I got pretty close I think but not the right answer

|(4 - 2e^p*ln(2)) - (4 - 2e^ln(2))| = 8

|-2e^(p+1) + 2(2)| = 8

|-2e^(p+1) + 4| = 8

|-2e^(p+1)| = 4

|e^(p+1)| = -2

Did I do something wrong up to this point? The ||'s are confusing me

[paper depot]

|(4 - 2e^p*ln(2)) - (4 - 2e^ln(2))|
hold on, where did this come from

[bleak prawn]

Its |A_p - A_1|

[bleak prawn]

I put || because we dont know which is larger, right?

[paper depot]

$|A_p - A_1|$ is somewhat questionable notation.

[twin meteor]

Ann

[somber portal]

i'd use dist(Ap, A1)

[paper depot]

remember these A's are points (like, points in R^2) and not merely x-values. they come with y-coordinates too!

[paper depot]

or just $A_1A_p$ lol

[twin meteor]

Ann

[bleak prawn]

Like multiply them?

[paper depot]

no!!!

dear god no

[somber portal]

no, same as you write a segment AB

[bleak prawn]

Okok sorry

So what actually are 4 - 2e^p*ln(2) and 4 - 2e^ln(2)

The x coordinates? Or the y?

[paper depot]

good question. maybe you could show your work for how you got these?

that and also the fact you left e^ln(2) unsimplified is somewhat alarming tbh

[bleak prawn]

g_p = 4-2e^px

I just subbed in p = 1 for A_1

And x = ln(2) for both

[pallid forge]

g_p(x) you mean

[bleak prawn]

Ah yeye

So its the y coordinates right

Wait no its x

Right?

With x = ln(2) subbed in

[paper depot]

the points lie on the line x = ln(2)

this means their x coordinates are both ln(2)

i am saying things that might sound obvious or trivial but this needs to be said

[dreamy monolith]

$$hi$$

[twin meteor]

Benjamin

[dreamy monolith]

Need help?

[paper depot]

!occupied

[vocal sleet]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[dreamy monolith]

Sorry

[bleak prawn]

So the coordinates are (ln(2), 4-2e^p*ln(2))

And (ln(2), 4-2e^ln(2))?

[paper depot]

yes, but you should really simplify these before doing anything else with them

[bleak prawn]

Yeye

So I simplify, put then in the distance equation, set them equal to 8 and solve for p, right?

[paper depot]

yes

[bleak prawn]

Awesome!!

Thank you everyone

❤️

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[vocal sleet]

Please don't occupy multiple help channels.

[elfin quarry]

Is this value correct

[crimson jetty]

bruh what

keep your old channel

[elfin quarry]

Sorry I am new mam

How can I do I am a kid

Only 13 year old

[paper depot]

i pinged you in your old channel #help-39

[elfin quarry]

Sorry mam 🤕

[paper depot]

go back there

closing this one

.close

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[nocturne basin]

can you guys help me with the 17th problem

[nocturne basin]

its of binomial

[vocal sleet]

@nocturne basin Has your question been resolved?

[nocturne basin]

@reef grove

oh sorry my bad

<@&286206848099549185>

[dreamy monolith]

Need help buddy

[dreamy monolith]

Buddy

[dreamy monolith]

Buddy??

[nocturne basin]

lol

yeah

[dreamy monolith]

With what buddy

[nocturne basin]

this

[dreamy monolith]

Did you read the problem

[nocturne basin]

i did

[dreamy monolith]

Good

Did you understand it?

[nocturne basin]

yes id id

[dreamy monolith]

Then what’s the problem buddy

[nocturne basin]

how do i solve it

[dreamy monolith]

So you don’t understand the problem

[nocturne basin]

pls tell

[dreamy monolith]

Try to visualize the problem

This discord doesn’t want me to send the answer out immediately

Even tho I have it

Are you visualizing it?

[nocturne basin]

i am trying

[dreamy monolith]

Really, try to imagine it in your head

[nocturne basin]

okayy

okay i am imagining

[dreamy monolith]

What do you see

[nocturne basin]

i see the problem in my head

i dont understand how will imagning the problem help

[dreamy monolith]

Are you doubting me

[nocturne basin]

no never

never

i just want to know how to solve this problem

cause i have been trying for a long time

[versed rune]

when you find a solution pls ping me as well

[nocturne basin]

lmao

see i have the solution

but i dont understand it

its pretty direct

in the book

if you guys can

pls tell

i am sening

sending

[versed rune]

i know this is one of those weird questions where you have to integrate somewhere

but idr why

[nocturne basin]

yeah yeah

just see the solution if you can understand it

do tell

[versed rune]

,rotate

[nocturne basin]

sorry for bad iamge

[versed rune]

that's all??

[nocturne basin]

no no

[versed rune]

ohok

[dreamy monolith]

Visualize it buddy

It will make sense

I promise

The integral is simple

The summation, that is

[lime urchin]

hi

[versed rune]

this isnt very helpful unfortunately

[dreamy monolith]

You aren’t doing it right man

Once you can visualize it well

[lime urchin]

man we aint so good at binomial to visualize it and all that

[dreamy monolith]

You can solve them so easily

[nocturne basin]

i am not able too

[dreamy monolith]

Let me guide you

[nocturne basin]

pls give me the powers

yes

[dreamy monolith]

You have to put your trust not in god, but mathematics

[versed rune]

you're waffling

[nocturne basin]

bro

[lime urchin]

he is an athiest

lmfao

[dreamy monolith]

Before you sleep, stare at the problems you need help in

I’m Jewish

You will see them in your dreams

[nocturne basin]

man just tell how to solve it

[dreamy monolith]

Trust the process 😔

[nocturne basin]

i am good

[dreamy monolith]

Do it buddy

[nocturne basin]

did you get the solution

like could you understand it

[dreamy monolith]

Ye soft

Ofc*

[lime urchin]

hey benjamin ill send you a question try to solve it aight?

[dreamy monolith]

The answer is set forth in the book. It is unto thee to dissect it and comprehend it

[dreamy monolith]

Okay mate

[nocturne basin]

man

go fight somewhere else

pls let me understand in peace

[dreamy monolith]

That is it young one

IN PEACE you will find it

[lime urchin]

i was kidding please help my boy hericulum here

[dreamy monolith]

The process seemeth hard, but once thou canst visualize the mathematics and apply the summation aright, the answer doth arise.

[nocturne basin]

<@&286206848099549185>

pls help with this

[dreamy monolith]

Thou needest none other. Put thy faith in me and let me aid thee in visualizing it

It will all work

[stray shard]

What's even going on? (wtf?)

[nocturne basin]

dont ask

jjust ignore this guy

[dreamy monolith]

I’m trying to help buddy visualize the problem, but he just wants the answer

[nocturne basin]

help me with the problem please

[dreamy monolith]

Look

Let me help you fr

What are you stuck in

[stray shard]

Anyway, let's try this from the top again. Can you describe your thought process?

[spare blaze]

have you tried to write it in summation form?

[nocturne basin]

general term

right?

[nocturne basin]

i was thinking how to solve

the part of

1 + 1/2

+1 /3

and so on

but not able to

i am not able to progress much

[stray shard]

1 is equivalent to 1/1

[nocturne basin]

yeah

i have sent the solution of the problem from behind the book

[spare blaze]

its a double summation

[nocturne basin]

if that helps a little

[nocturne basin]

hmm

how

like

its insinde

one part

okay i get it

[spare blaze]

i will write it

[nocturne basin]

okay ty

[versed rune]

I understand why the integrated
1+x+x^2...
when you integrate it it becomes
x + x^2/2 + x^2/3... and the limits have to be 0 to 1 to match with the original qs

then we know that C0-nC1+nC2-nC3... becomes zero, and the latter part of the expression becomes (1-x)^n because we know that

sum of (-1)^r nCr*x^r = (1-x)^n

the 1-x in the denominator is taken out of the summation for this step

(1-x)^n /(1-x) = (1-x)^n-1 and this is integrable to give the answer

that's waht i understood frmo the soln

[nocturne basin]

hmmm

[spare blaze]

$S=\sum_{r=1}^{\inf} {nC_r} \cdot {\sum_{k=1}^{r} \frac{1}{k}} \cdot (-1)^{r-1}$

[twin meteor]

therealtdp

[nocturne basin]

yeah

this makes sense

dam

[versed rune]

im gonna fail jee

[nocturne basin]

🤣

you also jee?

like preping for jee

[versed rune]

yes

[nocturne basin]

pro

2026?

or already given

[spare blaze]

i wrote mine 2 months ago

[nocturne basin]

woowow

how did it go

[spare blaze]

it went pretty good

[nocturne basin]

pro

i hope you get into an iit

with good rank

[spare blaze]

thank you <3

[nocturne basin]

and you too

also get into an iit

thanks for helping

[versed rune]

this year

[nocturne basin]

dam

how did yours go

[versed rune]

not so good

[nocturne basin]

doesnt matter

did you pass

cutoff

[spare blaze]

do you take coaching from any institute @nocturne basin

[nocturne basin]

yea

i am in allen

currently

[spare blaze]

hmm nice

[nocturne basin]

you?

[spare blaze]

i studied my 11th in allen

[nocturne basin]

ohh

[spare blaze]

in 12th our school cut partnership with allen and made their own teaching institute for jee

[nocturne basin]

ohh

ic

damn

[spare blaze]

allen is good 👍

[versed rune]

yep

[nocturne basin]

woow so pro

alright guhys good luck for advanced

[spare blaze]

[nocturne basin]

you will do pro

i belive

by ebye i have to goo

[spare blaze]

close the channel before you go

[nocturne basin]

yea

.close

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[pulsar marsh]

hey there,
whenever you want to find an angle in a triangle you would do "shift" tan\cos\sin and the rrst
but, the thing is, mr prof. told me to write down in the notebook "arc" tan\cos\sin or something like this
i was too shy to ask what was that word... was it "art"? or "arc"? pllease correct me

[viral copper]

arc

[pulsar marsh]

thanks a lot!!! yall are the best 🩷

[twin meteor]

neon

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[sinful nacelle]

im getting that a line meant to bisect 2 lines is equal to one of the lines itself aaaa

[obsidian stream]

I'm not really sure what it means exactly by inclined equally, but the way I interpret that, isn't your equation LA = LB only true if A and B are unit vectors?

[sinful nacelle]

the angle between them should be the same right?

why wouldnt they be?

wait a minute

i havent taken their magnitude into account

damn

yes youre right woops thank you

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[trim pecan]

can someone help me

[drifting hound]

May i get helped too?

[trim pecan]

with part b

[eternal obsidian]

What’s the query?

[trim pecan]

i dont understand what they want i thought

its 0.00679

in meters

[outer warren]

note the units

of 10^-2 m (essentially cm)

[vocal sleet]

@trim pecan Has your question been resolved?

[trim pecan]

i tried entering both though it keeps marking it wrong

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[strange helm]

yo could anyone explain how a zener diode regualtes voltage

[tidal dock]

idk if a math server can help you much with that

[strange helm]

anyone whos doing electronics could

[tidal dock]

try #old-network , there's an electrical engineering channel there

[strange helm]

yea but like how does it regulat the voltage

if it's rated 10V , applying something higher makes the voltage difference still remain 10 v across its ends - how does it do that

and even in the application of rectification after filtering normal diode rectification

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[little meteor]

Is equivalence just meant that their distance is 0?
so the only solution would be x1= x2 and y1=y2?

[paper depot]

you say "the only solution" but that sounds a bit weird

like, yes, two records are considered equivalent by this distance function if their x and y coords match

[little meteor]

okay cheers!

yeah only solution wasn't the riht word

thank you though

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[half mountain]

Anyone know how I would solve the integral:

1/(e^(x/2) + e^(x/3))

Im so lost

[uncut holly]

LMAO I did this same integral this week

Rewrite in terms of e^x, do a u-sub

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[haughty mural]

i need help with the steps

[outer warren]

start with half life formula

[haughty mural]

i know i use p=f(t)=ab^t using points 0,100 and 1700,50 right?

right?

ok

thanks for the help ig

.close

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[buoyant bone]

Hey

[buoyant bone]

could someone help me out with this

[vocal sleet]

@buoyant bone Has your question been resolved?

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[idle meteor]

Hi

[idle meteor]

This is an onoptimal experiment

Whats meant with that?

The course is about statistics but what does the map mean?

<@&286206848099549185>

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@idle meteor Has your question been resolved?

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[surreal niche]

I have a question but its in swedish ill explain it when i get someone to help

[surreal niche]

It says: The square ABCD has its corners on a circle with a radius of 5 cm. What fraction of the circle's area is covered by the square's area?

[surreal niche]

So i got it by simply extending the 5 cm to 10 cm so i get 1 triangle out of the square.

[cunning remnant]

You got the diagnol of the square

You can find the side lengths now

[surreal niche]

Yes and then i do the area for the triangle which is (10 x 10)/2

= 50 cm^2

[tawny salmon]

thats wrong, here the base and height are not same

[surreal niche]

Now for the circle, its 5^2 x pi

[tawny salmon]

the biggest side is 10, the other 2 arent 10

[surreal niche]

Yeah thats where im stuck

I dont understand how the other 2 arent 10

[tawny salmon]

its not an equilateral triangle

one angle is 90 degrees

[surreal niche]

Yh but how long is it then? Is it 5?

[tawny salmon]

do you know the pythagorean theorem?

[surreal niche]

Yes

[tawny salmon]

its a right angled traingle

the 2 smaller sides are equal as they are the sides of a square

[surreal niche]

I know its supposed to be simple but how i personally look at this is that the 5 cm is half of the leg. I know personally doesnt matter and what is correct is correct but my eyes look at it like that. Its a simple matter i know

I dont know how to explain what i mean

[tawny salmon]

could you maybe draw a figure to show what you think

[stray shard]

The circle area is r * r * π, and then the square area is s^2.

A/25π, if you staged the formula for the area it covers.

[surreal niche]

Im on my phone right now and dont have my computer cuz its dead and i wish i could show u what i mean

Its charging

[cunning remnant]

Remix a blank picture

Or something

[cunning remnant]

Remix this itself

[cunning remnant]

2x^2 = 100

so x = sqrt(50)

[stray shard]

It's actually a perfect square when I read the question.

All angles are 90 for it.

[cunning remnant]

The question says it's a square

[cunning remnant]

So the side length of the square is 5 times square root of 2

5sqrt(2)

[stray shard]

That's radius, not diameter.

[cunning remnant]

???

Diameter is 10

[stray shard]

Yeah and you only got a quarter of the square.

[cunning remnant]

what

Wdym

[tawny salmon]

@surreal niche do you agree that this is our traingle? x is side length of square

[stray shard]

It's 10 squared for the whole square with Pythagorean theorem.

[surreal niche]

Yes

[tawny salmon]

now apply pythagorean theorem to this triangle

[cunning remnant]

what are you saying

We are finding the side length of the square

[stray shard]

All sides are equal and if a^2 + b^2 = c^2 and b = a, 2a^2

[cunning remnant]

That is what I did

[stray shard]

Actually I misread your statement.

[cunning remnant]

Ahh no problem

[stray shard]

You meant that 5 is half the square root of 2 times a side.

[surreal niche]

Sorry guys but can i just quickly ask what program do yall use to draw on computer. I dont use computer that much

[cunning remnant]

what no

[surreal niche]

An illustrator

[stray shard]

Use some vector-based software. It's pretty good.

[cunning remnant]

I don't have a computer, can't help sorry, I just use discord built in remix thing

[tawny salmon]

i just use annotate on a screenshot, you can google and find many options

[surreal niche]

That is what i meant

Wait

I figured out my issue

I see now

[tawny salmon]

thats correct too

[surreal niche]

Or no

[tawny salmon]

u can apply pythagorean theorem on the smaller triangle now

[surreal niche]

Look if both are 5 wouldnt one of the leg/catheter (we call it "katet" in swedish)

Be 10

[tawny salmon]

you mean AB would be 10?

[surreal niche]

Yes

[stray shard]

It is equal to the diameter.

[tawny salmon]

no thats incorrect

AB^2 = 5^2 + 5^2

ab isnt just 5+5

its sqrt(5^2+5^2)

[surreal niche]

What isnt sqrt of that just 5 + 5

[tawny salmon]

nope

[stray shard]

Apparently he took the approach to √50 by calculating the radius as 2a^2 = c^2.

That kinda works too. The answer is all the same anyway.

[surreal niche]

Ok so 1 triangle would be 50 cm^2?

[cunning remnant]

You can't cancel out squares like that when they're in addition

[stray shard]

No!

[cunning remnant]

What triangle

[surreal niche]

Half of square in the picture

[stray shard]

Since multiplying two square roots together cancels out the square root, we effectively have 50 cm^2 as the whole square.

For the triangle, you would have divided it in half to get 25 cm^2.

[surreal niche]

This is what i understood from thus

[tawny salmon]

yes thats the side length of the square.

[cunning remnant]

That is equal to square root of 50

Which is the side length of the square

[cunning remnant]

5^2 + 5^2 = 25 + 25 = 50

[surreal niche]

Im sorry for making it hard when its not supposed to be. Im just trying to simplify it and make it easier in the future

[cunning remnant]

We are trying to make it simple for you

[stray shard]

This calculation doesn't exactly need the precise numbers. Just note it as √50.

[cunning remnant]

Or simplify it as 5sqrt(2)

[stray shard]

Pay attention to the answers too, life is too short for precise numbers.

[surreal niche]

Ok so next time i see something like this im not meant to think of the sides as the radius

[cunning remnant]

Yes

[cunning remnant]

Remember that radius is always form the centre of the circle

[stray shard]

One thing to be mindful about calculation is that every time you do the same calculation, the result is exactly the same. It is replicable.

[surreal niche]

Why im asking is if it goes from one of the corner to the middle it looks like its just half of the side of the square

Thats me thing

[tawny salmon]

its not half the side of square

[cunning remnant]

It's not half

It's half the diagnol but not half the side length

[surreal niche]

Ok i got that now so can it be interpreted as half of a triangle? Like the one i "drew"

[surreal niche]

.

[cunning remnant]

Yes

[stray shard]

It really isn't. If you paid attention in class, the usual lengths per vector assuming if each was the length of 1, the diagonal is the square root of 2.

[cunning remnant]

That's correct

[surreal niche]

💔i promise i pay attention its just im trying to simplify

[stray shard]

1 ≠ √2

Square root of 2 is approximately 1.414 IIRC?

[cunning remnant]

Yes

[surreal niche]

Ok i get it now

[stray shard]

The diagonal is simply a bit longer than the side and therefore wouldn't count as half a side.

[surreal niche]

So it would be 50 cm2 for the whole square

One triangle is 25

[stray shard]

Correct.

[cunning remnant]

Yes

[surreal niche]

Ok and now circle is just 5^2 pi

[cunning remnant]

Yea

25pi

[surreal niche]

So now its 50 / 5^2 pi

Thats the answer

[stray shard]

You should simplify it to that and then stage it up.

50/25π

More simplification here to be done.

[surreal niche]

Yes exactly

[stray shard]

It will converge to the correct answer you are looking for.

[cunning remnant]

Isn't it supposed to be 25pi/50

Oh wait

Nvm

[stray shard]

No, the circle is 100%.

[cunning remnant]

Yea I got it

[surreal niche]

Yh the whole is the circle

[cunning remnant]

I tripped

[surreal niche]

No its cool it happens

Well i appreciate u guys so mucj

How do i close this

Should i just leave?

[stray shard]

Dot command and close.

Then you can carry on your day.

[cunning remnant]

type ".close"

[surreal niche]

.close

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[mortal latch]

Hello

[mortal latch]

can someone help me find the answer to this

the answer I kept is wrong

[gaunt pelican]

do you know what factorials are?

[mortal latch]

yes

[outer warren]

you seemed to have multiplied instead of dividing

[mortal latch]

I get a error in the calculator

if I out this equation

[outer warren]

show what you're putting into the calculator

[mortal latch]

okay one second

I get syntax error

[outer warren]

put a multiplication symbol between the two 3!

[mortal latch]

it workedd

if it’s in a fraction does it divide the two factorials

or three in this case

I got a error for this too

the first one

and I added a multiplication sign

[outer warren]

although the end result is relatively small, the intermediate calculations are too big for calculators to handle

you can first apply properties of factorials to reduce that to something it can actually handle

[somber portal]

you're supposed to simplify before brute-force computing them

[outer warren]

e.g. consider
10!/8! = 10 * 9 * 8!/8! = 10 * 9

[mortal latch]

oh right we took this

[mortal latch]

is this just an example or it’s how to solve it

[faint quarry]

That's how you solve it, you expand the factorial until u can cancel it out through division

[outer warren]

that's an example of how you'd solve it

try applying a similar idea to what you have

[mortal latch]

I know we’re supposed to subtract the number

but there’s no substraction in this equation

[outer warren]

wdym

are you referring to the third one?

split the fraction and calculate each one individually

[mortal latch]

No the first one

[outer warren]

wdym subtrtact the number

there's no subtraction anywhere

oh

[mortal latch]

I learned this formula nPr there’s only 2 factorials and u have to substract the n- r

[outer warren]

are you refering to
$$\combo{n}{k} = \frac{n!}{(n-k)!k!}$$

[mortal latch]

so how does it apply here

[twin meteor]

ℝαμOmeganato5

[mortal latch]

yes but without the k

is it combination?

[outer warren]

some places use k, others r

[mortal latch]

wait this is right the order is different

[outer warren]

same thing

[outer warren]

its multliplication

order doesn't matter

so identify your n and r value

[mortal latch]

I meant that it’s not in the formula but nvm it is

[outer warren]

and the calculator can actually calculate it if you input it this way

[mortal latch]

so yes there’s n and r

that’s two factorials

600 is a error too

[outer warren]

so identify your n and r value

[mortal latch]

there’s theee values

three

that’s what I’m confused about

[outer warren]

r! and (n-r)! are linked

[mortal latch]

it doesn’t apply to the formula

[outer warren]

its symmetrical

[mortal latch]

OH SO ITS

600-10

[outer warren]

so doesn't really matter which one you take as r or n-r

yes, so you can use that as your r value

600-590 for r would work too

[mortal latch]

am I supposed to multiply

the values between 600-590

[outer warren]

including 600, excluding 590, yes

you could do that

its more tedious for the second question

but i think they just want you to apply that nCr definition

identifying that n=600 and r=10 or 590
you can just plug that into your calc

[mortal latch]

It says my answer must be a number

but the number is insane it’s like a billion

how am I supposed to write that

ugh

it doesn’t look right

[outer warren]

don't forget to divide by 10! as well

the number will be huge

use scientific notation

as the question states

[vocal sleet]

@mortal latch Has your question been resolved?

[vocal sleet]

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[eternal obsidian]

We need to find the solution to each and sum them up here, I have a problem with the first one

[bronze osprey]

don't ask that one yet

you're asking too many questions at one time

[eternal obsidian]

makes sense

[bronze osprey]

right, so for the 3rd question, we need to use Fermat's little theorem, which states that 3^5 = 3 mod 5

so we just need to figure out what 43^43 is mod 5, so 3^43 mod 5 or (9^21) * 3 mod 5

then you also need to find 3^(43^43) mod 8 using similar reasoning, where note that 9 = 1 mod 8

then use the Chinese remainder theorem

[bronze osprey]

also have you tried searching these problems up online?

[eternal obsidian]

Yeah

[bronze osprey]

https://approach0.xyz/search/

[bronze osprey]

or try searching here then

[eternal obsidian]

Sorry for not specifying earlier but I only have a problem with the 1st one

[bronze osprey]

oh wait I didn't read

ah I see

[eternal obsidian]

I just wrote it

[valid idol]

Hi

[eternal obsidian]

sup

[valid idol]

Can u work.my sums

[eternal obsidian]

Sure

[valid idol]

Shall I send u questions

[eternal obsidian]

Yeah send them in another channel

#help-4

[bronze osprey]

what have you tried so far with q1 then

do you see how 31C1 - 31C2 + 31C3 - .... + 31C1 = 1?

[eternal obsidian]

Yeah I do

This isn’t any basic question so, try advanced approaches

It is meant to train for International Math olympiad

this is the basic question sheet in that well

[bronze osprey]

https://approach0.xyz/search/?q=OR content%3A%24\binom{31}{1}-\left(1%2B\frac{1}{2}\right)\binom{31}{2}%24&p=1

found it

https://artofproblemsolving.com/community/c6h1550045p9421779

this thing can search all of AoPS and Maths Stack

[eternal obsidian]

well it is 1/n no wonder

And using induction isn’t really an option because we are solving it here not prooving

[bronze osprey]

I think just notice the pattern, try a smaller case

like try what happens when n = 3 or something small

[eternal obsidian]

Makes sense

But that’s just like proving collatz conjecture by.. computing the results

well thats a really good approach

I really appricieate it!

[bronze osprey]

yeah good luck on your IMO training!

I didn't get that far at your age

[eternal obsidian]

Especially because we re supposed to solve it in a time bound enviornment

[eternal obsidian]

Idk man I qualified the nationals

went through the national team training camp

with 50 others

and then they choose the top few of them

[bronze osprey]

oh you must be in the maths olympiad Discord server right

[eternal obsidian]

so i wasnt chosen sadly

[eternal obsidian]

it was last year

in my 11th grade

[bronze osprey]

no, as in

there's a Discord server you can join that is focused on IMO and other olympiads

[eternal obsidian]

Now I am just preparing for JEE Advanced

[eternal obsidian]

Not really into it now

[bronze osprey]

ah okay I see

then yeah JEE Adv now for you, I see

[bronze osprey]

you did mains already right?

[eternal obsidian]

I got a national rank of 3000 in mains, the first level

yeah mains

So you are from India too?

[bronze osprey]

ah that's impressive, well I'm not Indian so

[eternal obsidian]

Oh alright

[bronze osprey]

just know too much about India I guess

haha

[eternal obsidian]

Nice

[bronze osprey]

yeah trying to qualify for India's IMO is so exhausting

RMO is a big hurdle from what I remember other people have said

and then India doesn't historically do well in the IMO...

cause too many people focus their energy on JEE naturally

[eternal obsidian]

Damn is it so?

[eternal obsidian]

It is more rewrding tbh

rewarding*

[bronze osprey]

wait maybe it's the INMO

RMO doesn't look bad

like the AMC

[eternal obsidian]

Idk any of those acronyms

but we can talk this in dms imma share the next problem

[bronze osprey]

can you close this and ask in a new channel?

[eternal obsidian]

sure

.close

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[obtuse lodge]

I don‘t know where I went wrong but there shouldn’t be a remainder ;((

[obtuse lodge]

Nvm got it

.close

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[obtuse nest]

Hello math fellows, so here I am with another question.
Assuming a set A, I need to find out if it's linearly dependent or not. My question is if I need to limit my process to scalar multiplication, and combination(addition) of elements in order to find if an element is linearly dependent to other or not, or I'm allowed to use more operators?
Let say A={1,sin(x),cos(x)}, so if limiting the process to use only scalar multiplication and addition, then it's linearly independent.
But I can tell I can find 1 as sin^2 + cos^2, but sin^2 is not a scalar multiplication?!.
Or even find cos = sqrt(1-sin^2), but it advances two results, so it might not be accepted?
Thank you for your time and consideration

[hard atlas]

only linear combinations

so only scalar mult and addition

[obtuse nest]

Much appreciated sir. Thank you for your time and consideration.
Math bless us all 🍻

.close

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[wild helm]

guys what derivation and how to do it?

[gaunt sparrow]

Assuming you mean derivation as differentiation, well you usually follow a whole course to learn about it. It’s not really doable to explain everything here if we don’t know your background.

There are plenty of resources online though, see Khan Academy or Paul’s Online Notes.

[cunning plaza]

https://tutorial.math.lamar.edu/classes/calci/DerivativeIntro.aspx

[vocal sleet]

@wild helm Has your question been resolved?

[gentle nimbus]

just look at standard definition and subsequent formulae based around it

just so you know, you need to have familiarity with limits

[vocal sleet]

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[bitter delta]

$[ x^3 - y^3 ]

[twin meteor]

Paul04
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[bitter delta]

how can I figure out what that equals to without having the formula memorized?

[pallid zenith]

you can guess

[bitter delta]

is there not a better way

[pallid zenith]

guessing is the way if you dont know

[bitter delta]

so there's no method?

[pallid zenith]

guessing is the method, you can call it something fancy if you like

[hard atlas]

well how much effort do you want to go through. you can always rediscover it

[bitter delta]

yeah this

[pallid zenith]

you may try guessing some forms

[bitter delta]

how did the person who discovered it discover it

[pallid zenith]

probably guessing lol

[bitter delta]

😭

[pallid zenith]

(x-y)(ax^2 + bxy + cy^2)

[outer warren]

x=y is a trivial zero

[pallid zenith]

maybe you start here

[hard atlas]

well one option is that they randomly multiplied out other stuff and saw that it simplified nicely

[outer warren]

and then you can do long division

[pallid zenith]

yea

[hard atlas]

otherwise you can do polynomial division

[pallid zenith]

oh i forgot that division exists

[hard atlas]

for y=1 its the geometric sum which is also well known

[bitter delta]

https://math.stackexchange.com/questions/1172119/how-to-prove-x3-y3-x-yx2xyy2-without-expand-the-right-side

what does this mean?

[pallid zenith]

you need polynomial division

https://www.mathsisfun.com/algebra/polynomials-division-long.html

[hard atlas]

do you know how to divide x^3-8 by x-2?

[bitter delta]

nah, I forgot

I could figure it out though I've done it before

okay so basically x^3 - y^3 / x-y?

that gives x^2+xy+y^2 and you bring back the x-y so you get (x-y)(x^2+xy+y^2) = x^3-y^3?

[hard atlas]

yes

[bitter delta]

thanks everyone

.close

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[tired olive]

if n(A) = 5 and n(B) = 10, then find maximum and minimum possible values of n(A∩B) if

a) n(U) = 8
b) n(U) = 20

[tired olive]

i don't understand the first case

how can the max value be lower than the min value

it just twists my brain

[hybrid flicker]

what is U?

it feels like there's some context lacking indeed

because if "U" is A U B then both cases' premise are nonsense

[tired olive]

union

oh sorry

that's universal set

[hybrid flicker]

then a) still doesn't make sense indeed

you can't have more elements in B than in the universal set

[tired olive]

just a sec

that's the solution the teacher gave

i can link the video but its in hindi

if you want

i get how he solved it

but just don't understnad why the maximum value is less than the minimum

[hybrid flicker]

I still don't get it

[tired olive]

for example.

if there are 5 people who speak french and 10 people who speak spanish, and there are total of 8 people then 7 people must speak french and spanish both

[hybrid flicker]

ok but

"if there are 5 people who speak french and 10 people who speak spanish, and there are total of 8 people" that alone is nonsense

like I get they applied the n(A∩B) + n(A U B) formula

[silk osprey]

yea it doesn’t make much sense to allow for elements of B to not be in U

U is supposed to contain everything

[tired olive]

thats not elements

its number of elements

[hybrid flicker]

but

there's a total of 10 elements in B

[silk osprey]

i’m aware

[tired olive]

wait ill a draw a ven diagram

[hybrid flicker]

(the number of elements in B is 10)

but there's supposed to be 8 elements in the universal set

(the number of elements in U is 8)

how can there be more elements in B

than in the universal set

[tired olive]

well now that you say that

im a bit confused too

yknow what the world is fucked anyways , questions can be fucked too

how bout we leave it at that

.close

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[mortal latch]

can someone help me with answers I have a assignment due in a hour

[hybrid flicker]

maybe remove the comma?

[mortal latch]

I TRIED IM LOSING MY MIND

WHY IS IT WRONG

sorry didn’t mean the caps

[flat whale]

commas aren't numbers

[flat whale]

show what you tried

[mortal latch]

wait actually removing the commas worked sorry guys

[hybrid flicker]

its alr

[mortal latch]

I can’t figure out these three

[hybrid flicker]

is it just asking you to plug those in the calculator?

[mortal latch]

The calculator won’t take it it’s too big

and the scientific notation doesn’t work

[hybrid flicker]

so think about another way for the calculator to give those numbers to you

does the expression $\frac{n!}{k!(n-k)!}$ call something to mind?

[twin meteor]

rafilou is not not born in 2003

[mortal latch]

I tried this

[flat whale]

simplify before plugging into a calculator

[hybrid flicker]

even if that, the second one is gonna be hellish

oh well

[mortal latch]

what should I calculate exactly

also is this the right way to find the answer

the scientific notation

[vocal sleet]

@mortal latch Has your question been resolved?

[mortal latch]

No

[vocal sleet]

@mortal latch Has your question been resolved?

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[onyx fjord]

why 13?

[vocal sleet]

@onyx fjord Has your question been resolved?

[faint pumice]

they are using the addition identity of sine, where, in this case 12sin(2x)+5cos(2x)=Rsin(2x+θ). R, the amplitude, is found by sqrt((12^2)+(5^2))

which equals 13

[flat whale]

do you know calculus?

[onyx fjord]

yeah

[flat whale]

oh that works too

[faint pumice]

the amp is the max

[faint pumice]

would you just derive it and get the max from that

[flat whale]

derivatives is easier i suppose.

[slow mural]

cool

[flat whale]

but memorizing the amplitude/phase formula also works.

[onyx fjord]

do I need to match 0?

after derive

[faint pumice]

wym

[onyx fjord]

I saw this formula but I cant understand

[onyx fjord]

If I derive do i need to match 0?

[flat whale]

what's confusing

[faint pumice]

whta is the phase amp formula

[onyx fjord]

√(a² + b²)

[flat whale]

phase = theta. amplitude = R

oh it's called Harmonic Addition Theorem
https://mathworld.wolfram.com/HarmonicAdditionTheorem.html

[onyx fjord]

.

[faint pumice]

oh

yes sorry

you can find the derivative and set it equal to 0

[onyx fjord]

ok

I got it

[faint pumice]

did you get 13 ?

[onyx fjord]

yes

[onyx fjord]

but I find by this

I cant find by derivate

i dont know how complete this kkk

[faint pumice]

yeah neither do i 😹

i like to think im decent at math but lowk i cant do much hahaha

[onyx fjord]

kkkkkkkkkk thanks guys

how can I finish this doubt

[faint pumice]

what class is this for

[old sigil]

If you let a = sin 2x and b = cos 2x, then 10a = 24b and a^2+b^2=1

substitute b = 5a/12 into the 2nd equation to obtain a^2+25a^2/144 = 1, which simplifies to a^2 = 144/169 and a = 12/13(since a is positive)

from there you can also find b = 5/13, and the rest should be easy

[onyx fjord]

I'm studying for MEXT, but this is from ITA (INSTITUTO TECNOLOGICO DA AERONAUTICA) this is like a undergraduate program from brazil

[onyx fjord]

ohh I didnt see this

[faint pumice]

ah okay

i see

very nice, i pray you do well !

[onyx fjord]

thank

[old sigil]

using sin^2+cos^2=1 is a very common pattern, when you're given an equation that treats sine and cosine as seperate variables
it's nice to keep in mind

[onyx fjord]

thanks for advice

[onyx fjord]

.close

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[stone timber]

I'm ngl i dont know where to even start on this

It says to verify the trigometric identity

[inner osprey]

probably identify first if you want to use left to show right or use right to show left

[stone timber]

how do know which one is easier to start at

i have done the verifying trigometric identites before, just not with the double angle stuff

would i multiply the top and bottom by the conjugate in this example if i started on the left?

[blazing wigeon]

Usually I try breaking up the double angle functions first and see what’s left. So I would use the cos(a+b) for the cos3x

[stone timber]

so would that be cos(2x+x) = cos3x

[blazing wigeon]

Yes

[stone timber]

can you tell me if this looks alright

oh i meant - cosx on the bottom too

• cosx on the bottom 😭

sorry

[blazing wigeon]

Looks good so far

[stone timber]

now do i plug in cos2x= cos^2x - sin^2x?

[blazing wigeon]

Hard to say, pick one of the cos2A but definitely use the 2sinacosa for the sin2x

[stone timber]

uhhh

i’m getting lost atp😭

[blazing wigeon]

Make sure to keep the parentheses around the cos2x and you can divide out a cosx from the whole fraction

[stone timber]

sorry girl you lost me

you mean around the

1-2sin^2x?

[blazing wigeon]

Yes

Divide out a cosx from every term

And things will cancel out

[stone timber]

do you think you could write out what it would look like after you divide by cosx

[blazing wigeon]

$\frac{1-2sin^2x-2sin^2x-1}{1-2sin^2x-2sin^2x+1}$

[twin meteor]

mari