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[crystal lion]

for part c, i understand why the probability that the number of times the dart first hits the target is (1-0.1 = 0.9) but i dont understand why u would then do (0.9)^5 x 0.1

for part d, i dont understand why u would use a uniform distribution

[tired walrus]

a lot of misunderstandings from you here

(d) is not a uniform distribution at all

P(F=5) is 0.9**^4** * 0.1 because it is the probability of the specific sequence "fail throws 1, 2, 3, 4, then succeed on throw 5"

[crystal lion]

is this not a uniform distribution table?

[tired walrus]

no it's not

uniform would have the same number across the entire P(...) row

[crystal lion]

okk

[tired walrus]

uniform means each outcome is equally likely

yours are very much not

[crystal lion]

ohh okk

[crystal lion]

yep, im just kinda used to seeing that table with uniform distribution

thanks!

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[subtle path]

how to solve this type of equations using bezouts theorem/identity for example
a+b=360 / gcd(a,b) = 18 where a and b are two positive integers.

[onyx peak]

i would write d = gcd(a,b) = 18 and then a = a'd, b = b'd (so a' and b' end up being coprime)

[subtle path]

yep ok

it leads to something like this :
a' + b' = 20 / gcd(a',b') = 1

but then i'm stuck

[onyx peak]

There will be many sols

Are you trying to count them, find them all, or what?

[subtle path]

find them all

[onyx peak]

Then probably enumerate a' and b' one by one

[subtle path]

but what's the reasoning after this

[atomic moon]

Seeking the one that works

[subtle path]

how do I do that

[onyx peak]

a' = 1, b' = 19 works
a' = 3, b' = 17 works
...

a' = 2, b' = 18 doesnt work btw, because they arent coprime

[subtle path]

aint there a way that gives me all possible solutions logically

[subtle path]

yep

[onyx peak]

There is one

If a' and b' arent relatively coprime, then there is some prime p dividing them

but then that prime p has to divide a' + b'

so it has to divide 20

[subtle path]

ok

[onyx peak]

so a' and b' cant be multiple of any prime dividing 20, that is 2 or 5

[subtle path]

so all numbers not multiple of 2 or 5 ?

are the a' and b' we are looking for ?

[onyx peak]

I believe so, im not sure about the proof in the other direction yet

We still need to prove that if a' and b' aren't multiples of 2 or 5, then they are valid sols

[subtle path]

ig it aint that helpful, it just helps me filter through all integers below 20 to test if they work

so i should prob list them like you said before

[onyx peak]

Oh, it does actually work

[subtle path]

hmm ?

[onyx peak]

so the sols are all pairs (a', 20-a') where a' isnt a multiple of 2 or 5

then multiply those pairs by 18 and you get the original a,b pairs

[subtle path]

yep

that's what i was thinkin

ok thank you very much !

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[fresh wave]

Could I have some help here please?

[fresh wave]

This is what I’ve done so far

[severe canyon]

So far everything is correct

Make sure to write the new upper and lower bounds

[coral zenith]

Yeah after that you can just expand the square and integrate it

[fresh wave]

Ohhh ok

Thank you!

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[flint haven]

Question no. 9 Why is my solution wrong?

[tired walrus]

"AC bisects angle BAD" only implies AB+AD=AC if vectors AB and AD are known to have equal length to begin with and even then you could be off by a scalar multiple.

so your solution is kinda cooked from the very very start.

[flint haven]

So you mean the equation AB+AD=AC is wrong?

[tired walrus]

equation, and yes.

[flint haven]

Can you give me the best way to kickstart?

[tired walrus]

angles BAC, CAD, BAD are pi/3, pi/3, 2pi/3 resp.
take the length of vector AC as 1 for convenience and write down the lengths of the others.
then work out vectors BA and CD in terms of any of AB,AC,AD and solve from there.

[flint haven]

Yes, I want to know the way to work out vectors BA and CD in terms of any of AB,AC and AD

[stuck juniper]

What jee shift is this dawg

[tired walrus]

probably the best way would be to draw a diagram but at least try to be careful about how you do it, so that it doesn't end up a parallelogram

[flint haven]

Ok thanks

[flint haven]

Dpp

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[frigid laurel]

what am i supposed to do with pi

[winter lava]

tg is tangent?

[frigid laurel]

yeah

[brisk lion]

ugh systems of equations with trig...

[winter lava]

x = y+pi, try making a substitution using this

[frigid laurel]

ight

[tired walrus]

so x = pi+y

wait but what's the full question

[frigid laurel]

this

[tired walrus]

what does it say to do

[frigid laurel]

like solve for x and y?

[tired walrus]

you could maybe share the entire page that this was on

cause rn it looks a little strange to me

[frigid laurel]

the book is in macedonia

macedonian*

[tired walrus]

i speak bulgarian, that might be close enough

[frigid laurel]

in translation it legit just says solve these system equations

ight then

[tired walrus]

ok

i mean, you know that the tangent function (⁨tg⁩, or ⁨tan⁩ internationally) is periodic with period equal to pi, right?

[tranquil pine]

indeed i get 0=3

[frigid laurel]

?

[tired walrus]

$\tan(t + \pi) = \tan(t)$ for all real $t$

[soft zealot]

Ann

[frigid laurel]

i'm confused

how does this work

[brisk lion]

have you ever seen a graph of the tangent function before?

[frigid laurel]

sadly not

[winter lava]

so tg (y+pi) - tg y = 3

[frigid laurel]

yeah i got that

[winter lava]

but the period of tg is pi, so tg(y+pi) is the same as tg y

[frigid laurel]

the formula just confuses me

i get it now

wait so now i'm at tgy-tgy=3?

[winter lava]

yeah

so 0 = 3

[frigid laurel]

so the equation is not solvable?

[winter lava]

that means no solutions

yeah

[frigid laurel]

fair enough

thank yall ❤️

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[violet forge]

HEY
Can soemone give em help with this question

[violet forge]

<@&286206848099549185>

[halcyon ether]

!15m

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[wild venture]

Yeah

[shell condor]

Why are you asking for help, we should be asking you

[halcyon ether]

how tf BAC=90

[violet forge]

just is

[halcyon ether]

My eyes is disagree

[violet forge]

bad diagram

can anyone help

[halcyon ether]

give us a min

oh okay

damn 4 min, I'm getting old

[violet forge]

hmm

[halcyon ether]

Let H be the leg of the height from A

you make the diagram

[violet forge]

yup done

[halcyon ether]

Can you send it here

[violet forge]

umm i dont have my phone but ik what u r saying

line down from A perpendicular to BC

[halcyon ether]

Okay now use Thales Theorem or BPT

[violet forge]

well on which triangles

[halcyon ether]

Can you figure it out?

[violet forge]

wait
idk why it is taking time
i was doing it differently earlier
proved that triangle BDG and EFC were similar

[halcyon ether]

They are

hm

[violet forge]

hmm

[halcyon ether]

okay that's might be better than mine

Oh wait

Fuck this diagram

why is that it looks like isosceles triangle while it isn''''t

But doesn;t look like right triangle when it is

[halcyon ether]

nvm they aren't similar

The diagram tricks you

[violet forge]

no it doesn't
they are similar
as angle F= G =90
and DG =EF
and AE/EC = AD/DB
and angle BDG = angle CEF

[tiny kraken]

Hahaa they don't

This's my alt

I meant cupcake is my alt

[violet forge]

well am i triping

this is a 2 marks question T^T

[tiny kraken]

You are assuming BDG=CEF

[violet forge]

ahh its taking painfully long
what approach should i take

[tiny kraken]

I only know one way so I will show you that one

[violet forge]

plz

[tiny kraken]

Continue with this

Now EF/AH=FC/FH, right?

[violet forge]

nope it is going to be
EF/AH = FC/HC

what next

[tiny kraken]

Oh yeah lol

Do that

Again for the left triangle

[violet forge]

well now i know that ▲AHB is similar to ▲BGD and ▲FEC
and hence ▲BGD is also similar to ▲FEC

told you earlier too

[tiny kraken]

No 😭 , FEC is similar to HAC

If they are similar

[violet forge]

oh yeah sorry

but both triangles are equal no ?

[tiny kraken]

No

[violet forge]

oh

[tiny kraken]

That requires the triangle to be isosceles

[violet forge]

it has to be

as there is a square inside it

you forgot that ?

[tiny kraken]

It don't have to

[violet forge]

yup

[tiny kraken]

You can form a square within any closed curve I think

[violet forge]

ahhh
id think so
it is a triangle with one angle 90

[tiny kraken]

Okay so, EF/AH=CF/CH

DG/AH=BG/BH

Now multiply them together

[violet forge]

its not looking prommising
(DG x EF) / AH^2 = (CFxBG)/(CHxBH)

hello ?

[halcyon ether]

Yes

DG=EF=GF

Since it's a square

Now it look promising

AH^2=CH•BH, this's a common fact in right triangle

Try to prove it

[violet forge]

did it

turns out i was right all along

T^T

alright

have a Good Night
Bye

[halcyon ether]

The triangles are never similar bruh

K baii

[violet forge]

yes they were

i proved them to be similar

try

angle - C = BDG

and same for the angle B= CEF

trust

well how to close this channel

?

got it

.close

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[halcyon ether]

No

They aren't damn it

[violet forge]

yes they areee

try harder

you aer just not smart enogh

*slips out the window with proud

[halcyon ether]

Whatever, I don't have enough energy for this

[violet forge]

okay

get some coffee

and tryyy

byee

*slips again out of the window with proud

[north jasper]

u can find this is oswal questionbank it has a good solution there incase u have it

[violet forge]

well i need picture
my azz kinda broke to afford

[north jasper]

🥲 gave it away after tenth

[violet forge]

np its okay 😅

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[silk fjord]

That is wrong though

answer key states

why?

I though you could do that with triangles

[opal plinth]

In the first step, you use the area of GDE to find that of GHE, where GD and GH are the bases and E is a common point

In the second step, you try to use the area of EGH to find that of EFJ

[silk fjord]

oh wait yeah i see that

[opal plinth]

EG and EF are the bases, but the triangles don't have the same height

[silk fjord]

it works with DEF then right?

but idk the ratio there

[opal plinth]

No, still not the same height

[silk fjord]

DJE and EJF are same height?

just as DEF and DFG are

[opal plinth]

Oh ok I thought you were going with EGH

Are you supposed to solve this in two steps?

[silk fjord]

its a competition problem

solve however

as long as its right

im just making solutions

so im typing out the steps

[opal plinth]

Ok then my first thought is to find the areas of EGH, EFD, DEH, and DFG, and then you have a system of 4 equations and 4 variables

I haven't checked if it works

[silk fjord]

I see, let me try that

I get a system like this

sorry didn't format it

CAS gives this answer

which is no help since D is unknown

is this what you meant?

[opal plinth]

Right... I suspected that wouldn't be enough

I haven't looked at a geometry problem in a while

[final saddle]

@silk fjord Has your question been resolved?

[opal plinth]

Since there is no other constraint, you could just coordinate-bash it by choosing an appropriate coordinate system

G = (0,0), E = (7,0), D = (0,8)

Find the coordinates of J, then drop a perpendicular from J to GE giving point K, then find the area of KEJ and KFJ and subtract

I'm sure there's a slicker way to solve, but this should work just fine

@silk fjord

[silk fjord]

I see, that makes sense

thank you!

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[fallen valve]

how can i show fomrally that k(theta) must be constant?

[fallen valve]

it inutituelvy makes sense since it takes avlues in integers

so there must be a "jump" if not constant

[wet crater]

<@&268886789983436800>

[desert mantle]

that seems like not enough context

what is g

[final saddle]

@fallen valve Has your question been resolved?

[fallen valve]

Sorry yeah, g is

g is essentially a bvranch of arg(z)

on C \ {0}

im trying to use IVT to show that it is constant

i.e. smthing like k: (0, 2pi) \to Z, however I think IVT only applies when co domain is R

which is part of reason why I am stuck

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@fallen valve Has your question been resolved?

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@fallen valve Has your question been resolved?

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[soft pewter]

In a large wildlife reserve, a population of foxes and wild turkeys is observed. Today, there are 1200 wild turkeys, and it is predicted that this population will decrease by half every 10 years. At present, there are 300 foxes, and it is predicted that this population will increase by 2% per month. Based on this information, in how many years will there be the same number of foxes as wild turkeys?

[soft pewter]

a = 1200

b = 1/10

c = 1/2

f(x) = 1200(1/2)^(1/10)(10)

?

is c = -1/2 or 1/2

<@&286206848099549185>

[crisp mason]

remove that 10 from the turkey equation

[soft pewter]

alright

[crisp mason]

positive 1/2 because its decreasing

negative would give negative outputs, and you cant have negative turkeys lol

[brisk sierra]

Since the problem mentions animal populations, I will assume we're considering continuous growth and decay. Given this, you can use the following two formulas for turkey decay and fox growth respectively. They come from the half-life decay and continuous growth formulas: f(x)=1200\left(\frac{1}{2}\right)^{\frac{x}{120}} and f(x)=300\left(e\right)^{0.02x}

These formulas use months as opposed to years for the input variable gradations; that's what would make the most sense to me if I were to solve the problem.

[brisk sierra]

Desmos or a LaTeX editor can format the equations properly (sorry 😅 )

[soft pewter]

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[warm python]

Silly question. But this is basically saying if B subste A then B/I is a subset of A/I?

[worldly mesa]

Ideal not set

[warm python]

it says it's an inclusion preservinbg bijection

[warm python]

so what is being preserved

[worldly mesa]

The inclusion

[warm python]

of what

The inclusion being preserved is of subsets , right

[worldly mesa]

Yeah...

[warm python]

Assuming $I \subseteq B \subseteq A$

[soft zealot]

wai

[worldly mesa]

Then B/I subset A/I yeah

[warm python]

cool

and is obvious too

thanks

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[chrome ermine]

im confused

[tiny gorge]

when in doubt, rewrite everything in terms of sine and cosine

[chrome ermine]

i did

i have 1 +cot^2x * tanx

so then

[tiny gorge]

that doesn't sound like sines and cosines

[chrome ermine]

1+ (cos)^2/(sin)^2 * sin^2/cos^2

if i js directly converse to cosines

i get 1/ sinx* cos x

is that right

[tiny gorge]

nope

[chrome ermine]

oh

[tiny gorge]

oh wait it is right

[chrome ermine]

ok

uh now what

[tiny gorge]

try writing it as the sum of two fractions

[chrome ermine]

oh nvm i got it

i js used u=cotx

[tiny gorge]

one with sin(x) in the denom and the other with cos(x)

[chrome ermine]

i used u=cotx and then did usub

and got -ln|cotx|+c

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[lilac moat]

definitely the quickest way

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[night edge]

Hello, I'm mostly struggling to understand graphs right now, is anyone able to epxplain them? It's a packet and I'm genuinely unsure of what to do.

[tired walrus]

when you say graphs do you mean like y=f(x) graphs or like node-and-edge graphs

[night edge]

These are questions 1-4 and im still unsure after asking a ton of friends

[tired walrus]

ok so function graphs

[night edge]

Functions, yeah

sorry

[tired walrus]

ok. do you know what symmetry is

[night edge]

I'm pretty sure it's just how matching(?) Something is

so like, line etc

?

[tired walrus]

mmm kinda but not quite.

[night edge]

Oh, alrigjt

[tired walrus]

we say a shape is symmetric if you can for example reflect it or rotate it (by less than a full turn) in such a way that you end up with the exact same shape

have you seen snowflakes for example?

[night edge]

Yes! They're sort of the same on each side(?)

[tired walrus]

these things have rotational symmetry because you can rotate a snowflake by 60° and have it still look the same as when it started.

line symmetry is specifically about 2D shapes being the same when reflected.

[night edge]

The stuff we do on mcgraw hill is sort of similiar with the mirror tool, would that be the best way to really remmember it?

[tired walrus]

for example the shape of the uppercase letter Y (when printed) has vertical line symmetry

[tired walrus]

i guess kinda. a mirror-symmetric shape is one that can be UNAFFECTED by the mirror tool

[night edge]

Ohh alright

So 3-4 wouldn't be symetrical due to not being straight (to an aspect)?

[tired walrus]

mm nope

yeah ok so like

https://www.youtube.com/watch?v=rNURPkUEvg0 idk how best to explain a basic concept like this, but here's a video i found

[night edge]

ohh alright I'll go watch that

[tired walrus]

||probably easiest to simply say #1 and #3 have vertical line symmetry while #2 and #4 don't.||

[night edge]

i apologize, I'm really mind-blanked right now

[night edge]

Ohh thank you!

Im assuming because 3-4 are curved?

I think that just about answers what it can, hsve a goodnight!

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[velvet robin]

Hello, I would like assistance in a problem based on functions

[velvet robin]

Aside from the fact the the input is in a way related to it's mod 3, i cannot discern anything else

I have googled and tried AI, none of which provide the correct solution

[tired walrus]

!noai

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[tired walrus]

try thinking about the ternary representation of n

your f has a particularly simple description in terms of that

[velvet robin]

I have stated that the input depends on it's mod 3

I have not been able to proceed further

[tired walrus]

i am not talking about just mod 3

i am talking about the base-3 representation of n

do you know what this means y/n

[velvet robin]

No

I am in high school

This problem is from an olympiad prep book

[tired walrus]

don't olympiads sometimes have questions related to nondecimal base expansions

[velvet robin]

I have not encountered any so far

[tired walrus]

can i see the table of contents of the book

[velvet robin]

The book is called pathfinder and it is available online

[velvet robin]

Anyhow i am still very much stuck on this question

[tired walrus]

section 6.10, scales of notation

give that a read @velvet robin

[velvet robin]

Alright

[tired walrus]

what part of the book is this q from btw

[velvet robin]

Chapter 5

[tired walrus]

huh.

let me read through that

i see this on page 5.23 referencing binary expansions

so i guess it's kosher to look into future chapters here

[velvet robin]

It is

[tired walrus]

aight then yeah this is another one of those that requires look-ahead into ch6

[velvet robin]

I can't find the scales of notation though

[tired walrus]

page 6.35

[velvet robin]

Alright

Thanks

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[compact scaffold]

Anyone has any tips related to learning maths ? I wanna progress quick in maths but I'm in 9th grade(going to 10th in April) I wanna really understand maths, not just in the textbook way

[leaden moon]

a wise man once said:

99% people who learns math memorizes the rule, the 1% who actually is good at math actually understands why the rule works

[half raptor]

Yoooooo it’s uuuuuuu

[leaden moon]

what?

[half raptor]

Damn u forgot abt me

Oh well

[leaden moon]

remind me again please?

[half raptor]

My advice is to grind comp maths

By yourself

[leaden moon]

i don't remember helpee's name lmao

[half raptor]

You will experience pain but in the end you will understand

[leaden moon]

who the hell are you

[half raptor]

You know

[leaden moon]

.

[hollow stump]

Try to look for connections in your different maths classes. Often we think about using specific rules or such, but if you can first solve it a way you understand and try to figure out why the rule is being implemented you will understand it better. Also the use of tools like desmos and such to visualise always helped me

[leaden moon]

-# ||how did you know it was me who's a wise man lol||

[compact scaffold]

Hmm, I'll try abiding by this

[half raptor]

I was one of the person which we played gay roulette in discussion 2

[leaden moon]

oh shi

i remembered that thing

[half raptor]

Exactly

[leaden moon]

:(

[compact scaffold]

Any book recs to build a solid foundation?

[brisk lion]

in what topic(s)?

[half raptor]

Aops books are always good

[leaden moon]

https://books.google.com.vn/books?id=47UaDAAAQBAJ&lpg=PP1&pg=PP1&redir_esc=y#v=onepage&q&f=false

geometry

i only know that book which @ionic venture sent me

[compact scaffold]

Well right now, maybe proofs

[ionic venture]

Yo

[brisk lion]

How to Prove It by Velleman.

[leaden moon]

wassup

[ionic venture]

Sup

[leaden moon]

sup

[compact scaffold]

No pre requisites needed right? In order to learn proofs

[hollow stump]

Once you get to calculus there are a bunch of options. I like the depth that Spivak goes into about showing the why behind things and he even proves the most simple things like how (a+b)+c=a+(b+c) in the book

[brisk lion]

not much.

[leaden moon]

i don't think 10 even reached calc yet

[compact scaffold]

Alright that's good

[hollow stump]

Fair enough, but it is a nice read if you like math and once you get to calculus

[leaden moon]

• Genius is one percent inspiration, ninety-nine percent perspiration
• 99% people who learns math memorizes the rule, the 1% who actually is good at math actually understands why the rule works

[ionic venture]

Would you mind sharing me some topics in your curriculum?

[leaden moon]

thomas edison said the first sentence

2nd one is from a wise man

15 y/o

9th grade

yk who

[ionic venture]

I got this one fat big notebook

[spark swan]

art and craft of problem solving is a good book but unfortunately it's level is a bit higher

[ionic venture]

It's good

It gives you a distinction between problems and exercises

And a great mindset

that is to iterate

and explore

[spark swan]

[slim meadow]

This applies to all subject ngl and you might not need this, but keep grinding in once field of maths (ex. geometry) until you feel that you clearly improved, you switch to another field (ex algebra) while still doing geometry frequently
Note i didnt learn this from someone, its based on my own experience, and i improved a lot at geometry in just 1 week focusing on geometry (although i do feel like i got worse at other thing such as algebra so idk if this advice works)

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[silk fjord]

Honestly not to sure what to do here

If I draw quadrilateral ABCE then that is cyclic

And then interior angle E is 120°

And get a 30-60-90 right triangle by cutting the quad in half

But I don't see how that is useful

[opal plinth]

Make a diagram

[final saddle]

@silk fjord Has your question been resolved?

[silk fjord]

On mobile rn kinda hard 😭

Ig I can draw on paper

Sorry I have bad reception it's taking a while to upload

[opal plinth]

Mark the circle's center O, find the radius OE, draw OB and OD

The area is that of sectors BOE and DOE, plus that of triangles AOB and AOD

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@silk fjord Has your question been resolved?

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[unique ore]

L^(−1){(be^(−pa))(1/p​−p/(p^2+1)​)}

[glossy zephyr]

Inverse laplace?

[unique ore]

yup

[glossy zephyr]

,, \mathcal L^{-1}\left{ b\cdot e^{-pa} \left(\frac1p-\frac p{p^2 + 1}\right)\right}

i think

[unique ore]

no

[unique ore]

yup

[glossy zephyr]

Well, first of all, the b can come out

[unique ore]

yes

thats fine

my problem is e^-p𝜏×F(p)

[glossy zephyr]

Now, could you find the Inverses of both $\frac1p$ and $\frac p{p^2+1}$?

[unique ore]

yup

[glossy zephyr]

What are they

[unique ore]

Θ(t) and csn(t)

nesting it together is whats tripping me up

ℒ(Θ(t-𝜏)×f(t-𝜏))=e^(-p𝜏)×F(p)

[glossy zephyr]

Remember that the inverse laplace transform is also a linear operator

aka, it satisfies:

$$\mathcal L^{-1}{\alpha F + \beta G} = \alpha\mathcal L^{-1}{F} + \beta\mathcal L^{-1}{G}$$

[unique ore]

I guess

i mean yeah; threat it like coefficient

[unique ore]

ℒ(Θ(t-𝜏)×f(t-𝜏))=e^(-p𝜏)×F(p)

[glossy zephyr]

Divide the transform into two parts and do the corresponding shift for each

[unique ore]

"corresponding shift for each" -- wdym?

oh

-p𝜏 vs p

Θ(t-a)×[Θ(t)-csn(t)]

???

[glossy zephyr]

<@&268886789983436800>

thanks guys

[glossy zephyr]

You know that $e^{-pa}\frac1p$ is the L.T. of some function $f(t-a)$ times $\Theta(t-a)$

[glossy zephyr]

And this idea applies too for the other function once you do the distribution of e^-pa

Now the problem only becomes finding which $f(t)$ maps to $\frac1p$ or $-\frac p{p^2+1}$

[unique ore]

no; ℒ(Θ(t-𝜏)×f(t-𝜏))=e^(-p𝜏)×F(p)

F(p) = 1/p ???

why?

F(p) = (1/p-p/(p^2+1))

[glossy zephyr]

Try to divide it into two functions

$e^{-pa}\left( \frac1p - \frac p{p^2+1}\right) = e^{-pa}\frac1p - e^{-pa}\frac p{p^2+1}$

[glossy zephyr]

And since the ILT is a linear operator, you can solve the two function by themselves

[unique ore]

b×[ℒ^(−1){(e^(−pa))(1/p​​)}−ℒ^(−1){(e^(−pa))(p/(p^2+1))}]

[glossy zephyr]

Now your problem is a prime example of this property

You can see that you got $e^{-pa}\cdot F(p)$ here

[glossy zephyr]

You just have to find what f(t) relates to F(p) and do the correct shift

We call the shift the fact that we are doing f(t-a) here.

[unique ore]

b×[ℒ^(−1){(e^(−pa))(1/p​​)}−ℒ^(−1){(e^(−pa))(p/(p^2+1))}] = b×[Θ(t-a)×Θ(t)−Θ(t-a)×csn(t)]

[glossy zephyr]

not quite

[unique ore]

yeah

[glossy zephyr]

Its not "tau

[unique ore]

otherwise this would have been correct

[unique ore]

changed 𝜏 to a

[glossy zephyr]

Notice here that not only the step function is shifted, but also our function of interest

[unique ore]

ℒ^(−1){(b×e^(−pa))×(1/p-p/(p^2+1)​​)}=Θ(t-a)×[Θ(t)-csn(t)] --correct or not?

ℒ^(−1){(b×e^(−pa))×(1/p-p/(p^2+1)​​)}=Θ(t-a)×[Θ(t)-csn(t-a)] --correct or not?

ℒ^(−1){(b×e^(−pa))×(1/p-p/(p^2+1)​​)}=Θ(t-a)×[Θ(t-a)-csn(t-a)] --correct or not?

ℒ^(−1){(b×e^(−pa))×(1/p-p/(p^2+1)​​)}=Θ(t-a)×[Θ(t-a)-csn(t)] --correct or not?

[glossy zephyr]

Youre missing the b on all of them

but yeah

[unique ore]

ℒ⁻¹{(b×e^(−pa))×(1/p-p/(p²+1)​​)}=b×{Θ(t-a)×[Θ(t)-csn(t)]} -- incorrect
ℒ⁻¹{(b×e^(−pa))×(1/p-p/(p²+1)​​)}=b×{Θ(t-a)×[Θ(t)-csn(t-a)]} -- incorrect
ℒ⁻¹{(b×e^(−pa))×(1/p-p/(p²+1)​​)}=b×{Θ(t-a)×[Θ(t-a)-csn(t)]} -- incorrect
ℒ⁻¹{(b×e^(−pa))×(1/p-p/(p²+1)​​)}=b×{Θ(t-a)×[Θ(t-a)-csn(t-a)]} -- correct

[unique ore]

sry abt that

[glossy zephyr]

np

[unique ore]

is this correct now?

[glossy zephyr]

yea, last one

[unique ore]

ℒ⁻¹{(b×e^(−pa))×(1/p-p/(p²+1)​​)}=b×{Θ(t-a)×[Θ(t-a)-csn(t-a)]}

And so;

ℒ⁻¹{(b×e^(−pa))×(1/p-p/(p²+1)​​)}=b×{Θ(t-a)×[Θ(t)-csn(t-a)]} -- ❌ (my first attempt result)
ℒ⁻¹{(b×e^(−pa))×(1/p-p/(p²+1)​​)}=b×{Θ(t-a)×[Θ(t-a)-csn(t-a)]} -- ✔️

because
ℒ(Θ(t-𝜏)×f(t-𝜏))=e^(-p𝜏)×F(p) not ℒ(Θ(t-𝜏)×f(t))=e^(-p𝜏)×F(p)

I think I get now

thx a lot @glossy zephyr

[glossy zephyr]

no problem

Do ⁨.close⁩ if you dont got any other doubt

[unique ore]

.close

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[raven hawk]

can someoe explain how to set this up

[raven hawk]

i understand for Mary i will use a linear equation, and for Lisa an exponential

9-6pm is 9 hours ... but my question is , how do we start forming this

does 9-10 represent x = 0 or x=1 , for ax+b, and b(a)^x

i am always confused with the proper number i should use in the exponent

[glossy zephyr]

You can realistically just generate both as a geometric and arithmetic sequence.

[raven hawk]

ok, so its a1+(n-1)*diff

and a1*r^(n-1)

[glossy zephyr]

They both start at 2, so a1 is 2 for both.

[raven hawk]

mary is 2 + (9-1)*diff

oh i see now

= 512

[glossy zephyr]

Consider that Lisa is the one with the arithmetic sequence

[raven hawk]

right, so 2 + 8d = 242 , common diff = 30

and for the other the r = 2

and ill just list them out

question, when do you know the arithmetic sequence is a better fit instead of using a*x+b (general linear formula)

same for geom seq and exponential

[glossy zephyr]

They are basically the same, its just that for our problem, listing the sequences is realistic and easier

[raven hawk]

did you notice any cue in the wording that made it easier

[glossy zephyr]

The fact that there is only 9 hour intervals

If per se, it was 1000 intervals

[raven hawk]

because even though i knew the arithmetic and geom sequence, it went over my head

[glossy zephyr]

you should obviously use the continuous version

[raven hawk]

but even with 1000 intervals , would i just do n-1 ?

so 999

[glossy zephyr]

Youd have to generate all the terms and compare each other

Meanwhile, using the linear and exponential, you just have to be able to find the root(s).

[glossy zephyr]

Here b would be like a1, so 2
and a would be the slope from the point (0,2) and (8,242)

Thing is, you probably wont get an integer value (no idea if you actually do), because we are discretizing the function, you will have to find the x-value of the root, and test for the floor and ceiling of that value.

[raven hawk]

make sense

but cant i just set them as equals ad solve for n ?

2*2^n = 2+30n

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[vapid bloom]

Write and solve a system equation to answer the following questions. ```

I naturally thought through like
300g 10% + 300g 12% = 600g 11%
300/10=30
570g 10% + 30g 12% = 600g 10.1%
480g 10% + 120g 12% = 600g 10.4%

but how would you actually make this into a system of lin eqs?

[tired walrus]

call the amounts of weak and strong solutions x and y respectively

then you will have two equations based on:

• the volume of the mixture
• the total alcohol content in the mixture

[vapid bloom]

i seeeee thanks

.close

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[quartz swift]

guys, here why do we use well-ordering principle, if gcd is supposed to be the greatest of all divisors??

[trail mango]

wanting the smallest element of S and the word 'greatest' being in the theorem to prove are kinda independent

[quartz swift]

hmm, i didnt exactly get u, could u pls elaborate

[trail mango]

why do we use well-ordering principle
if gcd is supposed to be the greatest of all divisors
what do these have to do with each other?

[quartz swift]

i meant to ask , in this proof, if we are considering g, then why are we using well ordering principle that gives u the smallest element in that set, when g is supposed to be greatest common divisor

[winter lava]

S is not the set of common divisors of a and b

[quartz swift]

how??

[winter lava]

for one thing, S is infinite

[trail mango]

also i love your aesthetic

[quartz swift]

so if we go according to this theorem, shouldn't there be values smaller than g that also satisfy this condition

[quartz swift]

thanks bro : )

[winter lava]

values smaller than g that are in S?

[quartz swift]

yeah

like , d for example , wrt this definition

[winter lava]

you're saying if some number d divides a and b, then it should be in S?

[quartz swift]

hmm ye

[winter lava]

d may divide ax + by but that's not enough to put it in s

[quartz swift]

why tho?

[winter lava]

because it has to be equal to ax+by, not simply divide it

[quartz swift]

but theres no condition for x n y besides them being integers, so why is it not possible for d to be equal to ax+by for a different x n y

[winter lava]

it may or may not be

[quartz swift]

im sooo sorry for asking so many questions, im just a bit confused with it

[winter lava]

it's okay

[quartz swift]

how can we prove it?

[winter lava]

there actually is a common divisor of a and b which is in S

but not all common divisors of a and b are in S

so you should not try to prove it, because it's not a true statement (that no common divisors of a and b are in S)

[quartz swift]

huh??

[winter lava]

oh you asked how we can prove something, i thought that's what you meant

i'm not sure what you want to prove

[winter lava]

if we back up to here,
it is possible for d to be equal to ax+by

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@quartz swift Has your question been resolved?

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[quartz swift]

.close

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[candid pulsar]

hopefully translated right

do you need help with all sections?

[final saddle]

@vale onyx Has your question been resolved?

[wary siren]

does OP neeed help?

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[candid mulch]

Let ((G,) ) be a group. Assume ((H,)) is a subgroup of ((G,)), and ((J,)) is a subgroup of ((H,*)).

Suppose there are finitely many left cosets of (G) wrt (H) and (), and there are finitely many left cosets of (H) wrt (J) and (). Then there are finitely many left cosets of (G) wrt (J) and (*). Moreover, ([G:J]=[G:H][H:J]).

[soft zealot]

afi

[candid mulch]

Suppose [G/H={x_iH:1\leq i\leq m},\quad H/J={y_kJ:1\leq k\leq n}.] Let (g\in G) be arbitrary. Since (G) has partition (G/H), we must have (g=x_ih) for some (i) and some (h\in H). Then since (H) has partition (H/J), we must have (h=y_kj) for some (k) and some (j\in J). Thus, [g=x_i(y_kj)=(x_iy_k)j;] i.e., (g) is in the left coset ((x_iy_k)J) of (J) with respect to (x_iy_k). This implies [G=\bigcup_{1\leq i\leq m,1\leq k\leq n}(x_iy_k)J]

Let (f\in (x_{i_1}y_{k_1})J), so (f\in x_{i_1}H), and hence (f\in x_{i_2}H) iff (i_1=i_2). Furthermore, we can write (f=x_{i_1}y_{k_1}j) for some (j\in J). If (h\in H) is such that (f=x_{i_1}h), then (h=y_{k_1}j); i.e., (h\in y_{k_1}J), so (h\in y_{k_2}J) iff (k_1=k_2). Therefore, ((x_{i_1}y_{k_1})J=(x_{i_2}y_{k_2})J) iff (i_1=i_2) and (k_1=k_2), so each ((x_iy_k)J) through (1\leq i\leq m) and (1\leq k\leq n) are disjoint. Thus, we have the partition [G/J={(x_iy_k)J:1\leq i\leq m,1\leq k\leq n}.] It follows that [[G:J]=mn=[G:H]\cdot[H:J].]

[soft zealot]

afi

[candid mulch]

Could anyone please check my work?

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[jolly zenith]

Hello

[tranquil pine]

!da2a

[final saddle]

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.

[tranquil pine]

also, nohello

state your question

[jolly zenith]

Hello, here's my question

If I were to my differentiate e^(integral of f with respect to x)

with respect to x

what should I get?

I've been getting e^f

The answer should be f times e^f

[stone wagon]

is the integral definite or indefinite?

[jolly zenith]

indefinite

[stone wagon]

$e^{\int f(x) dx}$

[soft zealot]

artemetra

[stone wagon]

like that right

[jolly zenith]

yes

[stone wagon]

yeah it's chain rule

let F(x) be such that F'(x) = f(x)

i.e. F(x) = integral f(x) dx

[jolly zenith]

Yes, I got it

Thank you

[stone wagon]

if you don't have any more questions left, write ".close"

[jolly zenith]

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[old quarry]

can someone check my work for this

[final saddle]

@old quarry Has your question been resolved?

[stone wagon]

looks good to me

[old quarry]

okay thank you

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[crystal pawn]

here’s my ideas

i want to try the first idea but how could i get it correct

[leaden moon]

you learned variables yet?

[crystal pawn]

like x?

[leaden moon]

hold on a second

[crystal pawn]

or just any letter 😂

[leaden moon]

i think i don't need variables for this one

i can just explain this intuitively

[crystal pawn]

alr

[leaden moon]

anyways uhm

so you calculated that 20k/149 is definitely over 100

[crystal pawn]

yes

[leaden moon]

so that means that the school can definitely get the discount on the laptops

so apply discount on the prices of laptops

[crystal pawn]

yes

[leaden moon]

and divide again to get a final answer

[crystal pawn]

you also have the answers over 100 so

i assumed to

[leaden moon]

yea

[crystal pawn]

can you show it i like it visually 😭

[leaden moon]

wha?

it's common sense in real life?

[crystal pawn]

like this

[leaden moon]

oh i see i see

[abstract bramble]

the problem with these ideas is that you dont apply the discount to all of the laptops

[leaden moon]

Since the school definitely can buy at least 100 laptops, the school can choose to use all their money (with their discounts on)

We calculate the price of one laptop after applying discount: $$149(100% - 7.5%)$

After that, for the school to buy the most amount of laptop, we divide $$20000$ with the price of one laptop after discount to get the final amount

[crystal pawn]

ohh

[soft zealot]

1 divided by 0 equals Infinity
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[crystal pawn]

alr

thanks guys

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[restive sinew]

How do i solve it?

[tired walrus]

can you translate this from hindi

not clear what needs to be found

[daring lion]

If a, x, y, z, b are in an Arithmetic Progression (AP), then x + y + z = 15.
If they are in a Harmonic Progression (HP), then $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3}.$
Show that: a = 1 and b = 9.

[tired walrus]

"they" = all five numbers mentioned?

i uhhh wait hold on.

[wild venture]

Claim that a is 1 and b is 9, then take common difference to be d, then show it satisfies both equations for same value of d

[tired walrus]

the x, y and z that make the AP aren't the same x, y, z that make up the HP, are they?

[daring lion]

I think yes

[tired walrus]

what a shitty wording then

so really it should be

$a,x_1,y_1,z_1,b$ are in AP; $a, x_2, y_2, z_2, b$ are in HP; $x_1+y_1+z_1=15$ and $\frac{1}{x_2} + \frac{1}{y_2} + \frac{1}{z_2} = \frac53$. show that $a=1, b=9$

[soft zealot]

Ann

[daring lion]

I meant they are the same x,y,z

And I think I should leave it OP isn't responding

[tired walrus]

how can they possibly be the same? can you even name an example of a 5-term sequence that is both AP and HP at once??

[daring lion]

you are right

[daring lion]

yes

[wild venture]

a,a,a,a,a

[tired walrus]

and if we also ask for it not to be constant?

[wild venture]

That is a different story

[ebon agate]

I mean "if" is mentioned

[final saddle]

@restive sinew Has your question been resolved?

[muted prairie]

Find the average of a and b and the average of 1/a and 1/b

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[plain rain]

,tex
Hi. Suppose we have the non-empty set A, subset of $\mathbb{Z} $, and an equivalence relation, "R", that states: "$a,b \in A \implies \left( aRb \iff a-b = 2n, n\in\mathbb{Z} \right ) $ ". If we denote the equivalence class of element a, as $K_a $, then my textbook says that we have $$K_a = { a+2n | \ n\in\mathbb{Z} } $$ But I don't understand why that's the case. Dont rlly know what to ask exactly

[soft zealot]

fijokazż

[plain rain]

my guess is they took the equation a-b = 2n and isolated a on the left side, but i dont get it

[vital crag]

K_a is either all odd numbers or all even numbers depending on the value of a mod 2

[plain rain]

true

[vital crag]

if b satisfies aRb, then do you see why b is in K_a

[plain rain]

if a is even then b is too, and a being odd implies b is odd

so b is in the same equivalence class

or just multiply by -1 lol

i struggle understanding how to describe equivalence classes as sets in general tbh

[opal plinth]

That's pretty much it yes, what don't you get about it?

[vital crag]

what part of {a-b = 2n, n in Z} to {a + 2n : n in Z} do you not get

[plain rain]

it feels unsatisfactory to me

[plain rain]

so a-b = 2n is also describing an equivalence class?

[opal plinth]

b is in the equivalence class of a if b = a - 2n, for any n in Z

[plain rain]

that makes sense

[opal plinth]

That can be rewritten b = a + 2n of course, since n is in Z

[plain rain]

very true

thats it then?

i get it now

thank you

,close

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[velvet pivot]

Can I please get help understanding how they got L1/R1=L2/R2? I know that the whole point of all of those steps was to prove that, but I don't understand their explanation on how that was reached algebraically.

[velvet pivot]

Here is more context about the problem.

[velvet pivot]

None of the algebra here makes any sense to me, I don't get how you get the steps after each other by doing the steps like multiplying the numerator and denominator by K being equal to L2/R2, or where L1/R1=1*L1/R1 even came from.

I know what all of the individual terms mean, and I know claims 1-6 and how to get them, but I don't understand anything about claim number 7, the explanation inside of it doesn't make sense to me.

I know that any arc over its corresponding circle's radius is equal to that of any other circle's arc over its own radius because all circles are similar and their corresponding parts are equal in ratio to each other.

L1/R1=L2/R2 itself makes perfect logical sense, and I get what that means, but I don't know how that relates to the previous claims in terms of getting that equation.

And here's the Khan Academy article itself for additional context: https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-radians-intro/a/arcs-ratios-and-radians

I'm having immense trouble interpreting the reasoning behind this.

[rose brook]

average physics math

[velvet pivot]

How was the very first step l1/1=1*l1/r1 achieved? Why are we dividing L1 by its own radius?

[cosmic warren]

length of $l_1$ is $2\pi r_1\cdot\frac{\theta \text{ radians}}{2\pi \text{ radians}}$, while the length of $l_2$ is $2\pi r_2 \frac{\theta \text{ radians}}{2\pi \text{ radians}}$ Divide both by $r_1$ and $r_2$ respectively and you end up with $\theta$ in both cases. hence $\frac{l_2}{r_2} = \frac{l_1}{r_1}$

[neat meteor]

} after radians

[velvet pivot]

I don't get how the first step was achieved, I feel like it's similar to writing down 10 steps in a two-column proof on trying to prove that two triangles are similar and then on the 11th step saying something completely irrelevant but true that is the end statement saying that they're both similar. It doesn't make sense to me.

How is the length of L1 2 pi radians times theta radians over 2 pi radians?

[cosmic warren]

well, the perimeter of the entire circle is $2\pi r_1$

[soft zealot]

phoenixperson

[velvet pivot]

Yes

[cosmic warren]

but you only want the fraction subtended by the arc $\theta$

[soft zealot]

phoenixperson

[velvet pivot]

Which is equal to theta degrees over 360 degrees all times 2pi times radius

[cosmic warren]

yup. and 360 degrees is $2\pi$ radians

[soft zealot]

phoenixperson

[velvet pivot]

i haven't learned radians yet

Does knowing what this means matter?

[cosmic warren]

oh well then the only difference with degrees is just that when you divide by $r_1$, the value you're left with is $2 \pi \frac{\theta \text{ degrees}}{360 \text{ degrees}}$

[soft zealot]

phoenixperson

[velvet pivot]

I don't get why

[soft zealot]

phoenixperson

[velvet pivot]

And I don't even get why we're dividing by the radius

[velvet pivot]

Could i move on without knowing this stuff?

[cosmic warren]

cause you're multiplying $2\pi\r_1$ times the fraction the angle subtends of the arc, which is $\frac{\theta \text{ degrees}}{360 \text{ degrees}}$. Sowhen you divide that by $r_1$, what's left over is $2\pi \frac{\theta \text{ degrees}}{360 \text{ degres}}$

[velvet pivot]

I'm so dumb omg, it's just substitution!!!

This is why they're doing this:

It's because these two images are already given.

[soft zealot]

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

[velvet pivot]

Tysmmm!!

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[rain sentinel]

I'm not sure how they get the sine and cosine coefficients

[vital crag]

They just plugged y' and y" into the differential equation

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@rain sentinel Has your question been resolved?

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[rain sentinel]

.reopen

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✅ Original question: #help-36 message

[mint orbit]

they got these by matching it up to the particular piece

but, they look backwards

[rain sentinel]

isn't that what i've highlighted tho?

[mint orbit]

so you have a guess of the form of the particular part of the solution

then it has to satisfy the differential equation

so it looks like the part you highlighted is just ... im not sure

you mean in orange?

you just highlighted some terms ? I'm not sure what they mean

[rain sentinel]

i highlighted all the terms with cos in orange

[mint orbit]

hmm, no, i think you missed some

[vital crag]

If Acos(x) + Bsin(x) = Ecos(x) + Fsin(x) then A=E and B = F

[mint orbit]

this is the ultimate answer to the question

but youve made some mistakes here

I could write it out, if you want

[rain sentinel]

yes okay

[mint orbit]

okay

give me a sec

@rain sentinel

$(-64 \alpha+ 32 \beta + 3 \alpha )\sin 8x + (-64\beta + 32 \alpha + 3 \beta )\cos 8x = 1 \cdot \sin 8x + 0 \cdot \cos 8x$

[soft zealot]

jan Niku

[rain sentinel]

i think i understand now

Thanks

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[plush girder]

can someone tell me what i got wrong?

[final saddle]

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

[plush girder]

i need help on b

i used the formula for finding the volume of a sphere which is 4/3pir^3

[mint orbit]

these numbers dont make sense

lets say the reference radius is r*

so, one sphere has radius 5r*

and the other 6r*

what is the volume of the sphere with radius 5r*?

[hollow dove]

don't the constant terms cancel out in ratio

[final saddle]

@plush girder Has your question been resolved?

[hearty zephyr]

Not sure what you did but i didn't get 523.33

oh, you used 3.14, not pi.

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[soft zealot]

Renato

[tired walrus]

!status

[final saddle]

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

[gentle zephyr]

need help with b

[tired walrus]

ok, so what's your status on b then

[gentle zephyr]

no

how many numbers from 0 to 100 are congruent with 2 mod 3

2 5 8 11 14 17 20 23 26 29 32 35 38 41 44 47 50 53 56 59 62 65 68 71 74 77 80 83 86 89 92 95 98

14 is taken

[tired walrus]

ok well you listed them all out

there's 32 of them not counting the occupied value of 14

[gentle zephyr]

yes so f(4) has 32 possible

[gentle zephyr]

32 x 99 x 98 x 97 x 96?

unless I missed something

[tired walrus]

use * not x for multiplication

anyway why are there 5 factors though

f(5) is fixed

there are only 4 values that you have any choice on

[gentle zephyr]

32 choices for f4

f5 fixed so no choice

[tired walrus]

f(4) not f4.

do not write like your symbols are coming from the bargain bin!

[gentle zephyr]

f3 has 99 choices

[tired walrus]

f(3) not f3.

[gentle zephyr]

yes

[gentle zephyr]

f(2) has 98

f(1)has 97

32 x 99 x 98 x 97

[gentle zephyr]

interesting. is it me or am I getting better at counting

[tired walrus]

use * not x for multiplication

[gentle zephyr]

I think this exercise was too easy this time

[tired walrus]

you are improving, but this improvement shouldn't come at the expense of becoming sloppy in your notation again

[gentle zephyr]

I appreciate the help

and I appreciate the corrections

ty

.solved

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[deft kiln]

Can anybody help me out with this question. To find the area

[final saddle]

Please don't occupy multiple help channels.

[deft kiln]

like find the area of this.

[real geode]

do you need to find the complete area of the shape?

try breaking it into diffferent shapes and find the area of each shape and finally add them all together

like you can break the given shape into a trapezium and a semi-circle. find their areas seperately and add them to get final area

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@deft kiln Has your question been resolved?

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[pulsar axle]

how come i can put i^2 * i^2 in my calc but not i^4

[formal trail]

can you show pictures of each?

[pulsar axle]

im on my laptop

but i^4 just says error

[brisk lion]

without additional information it's hard to say.

[steady locust]

something to do with the way the calculator is programmed

maybe i^2 is preset to = -1, but they have not programed i^n in general

[pulsar axle]

so i just have to write out all the is

[brisk lion]

perhaps that model is hardcoded, yeah.

[formal trail]

what calculator is this?

[pulsar axle]

991-ES plus

[steady locust]

you should be able to simplify i^n without using a calculator

you know i^2 = -1 and i^4 = 1, so, e.g., i^7 = i^4 * i^2 * i = 1 * (-1) * i = -i

[rugged merlin]

I have the same calculator. Ill fact check this real quick lol

[steady locust]

i.e. taking n mod 4

[formal trail]

from reading the calculator manual it seems like they only allow hardcoded exponents (from the dedicated ^2, ^-1, and ^3 buttons) and not the arbitrary exponents button (probably because they didn't want to implement non-integer exponents)

[brisk lion]

would be weird if i^3 was one of the allowed ones though but not i^4.