help-39

[brisk steeple]

which part

[nocturne night]

the x is already in the denominator, so you don't have to multiply it

[brisk steeple]

Multiply the numerator by what you multiply the denominator by to ensure that there is no change in the value of the fraction.
Then use the fact that a/x - b/x = (a-b)/x

[sinful snow]

I don't understand

I'm multiplying 6y by 2x

[brisk steeple]

answer

[sinful snow]

i

The very start of the question

[brisk steeple]

my bad

[brisk steeple]

you can also do that because a/b - c/d = (ad - bc)/bd, but taking a common denominator of lesser complexity is preferable

[lusty jacinth]

We multiply 6y by just 2 because we are only trying to make the bottoms match, and the bottom already has an x, so to make the bottom match, we only need a 2. Because we are multiplying the bottom by 2, we do that to the top.

[sinful snow]

So would this apply for every question?

Say if it was 5x multiplied by 2y it would just be 2(5x)

[lusty jacinth]

5x *2 is 2(5x), but I have a feeling you mean somthing else, can you clarifiy?

[sinful snow]

I'm just saying if it was another equation would I ignore the y like I ignored the x

[lusty jacinth]

If you had something like $$\dfrac{5x}{y}+\dfrac{3}{2y}$$
then yes, you would multiply by only 2 on the top. However if you had
$$\dfrac{5x}{5}+\dfrac{3}{2y}$$
You would multiply by the full 2y because there is no y on the bottom of the left fraction.

Does this make sense?

[jolly parrot]

Morgan

[sinful snow]

Ohhhh!!!

Yes I get it now

Thank you so much I was gonna quit studying cause of this lol

[lusty jacinth]

Glad I was able to help! Good luck studying!

[pearl pond]

@sinful snow Has your question been resolved?

[pearl pond]

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[empty oriole]

I am struggling to see how he got that formula for q
(3)

[sharp vigil]

https://math.stackexchange.com/questions/141774/choice-of-q-in-baby-rudins-example-1-1

[empty oriole]

Oh, I could not find one

thanks

[empty oriole]

how do I see those answers for future reference? I only find questions related to this, but can't find the answer like in your link

.close

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[sharp vigil]

all of the answers should be available by scrolling down

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[lament storm]

This should be half life formula, im just not sure how to structure it or break it down?

[bitter herald]

Do you know whats the half life formula

[lament storm]

Its like A0(1/2)^t/h

[bitter herald]

Yea

So lets plug in what we know

'...has lost 25% of its carbon14'

This statement relates to the amount at time t

Meaning,
[
A(t) = A_0\8{\412}^{\5th}
]

[bitter herald]

How do you modify A(t) to describe this statement

[lament storm]

OH

[lament storm]

thats what im not sure about

[bitter herald]

So like

We know it starts with A_0 right

But then, that A_0 loses 25% of its value

What remains after it loses that much?

[lament storm]

75%?

[bitter herald]

Yeah

75% of A_0

[lament storm]

oh is that the answer?

[bitter herald]

no

Im saying A(t) is gonna be 75% of A_0

[lament storm]

OH

[bitter herald]

Now do you know what h is, based on your problem?

[lament storm]

ohh ok

[lament storm]

5750

[bitter herald]

Yeah

Can you write your equation now with those values

Ok I gotta go now. Hopefully another helper picks up the slack

[lament storm]

thank you!

[hollow barn]

For this, the simplest way to think about it (since there's no concrete values) is to take A(t) = 0.75 and A_0 = 1

Since this would mean we start with 100% and are now at 75%

After some time t

[pearl pond]

@lament storm Has your question been resolved?

[lament storm]

Oh the A_0 also has a value i should plug into it?

[hollow barn]

Well you aren't given any specific starting values, so we can assume the starting amount is 1 unit

Since it correlates to 100% starting

And then to have 75% left means we have 0.75

You could also start with 100 and then A(t) = 75

Just needs to be consistent

[lament storm]

so A(0.75) = A_1 (1/2) ^0.75/5750

?

is that correct

[hollow barn]

No no A(t) = 0.75

We don't know what t is yet it's what we want to find

[lament storm]

OHHH

[hollow barn]

Your setup should be $0.75 = (1)\bigg(\frac{1}{2}\bigg)^{\frac{t}{5750}}$

[jolly parrot]

TestTickler

[hollow barn]

Then you should be able to solve for t

[pearl pond]

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[sharp galleon]

Yo guys

[sharp galleon]

Whats the difference between invariant lines and invariant points in matricies

With invariant lines i js sub in mx right for y

[pearl pond]

@sharp galleon Has your question been resolved?

[pearl pond]

@sharp galleon Has your question been resolved?

[sharp galleon]

,close

.close

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[north talon]

Determine all finite, nonempty sets of positive integers $S$ such that for any elements $a,b$ of $S$, there exists some $c \in S$ such that $a | b + 2c$

[jolly parrot]

Copter

[north talon]

clearly if S = {a} works and if |S| = 2 then the only set that works is S = {n,3n}. I need help on the case where S >=3?

i dunno if thats possible

[pearl pond]

@north talon Has your question been resolved?

[north talon]

<@&286206848099549185>

[pearl pond]

@north talon Has your question been resolved?

[sharp smelt]

$a\mid(b+2c)$?

[jolly parrot]

what a wonderful world(wai)

[north talon]

yea

[sharp smelt]

Unless I'm tripping , this implies $a\mid b$ and $a\mid 2c$ no?

[jolly parrot]

what a wonderful world(wai)

[sharp smelt]

so $a=2$ , but we discard this as. then the set is infinite

[jolly parrot]

what a wonderful world(wai)

[sharp smelt]

uh @north talon you still here?

[north talon]

no

[sharp smelt]

oops

right

[north talon]

2 | 1 + 3, but doesnt divide either

[north talon]

my work for |S| = 2, idk otherwise

[north talon]

can anyone help?

[cursive wraith]

how did you go from b | 3a to the 3-adic valuation of b bigger or equal to nu_3(a) + 1

[pearl pond]

@north talon Has your question been resolved?

[pearl pond]

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[quick venture]

i dont understand where they got n from

[fleet ridge]

Its due to the periodicity of sin and cos

[quick venture]

so how do i actually solve cos(sin(x))=0

[fleet ridge]

Well if cosu=0, how would you solve that?

[quick venture]

u=cos^-1(0)

[fleet ridge]

Yes. True. But what would be the values of u?

[quick venture]

1/2pi

[fleet ridge]

Thats correct. But notice how in the full graph of cosx, there are many more points

So how would you collectively represent all the points?

U there @quick venture

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[thorny radish]

you should consider the largest two elements and do some cases of what class mod 4 they are

[thorny radish]

i don't know why this works but it does

well first i should clarify that for bookkeeping since all elements have the same ν₂ we can assume they are all odd

.close

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[fading nexus]

For e how will I find the greatest possible width and length?

[fleet ridge]

Assuming you know the value of x, you can just subtract 2x from the original rectangles dimensions

[fading nexus]

Oh I was over thinking it ty

[pearl pond]

@fading nexus Has your question been resolved?

[fading nexus]

For the life of me I can not figure out the window for the volume equation to graph even when I put the one my teacher put it does not work. I tried multiple version of the equations just changing the () to see if that is where I went wrong but it’s the same no matter what u do

[fleet ridge]

Im not getting what your saying. Could you maybe elaborate it a bit?

[fading nexus]

So I’m trying to put this volume equation in my calculator to find the maximum so it should be like a hill shape

but it’s just a straight line

No matter what window I ajust to

[fleet ridge]

1 minute I'll try desmos

This is probably why

@fading nexus the maximum is so abysmal high. Thats why it looks like a straight line in your calculator

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@fading nexus Has your question been resolved?

[pearl pond]

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[barren arrow]

can someone help me understand modulo arithmetic for competitive programming?

[barren arrow]

i dont understand the inverse thing

modulo inverse

i dont know if this is the correct channel to ask this question or now

not*

[plush bramble]

can you just show an example or problem of modulo inverse

[barren arrow]

should i send you the problem statement? or the formula i am using

[flint basalt]

if you have a specific problem that you were working through then send that

[barren arrow]

https://codeforces.com/problemset/problem/1731/B

[flint basalt]

send whatever you have regarding that problem

[barren arrow]

i am working with this problem

[flint basalt]

and please send a screenshot instead of a link or a pdf

[barren arrow]

so the formula i derived for this question is n * (n + 1) * (4 * n - 1) / 6

[plush bramble]

and the formula you were taught

[barren arrow]

but i dont understand the modulo part in this question

[plush bramble]

example

so in your problem instead of the 12 in the example, you have a much bigger number

[barren arrow]

yeah

1e7

[barren arrow]

after that??

[plush bramble]

uh no

[valid nova]

.pin

[plush bramble]

10^9 + 7 is not 1e7

[valid nova]

oh i am not a helpful 😭

[barren arrow]

1e9 + 7

sorry

[flint basalt]

.pin

[plush bramble]

so do you uderstand this then

[barren arrow]

yeah

is that it?

[plush bramble]

alright i don't understand what you're asking for then since you said you don't understand the modulo part.

[barren arrow]

with these test cases i am getting the wrong answer

for the last one

[plush bramble]

you haven't shown your work for the formula so no way telling it's right or wrong

[flint basalt]

i think that's because it overflows the long type

assuming this is java, long has a maximum limit of 2^63 - 1

[flint basalt]

,rccw

[barren arrow]

yeah, and when i multiply it overflows

i know that we have to use some modular properties in this question to prevent overflow but i dont seem to understand which one

read an article as well

didnt get what this means

[plush bramble]

you should learn some of these then

from https://en.wikipedia.org/wiki/Modular_arithmetic

[flint basalt]

hmmm

so it looks like you should multiply by the mod inverse

dividing by 6 should be equivalent to multiplying by 166666668L % MOD

[barren arrow]

how did you get this?

[barren arrow]

okay

[lusty jacinth]

<@&268886789983436800>

[barren arrow]

😭

[flint basalt]

so we need a number $x$ such that $6x \equiv 1 (\mod 10^9 + 7)$

[jolly parrot]

Krish

[flint basalt]

<@&268886789983436800>

[barren arrow]

okay

how to solve this?

x is the modulo inverse, right?

[flint basalt]

so basically it's saying that 6x and 1 give the same remainder when divided by 1e9 + 7

[plush bramble]

last sentence here

links to https://en.wikipedia.org/wiki/Modular_multiplicative_inverse#Extended_Euclidean_algorithm
and
https://en.wikipedia.org/wiki/Bézout's_identity

[flint basalt]

so you can set it up like $6x = 1 + k(10^9 + 7)$

[jolly parrot]

Krish

[barren arrow]

i will look up these articles, thank you so much

[barren arrow]

should i read the wikipedia article or are you explaining something?

[barren arrow]

okay

[flint basalt]

go ahead and read the articles reimann sent, it'll probably explain it better than i can

[barren arrow]

okay, thank you so much for the help!

[plush bramble]

you can probably find the algorithms that implement them elsewhere. wiki just explains what the code is doing and where it comes from

[pearl pond]

@barren arrow Has your question been resolved?

[pearl pond]

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[barren arrow]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[barren arrow]

i have read the wikipedia articles, but can you also continue what you were explaining

because tbh i understood very little

[flint basalt]

sure

[flint basalt]

does this part make sense so far?

[barren arrow]

yeah

[flint basalt]

if you set x equal to that number 166666668, then for k = 1 you get:

$6(166666668) \equiv 1000000008$

[jolly parrot]

Krish

[flint basalt]

which then gets you to $1000000008 \equiv 1$ when simplifying, since you have $1000000008 = 1 + 1(1000000007)$

[jolly parrot]

Krish

[flint basalt]

so that 166...8 is the mod inverse of 6

and if you're wondering why we chose k = 1, it would give the smallest positive solution which therefore would give the smallest positive mod inverse

for k, only values where k equiv 1 would work, so 1, 7, 13, etc

so you theoretically could choose k = 7 but it would just be a larger number

[barren arrow]

okay i understand this but how do we calculate the value of 166666668 in programming? is it (1 + 1e9 + 7) / a?

[flint basalt]

yeah i think you can do that

[barren arrow]

thank youu so much 🥹 you explained well

also is there anything else that would help me solve problems like these in future?

[pearl pond]

@barren arrow Has your question been resolved?

[flint basalt]

uhh i mean it's different for every case but i definitely recommend looking through those articles and fully understanding it or watching videos on it because it's definitely a tough topic

[barren arrow]

i looked it up, and i think i understand what is needed to solve a problem by now.

[barren arrow]

are you into competitive programming as well btw?

[flint basalt]

not much really, but i am going for a degree in math & comp sci right now

my moms worked as a software engineer for decades so i've learned a lot of java programming from her

[barren arrow]

woww, thats great!

[flint basalt]

you as well keep it up! you're doing great right now

[pearl pond]

@barren arrow Has your question been resolved?

[pearl pond]

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[woven river]

they really need to specify "harder" huh

[woven river]

current progress

how am I supposed to get f'(1/2) in the end

I will continue work on this, please ping to call me back, thanks

[plush bramble]

You can pick c to be 1/2 in Taylor's expansion

And x = 0 and 1 as well

[woven river]

a

[woven river]

wait x can be fixed too?

[summer kernel]

the Rolle theorem is not needed

[woven river]

ok, I just wrote what I thought at that time

ok ok

[summer kernel]

x can be fixed as long as x ∈[0,1]

taylor expansion holds for any point x

[woven river]

I thought taylor theorem said "for every x"

update

I can sense im getting closer here

I can see A/4

I can see f'(1/2)

[summer kernel]

let x = 0 and x = 1

you'd better write the remainder

[woven river]

isn't the remainder the one with f"(c)

[summer kernel]

ok, i miss that

c is between 1/2 and x

x can be smaller than 1/2

[woven river]

yea but u lost me at picking a specific point for x

I thought we are supposed to generalize it by saying for every x

[summer kernel]

for every x (in this problem, x ∈[0,1] by the domain of f), the taylor expansion holds, so you can pick x=0 or x=1

[woven river]

hmm I see

[summer kernel]

problem 6 gives two special points f(0)=f(1)=0, so just let x=0 and x=1 and find their relationship

[woven river]

I found this

seems I made a mistake somewhere, can u help check if 1 and 2 are correct already?

they are both derived from taylor expansion

[summer kernel]

wait

ah, you should bound |f"(x) | ≤ A later

[woven river]

wdym later

why

is there smth wrong

bro I just realized

what u mean by doing it later

soo annoying

[summer kernel]

if you bound by A first , ①-③ A/4 disappears

[woven river]

then we will use triangle inequality

[woven river]

yea...

[summer kernel]

that's right

[summer kernel]

yeah

[woven river]

I will be damned in the exam

[summer kernel]

haha, good luck

[woven river]

I think the main point to note is to pay attention to every detail on the ending

like there's a reason the absolute sign is there

great, just great

alr thanks so much vro @summer kernel , u have been helping alot, I will take a break now, cya

.solved

[pearl pond]

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[swift spindle]

Is this the right track?

[stuck berry]

yes

assuming you did partial fractions correctly

[pearl pond]

@swift spindle Has your question been resolved?

[broken moth]

$A=\frac{6}{6+1}$
and
$B=\frac{-1}{-1/2-3}$

[jolly parrot]

Roufaid

[broken moth]

not what you did

you should have used heaviside correctly

[stuck berry]

I'm pretty sure his partial fractions is correct after calculating it myself

[swift spindle]

uhh

so

hm

yea it was ok i got it thx

.close

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[ornate vine]

i have the function
y = 2x + sqrt 1-x
I'm trying to find the function cutting points with the X axis
I inserted 0 into the 0 and got this function:

0 = 4x^2 + 1 - x

now I did the 5 3 in the calcuator and got 2 points, which one is correct?(the answer page says only 1 is correct which is why im asking.)

[sharp sleet]

Hi

can i ask my doubts here.??

[ornate vine]

yeah

i mean

#help-14

is free

[sharp sleet]

i need help regarding jacobian matrix

[spare lark]

Open a new one

[ornate vine]

not here then

yeah open a new one

[sharp sleet]

open.?? what a channel.??

[ornate vine]

send your message in this channel please #help-14

[spare lark]

Yep right above the occupied help channel section there are free ones

[crystal dew]

yes. please read #❓how-to-get-help

you were just a little too late in claiming the channel

[ornate vine]

feel free to help me guys

[sweet junco]

consider extraneous solutions

[smoky musk]

You plug them back?

[spare lark]

How did you get that after y = 0 ?

[ornate vine]

squared them

[ornate vine]

gotta use google translate 1m

[sweet junco]

I assume 2x=-sqrt(1-x) and oh they got it wrong

[sharp sleet]

thanks ive sent it

[sweet junco]

2x+sqrt(1-x) squared is NOT equal to 4x^2+1-x

[ornate vine]

??

[sweet junco]

what you can do is
2x+sqrt(1-x)=0 -> 2x=-sqrt(1-x) -> 4x^2=1-x -> 4x^2+x-1=0

[sweet junco]

(a+b)^2=a^2+b^2+2ab, you're missing 4xsqrt(1-x)

[ornate vine]

nvm

I see where I got it wrong

thank you

you helped alot

.close

[pearl pond]

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[neat scroll]

This is the diagram I drew .

[wicked edge]

,rcw

[wicked edge]

What are you trying to find?

[neat scroll]

To prove

[wicked edge]

Prove what?

[neat scroll]

Wait

Questions didn't come

29th questions

[neat scroll]

Mentioned in pic

[wicked edge]

,rcw

[neat scroll]

I think u didn't understand how to solve u want me to share a pic of solution?

[wraith jacinth]

-# what have u tried to do so far?

[wicked edge]

Find cosine of A.

[wicked edge]

No, I am good.

[neat scroll]

Ntg am stuck

[wicked edge]

Given AB and AC, you can use which trigonometric function to solve?

[neat scroll]

1/2 so a is 30?

[wicked edge]

Correct.

$\sin{\theta} = \frac{1}{2}$

[jolly parrot]

Mercury (ヤフォダ)

[neat scroll]

Cos

[wicked edge]

Which angle A is?

[neat scroll]

Oh yes

[neat scroll]

Acute

[wicked edge]

Now just use the definition of angle sum.

[wicked edge]

No, the exact degree.

[flat falcon]

hi

How old r u guys

[verbal whale]

!occupied

[pearl pond]

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).

[wicked edge]

Not important.

[neat scroll]

90 + 30 + x = 180
So x is 60

[verbal whale]

!redirect

[pearl pond]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[wicked edge]

Alberto, you are abusing factoids.

[light helm]

if by a you mean A in your diagram
that isn't 30°

[wraith jacinth]

yup!

[neat scroll]

Thanks k understood

[wicked edge]

Total angle in a triangle is?

Then just work out the last angle, Q.E.D.

[wraith jacinth]

-# also for future refrence it might help to know that this is one of the most common right angle traingles-

[flat falcon]

can i help

?

[neat scroll]

It is

[neat scroll]

180

[neat scroll]

Wt is qed ?

[smoky musk]

Quod Erat Demonstrandum

i.e. the proof is finished

[neat scroll]

Hey I have another method . Can u explain that ?

[vestal tapir]

it's like lmao

[light helm]

that angle isn't 30°

[vestal tapir]

joy and relief

[neat scroll]

How ?

[wraith jacinth]

i assume pythagerom therom and then u find each aaangle respectively?

[neat scroll]

But y is that angle a n 30 ?

[light helm]

just for clarity, i'm saying that isn't 30°

[wraith jacinth]

well because sin^-1 (1/2) = 30

[wraith jacinth]

as mercury said earlier

[neat scroll]

If we take cos a ?

[light helm]

cos(what) = 1/2

[neat scroll]

Yes

[light helm]

that wasn't a yes/no question

i'm asking you cos(what angle?) gives 1/2

[wraith jacinth]

look at angle C

imagne doing sin(c) what should angle C be as the outcome?

[wicked edge]

You are doing Cos(A) = x/2x.

Then Cos(A) = 1/2.

By solving, you said A = 60. Not 30.

30 is C.

And in that case, you use sin.

Which yields sin(C) = x/2x.

Follow by Pluto and Omeganto's statements, it is important to not mix up these angles, because it can cause horrendous outcome in your proof (No Q.E.D.).

[wraith jacinth]

-# also small tip u can alway check ur solution by for example using both sin and cos if u know that cos^-1(1/2) = 60 then you know that sin^-1(1/2) should give u 30 and u can check like so and vice versa-

[wicked edge]

Great choice.

[neat scroll]

It will be x/2x
1/2

Yes

So wt should my next step be ?

A

Hey

I was saying th method

[light helm]

missing ^-1, or swapped those values around

[wraith jacinth]

-# ah oop-

[neat scroll]

Can anyone start fro starting I got con6

[light helm]

fro?
con6?

[neat scroll]

Wt?

[wraith jacinth]

u mean repeat from the start?

[neat scroll]

Yes

[neat scroll]

From

Confused

[wraith jacinth]

okay so lets starts from the start simplfied-

[neat scroll]

Yes

[wraith jacinth]

so we have this diagram here correct?

[neat scroll]

No

X and 2x is swapped

[wraith jacinth]

apologies ur right -

okayyy there we go-

[neat scroll]

Yes

[wraith jacinth]

okay so first things off we need to define our angles-

since the question tells us to prove that BAC = 2ACB

so we need to find each angle in this traingle-

well angle ABC is pretty east thats just 90 right?

[neat scroll]

U swapped 2

[neat scroll]

Yep

[wraith jacinth]

okay so if we find one angle we can find the other correct?

[neat scroll]

Understood till here

[wraith jacinth]

choose an angle!

[neat scroll]

Yes

[neat scroll]

A

[wraith jacinth]

okay soo what sides DO we have in comparsion to angle A

[neat scroll]

Ab and ac

[wraith jacinth]

okay if we have AB and AC which trigemetric function can we use??

[neat scroll]

Cos ?

[wraith jacinth]

YUP EXACTLY!!

[neat scroll]

Cos a =1/2?

[wraith jacinth]

YES!!

[neat scroll]

A =60?

[wraith jacinth]

THATS CORRECT WELL DONE!!

[neat scroll]

C = 30 ?

[wraith jacinth]

YUP!! buttttt- lets make sure.. to make sure we are 100% correct!!

in comparsion to C whatsides do we have?

[neat scroll]

Can u explain this method pls

[wraith jacinth]

,rcw

[neat scroll]

Bc and ac

[wraith jacinth]

so which function would we use?

[neat scroll]

But we dk bc

[wraith jacinth]

but we know AB right?

[neat scroll]

Cos

[wraith jacinth]

that would be true if we have BC

[neat scroll]

Oh yes so we will take sin

OK I got this method

[wraith jacinth]

yup does it makes sense now- anyway this is ONE method to solve it!! i can also explain the method you sent

[neat scroll]

Yes pls

[wraith jacinth]

so consider line AC

[neat scroll]

U r starting from. Start ?

[neat scroll]

Yes

[wraith jacinth]

yes for the other method-

[wraith jacinth]

lets put a point in the midpoint of line AC and call it D

[neat scroll]

Ok

[neat scroll]

Ok

[wraith jacinth]

now draw line DB

[neat scroll]

Ok

[neat scroll]

Got It

[wraith jacinth]

now we know AD = AB correct?

[neat scroll]

Yes

[wraith jacinth]

now since angle A is 60 degrres and traingle ADB is issocles that means that means that all angles are 60

do that means traingle ADB is equalateral-

[neat scroll]

How do we know a is 60?

[wraith jacinth]

so ADB becomes equlateral and BDC becomes isosles

[wraith jacinth]

1 sec- i am trying to use the same proof u sent and i cant read well

[neat scroll]

Ok

[wraith jacinth]

apperently its using a semi circle-?

[neat scroll]

Yesss

[wraith jacinth]

okay so u draw circle with center D and we use X as the radius

now because X is the radius u will hit point A and C on the circle

OHH I SEE-

OKAY OAKY i get it now-

[neat scroll]

Yyy will we do that ?

Reason?

[wraith jacinth]

you will see in a second-

[neat scroll]

Ok

[wraith jacinth]

okay so look at this circle

the radius here = X right?

[neat scroll]

Yes

[wraith jacinth]

(ignroe the fact ttat i drew it a little badly)

[neat scroll]

Yes

[wraith jacinth]

now notice how ANYTHING u draw on it IF it has an angle of 90 it will be a point on the circle?

[neat scroll]

How ?

[wraith jacinth]

U can imagne moving this extra points around and It will always be 90 degrees

[wraith jacinth]

because its a right angle traingle where the hyptonuse is the Diamater-

so both sides add up to 2x

and remeber how the equation of the circle is basssicly

x^2 + y^2

pythagerom therom is bassicly

c^2 (diaamter = x^2 + y^2

bassicly it uses the fact that ANY point on the circle connected tothe DIAMATER will always be 90 degrees-

[neat scroll]

Bassicle?

[wraith jacinth]

basically*

[wraith jacinth]

this is called Thales's Theorem i belive

[neat scroll]

Connected to diameter or edges of diameter only ?

[wraith jacinth]

now its connected to the poitns of the diamater and ANY point on the circle-

[neat scroll]

It applied always ?

Applies *

[neat scroll]

Can u help me connect this to our q ?

[wraith jacinth]

oayk i will show u how this is applied in our Q

[wraith jacinth]

we use this FACT to prove that DB = x

because reember if the circle we have touches point A it means the radius = x so it toch B too means DA = X

and from that fact we knwo that ADB is an equliateral traingle-

does that make sense?

[neat scroll]

Okay

[neat scroll]

Can u once fraw circle In our tri diagram

[wraith jacinth]

notice how ABC is always 90 degrees?

[wraith jacinth]

sure!

[neat scroll]

Ohhhh

Yesss

I visulaizwd It thanks

[wraith jacinth]

DO YOU SEE IT-

SO BECAUSEEEEEEE ABC is 90 DEGREES-

WE CAN CONCLUDE THAT DB = X

does that make sensee????!!!

[neat scroll]

How ?

[wraith jacinth]

REMEBERRR D is the center and X is the radius!!

[neat scroll]

We knew this before circle itself right?

[wraith jacinth]

yes yes!

[neat scroll]

Ohh yaaa

[wicked edge]

Why angle B is written as "b" here?

[neat scroll]

I got it now

[wraith jacinth]

-# becasue i am dumb

[neat scroll]

Same same bro

[wraith jacinth]

1 sec lemem fix it-

[neat scroll]

No u r nt u r genius

[wraith jacinth]

and from there u prove that ADB is equalateran and that angle A = 60

and the rest is history-

alsthough using sin and cos is so much easeir 😭

here^

[neat scroll]

But how will i prove agle c is 30?

Oh leave I got ittt

[wraith jacinth]

well traingle ABC all angles must add up to 180-

[neat scroll]

Thanks

[wraith jacinth]

yeah you get it!

ALRIGHT GOODLUCKKK! CYA!!

-# dont forget to .close when ur done

[neat scroll]

Cya ?

[neat scroll]

Next time can I dm u for solutions ? U explained so easily?

[wicked edge]

Using the properties of the perpendicular bisector and by the definition of circle here is alright, but isn't is quite tedious?

[wraith jacinth]

welllll u can ping me here and if i am online i will respond!! its better to ask here because i can make mistakes and other helpers bieng there to help when i do one its good also if i ever have to leave or i am busy someone else can respond!! but feel free to ping me and i will try to come if i can!!

[wraith jacinth]

the only reason i explained it tha why is cause OP asked to use the proof they had^ but iagree its quite tedious

[wraith jacinth]

^ @wicked edge

[wicked edge]

Yeah, OP should stick to sine and cosine, although it is correct, but a shorter way is much better.

[neat scroll]

Sorry for the trouble . I jst wanted clarity from all method so

[wraith jacinth]

no problemm glad i could help!!

[wicked edge]

Great help from Pluto anyways.

[neat scroll]

Yes I will but I wanted to learn new concepts so

[wraith jacinth]

thank youu you too!

[wicked edge]

Yeah that one is good, but the circle is no need here right, a bit redundant.

[sharp delta]

What was the question

[wicked edge]

(In my opinion), since you can just define the midpoint on hypotenuse, then reasoning that by the definition of the perpendicular bisector in right triangle, ->.

[neat scroll]

Question 30

I drew this diagram

[wraith jacinth]

,rcw

[neat scroll]

@wraith jacinth

[wraith jacinth]

hi hi i am here!!

[neat scroll]

Yes

[sharp delta]

What is the angle of other part of B

[neat scroll]

I am solving 30th now

[neat scroll]

It's x

[wraith jacinth]

okay so first things off it tells us that B = C which is equa to 2A

from that fact alone we can find each angle in the traingle

because well they all add up to 180

[neat scroll]

Yes

[sharp delta]

Use trigonometry I suggest

[neat scroll]

Oh ta

5x =180
X =36?

How does this helps as it's nt standard angle

[wraith jacinth]

okay lemme see-

[wraith jacinth]

i belvie that is true-

so knowing that we know which angle = 36?

and which angles are equal to 2(36)?

[neat scroll]

B and c

[wraith jacinth]

so b and c are = 72 correct?

[neat scroll]

Yes

[neat scroll]

Y3s

[wraith jacinth]

okay so we know that AC = AB correct?

actally there is 2 methods to prove this

[neat scroll]

Yes

[wraith jacinth]

(we can litterly prove this by angle if you would like)

[delicate temple]

Hloo guys

[wraith jacinth]

hi hi!!

[delicate temple]

Whats the problem?

[wraith jacinth]

^

[wraith jacinth]

-# i am trying to kesar to solve it ;3

[neat scroll]

Yes hiwn?

[wraith jacinth]

well can u find angle ADC?

[delicate temple]

O

[neat scroll]

180

[wraith jacinth]

okay now find angle BDC

also wait sorry-

i wrote slight mistake

tis 36 NOT 32

[neat scroll]

Yes it's 36

[wraith jacinth]

okay so knowing this try to find andlge BDC

the one in blue!

[neat scroll]

72

[wraith jacinth]

yup so we know that traingle ADC is isosles traingle correct?

[neat scroll]

Yes

[wraith jacinth]

now try to find angle ADB

[neat scroll]

108

[wraith jacinth]

okay so knowing all of the angles can u think how we can prove that AD = BC?

-# hint ||think of isosles traingles||

[neat scroll]

Ohh with hint I wad able tooo

Ad =bd
Bd =bc
Ad =bc

So thansk

Byeee

[wraith jacinth]

YUP well DONEEE!!!

[neat scroll]

Thanks uuuu

[wraith jacinth]

BYEEEE

[neat scroll]

Loves u 💓

Byeeee

[wraith jacinth]

-# dont forget to .close!!

[neat scroll]

.close

[pearl pond]

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[pearl pond]

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[quick venture]

hi for this integral

[timber cape]

hmmm

[quick venture]

why isnt it what i've written in red?

[torn flicker]

um

lcm should be (x + 1)^2 (x -1 ) right?

[rose lark]

because u can see that the bottom part already has x+1 squarednterm

[torn flicker]

so you just need to multiply by x - 1

[fleet ridge]

When you multiply everythign with "(x + 1)^2 (x -1 )", the "(x + 1)^2" gets cancelled

[timber cape]

because the deno has (x+1)^2 not (x+1)

[past perch]

the fractions when reformed need to have the same LCM (or denominator) as the LHS.

the B term already has (x+1)^2, so to match the lhs it only needs an (x-1)

to contrast, the A term has only (x+1) thus it needs one (x+1) AND one (x-1) to give the required LCM

remember that when you're taking an LCM you're not just multiplying the numerator, but also the denominator

1/2 + 4/5 becomes (5+8)/10 because the 1/2 becomes 1*5/2*5=5/10 (same for the other fraction and then addition is possible) and thus stays the same overall ratio

if you multiply the B term like you did in red you're basically doing numerator times (x-1)(x+1) AND denominator too, which becomes redundant

[quick venture]

excellent explanation thanks

[timber cape]

-# sam beam for a reason

[pearl pond]

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

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Please don't occupy multiple help channels.

[errant solstice]

<@&268886789983436800>

[sterile python]

<@&268886789983436800>

[pearl pond]

Channel closed

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[sterile python]

Fast one Chiaki

[timber cape]

damn

[errant solstice]

speed doesn't matter when... never mind.

[sterile python]

You showed me your speed

[errant solstice]

I get sniped all the time. speed matters not.

[fading cove]

yo guys can u helm me with thi s math question

[wraith jacinth]

#❓how-to-get-help

[sterile python]

Sniped?

[errant solstice]

in help channels.

[wraith jacinth]

-# this channel is already occupied open another channel

[sterile python]

I do not understand that term... :((

[errant solstice]

not even getting sniped sometimes. just straight up ramboed into a channel I'm helping.

but never mind. not a rule anyway. and as long as helpees get the help they need, I suppose.

[prime bramble]

[cinder flower]

[sterile python]

Well, what a pity, but good luck for next time

[light helm]

This is not the channel you're looking for.
Move along.

[errant solstice]

sorry, got a little carried away.

[sterile python]

Omeganto

[pearl pond]

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[steady dagger]

So basically i was working on application of shm using complex numbers
Yk
X= xmax* e^itheta
I was required to find the angle,
Which i got by equating both sin and cos imaginary part with sin and real part with cos
Cos inv gave 17.58degrees
Sin inv gave -17.58 degree
Tan of both RESPECTIVE angles
Gave -17.58
Which one to write? Im confused
P.s attaching the yt video ss i got here, in which the teacher gets 17.58 if u calculate sin^inv....
HELP

[steady dagger]

The second picture is ss from yt, i was looking for solution but i found that the value is same in magnitude but my -tive when I take tan

Which one to

Write can anyone help

[feral olive]

Do you know the unit circle

[steady dagger]

Yes

[feral olive]

Pretty much the absolute value of the angles you are getting is the reference angle

You need to find the quadrant of the unit circle where cos is positive, sin is negative and tan is negative

[steady dagger]

I get it the reference angle is always positive but im unsure how to state that In exam, like what do i write?
I know that angle must be in 4th quadrant

[feral olive]

Idk how it works for u but in this situation i just draw the circle with the CAST rule in the quadrants and circle which quadrant corresponds with the results

[steady dagger]

Or instead of taking tan i could like completely neglect imaginary sin and go with cos inverse only to find the angle

[terse blaze]

hi

[steady dagger]

Wassup

[terse blaze]

nothing much

[feral olive]

Why? Cos doesnt narrow it down theres still two quadrants that are positive and negative

Or are you saying to take the real value only since its SHM

[steady dagger]

Im confused what to do in the situation book has answer by only taking cos which is positive 17.58 and i use arg formula tan inv ( y/x)
So should I always do like
Mod of tan inv or smthing in exams to get positive angle or what

[steady dagger]

Yes

[steady dagger]

Cuz how would I know if angle will be positive or negative

[feral olive]

I guess you would only use cos since that marks the position of the object

[steady dagger]

Aha so no need to use arg formula

Why is ts so hard

[pearl pond]

@steady dagger Has your question been resolved?

[pearl pond]

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[acoustic tangle]

For part a I tried using this method

17/4 =4.25

Round up to 5

So the fifth value which is 63

But the markscheme is 61

[pearl pond]

@acoustic tangle Has your question been resolved?

[dense jasper]

Seems like they're using the $\frac{n+1}{4}$ convention instead. Not really a more satisfying answer than that.

[jolly parrot]

Civil Service Pigeon

[acoustic tangle]

Oh okk I’ll keep that in mind

[dense jasper]

[pearl pond]

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[pearl pond]

@acoustic tangle Has your question been resolved?

[pearl pond]

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[remote saffron]

What should be the thinking process for solving this type

[brisk steeple]

,rotate

[brisk steeple]

(3) is easy. You can just input the values.

[fleet ridge]

I think because there is a lot of sin^2 and cos^2, you'll look to use the identity sin^2x+cos^2x=1

[remote saffron]

Should I covert into cos² using sin²x= cos²(90-x)

[fleet ridge]

For which question?

[remote saffron]

3

[jolly parrot]

Ultimator

[wraith hinge]

same for cos

[fleet ridge]

This is probably the best way

[brisk steeple]

Interestingly, the value of (4) is 2

Just observe that pi/8 = pi/2 - 3pi/8
So cos^2 (pi/8) + cos^2 (3pi/8) = cos^2 (pi/2 - 3pi/8) + cos^2 (3pi/8) = sin^2 (3pi/8) + cos^2 (3pi/8) = 1

[remote saffron]

I see

But what about the last 2 terms?

[brisk steeple]

cos^2 (pi/8) = cos^2 (7pi/8) and cos^2 (3pi/8) = cos^2 (5pi/8)

[remote saffron]

How

[fleet ridge]

Using this

[brisk steeple]

Use the fact that cos(pi-x) = -cos(x)
Now square both sides

[pearl pond]

@remote saffron Has your question been resolved?

[pearl pond]

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[pearl pond]

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[grand mountain]

@dense jasper

[grand mountain]

I just wanna make sure this is the correct answer for SVD right

[plush bramble]

!noping

[pearl pond]

Please do not ping individual helpers unprompted.

[grand mountain]

My bad

[wicked edge]

What is the original question?

[grand mountain]

Find SVD

https://youtu.be/wHAJzemKQW4?si=0MQRx0M-SEUgLXoj

[pearl pond]

@grand mountain Has your question been resolved?

[pearl pond]

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[inland laurel]

How to solve this paragraph

Let's solve this together

[honest oyster]

Did you make any progress

[inland laurel]

yeah

i am sending it in a min

we want a smart approach this is not a smart one

[honest oyster]

I think it's better if we first find out which value f(x) will output for any given input

So comparing x-[x] and -x-[-x]

[tame prism]

What is the question??

[honest oyster]

10 and 11

[tame prism]

What is the question

[inland laurel]

^question

[tame prism]

So basically what is the trouble u are faciny

[inland laurel]

like how to take the first correct apporach

[tame prism]

So basically I think creating 2 cases is the approach x is -ve and x is +ve

[pearl pond]

@inland laurel Has your question been resolved?

[cyan heron]

Are the answers B and A?
If yes, i recommend drawing the graphs of these functions as a hint
If not, ive probably done something wrong

[pearl pond]

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[sharp smelt]

I'd like to solve this using the trapezoidal rule first

so $\frac{1}{2} (f(0)+f(1))$?

[jolly parrot]

what a wonderful world(wai)

[sharp smelt]

so $\frac{1}{12}\left(\frac{1}{12}+ \frac{1}{13}\right)$

[jolly parrot]

what a wonderful world(wai)

[sharp smelt]

As for simpson's rule

I'll have to consider the lagrange polys of the 2nd, 4th and 8 degrees?

[smoky gull]

-# all those names bro 😭

[sharp smelt]

this is for p=0

uh

$\int_{0}^{\frac{1}{2}} \frac{ (x-0.5)(x-1)}{0.5} \cdot \frac{1}{12} + \frac{(x)(x-1)}{-0.25} \frac{1}{12.25} + \frac{(x)(x-0.5)}{0.5} \frac{1}{13} dx$

[jolly parrot]

what a wonderful world(wai)

[sharp smelt]

as obtained by interpolating the function using lagrange interpolation

I'll do the one for 5 point as 9 is too painful

or am I supposd to use the formula

[pearl pond]

@sharp smelt Has your question been resolved?

[humble lintel]

I'm not sure what you are doing here, Why do you have all these terms inside of the integral

[sharp smelt]

Because I need to integrate the approxmating polynomial for simpson's rule right

[humble lintel]

Okay, but do you understand what SImpsons rule does

[humble lintel]

And how that is different from Riemann integration

[sharp smelt]

isn't

right

[humble lintel]

What is osm?

[sharp smelt]

typo

I meant to type isn't

[humble lintel]

What do you mean by isn't

[sharp smelt]

It's the same

is it not

like we're approximating the function via a polynomial

[humble lintel]

I don't really see them as the same thing, but moving past that for a moment, you take a Riemann sum to replace an integral

So I don't see why you are trying to change your integrand doing this instead of just applying the result

[sharp smelt]

so use the sum instead

[humble lintel]

Well, you can do what you are doing now instead if you want, but this is a bad idea because you are essentially reproving the formula for no good reason in the middle of the calculation.

[sharp smelt]

One more question

for the 5 and 9 point cases

would I use simpson's composite 1/3rd rule

or the composite 3/8 rule

[humble lintel]

I don't know what terminology your class uses. When I think "Use Simpson's rule" I think use the 1/3 rule by default and this phrasing is ambiguous

Frankly, if I were in your position, I would just do and submit both. It really isn't that much extra work

[sharp smelt]

noted

thanks!

.close

[pearl pond]

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[sharp smelt]

.close

[pearl pond]

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

[iron schooner]

why is angle A2 = to 20 degree I dont understand, and what is angle O3?

[brazen hare]

OA AND OB

are isoceles

so angle B and angle A will be same

[iron schooner]

so angle O3 is 30 degrees and angle A2 is 50 degrees?

[brazen hare]

noo

look at OAB

B1 and a(1+2) are equal

[iron schooner]

could you visually show me a image?

[iron schooner]

angle A 1 is 30 degrees

and angle B 1 is 50 degrees

those angles are both in different triangles

[brazen hare]

?? triangle OAB?

[iron schooner]

now what do we need to do?

[sacred python]

what are you guys trying to find

i think i can help

[iron schooner]

my question: why is angle A2 = to 20 degree I dont understand, and what is angle O3?

[sacred python]

is o2 given

[iron schooner]

based on this image below

[sacred python]

so we dont know o2

[iron schooner]

no we dont

[sacred python]

angle c is 30

[iron schooner]

how is angle c 30 they never said anything is parallel?

[sacred python]

in triangle coa it is isosclelene

[iron schooner]

ohh

yess

[sacred python]

similarly angle a is 50

[iron schooner]

yea i can see that

[sacred python]

so angle a2 = 50-30

[iron schooner]

ok so angle C1 is 50 degrees

hold on

30*

[sacred python]

yes

exactly

you see in triangleoab\

oab is isoscelene

[iron schooner]

wait now what is after that?

[sacred python]

so angle a=b=50

[iron schooner]

= sides opp = angles right?

[sacred python]

ya

[iron schooner]

so its angle A2 of the whole angle A?

[sacred python]

so angle a2=50-a1

no

look

b=a

[iron schooner]

show me a image?

[sacred python]

i dont have one

its just the thing you gave

[iron schooner]

ok wait lets do it step by step then hold on

[sacred python]

dude you dont need steps its just mental math

we now know a and b dont we

[iron schooner]

yes

[sacred python]

yes now find o3

using angle sum property it will be 80

[iron schooner]

wait

[sacred python]

joey are you confused

[iron schooner]

its angle A 1 = 30 degrees

[sacred python]

ya

[iron schooner]

not the whole angle A

[sacred python]

no A=50

so now a2 is 20

[iron schooner]

look at the given

its says A1

= 30 degrees

[sacred python]

yes a1+a2=a

[iron schooner]

but we dont know a2?

[sacred python]

30+a2=50

we know a1 and a though

[iron schooner]

whats the reason for that?

[sacred python]

now a2 is 20

[iron schooner]

reason?

[sacred python]

cus a1=30 and a=50

no formulas

just simple logic

[iron schooner]

I need a reason for that?

[sacred python]

look at the image once

[iron schooner]

we cant just pull it out the air?

30 + a 2= 50?

whats the reason

[sacred python]

dude just look at the picture

we know that a=50 right

we just found it out

[iron schooner]

= angles opp = radii?

that surely cant be the reason

[sacred python]

dude forget about the circle

[iron schooner]

im looking for the reasons

[sacred python]

focus on the triangle

oa and ob are radii

[iron schooner]

even without the circle it would be = angles opp = sides

which cant be the right answer bc it 50

[sacred python]

oa = ob

[iron schooner]

lit could be a2 = 50.

[sacred python]

no

listen

[iron schooner]

lets us find A2 first

[sacred python]

wait

for that we need to prove oab is a isosceles triangle

now tell me looking at the pic if it is or not

so then obviously a1+a2=b