help-38

[severe matrix]

alright

[glad grotto]

I just gotta do some more practice questions

[severe matrix]

okay keep it up

goodluck

[glad grotto]

Ty

Can u do the last 1?

If not it's fine

[severe matrix]

alright i will try

[glad grotto]

Alr ty

[severe matrix]

bcs it contains 'cot' 🥴

[glad grotto]

😭

[severe matrix]

OK, so the degree I chose is not random, but rather a degree that has the same value as the result we want to find, like this, tangent theta is equal to 1, I'm looking for the value of tangent which has a value of 1, and has a POSITIVE value.. i hope you got it

[glad grotto]

I see

Thank you

I'll review that

.close

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[raven plume]

matrix/linear algebra

[whole coral]

So many

[bright quarry]

do you expect people to do your entire homework for you?

[rose jay]

its just 2 problems amr

[bright quarry]

look at all the parts

[raven plume]

i sent 3 problems...

theres 13 in the whole set i did most of them lmfao

these are the ones im stuck on

[whole coral]

Well then, have you tried anything? Do you have any idea for any of them?

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[fluid shale]

can someone explain to me why v >u here

[fluid shale]

v is the velocity of the rocket relative to an external observer and u is the velocity of the gas relative to the velocity of the rocket

this is about the rocket equation

M_0 is the intitial mass of the rocket and m is the total amount of gas outputed by the rocket

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@fluid shale Has your question been resolved?

[fluid shale]

OOOOH I GET IT

is it assuming v_0 = 0 ?

i guess there is no initial velocity

and then v(t) = u ln ( M_0 / M_0 - m (t))

so v (t) / ln ( M_0 / M_0 - m (t)) = u

and if that number is bigger than 1 then v would neccessairly have to be bigger than u

to compensate for that

is my logic correct that they assume v_0 = 0

[trim jolt]

@fluid shale Has your question been resolved?

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[twin yoke]

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[twin yoke]

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[bright quarry]

man just use chain rule

[twin yoke]

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[bright quarry]

but ok you can use product rule

[twin yoke]

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[bright quarry]

$\frac{d}{dx} f(g(x)) = f’(g(x)) * g’(x)$

[solid kiln]

knief

[bright quarry]

yea

[twin yoke]

One message removed from a suspended account.

[bright quarry]

much simpler to use chain rule of course

just 2cos(2x)

[twin yoke]

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[bright quarry]

this is the chain rule

[twin yoke]

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[wary basin]

It's not part of the chain rule

You find the derivative though

[twin yoke]

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[bright quarry]

i was just showing you what the chain rule is

[twin yoke]

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[bright quarry]

✨ khan academy ✨

[bright quarry]

cardinality go brr

[twin yoke]

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[bright quarry]

anyways

just use product rule

[twin yoke]

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[bright quarry]

unfortunately you’re wrong

[twin yoke]

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[bright quarry]

why the -2sinx

[twin yoke]

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[bright quarry]

yea

[twin yoke]

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[bright quarry]

bulgogie

[twin yoke]

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[bright quarry]

nothin

[twin yoke]

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[bright quarry]

because bulgogie

[twin yoke]

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[bright quarry]

because why not use a trig identity

isn’t cos2x nicer

[twin yoke]

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[bright quarry]

this is 2(-sin^2 + cos^2)

[twin yoke]

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[bright quarry]

you didn’t edit it

it’s supposed to be cos

[twin yoke]

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[bright quarry]

https://tenor.com/view/zyzz-brah-zyzz-pose-yes-yes-brah-gif-25866912

[twin yoke]

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[bright quarry]

zyzz

[bright quarry]

yuh

[twin yoke]

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[bright quarry]

nope

why the extra factor of 2

[twin yoke]

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[bright quarry]

but also rewrite in terms of cos(2x)

[twin yoke]

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[bright quarry]

cos^2 - sin^2

[twin yoke]

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[bright quarry]

sure

[twin yoke]

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[bright quarry]

$-2\sin^2(x) + 2\cos^2(x) = 2(-\sin^2(x) + \cos^2(x)) = 2\cos(2x)$

[solid kiln]

knief

[twin yoke]

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[bright quarry]

perhaps exactly what i’ve written

[twin yoke]

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[bright quarry]

💀

[twin yoke]

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[bright quarry]

you don’t get why cos(2x) = cos^2 - sin^2?

consider cos(x+x)

angle sum for cos

[twin yoke]

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[bright quarry]

cos(x+x) = cosx cosx - sinx sinx

= cos^2 - sin^2

[twin yoke]

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[bright quarry]

double?

[twin yoke]

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[bright quarry]

cos(u+v) = cosu * cosv - sinu * sinv

suppose u = x and v = x

[twin yoke]

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[bright quarry]

^

[bright quarry]

it’s angle sum mate

[twin yoke]

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[bright quarry]

yes

hence why i showed you

goodnight

[twin yoke]

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[bright quarry]

what

[twin yoke]

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[bright quarry]

yuh

now goodnight

[twin yoke]

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[bright quarry]

💤

[twin yoke]

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[bright quarry]

😴

[twin yoke]

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[bright quarry]

why sinx

[twin yoke]

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[bright quarry]

huh

[twin yoke]

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[bright quarry]

oh

[twin yoke]

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[bright quarry]

it says in the second part

[twin yoke]

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[bright quarry]

oh this is another question lol

[bright quarry]

you didn’t differentiate it

[twin yoke]

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[bright quarry]

man this is so evil

[twin yoke]

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[bright quarry]

😭

[twin yoke]

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[bright quarry]

not teaching you chain rule

[twin yoke]

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[bright quarry]

it’s arguably the most important rule

they’re making you differentiate cos^2 and sin^2 using product rule too

😭😭

instead of power and chain rule

[twin yoke]

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[bright quarry]

do yourself a favor and learn it

it’s so easy

[twin yoke]

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[bright quarry]

goodnight

[twin yoke]

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[bright quarry]

you’re welcome

[twin yoke]

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[paper estuary]

H

[shrewd ridge]

,rcw

[prime lynx]

Limit as x tends to 0 probably

[paper estuary]

I don't know

[brittle fiber]

hello

[paper estuary]

Can you ilucidate

[brittle fiber]

what

[tardy hemlock]

Might use the Taylor expansion of sqrt(1 + x) around x = 0 if you had that already

[paper estuary]

I don't know taylor expansion

[feral mist]

Use binomial expansion

you can say that it is equal to (1-x)^1/2*(1+x)^-1/2

[tardy hemlock]

I'm not entirely sure what the exercise wants because I don't have any context on your course. Maybe it's enough if you just show that f(0) = g(0), f'(0) = g'(0) and f''(0) = g''(0)

[feral mist]

expand both the brackets and multiply the 2 brackets

[paper estuary]

Okay

[paper estuary]

How do you put zero

Are you there

[tardy hemlock]

What do you mean?

You plug 0 into x

[paper estuary]

Yes

[tardy hemlock]

$\sqrt{\frac{1 - 0 }{1 + 0}} = 1$

[solid kiln]

Kepe

[tardy hemlock]

$1 - 0 + \frac{0^2}{2} = 1$

[solid kiln]

Kepe

[paper estuary]

Which grade are you

[tardy hemlock]

The derivative of the LHS is $-\frac{1}{2 \sqrt{\frac{1-x}{1 + x}} \cdot (1+x)^2}$. Plugging in $0$ gives $-1$.

[solid kiln]

Kepe

[tardy hemlock]

The derivative of the RHS is $-1 + x$. Plugging in $0$ gives $-1$ too

[solid kiln]

Kepe

[tardy hemlock]

And the same for the second derivative.

But as I said, it really depends on what this exercise wants from you and what you previously covered in your course

[paper estuary]

Hmm

Cool

You are good at math

[tardy hemlock]

But again, you should give more context

What have you done previously in class?

What topic are you currently on?

[paper estuary]

In binomial expansion

[tardy hemlock]

Then follow this

[paper estuary]

Okay

[tardy hemlock]

What I provided are just two alternatives then

[paper estuary]

Yes

Are you studying

[tardy hemlock]

Yeah one could say so

In high school and in uni in parallel

[paper estuary]

I'm 11rh

[tardy hemlock]

Cool

[paper estuary]

I mean 11th grade

What about you

[tardy hemlock]

I started 12th recently

[paper estuary]

Fire

Do you have any advice because I memories sum

[tardy hemlock]

Advice on what?

[paper estuary]

Math

[tardy hemlock]

Like generally?

[paper estuary]

Yes

[tardy hemlock]

It's best if you really like it, hopefully even love it

[urban thistle]

hi

[tardy hemlock]

Then everything else comes naturally

Practice a lot

[paper estuary]

Do you follow youtube channel to get help

[tardy hemlock]

No

If I want to look up a school topic, I look it up in the book

For uni topics, I look them up in the lecture notes

Or in books

[paper estuary]

No internet and gained so much knowledge

[tardy hemlock]

Well I have internet lol

I just don't really use videos

[paper estuary]

Why don't you accept my friend request

[tardy hemlock]

Oh I didn't notice it

[paper estuary]

Bruh

Are you from America

[tardy hemlock]

No

I'm from Germany

[paper estuary]

I'm Indian

[tardy hemlock]

Oh, cool

You must know of Ramanujan, right?

A pretty good mathematical inspiration

[paper estuary]

Yes

Actually we are related

[tardy hemlock]

What do you mean?

In what way?

[paper estuary]

We are Brahmins

[tardy hemlock]

Oh

I don't know much about the hindu system

[paper estuary]

There is a movie about him

[tardy hemlock]

Yep I watched it!

You too?

[paper estuary]

Good right

[tardy hemlock]

Yeah it's absolutely great

Maybe we will even watch it in class

My teacher has it

[paper estuary]

He is a treasure

[tardy hemlock]

Very mysterious

He described to his friends that he saw a bloodstream

Then a hand drew elliptic integrals into it

The next day, he would remember it and write it down

[paper estuary]

The man who knows infinity

You it's all faith in God

[tardy hemlock]

He attributed everything to his goddess Namagiri

[paper estuary]

Yeah

Do you know

That he found formula for partition

[tardy hemlock]

Yeah, together with Hardy

[paper estuary]

German people are smart

[tardy hemlock]

Probably came up with it just like that again, haha

Without a proof

[paper estuary]

Johannes Kepler

[tardy hemlock]

Well Gauss and Euler would probably be the biggest I would name

Anyways, I gotta go now since I have to give a presentation to my teachers soon

[paper estuary]

About what

[tardy hemlock]

It's because I'm in university at the same time as high school

And they wanted me to present any topic I choose

I chose the jordan normal form

It's a nice form you can get matrices into

[paper estuary]

Do you know aryabhatta

[tardy hemlock]

Yeah

He's ancient

[paper estuary]

He discovered zero

[ornate garden]

do yall like n||u||gge||ts||

[tardy hemlock]

Later, the naturals were formalized by Peano

There are now if I remember correctly 5 or 6 peano axioms

Basically describing what the set $\mathbb N$ should be

[solid kiln]

Kepe

[tardy hemlock]

Well anyways, I'm gonna go now, cya

[ornate garden]

bro why u ignoring mme

[paper estuary]

I'm practicing math

But Im not that good

How about you

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@paper estuary Has your question been resolved?

[paper estuary]

Yes

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[brittle fiber]

.

[balmy hornet]

I put DNE since at X = 1, it approaches a local maximum but dne

Then I put 4 because at X = 6 there is a local maximum

Oh I got it

I need to include the absolute maximum since it's in the middle

.close

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[slim temple]

This is a problem related to Taylor Series, Im trying to find the Taylor Series for
x² . ln(x) with x₀ = 1
Doing the derivates, i found that the Series is divided into some sections [1st Image]. Them being: Two derivatives which dont seem to form any clear pattern

1st 2x . ln(x) + x
2nd 2 ln(x) + 3

All the following have a (2(2(n-3)!) . (x^(n-2)) pattern, starting from n = 3, alternating between positive and negative, which, after a lot of time, i was able to figure as the next sum [2nd image]

From there, I replaced all corresponding x₀ with 1. [3rd and 4th image] to be able to form the polynomical form, but it didn't work.

[slim temple]

I'd like to know if i did something really wrong

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@slim temple Has your question been resolved?

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@slim temple Has your question been resolved?

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@slim temple Has your question been resolved?

[slim temple]

.close

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[grim mauve]

I need help with this problem, deals with Parabolas.

Consider the function f(x)=(2x)^2 - 2

Describe each function as a transformation of the quadratic parent function.

I am having trouble about horizontal stretching or compressions.

I thought it is horizontal compression by 2 but the key says horizontal compression by 1/2. I am confused.

[dull temple]

it gets squished

the terminology can be a little confusing

[trim jolt]

@grim mauve Has your question been resolved?

[grim mauve]

Yeah but if it gets squished, shouldnt it be get squished by a positive number? If it gets squished by a fraction shouldn’t have it gotten wider?

[dull temple]

again this is just the wording being weird

[grim mauve]

Ok

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[daring brook]

on the interval [0, 2pi] this dip would be a local minimum right? not an absolute one?

[daring brook]

i dont think there's an absolute minimum here but i wanna make sure

[wicked badge]

ye

[daring brook]

because at x = 0 the y value is higher

[wicked badge]

well assuming this is a normal polynomial cubic function

the absolute minimum would be at y=-inf

because all polynomials are continuous

[daring brook]

even with the interval?

[wicked badge]

not with the interval

[daring brook]

its constrained to the int

but thx

[wicked badge]

wait

how is the question formulated

[daring brook]

it says find the absolute minimum and maximum values on the interval [0, 2pi]

[wicked badge]

then yeah that would be the absolute minimum and maximum

only for that given interval

[daring brook]

so no abs minimum?

[wicked badge]

abs minimum on that interval would exist

[daring brook]

but wouldnt it be local

[wicked badge]

you are not looking at the entire graph

you are looking at the interval

[daring brook]

its constrained to the interval

[wicked badge]

pretend everything outside of that interval does not exist

[daring brook]

oke

[wicked badge]

look

[daring brook]

the lowest y value is at x = 2.25 or so

and after the 2nd lowest is at x = 0

[wicked badge]

your view is constrained to this about

who is to say that to the left of this

this wouldnt happen?

so yeah you kinda just have to assume its an absolute min

[daring brook]

why can we assume O.o when it could go in either direction

[wicked badge]

its just that you have no information so you assume from what you are given

dont get fixed on the semantics :p

[daring brook]

welp

oke thx

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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[proven chasm]

Find all values of c such that f(x) has 2 integer roots, where f(x)=9x^2-12x+c

[proven chasm]

I used the the quadratic formula to get that

$\frac{\left(12\pm\sqrt{144-36c}\right)}{18}$ has to be an integer

[solid kiln]

TimesZeroed

[proven chasm]

Only way for this to be true is if

$\left(12\pm\sqrt{144-36c}\right)\operatorname{mod}18=0$

[solid kiln]

TimesZeroed

[proven chasm]

12 = -6 (in mod 18)

and then I added that to the other side

$\pm\sqrt{144-36c}\operatorname{mod}18=6$

[solid kiln]

TimesZeroed

[proven chasm]

Now I don't know what to do

If I square both sides, it doesn't work

[nimble stone]

the sqrt simplifies to 6sqrt(4-c) if that helps at all

[proven chasm]

thanks

so

$\pm\sqrt{4-c}\operatorname{mod}18=1$

[solid kiln]

TimesZeroed

[proven chasm]

Checking on desmos, the first solution is x=3

This is a pre-calc question

so I think I am overcomplicating it

[slim temple]

ill try to aproach it from some other way to see if i can help, but this seems like this should be easy, yeah

[proven chasm]

I am going to post the exact problem

170. Using the discriminant:

For what values of c will the equation

9x^2-12x+c=0

have

a) no real roots

b) one rational root

c) two real roots

d) two integer roots

[slim temple]

the first should be a c > a with a being some real positive number

the second is c = a

[proven chasm]

no

the first one is

c>4

ohh yeah

[slim temple]

yeah, thats a real positive number

[proven chasm]

my bad

yeah

b is

c = 4

[slim temple]

yeah

[proven chasm]

and c is

c < 4

idk what to do for d

[slim temple]

ill do some math for d), but maybe it has something to do with previous answers, ill check

[proven chasm]

wolfram alpha claims

these are the integer solutions

but I don't know if it produces

2 integer solutions

or just one

yeah

it just produces one integer solution

or only guarantees one integer solution

[slim temple]

this is what i could advance with, it may lead to something i guess

i missed a /18 on the whole root

[proven chasm]

so

6n

?

sqrt (discriminant) = 6n

[slim temple]

oh, sorry, i got it wrong, still i had a good idea-

[proven chasm]

yeah

anyway

so you got

sqrt(stuff) = n

ohh

that's nice

you just need 4-c to be a perfect square

nvm

[slim temple]

it must be (1+3n)/3, sooo, like, 1/3, 4/3, 7/3 for it to make up an integer while adding with 2/3

[proven chasm]

c?

c has to be an integer

[slim temple]

no, the roots with c must be

c could be any real

[proven chasm]

yeah

but

in the quadratic formula

you get

nvm

your write

actually

yeah c has to be an integer

in the quadratic formula

you get

sqrt(144-36c)

you factor out the 36

6sqrt(4-c)

ohh

c can have a denominator of 36 at max

my bad

[slim temple]

yeah, i was doing exactly that, i ended with

sqrt(4-c) = 1+3n

so infinite answers, now to find the pattern it follows

you ^2 both sides and gives -c = 9n^2 + 6n - 3 * with n being any integer

i think that would do

sorry, its minus c

sooo, basically c = -9n^2 - 6n + 3 , n = any integer.

it also makes sense, since that formula "mostly" gives negative numbers, and the answer as sqrt(4-c), which cant have any c > 4

[proven chasm]

so

-9n^2 - 6n + 3

this is the solution?

[slim temple]

i think, yeah

[proven chasm]

alright I will check it

but thank you so much

https://www.desmos.com/calculator/91ptpem9dx

I've just been

searching for value

by repeatly clicking on a->

[slim temple]

mfw i dont know how to use mod because my country education doesnt use it at all

[slim temple]

lol

[proven chasm]

dang

anyway

Can you explain your solution

[slim temple]

yeah, ill try to send it in LaTeX, wait a bit

[proven chasm]

thanks

https://www.desmos.com/calculator/ardks5rgos

here is my solution finder

it's very inefficient

but it is no longer manual

I haven't found any below 2000

Okay

I checked your formula

and it seems that it only gives value where there is one solution that is an integer

the other is 1/3

so I think the answer is that there aren't any

[slim temple]

i also find a problem, ill end the latex thing and check on my logic

sorry for the mess

$\frac{12 \pm \sqrt{144-36c}}{18}=x_{1},x_{2\in}\in \mathbb{Z}\ \frac{2}{3} \pm \frac{\sqrt{144-36c}}{18} = \mathbb{Z}\Rightarrow \pm \frac{\sqrt{144-36c}}{18} =\frac{1+3n}{3}\ \pm\frac{\frac{\sqrt{144-36c}}{6}}{3}=\frac{1+3n}{3}\Rightarrow\pm\frac{\sqrt{144-36c}}{6}=1+3n\ \ast \sqrt{144-36c}=\sqrt{12^{2}-6^{2}c}=\sqrt{(2.6)^{2}-6^{2}c}=\sqrt{(6)^{2}(2^{2}-c)}=6\sqrt{4-c}\ \pm\frac{6\sqrt{4-c}}{6}=1+3n \Rightarrow \pm\sqrt{{4-c}}=1+3n\Rightarrow (\pm\sqrt{{4-c}})^{2}=(1+3n)^{2}\ 4-c = 9n^{2}+6n+1\Rightarrow -c = 9n^{2}+6n-3\Rightarrow c = -9n^{2}-6n+3$

[solid kiln]

zzz0nnn

[slim temple]

for example, n = 0, which is an integer, makes c = 3, which is one of the solutions

[proven chasm]

no

it is not a solution

[slim temple]

no? :(

[proven chasm]

one of the roots is an integer

one isn't

[slim temple]

oh, yeah, cause -1/3 would lead to 1/3

its half the solution (only the sum part)

[proven chasm]

I think the answer is

there are no such values

[slim temple]

if the substract and sum part share solutions then there are solutions

ill make the same logic, wait a bit

like, i divided the problem in sum and substract without realizing

[proven chasm]

oh

anyway

my very bad desmos thing

has found nothing

for c > -252

[slim temple]

its the same problem but with 2+3n instead of 1+3n

[proven chasm]

yeah

I don't think there are solutions

[slim temple]

with 2+3n its
c = -9n^2 - 12n

now to see if there are c's which equate for that

[proven chasm]

ok

doesn't work

[slim temple]

$-9n^{2}-6n+3=-9n^{2}-12n\Rightarrow \ -6n=0$

[solid kiln]

zzz0nnn

[slim temple]

so the only answer to check is n = 0, which we know is not answer

[proven chasm]

so no solutions right

[slim temple]

yeah, id swear

[proven chasm]

ok

I need to figure out how to prove this now

😭

[slim temple]

what i just sent you is the full proof, it just need the 2+3n version

i can do it if you want

[proven chasm]

Before you try

I think

proving that (-6 + y) mod 18 = 0

then -6-y mod 18 can't = 0

is eaiser

because of -6 + y = 0 (everything is mod 18 for now)

then y = 6

-6-6 = -12

which mod 18

is 6

nvm

Ok I got a better proof

if you use the quadratic formula

and simplify it

you get

$\frac{\left(2\pm\sqrt{4-c}\right)}{3}$

[solid kiln]

TimesZeroed

[proven chasm]

we let y = sqrt(4-c)

then

$\frac{\left(2\pm y\right)}{3}$

[solid kiln]

TimesZeroed

[proven chasm]

has to be true

when it is plus

this is only true when y is of the form

3n + 1

and when it is -

[slim temple]

idk what the fuck u doing but go ahead, slay > again im not into mod, i usually use other methods

[proven chasm]

I need to explain it

so that my teacher gets it

[proven chasm]

yeah this doesn't require mod

you have 9x^2-12x+c=0

right

[slim temple]

yeah?

[proven chasm]

can I explain my proof sorry

nvm

[slim temple]

$\frac{12 \pm \sqrt{144-36c}}{18}=x_{1},x_{2\in}\in \mathbb{Z}\ \frac{2}{3} \pm \frac{\sqrt{144-36c}}{18} = \mathbb{Z}\\ + \frac{\sqrt{144-36c}}{18} =\frac{1+3n}{3}\ +\frac{\frac{\sqrt{144-36c}}{6}}{3}=\frac{1+3n}{3}\Rightarrow+\frac{\sqrt{144-36c}}{6}=1+3n\ \ast \sqrt{144-36c}=\sqrt{12^{2}-6^{2}c}=\sqrt{(2.6)^{2}-6^{2}c}=\sqrt{(6)^{2}(2^{2}-c)}=6\sqrt{4-c}\ +\frac{6\sqrt{4-c}}{6}=1+3n \Rightarrow +\sqrt{{4-c}}=1+3n\Rightarrow (+\sqrt{{4-c}})^{2}=(1+3n)^{2}\ 4-c = 9n^{2}+6n+1\Rightarrow -c = 9n^{2}+6n-3\Rightarrow c = -9n^{2}-6n+3\ \ - \frac{\sqrt{144-36c}}{18} =\frac{2+3n}{3}\ -\frac{\frac{\sqrt{144-36c}}{6}}{3}=\frac{2+3n}{3}\Rightarrow+\frac{\sqrt{144-36c}}{6}=2+3n\ \ast \sqrt{144-36c}=\sqrt{12^{2}-6^{2}c}=\sqrt{(2.6)^{2}-6^{2}c}=\sqrt{(6)^{2}(2^{2}-c)}=6\sqrt{4-c}\ +\frac{6\sqrt{4-c}}{6}=2+3n \Rightarrow +\sqrt{{4-c}}=2+3n\Rightarrow (+\sqrt{{4-c}})^{2}=(2+3n)^{2}\ 4-c = 9n^{2}+12n+4\Rightarrow -c = 9n^{2}+12n\Rightarrow c = -9n^{2}-12n\ \ -9n^{2}-6n+3=-9n^{2}-12n\Rightarrow \ -6n=0$

this is the whole thing, kinda messy, you just add that n = 0 is the only option and it doesnt work in neither of the formulas

and thats the whole proof

[solid kiln]

zzz0nnn

[slim temple]

we could also have gone for the geometric proof, which would have been far easier but just needed to write a lot

[proven chasm]

cool

can I show you my proof

it's a lot a shorter no mods

nvm it's fine

thank you for the help

.close

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[marsh forum]

Prove that $gcd(0,0) =0$

[solid kiln]

Mr bean is not $\R \setminus \Q$

[marsh forum]

The definition of gcd(a,b) is as follows, for a,b \in \Z

[night patio]

i feel like this depends on which definition you choose to take on

whether if a|b is written as b = ak or b/a = k (personally i dont like the latter definition but idk whats conventional definition elsewhere)

[marsh forum]

Let $gcd(a,b)=k$ . Then $k \mid a; k\mid b$.
\
if $l \mid a \land l\mid b \implies l\mid k$

[solid kiln]

Mr bean is not $\R \setminus \Q$

[marsh forum]

so $0=kk_1; 0=kk_2$

[solid kiln]

Mr bean is not $\R \setminus \Q$

[marsh forum]

It's obvious that $0=k \lor k_1,k_2=0$

[solid kiln]

Mr bean is not $\R \setminus \Q$

[marsh forum]

now we know that $0=ll_1; 0=ll_2$ so $l ;l_1l_2=0$

[solid kiln]

Mr bean is not $\R \setminus \Q$

[marsh forum]

What do I do now?

[marsh forum]

We use $b=ak$

[solid kiln]

Mr bean is not $\R \setminus \Q$

[night patio]

ok idk why it took me so long to realise but

like u agreed, 0 = D*0, where D is just any positive integer + 0

[marsh forum]

so $D \in \mathbb{W}$

okay

[solid kiln]

Mr bean is not $\R \setminus \Q$

[night patio]

so then by our previous definition we have D = {0, 1, ...}

every number within that set also divides into 0

so gcd is just 0

[marsh forum]

oh right

that make intutive sense, ofcours, but then we'd have have prove that 0 is the unique natural number with this proerty

Probably not required this point. I'm learning proofwriting , not Number theory (:vomit:)

thanks dootud!

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[vernal bronze]

help

[vernal bronze]

substitution method

[spring summit]

it gave u what x is

[vernal bronze]

no

[spring summit]

x = 4y - 17

[vernal bronze]

oh ya

[bright seal]

you have to substitute x=4y-17 into the equation above

Find y

And then plug this y in x=4y-17

🍫

[vernal bronze]

im not sure on how to

can u solve it

[bright seal]

Yes

Wait

y=5

3(4y-17)-y=4

I find the y

12y-51-y=4 -> 11y=55 -> 11y/11=55/11 -> y=5

Clear ?

[vernal bronze]

ya

thanks

[bright seal]

👍

[vernal bronze]

this one

you know what

I get it

help

help

[vernal bronze]

this

help

#help

@bright seal

hello

#❓how-to-get-help

<@&286206848099549185>

<@&286206848099549185>

[spring summit]

sub in the equation for y = x + 9 into 4x + 7y = 19

[vernal bronze]

solve it for me

write me the steps please

🙏

[spring summit]

4x + 7(x+9) = 19

[bright seal]

Hi

[vernal bronze]

i messed up

i substituted the x instead of the y

[spring summit]

works too

x = y - 9

4(y - 9) + 7y = 19

[vernal bronze]

can u solve it

like fully

[vernal bronze]

whats after

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

can u solve it please

[pulsar dust]

do not ping helpers multiple time you can ping only once

[vernal bronze]

oh ok

[pulsar dust]

as for this you have y=x+9 substitute y in 4x+7y=19

[vernal bronze]

then

[pulsar dust]

what did you get

[vernal bronze]

wait

4x+7x+63=19

[pulsar dust]

what is this equal to?

[vernal bronze]

4x+7x= 11x

then what

[pulsar dust]

subtract 63 from both sides in this

[vernal bronze]

the.

then

[wraith hinge]

i will teach

you bro

[vernal bronze]

teach me plz

[wraith hinge]

yeah'

we use elimination

method bro

[vernal bronze]

no

it has to be substitution method

[wraith hinge]

now

multiply

1 min

@vernal bronze

Substitute y into it ig

$4x + 7y = 19\ y = x + 9$

[solid kiln]

Sukiyaki

[wraith hinge]

You can

$4x + 7(x+9) = 19$

[solid kiln]

Sukiyaki

[wraith hinge]

Then solve it from there.

so it would become $4x + 7x + 63 = 19$

[solid kiln]

Sukiyaki

[vernal bronze]

then

[wraith hinge]

x = -4

do you get what i did?

Because you having y = x+9 essentially tells you that

[vernal bronze]

till her yes

[wraith hinge]

7 MULTIPLIED by Y.

so instead of writing y

what can you write?

[vernal bronze]

I understand

but

[vernal bronze]

after this

[wraith hinge]

After this, you sum it up.

and make the variable on one side and numbers on another.

That would make it easiest to solve.

[vernal bronze]

ok

[wraith hinge]

so, to remove 63 on the left side, what do we do?

-63 right?

[vernal bronze]

yes

[wraith hinge]

but you have to -63 on the other side as well.

[vernal bronze]

yes

[wraith hinge]

is y = 55/8

well once you do that,

x = -4

i think lemme check

[vernal bronze]

the teacher sent these are the answers

[wraith hinge]

right lemme

write it

1 sec

[wraith hinge]

that wouldn't be right..

hm...

that isnt right.

well if you substituted the following numbers into your calculator,

you'd actually get...

,calc 4(-(17/11))+7((82/11))

[solid kiln]

Result:

46

[wraith hinge]

y = 5 bro

wait a min

Yes.

y = 5 is correct.

okay, ill explain it in 1 text.

is this right sukiyaki

So, first you have two equations

$4x+7y=19$ - (1)
$y=x+9$ - (2)

So because you already have y, you can substitute into the first equation labelled 1. Now you'll have this; $4x + 7(x+9) = 19$

Of course when you expand it you'll get
$4x + 7x + 63 = 19$
Now, remove 63 from the left side, you should get
$11x = -44$
thus, x=-4.

Now substitute x = -4 into second equation,
$y = -4 + 9\ \therefore \text{y = 5}$

[solid kiln]

Sukiyaki

[wraith hinge]

ywah x = -4 right

Yes.

the answer key is wrong

This is called elimination and substituion right?

Well...

You could perform elimination too

we call it like that

i did elimiation

actually, this is purely substitution

what i did.

i should learn pure substitution

because i found x by substituting y into (1) and y by x into (2)

[wraith hinge]

well if you did elimination, it would be longer imo.

i learnt from

https://www.youtube.com/watch?v=0rAhyj4LYnk

But you could take 7y = 7(x + 9)

Ok

alright, notice that

3x + 2y = -6
2x + 5y = 7

[vernal bronze]

ya the teacher is wrong

[wraith hinge]

well, this would leave you a x + 3y = ...

[vernal bronze]

i sent her an email

[wraith hinge]

and you'd have to substitute here and there.

[vernal bronze]

how to do this

[wraith hinge]

,rotate

[wraith hinge]

let me see.

simply substitute x into the first equation.

hm

ans is 0

?

Yea, i got that as well which i found it weird lol

[vernal bronze]

but

how ?

[wraith hinge]

y = 0

y = 1-y -1
2y = 0
y = 0.
x = 1

yep, i'm p sure it's simply y = 0, x = 1

yeah

he is correct

[wraith hinge]

yes, this works

or simply, you know what i'd do?

y = x - 1 right?

yeah

well yea lets not do my method

it'll get confusing

[wraith hinge]

Anyhoo, this is how you can do it.

[vernal bronze]

how did the x become 1

[wraith hinge]

well you substitute y = 0 into x = 1-y

Well instead of me just giving you answers, let me see if you understand.

I'm going to give you two equations, see if you can solve it.

2x + 4y = 12
x = 4 + y

O

k

bro

[vernal bronze]

y=1

oh

[wraith hinge]

are both of you friends? @wraith hinge

[vernal bronze]

no

[wraith hinge]

o..

no

okay

we are stranger to each other lol

anyways @vernal bronze are you able to answer it?

[vernal bronze]

2/3

y= 2/3

[wraith hinge]

okay

lemem check LOL

which question are you doing

ye y = 2/3

So do you roughly get what you're doing now?

[vernal bronze]

ya

thanks

[wraith hinge]

good

[teal locust]

hi

[wraith hinge]

hi

you can close the channel and let others use it if youre done @vernal bronze you can text .close

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@vernal bronze Has your question been resolved?

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[little glen]

Isn't F'(x) just f(x) ?

so the integral is just 2f(x)e^x

= 2 sinx

after integrating -2cosx | 0, 1

= 2(1 - cos1)

[hollow tangle]

how?

[zinc ginkgo]

Yes

[little glen]

Differentiation of a definite integral

I think it was called Leibnitz rule or something

[hollow tangle]

oh F and f are different

[zinc ginkgo]

Did you try plugging this in

[little glen]

this thing is definitely greater than 1

but none of the options' upper bounds are > 1

[zinc ginkgo]

,calc 2(1-cos(1))

[solid kiln]

Result:

0.91939538826372

[little glen]

HUH

[zinc ginkgo]

No

[little glen]

OH LORD IM DUMB

sorry

but how would I do that without a calculator

im supposed to solve this without a calculator

no tables, no nothing

[zinc ginkgo]

,calc 2(1-cos(1))*360

[solid kiln]

Result:

330.98233977494

[zinc ginkgo]

Do you know Taylor series for 1-cos

[little glen]

i know the taylor for cos

I think it goes like 1 - x^2/2! + x^4/4!

[zinc ginkgo]

Yea plug 1 into that expansion until you get the accuracy you need

[little glen]

oh alright

i think till x^6/6!

since I get 720 that way

alright, tysm! @zinc ginkgo

[hollow tangle]

bro is this jee?

pls say no

[little glen]

(i thought it was increasing in (0, pi/2) xDDD)

[little glen]

it is

[little glen]

why'd you ask

[hollow tangle]

im in 11th

[little glen]

that question was a piece of cake compared to some other ones

[hollow tangle]

thanks for the motivation!

[little glen]

this one took me a solid 10 minutes to solve

[hollow tangle]

this is func and relation chapter right?

our teacher was telling us that 11th one is super ez

but 12th one

💀

[little glen]

integration

don't tell me you have never seen an integral S before...

[desert karma]

Find f'(ø)

[hollow tangle]

i have

cause of phy

but our one is not that hard

mostly just integrals trig or power rule

[little glen]

it's solved, thank you anyways mate :)

.close

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[past cargo]

how to prove the forward direction

can i start with saying that if circumcenter and incenter are the same point then all perpendicular bisectors are also angle bisectors?

[prime lynx]

,w incenter

[solid kiln]

Results provided by WolframAlpha

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Requested by ishigamisenku_

[prime lynx]

Is it like centroid and orthocenter

[past cargo]

it is

but different centers

centroid is the one for medians

this one are centers for perpendicular bisectors and angle bisectors

[prime lynx]

What are you required to use to prove it?
Construction?
Vectors?

[past cargo]

no vectors

this one doesn't require constructions so not that either

[prime lynx]

So plain geometry

[past cargo]

yes

[past cargo]

I mean if i can say this

then i think i am done

[prime lynx]

Yeah I think that works

@past cargo

[past cargo]

are u sure?

I think it works but idk how to prove it

just seems obvious to me lol

maybe i'm wrong

[prime lynx]

You can try assuming the triangle is not equilateral

[past cargo]

how do i prove this?

[prime lynx]

You can use the angle bisector theorem

I'm not sure if that's the name but I will send a picture

[past cargo]

i don't think my claim is riht

just because centers are the same doesn't mean they have to be the same line

since it's just one point

[prime lynx]

https://byjus.com/question-answer/the-perpendicular-bisector-and-the-angle-bisector-of-an-equilateral-triangle-are-the-same-and-6/#:~:text=The perpendicular bisector and the,they pass through the circumcentre.

[past cargo]

yeah that's the forward direction

not the reverse

if the centers are the same then that doesn't mean that those lines have to be the same either

i guess i know how to solve this though

if X is the incenter of ABC then it is equidistant from AB, BC, and AC

if X is the circumcenter then XA = XC = XB