help-0

[surreal meteor]

it would just go away, has no effect on the result of the definite integral, no constants do
$F(x)=2x+5$

$\int_0^1 F(x)dx= (2(1)+5)-(2(0)+5)...$

[ocean seal]

SimonWin

[surreal meteor]

How does a constant not effect the definite integral?

[jagged cobalt]

those 5's cancel out

have no effect

[surreal meteor]

Oh..

I see..

[jagged cobalt]

all constant just cancel

[surreal meteor]

I never thought of that

[jagged cobalt]

thats why the +C doesnt matter

[surreal meteor]

But yes makes sense wow

Yea..

Wow..

hmm

Wtf

So, what you did there was amazing, but i'm guessing i would've made my life much easier if i just picked u=6x+1

or would i arrive at the same

[jagged cobalt]

if youre doing a definite integral, i dont think theres any difference in hardship

with an indefinite one, you might get confused but it would still be right

[surreal meteor]

yeah since the constant cancels out

so i never thought when you took the definite integral the cancel didn't matter

but makes sense that when you do F(b)-F(a)

all constant terms will have mixed signs and cancel out

[jagged cobalt]

yup

[surreal meteor]

i just tried u=6x+1

just for fun

and practice

and seems i'd arrive at the same in the end hahaha

6*3x^2+6

[jagged cobalt]

yup

[surreal meteor]

Yeah hahah.. 😂

You called it.

But how would I arrive at like the same answer as wolfram? 😂

Without having to factor out a constant

Because honestly when im sitting in a 4 hour exam sweating i'm realistically not going to see that 😂

Perhaps it was better just letting u=x^2

like that Q guy said haha

And for some reason (i am probably calculating it wrong) but I am getting the wrong definite integral

all the way below^^

[jagged cobalt]

you can get to that directly by recognition of the fact that:
$$\int {\frac{x}{3x^2+1}} dx=\frac{1}{6} \int{ \frac{6x}{3x^2+1}} dx$$
Then as Q had said you can recognise the numerator is the derivative of the denominator after applying this factorisation, and that:
$$\frac{d}{dx} ln(f(x))=\frac{f'(x)}{f(x)}$$
therefore
$$\frac{1}{6} \int{ \frac{6x}{3x^2+1}} dx= \frac{1}{6} ln(3x^2+1)+C$$

[surreal meteor]

$\int_0^1 x(3x^2+1)^(-1)$

[ocean seal]

SimonWin

AℤØ

[surreal meteor]

Where yall find rules like that? 😂

[jagged cobalt]

1/6 ln(24) - 1/6 ln(6)

[surreal meteor]

This all i got man

[jagged cobalt]

its basically this one just applied more generally

[surreal meteor]

I see

if it was a above

a/a = 1

And then it would just be ln|ax+b|

Right?

$\int \frac{a}{ax+b}dx=\frac{a}{a}ln|ax+b|=ln|ax+b|$

[jagged cobalt]

they factor out 1/a so its a/(ax+b) then apply the reverse of the derivative i gave here

yeah

[surreal meteor]

Is that right? 🤣

Damn

[jagged cobalt]

no

you never integrated, also

[ocean seal]

SimonWin

[surreal meteor]

Why is it not right 😢

[jagged cobalt]

its just 1/a

not a/a

[surreal meteor]

I thought i was onto something

[jagged cobalt]

the 1/a is factored outside the integral

$\frac{1}{a}\int \frac{a}{ax+b}dx=\frac{1}{a}ln|ax+b|+C$

[ocean seal]

AℤØ

[surreal meteor]

shoudlnt it be 1 in the nominator then?

[jagged cobalt]

where?

[surreal meteor]

or wait no

No sorry i think i got confused

I thought you took the nominator out

[jagged cobalt]

no problemo

[surreal meteor]

But that's definitely not it

[jagged cobalt]

no it was 1, but by factoring out 1/a it becomes a

[surreal meteor]

Oh!

I see

because a/a = 1

So, you have just written a * 1/a as 1

[jagged cobalt]

pretty much

for convenience

[surreal meteor]

Yea okay, interesting!

Well thanks for all the help and patience! ❤️

[jagged cobalt]

no worries

[surreal meteor]

I always appreciate it!

Have a lovely day! 🫶

[jagged cobalt]

you too

[surreal meteor]

.close

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[onyx sparrow]

Can someone help with a step by step process on how to solve this?

[fickle heath]

Find the area of FGH, and then find the area of GPH

That should be a good starting point

[onyx sparrow]

So do I find the area of FGH and then find all the possible locations for P so that the other triangles can have half the area of FGH?

[fickle heath]

So that GPH has half the area of FGH, yes

[onyx sparrow]

Ooohhhh okay

[fickle heath]

Then you worry about FPG and HPF

[onyx sparrow]

Okay I think I got it, let me try doing it

Thanks

I solved it, thank you so much!

:D

.close

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[past violet]

Hi again, I am given a group (G, *) and e=1. assuming x*y*z=1, is y*z*x=1?

[past violet]

My thought process behind this is:

$e=x*x^-1$ where $x^-1$ is the inverse of x

[ocean seal]

sup sup

[past violet]

damn m formatiings doghsit

[exotic canopy]

x^{-1}

[past violet]

same goes for y and z

[exotic canopy]

only in an abelian group

or actually

[past violet]

with that said, if we take e=x*invx

that means that y*z is the inverse of x

[exotic canopy]

yeah that makes sense

[past violet]

with that said, either one of them can be invx and the other is e

same case for y and z

so in total, 6 cases

thats what i thought, but would that be a mathematically correct proof?

[exotic canopy]

you are correct then

[exotic canopy]

is the task to prove that?

[past violet]

task explicitly asks: if A, does B happen?

A and B being the conditions i gave

sadly stuff is in german so i cant just post

[past violet]

i mean, i thought about that because its commutative

but what if its not

[exotic canopy]

i feel like your argument holds and i can't seem to find a mistake so far

[past violet]

can i also send a picture of how i did the first 2 cases with X as the "real" variable

because i just started uni and i struggle to express everything in math notation

stuff is a lot more strict compared to highschool

[exotic canopy]

go for it

not sure about the (y ist x^-1 und z ist e) part

y*z can be x^-1 while neither y nor z are x^-1 and e

[past violet]

thats actually true

[exotic canopy]

you should just go off the argument that inverses are commutative

[past violet]

i didnt think of that

[exotic canopy]

$x*x^{-1} = x^{-1}*x = e$

[ocean seal]

artemetra

[past violet]

i get your reasoning

but

[exotic canopy]

even in non-abelian groups

[past violet]

how would that work if e=y*invy

[exotic canopy]

wdym

let's look at just $xyz=e$\
we can left-multiply by $x^{-1}$\
$x^{-1}xy*z=x^{-1}e$\
$yz=x^{-1}$\

[ocean seal]

artemetra

[exotic canopy]

does that make sense?

[exotic canopy]

and by this fact we essentially prove this thing

[past violet]

i get the idea, i was thinking about it this way: what if y is the "real" variable and x and z were the ones complementing it

i feel like that doesnt make too much sense

but i cant explain it

[exotic canopy]

i'm sorry but i am kinda not understanding what you mean by y is the "real" variable

lol

[past violet]

ah wait i was

still thinking of the

"either one can be inverse and the other is neutral", idea

[exotic canopy]

yeah that's wrong lol

i have counterexamples if you need

[past violet]

no i thought off one right of the top of my head when you mentioned it

it makes sense

an inverse can be a "product" of two other elements

doesnt have to be itsself and e

[exotic canopy]

6*1/3*1/2 = 1

[past violet]

exactly

[exotic canopy]

5+(-2)+(-3)=0

so yeah

thoughout your entire proof, you should consider y*z as one thing

it's there to spook you off 👻

[past violet]

we can also apply the same rule for x*y i guess?

since its commutative?

[exotic canopy]

wdym

if x*y=e, then y=x^-1

[past violet]

no

x*y=inv z

[exotic canopy]

yes but you don't have to do that

if you prove it just for x, you are done

like, completely

you don't even have to write without loss of generality

and you don't need all 6 cases

[past violet]

thats what i dont clearly get

how am i done just with x

wait lemme think about this

[exotic canopy]

because that's what the question is asking you do show lol

[past violet]

wait fuck

[exotic canopy]

x*(y*z)=1 implies (y*z)*x=1. that's it

all you have to show

you are overthinking it too much

[past violet]

it doesnt say that though

[exotic canopy]

what does it say

[past violet]

its x*y*z = 1 implies y*z*x=1

no parentheses

wait theyre associative

fuck

[exotic canopy]

yes

lol

[past violet]

i completely forgot

i was taking every single case assuming associativity wasnt even there

[exotic canopy]

i put them there for you to make it more clear that y*z is one thing that you shouldn't separate

[past violet]

if i just do it for 1 then its for all of them becasue theyre the same damn thing

[exotic canopy]

yes

[past violet]

yea thats too dumb of a mistake

lol

[exotic canopy]

personal question: did you sleep well last night?

[past violet]

kind of?

idk

[ivory pivot]

uuu very intimate

[exotic canopy]

lol

[past violet]

its my first ever excercise with group thery

[exotic canopy]

understandable

it gets intuitive after a while

[ivory pivot]

you ll never sleep as before when you start group theory

[past violet]

will group theory have me contemplating my life?

[ivory pivot]

yes it will

[exotic canopy]

yes

[past violet]

cool element of surprise

i gotta close this though, thanks a lot

[exotic canopy]

np

[past violet]

.close

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[thick lynx]

Find all pairs (a, b) with lcm(a, b) = 1500 and gcd(a, b) = 150

[thick lynx]

(I used a = 2^n * 3^m * 5^k, b = 2^n' * 3^m' * 5^k')

So I got these pairs

But their lcm is 300, so there is something wrong

Their gcd is 150, that's correct

@ivory pivot this is the problem from before

[ivory pivot]

it s becoming a nasty situation

$1500=10015=5^22^253$.. like you said

[ocean seal]

everg

[thick lynx]

yeah

50 = 2 * 5^2

So these max and mins should be correct

It's probably in the final step

[thick lynx]

Having these max and mins above, how would you construct the pairs?

[ivory pivot]

here the max should be 3...its actually 2

5-max

i have to go now 😭

[lone heart]

@thick lynx Has your question been resolved?

[thick lynx]

Thanks!!

.close

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[thick lynx]

.reopen

[lone heart]

✅

[thick lynx]

I have these pairs

But the gcd of the first would be 750, for example

the lcm 1500, correct

What's wrong?

[cosmic acorn]

Entschuldige die random frage, was isn das für ein thema?

oh english server

bruh

What is the topic of this question? I am interested.

[thick lynx]

elementary number theory (elementare Zahlentheorie)

[cosmic acorn]

Interesting (interessant xD)

[thick lynx]

Resolved it

.close

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[hidden plaza]

I am stuck on exercise 4 part b. I proved that the spaces are separate, but I can't find the linear decomposition ...

[lone heart]

@hidden plaza Has your question been resolved?

[tacit arch]

Show your proof that they're separate

[hidden plaza]

Let $f \in W_0 \cap \mathbb{R}_1[X]$. Since f(0) = 0, we have $f(x) = ax + b$ for some $a, b \in \mathbb{R}$. Since f(1) = 0, we have $a+ b = 0$. This means that $f$ must be the zero polynomial since otherwise it would not satisfy $f(0) = 0$. Therefore $W_0 \cap \mathbb{R}_1[X] = {0}$

[ocean seal]

mechap

[tacit arch]

That's almost the entire proof

You just need to identify that only linear function in W0 is the 0 polynomial

Oh you did that basically

The remaining is showing any continuous function can be written as a sum of W0 and R1[x]

[hidden plaza]

yes and this is where it gets tricky, because if I have f(x) = g(x) + p(x), p(x) always has a constant while g(x) must be the zero polynomial.

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[hidden plaza]

.reopen

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✅

[tacit arch]

W0 does not have to be a polynomial

Just continuous

e.g. sin(pi * x) is in W0

[hidden plaza]

but f has to be a polynomial

wait no

how do I know that I cover any continuous function in my sum ?

[lone heart]

@hidden plaza Has your question been resolved?

[tacit arch]

Maybe you could show they're orthogonal complements of each other

[hidden plaza]

I don't think it implies the sum covers all the space

Does it ?

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@hidden plaza Has your question been resolved?

[lone heart]

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[thick lynx]

Divide 15! + 200 with remainder by 182 without a calculator, 182 = 13 * 14.

[thick lynx]

We have that a = pm + q after the lemma of division of a by m with remainder

But that still doesn't help us in determining p and q

[tacit arch]

divide 15! + 200 by what?

[thick lynx]

by 182 = 13 * 14

[tacit arch]

oh

,w prime factorization 182

[ocean seal]

Results provided by WolframAlpha

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Requested by rie.mann

[tacit arch]

15! has all those as factors

[thick lynx]

Yeah, but we have a + 200 there

[upbeat gorge]

Then just get the remainder of 200 divided by 182

[tacit arch]

(a+b) mod n = a mod n + b mod n

[thick lynx]

So we have [\frac{15! + 200}{13 \cdot 7 \cdot 2} = 15 \cdot 12! + \frac{200}{13 * 14}.]

But 200/(13 * 14) is not in N

[tacit arch]

what

you're looking for the remainder

[upbeat gorge]

We’re dealing with remainders, not quotients

[thick lynx]

Not only

[tacit arch]

5 / 2 = 2 * 2 + 1, 1 is the remainder

[thick lynx]

It says "divide with remainder"

So we're looking for the whole thing, qm + r

[tacit arch]

you did the first part already

200 = n * 182 + r

find n and r

n in N, and 0 <= r < 182

[thick lynx]

Oh

[tacit arch]

you combine this form

[tacit arch]

with this fact

[thick lynx]

To find n and r?

[thick lynx]

How do we use this for 200 = n * 182 + r?

[tacit arch]

no

that's the second term. you should be able to do that without number theory

[hidden plaza]

Oops I forgot to close my question --'

[tacit arch]

think easier

[thick lynx]

Yeah, n = 1, r = 18

[tacit arch]

great you're done

[thick lynx]

So we're done, aren't we?

Yeah

Thanks!

.close

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[tacit arch]

plug that in there and finish the arithmetic

[thick lynx]

.reopen

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✅

[thick lynx]

Into 200?

But then won't we again have a quotient and not a remainder

[tacit arch]

wot

[thick lynx]

$= 15 \cdot 12! + \frac{182 + 18}{182} = 15 \cdot 12! + 1$, remainder $18$, right?

[thick lynx]

nvm

great

Thanks

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[jaunty knoll]

help

how do i solve this?

[slow haven]

is this relating to a course you've been doing? 🙂

[jaunty knoll]

idk

i am new

[slow haven]

haha, where did this question come from ? 😄

was it hwk or just you exploring maths

because i can think of one way to do this question but im not sure if you know the background to it

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[jaunty knoll]

what

[slow haven]

using a 2d rotation matrix

not the technical term for it..

does this look familiar to you or nah? 😄

[jaunty knoll]

wait

how do i create a channel

oh

[ebon sparrow]

#❓how-to-get-help

but you already made it by "what"

!original

[lone heart]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[jaunty knoll]

:C

[jaunty knoll]

i am so confuse

@lone heart

so

how do i solve this mathematical question
?

[ebon sparrow]

for E, locate the point on the x-axis and the point on the y-axis

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@jaunty knoll Has your question been resolved?

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[frail ravine]

answer key says i’m wrong, but i can’t find where i messed up 😦

[jagged cobalt]

one moment

what does the answer key say?

[frail ravine]

161

same question from the textbook + answer key

[jagged cobalt]

idk where they got that

i did it again using 0.5*b*h and you get the same thing

[frail ravine]

oh i see

[jagged cobalt]

(though ig its the same thing)

[frail ravine]

hodder tripping

[jagged cobalt]

161 is way out

[frail ravine]

alright thank you!

.close

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[brazen basin]

How can I simplify
cos(-x) + sin (-x) / cot (-x)
To a "single trigonometric function with no fractions"?

[brazen basin]

This is what I've done so far, but I don't know how to simplify it further

Initially I was thinking cos(x) + sin(x)tan(x) but that isn't right

[royal socket]

try continuing from the bottom right

cos(x) + sin^2(x)/cos(x)

[lone heart]

@brazen basin Has your question been resolved?

[brazen basin]

Yes but what do I do from there?

[royal socket]

Make a common denominator and see what you get

[brazen basin]

Oh

1/cos

Thank you!

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[alpine sable]

gödels argument pretty much relied on the idea that the set of characters is countably infinity and the set of proofs is uncountable, yeah?

[alpine sable]

like the the set of all possible proofs is the power set of all possible rules

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[amber violet]

Help

[tacit arch]

The relationship uses the fact lambda is an eigenvalue

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[lavish pier]

where do they get 3p/10 from??

.close

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[dusk umbra]

can anyone explain how bounds of infinitesimal change in z causes the delta function to disappear?

delta function in question

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@dusk umbra Has your question been resolved?

[tacit arch]

Show the entire page that includes all things referenced

[fleet rune]

guys help @everyone

[tacit arch]

!occupied

[lone heart]

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[dusk umbra]

sorry for colors, trying to highlight things as i read and parse them

[tacit arch]

They integrated it out

[dusk umbra]

oh yeah that makes sense

lmao

silly

thanks

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[waxen turtle]

I have no clue

I think A is one of them judging off those arc thingies

I'm not sure at all

I guess C so a bisector could go down the middle idk

And that would be perpendicular

Is that correct?

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[waxen turtle]

It's C right?

Since it's the bisector if we know one of those then the other will like be proven too idk

[finite flax]

it is not C

[waxen turtle]

😭 😭

[waxen turtle]

Why

[finite flax]

because what we need doesn't require DA congruent to BD

[opal stone]

it is A?

[finite flax]

which is not required

[waxen turtle]

Is it E

Cuz like

Idk 😭

[waxen turtle]

Oh yeahhh

[finite flax]

if this is the angle bisector and perpendicular bisector, then AD is congruent to AB

[waxen turtle]

So just A

Just A right fr

@finite flax

[finite flax]

yes.

[waxen turtle]

Ty

I misread B earlier😭

But I get it frfrfr

[finite flax]

if DA were congruent to BD, this would be an equilateral triangle.

which is not required.

[waxen turtle]

OK LAST PROBLEM

[finite flax]

we just need an isosceles for angle bisector to be perpendicular bisector.

[finite flax]

which homework module is this?

[waxen turtle]

Idk what a homework module us

Is

[finite flax]

what is the website called

[waxen turtle]

But this is the second to last unit in integrated math 1

[finite flax]

or the applet

[waxen turtle]

On Khan academy

Like proving congruence or something

Some questions are soo annoying (proofs 🥶 )

[finite flax]

looks like Zoe claimed that step 2 was given when it wasn't. (it's true though)

by vertical angle theorem

[waxen turtle]

Isn't it just the initial thingy rearranged fr

Idk tho

[finite flax]

I don't know what you mean

[waxen turtle]

So is its A right

[waxen turtle]

The letters fr

[finite flax]

"fr"?

[waxen turtle]

For real

😭 😭

But anyways it's A right cuz the reason isn't the correct reason

For step 2

[finite flax]

the problem is that Zoe assumed a congruence between angles was given when it wasn't. However, they are congruent by Vertical Angle (VA) Theorem.

[waxen turtle]

So it's A yes (plz confirm)

[finite flax]

and you should know what that is

so you can pass your test.

because it's really really

[waxen turtle]

( 🙏 )

[finite flax]

really really

easy

to understand

[waxen turtle]

So in other words

It's A

[finite flax]

yes.

it's A.

[waxen turtle]

WOOOOOOOO

Omg I hate geometry

[finite flax]

you should look at your notes

[waxen turtle]

Prob my second least favorite subject in math

[finite flax]

they help

[waxen turtle]

I don't have those

[finite flax]

why not

[waxen turtle]

Idk

Anwyays

[finite flax]

what do you do during class when everyone else is taking notes

[waxen turtle]

I have to do another geometry unit to get mastery for IM1😔 😔 😔

[waxen turtle]

Uhhh this is just online

Khan academy

Just

Watch vids

Then do the

Practice

I just skip to taking quizzes cuz it's faster

Mastery

Anyways

[waxen turtle]

Omg do u think I can do linear algebra after integrated math 3

[finite flax]

I teach high school math

and yes

[waxen turtle]

WHAT

[waxen turtle]

Omg like a real teacher

[finite flax]

but

you won't really have anything fun to do with it until you know calculus

and/or differential equations (which involves calculus)

[waxen turtle]

Calculus looks easy fr

but geometry 🥶

[finite flax]

I think a big part of the satisfaction some people feel in geometry comes from knowing they can prove things that look complicated at the outset

You will find more and more that your ability to solve a problem is going to depend a lot on recognizing what kind of problem you are dealing with and what tools you have to solve it

The most fun problems are the ones you have no idea how to classify, so you don't know right away how to solve it

[waxen turtle]

I hate proving things 😔

But it's ok

[finite flax]

yes, it is ok

[waxen turtle]

I will become 'him'

[finite flax]

?

I don't know that meme

[waxen turtle]

The one above all

[finite flax]

who is that supposed to be

[waxen turtle]

'him'

You use it when you think someone is overconfident

"bro thinks he's him"

[finite flax]

that must be a new one

[waxen turtle]

Fr

OMG r u like a real teacher

Like

An old person idk

[finite flax]

I'm not that old 👀

[waxen turtle]

Anything 30 and above is ancient

23+ is old

[finite flax]

>:P

[waxen turtle]

BYEYEYEY

.close

[finite flax]

laters

[lone heart]

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[little berry]

Why is it that the function $f\left(x\right)=\frac{x}{x^{2}+1}$ has two horizontal asymptotes, while the function $f\left(x\right)=\frac{1}{x^{2}+1}$ only has one?

[ocean seal]

Shinutsi

[little berry]

They have the same denominator

and afaik, we get the H.A. by equating the denominator to zero

[vale wigeon]

are you sure about this

[little berry]

about which part

[vale wigeon]

what are the two horizontal asymptotes of y = x/(x^2+1)?

[little berry]

oh wait I forgot the \sqrt

it should be y = x/\sqrt{(x^2+1)}, my bad

[vale wigeon]

no, that's for vertical asymptotes.

[little berry]

$y = \frac{x}{\sqrt{(x^2+1)}}$ and $y = \frac{1}{\sqrt{(x^2+1)}}$

[little berry]

ohhhhh

[ocean seal]

Shinutsi

[vale wigeon]

horizontal asymptotes, if they exist, have their $y$-coordinates equal to $\lim_{x \to +\infty} f(x)$ and $\lim_{x \to -\infty} f(x)$.

[ocean seal]

Ann

[little berry]

ohh yeahh

i switched things up

mb

thanks

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[faint monolith]

i think you use the remainder theorem (vaguely remember it and definietly forgot how to do it) like the x-a thing

other than that im completely stuck

[vale wigeon]

no

you in fact do not use "the x-a thing".

this is not about divisibility of polynomials at all.

do you know what an odd number is?

[faint monolith]

yeah

[vale wigeon]

ok, can you tell me in your own words what an odd number is

[faint monolith]

cant be divided by a even number without turning into decimals

[vale wigeon]

formally correct, but clunky and unworkable.

an odd number is a number expressible as 2n+1, where n is an integer.

[faint monolith]

oh

thats cool

[vale wigeon]

are you telling me this is your first time hearing about the proper definition of odd

[faint monolith]

yeah ive never seen that 2n+1 lol

[vale wigeon]

bruh what

wait, what level of education are you at

[faint monolith]

algebra 2/precalc doing it at the same time

[vale wigeon]

so like... highschool, or thereabouts

anyway, ok.

[rare gale]

here's how to start: let your first odd number be 2n + 1, the next one 2n + 3, then 2n + 5, see if you can figure it out from there

[vale wigeon]

do you know what consecutive odd numbers are

oh, spoiled.

[rare gale]

oops, sorry

[faint monolith]

oh

thanks i guess

[rare gale]

also, what do you think you need to arrive at to complete the proof? what would a remainder of 11 when divided by 12 look like?

in terms of a simple algebraic expression?

[rare gale]

^ ?

[faint monolith]

wdym

[rare gale]

well it's a bit hard to start off on this proof unless you know where you are heading

[vale wigeon]

do you know about remainders in ordinary number division

a = bq + r

[rare gale]

how will it look like to end up showing an expression has that remainder of 11 when divided by 12?

[vale wigeon]

this sort of shit

[faint monolith]

um actually never seen that

[rare gale]

example...if i am dividing 35 by 12 i get 2 and remainder 11

[faint monolith]

11/12??

[rare gale]

so i can simply write 35 = 2 * 12 + 11

[faint monolith]

oh

[rare gale]

more generally, ANY number n that has remainder 11 when divided by 12 will look like this: n = 12q + 11 where q is some integer

so basically what you want to show is that:

$(2n+1)^2 + (2n+3)^2 + (2n+5)^2 = 12q + 11$

[ocean seal]

Soosh

[faint monolith]

ohh

[rare gale]

for some q, so you would just manipulate stuff around algebraically

[vale wigeon]

you would want to manipulate (2n+1)^2 + (2n+3)^2 + (2n+5)^2, to be more specific.

and get it to look like 12 * (something) + 11

[faint monolith]

do i expand th ebrackets

[rare gale]

probably helps

i assume you mean the parenthesis

see if you can do the problem from here

[vale wigeon]

yes, do that.

[faint monolith]

i have to gtg rn but thanks for the help guys

[rare gale]

i didn't pick 35 on purpose as an example but i guess it was a relevant choice 😅

[faint monolith]

.close

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[gaunt solar]

Is there any way to find all the cube roots of a 2x2 identity matrix?

[gaunt solar]

I've found (cos(2pi/3), sin(2pi/3), -sin(2pi/3), cos(2pi/3)) and (0,1,-1,-1), but I'm not sure if there's a way to generalize the solutions

[lone heart]

@gaunt solar Has your question been resolved?

[gaunt solar]

.close

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[floral ermine]

How do I solve this

[violet urchin]

well lets go option by option

where is the vertex of the parabola?

[lone heart]

@floral ermine Has your question been resolved?

[slate rock]

@floral ermine

[floral ermine]

it's

(1,-9)

[violet urchin]

right

so first, where would the axis of symmetry be?

[floral ermine]

1

[violet urchin]

right

[floral ermine]

so the first option is correct

[violet urchin]

so is the first option correct?

yeah

[floral ermine]

yup

[violet urchin]

now is the parabola facing upwards or downwards

[floral ermine]

i don't know really

downards im assuming

[violet urchin]

well

[floral ermine]

since y = a|x-h|^2 +k and a is negative

?

[violet urchin]

is the coefficient of x^2 negative or positive

[floral ermine]

positive

[violet urchin]

is it?

look closely

[floral ermine]

yeah

[violet urchin]

try expanding out $-3(x-1)^2-9$ by hand

[ocean seal]

大野雄大 👻

[floral ermine]

oh wait

is

is h supposed to be x here

oh yeah

i forgot

[violet urchin]

?

do you know what the coefficient of x^2 is

[floral ermine]

no 😭

[violet urchin]

if we expand this out, we get $-3(x^2-2x+1)-9=-3x^2+6x-12$ right

[ocean seal]

大野雄大 👻

[floral ermine]

we didnt take that yet

in school

year 11 but my school is hard AS FUCK

[violet urchin]

you dont know how to multiply (x-1)(x-1)?

[floral ermine]

OH WAIT

I'M

SO

STUPID

Sorry

so i need to expand it?

right

one moment

[violet urchin]

yeah

[floral ermine]

-3x^2+6x-12

thats what i got

[violet urchin]

right

so now is the coefficient of x^2 positive or negative

[floral ermine]

negative

don't go crazy

i'm just dumb

whats the coefficient of x^2

so i can understand why

[violet urchin]

its the number in front of x^2

[floral ermine]

ahhh

okay

+1 info

so its positive

so its facing upwards

[violet urchin]

well the number in front of x^2 is -3 isn't it?

[analog arrow]

whats the coef of x^2 in (x^2-x^2)

[floral ermine]

oh i thought you meant infront of x^2

yeah its -3

[floral ermine]

x^2

[violet urchin]

0 💀

[floral ermine]

0

[analog arrow]

yes

[floral ermine]

yeah idk

anyways

its -3

[violet urchin]

anyways

right

[floral ermine]

so yeah negative

[violet urchin]

its -3 so its negative

which means its facing which way

[floral ermine]

down

[violet urchin]

correct

which means teh vertex is the maximum or minimum

[floral ermine]

max

bc its the top

top top

[violet urchin]

right

[floral ermine]

yeah

[violet urchin]

so then that's 3 isn't it?

[floral ermine]

going back here

[analog arrow]

ye

[floral ermine]

it opens upwards

is wrong

[analog arrow]

yes

[violet urchin]

yes

[floral ermine]

it opens downards is right cuz we just said that

[violet urchin]

it opens downward

[floral ermine]

vertex is -1, -9 wrong

[violet urchin]

like we said

[floral ermine]

it's 1, -9

[violet urchin]

correct

[floral ermine]

vertex is highest point yup

now

the 5th one

i dont get

cuz its right

y is -9

[violet urchin]

well not quite

[floral ermine]

why not

[violet urchin]

because if the maximum y value is at the vertex

and the vertex is at y=-9

[floral ermine]

OH

[violet urchin]

then how would we get values of y higher than -9?

[floral ermine]

so y CANT be greater than 9

[violet urchin]

right.

[floral ermine]

and xi s the symmetry

there w ego

now i understand

i wanna give you a big kiss

thanks folks

wait lemme see before i go

this is

the same thing

right

except i have to remove

[violet urchin]

well ok

[floral ermine]

the 2 from x-h

right

[violet urchin]

lets just check this

is there any reason the domain wouldn't have any of the real numbers?

[floral ermine]

nope

cuz you have

+3 right

[violet urchin]

+3?

[floral ermine]

nah

[violet urchin]

mhm

[floral ermine]

the domain is

[violet urchin]

so we know that 2 is wrong

[floral ermine]

3 here right

[violet urchin]

and 1 is right

[violet urchin]

the domain is 3?

[floral ermine]

what we took

is

h, k

x, y

h = x and k = y

in this atleast

[violet urchin]

do you know what domain means?

[floral ermine]

yeah

damn really hurt

domain is the set of possibilities u can put in x basically

right?

[violet urchin]

yeah its the values that x can take

[floral ermine]

here x can takeanything

theres no restrictions

so yeah its right

[violet urchin]

right

so 1 is right, 2 is wrong

[floral ermine]

yup

[violet urchin]

now what do we know about the range?

[floral ermine]

its 5

[violet urchin]

the minimum or maximum is 5?

[floral ermine]

i dont really know

maximum probably

[violet urchin]

well

what is the coefficient of x^2

positive or negative

[floral ermine]

positive

[violet urchin]

right

so is the parabola facing up or down

[floral ermine]

up

OH

OHOH OH

brodie turned me into santa

holy shit

[violet urchin]

right, so is the vertex the minimum or maximum

[floral ermine]

minimum

fo sho

[violet urchin]

if its facing up

[violet urchin]

yes

[floral ermine]

so 3rd and 4th are wrong

5th is right

[violet urchin]

exactly

[floral ermine]

dude

your gonna be the reason i dont fail ym exam

do you have time or nah

[violet urchin]

uh

how much time

[floral ermine]

well

any amount

💀

[violet urchin]

sure

[floral ermine]

alright

awesome

now we got

this bad boy

[violet urchin]

well

how do we calculate distance?

or actually we dont need tha i misread

that solution looks right

[floral ermine]

it is right

but

idk how

[violet urchin]

oh

ok so basically

[floral ermine]

hollon what

we went into physics

😭

[violet urchin]

ok so

do you know the pythagorean theorem? sorry i have to check

[floral ermine]

yeah

hurt my feelings

[violet urchin]

sorry sorry

i have to check

[floral ermine]

a^2 + b^2 = c^2

watch this be wrong

[violet urchin]

so if we have two points (a,b) and (c,d)

[violet urchin]

good

[floral ermine]

got me scared

[violet urchin]

if we have these two points

[floral ermine]

yeah

[violet urchin]

then we can make a right triangle

with the hypotenuse being the line between them

[floral ermine]

UH OH

HOLLON WHAT

i dont know how to make triangles

[violet urchin]

uh

lemme open desmos

ill show you

[floral ermine]

alr

[violet urchin]

look at this

[floral ermine]

yeah

[violet urchin]

we took the points (1,2) and (4,5)

and we made this right triangle with the horizontal distance and vertical distance and the line between them

does that make sense

[floral ermine]

not really

wait

yeah maybe

it does

[violet urchin]

alright so

we want to find the distance between the two points

which is equal to the length of the hypotenuse right?

[floral ermine]

uh

yeah

it is

[violet urchin]

and we can pretty easily find the lengths of the two legs

since they're just 4-1 and 5-2

right

[floral ermine]

i

dont know

how to do that

but

im assuming its just a right triangle

so ill just connect them

to a right angle

easy

[violet urchin]

hmmm

i guess that should work

so then lets say the distance between teh points is d

then we know that d^2=(length of the horizontal side)^2+(length of the vertical side)^2

[floral ermine]

AAAH

WHAT

[violet urchin]

its just the pythagorean theorem

[floral ermine]

oh

yea!

[violet urchin]

so then d^2=(4-1)^2+(5-2)^2

and that simplifies to 3^2+3^2=18

and then we have d^2=18

and $d=\sqrt{18}$

[ocean seal]

大野雄大 👻

[violet urchin]

i know that was a little fast but did you get it?

[floral ermine]

not quite

this is too much for my tiny brain

can we go through a different route

[violet urchin]

well i mean

for you to understand the distance between points

you kind of need to know how distance between points works

[floral ermine]

waitttt

[violet urchin]

i can give you a khan academy article/video if that would work better

[floral ermine]

I KNOW WHY

I STUDIED THIS BEFORE

its just a law that we took

when we have

2 lines and something left out in the middle

we add a line in the middle so we have them parralel

smt like that

does that work

[violet urchin]

uh

can you explain to me how that would help you find distance?

im not following sorry

[floral ermine]

it wouldnt

its fine

but

we have this law right

hold on let me draw

it wouldnt help u find the distance

but

if we have

2 lines

and left out space in the middle

we add a line

thats just the locus

of those 2 lines

the distance between them

thats how my school takes it atleast

[violet urchin]

so youre saying that if you have two parallel lines

the line that's parallel to them and halfway between them is equidistant?

[floral ermine]

yup

that'd be the locus

for the 2 lines

[violet urchin]

yeah that works for this problem i suppose

[floral ermine]

yeah well

for

THAT problem

[violet urchin]

its not super generalizable but it works well enough here

[floral ermine]

this is different