help-19

[narrow crypt]

u can use derivative to study the behavior of a function

[daring frost]

Just write it in form a(x-x_1)(x-x_2) and then you know?

[narrow crypt]

ye

i sent op on the journey to find how that method works already

[mystic nova]

ofc I know, but sometime we have irrational roots

That's painful to factor

[narrow crypt]

completing the square > quadratic eq

[tepid beacon]

math is such an adventure 💀😭 , thanks everyone

.close

[odd edge]

Channel closed

Closed by @tepid beacon

Use .reopen if this was a mistake.

[narrow crypt]

u got it??

[tepid beacon]

yes

[odd edge]

Available help channel!

Send your question here to claim the channel.

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

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

[edgy holly]

How to simplify...

[edgy holly]

componendo Dividendo and cubc both sides

[mystic nova]

It looks like $(a+b)^3$ expansion

[clever fjord]

Alexis_Fx

[edgy holly]

Also tried to form perfect cube

[nocturne belfry]

[edgy holly]

not helping, I dunno how to foorm tho

[narrow crypt]

[ 90\sqrt{6} = 90\sqrt{3}\sqrt{2}]

[clever fjord]

k

[narrow crypt]

by guessing its prolly smth like ${(a\sqrt{3} + b\sqrt{2})^3}$ = [3a\sqrt{3} + 2b\sqrt{2} + 3a^2b\sqrt{2} + 3ab^2\sqrt{3}]

[edgy holly]

the term with $\sqrt{2}$ and $\sqrt{3}$ are the $a^3+b^3$ and $a^3-b^3$, and rest are $3a^2 b$ and $3ab^2$

[clever fjord]

Double_mytrouble

[edgy holly]

I dunno what to do next

solving it becomes complicated

[clever fjord]

k

[narrow crypt]

hmm its not actually perfect ye

lemme try

[edgy holly]

umm, no solutions are coming

it could also be $(a\sqrt{2}+b\sqrt{3}+c)^3$ type

[clever fjord]

Double_mytrouble

[narrow crypt]

fair

[waxen talon]

You should note that $\sqrt{6} = \sqrt{2}\sqrt{3}$

[clever fjord]

1 divided by 0 equals Infinity

[mystic nova]

already did man

[edgy holly]

how does that help

[waxen talon]

Then i think like $a = \sqrt{2}$ and $b = \sqrt{3}$

[clever fjord]

1 divided by 0 equals Infinity

[edgy holly]

?

[tacit haven]

offtopic but whys the file name whatsapp_image

[edgy holly]

are you suggesting to transform the equation taking a and b as variable?

[edgy holly]

I sent it to my whatsapp and downloaded it so I can post it from my laptop

I can't take pic from my laptop

[edgy holly]

or atleast clear ones

[waxen talon]

Yeah

[tacit haven]

,w expand (a + bsqrt(2) + csqrt(3))^3

[clever fjord]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by copterforgor

[narrow crypt]

💀

[edgy holly]

bro 💀

[tacit haven]

m-match coefficients...

LMAO

[edgy holly]

I ain't solving that shit

[tacit haven]

fair

[edgy holly]

<@&286206848099549185>

[desert jolt]

Are m and n integers?

[edgy holly]

yeah

final answer is a positive integer between 00 and 99

[lean yew]

oh my god

this is gonna be ugly

[lean yew]

but this essentially is what you need

[narrow crypt]

i give up

[mystic nova]

I don't think so, the denominator is an integer which means the numerator must somehow cancel it out and I don't think it can with that

[lean yew]

i believe once you've done all the stuff and did some rationalization

it should work

[narrow crypt]

holy

WA can solve it

[narrow crypt]

it is indeed this

[edgy holly]

This is a olympiad question and time given for it is 6 minute....

[hidden token]

Were there options?

[edgy holly]

I don't think comparing coefficient is the way

[edgy holly]

no

olympiad exams are subjective

[narrow crypt]

there must be a way..

how is the solution so cute

but this thing so ass

[edgy holly]

average olympiad moment

Should I close it?

[narrow crypt]

wait

[desert jolt]

this is an oly question?

doesnt look like one

[edgy holly]

yeah, first level of Indian olympiad easiest level...

here its like IOQM-->RMO-->INMO-->IMOTC-->PDC-->IMO. IOQM has 3 sections, easy medium and hard, this is the easy one

[narrow crypt]

ye i cant solv eit

[mystic nova]

I thought "I will solve this last problem and then sleep" and now I'm still trying

I give up

[narrow crypt]

lmao

the so-called "6 minute"

[brittle plinth]

cố lên em

a ngủ đây

[odd edge]

@edgy holly Has your question been resolved?

[desert jolt]

which year PRMO was this

interesting question

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[random pumice]

Problem 1.7 (Greece Junior MO 2015). Find all values of the real parameter a, so that
the equation x2 + (a−2)x−(a−1)(2a−3) = 0 has two real roots, so that one is the
square of the other.

[random pumice]

Progress: D>0 and vieta

[narrow crypt]

!show

[odd edge]

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

[random pumice]

Wht to do next

[shut nymph]

Solve the inequation

[random pumice]

Then

[shut nymph]

then you're done

oh wait

[random pumice]

?

[shut nymph]

i didnt see the second part of the problem lol mb

First of all i think you made a mistake in your polynomial with a

I get 9a^2 - 24a + 16

So you can start by giving the condition you found on a for there to be 2 real roots, then work from there

[random pumice]

I think i solve this problem

Subs a = 2-n2-n

Use rational root theorem

(r+1 and r-1/2 are roots

[odd edge]

@random pumice Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[errant inlet]

For this question, Im having trouble finding the solution for x(0). I understand how to find the other missing eigenvector (In this case its [1,1,1]), and my understanding of how to find x(t) is $$x(t) = \Phi (t) \Phi (0)^{-1} x(0)$$. Yet, with the way the fundamental matrix is created here, its impossible to invert and thus, obtain $$\Phi (0)^{-1}$$. Im not sure on what to do next here

[clever fjord]

Emperor Voldy

[odd edge]

@errant inlet Has your question been resolved?

[odd edge]

@errant inlet Has your question been resolved?

[sharp oak]

x(t) = Σ C(eigenvector)e^(λt)
Or something like that

Oh, that's only if the matrix has n linearly independent eigenvectors though

[odd edge]

@errant inlet Has your question been resolved?

[errant inlet]

What if they aren't linearly independent?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[sharp dew]

How do i solve this? Super lost

[sharp dew]

honestly failed this class like 42 times i cannot.

I respect and hate math

[nocturne belfry]

sorry whats the question?

[sharp dew]

SORRY height of the tree

3. Find the height of the tree in the diagram below to the nearest foot.

[nocturne belfry]

sure, what have you tried

[sharp dew]

i calculated all the angles, Dont know how to get the full base since i only have the base of one triangle

also just whole i have help, Is the angle behind 75 just 105?

[nocturne belfry]

it is, yea

[nocturne belfry]

this is fine, you can name it something else

[restive fern]

Well, for a 45 45 90 triangle the opposite and adjacent must be the same so we just find the value where decreasing the adjacent by 20 gives us a 75 15 90 triangle, that's the best I can do because I suck at trig

[sharp dew]

Just got an idea, Do i calculate the hypotenuse of the smaller triangle, then get the height?

since i have 3 angles and a side?

[restive fern]

Could

Use law of sines or something

[sharp dew]

i consider myself mathematically illiterate, i'll be back with an attempt lol

[restive fern]

Lol

[nocturne belfry]

you draw all the angles, and label all the important quantities

then you identify how many unknowns you have, and search out that many equations

[nocturne belfry]

yea thats what i get

I got $20 + x = h$

[clever fjord]

jan Niku

[nocturne belfry]

and $\frac{x}{\sin 15} = \frac{h}{\sin 75}$

[clever fjord]

jan Niku

[nocturne belfry]

by substitution, $h = \frac{20}{1 - \frac{\sin 15}{ \sin 75} }$

[clever fjord]

jan Niku

[odd edge]

@sharp dew Has your question been resolved?

Channel closed

Closed by @sharp dew

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[glossy marsh]

back at it again

[glossy marsh]

asking the same dumb question

(me)

[steep mantle]

?

[glossy marsh]

crap im sorry im forgetting how to word it

[steep mantle]

take your time

[glossy marsh]

#help-31 message

copied: yall this is a hobby thing ive already talked abt here months ago but im trying to make a cheat sheet where (a√5 + b) can be expressed as n1(√5 + 1)^n2

for example (4√5+9) = ((√5+1)^6)/64

i have no way of telling if (√5+4) can fit this pattern

aside from brute force

so far it has its own category

(√5-7) = -(√5+4)8/((√5+1)^2)

[glossy marsh]

this is the question

everyone reading this

oh wow theres very few help channels rn
well ig that makes sense
it aint school season

[steep mantle]

n_1 at the start allows you to just consider the ratio between a and b

[glossy marsh]

what

[steep mantle]

im p sure you can expand it into a sum of binomials that represent a and b, and using some algebra you can simplify it

[glossy marsh]

i did graph it but it just made shit harder

with n1 being x and n2 being y

was (√5 + 4) = n1(√5 + 1)^n2

to solve for the 4 one

oh

ohh

wait whyd you delete that

i thought you meant between n1 and n2 and i was like no?

[steep mantle]

so like, $\frac{a}{b}=\frac{\sum_{i=0}^{\floor{\frac{n_2-1}{2}}}5^{\floor{\frac{n_2-1}{2}}}\binom{n_2}{2i+1}}{\sum_{i=0}^{\floor{\frac{n_2}{2}}}5^{\floor{\frac{n_2}{2}}}\binom{n_2}{2i}}$

if im not mistaken

[glossy marsh]

i
i dont know calculus

[clever fjord]

skissue.in.a.teacup

[steep mantle]

theres no calculus here

[glossy marsh]

what is the sigma

[steep mantle]

sum

[glossy marsh]

i dont know how to read that

[glossy marsh]

oh right

[steep mantle]

actually hold on lmao

is n_1 a real number?

[glossy marsh]

sum of... i = 0 and (n2-1)/2???????

whAT

[steep mantle]

sum of f(i) from i=0 to floor((n2-1)/2)

[glossy marsh]

n1 or x is any positive intiger

as is n2 or y

[glossy marsh]

the hell

[steep mantle]

if n1 is a real number you literally can just set n2=1 and n1 will do the rest

[glossy marsh]

no

[steep mantle]

ie $\sqrt{5}+4=\frac{\sqrt{5}+4}{\sqrt{5}+1}(\sqrt{5}+1)^1$

[glossy marsh]

if n2 = 1, n1 cant be an intiger

[clever fjord]

skissue.in.a.teacup

[glossy marsh]

i checked

[steep mantle]

does n1 have to be an integer?

[glossy marsh]

i brute forced sqrt5 + 4 a lot

[glossy marsh]

yea i just said that

[steep mantle]

didnt see that

[glossy marsh]

right here
well ok it dont need to be positive

sometimes these things get a random negative

but when i plugged in the equasion,

it showed no negative x
and i just assumed

i realise now that thats wrong

idk where it went wrong

but x might be negative

[steep mantle]

a lot of a,b arent possible to represent like that then

[glossy marsh]

basically im factoring (√5 +4) with (√5 + 1) and √5 as potential factors

[steep mantle]

also

[glossy marsh]

i will be doing this for everything on my dumb cheet sheet

[steep mantle]

"say for example (√5+3) = ((√5+1)^2)/2"
you used n1=1/2?

[glossy marsh]

oh shit youre right

i expressed it wrong

i guess um

[steep mantle]

n1 is rational then?

[glossy marsh]

the only irrational numbers allowed are sqrt5 and sqrt5 + 1

[glossy marsh]

yes

[steep mantle]

then this should work if i didnt crunch the numbers wrong

[glossy marsh]

N2 needs to be an intiger, and n1 is rational, and not 0
thanks for pointing that out

[glossy marsh]

i cant read it but thank you

[steep mantle]

honestly this is so arbitrary, whats the cheat sheet for?

[steep mantle]

tldr you make the binomial and split the odd terms and even terms

[glossy marsh]

i stopped doing math at precalc

everything else i know is bc i researched for fun

[steep mantle]

everything i used should be precalc

[glossy marsh]

god

[steep mantle]

every terminology i used should be precalc, the usage might not be specifically

[glossy marsh]

i dont know how to read this

[steep mantle]

google sigma notation

[glossy marsh]

😭

oh my god this is so dumb um

so its uh

n plus itself times the thing next to it capping at the thing atop it?

i hate this

i got it yay

although im only 2 minutes in :devestation: https://youtu.be/xavgv1m9feE

oh wait what

no but

but

no

he did it wrong i think

5
Σ n^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2
n = 1

OH

OH

6
Σ 2^n =
n = 1

does not mean 2^1 + 4^2 + 6^3
no
bc only n is changing and 2 is not n

oh god

wait

is the top "until n is 6" or "do it 6 times" @steep mantle

[north sparrow]

from n=1 until n=6 by increment of 1

[glossy marsh]

k thanks

what if n = 2

than itd be 3 things adding?

[merry finch]

if it says n = 2 then it does it 5 times

it'll add up when n = 2, n = 3, n = 4, n = 5, and n = 6

[glossy marsh]

what

[merry finch]

$\sum_{n=2}^6 f(n) = f(2) + f(3) + f(4) + f(5) + f(6)$

[clever fjord]

frosst

[glossy marsh]

the fuck

[merry finch]

$\sum_{n=2}^6 n^2 = 2^2 + 3^2 + 4^2 + 5^2 + 6^2$

[clever fjord]

frosst

[merry finch]

here f(n) = n^2

[glossy marsh]

ok so i think i can start to apply this

[glossy marsh]

@steep mantle

is this supposed to be (√5 + 4) = n1(√5 + 1)^n2

[loud mural]

did you guys try factoring it?

[glossy marsh]

adam fuck off

[loud mural]

??

why the attitude

[steep mantle]

no its general

[glossy marsh]

wym

(a√5 + b) = n1(√5 + 1)^n2?

[loud mural]

"wym" refine your choice of words

[glossy marsh]

@short seal maybe you could be of help

[loud mural]

@short seal can you help

[glossy marsh]

have i done this correctly so far

i dont know what the L brackets mean nor how to apply them

[steep mantle]

L brackets are floor

[glossy marsh]

i cant figure out how to express this in desmos

[quasi sparrow]

https://help.desmos.com/hc/en-us/articles/33685875217677-Summation

[glossy marsh]

i think if there is a part explaining how to express floor like that its this but eh

i think it means a and b

[quasi sparrow]

Yes try googling floor desmos

[glossy marsh]

this is stupid

[bitter lodge]

i think im nearly at a proof to say that not all values of a/b are possible

[glossy marsh]

i know i can have floor(input) but i dont know enough abt this to apply it

[glossy marsh]

not all combinations are

[bitter lodge]

is that not the goal of this help channel ?

[glossy marsh]

well that and like a has to be rational and b needs be a whole number but can be 0 or negative

[glossy marsh]

the goal is to find out how to tell how to factor all instancesx of (a√5+b) considering (√5+1) is a factor and so is √5

although i am pretty freaking sure (√5+4) cannot be expressed as such

but i can only make progress via brute force

[glossy marsh]

im still confused bc i cant read this but ive made progress

skissue made that equasion

[glossy marsh]

i have a cheat sheet

a um

[quasi sparrow]

That's not informative

[glossy marsh]

my brain wont retreive the word

[glossy marsh]

i know i fergot the word

spreadsheet

[quasi sparrow]

That's not informative either

[glossy marsh]

i have a spreadsheet of different values of a and b for this

[quasi sparrow]

Can you state your goal if it's not this

[steep mantle]

cant you like a=1 b=0

[glossy marsh]

stop it i know im tryoing im just slow

i know that

im trying to get the information out

i know that

[quasi sparrow]

Feel free to type only when you've understood what you're trying to say

[glossy marsh]

id show it here but adams here

[quasi sparrow]

Adams?

[glossy marsh]

adam is here

[quasi sparrow]

Is that supposed to mean something

[glossy marsh]

dont be a dick

im trying my best

[glossy marsh]

are you stupid

[quasi sparrow]

Alright this wasn't fun

<@&268886789983436800>

[glossy marsh]

oure telling me to not talk unlress i can be 100% cohesive but you have no idea how names work

[meager juniper]

@glossy marsh I'm going to need you to calm down by a lot please.

!vol

[odd edge]

Helpers are just people volunteering their time to help you. Be polite and patient.

[glossy marsh]

riemans being rude though
and why is the bot a she they what

[meager juniper]

let me read up and check

I've read the above, and riemann isn't reading as rude, just a bit blunt. It's a tendency with people involved in math to state things really quite bluntly, mostly because clarity of communication is valued. I don't doubt that he was being sincere and trying to clarify things with you.

Please don't take it personally

riemann, please be a little more charitable in the future. This sort of back and forth can be avoided.

[quasi sparrow]

Alright man I get called stupid and I'm the one being asked to change?

[meager juniper]

The particular message where I feel you could have been a little nicer on was prior to that.

but it's just a minor thing

[quasi sparrow]

Wut

[meager juniper]

This one.

Like, I don't doubt that you meant well.

[carmine idol]

i'm seeing two people prioritizing two different conversational norms

[meager juniper]

Same!

I get it!

[quasi sparrow]

I don't know how that addresses what I said here

[meager juniper]

I'm asking you both to change, and not by the same amount?

OP is clearly in the wrong here.

anyway, this is an interesting problem.

[glossy marsh]

im not saying anything bc i dont know how to appropriately acknowledge this

[carmine idol]

yeah i think so too

might need to talk about fields and stuff

[meager juniper]

I wonder if there's a closed form for the summation of every other binomial coefficient.

[carmine idol]

I think there is

just expand $(1+1)^n+(1-1)^n$

[clever fjord]

Element118

[glossy marsh]

what 😭

[meager juniper]

This is just an informal warning, I'm not even noting it down. It's really not a big deal.

[carmine idol]

yeah i also think it's not such a big deal, after all, this place is to get help

[glossy marsh]

regardless if i persist with "but they did blah and its ok i did blah" thatd make shit worse

thats all

[meager juniper]

With help threads, sometimes frustrations can mount, and all that's really required is a reset, and that's fine.

[glossy marsh]

why are you oganession

[carmine idol]

I'm thinking we can make use of the norm in $\mathbb{Q}[\sqrt{5}]$ (assuming we haven't yet)

[clever fjord]

Element118

[carmine idol]

i used to be, and it stuck around

it might be time for a rebrand

[meager juniper]

oh, so the summation is apparently 2^(n-1)? Even on odd rows?

[glossy marsh]

thar be this
graphing it made brute force harder
#help-31 message

[meager juniper]

that makes a certain amount of sense.

[glossy marsh]

i dont list obvious factors, like (2√5 + 2) in the cheat sheet

im debating not listing thngs that include sqrt5

[carmine idol]

but basically the norm in $\mathbb{Q}[\sqrt{5}]$ is $a+b\sqrt{5}\mapsto a^2-5b^2$, which for $\sqrt{5}+4$ is $16-5=11$, since 11 is prime, $\sqrt{5}+4$ only has units and associates as factors.

[clever fjord]

Element118

[meager juniper]

slick

[glossy marsh]

no idea what this means

[carmine idol]

don't worry, we'll get there at your pace

The norm is $(a+b\sqrt{5})(a-b\sqrt{5})$. Try expanding that.

[clever fjord]

Element118

[glossy marsh]

also this is the main reason im doing this
lots of shit youd have no idea you could simplify
like how am i sposed to know (3√5-7) = -32/((√5+1)^4)

[glossy marsh]

i got major brainfog im sorry ppl
what do you mean by norm

[carmine idol]

okay sorry i didn't cover the prerequisite field theory

[glossy marsh]

i mean like fractions and square roots involving asqrt5 + b

in really large things

[carmine idol]

so what happens is that $\mathbb{Q}[\sqrt{5}]$ is a field containing $\mathbb{Q}$. It's a two dimensional field extension of $\mathbb{Q}$.

An analogy is how $\mathbb{C}$ is a two dimensional field extension of $\mathbb{R}$

[clever fjord]

Element118

[glossy marsh]

😭

[carmine idol]

two dimensional, as in, it's also a two dimensional $\mathbb{Q}$ vector space

[clever fjord]

Element118

[carmine idol]

okay maybe i didn't recurse far enough with the explanation

let's try again

[meager juniper]

I don't think nicole is familiar with "fields" or "vector spaces" perhaps

[carmine idol]

yeah skill issue on my end

[glossy marsh]

probably not no

[carmine idol]

You know about rational numbers right

we can add subtract multiply divide-by-nonzero

[glossy marsh]

yea

[carmine idol]

we call this a field

[meager juniper]

any set of numbers where we can do all of thing things you'd normally expect out of numbers (add, subtract, multiply, divide) and get back the same sort of number, we call a field.

So rationals are fields, because when you apply these operations to them, you always get back other rational numbers.

[carmine idol]

[okay vector spaces might be important, maybe we cover that too]

[meager juniper]

But integers are not a field, because if you divide two integers you might not get an integer back

[glossy marsh]

weird ok

[carmine idol]

so if you have a field $\mathbb{Q}$, we can get a vector space $\mathbb{Q}^n$ (read: n-tuples of rationals)

[clever fjord]

Element118

[meager juniper]

we care about fields, because if we have something that isn't exactly the numbers as we're used to them, but they have the above behaviors, we can do algebra on them the way we would normally expect.

[carmine idol]

Example time:
$(1/2, 1/3)\in\mathbb{Q}^2$

[clever fjord]

Element118

[carmine idol]

two rationals, it's a vector from $\mathbb{Q}^2$

[clever fjord]

Element118

[glossy marsh]

so like
shit im confused as to how my thing works now

[carmine idol]

now, with vectors we can do cool stuff, like adding entry-wise

[meager juniper]

I'll let Og explain.

[glossy marsh]

what 😭

[carmine idol]

$(1/2,1/3)+(1/4,1/5)=(3/4,8/15)\in\mathbb{Q}^2$

[clever fjord]

Element118

[glossy marsh]

i think i need to go eat and get water

[carmine idol]

yeah go get some water

stay hydrated

Also, you can multiply by a rational multiple, term-wise:

$6\times(1/2, 1/3)=(3,2)$

[clever fjord]

Element118

[carmine idol]

huh, to talk about norms i might need matrices and determinants to explain why they work

[meager juniper]

yeah, it's pretty difficult to go from high school math directly to field extensions

[carmine idol]

lemme see if there's a shortcut

i can basically replace it with an assertion that $a+\sqrt{5}b$ and $a-\sqrt{5}b$ are in some sense "algebraically indistinguishable"

[clever fjord]

Element118

[carmine idol]

in some sense, we are just treating $\sqrt{5}$ as a symbol instead of the usual value of $2.236$-ish

[clever fjord]

Element118

[carmine idol]

!xy

[odd edge]

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.

[carmine idol]

though

[carmine idol]

because the main aim might be this

[odd edge]

@glossy marsh Has your question been resolved?

[glossy marsh]

well i mean i know high school math
and a bunch of random other shit
i know lambda calc
and todat learned how to use sigma

[meager juniper]

knowing about lambda calculus but not knowing about summation notation is a pretty odd progression, ngl

[glossy marsh]

no this is a whole new project at this point

[glossy marsh]

yea

i learned lambda calc from this and the guy it coulda done a better job explaining why the false and true function work
https://www.youtube.com/watch?v=RcVA8Nj6HEo

had to watch it 3 timrs and still didnt get it

[bitter lodge]

ok, heres my elementary approach, skipping over a lot of algebra and etc, just giving an overview \
we set up the equation $\sqrt{5} + q = a (\sqrt{5} + 1)^k$, where $a, q \in \mathbb{Q}$ and $k \in \mathbb[N}$ \
when you do binomial expansion on the RHS, half of the terms will be a multiple of rt5, the other half will be rational\
now, in our equation we already know the coefficient of rt5 from the LHS, its 1. this allows us to find the value of $a$ in terms of $k$ \
this in turn gives us an expression for $q$ in terms of $k$

as a quick aside, i've done the math with a variable $x$ in place of the $5$ and that's what ill be using from hereon. the only conditions are $x > 1$ and non-square. i also define $r = \left(\frac{\sqrt{x} + 1}{\sqrt{x} - 1}\right)^2$ for convenience
\begin{align*}
q_{2k} &= \frac{\sum \limits_{i=0}^k {{2k} \choose {2i}} x^i}{\sum \limits_{i=0}^{k-1} {{2k} \choose {2i+1}} x^i} = \sqrt{x} \frac{r^k + 1}{r^k - 1}\
q_{2k+1} &= \frac{\sum \limits_{i=0}^k {{2k+1} \choose {2i}} x^i}{\sum \limits_{i=0}^k {{2k+1} \choose {2i+1}} x^i} = \sqrt{x} \frac{r^{k + \frac{1}{2}} - 1}{r^{k + \frac{1}{2}} + 1} \
\end{align*}

at this stage, i did a ton of algebra that couldve been avoided, anywho the result was $q_{2k}$ is a decreasing sequence with limit $\sqrt{x}$ and $q_{2k+1}$ is an increasing sequence with limit $\sqrt{x}$. these results are enough to rule out a lot of possibilities for $q$ (and thus prove that not all values of $q$ are possible) \
you can go one step further and find the inverse functions of the above to determine if any particular value of $q$ is achievable

\begin{align*}
\text{for } q > \sqrt{x} \text{, if } &\frac{\log \frac{q + \sqrt{x}}{q - \sqrt{x}}}{\log \frac{\sqrt{x} + 1}{\sqrt{x} - 1}} \text{ is an even integer, or}\
\text{for } q < \sqrt{x} \text{, if } &\frac{\log \frac{ \sqrt{x} + q}{\sqrt{x} - q}}{\log \frac{\sqrt{x} + 1}{\sqrt{x} - 1}} \text{ is an odd integer}
\end{align*}
then $q$ is achieved with that integer

[clever fjord]

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

[glossy marsh]

cant find a subtraction nor pred function that work but i think its half skill issue half bad resources

[bitter lodge]

the problem is fun and id recommend you try to go through it yourself, you can use that overview as a guide for hints / to check

if a mod or whoever has a problem that thats just the answer laid out then feel free to delete, imo this is the kind of instance where its reasonable to do so

[glossy marsh]

well im glad you like my dumb problem
i still cant do like 3 or 4 kinds of functions being used here

[bitter lodge]

just clarify at the bottom by "q is achieve by that integer" i mean k = that integer will give that q (with the appropriate value of a)

[carmine idol]

alright then, if you want i can try to bring you towards the field theory that you can use to check how to factor numbers of the form $a+b\sqrt{5}$

[clever fjord]

Element118

[carmine idol]

but it's definitely going to take a while

[glossy marsh]

well thanks

um

i cant think words

oh right yea
was going to say i dont think ill need this nearly as frequently as i used to

... maybe ill make a rotatable icosahedron next. :3c

[carmine idol]

ah okay

[glossy marsh]

dont think i said it but i did get water and coffee and took care of myself and stuff

yay

[carmine idol]

yay

[glossy marsh]

@carmine idol @meager juniper @bitter lodge ill delete this in a sec but
the colored regions are known to be of the same family of things, the red ones definately fit the rule, the gold ones fit the rule if we include sqrt4 + 5, and the others i havent pried into yet
the dark ones have common factors and as such arent worth listing, and grey is "i havent gotten around to it yet"
the red ones seem to have this strange property where asqrt5 + b is the reciprical of asqrt - b times or devided by a perfect 2, though this is too compressed to see the equasions

i was surprised the red did that, but even more surprised to fail to find gold NOT doing that

[carmine idol]

hmm

[carmine idol]

in your investigation, you see that $a+b\sqrt{5}$ and $a-b\sqrt{5}$ seem to behave similarly right

[clever fjord]

Element118

[glossy marsh]

well yea

[carmine idol]

great we have enough to talk about norms

[glossy marsh]

sometimes the result will randomly become negative though

if b is negative

like um
(√5-1) = 4/(√5+1)

[carmine idol]

so if $c+d\sqrt{5}$ is a factor of $a+b\sqrt{5}$, then $c-d\sqrt{5}$ is a factor of $a-b\sqrt{5}$.

Multiplying them out:

$(c+d\sqrt{5})(c-d\sqrt{5})$ is a factor of $(a+b\sqrt{5})(a-b\sqrt{5})$

[glossy marsh]

reciprical x or / perfect 2

[clever fjord]

Element118

[glossy marsh]

i followed until the last line

and

and

its only true for the red ones and im not sure why but i think i know how to figiure that out

[carmine idol]

it follows this logic:
5 is a factor of 15
7 is a factor of 21
(5 times 7) is a factor of (15 times 21)

[glossy marsh]

ok i understand now

[glossy marsh]

yea i got it

just took a sec

[carmine idol]

yeah, now multiply out:

$c^2-5d^2$ is a factor of $a^2-5b^2$

[clever fjord]

Element118

[glossy marsh]

i cant prove the reciprical thing, the property is just aperent, though i do think its easily provable

[carmine idol]

(of course, I'm sweeping away some details, but that's the main idea you can use to help check if you can factor $a+b\sqrt{5}$)

[clever fjord]

Element118

[glossy marsh]

i think honestly i need to brute force a bit more to get the easy ones down and see which of the colored regions are the same in the small scale so i can get more of the chart filled out

what i was doing is far faster than trying to make a formula for it

despite sucking

i um

i guess ill start working on things we can assume as truths for shortcuts in any proofs i might be able tio use

i should have a proof list

(√5+1)(√5-1) = 4 aint listed anywhere in me notes

though that is an easy one

i need some special sign
bc i want to say (√5+1) = (4√5-9) but thats not true

"=" is wrong

they both fit the pattern

but its like saying 15 = 12 bc they share the factor 3

i might make a sign

work in progress idfk

its dumb

@carmine idol should i organize this so instead of pairing (√5+1) with (√5-1) i just make the counting order go negative like a numbre line

im considering it

ykw ima make a copy once ive made more progress on the redorange chunk

the bottom 4 oranges are new

the only patterns i can see in the color groups behavious is factors

which make repeating patterns bc yea

ok uh

i did that

has pros and cons

im actively trying to color it correctly

[carmine idol]

What's this table for by the way

[carmine idol]

are you just trying to factor numbers of the form $a+b\sqrt{5}$ or is there some organisation I'm not aware of

[clever fjord]

Element118

[glossy marsh]

yea

im tired

uh

uh

basically if i use the factors in the ruleset to convert one equasion into another they get the same color

if i find that 2 colors express the same group, they just become one color

left and right is the a that sqrt5 is multiplied by

up and nown is how much b

[meager juniper]

so just to confirm, you're looking for all numbers a + b sqrt(5) which are obtainable from c (1 + sqrt(5))^n, where c is some rational value?

[glossy marsh]

all instances of asqrt(5) +b where a and b are whole numbers have a block in the chart
except the chart is very small rn

[meager juniper]

and n is a postive integer?

[glossy marsh]

no n can be negative

likew half of these are devided by (sqrt5 + 1) at least once

i reorganized it
so now +b and -b are no longer paired
i need a big ass screen for this

i do need to color it more

black is when theres an obvious whole number factor

color groups are when the factor difference shared between them includes (sqrt5 + 1)

theres a few goals
one is to prove if color groups are the same as eachother
the point is to make it so i can simplify instances of asqrt5 + b that show up in huge equasions in irregular ways ig

[glossy marsh]

if i can find how/if (√5+4) can be expressed with (√5+1), i can do that for (3√5+1) bc (3√5+1) = (√5+4)(√5-1)

and then the gold section would also be red

itl probably be more than good enough if i can do this for every instance where a and b are between -16 and 16
but i dont need to express negative a because then thered be symetry due to the double negatives

red is probably going to be symetrical in the +-b bc thats whats been happening

yall were typing

[carmine idol]

I feel this might be similar to the idea of ideals.

(for any other people on math server who knows what this means so they can continue: $\frac{1+\sqrt{5}}{2}$ is a unit in $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$)

[clever fjord]

Element118

[glossy marsh]

i only started coloring the graph in things that werent greyscale 2 days ago
didnt touch this in almost a year prior to that

this is quite a bit more detailed now

[meager juniper]

let z = sqrt(5) for brevity, so we have (a + bz) (c + dz) = ac + 5bd + (ad + bc)z, additionally, we have (a + bz)/(c + dz) = (ac - 5bd + (bc - ad)z) / (c^2 - 5d^2)

If we imagine an (x, y) plane where we plot 4 + 5 sqrt(5) on the point (4, 5) (for instance) then we can figure out the geometry of multiplication. Because we can freely scale this value up or down, all we need to worry about is the direction of this vector.

So instead, let's move from cartesian coordinates to polar (r, t), and really all we care about is t.

We have x = r cos t, and y = r sin t, and we are assuming r is irrelevant, so it will be treated as 1. (a + bz) (c + dz) = (cos t + z sin t)(cos s + z sin s) = cos t cos s + 5 sin t sin s + z (sin t cos s + cos t sin s)

So atan2(sin t cos s + cos t sin s, cos t cos s + 5 sin t sin s) is the resulting angle. We can clean this up a little bit by using some trig identities. cos t cos s + sin t sin s = cos(t - s), sin t cos s + cos t sin s = sin(t + s), so we have atan2(sin(t + s), cos(t - s) + 4 sin t sin s)

However, this is not a very nice expression, and the geometry it induces doesn't seem to be anything that I'm familiar with.

[glossy marsh]

just multiplying a "(asqrt5 + b)" with sqrt5 + 1 and removing the factors (always perfect 2s????) i get another different (asqrt5 + b) and thats how ive been doing this

or deviding

[carmine idol]

yeah there are always 2s there, and there are good reasons why

involving field theory

[glossy marsh]

alr

ive tried brute forcing a = 1 b = 4 last year and didnt record most of what i found mostly bc it was almost exclusively giant numbers

bc i went big

trying to make it work and fit into the red

i doubt itl happen

[carmine idol]

yeah, there's a whole class of massive numbers. No smaller ones though.

[glossy marsh]

?

red so far is the biggest group
and red is the stuff that can be made of only sqrt5 + 1, sqrt5, and n

i havent even found all of the small red

though i have not explored the other colors nearly as much and am just yesterday and today doing the oranges

[carmine idol]

You know, you can write an algorithm to do this for you

if you want it faster

[glossy marsh]

i know that

though i dont know the math to do so
so im doing it it this way

well not yet

id like to know but i also think itd be faster to do it this way than learn

bc itd be slow that way too

there are lots of boxes

technically infinite but im doing the 16x32 rn

bc that should be enough

and bc the page isnt giiant yet

and at that point a ton of the expansion would be factors of these

and if i have say (2sqrt5 +35)
i could express it as ((sqrt5 +5) +(sqrt5 +5)+25) and then disect those
or something

wouldnt help all cases

but itd be useful

i realise now i do need to solve for everythings xsqrt5 equivilent

bc i do need list those

im sorry yall

[odd edge]

@glossy marsh Has your question been resolved?

Channel closed

Closed by @glossy marsh

Use .reopen if this was a mistake.

[faint knot]

that sucks

[odd edge]

Available help channel!

Send your question here to claim the channel.

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

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

[undone grove]

Q 6 idk how to disprove an answer

[lean yew]

i see 12 and 5 and my mind immediately thinks of pythagorean triples

[lean yew]

wait there's more, right?

can you show the full question?

[undone grove]

O mb

,rotate

[lean yew]

ah ok

[wooden python]

if t = tan(A/2), solve for t the equation 12tan(A) = 5, 180<A<270

this one?

[undone grove]

Ye

[wooden python]

btw what is this

[lean yew]

huh, they asked for the value of t. it seems like you got the value of t already?

[wooden python]

what's an equals sign doing in the numerator of a fraction

[lean yew]

oh wait i just noticed

[undone grove]

Ye mb mistake

[undone grove]

.

[lean yew]

wait disprove an answer?

[undone grove]

Ye something doing with tan has to be positive I think

Bc a has a domain

I think it's called

[lean yew]

so have you gotten a value of t yet? seems like you have

[undone grove]

Ye

1/5 -5

[lean yew]

seems to me this is in radians

[undone grove]

O

[lean yew]

oh i'm mistaken i think

[wooden python]

did you miss a comma here

[lean yew]

i'll let Ann explain, since if it's not i don't think i get it

[undone grove]

Ye

[wooden python]

so you got the values of t as -5 and 1/5?

[undone grove]

Yes

[wooden python]

ok let me just double-check this real quick

,w simplify 2*(-5)/(1-(-5)^2)

[clever fjord]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ann_dec

[wooden python]

,w simplify 2*(1/5)/(1-(1/5)^2)

[clever fjord]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ann_dec

[wooden python]

ok we're good to go those t-values are correct

[undone grove]

Ye

[wooden python]

now remember t = tan(A/2) and A lies between 180° and 270°

between what two values does A/2 lie?

[undone grove]

Pi and 3pi/2 ?

[wooden python]

i didn't ask you to convert anything into radians

[undone grove]

O oops

[wooden python]

and even if you did, that would be wrong still

bc pi and 3pi/2 are just 180° and 270° resp

(180°, 270°) is the interval in which A itself lives.

in what interval does A/2 live?

in degrees.

[undone grove]

90, 135

[wooden python]

ok

so 90° < A/2 < 135°

so, in what quadrant does A lie?

1st, 2nd, 3rd, or 4th?

[undone grove]

2nd

[wooden python]

good

and what sign is tan in the 2nd quadrant?

[undone grove]

Ooo negative

[wooden python]

ok

do you now see which of your two values of t is the one you want

[undone grove]

O thanks I thought I had to sub back in but clearly not

And I didn't find the region of a/2

.close

[odd edge]

Channel closed

Closed by @undone grove

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[sick pier]

the answer should be just pi/4 right?

[sinful grove]

Mhm

I’m assuming this is the function inverse and not some sort of fibre

[sick pier]

yea inverse

[sinful grove]

Some high school math like to do that for some reason with this notation

[sick pier]

wdym some sort of fibre

[sinful grove]

taking the preimage of {pi/4}

[mystic nova]

Assume this is arcsin(x) the output should only be in [-pi/2,pi/2] so, yeah

[sinful grove]

So what I’m saying is, yes that is correct. But I’m only complaining how sometimes it’s not clear with lower level math questions

They sometimes abuse notation

[sick pier]

sooo, 3pi/4, pi/4 +2*pi*n and so on are technically right?

[sinful grove]

No

Only within [-pi/2, pi/2] (which is only pi/4) if this is the actual sin inverse theyre using and not some sort of abuse of notation

[sick pier]

i think we're abusing the notation cuz teacher did the same

this what teacher did

[sinful grove]

Welp as I knew it would…

[sick pier]

welp

[sinful grove]

Okay so in that case these are right

[sick pier]

🙂

[mystic nova]

"In mathematics, a function from a set X to a set Y assigns to each element of X exactly one element of Y" so a function should only have one output for each input

[sinful grove]

Theyre abusing notation and taking the fibre of pi/4

[sick pier]

erm

[sinful grove]

Do you see what the “full solution set” is?

[sick pier]

heh

[sinful grove]

Youre solving sin(x) = 1/sqrt2

[sick pier]

yep

[sick pier]

this is the lecture with the workings and lecturer solving it ^

welp

[sinful grove]

!status

[odd edge]

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

[sick pier]

5

[sinful grove]

What is the question?

[sick pier]

why is the answer not just pi/4?

[sinful grove]

It seems your teacher is using the notation sin^-1(y) for solving the problem of finding all x such that sin x = y. This solution set will be infinite and pi/4 is just one solution.

However youre correct if sin^-1 represented what it actually does in math, which is an inverse function. Then pi/4 would be the only answer by how it’s defined

[sick pier]

hmmm

thank you!

if i pass my exam, i'll boost the server!

:)

[sinful grove]

[sick pier]

.close

[odd edge]

Channel closed

Closed by @sick pier

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[round pivot]

Hiii

[dense moth]

iiiH

[wooden python]

hello, do you have a question to ask?

if you do, then ask it; otherwise, if you just wanted to chat, there's #discussion and #chill.

[sturdy cape]

Damn, you beat me to it; I was about to say "Not the relevant place to do this - these channels are for math help"

[wooden python]

!occupied

[odd edge]

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).

[sturdy cape]

[use an OPEN channel to ask this]

[narrow crypt]

https://tenor.com/view/thats-so-epic-gif-6804949165771866469

[odd edge]

@round pivot Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[odd edge]

Available help channel!

Send your question here to claim the channel.

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

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

[sonic juniper]

I need someone if I'm doing it correctly, its about operations on functions

[wooden python]

your pictures are a bit blurry and i can't read the top of the first one, can you re-upload?

[sonic juniper]

sure

its kinda messy 'cause im not 100% sure about my answers

[odd edge]

@sonic juniper Has your question been resolved?

[graceful gate]

hello guys

[sonic juniper]

hii

[graceful gate]

how can I improve my level in maths? I am grade 8 and I want to be intellegent

[sonic juniper]

I personally not a mathematician or sort of math wizard, im weak in math but most people say is to be familiarize with the formulas and how you solve different kinds of equations

just be familiarized with the lesson

[sonic juniper]

^

[sonic juniper]

<@&286206848099549185>

[odd edge]

@sonic juniper Has your question been resolved?

[sonic juniper]

<@&286206848099549185>

[inner berry]

Mhm?

[sonic juniper]

I need help in checking my answers

[inner berry]

kk

[sonic juniper]

its about operations on functions and im not 100% if my answers are correct

[inner berry]

I love your keyboard

[sonic juniper]

oh thank u 😅

[inner berry]

ur welcome

you have correct answers and incorrect

[sonic juniper]

hmm

[inner berry]

May I see where u got the questions from

[sonic juniper]

sure

[inner berry]

Which grade are u or how old are u

[sonic juniper]

i'd like to keep my grade confidential but i'm 16

[inner berry]

1. ✅

2. ✅

3. ✅

4. ✅

5. ✅

6. ✅

7. ✅

8. ✅

9. ❌

little one

did u take domains and restricted values?

[sonic juniper]

im not familiar with that yet

or we maybe haven't discussed that yet

[inner berry]

oki

10. ???

11. ✅

[sonic juniper]

from 10-12, idk what I have been doing T_T

i've tried watching tutorials on how but that's what I can do T_T

[inner berry]

12. ✅

Try to understand harder examples

Or try to solve it again

Maybe u miscalculated smt or skipped a step without knowing

[sonic juniper]

I've actually asked for help here recently, and that's what we have discussed

I solved this with assistance

and I tried practicing after since the person who helped me needed to go 😭

[inner berry]

10. ✅

You got all correct except 9

Try solving it again

[inner berry]

ooooo

[sonic juniper]

rlly? I have only 1 mistake?

[inner berry]

practice means ur determined

[inner berry]

mhm

[sonic juniper]

I swear I was not on my right mind when I was solving these

[inner berry]

oop

[sonic juniper]

I just remembered what the person told me to do in solving these and watched a few

and was confused mainly from 9-12

[inner berry]

mhm

ye the hard ones

[sonic juniper]

can u assist me in no. 9?

[inner berry]

Mhm

[sonic juniper]

=(-x^2 - 1 - 2x)(x +5)

[inner berry]

Mhm

[sonic juniper]

that what I only know what to do..

[inner berry]

:>

[sonic juniper]

oh wait

[inner berry]

mhm

[sonic juniper]

i added a subtraction by accident

[inner berry]

I told u

😭

[sonic juniper]

okay wait let me try re-solving this

stand by

[inner berry]

pets

I’m waiting dw

[sonic juniper]

okey, thank u

[inner berry]

no problem

[sonic juniper]

=(-x^2 - 1 - 2x)(x +5)
im stuck here ;)

[inner berry]

$(-x^2 - 1 - 2x)(x +5)$

[clever fjord]

someone

[inner berry]

this?

[sonic juniper]

yup

[inner berry]

Simplify

I’m here dw

[sonic juniper]

im not sure but is it = -x^2 - 4 - 2x?

[inner berry]

Try again

Try to listen to smt calming instead of this

so u can focus

what’s ur name chomp

[sonic juniper]

u can call me lou

[inner berry]

Lulu u stuck or u solving?

[sonic juniper]

im stuck..

[inner berry]

show me

[sonic juniper]

im just trial and erroring it

[sonic juniper]

but im stuck here

[inner berry]

Think harder

can u think or ur kinda blank?

[sonic juniper]

let me try it again

i did the other others so maybe i can do this too

[inner berry]

Mhm

[sonic juniper]

okay im blank

[inner berry]

flicks head

try doing some pushups it’ll focus blood into ur brain it helps

[sonic juniper]

I think i've had enough exercise today

plus its night time here

[inner berry]

and still blank

ah

makes sense

[sonic juniper]

yea..

[inner berry]

I mean u still did the others correct

lemme help

$(-x² - 1 - 2x) - (x + 5)
-x² - 1 - 2x - x - 5 = -x² - 3x - 6$

[clever fjord]

someone

[inner berry]

No

Wrong

(-x² - 1 - 2x) - (x + 5)
-x² - 1 - 2x - x - 5 = -x² - 3x - 6

like this

Simplification after ur step

[pastel steeple]

make a 3x2 table if you need to

[inner berry]

@sonic juniper lulu

[sonic juniper]

??

[inner berry]

(-x² - 1 - 2x) - (x + 5)
-x² - 1 - 2x - x - 5 = -x² - 3x - 6

It’s a subtraction

so u just put it out

Expand

And then u solve

Combine like terms

[sonic juniper]

how did u get that

[inner berry]

😭

didn’t u get it too

the one ur stuck on

It’s subtraction

So u just take off the brackets

[sonic juniper]

how did u get = -x² - 3x - 6?

[inner berry]

Combine like terms

combine the stuff that are like eachother

like

[pastel steeple]

i assume you're missing the negative sign turning the +5 in the parentheses into a -5

[inner berry]

numbers

[inner berry]

are u sure?

[pastel steeple]

i said assume didnt i

[inner berry]

Where’s the positive 5

[pastel steeple]

can you show what you got for the subtraction @sonic juniper

[inner berry]

the 5 is negative

[pastel steeple]

i said +5 in parentheses

[inner berry]

That’s correct

[sonic juniper]

are you familiar with the FOIL method?

or any of u

[pastel steeple]

have you done the subtraction correctly first

[inner berry]

Mhm

[sonic juniper]

First Outer, Inner Last is the meaning

[inner berry]

Ye ik

[sonic juniper]

so do I have to do that? or just simply combine like terms ?

[inner berry]

combine em

[pastel steeple]

what do you currently have on your paper for this problem?

[sonic juniper]

-x² - 1 - 2x - x - 5 = -x² - 3x - 6 this one then I have to solve like he said

correct me if im wrong

[pastel steeple]

ok good so that's g - f

[inner berry]

mhm

[pastel steeple]

the question asks you to find (g - f)(x)

whats the last step then

[inner berry]

To find (f+g)(x), add f(x) and g(x)

[pastel steeple]

what the hell are you talking about

[inner berry]

what

[sonic juniper]

isn't it g - f?

not f+g?

[pastel steeple]

we are on number 9 right?

[sonic juniper]

this one

[inner berry]

Sorry

[sonic juniper]

its okayy

[inner berry]

I meant it more general form but mhm

u can do it

u did the rest

try

[sonic juniper]

[inner berry]

CMONN DONT LET ME DOWN

[pastel steeple]

multiply by x

[inner berry]

she’s doing it dw

u stuck lulu?

[sonic juniper]

hold on

[inner berry]

oki

[sonic juniper]

i have -x^3

am in the right path?

[inner berry]

^3 or 2

[sonic juniper]

^3?

[inner berry]

maybe u use a different method

[sonic juniper]

i combined like terms

[inner berry]

😭

[sonic juniper]

AM I IN THE WRONG PATH

[inner berry]

SHOW ME WHAT U WROTE

[sonic juniper]

I followed this

so

(-x^2)(x)-(1)(x)-(2x)(x)-(x)(x)-(5)(x)

IDK

IM SO WEAK IN MATH

[pastel steeple]

great now simplify it by combining like terms

[sonic juniper]

thats what I did pero im stuck on the 2nd part

i already got -x^3

[inner berry]

(x + 5) + (-x² - 1 - 2x)

This is what I got for f and g

Wait

[pastel steeple]

then (-1)(x) = -x?

[inner berry]

wait wait

[sonic juniper]

yes

[odd edge]

@sonic juniper Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[lone ether]

if 0.25m is 1/4th of one meter why do I need 16 tiles to make up 1 meter?

[late sinew]

add and multiply

ab=1, a=b

solve it

[nocturne belfry]

its not 1 m, its 1 m^2

you dont need something 1 m long or 1 m wide, you need something 1 m long AND 1 m wide

[lone ether]

wouldn't 0.25m be 1/4 wide and long

[nocturne belfry]

it is, yea

[lone ether]

If one tile is 1/4th of 1 m^2 why do I need 16 tiles

[nocturne belfry]

one tile isnt actually 1/4th of 1 m^2

[amber veldt]

(1/4 m)²=1/16 m²

[nocturne belfry]

we need to make a shape thats 1 m long, and 1 m wide

so start with 4 of your tiles in a line

this shape is 1 m wide, but its only 1/4 m tall, yea?

[amber veldt]

you are confusing (1/4 m)² with 1/4 m²

[lone ether]

I think I see it I'm only adding 4 tiles on the length to get 1 but I forgot I need to do the same with the width

[nocturne belfry]

yea, and then you have to fill it in

in the end you need 16 tiles to do that

[lone ether]

.close

[odd edge]

Channel closed

Closed by @lone ether

Use .reopen if this was a mistake.

[lone ether]

thanks y'all

[odd edge]

Available help channel!

Send your question here to claim the channel.

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

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

[hazy falcon]

Hi need help

[mystic saffron]

what do you need help with?

[odd edge]

@hazy falcon Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[noble dome]

Let ( f : A \to B ) be a function. If ( f ) has a left inverse, then it is unique.\

\textbf{Proof:} Suppose there exists ( g: B \to A ) such that ( (g \circ f)(x) = x ).
Also suppose there exists ( h: B \to A ) such that ( (h \circ f)(x) = x ).
Then, ( (g \circ f)(x) = (h \circ f)(x) ).

[clever fjord]

licentia

[odd edge]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

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

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

[shrewd trellis]

https://cdn.discordapp.com/attachments/1366639984374710365/1396475826492674110/image.png?ex=687e38f6&is=687ce776&hm=28bb1a08f12bee4f79561307bd922db7cc367f62c7d42ad0b63afda1c6d6a57e&

[shrewd trellis]

do u know why tan and tan inverse gets canceled out?

[latent scaffold]

Because tan inverse is the inverse of tan

[shrewd trellis]

What is tan tho, I still havnt understood

Tangent?

[wooden cypress]

It's a trig function but it doesn't actually matter what it is for this to make sense

tan^-1 is just defined as the inverse of tan

[shrewd trellis]

What about the reciprocal trig functions

Cotan
Cosec

Sec

I don’t know what they mean

[latent scaffold]

They're the reciprocals of the usual trig functions.

[shrewd trellis]

Ik

[latent scaffold]

They do carry geometric meaning, but not much more than sin, cos and tan do

[shrewd trellis]

So what do they mean

[wooden cypress]

For any number x, you can draw a right angled triangle like that

With that angle being of size x

then $\tan x = {a\over b}$, $\sin x={a\over c}$, $\cos x={b\over c}$

[clever fjord]

depression

[wooden cypress]

If you want to actually compute one of the functions, it's a bit trickier, so you would just use a calculator

But that's where the functions come from

[shrewd trellis]

“Compute on if the functions” do u mean calculate without calculator

[wooden cypress]

Yeah

It can be done but it's a pain

My point is it doesn't matter what the functions are exactly

You can still do all the math with them, without acutally having to worry what $\tan^{-1}\left({8\over10}\right)$ is exactly

[clever fjord]

depression

[shrewd trellis]

I’m talking about the reciprocal trig functions