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@edgy mantle Has your question been resolved?

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[hot sleet]

been stuck here for 3 hours and no YouTube tutorial has helped me lmao

would converting 1-cos to 1-sin(x/2) would do the trick?

[tacit arch]

are you allowed to use substitution? w = 1/x, then yea this could work

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[wheat pendant]

I have some questions regarding the differential equation problem.

[wheat pendant]

However, when i asked wolfram to solve it gave a very different result.

What did i missed on this one?

[lone heart]

@wheat pendant Has your question been resolved?

[wheat pendant]

<@&286206848099549185>

[tacit arch]

the previous -2 became a +2

[wheat pendant]

OH MY GOD

but using the reduction order doesn't require that term

isnt it?

since y1 already present which i need to only get y2 to get the general solution

[tacit arch]

i don't know the reduction of order technique you learned

this is the one i'm familiar with https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx

which requires you to solve for a v(t) that equals to y_2(t) = v(t) * (known solution)

[wheat pendant]

okay thanks for the info

ill try again

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[timber jungle]

Task in text form: Given a rectangle ABCD inscribed in a circle ω with center O. The line ℓ
passes through O and intersects the segments BC and AD at points
E and F, respectively. The points K and L are the points of intersection of ℓ and ω, with the points K, E,
F, L lying in that order on the line ℓ. The lines tangent to ω at points
K and L intersect the line CD at points M and N, respectively. Prove that the points E, F, M, N lie on the same circle.

S is the center of circle which existance should be proved, figured it out in geogebra just to make drawing a proof a bit more convenient
what I figured out so far:

1. obviously LN and KM are parallel
2. using the Thales's theorem I can prove, that if I draw the circle with the diameter MN and center S and draw parallel to LN going through O, it will also go through S.
   So I can tell that ES=FS and MS=NS, so now I only need to prove that ES=MS or FS=NS.
3. Through geogebra I figured out that ∠OSE = ∠CME - ∠EMK = const for the same rectangle

And after this I couldn't come up with anything that can help me.
I can send the geogebra file here if somebody needs it and if I'm allowed to.

[tiny sky]

could u send me a geogebra link?

[timber jungle]

actually, I don't have an account, I can only send file. I don't know if I'm allowed to send it in this channel, I could send it to you in DM

[tiny sky]

ah nw it's fine

[timber jungle]

so, do you want me to send it in DM?

[tiny sky]

it doesn't take that long to draw up in geogebra

also idk how geogebra files work lol

[timber jungle]

I just click open and there is local file button

[tiny sky]

i have to go now, i'll think abt this when i'm back

[timber jungle]

okay 👊

[tiny sky]

i'd like to preface this by saying that i really really suck at olympiad geometry

but just an observation

CMKE and NBFL are cyclic

[timber jungle]

me too, the worst thing I can imagine is geometry and probability

[tiny sky]

so can we use that to angle chase DNF etc.?

[timber jungle]

ye, I figured it out too, but couldn't come up with idea how to use that

[timber jungle]

can't tell if I already tried that, but I'll try this anyway

thx for idea

[lone heart]

@timber jungle Has your question been resolved?

[tiny sky]

ok so if u angle chase correclty

you get that the question is equivalent to proving that AL = BK

which is true

[timber jungle]

but... isn't AL almost always not equal to BK because of the angle of line l?

for example on the drawing I made AL is longer than BK

Maybe I'm missing something and I made a bad drawing

[tiny sky]

oh wait i've might not have the right labellings

oh yeah my points are labelled badly lol

lemme relabel my points 1 sec

right ok so i'll walk u through my thought process

[tiny sky]

first piece of advice is to generally not add in circle EFMN

sometimes adding in the circle can be useful but here i think it clutters the diagram

so there's many ways to show something is cyclic

for example, if we could prove NFE + NME = 180 we'd be done

now think about why that is equivalent to proving that NFD = EMK

did that make sense @timber jungle ?

[timber jungle]

no, but I'm trying to understand this

Okay, my guess is that if those angles are equal, that means that △NDF~△EMK and that should mean somehow that △NLF~△CME. If △NLF~△CME, I wrote few equations and I proved that NFE + NME = 180 @tiny sky

But now I don't really understand how to prove that △NLF~△CME and the original statement that NFD = EMK, but I'm on my way to understand this

okay, I managed to prove that △NLF~△CME. Now everything you said makes sense @tiny sky

The only thing left is to prove that NFD = EMK

[tiny sky]

i mean it's actually a lot more simpler

DFE + FEC = 180

FEC = KMC

so then look at that and realise that that's equivalent to NFD = EMK

[tiny sky]

anyway once we have this, NFD = NLD = angle subtended by NL

[tiny sky]

EMK = ECK = BCK = angle subtended by BK

so it suffices ot show BK = DL

which is true by symmetry/drawing in a few lines to show that 2 triangles are congruent

hope that all made sense @timber jungle

[timber jungle]

welp, I'm processing it :D
Rn I'm trying to understand why you said this:

that's equivalent to NFD = EMK
I've used the fact that FEC = KMC in writing some equations, but that didn't lead me directly to saying that NFD = EMK

[tiny sky]

well then DFE + KMC = 180

KMC = EMC + KME

DFE = NFE - NFD

so EMC + NFE + (KME - NFD) = 180

so suffices to show KME - NFD = 0

[timber jungle]

okay, now I understand how you proved that KME + NFD and proving that 2 triangles are congruent

that makes sense

so how would you prove that NFE + NME = 180 because of that?

[tiny sky]

^

[timber jungle]

man, you're writing in an occupied

[ancient lynx]

oops

[lone heart]

@timber jungle Has your question been resolved?

[timber jungle]

I thought I understand, but I don't. So from your equation to prove that KME = NFD, EMC + NFE should be equal to 180, so that we can get KME - NFD = 0
But I can't find when did we get this information and if we have this info, we are already done because EMC = NME
And I don't really understand the part with the triangles, what are you trying to prove with those triangles? And am I getting this right, you proved that △DFL = △BKE?
Sorry to bother you for such a long time, it's just difficult for me to understand this

[tiny sky]

we're not doing anything with triangles?

[tiny sky]

we start with this

[timber jungle]

okay, seems like I'll just need to try to rethink it a hundred times more tomorrow until I get it. Anyway, thank you very much for helping me out, at least I'm not stuck anymore @tiny sky

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[round zealot]

Ik that the sqrt of 65 is about 8, but with the negative in front of the sqrt symbol, if I keep -8 it won't work because -8 squares is positive 64

for 22

<@&286206848099549185>

<@&286206848099549185>

[tacit arch]

chill your pings

[round zealot]

ok

[round zealot]

help

[tacit arch]

just wait

[round zealot]

ok

[strange schooner]

First, that square root is closer to 9 than it is to 8

[round zealot]

how

[strange schooner]

And second, fairly sure that - was there just so you had to write -9 instead of 9, from the looks of it, you haven't taken complex numbers, which would be needed to get a negative result when squaring

[round zealot]

also guys i ment question 22

??

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[round zealot]

.reopen

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✅

[round zealot]

.close

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[strong night]

Hi

[strong night]

Can i get help on these two plz. Im very confused

Plz @ me when responding. Thanks

<@&286206848099549185>

[smoky crane]

.reopen

[pseudo ice]

For the first, you might want to consider difference of cubes…
For the second, consider that as x is going to infinity, you can consider x positive, and compare the numerators and denominators

[strong night]

I thought of this

Still cant do it @pseudo ice

😭

[pseudo ice]

Of which one? Both of them?

[strong night]

Workin
On first rn

Rlly need help wit it. Been doing it for a while

@pseudo ice

Im tryinh to factor an x from numerator right?

[pseudo ice]

If you can make a factor of x appear in the numerator appear, yep

Right, you’re happy that the general difference of two cubes is
[
(a - b)(a^2 + ab + b^2) = a^3 - b^3
]

[ocean seal]

@pseudo ice

[strong night]

I did that but then do i like multiply it out?

[pseudo ice]

Well, we have cube roots, and somehow want to get rid of them

[strong night]

My teacher recommended thid

This

And said to use the formula

Idk where to go from here

[pseudo ice]

Yep, so notice how our numerator is (a - b) right now, if we multiplied it by something, we would turn it into (a^3 - b^3)

[strong night]

Ohhh

Cube top and bottom?

Is that what ur saying? @pseudo ice

[pseudo ice]

Noooo

[strong night]

Oh

@pseudo ice

[pseudo ice]

What would you have to multiply (a - b) by, to get (a^3 - b^3)?

[strong night]

Idk @pseudo ice

☹️

Brainfart

[pseudo ice]

Awwww

Do you at least see it now?

[strong night]

No

@pseudo ice

Oh im dumb

Its the other part of the formula

Asquared plus ab plus b squared

Y do we want to multiply by this tho? What is being accomplished by getting a cubed minus b cubed?

[pseudo ice]

Yep, you’d need to multiply by that, but of course if we multiply by it, we also need to divide by it too

[pseudo ice]

And work out what a^3 - b^3 is

[strong night]

Im confused y we need that

[strong night]

This

[pseudo ice]

Did you work out what it turns into?

[strong night]

Not yet lemme do that so multiply top and bottom? @pseudo ice

[pseudo ice]

Yep, multiply top and bottom by that a^2 + ab + b^2

[strong night]

K one second

[pseudo ice]

You know the numerator becomes a^3 - b^3, so then check what that is with your choices of a and b

[strong night]

So top is x?

Lemme do bottom rq

@pseudo ice

[pseudo ice]

You don’t need to really expand it out, just leave it as x(a^2 + ab + b^2) and then cancel the x in the denominator with the x you get from the numerator

[strong night]

Ohhhhh

Ur right

Can we do second one together ur rlly good help btw

@pseudo ice

[pseudo ice]

Sure

[strong night]

Thx alr lemme just right answee for first one

@pseudo ice ok moving onto next one

Not rlly sure where to start ngl

Divide everything by highest power of x?

[pseudo ice]

Well, I mean, you could, if you’re careful about it-

As the limit of x is going to infinity, we can safely assume x is positive

Now, if x is positive, what’s sqrt{x^2}?

[strong night]

I want to do the same steps my teacher would do

This is all the notes @pseudo ice

[pseudo ice]

,rccw

[strong night]

x

[pseudo ice]

Cool, so what’s the “highest power” of the denominator?

[strong night]

I think i would need to divide everything right?

So would i divide by x or x squared @pseudo ice

I see the highest power is x squared but it rlly is just x

@pseudo ice

Is this right?

[pseudo ice]

Yep, the highest power is basically x, so divide both top and bottom by x

Then use the fact you have x as sqrt{x^2} to simplify the insides of the root

[strong night]

Wait can i take out a term if root?

@pseudo ice

Under root

I kinda forget this

[pseudo ice]

Well, yea, you can either factor x^2 from the root and pull it out, or equivalently divide the insides of the root by x^2 and the numerator by x

[strong night]

@pseudo ice can i take out 9x squared?

[pseudo ice]

You could if you wanted to, sure

[strong night]

Wait but i have a question

Can u always take out a term if it is able to from under a square root. For example: root of 9-5 is 2 but then if u favtor out 9 first then its 3 times root of negatuve 5

@pseudo ice

I always get confused on this

[pseudo ice]

Well, if you factor 9 out, it’s 3 * sqrt{1 - 5/9}, which works out as 3 * sqrt{4/9}

And of course sqrt{4/9} is 2/3

[strong night]

Wait can u say tht more simply @pseudo ice

Like clearly

Where do u get 5/9 from

And 4/9

I dont think i can factor is out @pseudo ice

[pseudo ice]

Because you're factoring the insides, you first need to factor a 9 from 9 - 5

And as 9 - 5 = 9(1 - 5/9), as you have to divide the "other term" by 9

[strong night]

O ya. I think i have had too much math for today. After this question i will stop

So i will go ahead and factor the x on’y

@pseudo ice o wait but if i factored 9 it would be just a negative under tje root

O wait no

Nvm

1

• whatever

-*

[pseudo ice]

Remember that if you factor, you have to divide everything by what you're factoring out

So like if you factored an a from a^2 + 3a, you get a(a + 3)

so the a^2 gets divided by a, and the 3a gets divided by a too

[strong night]

Is this right?

So far?

Or did i make a mistake

@pseudo ice

[pseudo ice]

Not quite, no

[strong night]

@pseudo ice oh

Damn

[pseudo ice]

New question, can you factor an $x^2$ from the expression $9x^2 - 5x$? bear this in mind

[ocean seal]

@pseudo ice

[strong night]

Oh no u can’t

@pseudo ice

So i cant take anything out of root?

@pseudo ice

@pseudo ice im stuck

Can u explain what and why ur supposed to do

[pseudo ice]

You can either factor out an x^2, or a 9x^2, out of the root, because you know you can square root those

[strong night]

But u said u cant factor

[pseudo ice]

You can factor an x^2 from this expression, it just doesn’t look as nice and you will have fractions in it

[strong night]

X squared

[strong night]

True

Ok i will do that rn

Ill do one step at a time and then send to y

U

Like this?

Sorry x

Not x squared

Outside

Wait a minute

I think i screwed it up

[pseudo ice]

Yep, better

[strong night]

Is this right?

[pseudo ice]

Yep, with x on the outside of the root, it is

Now, if you factor an x from the numerator too, that x + 3, in a similar manner

[strong night]

How do u factor an x from numerator

There is one x

[pseudo ice]

Alright, new move then-

[strong night]

Dont i divide by highest x?

Power

[pseudo ice]

Rewrite this as $\frac{ (x + 3)/x }{ \sqrt{9 - \frac5x} }$ and work out that $\frac{x + 3}x$ in the numerator

[ocean seal]

@pseudo ice

[strong night]

How does the x move to numerator @pseudo ice

[pseudo ice]

You are aware of how to manipulate fractions, such as the fact that $\frac{a/b}c = \frac{a}{bc}$ right

[ocean seal]

@pseudo ice

[strong night]

Ya u just multiply by b

Top and bottom

@pseudo ice

[pseudo ice]

Well, going backwards is dividing both numerator and denominator by b

[strong night]

Are we able to do this way @pseudo ice ?

Like divide by x highest power?

My teacjer will mark us like this

[pseudo ice]

It’s literally what we are doing, the highest power of both the numerator and denominator is x

You just need to know how to work with roots and how they simplify

[strong night]

Does this work?

@pseudo ice

[wintry shore]

,rotate

[pseudo ice]

There was an x outside the root which you’re forgetting about

[strong night]

It cancels?

Doesnt it?

[pseudo ice]

It cancels with the x you’re dividing by

[strong night]

Yes so no need to write?

@pseudo ice

[pseudo ice]

Correct, don’t have the /x for the denominator

[charred pilot]

\frac{1}{\sqrt{9-\frac{5}{x}}\cdot\frac{x}{x}+\frac{3}{x}

[strong night]

Wait like what i have rn is correct or no? @pseudo ice

[pseudo ice]

Now it is

Because your original was

[strong night]

Oh yes

Okay ic

What do i do now

@pseudo ice

[charred pilot]

Sorry not trying to take up space

[pseudo ice]

do you wanna take over explaining, by any chance?

[charred pilot]

Yeah I was trying to format it for them

$\frac{1}{\sqrt{9-\frac{5}{x}}\cdot\frac{x}{x}+\frac{3}{x}$

[charred pilot]

Can you by chance check my code

[strong night]

@pseudo ice am i close to finishing?

[charred pilot]

You're 1-2 steps away

[strong night]

Alr

[pseudo ice]

You’re missing one closing } after the one for the sqrt

The one so you have the denominator of the first \frac

[charred pilot]

I tried that in the latex test channel and its still messed up

[pseudo ice]

(Tl;dr “what happens to this as x gets larger and larger?”)

$\frac{1}{ \sqrt{9-\frac{5}{x}} }\cdot\frac{x}{x}+\frac{3}{x}$

[ocean seal]

@pseudo ice

[strong night]

Closer to 0?

O wait closer to

…

1/3???

[charred pilot]

\frac{1}{ \sqrt{9-\frac{5}{x}} }\cdot(\frac{x}{x}+\frac{3}{x})

[charred pilot]

Yes

[pseudo ice]

Put it in an equation environment?

[strong night]

Yay

Thx @charred pilot And @pseudo ice

.close

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[rose sigil]

there are some typesetting errors here

uniformly convergent on R, i'm guessing?

[lone heart]

@dim junco Has your question been resolved?

[rose sigil]

if so then yea looks fine

[ocean seal]

Architect

[rose sigil]

not really. you can just take whatever term in the tail (say the first one)

[ocean seal]

Architect

[rose sigil]

for any k, x^k/k! is unbounded

by taking limit as x goes to infinity

so just set k = n + 1

this sum will be greater than x^(n+1)/(n+1)!

(if x > 0)

yep

[ocean seal]

Architect

[rose sigil]

no

e.g., $\sup_{x\in R_{\geq 0}} 1/x = \infty$ but $\lim_{x\to\infty} 1/x = 0$

[ocean seal]

:bending_skull:

[rose sigil]

the sup of a function over a set is like the "largest" (but not quite that, there might not be a largest even if there is a sup) value the function takes over the set

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[rose sigil]

what i mean by that is e.g. $\sup_{x\in (0,1)} 2x = 2$ but $2x$ isn't ever 2 for any $x\in (0,1)$

[ocean seal]

:bending_skull:

[rose sigil]

but largest is a good intuitive way to think of it

[lone heart]

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[fervent rampart]

Yo hello mu friends

[fervent rampart]

i have a question

how can i learn mathematic?

Beginner to Professional

[livid sage]

what is your current level of mathematics education?

[fervent rampart]

Hmmm

i should low

What do you think the about

[livid sage]

you can answer this by listing the last few math courses you have taken or the current math course you are enrolled in

[fervent rampart]

Wait

i know basic of math

thats all

i need so much

[livid sage]

https://www.khanacademy.org/math may be a useful resource

[fervent rampart]

Oh

the website is so cool bro

thank you so much

for your help

.close

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[tawny imp]

bro 💀

[gritty bramble]

cooked

[manic pivot]

i went in here to ask for help w actual math probelms theres no way u need help w this wtf

is this not common sense like u could have never taken gemoetry and still understand this i would think

[tawny imp]

if its so easy then you can help me

🤷‍♂️

[manic pivot]

no u j have to go thru it it takes like 5-10 min for me to do that easy but a waste of time for me

im sure theres a lot of nerds in here happy to help

[dense vapor]

<@&268886789983436800> this dudes being an arsehole

[tawny imp]

so its too hard for you to solve?

[wide tartan]

don't be rude to other users please

[manic pivot]

the person i said that to wasnt offended i didnt really say anything bad just sorta banter no need to be so sensative what u consider rude isnt subjective and if that guy i said it to has no issue w that i said then i wasnt being rude

[wide tartan]

[tawny imp]

https://tenor.com/view/cat-cat-meme-funnt-cat-black-cat-sleepy-cat-gif-16409614852005252877

[wide tartan]

don't tell people in the help channels that the thing that they're struggling with is easy.

is that better?

[manic pivot]

like im not trying to be problematic i j joined for help w a math problem and saw an easy problem and gave him some shit for it but nothing serious its really sensative and sad that society is so sensative that telling someone that a literal math problem is easy offends someone i wont say it and ill follow ur communist rules to not get banned but u should stand up for whats right and moderate fairly.

[tawny imp]

https://tenor.com/view/cat-cat-side-eye-side-eye-cat-food-cat-food-side-eye-gif-3423883768828421583

[dense vapor]

Oh good lord 🤣

[manic pivot]

like the person i said it to is literaly putting cat memes at it he doesnt care so im just confused why u as a mod felt the need to say anything about it

[manic pivot]

and even more confused why this guy felt the need to step in

[dense vapor]

https://tenor.com/view/getherjade-get-her-jade-rpdr-gif-9479342

[tawny imp]

congratulations, you just spent the 10 minutes that you said it would take to solve that problem on arguing with a highschooler

good job

https://tenor.com/view/cat-kitty-eepy-cat-eepy-kitten-eepy-gif-8186911474993351997

[manic pivot]

like ik im making a big deal abt it but theres no way that u guys responding to it dont see my point thats why i argued abt it sm baffels me that ppl r like this

[wide tartan]

can we keep it on-topic please - I think I'm done here

[manic pivot]

the topic was my comment and discussing weather or not it was wrong so isnt that on topic?

[manic pivot]

i dont appreciate how u as a mod are just ignoring my feedback on what happened

[wide tartan]

okay, dm @fierce herald if you have issues with the way I carry out the role.

can we please get back on-topic now?

[tawny imp]

yeah

[tawny imp]

i was thrown this question in my math with no explaination on how to solve it

[manic pivot]

i dont have big enough issues of how u carry out the role to message mod mail its just not okay how u do this and dont care about being unfair and when i call it out u laugh at it but anyways since u wanna be like that then please answer this at least mod person how do i make a thingy thing to ask my math question?

[wide tartan]

it's laid out in #❓how-to-get-help

[tawny imp]

can you just quit the yappin and go do something useful? you have a lot better to do then argue with a high schooler yk

[manic pivot]

taking a few mintues to argue with someone while im doing a math assignemnt is considered useful to me since it doesnt take that much time and i do it to try and understand where lower intelligent indivisuals are coming from sometimes i like to understand others povs cus for ex i genuinely cannot fathom how someone that has nothing to do w me or u can look at me basically telling u that ur math problem is easy and then be like OMG DID THIS PERSON JUST CALL A MATH PROBLEM EASY AND MAKE A JOKE OMGGG I NEED TO TAG A MOD THAT IS SO WRONG like its crazy anyone is like that i was arguing to try to understand their pov but they just ignored all the logic i was spitting back at them becasue i dont even think they understood why they got upset themselves

[tawny imp]

erm

"considered useful"

https://tenor.com/view/cat-underwater-gif-922906369727670801

[subtle birch]

what the sigma

[wide tartan]

please just drop it. this isn't the place to argue with me, there are other avenues for that. I'm gonna just mute you if you continue to say off-topic things.

[tawny imp]

what have you accomplished by arguing with me

[urban tree]

W

[manic pivot]

excuse me but jmao is the one arguing w me and im responding so tell that to him to please this is not me arguing with you just standing up for u only enforcing rules on me and not jmao

[azure needle]

Please mute 'em

[tawny imp]

yeah im done

[wide tartan]

come back later.

[wide tartan]

for reference for people coming in: here is the original problem, no progress has been made yet

[tawny imp]

<@&286206848099549185>

[latent ore]

Ok

[tawny imp]

ignore everything in between the problem i posted and smay's last message

[rare gale]

are you familiar with what a tangent is and some of the properties of tangents?

[tawny imp]

not really no :/

my math program tends to explain things very simply and make hard problems

[rare gale]

well in short, a line or line segment "is tangent" to a circle if it intersects it at a single point

[tawny imp]

alr

[rare gale]

so you can think of that as just grazing the outside of the circle, but not passing through the interior area

[tawny imp]

so like the line AD or AB?

[rare gale]

right, we can say those line segments are tangent to the circle

[tawny imp]

okay

[rare gale]

one of the key properties with tangents is that if you draw a radius of the circle from the point where they hit the circle, the radius and the tangent line \ segment will form a right angle

[exotic belfry]

do you really expect helpers should know what (for example) "Axiom 4 (Lesson 7)" means?

[rare gale]

so for example you can see segment AB here is part of a line that intersects the circle at a single point

and if O is the center of the circle, then angle ABO will be a right angle

[tawny imp]

huh ok

[tawny imp]

he seems to understand it pretty well so yeah

[rare gale]

so summary so far:

1. tangents intersect a circle at a single point, "grazing the outside"
2. they form right angles with some radius of the circle at the point of tangency

btw in that problem you showed, are you also supposed to change the order of the different steps?

because AB ~= AD doesn't really seem like the first step in the proof

[tawny imp]

its just a drop down menu to select which one defines the step best

[rare gale]

confusing because for the AB congruent to AD step i would think you would have to establish that BE congruent to ED first, but that step is further down

reposting so don't have to scroll up

so anyway, you chose circle definition for step 2, why was that? can you maybe show list of available options and think a bit more \ reconsider what you think it might be?

[fickle heath]

I'd assume the "tangent segment theorem" has something to do with that

[rare gale]

so if you don't know what any of these options mean, you probably won't be able to answer the question. so if that theorem sounds unfamiliar, can you look it up in your book or wikipedia as you want and see what it means?

what does tangent segment theorem state?

[fickle heath]

(the question is kinda terrible because of it, since the proof of that theorem - if I understand what it is - uses triangle congruency...)

[tawny imp]

i think im just gonna skip the question

i did crap on the first half of the lesson so i just gotta pray for higher than 80

[exotic belfry]

***so if you don't know what any of these options mean, you probably won't be able to answer the question. ** thats what i meant before. anyway.

[rare gale]

why? because you don't wnat to bother looking through your book on what the different names mean? 🤔

[tawny imp]

i dont have a book its video lectures

anyways i skipped the question and i got a 66 😭

technically i passed

[rare gale]

really, introductory math classes with no book? sigh

that's pretty atrocious

do you take notes while watching the lectures?

[white widget]

i dont know what i need to do after having found the dimension of the big rectangle (4x+1)(5x+2)

[rare gale]

!occupied

[lone heart]

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).

[white widget]

oh sorry i though it was a general help channel i didnt know it was for one person at the time

[lone heart]

@tawny imp Has your question been resolved?

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[dull lintel]

!status

[lone heart]

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

[velvet light]

1

[dull lintel]

Replace A with B-d
And C with B+d

Then use sine sum and difference formulas

[velvet light]

thanks

.close

[lone heart]

Channel closed

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Use .reopen if this was a mistake.

[lone heart]

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Send your question here to claim the channel.

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• 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>.
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[solemn geyser]

hi

[solemn geyser]

i am not really familiar with the english vocabulary for structures so please excuse me

i found that A5 , the group of even permutations of 5 elements is of the order of 60

and even though 6 devise 60 there exist no sub group of the 6th order

but couldn't we consider the group generated by (12) and (345)

since they can commute with each other we can consider it isomorphe with Z/3Z x Z/2Z

[wide tartan]

(12) is not an even permutation

that does not mean that A5 doesn't have a subgroup of order 6, though.

you can consider <(123), (12)(45)> and that's a subgroup of A5 with order 6 👍

[solemn geyser]

thanks!

but then since we've got cauchy's theorem that state that for every prime divisor of the order of the set there exist a sub group of that order

this means that there are non prime divisors of this order where there doesn't exist a subgroup of that order

[wide tartan]

there are some examples even in A5 yeah

15, for example

[solemn geyser]

oooooh i am dumb

but is there a way to know such a sub group

[wide tartan]

this was actually one of my past exam questions :0

[solemn geyser]

lol

[wide tartan]

you mean a way of telling whether there is one? there are no subgroups of A5 of order 15 is what i meant

[solemn geyser]

i understood that but i am not sure why, i think it has to do with how permutations commute with only those of disjoint cycles

then we can't have two permutations that generate a group of the order 15 if we don't have a permutation of the 3rd order that can switche with another of the 5 order

but why is them being able to switch so important

[wide tartan]

well, it's not incredibly important, except for the fact that the only group of order 15 is cyclic

for 6, there are non-cyclic groups of order 6, and we were able to find it in A5 by taking a group generated by two elements that was not isomorphic to Z/6 (notice how in my example the generating elements were, in fact, not disjoint permutations)

but for 15 we can't do that.

[solemn geyser]

yes but why can't we find a permutation of order 3 and another of order 5 and then see what they generate(in A5 we can't because there necessarily exist some elements that are in the cycle of the 3rd order permutation and the 5th order one at the same time)

this means that it's because the two permutations can't switch position with one another

[solemn geyser]

oh yeah you're right

now i am more confused lol

[wide tartan]

well, Z6 is not the only place to look for subgroups of order 6

in fact, there are no cyclic subgroups of order 6 in A5

[solemn geyser]

yes i agree with that

but the subgroup doesn't have to be cyclic

[wide tartan]

that's right

so what you might do, is see if there are any other isomorphism classes (the example that i gave was isomorphic to S3)

[solemn geyser]

what i understood is that the cycles being disjoint doesn't matter at all

[solemn geyser]

????

[wide tartan]

check it!

[wide tartan]

well, we do get a statement that if the cycles are disjoint, like (123)(456), then the order of the group generated by the two is gonna be the product. but if they're not disjoint, then you can't say that much about the order of the group generated by the cycles

[solemn geyser]

i see ...

i still don't get how your example works out

no wait i see

same cardinal

but then why can't we do the same to 15

[wide tartan]

because the only group of order 15 is isomorphic to Z/5 x Z/3

[solemn geyser]

oooh yeah yeah i get it now Z/15 isn't a group i

[wide tartan]

Z/15 is isomorphic to Z/5 x Z/3

it is a group but those two are the same isomorphism class

and we already know that we can't find one like that in A5 (because Z/15 is abelian!)

so now there's nothing left to check

[solemn geyser]

yes

[solemn geyser]

but then why does your example work ,Z/6 is abelian too

i am sorry for asking so much

[wide tartan]

yeah, and there are no subgroups that are isomorphic to Z/6. but the key thing is that Z/6 is not the only group of order 6, where in the 15 case it was the case that once we were done checking that no subgroup was isomrophic to Z/15 we were already done

don't be sorry, i enjoy talking about it

[solemn geyser]

oooooh i get it now in the first case the subgroup didnn't have to be cyclic so we could consider S3 since it is also of the order 6 but this doesn't work for 15 since we don't have another group other than Z/15 to work with in the first place

[wide tartan]

yep

[solemn geyser]

i get it now

thank you so much

[wide tartan]

good luck on your studies! also check out #groups-rings-fields

[solemn geyser]

thanks , this seems like it would help!

.close

[lone heart]

Channel closed

Closed by @solemn geyser

Use .reopen if this was a mistake.

[lone heart]

Available help channel!

Send your question here to claim the channel.

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

[placid flume]

Solve tan(2x) + 1/sin(x) = 1/tan(x) + 1/sin(5x)

[placid flume]

i have actually tried bashing it out algebraically

i can confirm

it does not work at least not easily

there MAY be a geometric solution however i don't see how to find it

PLEASE help

[unborn plover]

oh wait that's not a trig identity

are you just looking for solutions?

[placid flume]

yes

what...

[slender forge]

$\frac{\tan(2x)}{1 - \tan^2(x)} = \frac{1}{2\tan(x)}$ and $\frac{1}{tan(x)} = \cot(x) = \frac{\cos(x)}{\sin(x)}$

[dull lintel]

tan(x) is generally not equal to cot(x)

[placid flume]

let him cook

[dull lintel]

sure

[slender forge]

omg i cant get it written

1/tanx = cotx

YEAH

i think thats right now

[placid flume]

but that's NOT equal to sinx/cosx is it?

[slender forge]

😭

omg

im still just learning texit

[placid flume]

same 😭

[ocean seal]

Quantie

[slender forge]

ok now

does this look right ⁉️

[placid flume]

YES

[slender forge]

im so unserious but i know how to solve this i just gotta get this shit right

also do you have any specifics for x

[placid flume]

no :(((

implied x is real

and that's abt all ur getting

[slender forge]

yeyeye

[placid flume]

you're gonna have to write it in the thingy thingy notation

yk what i mean

[slender forge]

lemme check so i havennt messed up

do you have the answers

[placid flume]

DID U SOLVE IT

[slender forge]

pi/6 and pi/3

[placid flume]

OMGGGGGG

[slender forge]

trig is always messy to write, so thats the solutions then

[placid flume]

wait there's pi/2 as well apparently

show ur working

show ur working

[slender forge]

makes sense

[placid flume]

i dont need the solutions i cld get them from wolfram alpha i need the working :(((

[slender forge]

not allowed to just give the solution so we will get there together

[slender forge]

you have this

[placid flume]

oh yh ok

[slender forge]

put in the terms in the equation

and simplify it

whatu get

[placid flume]

what replace the 1/tanx on the RHS with tan2x/2-2tan squared x?

[slender forge]

you can simplify it to sin and cos terms

wait this is easier

oh

[placid flume]

yeesh ka peesh what now

[slender forge]

rewrite 1/tanx to cotx and bring them to the left and right side

[placid flume]

so what tan2x-cotx + 1/sinx = 1/sin5x?

quantie helpppp me

[slender forge]

@placid flume do you follow this

[placid flume]

Yes

Nice use of compound angle formulae to get the cos 3x

[slender forge]

Thank

Thats the LHS

Do you know how to simplify RHS

[placid flume]

Not immediately no but instinct is to use multi angle formulae?

That feels wrong however

[slender forge]

Could maybe work with more work, but i used the reciprocal

[placid flume]

??

[slender forge]

sin5x - sinx) / (sin5x * sinx

[placid flume]

oh yh ofc

just combining the fractions

i gotchu

ahh u can cancel sinx from both sides as long as it isn't equal to 0

checking that gives the pi/2 solutions

[slender forge]

After simplifying sinx - sin5x you can do that

[placid flume]

how do u do that????

[slender forge]

The sines identity

[placid flume]

remind me???

currently i have -cos3x/cos2x = sinx-sin5x all over sin5x

[placid flume]

oh yh i've seen that

so in this case alpha is x and beta is -5x

and then i can re cancel ANOTHER cos 3x of both sides as long as i check when it is eqaul to 0

[slender forge]

Yeah

[slender forge]

Wait with the cancels until youve completely simplified

[placid flume]

wait so currently the RHS is 2sin(-2x)cos(-3x) all over sin(5x)

that simplifies to -2sin(2x)cos(3x) all over sin(5x)

and then i divide the cos(3x)?

[slender forge]

Yeahh 🦅

But over sinxsin5x

[placid flume]

wait didn't the sinx get cancelled out earlier?

[slender forge]

Wait for that

Now for the million dollar question

Can you simplify RHS more

[placid flume]

hold up lemme just quickly rewrite everything without cancelling sinx yet

wait is that a double minus on the RHS for urs??

[slender forge]

no

[placid flume]

wait the numerator of it is -2cos(3x)sin(2x) right???

i dont see the simplification

[slender forge]

what can you do with sin2x

[placid flume]

double angule

2sinxcosx

WHOAH

[slender forge]

yeah

[placid flume]

and that cancels w the sinx on the bottom

right??

which is -2sinx*cos3x/sin5x???

[slender forge]

yeah you can cancel now

😛

[placid flume]

wait no 4

[slender forge]

but wait for the cos3x

[placid flume]

so we have -4cosx*cos3x all over sin5x

[slender forge]

yeah!

and LHS is simplified

now multiply both with -1 to get rid of the annoying -

[placid flume]

yes i seeee

then NOW can we cancel the cos3x??

or is therer still smth brewing

[slender forge]

keep it

[placid flume]

cos so far we have cos3x/cos2xsinx = 4cosxcos3x all over sin5x

[slender forge]

yeah

how do you continue

[placid flume]

and we're not removing the cos 3x?

[slender forge]

wait imma actually check

maybe i couldve all this time, not that itd matter much anyways

ok you cann

sorry

as long as cosx =! 0 🗿 👍

[placid flume]

yyh yh check those values

[slender forge]

hahaha

so yeah you got that without cos3x

how do you conntinue

[placid flume]

we have 1/cos2xsinx = 4cosx/sin5x

cross multiply the denominators seems natural?

[slender forge]

yep

[split aspen]

@little marlin

[placid flume]

with sin(5x) = 4cosx*cos(2x)*sinx

[slender forge]

yepp

now what

[placid flume]

there's still the cosx term so i dont think i can fully turn it into a sin polynomial

i can turn cos2x into 1-2sin squared x

[slender forge]

thats true

but look closer

[placid flume]

i can't see what ur seeing

wait is this the angle identity again

[slender forge]

4sin2xcosx * cos2x

can you see it

[placid flume]

2sin(2x)cos(2x)

i haven't made a mistake there right?

WHICH GOES TO SIN(4X)

[charred pilot]

sin(5x)=2cos(2x)(2cosxsinx)

[slender forge]

yeahh

[placid flume]

so we're looking for values when sin(4x) = sin(5x)

[slender forge]

you go back to -2cos3xsin2x

[slender forge]

YEAH

[placid flume]

okok

nice CRAZY reduction

[slender forge]

hahaha

[placid flume]

alright final steps now?

[slender forge]

love hate relationship with trigonometry

[placid flume]

real 😭

[slender forge]

sin5x - sin4x = 0

recognise that?

[placid flume]

OMG IT'S THE ANGLE IDENTITY THING FROM BEFORE

[slender forge]

yessir

[placid flume]

with alpha being 5x beta being -5x

-4x

so that's 2sin(x/2)cos(9x/2) = 0

[slender forge]

just 4x, not -4x

wait its the same thing

its fine

[placid flume]

okok

[slender forge]

now just solve for each case

[placid flume]

YH SIN(X/2) = 0 AND COS(9X/2) = 0

[slender forge]

cos 9x/2 = 0 and sin x/2 = 0

yep

[placid flume]

DAMN

[slender forge]

that wasnt that hard was it

[placid flume]

ur acc insane

[slender forge]

$\frac{\pi(1+2\pi}{9}$ and $n \in \mathbb{Z}$

damnnit

[placid flume]

i get it tho yeah

solving the cases is trivial

[slender forge]

there we go

wait no

[ocean seal]

Quantie

[slender forge]

aha!

DAMNIT i give up on making it pretty

alright so yeah

thats it

[placid flume]

😭

thanks so much

journey of a q

.close

[lone heart]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[astral void]

Hello! Can someone help me understand primes (no not prime numbers, but f'(X) and rate of change kind of prime numbers)? I know the answer to both of these questions is 14, but I have no idea why or how we got there.

[azure needle]

Derivative?

[hushed locust]

f'(x) means the derivative of f(x), so you first need to take the derivative

[astral void]

Sorry if this is a dumb question, but what is a derivative?

[azure needle]

Kind of like the differentiated form of an expresion

[opal lintel]

I think it is first to derive and then replace the value

[azure needle]

Yes

Although that's not needed

[opal lintel]

yep

[azure needle]

Do you know what rate of change means

[astral void]

I know rate of change is the slope between two points

[sonic sedge]

hiii

can someone help me with a math question

[azure needle]

Now the equation is in slope intercept form

[lone heart]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[urban tree]

?

[azure needle]

Do you know what the slope is?

[astral void]

so the slope would be 14... is that what a derivative is? I guess I'm still a little confused on what a derivative is. I know you said it is the differentiated form of an equation, but I'm not really sure what that is either :/

[azure needle]

No

[astral void]

btw I really appreciate you guys helping me with this, I've been stumped for like an hour trying to figure it out by re-reading my class notes

[azure needle]

y=ax+b

a is the slope

Compare this with your question

@astral void

[astral void]

the equation is F(x) = 14x-16, so 14 would be the slope.

[azure needle]

Yup

[astral void]

So the derivative of a function is just the slope?

[azure needle]

Yes

Derivatives are in calc 1 or something

[astral void]

Yup, that's the class I'm in

Okay thanks guys! I appreciate it!

.close

[lone heart]

Channel closed

Closed by @astral void

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[lone heart]

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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.

[green osprey]

this gype of graph is ocnsidered an even graph right?

[hushed locust]

it's a graph of an even function

[green osprey]

yeah but why is it an even function

how would I decipher that if I was given nothing

I understand I can split it up into e^(1+x)(1-x)

but how would that give me anything

[hushed locust]

given the graph, you can see it's symmetric about the y-axis
given the function definition, you can find that f(-x) = f(x)

[green osprey]

so would I sub in -x to confirm

oh wait

I see what you mean

so f(-x)=f(x) means even function

then does -f(x)=f(x) mean odd function?

[hushed locust]

odd function means f(-x) = -f(x)

[green osprey]

ok thanks

.close

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[wispy matrix]

Graph the following system of linear inequalities.

−6x + 3y   ≤   −33
x − y   ≥   5
x   ≥   0
y   ≤   0
I was given this problem, I know it's unbounded but I can't figure out what area to shade, I'm thinking the whole thing will be since one linear inequalities is less than, and the other is greater than. Also since the bottom two linear inequalities are false do I not graph them?

[wispy matrix]

hope this helps

.close

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[alpine sable]

is this the right integral?

I'm not sure if I need the 9.8. pretty sure I do but taking 9.8(40,000) outside as a constant seems like a really big number

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@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[lusty sphinx]

Hey im trying to get some help with finding the domain on z(x)=sqrt (x/x-1). i got an answer of 0,1 U 1,infinty but im confused on what to do with the brackets rather they are open or closed brackets

[alpine sable]

@everyone im new

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@alpine sable Has your question been resolved?

[alpine stump]

can someone help me with a finance problem

When you reach retirement​ age, you would like to have enough money saved to be able to​ “pay yourself” an annual salary of ​$62​,000 per year for 20 years. To put this another​ way, your plan is to start your retirement with a large amount of money​ saved, and you will withdraw ​$62​,000 from these savings once a year for the next 20 years until all of your savings are depleted.
In the​ meantime, you are a​ 25-year-old new UIC​ graduate, and you plan on working for 40 years until you retire. To fund your retirement​ goals, you plan on investing some money in the stock market. More​ specifically, at the end of each year until you​ retire, you are going to put part of your paycheck into the stock​ market; you'll put in the same dollar amount every year for the next 40 years.
You are a pretty decent stock​ investor, and you think you can make a 12​% return on the market each year you​ invest, both until you retire and after retirement.
What is the annuity payment​ (to the nearest​ dollar) you need to put into the stock market every year for the next 40 years to fully fund your​ retirement?
Write your​ answer, without a dollar sign in​ front, rounded to the nearest whole dollar. ​ (If you do this​ correctly, it might be a smaller number than​ you'd think!)

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[candid dune]

hey

[lone heart]

Please don't occupy multiple help channels.

[candid dune]

i have a question for a math problem

[charred pilot]

What is it

[green shoal]

what is your question

[candid dune]

i need to find the distance to all the points

[candid dune]

so like A to D

A to B

D to C

B to C

[green shoal]

So do you know how to do this and need help or have never done this before

[candid dune]

well i know how to do it but im stuck with a to D

because D had (x,8)

@green shoal

[green shoal]

please get someone else I've got something that came up sorry

[candid dune]

its alright

could someone help me?

[charred pilot]

I would recommend setting variables to the coordinates of C and finding how each point connects

For example, C is (x+12,0)

[candid dune]

(12x,0)?

[charred pilot]

No

Do you see how I got that point

[candid dune]

i dont know how sorry

sorry btw my english is bad

[charred pilot]

The y value is 0 because it is on the x axis

[candid dune]

my math is in french

[candid dune]

ohh

[charred pilot]

The x value is the same distance from D as A is from B

The x value of A+12 is the x value of B

[candid dune]

alright alright wait

where is 12

[charred pilot]

14-2

[candid dune]

but why

[charred pilot]

That is how far apart B is from A considering the x value

[candid dune]

oh so how can i find the x,8

(x,8)

[charred pilot]

The distance formula is

\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

[ocean seal]

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

[candid dune]

alright ill write that

[charred pilot]

You know that the distance from A to B is the same as D to C

[candid dune]

yes i know

[charred pilot]

(Since it is a rectangle)

[candid dune]

but how is D the same distance from B as A

[charred pilot]

D to C is the same as B to A

Find the distance from B to A

[candid dune]

alright alright

will do

[charred pilot]

\begin{align}
\sqrt{(x-2)^2+(8-2)^2}\\
\sqrt{x^2-4x+4+36}\\
\sqrt{x^2-4x+4+36} = \sqrt{(a-14)^2+(0+6)^2}\\
\sqrt{x^2-4x+4+36} = \sqrt{a^2-28a+14^2+36}
\end{align}

[candid dune]

i got 14,42

for distance to A to B

[charred pilot]

14.42?

[candid dune]

yes

[charred pilot]

What is it in exact form

[candid dune]

14,42 units

wait wdym

[charred pilot]

Without the decimal

[candid dune]

14

[charred pilot]

No

Before you calculate the decimal

[candid dune]

ohh

208 but with the square root

[charred pilot]

So $\sqrt{208}$?

[ocean seal]

mari

[candid dune]

yes

but can u make sure that it was the right thing that i did

[charred pilot]

In a bit

[candid dune]

alright

[ocean seal]

mari

[charred pilot]

Okay

Ignore that

[candid dune]

alright

[charred pilot]

To find the distance from A to D, you have to find the x coordinate for D. You know that the distance formula from A to D is equal to the distance formula from B to C. Instead of putting x in the distance formula, put it in as 2+a for the A to D one and put it as 14+a for the B to C one

Ill check your work for the other while you set that up

[candid dune]

wait what do u mean by 2+a and 14+a

[charred pilot]

\begin{align}
\sqrt{(14-2)^2+(-6-2)^2}\\
\sqrt{144+64}\\
\sqrt{208}
\end{align}

[ocean seal]

mari

[charred pilot]

For the unknown points

Instead of putting just the variable x, put 2+a for the first one

And 14+a for the second one

[candid dune]

oh alright

[charred pilot]

\begin{align}
\sqrt{((2+a)-2)^2+(8-2)^2}\\
\sqrt{a^2+36}\\
\end{align}

[ocean seal]

mari

[charred pilot]

Like that, but do 14+a for the next one

[candid dune]

so (2+a - 2)^2 + (8-2)^2

[charred pilot]

\begin{align}
\sqrt{(x-14)^2+(0+6)^2}\\
\sqrt{x^2-28x+14^2+36}\\
\sqrt{x^2-28x+160}\\
\sqrt{x^2-28x+160} = \sqrt{a^2+36}\\
x^2-28x+160 = a^2+36\\
x^2-28x+124 = a^2
\end{align}

Actually keep the a for one and put an x in for the other

So we can find what one of them equals

[candid dune]

oh

wait so at the moment i have (6)^2 + (2+a-2)

[ocean seal]

mari

[charred pilot]

This is harder than I thought it would be

[candid dune]

wait i have an idea

we could find the slope for A to B

and use it to find like the distance

D to C maybe

By replacing

[charred pilot]

You already know what D to C is

[candid dune]

14,42x right

[charred pilot]

But yes we should find the slope of the line connecting A and B and find a slope perpendicular to it

[charred pilot]

Just use sqrt(208)

[candid dune]

alright

[candid dune]

but why

[charred pilot]

To not have to intermediately round, its a good habit, exact answer, easily manipulatable

[candid dune]

alright got u

[charred pilot]

\begin{align}
\frac{-6-2}{14-2}\\
\frac{-8}{12}\\
-\frac{2}{3}
\end{align}

[ocean seal]

mari

[charred pilot]

That's the slope from A to B

[candid dune]

wait i did a mistake, 14,42 is the distance between A to B , thats not the slope sorry

oh alright thanks

so -2/3

[charred pilot]

Now we need to find the slope perpendicular to that

[candid dune]

so 3/2?

[charred pilot]

Yes

So move up 3 units from A and to the right 2 units

[candid dune]

so its A to D and B to C

[charred pilot]

x is 6

[candid dune]

oh but how did u find that

[charred pilot]

D is (6,8)

[charred pilot]

This

But twice

C is (18,0)

[candid dune]

sorry to ask why so many times but why lol

[charred pilot]

Now find the last distance

[candid dune]

alright

[charred pilot]

Because that is how slope works, it is a change in y over a change in x. I am moving up 3 units and moving right 2 units for each interval. Since the y value for D is 6 more than the y value for A, I have to do this slope progression twice

[candid dune]

ah so if D'S y is higher value than A u do it twice?

[charred pilot]

No

If I do the progression once, I will find the midpoint between the two

How about this, try the progression once and tell me what point you get

[candid dune]

alright wil ldo once i find the distance between D and C

i found 208 again

the square root of 208

[charred pilot]

It's not 208

[candid dune]

what is it

[charred pilot]

I haven't done it but show your work here

[candid dune]

alright so i did (x2-x1)^2 + (y2-y1)^2

(18-6)^2 + (0-8)^2

good so far or no

[charred pilot]

You found the distance from D to C

We already knew it was sqrt(208)

Find the distance from A to D

Which will equal the distance from B to C

[candid dune]

alright

is it sqrt(52)

[charred pilot]

That sounds right

[candid dune]

alright so now i have to do it times two

[charred pilot]

Do what times two

[candid dune]

after doing the square root

vbecause

i need to find the air

[charred pilot]

What?

[candid dune]

like base x height

oh wait no

i dont do times two

[charred pilot]

The area is sqrt(208)xsqrt(52)