help-27

[fossil carbon]

if that’s the right graph

then uh

yeah

[maiden nexus]

Oh sorry

For 1

I picked C

For 2 I picked D

Those ones I was pretty confident

It’s these last 2

For 3 I picked D

And 4 D as well

<@&286206848099549185>

[fossil carbon]

C

and D

are the answers

actually

i’m not even sure anymore

the more i think about it

let me google smth

[maiden nexus]

🤣

[fossil carbon]

no i was wrong

it’s D and D

it would be C

if the highlighted bit

was concave

but it’s convex

is that all u needed help with?

actually

@maiden nexus

do u have this shit without ur pencil drawing

cause the last one might be wrong

[maiden nexus]

See and that why I’m confused

[fossil carbon]

but i can’t tell

with ur drawing

[maiden nexus]

No that’s how it was printed from my teacher

[fossil carbon]

well that last one

could either ber

[fossil carbon]

D or C

[maiden nexus]

That’s the answer that my teacher gave us

[fossil carbon]

okay

last one is 100% C

[maiden nexus]

But what about 3

[fossil carbon]

yeah i dont actually get that

lmao

because it’s decreasing

but the gradient

is getting closer to 0

[maiden nexus]

He says that he’ll get some wrong in purpose

[fossil carbon]

then

[maiden nexus]

But I just wanna make sure

[fossil carbon]

number three

i think is D

if i’m wrong

i apologise

[maiden nexus]

That’s what I think too thank you

.close

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[gentle urchin]

can someone PLEASE check my work as thoroughly as possible

[devout snow]

@gentle urchin Has your question been resolved?

[gentle urchin]

<@&286206848099549185>

[gentle urchin]

hiii can you check my work please if you know how

did you actually check though

im not sure that has anything to do with my question?

why do you have the helper role then?

<@&286206848099549185> can someone else take a look please

yeah but not really for this question

university

[boreal lance]

why not shouldnt it work here

[gentle urchin]

cause if you look at the graph its not really only curve on top of another

@misty crest sorry for the ping but if youre willing to could you take a look please

similar question to yesterdays

ok i believe you

thanks for your help

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[heavy current]

how's this proof?
#1260424130822541403 message

[devout snow]

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[mossy cave]

How can i fold a square to get another square that's half the area of the original square without folding the corners in

[solid osprey]

wdym "folding the corners in"

[mossy cave]

Like there's one way where u fold all four corners to the center

Not including that method

[upbeat marten]

just fold it in half?

[frosty cave]

I’m guessing we fold the edges which somehow gives half the area

[upbeat marten]

would that not work

[mossy cave]

U need to get a square

[solid osprey]

not a square

[upbeat marten]

oh

[frosty cave]

Square not rectangle

[solid osprey]

just fold till the sides are 1/sqrt 2 :p

[mossy cave]

Idk the measurements

Ur not allowed to use a ruler

[solid osprey]

ow

[rare mantle]

what is this for lmao

[mossy cave]

Ig ill just give up

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[lost laurel]

what does this even mean ?

[lost laurel]

Like I don't understand what non-leading variables are

[restive river]

i think that when you multiply said reduced matrix with whatever the variable matrix is to form the equations, every variable that gets multiplied with one of the leading 1's in each row is called leading

and the ones that never get multiplied with a leading 1 are non-leading variables

[rustic laurel]

a leading variable is the first variable in a row

[radiant anvil]

https://m.youtube.com/watch?v=VNMjJ92rAw0

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[errant pier]

hi so i have a problem,i did this expression,the result should be 0 at the end but i dont know what im doing wrong at the final steps..can some1 help me?

[uncut crow]

slay

[restive river]

slay

[uncut crow]

looks like you did get 0, if that's (-2/3 a^2) * 0 at the end there

[devout snow]

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[errant pier]

ohh this explains it

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[viral lynx]

I drew a lil diagram its probably all wrong

Cause idk what f looks like or l

l could have a positive slope, f could look like anything

B could be above or below the origin, A could be negative aswell

How am I supposed to do this without any numbers?

Omg 💀

They gave us the numbers in part a I didnt realize

🤡 🤡 🤡 🤡

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[lilac tusk]

Hello, I have a question!
Regarding the bolanzo-weierstrass theorem of sets in n dimensional space.

Is my reasoning in the proof correct?

Suppose S is an infinite set in Rn contained within some n-ball B(0;a).

Then S is contained within the n-dimensional interval (n-cell for short) [-a,a]×[-a,a]×...×[-a,a].

We can split this n-cell into exactly 2^n new n-cells(for example a line can be split in half, a square can be split into 4 new squares, a cube into 8 smaller cubes), composed of the cross product of n 1-dimensional intervals, of length 2a/2=a.

At least One of these n-cells contains infinitely many points of S, call this n-cell C_1.

Repeat the process and split C_1 into 2^n new n-cells, composed of 1-dimensional intervals of length a/2. Choose the n-cell containing infinitely many points of S, call it C_2.

We may repeat the process to the mth n-cell, C_m an n-cell, composed of n 1-dimensional intervals of length a/2^(m-1).

Since C_m is always a subset of C_(m-1) from the way we constructed them, and all C_k contain infinitely many points of S, if we consider the infinite set of all n-cells constructed in this way, there must be some point t that is an element of all n-cells C_k. And then we proceed to show t is an accumulation point, since for any n-ball B(t;r), we simply choose m sufficiently large s.t. C_m is contained within B(t;r), and since C_m contains an infinite subset of S, we are done and t is an accumulation point of S.

(I know the last part would have to be described better, but Im interested in if I can assert that t exists, since all n-cells C_k are all non empty subsets of one another, they must contain a common member t?)

[devout snow]

@lilac tusk Has your question been resolved?

[lilac tusk]

<@&286206848099549185>

[devout snow]

@lilac tusk Has your question been resolved?

[lofty lintel]

you can assert t exists by using the nested intervals theorem

so first justify that each [-a, a] that you start with follows this pattern, for example the first [-a, a] might go like this:
[-a, a] -> [0, a] -> [0, a/2] -> [a/4, a/2] -> ,,,

(make that a bit more rigorous and show that conditions (i) and (ii) hold)

[fossil carbon]

I just wanna appreciate how mind boggled i was when i read the first sentence of this question

[lofty lintel]

now consider C = C_1 ∩ C_2 ∩ C_3 ∩ …

and by construction, write each C_i = I_i1 x I_i2 x I_i3 x …

where each I_ij is a closed interval

by the nested intervals theorem, for each j, I_1j ∩ I_2j ∩ I\3j ∩ … is a singleton, call that element t_j

[lofty lintel]

thus C is also a singleton and contains the element t = (t_1, t_2, t_3, ..) and thats how you justify that t exists

[lofty lintel]

i think maybe asserting that there is such a sufficiently large m so that C_m is contained within B(t;r) needs more justification

Like of course each C_i is a cross product of intervals of length a/2^(i-1)

meaning for any x = (x_1, x_2, x_3, …) in C_i, we have that |x_j - t_j| <= a/2^(i-1) for each j.

thus |x - t| <= √((a/2^(i-1))^2 + (a/2^(i-1))^2 + … + (a/2^(i-1))^2) = √(n * (a/2^(i-1))^2) = √(n) * (a/2^(i-1))

so you really have to choose m large enough so that √(n) * (a/2^(m-1)) < r

i think the usual proof of the n-dimensional case is just generalized from the 1-dimensional case like this

[devout snow]

@lilac tusk Has your question been resolved?

[lilac tusk]

By sequence this means set of ordered pairs of form (1,$I_1$)?

[woven radish]

zigzag

[lofty lintel]

i mean sure i guess

technically a sequence is a function with domain N

but i really just mean I_1, I_2, I_3, …

or you can think about it as two separate sequence a_n and b_n

where I_n = [a_n, b_n]

[lilac tusk]

So i see how that theorem works

U sure its ok to use convergance and limits in point set topology ?

They werent introduced un tge book yet

I feel like there must be a nice way with sets only

[lofty lintel]

idk what you mean

this is in R

so you can use theorems about R

[lilac tusk]

I guess

[lofty lintel]

if you want you can prove the nested intervals theorem too

[lilac tusk]

Using R to prove Rn

[lofty lintel]

but i don’t see any way around using the nested intervals theorem with your proof method

[lofty lintel]

this is an alternate way to do the proof

[lilac tusk]

Yeah

[lofty lintel]

(and you still have to prove the 1-dimensional case if you do this)

[lilac tusk]

Gtg out now and will thoroughly read your method later

Consider this question answered

Thankyou!

.close

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[lofty lintel]

np

[devout snow]

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[stone stump]

this is an english server so you will have more luck by translating it into english

[pulsar sand]

Omelette du fromage

[devout snow]

@small solar Has your question been resolved?

[olive snow]

Que fait tu du deuxième au troisième équivalent ?

[woven radish]

RH

[olive snow]

Mais cos(x) c'est pas /2 aussi ?

Et sin par 2i ?

Après l'arc moitié

[woven radish]

RH

[olive snow]

Ok je vois

Un peu trop rapide pour moi xd mais ça fonctionne

[olive snow]

Tout depends du niveau auquel tu es, peut être une étape de plus pour ça

Et après éventuellement préciser la multiplication par i-z

Sinon il y a des français ici, c'est juste que tomber sur un d'entre eux est assez rare

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[exotic oriole]

complex numbers smh

[exotic oriole]

The complex number 2 + yi is denoted by a, where y is a real number and y < 0. It is given that f(a) = a^3 - a^2 -2a. Find a simplified expression for f(a) in terms of y.

imma send my workings, i dont know what i did wrong

The answer is -5y^2 + (6y - y^3)i

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@exotic oriole Has your question been resolved?

[exotic oriole]

<@&286206848099549185> 😩

[cold bone]

you did mistake in (iy)^3 expansion

[exotic oriole]

urgh i think i saw it

stupid mistake

i'll work it rq

thanks!

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[sterile cairn]

No idea

[chrome isle]

The given circle is the incircle
And there is a formula to find the inradius (radius of incircle ) = area /Semiperimeter

[devout snow]

@sterile cairn Has your question been resolved?

[sterile cairn]

if u use this formula does it always work?

I got 0.707 as the radius

but none of the options give smth that can sprout from that radius

[chrome isle]

It should since it was proved thousands of years ago

[sterile cairn]

oh but what did u get using the inradius

it can't be 0.707

right

?

[chrome isle]

Show ur work

[analog badge]

The formula is ( a + b - hyphothenuse ) /2 i think

[sterile cairn]

I used a program for it

maybe the program is wrong ig

on my calc

[chrome isle]

u just have to calculate the perpendicular and find area and semiperimeter

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[sterile cairn]

How do I do this?

[chrome isle]

Equate the two functions since they meet only at one point there is only one x they will satisfy it

[sterile cairn]

I got x^2+2x+k^2-2kx=0

then idk how to continue

[mystic scarab]

If the solution is one point, how must the discriminant (∆) be?

[sterile cairn]

I equalled it to 0 and got 4k^2-8kx-4=0

[chrome isle]

Since only one x satisfies this the roots must be equal

[mystic scarab]

The discriminant can't contain the variable x

[sterile cairn]

yea

thats y I didnt know

[chrome isle]

how does the discriminant have x in it

[sterile cairn]

lol

wait so what do I do then?

[chrome isle]

Whats your quadratic eq write it purely in terms of x

[sterile cairn]

got it, its x^2+(2-2k)x+k^2
so the discriminant of this is (2-2k)^2-4k^2

and then equal to 0 and so on

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[fading canopy]

I just wanna make sure im not crazy here, for the question (see picture) this desmos function (also see picture) will always give the correct solution for this problem type right?

Im a little new to desmos regression functions but this seems way faster to me than evaluating this by hand.

[fading canopy]

s=8 here is the correct answer, and I tried with a couple similar problems and a correct answer

i'd also be fine with a hand solution that is relatively quick rather than this as I trust doing it manually more. The problem with the normal hand method for me is that it is too slow

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[devout snow]

@fading canopy Has your question been resolved?

[fading canopy]

.close

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[raven trout]

I tried using this method my math teacher explained to us, I’m not sure if I even did it right.
Question: find the exact values of cos2(theta) -5=-6

[raven trout]

Would appreciate the help ❤️

[patent birch]

when you have cos(2theta), you can't just divide by 2

[unique marlin]

Yea the two is inside the cosine, so it would be divided after taking the inverse cosine

[raven trout]

So I would divide 2pi/3 by 2?

[unique marlin]

well it would be the inverse tangent of -1

then divided by two, yes

[raven trout]

Ok I think I get it now

Thx sm

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[raven trout]

This is another problem. Would appreciate the help double checking it’s all good. The main problem is f(g(2)) and using the functions to solve it

[winter patrol]

is there more context?

the question makes no sense

[raven trout]

Not really sorry, the section of the hw that contained that problem was just titled evaluating trig expressions

[winter patrol]

can you take a pic of the original

[raven trout]

Yeah lemme send it

It doesn’t have much context tho

Sorry abt the handwritting, our teacher had us write the problems

She didn’t rlly give us a handout

But it was just like plugging in 2 for every x

[winter patrol]

there's no info about f,g

just skip

[raven trout]

Alrighty

I appreciate the time you tried to help me

No worries

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[civic grail]

Is this correct? khan academy says im wrong. the instruction is rewrite the function by completing the square

[winter patrol]

,rotate

[winter patrol]

whats your final answer

you messed up your signs

[civic grail]

it just says rewrite the function so my answer is (x-9/2)² +134/7

[winter patrol]

you did
81/4 + 14
instead of -81/4 + 14

[civic grail]

ahhh i didnt notice that

so its -67/4?

[winter patrol]

no

now you're forgetting denominators

[civic grail]

yeah 😭

1 under 14 then common denominator?

ah okay there

-25/4

[winter patrol]

yes

[civic grail]

now i get (x - 9/2)² - 25/4

i dont know why theres this blank before the quantity in khan academy

oh

sorry im sleepy

okay i got it now

thank you

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[scenic surge]

For this DFA how am I suppose to move out of q_0

if it has a self loop

like that

if I input 1 or 0 it self loops...

oh wait these r NFAs

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[granite bough]

Hello

[granite bough]

I have an expression I need to turn into a quadratic? idk what it is called in English

the expression is x^2 + 4/3*x - 1

[buoyant laurel]

?

4/3x?

[granite bough]

They want me to write that expression in the form of (a-b)^2

i dont know what it is called

or it could be (a+b)^2

I am not sure where to begin with this particular expression

[granite bough]

yes sorry

[buoyant laurel]

$(x+2/3)^2-5/9$

[granite bough]

how did you arrive to that expression

[cold bone]

completing square

[granite bough]

thanks

[woven radish]

zephyr

[granite bough]

yeah I don't know how you got there, I have been doing a few completing the square problems that were much easier than this one

[buoyant laurel]

I tap wrong

[granite bough]

and they have been going okay but this one I have no clue where to begin

<@&286206848099549185> How would I begin completing the square of:

$x^2 + \frac{4}{3}x - 1$

[woven radish]

Neo

[granite bough]

Or rather where

[chrome isle]

This can be written as
x² + 2×2/3x + (2/3)² -1 -(2/3) ²

[granite bough]

I don't want the answer 😦

[restive river]

bring the constant to the other side

[warm breach]

source: mathisfun

[restive river]

add one

[chrome isle]

Think how can we turn the expression into the form a² + 2ab + b²

[granite bough]

I will try to do it a little bit longer and come back if I need more help

[chrome isle]

Ok

[granite bough]

Been trying but I can't see it yet

[chrome isle]

There is a a² term as x²
There is a 2ab term as 4/3x just think what b is here

[granite bough]

I was initially thinking among these lines but

okay

I think I get it now

What would I multiply 2 with to get 4/3

[chrome isle]

Yes that's your b

[granite bough]

ah

okay i will keep going (im very slow)

i should be able to solve it

[chrome isle]

np

[granite bough]

i managed to solve it but that was pretty difficult for me

i got it to

$(x + \frac{2}{3})^2 - \frac{13}{9}$

[buoyant laurel]

Oh yes

U got it

[woven radish]

Neo

[granite bough]

My calculations were pretty confusing and tedious, would there be a more efficient/less tedious way to do this? I found it difficult to "turn" the -1 to 4/9 without changing the overall value of the expression

[raw knoll]

uhm yes

[granite bough]

wait I just realised I added and subtracted 1 for no reason

[raw knoll]

so when you have an ax²+bx+c=0

you can change it to x²+b/a+c/a=0

[granite bough]

okay nevermind I understand how to make it more efficient as well now

[raw knoll]

yeah you just divide it by 2

[granite bough]

thanks for all the help, sorry if anyone found me rude to interact with

.close

[devout snow]

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[granite bough]

sorry my internet isnt the best

i didnt see anyone replied when i closed it

[devout snow]

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[granite bough]

Hello

[granite bough]

Where did I go wrong?

I think there are many places

Because my answer is way off

And is there a fool proof way to do this

Someone showed me this

would this be the fool proof way?

[granite bough]

<@&286206848099549185> I would appreciate if someone could just point out the first step that went wrong here

hello mathbowl

I'm from Sweden

Sorry my internet is bad, I’m not entirely comfortable sharing exactly where I study

[long sundial]

It should become $(x+1/2)^2-3-1/4, not +1/4$

[woven radish]

Max

[granite bough]

okay i will try to correct

i unfortunately still cant make sense of it, im going to come back to this tomorrow and just move on now

.close

[devout snow]

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[granite bough]

.reopen

[devout snow]

✅

[granite bough]

I am doing something wrong with the negative sign, there was another problem with -x^2 and I ran into a similar issue, though the original problem seems to have gone more awry than this one.

-26 should be positive

is this what i have done wrong?

is it because i broke out -1 at the start

that i am making these mistakes

if i simplify that expression i see that it is wrong

If I break out -1 at the start, how would I add or subtract?

[dreamy linden]

yes as it should be positive 26

[granite bough]

yeah but why should it be pos. 26

i get that it should be but i don't understand why

i see that it would be wrong if it isn't positive

[dreamy linden]

just open up the expression in third step and you will know

[granite bough]

yeah i did

i can see why it is wrong

i dont understand why it is right

it has something to do with breaking out -1

i must be missing some parenthesis as well

[mystic scarab]

Because -1 = 25 - 26

[mystic scarab]

You put the closed parenthesis in the wrong position

[dreamy linden]

yes -25 and -26 don't sum up to 1

[granite bough]

yeah that sounds right

where should i have put it

[granite bough]

i agree

[mystic scarab]

In your opinion?

[granite bough]

i dont know! that's why i am asking

this step was correct right?

[dreamy linden]

yes

close the bracket after -26

[granite bough]

if i want to add 25 into the inside of that parenthesis and also subtract at the same time

i dont know how i would do that

[mystic scarab]

I'm forcing you to think about it, otherwise you'll encounter the same issue in the future

[dreamy linden]

that will yield the orignal expression

[granite bough]

alright i will think about it

i have already but i will think some more

[granite bough]

idk if this is right or not but from what you helping me have said this is what I'd imagine it is supposed to look like

[granite bough]

but I don't understand why, and when I try to justify why then I imagine when you would move the -26 out of the parenthesis, since the parenthesis was being multiplied by -1, you would have to multiply -26 by -1

[mystic scarab]

That's exactly it!

[granite bough]

are you just saying that to be nice

because that sounds very foreign to me

but that would explain why I am having this issue in the first place

[mystic scarab]

It's the same as $-(a+b) =-a-b$, nothing different from this

[woven radish]

Alberto Z.

[granite bough]

yeah i guess im okay with that

i think my biggest issue was where to put the parenthesis?

[mystic scarab]

I just pointed out the fact that you found exactly why you were wrong with that sign

[mystic scarab]

Leave them where they are

[granite bough]

yeah but I put them there based on your guys' comments

that's not something I figured out

this was the original and previous problem

I think I have done something similar wrong here as well

except here I put the parenthesis in the righ tspot

no wait

nah i dont know what is happening

let me try to redo this problem

That's the right answer, but I don't know if I brute forced it or not

Did anything illegal happen here

[granite bough]

I still can't explain what happened here

[granite bough]

<@&286206848099549185> I just need someone to confirm that I didn't do anything illegal here

[opaque folio]

Yep

Nothing wrong with that

[granite bough]

Maybe a little long winded but I'm okay with that

[opaque folio]

Yeah

You’ll get faster

[granite bough]

Thank you Kiter007 and everyone else helping me

[opaque folio]

As you start to do the fractions in your head and finding the number ur adding and subtracting

[granite bough]

I wish you all only the best

.close

[opaque folio]

Your welcome

[devout snow]

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[devout snow]

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[lost crag]

how to prove, given A_0, A_1,... are sets

[lost crag]

after taking $a \in LHS$ idk how to prove it

[woven radish]

Ayanokoji (ALWAYS PING ME)

[static ember]

if a is an element of the left hand side, think of what that means. It means there exists an $i \in \mathbb{N}$, such that ...

[woven radish]

rbit

[lost crag]

can you show me the proof? this is unintuitive to me

[arctic field]

expand out the definitions of union and intersection

its purely a definition chase

[static ember]

no sadly I cant just show the solution

[arctic field]

are you the real rbit

[static ember]

I think so

[devout snow]

@lost crag Has your question been resolved?

[devout snow]

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[fervent pewter]

why is it lnv - ln (1-v)

[devout snow]

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[safe trellis]

Can someone help me understand this? Im not sure if im correct

[median musk]

It seems you are correct for question a

[safe trellis]

Oh, thank you so much. I wasn't quite sure why the calculator multiplie the 1/4 and √3/2.

[devout snow]

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[inland mesa]

can anyone give me a hint

i have no clue how to do this

[misty crest]

how would you solve a quadratic?

there’s your hint

[inland mesa]

ah ok ill try that then

[restive river]

quadratic.

[inland mesa]

but how would i get it into that form of ax^2+bx+c

[misty crest]

rewrite 2^2x+1

[restive river]

you need to smartly consider some term(s) as other single variable

[misty crest]

think of it as au^2+bu+c

[inland mesa]

2*2^2x

[misty crest]

mhm which is also 2(2^x)^2

[inland mesa]

so do i assighn it to a variable then

oh

[misty crest]

make sense

it’s a quadratic in terms of what?

what is the "u"

[misty crest]

^

[inland mesa]

2^x is the u?

[misty crest]

mhm

so then you’re really solving for 2^x

the same way you’d solve a quadratic

and from there you’ll have 2^x=some number

then you can solve for x

[inland mesa]

would it be best if i assighn a random letter to 2^x

[misty crest]

yea sure you can do that

like u

[inland mesa]

2g^2-5g-12=0

[misty crest]

or that yea

then solve for g

then go back after that to solve for x

once you have values for g

[inland mesa]

(2g+3)(g-4)

so G is -3/2 and 4

[inland mesa]

how

do i do 4 = 2^x?

[misty crest]

mhm

and the same for the other one

but you’ll notice that 2^x>0

and thus ≠-3/2

[inland mesa]

wait so if -3/2 = 2^x then how do i solve for x?

[misty crest]

log base 2 on both sides

but the domain of a log function is x>0

[inland mesa]

AHHHH ok

ty i understand now

[misty crest]

there is no power x such that 2^x<0

[inland mesa]

i see

thanks

.close

[devout snow]

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[devout snow]

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[untold pivot]

Im having trouble with (b)

[woven radish]

somethingwrong

[untold pivot]

but im not too sure what to do to show that the tangent the rest of the points exists

[devout snow]

@untold pivot Has your question been resolved?

[devout snow]

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[visual maple]

what was the formula for perfecting the square?

[visual maple]

$a(x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a}$

[woven radish]

pixel

[visual maple]

was it this?

[jagged harbor]

hopefully, yeah

[devout snow]

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[visual maple]

can all of these be solved using the quadratic formula

[eager nova]

Well, they are quadratic equations so…

[visual maple]

💀

i am sorry for wasting your time

[eager nova]

If I would have thought u wasted my time I would have not answered to you.

Just take the answer, no need to be dramatic

[mild sorrel]

https://tenor.com/view/lion-king-dramatic-gif-24607395

[devout snow]

@visual maple Has your question been resolved?

[tepid fulcrum]

hi pixel

im pixel

close the channel if you're done btw

[devout snow]

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[soft trench]

can anyone check my work (exercise 22 ch 3 of rudin)

[devout snow]

@soft trench Has your question been resolved?

[tiny panther]

This all looks correct

[devout snow]

@soft trench Has your question been resolved?

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[sweet musk]

Problem 530

[sweet musk]

I got answers that make sense but when I multiply √33.2812 and √77.888 they are not equal when they should be

[opaque folio]

Ur reading ur numbers wrong my fiend

[winter patrol]

actually your issue seems to be with rounding

[opaque folio]

Nope his issue seems to be the fact 74.88 turn into 77.88

[sweet musk]

What am I doing wrong

[opaque folio]

Messy working out

[sweet musk]

Oh thanks

I didn't notice

[sweet musk]

Ok yeah I see

[opaque folio]

Ur welcome

[sweet musk]

.close

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[outer pulsar]

hello!

[devout snow]

@outer pulsar Has your question been resolved?

[outer pulsar]

@fervent swallow

[outer pulsar]

.close

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[hard ocean]

How do I find P and D?

[heavy current]

P is going to be the matrix formed by taking the eigenvectors of A as its columns, and D is a diagonal matrix with the eigenvalues of A on the diagonal

[sand dove]

btw you're lacking one final eigenvalue with its eigenspace basis

[hard ocean]

Yeahhhh thats what i realized just now

Thats why it wasnt working out cuz my P matrix would be 3x2 when it has to be 3x3

[sand dove]

oh and btw P and D aren't unique

D has three choices here, it depends of how you arrange the eigenvalues on the diagonal

and P has infinite choices as there are infinite possible eigenspace bases, and so for each eigenvalue

[hard ocean]

I don't understand how is there infinite possible eigenspace bases?

@sand dove

[trail eagle]

The columns of P are given by the eigenvectors, and for any eigenvector any scalar multiple of it gives an equivalent eigenvector.

[hard ocean]

Ohh ok

That's cool

.close

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[soft thistle]

how do i determine the fractal dimension of a pseudo fractal?

[devout snow]

@soft thistle Has your question been resolved?

[hybrid pond]

are you familiar with box-counting?

[soft thistle]

what do you mean?

probably not

[hybrid pond]

alr

lemme explain then

first you wanna cover the fractal with a grid of boxes (or squares in 2D, cubes in 3D) of side length 𝜖

[soft thistle]

ok

[hybrid pond]

Then, for each grid size, count how many boxes contain part of the fractal. This gives you a number 𝑁(𝜖).

Plot the log of the number of boxes N(𝜖) against the log of the size of the boxes 𝜖

[soft thistle]

oh

thats gonna be a pain

[hybrid pond]

The slope of the line in this log-log plot gives you the fractal dimension

yeah

there might be an easier way

but this is the way i learned it

[soft thistle]

this is it

[hybrid pond]

first ig you could divide the image into a grid of squares with side length e. The grid can start with larger squares and progressively use smaller squares.

[soft thistle]

alright thanks

[hybrid pond]

that's gonna take a lot of time however

[soft thistle]

yea

[hybrid pond]

sorry i can't be much of help

i'm sure there is an easier way

[soft thistle]

.close

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[deft berry]

Is my limit proof correct? im not too familiar with [x] functions

[deft berry]

im stating the limit is 2 btw

[devout snow]

@deft berry Has your question been resolved?

[deft berry]

<@&286206848099549185>

hi 🙂

i dont think 2/[x] = [2/x] is valid by itself, but it should be valid in this context, or else I dont know how you could transform the function

I cover both cases (x greater than or less than zero)

really? im not too sure how the function works but if x=1.6, then [2/x] and 2/[x] should be different

ah I see, no problem

but again, I think it should be valid in this context, I just cant explain the why

no worries, thank you for helping

[devout snow]

Please do not ping individual helpers unprompted.

[deft berry]

?

[x] is an integer function, as in, it rounds to the nearest integer

in this case it rounds to the lowest integer

so I assume, as x approaches 0 from the left, x is -1, and so 2/[x] = [2/x]

but i dunno if that works

2/1.6 is greater than 1

its ok, thank you very much for trying

[deft berry]

<@&286206848099549185>

[deft berry]

.

[chrome isle]

x-1 < floor(x) <= x

[deft berry]

then my argument for x<0 id correct, but for x>0 is not?

as x would just go to 0

sure

[restive river]

Wholesome stuff

👍

[deft berry]

its still unanswered

[devout snow]

@deft berry Has your question been resolved?

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[deft berry]

.reopen

.reopen

[deft berry]

.close

[devout snow]

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[devout snow]

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[sage wave]

.

What does this mean

[cold bone]

"this" what?

any specific part?

[sage wave]

i just dont get the entire question at all

[cold bone]

you're supposed to perform the given operations in the gives matrices

[sage wave]

ohh

i get it

thank you

[devout snow]

@sage wave Has your question been resolved?

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[glossy basalt]

is there a reason the answer i came up with is a tiny bit different than what i need to show?

[devout snow]

@glossy basalt Has your question been resolved?

[glossy basalt]

<@&286206848099549185>

[glossy basalt]

<@&286206848099549185>

[scenic surge]

wont it be

(x_0)^2/a^2 + (y_0)^2/b^2 - (z_0)/c + (2x_0/a^2)(x - x_0) + (2y_0/b^2)(y - y_0) + (-1/c)(z - z_0)

which simplifies to

-x_0^2/a^2 + 2x_0x/a^2 -y_0^2/b^2 + 2y_0y/b^2 - z/c

You can move -z/c - x_0^2/a^2 - y_0^2/b^2 to the other side

2x_0x/a^2 + 2y_0y/b^2 = z/c + x_0^2/a^2 + y_0^2/b^2

you can sub in z0/c for x_0^2/a^2 + y_0^2/b^2

2x_0x/a^2 + 2y_0y/b^2 = z/c + z0/c ?

[glossy basalt]

oh that makes sense

i didnt even try to expand out

Also, do you know if this is possible or not? I feel like you lose the information to be able to find the gradient after

this is a separate problem

[devout snow]

@glossy basalt Has your question been resolved?

[flint spire]

it is possible, only because the two unit vectors are not parallel

you get the gradient by solving a system of 2 simultaneous linear equations

[devout snow]

@glossy basalt Has your question been resolved?

[devout snow]

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[frail furnace]

need help, pic on the right is what i tried idk what im doing

[olive snow]

Wrong proprety of exponents

A^(mn)= (A^m)^n

So its 3^x + (3^x)^2 = (3^x)^3

So letting u = 3^x

u + u^2 = u^3

I let you solve from here

[devout snow]

@frail furnace Has your question been resolved?

[frail furnace]

ooooh okay thanks

[devout snow]

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[left pulsar]

Can someone help me how to start this?

[topaz axle]

you multiply 2 numbers

there's some amount of pairs of face cards

and some amount of red non face triples

[left pulsar]

I got this one. It's 20C3 right?

[left pulsar]

how do I solve this?

[topaz axle]

4 × 4c2

[left pulsar]

why 4?

[topaz axle]

JKQA

[left pulsar]

A is considered a face card?

[topaz axle]

i'm not sure

[left pulsar]

is it is not considered a face card, how would it look?

[topaz axle]

3 × 4c2

[left pulsar]

okay thanks!

.close

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[devout snow]

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[visual maple]

how do i solve "b'

this is answer for 'a'

[tepid fulcrum]

what's the question

whether it's right?

[visual maple]

nono how do i solve 'b'

so 6b

state the set of values of...

also hi pixel

[analog badge]

So what is |f(x)| ?

[tepid fulcrum]

ohh

sorry

misread

[ancient sluice]

From graph ?

[analog badge]

You already graph it

[tepid fulcrum]

hi

[visual maple]

y= |4-x|?

[tepid fulcrum]

yeah

[ancient sluice]

You see when x element of [0,4] y assumes same value

Two pixel ?? Ah !

[visual maple]

well

im not sure what u mean

so how did u get [0, 4]

[ancient sluice]

From your graph

[visual maple]

|4-x| = 4 - |x|

[visual maple]

mmm what did u look at from the graph tho

[ancient sluice]

The overlaping region

[visual maple]

oh yeahhhhh

thank you

.close

[devout snow]

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[sweet token]

If the light intensity of a lamp (X) is inversly proportonal to the square of the distance (Y) between the lamp and a student who is studying bellow the lamp by 8 meters. If the intensity was X at this place. Then at what distance should the lamp be to achieve an intensity of 4X?

could I get some help in this?

[devout snow]

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[eager stone]

its tangent at A

[eager stone]

BA=BD and its asking whats the degree of ADB

everything else is drawn

[restive river]

🤑🤑🤑🤑🤑🤑🤑🤑🤑

[eager stone]

hmm i got how you found that but what do i do from there

[restive river]

cob = aoc

[eager stone]

okk

[restive river]

u need more?

[eager stone]

...i think yeah

[restive river]

dab = 90 - oab

[eager stone]

okok i think i can do from here

ty

.close

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[noble zinc]

this q is confusing me alot

[tepid fulcrum]

see if you can transform the last integral in some way

[noble zinc]

hm

can u do integrals on latex i wanna draw it out

how do i do it

[tepid fulcrum]

yes, use $\int_{a}^{b}f(x)$

[woven radish]

pixel

[noble zinc]

$\int_{b}^{a}f(x) = - \int_{a}^{b}f(x) = \int_{a}^{b}f(-x)$

i know this

[tepid fulcrum]

yes

[noble zinc]

wait

sorry

this

finally

so can i say

[woven radish]

Ishaan

[tepid fulcrum]

not exactly, how do you know that -f(x)=f(-x)?

[noble zinc]

oh yeah we dont

[tepid fulcrum]

maybe there is another way to get rid of the -x in the function?

[noble zinc]

hm

so $f(-x)$ = reflection in y

[woven radish]

Ishaan

[noble zinc]

so

do we make the integrals negative or smth

[tepid fulcrum]

have you learned u-sub yet?

[noble zinc]

nahh

[tepid fulcrum]

oh

well

[noble zinc]

i j tried to draw a sketch

of a random function f(x)

and then i saw that

if u swap the bounds and make them negative

it may work

$\int_{b}^{a}f(x) = \int_{-a}^{ -b}f(-x)$

oops

[tepid fulcrum]

u forgot the _

before the {b}

and {-a}

[woven radish]

Ishaan

[noble zinc]

thanks

is this right?

[tepid fulcrum]

yes

this is true

[noble zinc]

im tryna generalise it

omg acc?

i j drew a random sketch it looked right

oh nice so

[tepid fulcrum]

wouldn't know how to prove this without u sub tho

but you can just use it

[noble zinc]

hmm yeah ur right

and also

[tepid fulcrum]

i do have to go for dinner rn so maybe another person can help you

or if you can do it yourself from this point on

[noble zinc]

Generalisation 1: $\int_{b}^{a}f(x) dx = \int_{-a}^{ -b}f(-x) dx$
Generalisation 2: $\int_{b}^{a}f(x) dx = - \int_{a}^{b}f(x) dx$

[tepid fulcrum]

i think this is basically the key

to getting it

but yeah ill be back later

[noble zinc]

tysm

is someone able to verify this

i forgot the dx

[woven radish]

Ishaan

[gusty nexus]

I don't think first one is correct

I think you should swap -b and -a

[noble zinc]

r u sure

cuz if ur reflecting in the y axis

wont u need to swap them around

cuz the upper bound a value on right side of y axis is now on the left

so it becomes the lower bound

[gusty nexus]

let me check by creating a function

[noble zinc]

okay thanks

[gusty nexus]

yes, you are right

[devout snow]

@noble zinc Has your question been resolved?

[noble zinc]

.close

[devout snow]

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[proven harness]

Hi ! so im relearning algebra because i realized i still lack in it, for this question im still kinda confused, from where did b = 4/4 + 1/4 came from ?

[tepid fulcrum]

in the equation y=1/4x+b, they substitued the point (-3,1), which implies 1=-3/4+b

then they solved for b

they are using the equation y=1/4x+b, because earlier they found that the slope of the line is 1/4

[tepid fulcrum]

i think you mean b=4/4+3/4 right?

[proven harness]

yes

but what im pretty confused is with, isnt b already solved when y = 1/4x - 1/4, which means b = -1/4 ?

[tepid fulcrum]

in that case they just rewrote the equation x-4y=1 to get the slope of that function. they had to do this since we know that the line is parallel to x-4y=1 (parallel=equal slopes)

the line x-4y=1 doesn't go through (-3,1), so that's why they had to find a different b value

[proven harness]

ahh alright, so we find it by subtituting it to
y=1/4x+b
1=-3/4+b ?

[tepid fulcrum]

yes

[proven harness]

which means 1 + 3/4 = 4/4 +3/4 = 7/4 ?

[tepid fulcrum]

exactly

[proven harness]

ahhh thank you for the help

[tepid fulcrum]

you're welcome

[proven harness]

but how do we know that we need to search for b a second time ?

[tepid fulcrum]

there is a chance that the point does lie on the line given, but that doesn't happen often

because that would make the question a lot easier

so always just do it like this(this=finding b)

it's much more work first checking if the point lies on the line given

[proven harness]

ah i see, then if the question ask its parallel, then you need to find the other b that could include (-3,1)

[tepid fulcrum]

yess

[proven harness]

ightt

thank you sm, for the help

[tepid fulcrum]

np

[proven harness]

.close

[devout snow]

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[devout snow]

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[grim escarp]

i’m not sure if i’m doing this correctly. questions that have translations of other functions or stuff like this always confuse me

[grim escarp]

according to the answer key it’s correct? but my teacher puts incorrect stuff into the key on purpose so idk lol

[misty crest]

so what does this remind you of

for each increase in x by 1 g increases by 2

[grim escarp]

linear function?

[misty crest]

mhm

with slope of what

[grim escarp]

2

[misty crest]

right so g has a positive slope, and is increasing

why isn’t it concave up?

do you know

[grim escarp]

because it’s linear

[misty crest]

yup

the slopes of a linear function are constant

thus have a concavity of 0

[grim escarp]

mhm

i still don’t get the g(x+1) - g(x) stuff though

[misty crest]

🤔

[misty crest]

^

thus g(x+1)=g(x)+2

[grim escarp]

ohhhhhhh

ok i didn’t realize that

thank you!

[misty crest]

you’re welcome

[grim escarp]

can i ask another question here ?

[misty crest]

sure

[grim escarp]

okk

let me find it

for this stuff about sum of two functions and determining aroc of the new function - does it apply to difference of two functions?

i copied these notes from a question on ap classroom

[misty crest]

i think i saw this exact question yesterday

[grim escarp]

and then i tried to apply it to question 5

hahahaha yeah it’s ap precalc work

[misty crest]

where you can determine that its decreasing for the first half of the interval then it’s inconclusive for the second half

[misty crest]

makes sense

they were doing that too

[grim escarp]

i tried asking my teacher about it but he can’t even decide on an answer :/

[misty crest]

😹😹😬

yikes

[grim escarp]

:,)

[misty crest]

,rotate