help-27

[loud spade]

so basically C

[summer summit]

yes

[loud spade]

https://cdn.discordapp.com/attachments/903486480075354182/1503115240991101008/image.png?ex=6a022ca7&is=6a00db27&hm=b56ff89424b1050488bc299dd09da5042d945d2fc3648286c9e87042fee6b216&

hmm ok

[summer summit]

sorry was scrolling back up to check lol ty

[loud spade]

dw

uh

so

my exam is 7 am

its currently

11:11

do I sleep

[summer summit]

yes

[loud spade]

or do I take a mock

[summer summit]

sleep now, review everything early in the morning

you'd get the same amount of sleep regardless, but if you do a mock now you'll have a lower chance of remembering everything since you're sleeping in between

unless of course you dream of mock exams

[loud spade]

aight bet

thank you gang

.close

[devout snow]

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[summer summit]

of course! good luck with your exam!

[loud spade]

Thank you so much for your help!

@lunar harbor And obviously you too!

[lunar harbor]

[devout snow]

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[devout haven]

Does anyone have potential ideas for how to solve the upper bound? I've been able to prove the lower bound using sudakov to bound E max Z, and then bounding the packing number using K_eff. For the upper bound I've tried dudley, but I'm not able to meaningfully bound the covering number in terms of K_eff, only K.

potentially the help channels aren't quite right for this? Maybe I should go to #advanced-probability ?

[supple knot]

yea that's a good channel for your question

[devout haven]

okay I'll ask over there!

.close

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[devout snow]

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[vast ledge]

part c

this is my work

the last 3 values are where g(x) instrect the x axis

i got t= 2.83 to be correct

this the answer

what did i do wrong

[supple knot]

How do you know g(2.83) is the greatest of the local maxima

[vast ledge]

its the biggest number

i found out the critical nmbers for g'(x)

then plugged into g(x)

[supple knot]

Did you compare it to a = 6.203...

[vast ledge]

thats not a point

on g'(x)

[supple knot]

I see

How do you know g'(2.83) is zero

[vast ledge]

i checked on desmos

green is g(x)

black is g'(x)

the furthestmost point is 6.203

but thats on g(x)

so im confused

cuz i need to find wehre g'(x)=0

[supple knot]

My guess is there's a mistake in the equation
F'(t) = E(t) - F(t) = 0

It should have been what you have
F'(t) = E(t) - L(t) = 0

[vast ledge]

so the answer key is wrong?

[supple knot]

That's my guess yea

[vast ledge]

hol on

i see

the problem

e and l

are rates

so im wrong

i interepreted it wrong

[vast ledge]

.

[supple knot]

I see. Maybe they just have the one typo with F' = E - F instead of F' = E - L but the rest of their work is correct

[vast ledge]

ye

wait

ive confused my self

can you like walk me through the problem

from part c

like the whole thing

i dont know what im doing anymore

[supple knot]

Read the solution one line at a time until you get confused

[devout snow]

@vast ledge Has your question been resolved?

[vast ledge]

i get it

im good now

.closer

.close

[devout snow]

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[devout snow]

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[bronze void]

Use ftc for each

[celest lantern]

What have you tried?

Just add integration.

[bronze void]

You get similar answers which you can sub values in for the final integration, notice how for each section of the problem, we can rearrange and get a value for G(-10) and G(6)

[celest lantern]

What?

KB that is tedious.

[glass maple]

i got it

added the integrals and got 8

[celest lantern]

Correct.

[glass maple]

.close

[devout snow]

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[celest lantern]

No need for any weird substitution.

[devout snow]

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[desert seal]

Not sure how to to start this

[sturdy fiber]

,, let \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}

[woven radish]

Itsuki

[sturdy fiber]

after that just find r cross (2i-j+2k) using determinant method

[desert seal]

I’m confused on what the actual question is asking sorry 😭

[sturdy fiber]

basically we need to find all (x,y,z) such that they are in the same plane as (2,-1,2) but at a fixed perpendicular direction and magnitude to the plane from (4,2,-3)

[desert seal]

Ahh I see

Lemme try the question now

[sturdy fiber]

👍

(ok i gtg unfortunetly just ping @ helpers when u come back)

[desert seal]

.close

[devout snow]

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[devout snow]

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[civic crag]

I've done (b) which is Yes G is irrationational
but I want to know if this would be a good justification for (c)

[civic crag]

as opposed to this solution here (where it's used the cofactor expansion)

Here's the problem

[devout snow]

@civic crag Has your question been resolved?

[devout snow]

@civic crag Has your question been resolved?

[velvet geode]

yk u can ping the 'helpers'

@helpers

[devout snow]

Channel closed

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[civic crag]

.reopen

[devout snow]

✅ Original question: #help-27 message

[civic crag]

<@&286206848099549185>

Just want to know if it’s worth doing the extra step

[deft furnace]

Yes

[civic crag]

So this solution makes sense would be better then?

[deft furnace]

The first version basically says:

“” and “” because the matrix becomes block diagonal / extends nicely.

That is correct, but it skips the justification a bit fast.

The second version explicitly explains why:

only one nonzero entry exists in that row/column,

so cofactor expansion works immediately,

therefore , etc.

That’s cleaner and more rigorous, especially for a proof-based course. It shows you understand the determinant computation instead of just pattern-matching.

The conclusion is also correct:

⇒ only critical point is ,

leading principal minors alternate signs,

by Sylvester’s criterion the matrix is negative definite,

therefore is a strict local maximum,

and since the quadratic form goes to , it’s also the global maximum.

[celest lantern]

Is this AI?

Why it's not well-worded and not well-phrased?

[civic crag]

This is what ai said lol

[celest lantern]

!noai

[devout snow]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[deft furnace]

Atleast i am trying to help

[civic crag]

I appreciate it

[last parrot]

At least?

AI is helping, you are not even contributing

[deft furnace]

I am trying

[celest lantern]

You are trying?

[deft furnace]

Is it bad to try to help?

[celest lantern]

Sounds like AI trying.

[celest lantern]

Yes, you are not helping, and it's straightforwardly bad.

[deft furnace]

Did i said no?

[gloomy aurora]

It does not matter whether its 'atleast' or 'atmost'; server policy deems so. We'd rather have good inputs than inputs that were not thought out or original. I appreciate your interest and your dedication, but please if you were trying to help, put it in more credible forms of helping than pinging an AI to do so. If the AI content was viable, OP could have asked an AI himself, you didn't need to do it for him.

[deft furnace]

Chill out

Sorry

Cuz i did a good promot with ai

This is whyvi used it

Its always accurate

[gloomy aurora]

I feel my message is clear enough for me to not repeat my statement again.

[deft furnace]

Yes

[last parrot]

Cold

-# Hi Aza! Im Minh btw i changed nickname 😭

[true pollen]

This is wrong. AI is never "always accurate", no matter how good you are at prompting. Always double check any answer you get from AI, and don't use it in this server.

[deft furnace]

Yh insted of helping the guy u guys are blaming me for using ai to help

This gen is coocked tbh

If k know i would totally help

But he was just being ignored so i tried to attempt

[true pollen]

Informing is not the same as blaming.

[deft furnace]

I see it as a blame for me

[celest lantern]

You are not attempting.

Yup, and you deserve that blame.

[deft furnace]

I dont wana give something wrong from me

[true pollen]

!redir

[devout snow]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[true pollen]

I'd say the second solution is better only because it explains more what is actually happening.

I think both would be valid though.

[devout snow]

@civic crag Has your question been resolved?

[last parrot]

Well, true, but he is going against the rule though.

[devout snow]

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[clever sphinx]

Can someone help me please with b

Part a^

[faint gorge]

[ prpl G_Y(y) = \P(Y le y) = \P(T^3 le y) = \P(T le y^{1/3}) = F_T(y^{1/3}) ]

[clever sphinx]

Why is Y<=t?

[faint gorge]

This is what the cummulative func is by def

[clever sphinx]

I always thought it was P(Y<=y)

[faint gorge]

it's a variable name

[clever sphinx]

Hm

Oh ok

Can I use y instead of t

[faint gorge]

yes

let me adjust it

[clever sphinx]

Thanks

[clever sphinx]

And then I substitute in y^1/3?

[faint gorge]

yes

then differentiate

[clever sphinx]

Ohhh okay

Yeah I got it!!

Can we do another similar example please

[faint gorge]

hmm okay

I suppose Q2?

[clever sphinx]

Yeah

[faint gorge]

Try it and see

u can share ur work

[clever sphinx]

Also sorry this is really silly but what’s the diff between Y and y😭

[faint gorge]

Y is just a random variable, y is a specific value Y can take

Y=Y(omega) maps a random outcome omega to a real number y

[clever sphinx]

Oh okay!

Ive tried 2

[faint gorge]

for example T was machine life time and t is then the number of months for instance

[faint gorge]

yes

[clever sphinx]

Wait the markscheme says it’s 1-

[faint gorge]

wat

oh

Yes

carefu

[clever sphinx]

Ohh okay oops

[clever sphinx]

Thanks

I understand it now

[faint gorge]

\prpl Also notice that
[ X^2 ge frac{1}{y} implies X ge frac{1}{sqrt{y}} \text{ and } lp X le -frac{1}{sqrt{y}} rp ]

but we can omit the second case since $X ge 1$.

[clever sphinx]

Also

How do the limits change ?

[faint gorge]

of the integral?

[clever sphinx]

Like for the cumulative distribution function

[faint gorge]

Are you referring to iii

[clever sphinx]

No for 2I

Im writing the cumulative distribution function for y

So shouldn’t the limits change

[faint gorge]

If x>=1 then what does that imply for y

[clever sphinx]

y>=1

[faint gorge]

x>=1 => x²>=1 => 1/y>=1

[clever sphinx]

Hm

Yeah

Ohh

[faint gorge]

so what is y

[clever sphinx]

1>=y?

[faint gorge]

yes

and lower bound

x² -> oo so 1/y?

[clever sphinx]

0

[faint gorge]

yes

[clever sphinx]

So it’s 0<=y<=1

[faint gorge]

yes

[clever sphinx]

Perfect

[devout snow]

@clever sphinx Has your question been resolved?

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[devout snow]

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[slim smelt]

I don't understand the step involving forming a determinant from the second order differential equation.

Why does that work? How can you just ignore Ae^3x and Be^5x in the drterminant like that? Why is it valid? I don't remember studying this. Please explain it to me.

[supple knot]

There's a lot of theory with Wronskian involved. Here's a start
https://tutorial.math.lamar.edu/classes/de/FundamentalSetsofSolutions.aspx#Wronskian

[jagged harbor]

<@&268886789983436800>

[devout snow]

@slim smelt Has your question been resolved?

[slim smelt]

I'm still reading the notes

[slim smelt]

Honestly, man, I read this entire thing and I didn't understand a thing. I've never come across the term "Wronskian" ever before and it seems like too much theory for just one problem especially since it's never been mentioned before. I could, of course, go through the entire notes of this guy and understand it thoroughly, but that seems a bit excessive for just trying to understand one problem. Thank you, though. I've bookmarked that blog

[safe knoll]

i feel like they just converted the polynomial into vector form

Ax=0 , solve for A=0 cuz x cant be = 0 in this case

lemme try use my bad handwriting real quick

tho i dont understand the question fully , are we just supposed to make a DE out of the given solution?

[slim smelt]

Yeah

[supple knot]

yes math problems beyond high school is typically based on a lot of theory and if you want to learn the "why" behind doing the calculations, you have to learn the theory. if you just want to do the rote calculations then you don't need to understand the theory behind wronskians

[slim smelt]

Here's the rest of the solution but they just solved the determinant which I know how to do. I'm just stuck on how they created it out of thin air like that in the first place

[slim smelt]

I mean, fair enough. I was hoping there would be some easier explanation

[supple knot]

easy and math don't go together beyond calculus, for most people at least

[slim smelt]

Is there a particular book on ordinary differential equations that you can recommend for undergraduates that goes over alot of theory but starts from the basics (like only assuming calc 1 or 2 knowledge)

[supple knot]

there's plenty in #book-recommendations

[slim smelt]

or are these notes that you've given enough?

[supple knot]

enough for this problem yes

[slim smelt]

hmm, alright

.close

[devout snow]

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[devout snow]

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[clever sphinx]

Im doing q3

Im unsure on how to put everything together

[spiral raptor]

Assuming you got it correct, you can probably set a=1, and enter the values for b,c, and d appropriately

[clever sphinx]

Hmm

Do I need to take into account that it’s -b/a?

[spiral raptor]

Well a=1 right?

[clever sphinx]

yes

[spiral raptor]

So it becomes -b and not -b/a

[mild sorrel]

Just find everything in terms of a.

[spiral raptor]

Also, you can wrap the whole thing with k, so its k(x^3+bx^2+cx+d)

[fluid musk]

hey , hey does anybody here speak german , has time and could explain to me how parabolas(parabeln) work? I write the big exam tomorrow and I understand NOTHING OF PARABOLAS I need help and I really need to pass

[mild sorrel]

!help

[devout snow]

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

[clever sphinx]

So the polynomial is in the form ax^3+bx^2+cx+d=0 and I’m solving for a b c and d?

Im just confused about the signs really

If -b/a = -2/5 I sub in 2/5?

[frigid shale]

Basically just x cubed+ (sum of the roots) x squared - (sum of two roots) x + (product of roots)=0

[mild sorrel]

-b/a = -2/5 implies b/a = 2/5, so yeah, if a = 1, b = 2/5

[frigid shale]

From vesta's formula normal is x squared - (sum of roots)+ (product of roots)=0

[spiral raptor]

So let's take a out. The polynomial becomes a(x^3+b/ax^2+c/ax+d/a) now you should easily be able to solve it

[clever sphinx]

Thank you

[mild sorrel]

np

[frigid shale]

Actually signs came from normal quadratic equation, -b/a from sum of the roots, c/a is the product of the roots

Since cubic equation have 1 more roots, they will have sum of 2 roots in the middle

Becomes -b/a, c/a and -d/a

[mild sorrel]

The signs have already been taken care of in this work. The polynomial assumed was ax^3 + bx^2 + cx + d = 0, and from the relations, one gets (taking a = 1) that b = 2/5, c = -27/25 and d = -81/25
So the cubic equation is x^3 + 2/5 x^2 - 27/25 x - 81/25 = 0
Multiplying it by 25 does not extinguish any of its solutions, nor does it add any.

[devout snow]

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[devout snow]

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[spring oasis]

how do you even graph this shit?

x^2=y^2

[buoyant verge]

mmm

could you try some test points?

[spring oasis]

?

ah you parametrize it, easy

[buoyant verge]

Well, I ment like taking points like -1, -0.5 , 0, 0.5 and 1

[spring oasis]

r(t) = (t^2, t^2)

[buoyant verge]

plotting in geogebra leads to an "X" type shape

[spring oasis]

how

[buoyant verge]

oups

sorry

I see my error, I was testing for y^2

not y^3

[spring oasis]

yeah

[buoyant verge]

I mean, your graph is pretty accurate to what geogebra plots

[spring oasis]

what

[spring oasis]

.close

[devout snow]

Channel closed

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[spring oasis]

i will continue later but this graph looks like a butt crack, right?

[uncut crow]

sir this is a math help channel

[spring oasis]

srry

[buoyant verge]

A bit hahahahha, nice, im going to sleep, good luck with it

[devout snow]

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[unique pebble]

i need help with asvab math like alegbra and geometry

[devout snow]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[celest lantern]

Ask away!

[devout snow]

@unique pebble Has your question been resolved?

[devout snow]

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[junior heron]

i am completely stumped on this

[junior heron]

i barely understand affine transforms in general

how should i start here

[devout snow]

@junior heron Has your question been resolved?

[late mortar]

have you learnt it yet?

[junior heron]

yeah i understand that its like w = Pz so that you can convert a crazy matrix into a normal one to make a phase image aligned with the axes

i dont know how to show this question tho

idk where to start

[late mortar]

have you tried applying P to the line equation?

also I assume here we assume w=(u v) an z equals to (x y)?

[junior heron]

yes

[junior heron]

i dont even know how

the line isnt a matrix or a vector but P is

idk how to apply it

[late mortar]

well notice that the equation is basically [D E][u; v] +F=0

[junior heron]

HMMMM

true

so we've found w

so i need to apply p to this

wait is [D E] = P

hmmm

i dont know how to apply P to this still

[junior heron]

i tried this

i guess that's still a line because there's no exponents or anything so this is still a line

and pluggging in 0 for u and v results in the same thing

is that all it wants

[devout snow]

@junior heron Has your question been resolved?

[junior heron]

.close

[devout snow]

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[floral field]

Hello, does anyone know how to find angle A?

[celest lantern]

Send this to new channel please.

This channel is going to collapse soon.

[late mortar]

Yes except it’s x y and not u v on line 4

Sorry for the late reply btw I had PE class just now

And P inverse if it has one

[devout snow]

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[crystal dawn]

$(a + b)^2 = a^2 + b^2$?

[woven radish]

Hanako(x, y); ∂(fox)/∂x

[crystal dawn]

aren't you forgetting a couple of terms here?

[keen sundial]

hm

[topaz axle]

undeserving terms will be igniored

[keen sundial]

i tried again but i dont think im doing it right

[crystal dawn]

well you'll probably have to first remember what (a + b)^2 expands to

[keen sundial]

a^2 + 2ab + b^2

wheres (a+b)^2 here? we have z^2

[obsidian nebula]

write z as a+ b

[faint gorge]

a is the real part, b the imaginary part bruh

[crystal dawn]

I want you to look carefully at your complex number representation

just, hard stare at what you wrote about z

then think about your own question

[keen sundial]

well I copied the question correctly right

[crystal dawn]

I never said you did it wrong

but I want you to look hard at your z, then look at your question

[keen sundial]

ok I understan

d

how do i multipy those 2 in red?

[celest lantern]

Well, if you have the terms in those things, just let them be a and b?

Then sqrt a * sqrt b = sqrt(ab).

I don't think of anything else at this point, would you have to just put 2i to the side and then multiply with the square root of (ab).

Whereas a = 2+sqrt 2, b = 2-sqrt 2.

[keen sundial]

wdym?

[celest lantern]

Then it aligns with a^2 - b^2 form also.

Well, if you let $a=2+\sqrt{2}$ and $b=2-\sqrt{2}$.

[woven radish]

Mercury (ヤフォダ)

[keen sundial]

thats not what i did tho

[celest lantern]

Then would you have $2\sqrt{a} \cdot i\sqrt{b}$.

[woven radish]

Mercury (ヤフォダ)

[celest lantern]

Which can be simplified further into: $2i\sqrt{ab}$.

[woven radish]

Mercury (ヤフォダ)

[keen sundial]

look

i had square root over those terms

[celest lantern]

What did you do?

Well, isn't that's what I explained?

[celest lantern]

No no.

Then you have $2\sqrt{a}$ and $i\sqrt{b}$.

[woven radish]

Mercury (ヤフォダ)

[celest lantern]

The things inside that big square root.

[keen sundial]

how?

[celest lantern]

Look here.

Now, note the left term.

Do you see number 2 and the big square root next to it?

Then you can just let the values $2+\sqrt{2}$ inside that root be $a$.

[woven radish]

Mercury (ヤフォダ)

[keen sundial]

where did u get 2 + sqrt(2)?

wait nvm

ok yeah continue

[celest lantern]

Well, you got that?

[keen sundial]

yeah

but why do we do that

[celest lantern]

Let me explain.

Next, go to the left term.

We do the same, let $2-\sqrt{2}$ be $b$.

[woven radish]

Mercury (ヤフォダ)

[celest lantern]

So now, you have $i\sqrt{b}$.

[woven radish]

Mercury (ヤフォダ)

[celest lantern]

Right?

[keen sundial]

correct

[celest lantern]

Well then you have $2\sqrt{a} \cdot i\sqrt{b}$

[woven radish]

Mercury (ヤフォダ)

[celest lantern]

Can you simplify it?

[keen sundial]

$2i/sqrt(ab))$

[woven radish]

ø

[keen sundial]

i forgot how to

[celest lantern]

Correct.

No worries.

Now, do you see a and b, when a times b, do they looks like an identity?

$(2+\sqrt{2})(2-\sqrt{2})$

[woven radish]

Mercury (ヤフォダ)

[keen sundial]

that evaluates to 2+4sqrt(2)

[celest lantern]

Hmm, not quite.

Well, let me do this.

$(a-b)(a+b)$

[woven radish]

Mercury (ヤフォダ)

[keen sundial]

ah

thats

[celest lantern]

Is it flick an idea?

[keen sundial]

difference of squares

[celest lantern]

Correct.

[keen sundial]

a^2-b^2

[celest lantern]

Now can you redo?

[keen sundial]

redo what?

[celest lantern]

This.

Simplify it.

[keen sundial]

yea 4-2 which is 2

[celest lantern]

Now, $\sqrt{ab}$ will be?

[woven radish]

Mercury (ヤフォダ)

[celest lantern]

You just found out ab = 2.

[devout snow]

@keen sundial Has your question been resolved?

[devout snow]

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[dawn oar]

I feel dumb but I completely forgot how to do this

[runic prawn]

what's the question

[hasty cargo]

Err can you please state the full question, or give us a screenshot of it

[dawn oar]

OH my gosh duh im sorry i knew i was forgetting something

Segment DE is the perpendicular bisector of AC. Find the length of AE

[supple knot]

not enough information given

!original

[devout snow]

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

[supple knot]

did you try entering 14x + 12

[dawn oar]

can’t

It doesn’t work

[supple knot]

show the error

[supple knot]

ah my bad

i didn't notice the tick marks

these mean the lengths AE and EC are equal

[dawn oar]

yep I tried that it was wrong

[winter patrol]

you could apply the definition of bisector,
however the end result doesn't make sense

[supple knot]

!show

[devout snow]

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

[supple knot]

,w solve 14x+12 = 8x-6

[woven radish]

Results provided by WolframAlpha

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

Requested by rie.mann

[dawn oar]

oh

I got -30

somehow

oh no I got -3

but when I plugged it back in it was -30

[supple knot]

yes that's why it's important for you to show your work

[dawn oar]

For ar

Ae

Which is wrong

[dawn oar]

bruh my work is really messy

14(-3)+12
-42+12
-30

[supple knot]

make it not messy

[dawn oar]

there u go

It just gives me a thumbs down and says try again

could be a teacher error

[supple knot]

yes it is

move on to next problem and bring it up to your teacher

[dawn oar]

alright thankyou for the help

.close

[devout snow]

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[zealous stone]

Hi, I need help w/ geogebra
I'm doing some last minute stuyding of trigonometry, and my geogebra is betraying me

I'm trying to use cosinus to find the degree of a corner, but I'm not getting the answer in degrees

My input is acos(9/10)
My answer is correct (chatgpt said so), but I don't know how to get the answer in degrees

thanks in advance :)

[zealous stone]

Geogebra Classic 6 on mac by the way

CAS window

[knotty sage]

Either check the options or multiply the result by 180/pi

[zealous stone]

I checked options and it's set to give the answer in degrees

I'll try the pi thing though

oo lets go, it worked

Thank you so much!!!!!!

[knotty sage]

Note that it only works on angles

[knotty sage]

So sin(x [rad]) = sin(x * 180/pi [deg]))

[zealous stone]

Noted

.close

[devout snow]

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[devout snow]

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[sick delta]

somebody plssssss

[neon folio]

https://tenor.com/view/bad-bunny-covering-eyes-covering-face-crying-grammys-gif-2141802334945557270

[supple knot]

just copy paste the question here so people don't have to click different things to help you

[sick delta]

i will.

[devout snow]

@sick delta Has your question been resolved?

[devout snow]

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[untold ravine]

<@&268886789983436800>

[devout snow]

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[tepid mantle]

{x€N|3<x}

[faint gorge]

Do you have a math question

[tepid mantle]

Yes

[supple knot]

[faint gorge]

Very informative

[devout snow]

@tepid mantle Has your question been resolved?

[devout snow]

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[echo pilot]

Can someone explain what these mean?

[red fractal]

Distributive property

Means expansion

Because of the word distributive

[echo pilot]

I know what it is, just what makes each problem have, commutative, associative, or distributive

[red fractal]

Ah I see

[summer summit]

commutative: order doesn't matter
associative: grouping doesn't matter
distributive: multiplying across/expanding

ex. for number 1, it shows that the order does not matter

therefore it is commutative

[red fractal]

1st is because its +

[hollow vine]

⬇️

[red fractal]

Refer to this image above

[echo pilot]

So on question 2 what made it associative

[hollow vine]

sorry u mean question 4?

[summer summit]

question 2 is commutative

[red fractal]

Yeah 3x2 is the same as 2x3 you still get the same answer

[echo pilot]

No question 2 sorry

I typed associative

[red fractal]

As an example

[summer summit]

it's not associative because it is not changing the grouping.

its just changing the order in which it is written. the parentheses are around the same terms

[echo pilot]

Wait that’s so easy

Okay more questions

How would I do this stuff sorry about the lighting

Uhh

[summer summit]

same thing, youre just naming the properties

[hollow vine]

compare to this

[echo pilot]

The first one is associative?

[summer summit]

tried to make it look a little better. refer to original pic for the numbers that didn't generate properly.

sometimes ai is in fact useful :)

[echo pilot]

AI wasted water why I come here instead of asking ChatGPT

[summer summit]

don't worry, i'll balance it out by not drinking water for 1 year.

[echo pilot]

So the first one is associative

[summer summit]

are you starting at question 16 or 17?

[echo pilot]

16

[summer summit]

that is not associative in the first step

[echo pilot]

How

Associative is changing the grouping

[summer summit]

(try taking a look at the other properties first)

[echo pilot]

I really don’t understand

[summer summit]

it isn't explicitly stated in the image sent by the other helper, however this is a form of distributive property

$(a+b)(c+d) = a(c+d) + b(c+d)$

[woven radish]

Krish

[summer summit]

which eventually evaluates to ac + ad + bc + bd

[echo pilot]

It would be distributive prop of multiplication

[summer summit]

yes

[echo pilot]

And then next is also distributive

[summer summit]

well, also you don't have to state distributive property "of multiplication"

distributive property only applies to multiplication

[echo pilot]

Okay

The next is distributive?

[summer summit]

what are you distributing in that step?

[echo pilot]

Is it associative

[summer summit]

did any of the parentheses move around?

[echo pilot]

No

So commutative

[summer summit]

yes, and explain your reasoning?

something besides "because its not the other two"

[echo pilot]

Idk

[summer summit]

well, what does the property say?

[echo pilot]

Commutative

[summer summit]

i know the name, i'm asking the definition.

[echo pilot]

I don’t know

[summer summit]

look back at the image and tell me what it says

[echo pilot]

Which image

[summer summit]

^

[echo pilot]

It’s the same both ways I don’t know

[summer summit]

yes, and is that what your step is showing?

think of (a + b) as one term

[echo pilot]

Yes?

[summer summit]

right so that's why its commutative

does that make sense?

[echo pilot]

No

[summer summit]

https://www.youtube.com/watch?v=FSQVYBtTxJI

try watching this video

[echo pilot]

It’s rearranging

[summer summit]

yes, changing the order.

[echo pilot]

I got it
Why is the last one distributive

[summer summit]

like the last step?

[echo pilot]

Yes

[summer summit]

the last step is not distributive

or are you talking about the last one we did (the first line)

[echo pilot]

On number 16 AC+AD+BC+BD

[summer summit]

that is not distributive property

what is being distributed there?

[echo pilot]

I don’t know that’s what my sister said

It’s commutative

[summer summit]

hmm, that's wrong. going from CA+DA+CB+DB to AC+AD+BC+BD is not distributive

[summer summit]

yes

[echo pilot]

And the one above it would it be property of addition

Also for the one we are on

[summer summit]

which property of addition?

[echo pilot]

This one

[summer summit]

i know which one, im saying which property of addition. it has two different properties, associative and commutative

[echo pilot]

Commutative

[summer summit]

for the last line it's not the commutative property of addition

the commutative property of addition states something like a + b = b + a

[echo pilot]

Just commutative property?

[summer summit]

well no, you have to specify it

[echo pilot]

If it’s not addition then it’s multiplication

I don’t know genuinely

[summer summit]

yes but don't just use process of elimination, you should understand why

[summer summit]

did you watch the video?

[echo pilot]

I do not understand anything that is going on right now

My teacher is stupid and doesn’t teach

[echo pilot]

No

[summer summit]

go watch that, that's why i sent it

[echo pilot]

I don’t learn from videos I learn from visual and verbal explanations

Something my teacher lacks

Yo

.closr

.close

[devout snow]

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[devout snow]

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[keen sundial]

can someone prove this

[keen sundial]

Into vector form

this too

[sacred sedge]

wdym by prove?

like how can you go from parametric to vector?

@keen sundial

[devout snow]

@keen sundial Has your question been resolved?

[keen sundial]

Yea

[sacred sedge]

So you know how you can represent a line in vector form as $\vec{r} = r_{0} + t \vec{v}$ where v is the direction vector

[woven radish]

InterGalactic

[sacred sedge]

and $r_{0}$ has $<x_0, y_0, z_0>$

[woven radish]

InterGalactic

[sacred sedge]

with $\vec{v} = <a_1, a_2, a_3>$

[woven radish]

InterGalactic

[sacred sedge]

putting all together we have $<x_0, y_0, z_0> + t<a_1, a_2, a_3>$

[woven radish]

InterGalactic

[sacred sedge]

now upon adding them up, we get

$<x_0 + t a_1, y_0+t a_2, z_0 + t a_3>$

[woven radish]

InterGalactic

[sacred sedge]

so the x,y,z of our vector is

$$ x = x_0 + ta_1$$
$$y = y_0 + ta_2$$
$$z = z_0 + ta_3$$

[woven radish]

InterGalactic

[sacred sedge]

now for each of them if you isolate t

you will get that

$\frac{ x - x_0}{a_1} = \frac{y-y_0}{a_2} = \frac{z-z_0}{a_3}$

[woven radish]

InterGalactic

[sacred sedge]

@keen sundial

does that make sense

it is just expanding the vector form

to get the parametric representation of it

[keen sundial]

No I’m talking about

How do we represent a Cartesian into vector

.close

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[keen sundial]

“The span of two non zero non parallel vectors is a plane through the origin”

[keen sundial]

I don’t get the “plane through the origin” bit

[solid hull]

is it the plane bit, the origin bit or both

[keen sundial]

That whole statement

[solid hull]

alright

the idea behind it is that when two vectors are not parallel (in the 2d case at least - more precisely, if the two vectors are independent), then each vector will increase the span of the set of vectors

in the case of having two linearly independent (or non-parallel) vectors, this means that the combination of both vectors spans a 2 dimensional space

this is what constitutes a 'plane' in higher dimensions

the part about the origin is that due to the properties of linear algebra, the origin should not move in any way under the product of any transformation

so every plane should contain the origin

[summer summit]

the definition of span is the set of all possible linear combinations. the zero vector is in any span regardless of the question or values given to you

[keen sundial]

Wdym?

[solid hull]

A better way to explain it is as Krish mentioned, the 0 vector (or the origin) will always be present in some linear combination of vectors by setting the coefficients to 0

[keen sundial]

Also after we’re done with this, I need this explained too

[solid hull]

in the case of a 2-dimensional set of vectors ${v_1,v_2 }$ for example, for some linear combination $av_1+bv_2, \ \a,b \in \mathbb R$, $a=0,b=0$ corresponds to the origin

[woven radish]

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

[summer summit]

to add on to this, this is also the reason why spans are subspaces (if you have covered that), since every subspace must also contain the zero vector

[keen sundial]

Fair

[keen sundial]

What’s a subspace?

[summer summit]

basically a collection of vectors. theres three main rules of a subspace iirc, that the zero vector is in the set, if you add two vectors in a set the result should also be in the set, and any scalar multiple of a vector in the set should also be in the set

for example, R^3 is a subspace itself

another ex. a line y = 5x in R^2 is a subspace since it goes through the origin, but y = 5x + 1 is not, because (0, 0) is not on that line

[devout snow]

@keen sundial Has your question been resolved?

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[proper otter]

How to connect these to I tried the elipse formulla

[runic prawn]

rotated parabola

[burnt lark]

yeah, you could use a horizontally stretch parabola of the form $x = ay^2$, where $0 < a <1$

[woven radish]

TestTickler

[proper otter]

this the formula im using

[runic prawn]

rotate it

[proper otter]

how

[runic prawn]

rotation matrix

[proper otter]

?

[burnt lark]

i dont think theyve learned that yet

what functions are you allowed to use

[proper otter]

.

[burnt lark]

i guess you could play around with a radical function of the form $a\sqrt{x+b} + k$

[woven radish]

TestTickler

[burnt lark]

thats only for the top half

for the other half missing to connect just do the negative version, so something of the form $-a\sqrt{x+b}+k$

[woven radish]

TestTickler

[burnt lark]

probably want a higher a value since it needs to be a bit steeper

[proper otter]

im so confused

im super bad at math

[burnt lark]

since youre not allowed to use the rotated parabolas like x = y^2

you can do it this way

just mess around with the scaling on the radical function and also the height at which it needs to be to fit your graph

[proper otter]

what variables do i put im using sliders

[burnt lark]

of this form $a\sqrt{x+b}+k$

[woven radish]

TestTickler

[burnt lark]

mess around with a,b,k

[proper otter]

??

[burnt lark]

yeah

[proper otter]

whys it missing the bottom curve

oh

i have to make it

[cold obsidian]

Hi

[burnt lark]

yeah just combine it with the "negative" version of the function

[proper otter]

how to cutoff green

[burnt lark]

is that the equation for green?

try restricting the domain in the x-direction?

[proper otter]

ye

originally this was absolute value but i just halved it

and now im trying to do linear

[proper otter]

should i do linear or a mirror of absolute value

[burnt lark]

for what

[proper otter]

the brows

[burnt lark]

just do linears

unless you have no abs value yet

[proper otter]

i didnt

i just mirrored

it

before

after

[proper otter]

for the eye shape what should i do

[burnt lark]

idk man

if u could do a circle it would be easiest

[proper otter]

for this gap

i tried doing that sideways parobala

[burnt lark]

just do linear for that

[proper otter]

SCAM

[proper otter]

i did ax^2+bx+c

[burnt lark]

yeah you could do a negative one

where $ -1 < a < 0 $

[proper otter]

SCAMMM

<@&268886789983436800>

is ther anyway to make touch the corner of the purple

[burnt lark]

probably need two different abs value functions

[proper otter]

yoooooooooooo

[devout snow]

@proper otter Has your question been resolved?

[burnt lark]

nice bro

[proper otter]

i need help are you still available

[burnt lark]

one last thing because im bouta go to bed

what u need

[proper otter]

how can i connect that

[burnt lark]

so like what we did with the $a\sqrt{x+b}+k$

[woven radish]

TestTickler

[burnt lark]

but this time its in the other direction, so $a\sqrt{-(x+b)}+k$

[woven radish]

TestTickler

[burnt lark]

and you unite them in the same way we did the other

$-a$

[woven radish]

TestTickler

[burnt lark]

yeah just use sliders on that

should be able to get it

[proper otter]

it stay straight

wait nvm

thanks

[burnt lark]

ofc

im gonna head to bed now, gn and gl

[devout snow]

@proper otter Has your question been resolved?

[violet mauve]

Could i have some help with a trig question?

[celest lantern]

!occupied

[devout snow]

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).

[celest lantern]

Make another channel, please?

Try #help-2 .

[violet mauve]

Thank you!

[celest lantern]

No worries!

[devout snow]

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[keen sundial]

I assume this is wrong

[mild sorrel]

What even is (2 0)?

[keen sundial]

vec a

[mild sorrel]

you need to compute the magnitudes i guess

[coarse flume]

Ye the dot product is a magnitude not a vector

[mild sorrel]

|a->| represents magnitude

[keen sundial]

this was their solutionon

[trail eagle]

Yes. You need to compute the dot product along with the magnitudes.

[keen sundial]

how do we find magnitude again

[coarse flume]

Sqrt(x^2+y^2)

[sturdy fiber]

-# why is there a random lego brick on the side ;-;

[keen sundial]

bru I cant lie, no clue either

[coarse flume]

What = 2sqrt2 cos(theta)

You can't just write half of an equation

[trail eagle]

Yes, and so cos(theta) is not just the product of the magnitudes.

[coarse flume]

So write that the dot product=2sqrt2 cos(theta)

And then find the dot product another way

[keen sundial]

theta = pi/4?

@coarse flume @trail eagle

[coarse flume]

Yes

[keen sundial]

also what does he mean

[trail eagle]

The outside angle could equally represent the angle separating two vectors.

It's conventional to take the smallest of the two.

Hence why they specify in [0,pi]. Even then generally the "angle between two vectors" refers to the smallest such angle, and arccos will always yield that one anyway.

[keen sundial]

makes sense

so does this formula ( I assume its called the re-arranged dot product formula or if there is a specefic name let me know) always considers the internal angle?

[trail eagle]

It would work equally with either, since they would both have the same cosine.

But arccos always spits out some angle in [0,pi] anyway

So unless you’re looking for the bigger angle explicitly you’ll always get the “right” angle

[keen sundial]

ohhh makes sense

and whats this formula called? the one i mentioend in the brackets

[trail eagle]

I don’t think it has a specific name. As you said it’s just derived from the dot product definition.

[solid perch]

-# just as you wouldn't say that V = IR is Ohm's law but I = V/R, which you get by dividing by R on both sides, has a separate name.

<@&268886789983436800>

[keen sundial]

alright well thank you!

.close

[devout snow]

Channel closed

Closed by @keen sundial

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

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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.

[zealous bronze]

19 th question

[stuck hill]

,rccw

[stuck hill]

okay so i feel we can do this if we show EF is parallel to GH

[zealous bronze]

The diagram I drew is right

I am jst nt able to understand how is angle hae and angle gcf equal

[zealous bronze]

Yes but idk how to do next step

[stuck hill]

,rccw

[zealous bronze]

If we prove these angles equal then by sas
We can say that he and gf are parallel
And similarly opp side is parallel hence Its a parallelogram

[stuck hill]

i mean you can show AE is parallel to FC

because AE is parallel to OB and OB to FC

now also AE = OB and OB = FC

so AE = FC

so AE = FC and AE || FC

so AEFC is a parallelogram

got it?

similarly HAGC is a parallelogram

can you continue from here?

[zealous bronze]

But we jst know its parallel nt equal right ?

[zealous bronze]

No

[stuck hill]

why?

[zealous bronze]

But how will I prove hae is an line ?

We can jst assume it's an line

[stuck hill]

angle chasing i think will work here

oh i got it

so

[zealous bronze]

Wt is that

[stuck hill]

we need to show

HAD + DAO + OAB + BAE = 180

now HAD = ADO and BAE = ABO

[zealous bronze]

Had?

[stuck hill]

angles

now since HAD = ADO and BAE = ABO so i feel if we consider triangles AOD and AOB we can show this

[zealous bronze]

Wt can we show

[stuck hill]

from triangles AOB and AOB, since BD is a straight line we have AOD + AOB = 180
so (180-OAD-ODA) + (180-OAB - OBA) = 180

i feel we cna continue

if we cannot im really sorry

[zealous bronze]

How can u say bd is an straight line ?

How abt this method

Bd is nt a diagonal of parallelogram abcd

[stuck hill]

because it is a disgonal of ABCD

[zealous bronze]

It's nt mentioned in the question wont it be our assumption then ?

[stuck hill]

no

[zealous bronze]

We can't assume things by diagram right ?

[stuck hill]

because ABCD is a paralleogram

by definiiton

BD and Ac

are its diagonals

no need for a figure anyhow

[zealous bronze]

But o Is random point in parallelogram

[stuck hill]

oh im really sorry then

sorry.

[zealous bronze]

Like is my point wrong ?

Or wt

[stuck hill]

no im wrong

sorry

[zealous bronze]

Can u see this

[zealous bronze]

And explain me ?

[stuck hill]

nono my approach is wrong

[stuck hill]

,rccw

[stuck hill]

yes it seems to work, but im not exactly the best on geoemtry so ill wati for another helper

[zealous bronze]

Okayy

<@&286206848099549185>

[velvet geode]

wsgg

[velvet geode]

,rccw

[velvet geode]

,rccw

[velvet geode]

demn

[zealous bronze]

Wt

[velvet geode]

ion rlly do geometry tbh

mb gng

peace out ✌️🕊️ ☮️

[zealous bronze]

Mb

?

[velvet geode]

mb = my bad

kinda like sorry

[zealous bronze]

Oh😔

[velvet geode]

🥀

[zealous bronze]

<@&286206848099549185>

[mossy flicker]

What's the problem