help-38

[supple copper]

if this sounds like an integral that's because it is

[feral trail]

yeah lol i’m looking at that in my notes right now, i understand

[supple copper]

so let's take an arbitrary slice, at height x

also, upsidedown things is weird cos you need to start at 0 then go does x metres then do stuff

but there's nothing stopping us from putting the cone right side up and just flipping a negative sign in the density

if you dont want to do that that's okay as well

[feral trail]

i’d like to keep it as is for right now until i get more comfortable and not go through that mental gymnastics LOL

[supple copper]

that's okay all good

so now we need to know how big our discs are

given that we go down x meters from 0, what is the radius of the disc?

(use similar triangles)

[feral trail]

oh, i see. i remember my prof went through an example like that where he used the pythagorean theorem to find the radius of the disc like that?

[supple copper]

i dont think you need pythagoras here

similar triangles use scaling factors

which means you just need fractions

so like, if you go down halfway, ie x = 4.5, then the radius should be halved as well, ie r = 5.5

because the similar triangles created should differ by a scaling factor of 2 here

[feral trail]

okay, i understand what you’re saying, im just having a difficult time answering this LOL

[supple copper]

So think of it proportionally

If you go down x metres, what proportion of the whole way have you gone?

[feral trail]

x/9? because our depth is 9 meters?

[potent cove]

i mean it’s just linear variation so just integrate dm

and dm = pdV

p is density btw

dV is a volume element as you go along x

and ig you should take an infitessimally thick disk of thickness dx for that

[supple copper]

What’s crazy is that’s exactly what we’re doing

Already

[potent cove]

damn so

Looks like it’s done?

[supple copper]

Uh huh, so proportionally speaking, how much should the radius be?

[potent cove]

are they trying to find dV or smt?

[feral trail]

hold on the gears are turning

[wraith hinge]

since the raidus increases linearly from 0 to 11 m, shouldn't it be
r(x)=11/9x

[supple copper]

You’re not the op

If you have a question ask it in your own channel

I’m trying to help bianca here

[potent cove]

yeah ig he’s tryna help

mb im wrong

[feral trail]

i’m a little thrown off by the radius at the top being 11 meters, i’m trying to figure out how this relates in proportion to the x/9 i stated earlier

[potent cove]

not here to argue

[supple copper]

So if x is 0 what should the radius be

[feral trail]

11, because x at 0 is the top ?

[supple copper]

Yep

And what should it be when x = 9?

[potent cove]

i’m really curious if you are trying to find r(x) rn?

[supple copper]

We are

[potent cove]

hmm

[feral trail]

1?

[wraith hinge]

no shit sherlock , you could've just said nicely , I'd understand

[supple copper]

Nah it’s 0 because it’s a point

[potent cove]

i mean since any triangle formed in the cone will be similar

[supple copper]

There’s no radius to disc that’s just a point

[potent cove]

x/r is constant and equal to total height/total radius isnt it?

[supple copper]

I did say earlier to use similar triangles but they are struggling

[potent cove]

damn

[supple copper]

Guys you’re not helping

[supple copper]

By the way, let’s just try it upright

[feral trail]

okay!

[supple copper]

If I stand the cone upwards can you find the radius at some height?

[feral trail]

actually, can i show you my notes that i took of the example my professor provided? it’s got a cone shape and he used a triangle to determine the radius and im wondering if it would be useful at all. only thing is its the pythagorean thing i mentioned earlier

[supple copper]

Sure

[feral trail]

if you can’t read something let me know, thanks for taking the time to help^^

[supple copper]

Yeah okay so here Pythagoras helps because it’s a hemisphere

We have a cone so we don’t even need to think about the hypotenuse

[feral trail]

ohh right i completely forgot that it was a hemisphere, my bad

saw the disc and triangle and immediately assumed cone whoops

[supple copper]

Lots of things going on in this picture

But the blue part is the question info

Red part is what we want to find out

Green part are some extra bits of construction to help us find red

Do you see that the green and blue triangles are similar?

[feral trail]

yes, i do!

[supple copper]

So if I give you 9-x can you find the red ? Length?

[feral trail]

wait, so 9-x is the height of our green triangle right?

[supple copper]

Yes

[feral trail]

i see two separate arrows so i’m a little confused lol

[supple copper]

Oh sorry

The arrow in the middle is me saying use this to find the green triangle’s base and subsequently the ? Length

[feral trail]

okay, i don’t really remember how to find the length like that. i’m really tempted to use something with 1/2(b*h) but i don’t see a value for area we can use to determine the base’s value and it sounds inefficient to use that formula anyways, so im trying to remember what else i know

[supple copper]

What’s the relationship between a, A, b and B?

[feral trail]

A is double the size of a, B is double the size of b?

the green triangle is twice as big than the red triangle

[supple copper]

Well I haven’t actually given you that they are exactly double

[feral trail]

oh

[supple copper]

But proportionally speaking they are the same right

[feral trail]

yes

[supple copper]

As in, if I multiplied a by some factor k to get A (ie. ak=A)

Then this same factor will also scale b to B (ie. bk = B)

So we can say that k = A/a and k = B/b

In other words, A/a = B/b

Is that right?

[feral trail]

okay, yes

[supple copper]

so that answers this question

[supple copper]

Can you apply the same thinking to this problem? We have the bigger blue triangle and the smaller green triangle

[feral trail]

11/? = [(9-x) + x] / (9-x)

[supple copper]

Let’s use r instead of ?

I should’ve put r from the very beginning

[feral trail]

okay!

11/r = [(9-x) + x] / (9-x)

one second writing out the radius

[supple copper]

Awesome

So we have the radius in terms of x now

[supple copper]

So we’ve answered this question

Now we have the radius we can find the volume at each height

[feral trail]

it wouldn’t happen to be pi(r)^2 would it?

[supple copper]

Yes but replace r with the thing with x

And also we need to multiply by a little height we will call delta x

Δx

[feral trail]

ah yes yes okay

and for the mass of just the slice we multiply this by the density?

[supple copper]

Yep!

[feral trail]

like so?

[supple copper]

Yep

[feral trail]

and then the total mass is just the integral of that from 0 to 9?

[supple copper]

Yep

And turn the Δx into dx

[feral trail]

okay!! let me punch this into desmos and see if the answer is correct

[supple copper]

I think wolfram is usually better than Desmos at computing stuff

[feral trail]

ooo okay i’ll try that instead

i’ve never used wolfram before, does this input look okay?

oh it didn’t work hm

[supple copper]

What do you mean it didn’t work

[feral trail]

oh no just the way i punched it in to wolfram was wrong

i went back to desmos LOL here’s what i got

it’s finally right ☹️ thank you i really appreciate your help

.close

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[grim sparrow]

well done @feral trail

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[odd pollen]

may someone help me with basic grade 9 physics

[odd pollen]

using the circuit below calculate total current 1T. potential difference for 2nd resistor V2 and total power Pt show all work and use grass method

is what the question says btw

[hollow cypress]

,rccw

[rose inlet]

first off, what's given to you (all ur voltages and resistor values)

and what are you solving for

[hollow cypress]

theyre at #help-44｜stanley-🌲-v2-dans

[rose inlet]

oh]

i take it back

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[sleek imp]

The angle between two intersecting planes passing through one of the cylinder's radii is 60°. The areas of these intersecting planes are 360 and 720. If the height of the cylinder is 30, find the radius of its base:

[sleek imp]

I tried to do it through Pythagorean theorem but there was no answer like that

What else can I do for this one?

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@sleek imp Has your question been resolved?

[sleek imp]

<@&286206848099549185>

[floral lodge]

Not sure if you have the problem set up right, or if I’m just wrong.

[sleek imp]

I got the same answer but it isn't there as you can see on the second pic

[floral lodge]

Yeah that’s why I’m thinking you got the problem set up wrong

[sleek imp]

How else could it be drawn then?

[floral lodge]

Good question

[sleek imp]

I just assumed that it would look like this since the angle between is 60° and it's a cylinder

[floral lodge]

I don’t think you’re simply just finding the radius the the cylinder, but the radius of the plane being intercepted

[sleek imp]

Yeah but wouldn't those two be equal?

[floral lodge]

No since it’s at an angle

Kind of a bad picture

[sleek imp]

But wouldn't that be an ellipse?

[main sigil]

I don't really understand the question

[sleek imp]

Well, there is another translation for that
The bilateral angle between two sectional planes passing through one of the cylinder makers is equal to 60°. The areas of these sections are equal to 360 and 720. If the height of the cylinder is equal to 30, find the radius of its base:

But I don't think that changes anything

[main sigil]

can this be right?

the sketch

[sleek imp]

Wouldn't that pass through both radiis (like the heights)?

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@sleek imp Has your question been resolved?

[fossil sedge]

What is bilateral angle?

And where are you from? @sleek imp

[sleek imp]

Kazakhstan

[sleek imp]

Its just the angle between two sections/planes/cuts

[sleek imp]

.close

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[safe nacelle]

Let A be an nxn matrix. Let S be m-dimensional subspace of Rn with m<n.

Does any subspace of Rn have a basis? and if the rows of A are linearly independent the is it true that Rn is a subset of col(A)

[thorn bay]

i dont see how S is relevant

every vector space has a basis

[safe nacelle]

it was for a follow up but its not relevant here

[thorn bay]

for finite dimensional spaces, you can prove that with just lin alg

for infinite dimensions you need set theory

did you learn that?

[safe nacelle]

yes, but my lecturer marked my multiple choice answer incorrect when i said it was true

[thorn bay]

what was the question?

[safe nacelle]

^this is his memorandum

[thorn bay]

that is absolutely wrong

[safe nacelle]

right?!

[thorn bay]

write a proof!

prove that you're right

[safe nacelle]

i just want to know about (I) now as well. Because would it not be true if it was col(A) subset Rn

[thorn bay]

do you know any theorems connecting column space to row space?

[safe nacelle]

yup. row(A) = col(A) and if they linearly independent they span the subspace
is that what you mean?

[thorn bay]

not necessarily row(A)=col(A). you need more conditions

[safe nacelle]

right..
but how is Rn a subset of col(A)

[thorn bay]

I have to go, but, the hint is

dim row space A = dim col space A is the formula in general

see if you can justify equality here gl

[safe nacelle]

rip. okay thanks

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[wraith hinge]

why is tany plus minus?

even when the tany is positive all the way in the

0 < y < pi/2 and pi/2 < y < pi

im

fucking stupid

thats why

SHAKE MY HEAD

pi/2 < y < pi is another damn strand of tan function

which is negative

[lime sphinx]

they used a trig identity imo

[wraith hinge]

i asked about why they got plus minus

[lime sphinx]

then to solve for tan y using that trig identity you need to take square roots on both sides in the second to last line of what I posted

i mean i skipped some algebra to solve for tan y but yeah

[wraith hinge]

yeah but the purpose of plus minus i asked lmao tho i got it as i mentioned above it is because theta angle is between 0 and pi/2 and pi/2 to pi which is negative and positive

so it would chnge signs

so they got both of them

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[obsidian kraken]

Please help 😭😭😭

[mystic iris]

,rccw

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[solemn wagon]

Usually in these type of problems its nice to extend some side in order to simplify the problem into proving two lengths are equal instead of the sum

Id suggest extending MB by the length of ND

"upwards"

now we have to prove a triangle is isosceles

this channel is occupied please use the available help channels

[zinc ginkgo]

#❓how-to-get-help

[soft apex]

oh sorry

[solemn wagon]

@obsidian kraken

[obsidian kraken]

I tried it, but I couldn't find anything

[solemn wagon]

like this PD = NB and we have to prove that PM = CM

okay

Ill give you a hint

notice that triangles BCN and DPC are similar

can you see why?>

[obsidian kraken]

Hmmm

Let me realise it please

Okayyy

But, what can we do with angles?

[solemn wagon]

okay from the similarity we get <MPC = <NCB

call this angle a

[obsidian kraken]

Ohhhhh I got ittt!!!

Thank you very much!!!

[solemn wagon]

no worries

nice

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[old patrol]

i’m trying to solve for c and i got to how many cycles there are in 29 seconds, 5.8, but im not sure what to do with the 0.8 cycles that’s left

[old patrol]

i vaguely remember having to that that into the function but i’m not quite sure

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[jade parrot]

<@&286206848099549185>

[zinc ginkgo]

!15m

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[summer haven]

!status

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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

[jade parrot]

1

?

[jade parrot]

1

[summer haven]

okay

so what's your understanding of turning points

and how to find them

[jade parrot]

idk

[summer haven]

so are you familiar with derivatives and what they mean?

[trim lichen]

does the word "vertex form" sound familiar

@jade parrot

[hollow cypress]

alternatively completing the square form

[jade parrot]

no

[jade parrot]

no

[summer haven]

oh yeah you don't need them at all here mb

[trim lichen]

... ok what level of education are you at

[jade parrot]

secondary school

@trim lichen

[trim lichen]

$y = a(x-h)^2+k$

have you seen something like this before at all

[solid kiln]

Ann

[jade parrot]

yeah i think so

[trim lichen]

ok that is called vertex form in English usually

can you convert your quadratic into this form?

[jade parrot]

(x-4)^2+3

[trim lichen]

why (x-4)^2 tho

that gives you a -8x term upon expanding back and not +8x like you had

sign error?

[jade parrot]

Idk

[hollow cypress]

could you show your work as to how you got to this answer

[jade parrot]

I cba anymore

.close

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[waxen cloak]

Hey, I'm trying to prove a greedy algorithm is always optimal for this merge operation on an array: repeatedly merge adjacent $x$ and $y$ into $\max(x,y)+1$, goal is to minimize final value. At each step, pick the merge with smallest $\max(x,y)+1$.

I tried formulating an exchange argument but got stuck. Wrote this up cleanly here: https://math.stackexchange.com/questions/5063365/optimality-of-the-merge-the-smallest-greedy-algorithm

If anyone has ideas or sees a clean proof, would love to hear it.

[solid kiln]

Avighna

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[waxen cloak]

<@&286206848099549185>

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[unreal river]

Hi

@waxen cloak seems not optimal

For instance take $1, 1 - \varepsilon, 1 - \varepsilon, 1$

the greedy will begin with the middle 2

[solid kiln]

Cain

[unreal river]

instead of merging (1,1-eps) and (1,1-eps) and then merge (2,2) to get 3

the greedy will merge (1-eps,1-eps) and then (1,2-eps) and then (3-eps,1) and get 4-eps

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[waxen cloak]

okay im really sorry but i shouldve mentioned that all values are non-negative integers

in particular, 0 ≤ a_i and a_i is an integer

i'll edit this into the post

[solid kiln]

Avighna

under this constraint, $\varepsilon$ would have to be $1$ so the final answer would be $3$
```Compilation error:```Image processing timed out!```

[unreal river]

@waxen cloak it doesn't matter

still 1,1,1,1 would do the same

as greedy might choose the middle two first.

[waxen cloak]

Ok yeah, you're right. The version I was imagining always picks the leftmost pair in case of ties, which I now realize might matter.

This problem is from the Baltic Olympiad in Informatics, 2025 (Day 2, Problem 2), and- spoiler alert- my friend coded this greedy solution that worked for $n \le 2000$. I’m guessing his implementation broke ties by picking the leftmost option, which might be why.

It might seem like I’m just patching the algorithm to make it work (which I know can feel a bit annoying), but I’m genuinely curious whether this tie-breaking tweak is enough to make the greedy strategy provably correct. Or is there a counterexample even then?

[solid kiln]

Avighna

[unreal river]

still not

2,2,1,2

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[opal snow]

can smn help with this

[opal snow]

i need to find angle DCE the question mark

please

[stoic garden]

I do not think there is enough information.

You could rotate point E about point C, and none of the given information would be contradicted.

And, looking at your work, it doesn't look like you added in any extra information. This problem is likely impossible.

@opal snow

[opal snow]

its possible

a teacher wouldnt give us an impossible question dont u think

[grim saffron]

Is FCA a straight line

[opal snow]

thats what am saying its weird

[stoic garden]

It has happened more times than you likely think.

[opal snow]

it should be but it doesnt look like it

maybe a drawing error

[stoic garden]

It looks to me like the lines were drawn on a computer, after which the labels were hand-drawn. I doubt they meant to make FCA a straight line.

Thoughts? @grim saffron

[versed jasper]

pretty sure you can solve it without any trig

if im not going insane

[stoic garden]

#help-38 message

[stoic garden]

Are you assuming angle FCB measures 90 degrees?

[versed jasper]

mm perhaps

[grim saffron]

I think it's more likely that the diagram is drawn by hand, since the line FCA is almost a straight line. You'd think that if the angle is intentional, it would be exaggerated.

[versed jasper]

just noticed AF may not be a single line

[stoic garden]

Yeah, that seems to be the deciding factor.

@opal snow for now, I think it's best if you assume FCA is a straight line.

[opal snow]

yeah

but can u solve it if that was the cases

[stoic garden]

You certainly can.

[opal snow]

how

[stoic garden]

Angles ACB and FCB are supplementary. In other words, they form a linear pair.

[opal snow]

alr ty

[versed jasper]

that's how i thought of it

(wait image upload slow)

[versed jasper]

that was assuming AF was a straight line

[stoic garden]

This seems like an unnecessary overcomplication. Can I DM you a potentially faster way?

[versed jasper]

sure

im no geometry expert disclaimer

[opal snow]

so we should just find all the angles and add and minus 180

the nagles that are on af

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[stoic garden]

<@&268886789983436800>
This user's profile is full of sexual references.

[manic lagoon]

Ok

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[real herald]

My book legit just didnt provide a proof for this proposition

[real herald]

is there a reason this holds?

[thorn bay]

i hope so
in math, things usually are true for reasons

even not in math actually

what's the definition of lin dep and lin ind?

[real herald]

a linear combination is linearly independent if when the linear combination equals $0 \implies$ the coefficients must also be zero.

[solid kiln]

BOSS

[real herald]

or if a vector cannot be written as a lienar combination of the other vectors

oh

wait

i get it lmfao

@thorn bay

if its a basis by definition

anything else will have to be generated by a linear combination of those elements

so it wont be linearly independent

.close

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[thorn bay]

right because you could v - (linear combination of the other elements chosen to equal v) = v-v=0 even if coefficients are not 0, exactly what you said but im.spelling out the details

[real herald]

@thorn bay yeah thats what im getting at

.reopen

[trim jolt]

✅

[real herald]

@thorn bay do you know how to define maps between vector spaces?

sorta like homomorphisms preserving verctor addition and scaler multiplication?

this is from my lecture notes and i kinda just

[ionic pendant]

this is a standard proof in linear algebra texts. this textbook has a free pdf online: https://linear.axler.net/

[real herald]

confused the shit outta myself

oh tysm i love axler

[thorn bay]

yes that's called a linear function, in general
if the domain and dcodomain are the same it's called a linear transformtaion
if the domain is V over F and the codomain is F it's called a linear functional

[real herald]

like

[thorn bay]

and if the domain is a set of functions it's called a linear operator

[real herald]

This was not in the reading its part of a question

ok im gonna work them out then

[thorn bay]

linear transformations = linear functions, theyre cinnamonous

[real herald]

how much of this is needed to understand extention fields?

[thorn bay]

honestly, i'm bad at algebra

I like analysis and topology

so, I dont even know what extension fields are lol

[real herald]

ah nw

im learning lin alg to understand them so i was curious how deep i need to go

tysm for the proof help!

.close

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[twilit lion]

I need help on radicals

[trim jolt]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[twilit lion]

Does anyone know about radicals

[trim lichen]

you should just send your question

not ask "does anyone know about <topic>"

just send your question

[twilit lion]

Ok I’m sorry

I need help with both of them this was my check in and my teacher is learning the material with us so it’s tuff

I copied the work off what she was showing for review but I got no idea how to do this

[trim lichen]

ok right

so first off do you know what a square root is?

like just to ensure we got the basics of the basics down for this whole topic

[twilit lion]

Yes

[trim lichen]

ok

[twilit lion]

Like I know what a square root is but it’s just the showing of the work part

[trim lichen]

ok so do you know any properties of square roots

[twilit lion]

What do you mean

[trim lichen]

we're gonna need some of those for the work

sqrt(ab) = sqrt(a) * sqrt(b) for example

that ring a bell?

[twilit lion]

Yeah

[trim lichen]

ok let's see

do you understand what goes on with that numbers example

[twilit lion]

I think you separate the two doing square root of 9 and 5 multiply them with each other then because square root of 9 is 3 and 5 doesn’t the answer is 3 square root of 5

Is that correct?

[trim lichen]

when typing roots in plain text you can and should abbreviate them as sqrt(), so instead of "the square root of 5" you write sqrt(5)

[twilit lion]

My bad

[trim lichen]

this is more or less correct otherwise. not how i would have worded it but otherwise okay.

[twilit lion]

How would u have worded it

[trim lichen]

i would just say sqrt(9) simplifies to 3 and sqrt(5) stays as is

anyway,

[twilit lion]

Dang I’m dumb

I make it much harder for myself

[trim lichen]

now looking back at question 7 here, you've got a sqrt(50) and a sqrt(8). these are both simplified following the same logic as in the concept example: 50 is broken up as 25*2, and 8 is broken up as 4*2

from which you get $-5\sqrt{50} = -5\cdot 5\sqrt{2}$ and $3\sqrt{8}=3\cdot 2\sqrt{2}$

[solid kiln]

Ann

[twilit lion]

Ok I’m getting it more

[trim lichen]

the general strategy when faced with a root of a big number is to factor out as big of a perfect square as you can from it

as far as simplification goes

[twilit lion]

Yea I understand that concept

[trim lichen]

ok so we good on question 7 then yeah?

[twilit lion]

Ohh I’m starting to understand it now

Yep question 7 is all good

Is it ok if we move on to question 5

[trim lichen]

yes alright

you may have heard of such a term as rationalizing the denominator

[twilit lion]

Yes

[trim lichen]

ok yeah that's what is happening there

[twilit lion]

Is that multiplying the denominator

To the equation

My wording might be off

[trim lichen]

multiplying the top and bottom of the expression by the conjugate of the denominator

an equation is something that has an equals sign

[twilit lion]

Wait

Would the conjugate be 2

[trim lichen]

no, it is never just a raw integer

the conjugate of 2 - sqrt(8) is 2 + sqrt(8)

[twilit lion]

Ohh so it would be multiplied by 2+sqrt(8) top and bottom correct

[trim lichen]

yes

[twilit lion]

So conjugate is like opposite

Where the sign changes

[trim lichen]

the idea behind this move is to set up a difference of squares identity on the denominator so that the square root in there simplifies away

[twilit lion]

Ok

[trim lichen]

idk what went on with this three or the 2 just randomly hanging there

[twilit lion]

No the -2 was because originally in pencil I got the problem wrong

Pen is corrections

And idk abt the three I probably accidentally wrote wrong number in there

It should be a 2

[trim lichen]

it should be an 8 actually

let me rewrite the entire thing in a less messy manner

idk if this handwriting is yours or the teacher's but it is a bit sloppy

[twilit lion]

It’s mine 😂😂 I Know my handwriting isn’t good

[twilit lion]

Can you explain what you did here?

[trim jolt]

@twilit lion Has your question been resolved?

[dense grove]

Rationalize

[twilit lion]

I still need help with this one

I don’t understand how they got -2-2 sqrt(2)

<@&286206848099549185>

[dapper swift]

which part didn't you understand?

[twilit lion]

I guess after staring at it for 15 minutes I kind of understand the last part

The 8 simplifies to -2 and 4 simplies to -1 multiplying with 2 to get -2

[thorn bay]

you have pretty handwriting

[twilit lion]

So it will be -2-2 sqrt(2)

But I do need help with a couple more problems

If anyone can help with that

[trim lichen]

which line

[twilit lion]

I got a test so I’m cooked

[trim lichen]

sorry i had to disappear for a while

[twilit lion]

All good

[twilit lion]

I figured out the last one kinda

[twilit lion]

^

It was the last part

[dapper swift]

ah so yeah it's the distributive law

$\frac{8}{-4} + \frac{4 \cdot 2 \cdot \sqrt{2}}{-4}$ first

all good

[solid kiln]

south

[twilit lion]

All good

Can anyone help me with questions 2,4 and 8?

[dull temple]

2 and 4 are right and are roughly how I'd do them

for 8, start by factoring the stuff under the radical

like 24 = 2.2.2.3

which is 2³.3, so if you take the cube root you get 2.³√3

[twilit lion]

Alright thanks

[trim jolt]

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[opaque dust]

okay so i know how to do most of this question

im issue is i get the pi/3

but idk how to figure out that the next one would be at 5pi/3

(ik that if it was like 1 that hapepns every 2 pi, or 0 hapepns ever pi..

[dusky thunder]

!occupied

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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).

[dusky thunder]

do u know the unit circle

[opaque dust]

yes

barely

[dusky thunder]

notice how at 5π/3 and π/3, cos has the same value

[opaque dust]

i dont have it memorized to the point that i could remeber this tho

thats the isuse

[dusky thunder]

u only need to remember the first quadrant

the other quadrants are just reflections

[opaque dust]

like i said for hteo ther one since it = 0 i just knew the oher one would be within pi

is there any similar trick without remeebring it

[opaque dust]

i dont have time to remeber it before my test

any other way i couold do anything at all

[dusky thunder]

can u bring a cheatsheet?

[opaque dust]

nope :(

[dusky thunder]

https://youtu.be/75dMcyCUo2g?si=Xq0GxeiXqSA6mlHb

try this

[opaque dust]

i dont have time

[dusky thunder]

one way or the other, u at least need to know the first quadrant

[opaque dust]

i need something else sorry

[dusky thunder]

then no

[vital grail]

what do you want to remember? like the values of sin, cos, and tan? As in sin 90 = 1, etc

[opaque dust]

i know the 45 45 90 triangle

and the 90 30 60

like 30 60 90 is just 1, 2, sqr 3

[trim lichen]

those two triangles give you most of the 1st quad

you also need to remember the trig values at 0 and 90°, which for sin and cos are just 0 and 1 (but dont mix up which is which)

[opaque dust]

is there a way to fiure out how long untill the next solution, like i said eerlier when cos=0 i knew that the next one would be A total pi after. any way to do a similar thing for 1/2

[dull island]

can somone help me through this simple polynomial factoring " x^3-6x^2+11x-6 "

[trim lichen]

i think "how long until the next solution" is not a good phrasing

[trim lichen]

!occupied

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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).

[opaque dust]

sorry im just saying it the best i can

[trim lichen]

when solving cos(x)=a the two fundamental solutions are horizontal reflections of each other

for sin(x)=a they're instead vertical reflections of each other

[supple copper]

I have a trick for you

[vital grail]

yea i'm not sure what you're looking for here. I'll drop this just in case.
https://www.youtube.com/watch?v=PF2nmCVSUEs

[trim lichen]

oh yeah the trigonometric hand thing

[vital grail]

good luck

[opaque dust]

what

[supple copper]

Suppose they ask you what is sine 60

[opaque dust]

that one ik off the top of my head

[supple copper]

Or what is inverse sine of whatever

This is what you do

They ask you a fraction

Arcsin(a/b)

If the argument is positive then there’s 2 options, the top is positive and the bottom is positive

Or they are both negative and cancel each other out

If the argument is negative then either the top or bottom (but not both) are negative

So let’s try -1/sqrt 2

Sine is opp/hyp so the triangles look like this

[opaque dust]

also say there was no like domain or whatever

and i had to answer in general form, then id need to know how far between each one

so how do you do that?

[supple copper]

Then you can use Pythagoras to figure out which angle is 30 and which is 60

Or in this case they are 45 degree angles

For example these are 4 of the 8 ways these 2 numbers can be arranged

And of course the hypotenuse here is 1

Red are the other 4

[opaque dust]

yes i know all of this'

i think

[supple copper]

If you need cos or tan then maybe you need to flip the triangles

[opaque dust]

b ut like

in the case of one half

[supple copper]

Arcsin(1/2)?

[opaque dust]

i know that its cos = 1/2 and i know that means that its 60deg, or pi/3

but then idk what how to find how long untill the next one for gneral form

[supple copper]

Look at the red triangle

The angle is pi/3 as you said

But the blue one is the other way of drawing the triangle

It still has +1/2 as the adjacent side and the hypotenuse is always +1

The angle for the blue one is clearly 2pi - pi/3

[supple copper]

So the general solution will be 2kpi + this

[supple copper]

And 2kpi + this

[supple copper]

You need to recognise that there are 2 distinct solutions that aren’t really related to each other (their relationship depends on the trig function)

So I use the triangles to find both solutions for me

Then I use that to then add on 2kpi onto each of the 2 solutions to get all my solutions

[supple copper]

This corresponds to the 2 different ways you can draw each triangle

[opaque dust]

okay i think i mnaged to remeber the unit circle

so

how do i use it exactly

[supple copper]

I just explained it…

[opaque dust]

sorry im just overwlemed

[opaque dust]

also if i had sin= sqr 2

sorry

sin = sqr2/2

since sin is y/r

i look at when y ris sqr2/2 on the uniti circle?

[supple copper]

So sqrt2/2 is just 2/(2sqrt2)

Which is really just 1/sqrt2

[opaque dust]

so pi/4 3pi/4

[opaque dust]

this crrect tho

if i have the unit circle cheat sheet if im allowed i just look fro the y

i dont think im allowed but i thinkim close to remebering it good enough

[supple copper]

Well it’s just this without the minus signs

So it’s the ones up above

Or as you said, pi/4 and 3pi/4

[opaque dust]

okay

im cooked for my exam tmr

tty for the help to make me do a little bit better

if im not allowed a cheat sheet i lowkey might need ot load the image onto my calc 😭

idk if i will do that tho since i hate cheataing even tho its easy to get good grades

doestn feel right to cheat for my self (unless everyone is)

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@opaque dust Has your question been resolved?

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[trim jolt]

@lunar stirrup Has your question been resolved?

[lunar stirrup]

<@&286206848099549185>

[woven python]

!work

huh

the factoid didn't work 😢

just show your work

or

!status

[trim jolt]

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

[lunar stirrup]

didnt do anything

im trying to find the source of the flux

[lunar stirrup]

@woven python

[trim jolt]

@lunar stirrup Has your question been resolved?

[burnt sable]

Have you heard of Gauss's Law/ The divergence theorem?

@lunar stirrup

[lunar stirrup]

yeah

[lunar stirrup]

sorry

[burnt sable]

So we know that this is the expression for electric flux through a surface

But we also know from the divergence theorem:

So how do you think we can use these for this problem?

[waxen cloak]

Greedy gets 4

Both 2,2 and 2,1 give 3, so greedy will pick leftmost and will merge 2,2

It isn’t a counter example

It seems like someone else has claimed this help thread

Since my question got auto closed, i had gone to sleep

what can i do 😔

[burnt sable]

youd have to post it again in a free chat

[lunar stirrup]

uhh

i dont know this one

[lunar stirrup]

only this

[burnt sable]

Thats ok

all it's saying is that integrating F.dA over a closed surface is the same as integrating div(F) over the volume the surface encloses

for this problem you actually dont have to do any integration

Have you seen divergence of a vector field before?

[lunar stirrup]

oh ok

[lunar stirrup]

never

our teacher told us that the line causing this flux is the z axis

but im not able to explain how

[burnt sable]

ah i see the approach they're going for

[lunar stirrup]

yeah?

could u help

[burnt sable]

yeah so they probably want you to think about the symmetry of this situation

So since z doesn't appear in the electric field equation

if you fix x and y and move along z the electric field will be the same

[lunar stirrup]

ohh

what is z was there in the eqn

*if

[burnt sable]

then the field would change as we change z

but we can use this symmetry to give us a clue to solving that first integral

do you think integrating using (x,y,z) would be the best approach here?

[lunar stirrup]

for this one

its z axis right

which is an infinite line charge

so we can use gauss law right

[burnt sable]

yeah exactly

[lunar stirrup]

2k lambda/ r = a/r

ok got it

thank u

.close

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Do not open multiple channels

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[sudden violet]

hi, i'm having some problems solving an exercise. I've asked in stack exchange but I don't understand the solution they suggest me and I don't know how to prove it, as it says minimum/maximum is reached when x = 0 with a geometric interpretation.

[sudden violet]

the exercise is Given an ellipsoid
$$\frac{x^2}{a^2} + \frac{y^2 + z^2}{b^2} = 1$$
and a plane
$$Ax + By + Cz = 0$$
let $S$ be the set of points of the intersection. Find the farthest and nearest points to $(0,0,0)$ that belong to $S$.

[solid kiln]

diiegorgueez

[sudden violet]

and here's the stackexchange link: https://math.stackexchange.com/questions/5063457/minimal-and-maximal-distance-to-the-intersection-of-an-ellipsoid-with-a-plane

also the solution I've been able to get is this one, but that is assuming none of A, B and C are zero.

should I divide into cases A = 0, B = 0, C = 0, A=B=0, B=C=0, A=C=0 or something like that?

and also 1-\lambda/a^2 and 1-\lambda/b^2 have to be non zero here as well smh

<@&286206848099549185>

[trim jolt]

@sudden violet Has your question been resolved?

[trim jolt]

@sudden violet Has your question been resolved?

[trim jolt]

@sudden violet Has your question been resolved?

[lime sphinx]

https://www.desmos.com/3d/ntoe4ujdsg

[lime sphinx]

Your big solution seems to work in Desmos empirically, but it only produce a pair of maximal or minimal distance. The solution proposed on stack overflow utilize more geometric understanding to get the other pair of solution on the green circle

[trim jolt]

@sudden violet Has your question been resolved?

[sudden violet]

yeah, I understand it must be that solution but I'd like to know how to prove it non-geometrically and if I have to divide into cases or something.

[lime sphinx]

in theory lagrange multipliers should get you the four solutions, but the computations and case checking seems to be horrendous tbh

[lime sphinx]

this works unless B = 0 and C = 0 imo, but in that case Ax=0 implies x=0 and it’s much simpler to solve

you can check this with the desmos demo

[lime sphinx]

also if a = b it becomes a sphere but outside of that imo that's it

[sudden violet]

alright, thank you :)

[sudden violet]

also, is this enough to prove that ellipsoid \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 is bounded?

[lime sphinx]

if your definition of bounded is that the distance from the origin of any (x,y,z) in the ellipsoïd is less than some positive constant, yeah this seems ok even if it's a bit hard to read.

alternatively you could put the ellipsoïd inside a closed ball of radius max{|a|,|b|,|c|}

[sudden violet]

yeah i'm sorry my handwriting is too bad 🥲

nice then, thanks

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[real zodiac]

basically i need to use desmos to make a map of sorts and im stuck

[real zodiac]

the brief says "outline of the island. Gradient of curve at a particular point: Hint: Functions will join together more smoothly if they have the same gradient at their intersection point

im not sure how to do this

like if i have a quadratic and a sin function even if i make both the same gradient at the point i cannot further manipulate the graphs

like if i wanted the top part to be flatter i cant change that

idk what to do

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[real zodiac]

.close

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[thin sail]

؟

[thin sail]

i know i can plug in but how can by basics

how can i solve it only by basics*

[simple haven]

You have to think about the 4 ways this equation might be true

[thin sail]

im listening

[simple haven]

abs is a function that takes an input and makes it positive, right?

[thin sail]

100%

[simple haven]

So either you have (x+4) is positive, in which case abs gives you back (x+4) directly

Or (x+4) is negative, and abs gives you back -(x+4)

[thin sail]

Okay

and the same applys to the other side

right ?

[simple haven]

So for each abs we undo it by splitting it into cases.

Exactly

[thin sail]

i will try bro give me sec

[simple haven]

You should have 4 cases

||One doesn't make sense, two give you no result.||

||And the final case gives you the solution.||

[thin sail]

and i will make them equal how ?

i mean every equation equal to 2 cases of the other

like x+3 = y+4
x+3 = y-4

[simple haven]

Let's say you're handling the case where (x+4) is positive and (x+5) is negative. You just set them equal to each other. (x+4) = -(x+5)

[thin sail]

-y-4

okay and that apply to every case

[simple haven]

Yup

[thin sail]

i will try again

so it will be -4.5 after we eliminated the ones who are illogical

correct ?

[simple haven]

That is correct

[thin sail]

Yuppy

thanks bro

you are the best

[simple haven]

You're not too shabby yourself!

[thin sail]

how to close the ticket tho ?

[wraith hinge]

!done

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[thin sail]

.close

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[thin sail]

but i have another question what if we have a quadratic. ?

what we doin here

.reopen

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✅

[thin sail]

I tried to use mode 5 3 to solve it by calc

but what if I did it ;what I will do after that ?.

[thin sail]

is the factorized form equals = (abs(x)-5) (abs(x)+2) ?

if we using calc i can just product the root and ignore the absolute X

so it gonna be -10

right ?

!done

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[main sigil]

What's your definition of a tree?

[zenith venture]

a graph connected in the least number of edges possible

[patent cape]

Proving by contrapositive is straght-forward from the definition you gave.

[zenith venture]

wouldnt the question be saying that something like this is a tree though?

[patent cape]

There are two paths from upper left to lower right, for example.

[zenith venture]

oh

so is the contrapositive if graph G is not a tree, then there is not only one path between any two verticies

[patent cape]

Yes, i.e., if G is not a tree, then there exists two vertices u,v such that there exists more than one path between them.

[zenith venture]

how would I prove that?

but i need to use contradiction for this question

[patent cape]

Start from the definition of a tree. If G is not a tree, then there exists an edge ${u,v} \in E$ such that $G \setminus {{u,v}}$ is still connected.

[solid kiln]

ucheo

[patent cape]

We can use the same argument. For a contradiction, the hypothesis could be that there is only one path between any two vertices in G but G is not a tree.

Then you apply the argument I hinted at.

[zenith venture]

contradiction because there is only one path between {u,v} and hence graph G would be disconnected?

[patent cape]

Yes, since $G \setminus {{u,v}}$ is connected, then we can find a path between $u$ and $v$. But this path will not include the edge ${u,v}$ since it was removed.

[solid kiln]

ucheo

[patent cape]

So we found two paths between u and v, a contradiction.

[zenith venture]

ah ok thanks

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[bright jacinth]

guys

[bright jacinth]

how to solve for x and y

when we have 2 expressions

not equations?

3x+5y-3 and 8x-3y+6

x=?

y=?

[native shuttle]

3x+5y=3 , 8x-3y=-6

[real herald]

are they equal to zero?

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[coral phoenix]

anyone know how to do b

[fierce lake]

Take any three numbers say a, b, c

Since they are in an AP, therefore b = (a + c)/2.

Can you continue from here

[coral phoenix]

uh im not sure what to do after that

[fierce lake]

Uh okay

Well b is an integer right?

[coral phoenix]

yes

[fierce lake]

Then if (a+c)/2 is an integer

That implied a+c must be even correct

[coral phoenix]

yes

[fierce lake]

So now take cases
C1: both a and c chosen are even

C2: both a and c chosen are odd

a and c cannot be even and odd or odd and even respectively

[coral phoenix]

yes

[fierce lake]

And b will be automatically chosen

So you only have to vary a and c

[coral phoenix]

yes

[fierce lake]

From 1 to 10001 there are 5000 even numbers and 5001 odd numbers

So C1 + C2 gives 5000C2 + 5001C2

Divide that with sample space

I.e 10001C3

C1 and C2 mean cases and 5000C2 means combinatorial C

[coral phoenix]

wdym by combinatorial C

[fierce lake]

nCr

Number of ways to choose r objects from n given objects

[coral phoenix]

oh yeah

[fierce lake]

And voila

[coral phoenix]

ok uj

uh

[fierce lake]

You still have a problem?

[coral phoenix]

what does it mean by answer in the form of l/k where l and k are coprime integers

[fierce lake]

Co prime means that the gcd of the 2 integers must be 1

For example 3 and 4

So you gotta simplify the answer into an irreducible fraction

[coral phoenix]

ohhh

[coral phoenix]

ok wait

so u add the two cases

[fierce lake]

Yes

[coral phoenix]

then choose 2 from 5000

why do we add them both

[fierce lake]

Because it's an 'or' situation that's why

A and C can be either both odd, or both even

[coral phoenix]

hmmm

then i divide that by selecting 3 consecutive numbers

like the probability

[fierce lake]

Yes which is the sample space

[coral phoenix]

or just 3 from 10001

[fierce lake]

Yes 3 from 10001

[coral phoenix]

why not like cosequtive

consecutive

or is that already like included in the 5000c2 and stuff

[fierce lake]

As in?

[coral phoenix]

because u could select any two random even numbers

[fierce lake]

Yes

[coral phoenix]

but it doesnt guarentee its consecutive does it

nvm i just realised

[fierce lake]

Even numbers can never be consectuvie

[coral phoenix]

what artihemtic progression means again

[fierce lake]

Lol

[coral phoenix]

my bad

anyway thanks for the help

[fierce lake]

Np

[coral phoenix]

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[blazing geode]

any suggestion to calculate the sum of products between two numbers that draw from 1 to 10 and are different from one another.

[marble wharf]

can you write it down in sigma notation?

[blazing geode]

I think so?

no

It is not viable

[marble wharf]

wdym viable

[main sigil]

there is a way to do it without sigma notation if you'd like to try that

can you find a way to connect your question to the pic i sent above?

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[opal badge]

oops

.close

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[lean kraken]

I know that self inverse is property of klien group four

So I checked and I got answer D

But i didn't understand the upperline means in options A and B

[opal badge]

It means the residue class of the number

sometimes also denotes by brackets

[lean kraken]

SO if I multiply 1×1 i will always get 1 which is correct

[opal badge]

so like in mod 8, $\overline{2} = [2] = {2, 10, 18, 26, \ldots, 8k +2, \ldots}$

yup

[lean kraken]

3×3=9 and divided by 8 will gives 1

And yes 1 is Identity

[solid kiln]

Nanigov

[opal badge]

yeah

[opal badge]

well it should also include the respective negative integers but ygwim

[lean kraken]

7×7=49 and divided by 8 yes correct

Option B seems wrong

Residue class 5×5 =25 mod 40

Not getting Identity element

[opal badge]

correct

b) also has another more glaring issue tho

[trim lichen]

nope! klein 4 group lacks an element of order four, but {1,-1,i,-i} has them.

[lean kraken]

i×i=-1

And -1 is not identity element

[trim lichen]

oh it asks you for a NON klein 4 group

my bad.

everything is in order.

[trim jolt]

@lean kraken Has your question been resolved?

[lean kraken]

How D is in order?@trim lichen

I am saying it is not klien group four

[trim lichen]

when i said "in order" i didn't mean order like in group theory

i meant like everything is OK

[lean kraken]

What you mean everything is OK@trim lichen

I am straightforward saying D is incorrect and ABC are correct in simple words not in hypothyroidism

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[lean kraken]

.close

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[wraith moat]

How should I write out the midpoint sum? I forgot a lot about this stuff

Would it be pi/4 * f(pi/8) for the first interval?

[ionic pendant]

yes

[wraith moat]

• pi/4 * f(3pi/8)?

[ionic pendant]

yes that works

[wraith moat]

Ah okay its just A

Thanks

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[near leaf]

anyone explain this? how did he get so many 99s

[trim lichen]

this hundred was broken into 1+99

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[near leaf]

@tawny escarp