help-23

[floral sequoia]

Which casts a shadow of the horse onto the fence

[north dragon]

yea

[floral sequoia]

As the horse runs, the shadow also moves

[north dragon]

oh

ye

i understood the siutaion

[floral sequoia]

The question is asking what the speed of that shadow is, the moment the horse ran 1/8 of the circle’s circumference

[north dragon]

ye

[floral sequoia]

Which part still isn’t clear to you?

[north dragon]

how do i find the speed of the shadow

the more distance the horse travels, the shadow travels much faster

[floral sequoia]

This is calculus right?

[north dragon]

yeah

[floral sequoia]

Draw a line from the light source passing through the horse and hitting the fence

[north dragon]

uh ye

[floral sequoia]

Draw another line perpendicular to the fence from the light source

[floral sequoia]

The point that hits the fence is the shadow

[north dragon]

ye so a triangle

that goes out of the circle

[floral sequoia]

Since you already know the speed of the horse, you can find the position of that point as a function of time

[north dragon]

yea thats what i dont get how to do

[floral sequoia]

Let’s first look at what values in the triangle we already know

[north dragon]

we know the speed of horse n the angle made by it

so for the triangle we already have the base and the angle, we can find what we want

[floral sequoia]

Yeah, we have one side, and an angle, so we can make a function for the other side using trig

[north dragon]

how do we make the function

[floral sequoia]

Let me draw the diagram

[north dragon]

ye

tysm

[floral sequoia]

We’re trying to make a function to find AB

[north dragon]

yeah

[floral sequoia]

Which is tan(AOB)*r

[north dragon]

uh yea

[floral sequoia]

r isn’t given in the question

So we can assume that the answer would be the same for any value of r

[north dragon]

it isnt needed it seems

yea

its independent

so we can say 1=r/ab

r=ab

[floral sequoia]

?

No, the ratio between ab and r still changes based on the angle

[north dragon]

the angle will be pi/4

[north dragon]

oh i see

so we put in the values at last ?

[floral sequoia]

The angle is pi/4 at that moment yes, but we’re finding the speed, so we still need ab as a function of time

We can do this by first writing AOB as a function of time then plugging that in

[north dragon]

i see

so tan0 = ab/r

[floral sequoia]

No, that’s still our original equation, we need to find the way to replace 0 with some operation of time

[north dragon]

i actually have the ques solution

but i wasnt able to understand it

[floral sequoia]

The horse moves 20km/hr that’s 20/2pi*r of the circle per hour

[north dragon]

oh

so function of time as

ye i got it

[floral sequoia]

Oh okay

[north dragon]

so

what do we do nex

t

[north dragon]

i got what u mean

but then how do we progress further

.close

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[lime ocean]

how to do this

[lime ocean]

cos420

[quasi bison]

cosine is \textbf{periodic}; specifically $\cos(x+360\dg) = \cos(x)$ for any $x$

[flat frigate]

Ann

[quasi bison]

or to put it in words: cosine repeats itself every 360° (every full turn)

[lime ocean]

is that the same for sin and tan

[quasi bison]

yes

though with tan the repetition is actually twice as frequent

tan repeats itself every half-turn

[lime ocean]

so if they gave me the question with tan i would do same thing as i did with cos but sometimes the answer is postitive and sometime is negative

when do i know which one is which

[lime ocean]

also this would be cos780

?

[quasi bison]

no, you're overthinking it and also going in like

exactly the wrong direction

the angles 420° and 60° are a full turn apart

therefore their cosines are the same

[lime ocean]

whats the method do this questions bc i need to do alot of them for hw

is there no specific formula

[quasi bison]

theres no "method" and no "formula"

you are thinking too mechanically right now

[lime ocean]

then what do i do

[quasi bison]

the angles 420° and 60° are a full turn apart
therefore their cosines are the same

[winged flare]

i mean if you mean questions where large angles are given

[lime ocean]

yh

[winged flare]

• try to keep subtracting or adding 360 degrees until you get an angle between 0 and 360\
• if you dont get an angle between 0 and 90 degrees then use required formulae such as\
  $sin(x) = sin(180 - x); sin(180+x) = -sin(x)$\
  $cos(180-x) = cos(180+x) = -cos(x)$\
  $tan(x) = tan(180 + x)$ (that is, subtract 180 degrees from angle once again)

[lime ocean]

ok let me try apply on this question

i substratced 420-360

so cos60

[winged flare]

yup

[lime ocean]

which is -1/2

?

[winged flare]

no..

cos of angles between 0 and 90 would always be positive

same goes for sin and tan

[lime ocean]

so 1/2

[winged flare]

yeah

[lime ocean]

danggg

wow

[winged flare]

make sure you remember the actual values for the angles between 0 and 90 degrees

it would be given to you

[lime ocean]

yes thank you so much

you made it really simple

[winged flare]

here we change angle by 360 since all basic trig functions repeat 360 degrees

is that the only doubt you have

[lime ocean]

yh

[winged flare]

you could close using .close if youre done

[lime ocean]

.close

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[verbal vector]

how do u do part c? i think i got it wrong somehow

[verbal vector]

wouldnt g(f) being one to one immediately imply that g is one to one?

and from that you can conclude that f is also one to one

[dense prism]

Just assume its not one one and arrive at a contradiction

[safe radish]

@verbal vector Has your question been resolved?

[raven heart]

no, part (d) very much hints that's it's not the case

the f is onto part is important

think about what could happen if it isn't

[verbal vector]

if f:A-->B and g:B-->C that means the range of f is equal to the domain of g?

[raven heart]

no

all functions would be onto in this case, it wouldn't even make sense to give it a name

[verbal vector]

i thought B was the range

[raven heart]

not necessarily

f : A -> B only requires that range of f is included in B

[verbal vector]

ok i get it but how do i show part c?

[raven heart]

well what do you have to do to show f is one to one ?

cause that's your goal

[verbal vector]

f(a)=f(b) implies a=b

[raven heart]

yeah

[verbal vector]

we're given g(f(a)=g(f(b) implies a=b

[raven heart]

indeed

[verbal vector]

i cant assume g is one to one

[raven heart]

do you really need that ?

[verbal vector]

so i cant assume g(f(a))=g(f(b)) implies f(a)=f(b)

[verbal vector]

probabaly not

[raven heart]

that's not a direction you need

[verbal vector]

im not sure how else to do it

[raven heart]

what about the other way

f(a) = f(b) => g(f(a)) = g(f(b))

is that true ?

[verbal vector]

i guess so

[raven heart]

why so?

[verbal vector]

cuz then u get something like g(x)=g(x)

this is always true

[raven heart]

yeah g is a function so we're good

[verbal vector]

so we're done with part c now

right?

[raven heart]

how exactly are we done ?

[verbal vector]

idk at this point

maybe not

[raven heart]

[not a trick question, we've just been throwing ideas off the wall, that's not a coherent proof]

[verbal vector]

but how does this help tho?

[raven heart]

well we better involve g°f in the proof, if we're told g°f is one-to-one

[verbal vector]

from that we have g(f(a))=g(f(b)) implies a=b, what else?

[raven heart]

yeah right

[verbal vector]

we also have f(a)=f(b) implies g(f(a))=g(f(b))

[raven heart]

right

[verbal vector]

we supposed f(a)=f(b). then you apply g to both sides which gives g(f(a))=g(f(b)). and since we're given g(f(a))=g(f(b)) implies a=b
this means f(a)=f(b) implies a=b

this means f is one to one

[raven heart]

yeah exactly

gg

[verbal vector]

.close

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[bleak notch]

Prove that P(2n, n) is divisible by 2^n

[bleak notch]

I can tell I’m on the right track I just don’t know where to go next

[elder moss]

Hmm.. do you know how to find power of a prime number in n! ?

[bleak notch]

Uhh I don’t think so

I don’t know many properties with factorials haha I just know how they work 😂

[elder moss]

K

[bleak notch]

I don’t know if I can do this, but can I factor out the 2 of numerator and then leave it to be 2(k+1)!/(k+1)! ?

[cold ridge]

nope

[rugged arch]

no

[bleak notch]

Ok I figured not

[cold ridge]

(ab)! isnt equal to a! * b!

[bleak notch]

Good thing I didn’t do that 😂

[rugged arch]

from your inductive hypothesis, you have $\frac{(2k)!}{k!} = t \cdot 2^k$, right?

[flat frigate]

mmmm7

[bleak notch]

I’m not sure, what is the t? And the denominator looks different

[rugged arch]

t is some integer

[bleak notch]

Oh

Oh that’ll let it divide I see

[cold ridge]

I can't really read

[rugged arch]

$P(2n,n) = \frac{(2n)!}{(2n - n)!} = \frac{(2n)!}{n!}$

[cold ridge]

What does it say?

[flat frigate]

mmmm7

[cold ridge]

What do you have to check?

In the induction

Or prove

[bleak notch]

I’m proving P(2n, n) is divisible by 2^n

[rugged arch]

wdym?

yes if a is divisible by b then a = tb for some integer t

[bleak notch]

Ok yeah this is right then

[cold ridge]

Okay so

(2k+2)! = (2k+2)(2k+1)(2k)!

And (k+1)! = (k+1)k!

[rugged arch]

yeah so the inductive step requires you to show $\frac{(2k + 2)!}{(k + 1)!}$ is a multiple of $2^{k + 1}$

[flat frigate]

mmmm7

[rugged arch]

which is almost trivial

[cold ridge]

From there, we have that P(k+1) = (2k+2)/(k+1) * (2k+1) * (2k)!/k!

(2k+2)/(k+1) is just 2 because 2(k+1)/(k+1) = 2

So it becomes 2 * (2k+1) * ((2k)!/k!)

[bleak notch]

But the factorials don’t let me take out the 2 right?

[cold ridge]

Theres no factorial here

Let me write it for you on paper

[bleak notch]

Yeah hard to read it in one line 😂

Ping me when you reply so I see it!

[cold ridge]

@bleak notch

[bleak notch]

Oh you wrote out the first few factors of the factorial

[cold ridge]

Yes

It helps a ton

So it can relate to P(2k, k) which we have already proved is a multiple of 2^k

[bleak notch]

Yeah that makes sense

I’m still reading and understanding the last few lines one second

[cold ridge]

M is an integer that is supposed to represent any multiple

Random

[bleak notch]

How come we can do 2*(2k+1) and it doesn’t become 4k+2 if that makes sense

Wouldn’t you need to distribute that before you move it around

[cold ridge]

No because a * (b * c) is equal to b * (a * c) and that's equal to c * (a * b)

"Associative Property: The way in which numbers are grouped does not change the result"

[bleak notch]

Ah ok

I think that makes sense

[safe radish]

@bleak notch Has your question been resolved?

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[fleet zephyr]

I need help solving this using matrix math. I am told to do it by finding the matrix Q where each column vector is each eigenvector of A.

[fleet zephyr]

this is the work I did so far

But I get a det(Q) = 0?

btw v = <x(t), y(t), z(t)>

the eigenvalues are λ1 = λ3 = -2 and λ2 = 2

[safe radish]

@fleet zephyr Has your question been resolved?

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[solemn anchor]

Can someone explain Eulers method to me. I guess I'm just not understanding what exactly is happening.

[plucky elk]

Show a resource you're not understanding

[solemn anchor]

I might actually close this

I figured it out

[plucky elk]

Great

[solemn anchor]

.close

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[hoary seal]

I dont understand this slide

[obsidian oracle]

a random variable is a function that takes as inputs outcomes from a sample space

[hoary seal]

especially with the X and Y

[obsidian oracle]

and outputs numbers

[obsidian oracle]

well those are examples of random variables

you take a random event such as "tossing a coin"

consider its sample space S = {H,T}

and now, a random variable X for this event can be any map from S to the real numbers

so X is defined here by X(H) = 1 and X(T) = 0

[hoary seal]

Im confused from here

[obsidian oracle]

it's called a random variable only because it interprets a random event

[obsidian oracle]

can you explain what you're confused about?

[hoary seal]

why is tail 0 and heads 1?

it should be an even chance

[hoary seal]

I got confused with "can be any map from S to the real numbers"

[obsidian oracle]

It's not about probability

Imagine we're playing a game of coin toss

And I give you 1$ if you get heads

And 0$ if tails

[hoary seal]

so its more of an event?

or is an event

[safe radish]

@hoary seal Has your question been resolved?

[obsidian oracle]

no, it's a function that depends on the result of an event

it's an event interpreter if you wanna see it like that

[safe radish]

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[fair goblet]

How do i tell if a vector field is conservative
by looking at its graph
im unsure of the rules
can someone guide me

[plucky elk]

https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/16%3A_Vector_Calculus/16.03%3A_Conservative_Vector_Fields

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@fair goblet Has your question been resolved?

[safe radish]

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[junior bobcat]

ik thats not math but can someone help

[dark loom]

Can we get a translation?

Oh wait

[dark loom]

Thank you

[safe radish]

@junior bobcat Has your question been resolved?

[tardy mango]

discord.gg/chemistry

[junior bobcat]

i join the CHEM sever

and i dont have help

[safe radish]

@junior bobcat Has your question been resolved?

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[merry stag]

proving by induction here, the a shouldn't get a value right? what am i meant to do with it otherwise?

[plucky elk]

What does "the a shouldn't get a value" even mean

[merry stag]

i am unsure of what to do with a in my proof, the a /= 0 and a /= 1 implies a value is assigned to it at some point for the purpose of the proof

[plucky elk]

Yes the very beginning

You can assume a=3 and your proof will be nearly identical

[merry stag]

do i work with a value of a while proving n and n+1 hold?

[plucky elk]

Same as you would for 3

[merry stag]

so to be clear, if i do the proof with a = 3 is that only for the base case or does the proof for n and n + 1 hold with a = 3?

[severe pond]

you're misunderstanding what the base case is

a is an arbitrary real number not equal to 0 or 1

you don't assign it a value

maybe if they wrote for all a in R such that a \neq 0, 1 at the beginning instead of "where a is in R, a \neq 0, 1" it would make more sense

[merry stag]

i think that's what i was looking for clarification with, thanks, a stays fixed and i do induction over n, right?

[severe pond]

yes

[merry stag]

got it

.close

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[deep depot]

How to proceed ?

Any approach ?

Like maybe I've to prove p(45^(1/100))=p(46) ?

[severe pond]

i think you’re confused

you want to show p has a root

[deep depot]

Yeah a bit

[west inlet]

whats 1+1

[severe pond]

not that p’ has a root

[deep depot]

So I should consider a function?

[severe pond]

rolles theorem tells us that f’(c) = 0 for some c in the interval so clearly we want f’(x) = p(x) yea?

[deep depot]

Yeah

[severe pond]

so integrate p

[west inlet]

holy fuckin dorks

[severe pond]

you’re playing roblox mate

[west inlet]

ikr

[deep depot]

So let's say
h(x)= integral of p and analyse h

[severe pond]

holy fucking dork

[west inlet]

ok math wiz

[severe pond]

thanks

[west inlet]

zip it up when your done

[severe pond]

yes

[deep depot]

Let me try

[severe pond]

go play roblox

nice one

[deep depot]

,w 2323/101

[flat frigate]

Results provided by WolframAlpha

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

Requested by arnabpal

[deep depot]

Alright

It's right?

[severe pond]

yep

looks good

[deep depot]

Ohk thanks man

.close

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[analog arch]

I'm not sure for b) how the answer is 1 (graph theory)

[severe pond]

do you know what the degree of a vertex is

[analog arch]

how many edges a vertex has

[quasi bison]

and how many vertices in this graph have exactly 2 edges going into them?

[severe pond]

sure so how many vertices here are only connected to two other vertices

[quasi bison]

or better yet

which vertices here have degree 2?

[analog arch]

sorry how can I tell

just v1 seems like its connected to v2 and v3, so two edges there?

[severe pond]

yes

and it’s the only one

[analog arch]

oh

sorry idk why I overthinked it to be wrong

.close

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[carmine jay]

i got 21 but idk if it’s right
can anyone check? pls

[thin bridge]

show work

[turbid crag]

that is correct

[carmine jay]

its messy haha but i can write the solution again neatly

[carmine jay]

oh alright

thanks

[turbid crag]

actually — maybe i misunderstand but — in these cases we can just check the work, or should we enforce showing work?

[thin bridge]

in most cases, its better work is shown if it exists

[turbid crag]

ah, i see; my mistake then

i forgor the magic word

[thin bridge]

minor mistakes can be corrected
flawed logic may have been applied to reach the correct value
helper doesn't necessarily have to go through the process of solving it
etc

[ember socket]

(nice handwriting)

[turbid crag]

i see

[thin bridge]

looks good

[turbid crag]

++;

[carmine jay]

i checked the answer on multiple AI but they all gave different answers

[thin bridge]

AI is unreliable

[carmine jay]

seems like it

guys this one too

[mossy lotus]

you also got an answer for us to check for this or you want a hint?

[carmine jay]

just to check

[mossy lotus]

ok, so what you got?

[carmine jay]

oh sorry
i thought u were talking about the previous question i asked

my bad

[carmine jay]

i have no idea for this one

want a hint

[turbid crag]

there are two ways to solve this problem, the intended way and the fast way, which one do you want lol

[mossy lotus]

ok, a small hint would be ||connect EF||

[carmine jay]

i would love to know the both actually

[carmine jay]

ah i see

[turbid crag]

(and see what you can get from there)

[safe radish]

@carmine jay Has your question been resolved?

[carmine jay]

ok so area( tri CDE) / area(CBA)= 1/4…i can’t go further after this lol

idek if that’s helpful

[turbid crag]

explain — why?

[note for anyone who's noticing — im trying to fix this issue with my gender roles being completely screwed over, disregard it for now; my pronouns are they/them] edit: resolved! hooray

[carmine jay]

when two triangles are similar with corresponding sides then that ratio is valid so here it will be equal to (CE/CB)^2 which is 1/4 as CB=2CE

[mossy lotus]

and how are they similar? you got a proof for that?

[turbid crag]

++;

[carmine jay]

oh yea lmao
then how do we do it chat

[mossy lotus]

https://tenor.com/view/bruh-kanye-west-rapper-drop-shattered-gif-17839027

[carmine jay]

i just saw that theorem in my book and applied 😂

[mossy lotus]

whyd you claim them similar if you dont know lol

[carmine jay]

tbh i’m tired by that q, i tried it for hours last night. i don’t wanna put more brain to it now

[carmine jay]

i’m dumb

[turbid crag]

sure — feel free to .close it off

[mossy lotus]

take a rest, have a bit of fun and return to the problem with your head refreshed

that helps a lot more than people think

[turbid crag]

yep! may i suggest tetris

[carmine jay]

ok

thanks everyone

[turbid crag]

.close it off

!done

[safe radish]

If you are done with this channel, please mark your problem as solved by typing .close

[carmine jay]

.close

[safe radish]

Channel closed

Closed by @carmine jay

Use .reopen if this was a mistake.

[carmine jay]

.reopen

[safe radish]

✅ Original question: #help-23 message

[carmine jay]

sorry for reopening but i want a solution for this

i wanna move ahead of my homework lol(this is keeping me stuck)

[turbid crag]

the short solution: since it doesnt actually say anything about where D and G are, we can just move D to the midpoint of AC and G to A, degenerating the problem

[turbid crag]

under this case, you now can use this logic, and since tri CDE = 1/4 tri ABC, ADEB = (1-1/4) tri ABC, the rest should be easy

[mossy lotus]

thats not very mathematically rigorous

[mossy lotus]

this one I mean

[turbid crag]

i know (and perhaps i should make it clear that this is math comp logic lol)

[mossy lotus]

you got a nice pair E and F as midpoints, why not use them to form the similar triangles?

[carmine jay]

i don’t think the degeneration works here.
moving D to the midpoint of AC and G to A actually changes the geometry, because DE ∥ GF must still hold in the original figure, not in a collapsed version. right or am i wrong?

[mossy lotus]

have you heard of the mid point theorem?

[turbid crag]

you're right — but im applying the trick that, if it works for a general case, it must also work for a specific case, and so we can get the answer off of a specific case

[lone hornet]

You can just construct a line from B to AC (let's call point M) parallel to DE and show that CBM is similar to CDE, and AGF is similar to ABM (by two angles), with a linear scaling factor of 2

And then you immediately get that white area is 1/4 of the area of ABC

so purple is 3/4

then 60 * 4/3 = 80

[carmine jay]

makes so much sense
thank u so much

.close

[safe radish]

Channel closed

Closed by @carmine jay

Use .reopen if this was a mistake.

[safe radish]

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[old root]

I am up to checking a question where 3 = rootx^2+2x, and the current answers I have for x are -1 +- root 10.

Current working out (for -1 + root10) is:
3 = root(-1+root10)+2(-1root10)

I know this is solvable by a calculator, however, this does not mentally make sense to me. Is there a way to make sense of the question, or is this a question where I use a calculator and not think so much about the question?

[quick crater]

making sure, the question is
$$3=\sqrt{x^2+2x}$$

[flat frigate]

ihave<skissue>

[old root]

yeeep

[quick crater]

okay, and you got $x=-1\pm\sqrt{10}$ for the solution

[old root]

ye

[flat frigate]

ihave<skissue>

[old root]

p

yes

[quick crater]

and what are you asking specifically? you want your answer checked?

[old root]

wait nevermind stupid question

so sorry

.close

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[safe radish]

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[teal matrix]

22(ii)

[thin bridge]

what have you tried

[teal matrix]

I know that

Cot=cos/sin

And tan=sin/cos

Need help from here

[thin bridge]

you should make certain horizontal fraction lines longer and/or use () to clearly indicate the order of operations

[teal matrix]

Yes waiy

[teal matrix]

Thanks

[thin bridge]

would you be able to simplify something like
$$\frac{p}{\br{\frac qr}}$$

[flat frigate]

ραμOmeganato5

[teal matrix]

I cant

I forgot the basic rule

Thats why i came for help

pr/q?

[thin bridge]

yes

apply that to your question

[teal matrix]

sin²+cos²=1

[thin bridge]

but you don't have that present,

[teal matrix]

But

We have sin²

[thin bridge]

that identity doesn't really help here

[teal matrix]

Yeah

It messes up

[thin bridge]

note that you have a common denominator now

[teal matrix]

Bases are common

Yeah

[thin bridge]

combine your fractions
and consider what identities would be applicable to the numerator

[teal matrix]

We are soo cooked

[zealous tundra]

what is a^2-b^2

[teal matrix]

.close

[safe radish]

Channel closed

Closed by @teal matrix

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[strange cloud]

can anyone help me this? I also translated this (it deleted some vector symbols)

[strange cloud]

Looks like people don't like vectors very much

.close

[safe radish]

Channel closed

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[fickle mantle]

How do i find the o thing pls

[safe radish]

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[midnight salmon]

Can somebody help me with cardan's method of solving cubic plynomials?

[wary python]

yes?

[prisma wren]

!da2a

[safe radish]

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

@midnight salmon Has your question been resolved?

[midnight salmon]

pls explain with this eqn: $2x^3 + 5x^2 + x -2 = 0$

[flat frigate]

Professor Uchiha

[midnight salmon]

@wary python

[wary python]

have you ever heard of rational root theorem?

[midnight salmon]

whats that

i might have heard about it

but not exactly as "the rational root theorem"

no i have not heard of it

[wary python]

basically a smart way to guess the first root of a polynomial like the one we're solving

[midnight salmon]

hmm

could you explain it to me pls

[wary python]

The Rational Root Theorem is a rule that gives us a small list of all the possible rational numbers that could be a root (a solution) to a polynomial equation

Think of it as a set of rules for making the most educated guesses

@midnight salmon

[midnight salmon]

ah i see

so i can narrow down the roots of the polynomial

and since its a cubic polynomial

one of my roots HAS to be

a rational root?

[wary python]

Not necessarily

[midnight salmon]

why not

[wary python]

a cubic polynomial does not have to have a rational root

[midnight salmon]

dont irrational and imaginary roots occur in pair

[midnight salmon]

i dont understand

If there is a imaginary or irrational root, then they have to occur in pairs right?

doesnt that mean the cubic polynomial has to have a minimum single rational root?

@wary python

[wary python]

yes, we are guaranteed to have at least one rational root

like this
$2x^3 + 5x^2 + x -2 = 0$

[flat frigate]

aslx

[midnight salmon]

so defenitely one rational root right?

[wary python]

as coefficients ($2, 5, 1, -2$) that are all rational numbers

[flat frigate]

aslx

[midnight salmon]

and using the rational theorem we guess the roots?

ok so if we sucessfully guess the roots

cant we just

[wary python]

the rule of conjugate pairs guarantees that at least one of the three roots must be a rational number

you got it right

[midnight salmon]

factor $x-\alpha$ out of the cubic polynomial and then get the other 2 roots?

[flat frigate]

Professor Uchiha

[midnight salmon]

good ot know

[wary python]

you got it

[midnight salmon]

ah but is this cardan;s method?

[wary python]

related but like i shorten most of the part, if you use the original cardan's method things would be much complicated and long

cardan's method used the cubic formula

while we are currently using factoring

[midnight salmon]

cubic formula? Ig the way it was taught to me was through a series of steps

[wary python]

when all three roots are non-rational, or you just need the exact algebraic solution for any cubic (if the rational fails)

[midnight salmon]

Reduce the cubic equation from$x^3 + lx^2 + mx + n = 0$ to $x^3 + ax + b = 0$

[flat frigate]

Professor Uchiha

[midnight salmon]

something like this

[midnight salmon]

i was taught in a different way

[wary python]

hmm

[midnight salmon]

from this we take $y = u + v$ then compare the above equation with $y^3 = 3uvy -(v^3 + u^3)$

[flat frigate]

Professor Uchiha

[midnight salmon]

then we find $u^3 v^3$ and $v^3 + u^3$

[flat frigate]

Professor Uchiha

[midnight salmon]

then find a quadratic that corresponds to the above roots of $u^3 and v^3$

[flat frigate]

Professor Uchiha

[midnight salmon]

then solve for u^3 to get some $(\lambda) ^3$

[flat frigate]

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

[midnight salmon]

then u values would be

$\lambda , (\lambda) (\omega) , \lambda (\omega)^2$

[flat frigate]

Professor Uchiha

[midnight salmon]

this was how it was taught to me

[wary python]

yes you can use that but like i perfer to use (RRT + Division) when you do have a rational root

its easier to do

[midnight salmon]

i agree

but what if my dumb ahh cant guess the roots

?

• when i solve the quadratic for the cubic i gave you earlie

r

i get like 2 imaginary roots

whatt he fuck do i do after that

im genuinely confused

should i multiple by omega

or something

im sorry i typed that like i have asthma

[wary python]

Since we proved that our polynomial must have a rational root, you only have to check the 6 numbers on your $\frac{p}{q}$ list ($\pm 1, \pm 2, \pm \frac{1}{2}$). One of them has to work.

[flat frigate]

aslx

[wary python]

but if you got 2 imaginary roots

in this case

[midnight salmon]

how did you guess thos enumber?

[wary python]

probaly smth wrong

[midnight salmon]

oh

maybe ill check with a calculator

[wary python]

yes

it supposed to be the real ones

not imaginary

[midnight salmon]

but what if they are of the form $\lambda \omega$

[flat frigate]

Professor Uchiha

[midnight salmon]

thne i can multiple with $(\omega) ^2$ to get the rational root

no?

[flat frigate]

Professor Uchiha

[wary python]

following the rule $\frac{p}{q}$, where $p$ comes from the end of the equation and $q$ comes from the beginning

[flat frigate]

aslx

[midnight salmon]

oh ok

thanks

thank you @wary python

how do i close?

[wary python]

you are mixing up the final answer with wrong formula with this one tho

[midnight salmon]

this thread

[midnight salmon]

wdym

[wary python]

The idea of multiplying roots by $\omega$ (omega) is only part of Cardano's method. Those terms ($\lambda, \lambda\omega, \lambda\omega^2$) represent the three final solutions derived from that complex formula

[flat frigate]

aslx

[midnight salmon]

they are the solutions of u no?

[wary python]

Roots are solutions; you don't multiply solutions to get other solutions. You can't multiply $\lambda\omega$ by anything to magically turn it into the rational root $\lambda$. For our problem, we find the roots by factoring the cubic into three parts: $$(x - a)(x - b)(x - c) = 0$$ Since we are doing the easy method (factoring), $\omega$ is totally irrelevant.

[flat frigate]

aslx

[wary python]

you get it?

@midnight salmon

[midnight salmon]

uhh yes

factoring

but what if i dont know the rational root?

wait

is u^3 supposed to be rational

?

[low echo]

Hey

[wary python]

No, $\mathbf{u^3}$ is not necessarily rational. In Cardano's method, $u^3$ is a root of the auxiliary quadratic equation (the resolvent). Those roots can be irrational or even complex

[flat frigate]

aslx

[midnight salmon]

hi

[wary python]

okay are we done?

[low echo]

How are you?

[midnight salmon]

ok what to do if complex

[wary python]

im rlly sleepy rn

[midnight salmon]

im sorry

but can you just explain what if the roots of the resolvent is complex

[wary python]

its okay you are the last case of today

[low echo]

Are you in thailand

[midnight salmon]

lol k

[midnight salmon]

we'll talk somewhere else pls

[low echo]

Alright bye

[midnight salmon]

@wary python

[wary python]

The original cubic equation has three distinct, purely real roots

pretty amazed right?

[midnight salmon]

woah

hows that

possible

well never mind that

how do i solve for u then

i have u^3 = some complex stuff

how do i compute u

[wary python]

can i sleep 😭

[midnight salmon]

welp lol

mb

Thank you for helping me tho

[wary python]

i mean one more

[midnight salmon]

Have a good nights sleep

[midnight salmon]

??

[wary python]

in that case you probably have to use trigonometry

[midnight salmon]

cis theta thingy?

alright

thanks for the hint

[wary python]

okay gn

have a nice day for you

[midnight salmon]

gn lol

[midnight salmon]

i am also gonna sleep lmao

[wary python]

i mean you can do polar conversion or de moivre's theorem

but yeah

cya

[midnight salmon]

ah

ok cya

bye thanks

[wary python]

try .close

[midnight salmon]

mm sure thanks

.close

[safe radish]

Channel closed

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[safe radish]

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[fickle mantle]

can i get help understanding how to prove congruency and parallograms

[frail delta]

any example q you can show so i get an idea?

[fickle mantle]

ok wait

[frail delta]

hmm i see

so do you know congruence criteries or do i refresh them?

[fickle mantle]

uh

sss

as

sas

Sara

Rhys

rhs

[frail delta]

yep

and for parallelogram

[fickle mantle]

uh

opposite equal sides

two pairs of parallel sides

opposite sides equal

[frail delta]

yep

[fickle mantle]

something about th e diagn=onal

[frail delta]

the diagonals bisect each other

meaning they cut eachother in half at their point of intersection

and now for the 1st q

so have you tried anything else?

or could think of anything else here?

[fickle mantle]

that’s all I’ve done

[frail delta]

i see

so these points make a parallelogram right?

and from what you mentioned earlier, the opposites sides are equal

so can i say AD = BC?

[fickle mantle]

yes but how do i prove tht

ik how to prove parallel lines a little

[frail delta]

you want to prove that opposite sides are equal?

[fickle mantle]

i can’t prove things in general

[frail delta]

what exactly do you want to prove

1s

[fickle mantle]

uh congruency

and parallel

idk like how to prove it

what to do maybe

[frail delta]

proving is simple look

this is a parallelogram right?

[fickle mantle]

ic

[frail delta]

actually no wait not those 2

mb

[frail delta]

here consider AEB and CED

[fickle mantle]

for

[frail delta]

to prove congruency

so we can conclude that opposite sides are equal

[fickle mantle]

ic

[frail delta]

so a property of the diagonals is that they bisect each other

[frail delta]

so here AE = EC and BE = ED

AE isnt equal to BE btw

the point E cuts the diagonals in half

and now here

those 2 sides are equal

we got that

and now for the angle

angle AEB will be equal to angle CEB because they are vertically opposite

so now we got 2 sides and an angle

so with the SAS criteria we can show congruency and prove that AB = CD

[fickle mantle]

what topics help understand this pls

ohh

[frail delta]

didnt understand....

isnt congruency new probably?

[fickle mantle]

to me?

ig so\

[frail delta]

ye

[fickle mantle]

then what do the rest find

the sss

Sara

saa

[frail delta]

they are also congruence criterias

if you find out that all 3 sides of a triangle are equal

then you use the SSS criteria

[fickle mantle]

does sas help prove parallel

oh also how do I or I’ve tht something is perpendicular

[frail delta]

no

[fickle mantle]

oh

[frail delta]

perpendicular meaning at the meeting of the lines an angle of 90 degrees is formed

if i draw something like this (its very rough rn) it means that the 2 lines are perpendicular to each other

so now can you try attempting this?

first part has nothing to do with GH and FH btw

nor it got anything to do with congruency

[fickle mantle]

yeah and im still confused

does proving a parallelogram relate to congruency

[frail delta]

if you find out that one pair of sides is equal and parallel then its a parallelogram

as easy as that!

[fickle mantle]

how do i do tht

w sas?

[frail delta]

depends on the question

[frail delta]

for this one, forget congruency and just observe the diagram

[fickle mantle]

ok

[frail delta]

you making progress?

[fickle mantle]

i still don’t get it

[frail delta]

alright

no worries

[fickle mantle]

so congruency is h=just for congruency right?

[frail delta]

what?

congruency meaning the shapes are equal in size

[fickle mantle]

yes

so when proving parallelogram I don’t use congruency right?

or do i for some

[frail delta]

it depends on the question

atleast for that one you dont need congruency

[frail delta]

it purely depends on what info is given to you

you'll have to figure out accordingly, what to apply

[fickle mantle]

dang

could u give me examples

[frail delta]

of?

[fickle mantle]

when i use congruency

[frail delta]

um i have no idea but ill see

hard to find them randomly

but there might be some

[fickle mantle]

for this question

[fickle mantle]

.

can i say xf=fh

[frail delta]

you can use congruency of triangles in that one btw

[frail delta]

isnt 'x' the length of GF

?

might be wrong but from what i see

[fickle mantle]

yuh

honestly idk anymore

[frail delta]

so X isnt really a point

so you cant say XF

xf*

[fickle mantle]

gf?

[frail delta]

yea

[fickle mantle]

i still dk how to work it

[frail delta]

dont look at X then

look at the sides of the parallelogram

lemme show this one then

[fickle mantle]

ok

[frail delta]

these points make a parallelogram

and its given that line AD = DE

well its split into 2 parts

and acc to the property of a parallelogram, AD = BC

so since AD = DE and AD = BC, DE = BC

and they are parallel too because AE is a singular line

so one pair of equal and parallel sides

hence proved that DECB is a parallelogram!

you got it?

was that too much to take in?

[fickle mantle]

what does this have to do to do with it tho

i understand for making the first parallelogram

[frail delta]

your first subq says that you have to prove that DECB is a parallelogram

by seeing the diagram, this was the easiest method to prove it

[fickle mantle]

ok wait, how about AD = DE and AD is parallel to BC

hence DE is parallel to BC

is tht how you do it?

[frail delta]

well AD and DE is a single line afterall!

just that the point D is cutting it into half

so them being equal doesnt exactly have to do anything with it being parallel

but since it is a single line

thats why its parallel

[fickle mantle]

huh

[frail delta]

okay so making it simpler

AE is a line

[fickle mantle]

what about angle d is equal TO C

[fickle mantle]

yes

pls

[frail delta]

and there is a point 'D' on that line?

which is cutting the line into 2 pieces

[fickle mantle]

yes

[frail delta]

now well ignore the point for now

[fickle mantle]

ok

[frail delta]

wouldnt the entire line be parallel to BC?

[fickle mantle]

yes

[frail delta]

exactly

and thats why DE is also parallel because afterall, it lies on the line AE too

[fickle mantle]

ohh

[frail delta]

D is just a point on the line

[fickle mantle]

your so smart

ty

[frail delta]

oh im not but ty lol

[fickle mantle]

what about them being equal

[fickle mantle]

nah to me

cause this is so hard to get

[frail delta]

DE and BC?

[frail delta]

oh just practice more qs

[fickle mantle]

yes

bc without the d

it’ll b equal

i think

[fickle mantle]

my exam is tomorrow🥀

[frail delta]

AD and DE are equal right?

[frail delta]

dang

[frail delta]

^ well its given, nothing to prove in that lol

[fickle mantle]

ye

[frail delta]

and AD and BC are also equal right?

because its also given that ABCD is a parallelogram

[fickle mantle]

yes

[frail delta]

so BC and DE are equal then?

because they are both equal to AD

[fickle mantle]

yes

[frail delta]

so equal and parallel

so its a parallelogram

[fickle mantle]

ic

tysm

i got another question

for measurements

uh when two or more diff shapes are added does it only affect the sa

surface area

(going to brush my teeth rq)

[frail delta]

no it affects both, the volume and surface area

[frail delta]

?

[fickle mantle]

it means a dot

the point for decimals

[frail delta]

oh

crap

[fickle mantle]

thts what they teach here

[frail delta]

then mb

[frail delta]

wb this

you didnt understand it?

[fickle mantle]

yes

it’s almost time to go slepe\

[frail delta]

hmm

you know what this is?

or what exactly they did

[fickle mantle]

formula

uh

added the volume of the shapes

dirt one is sphere i think

second isa cone

[frail delta]

mhm

and same for the surface area

!

they added because the figure is a combination of a cone and a hemisphere

[fickle mantle]

ic

[frail delta]

and yk pi

[fickle mantle]

but some of the main formula isn’t there

[frail delta]

which?

[frail delta]

so which of this they missed?

[fickle mantle]

For sa

They missed the parts w the circle

[frail delta]

well nope they didnt miss anything

[fickle mantle]

I have to go

[fickle mantle]

Are u sure

[frail delta]

yes because pi * r * l is the formula for lateral surface area

and not the total surface area

[fickle mantle]

For the cone

The volume had h

But there's no h

In there

Can I DM u?

I gtg

[frail delta]

sure ig

[fickle mantle]

Oh

So it's the sa

[frail delta]

no thats the volume

they found the lateral surface area which is given by

[fickle mantle]

Ok nvm

Maybe I need to. Learn it better

[frail delta]

no dw

the pi * r^2 is coming from the base

and here the hemisphere is covering the base

you cant include that

so they just excluded the pi * r^2 part

[fickle mantle]

Ohh

What time is it for u

[frail delta]

so just directly add the rest

[frail delta]

midnight 1:30

[fickle mantle]

Dang

[elder moss]

Same

[fickle mantle]

So ur like 3hrs ahead of me

[frail delta]

lmao

[frail delta]

prob

[fickle mantle]

Will u be awake at 5

[frail delta]

no 💀

[fickle mantle]

It's 10:03 fo me

[fickle mantle]

Oh ok

[frail delta]

but there will be other ppl to help

dw

[fickle mantle]

That's fine

Gn

[frail delta]

gn

and gl for your exam

[fickle mantle]

Tysm

. close

.close

[safe radish]

Channel closed

Closed by @fickle mantle

Use .reopen if this was a mistake.

[safe radish]

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[cold pilot]

Hi

[cold pilot]

Carlos has two younger brothers.
The sum of their ages is 30, and the product of their ages is 720.
If Carlos is the oldest, how old is Carlos

How can I solve this?

[silver oar]

what have you tried so far

[lime dust]

You can't solve that unless I'm reading wrong.

!original

[safe radish]

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.

[cold pilot]

What if the problem it's in Spanish?

[lime dust]

No problem

I understand spanish

[cold pilot]

Carlos tiene 2 hermanas más chicas que ella, la suma de las 3 edades es 30 y el producto es 720, cuantos años tiene Carlos?

[lime dust]

What equations can you create from there?

[cold pilot]

I tried linear equations but I couldn't

[lime dust]

Tell me the equations.

[cold pilot]

{a+b+c=23
a(b)(c)=396

[lime dust]

why 23 and 396?

[cold pilot]

Teacher also said that we can use Least common multiple

[lime dust]

you said 30 and 720

[cold pilot]

Ohh right

Its another problem

[lime dust]

So: A+B+C=30

ABC=720