help-19

[zenith tartan]

its not clear to me

[zenith tartan]

,rccw

[ruby nacelle]

i dont know how a) can help me

[idle peak]

let f(x)=4x^3 - 2x^2 - 3x + 1
then do long division
if the remainder of the division is 0, hence proven

[zenith tartan]

they r asking help for c part i believe

[ruby nacelle]

i know how to do a

i just dont know how (a) can help me in (c)

[idle peak]

oh

lemme see

[ruby nacelle]

it doesnt miss a puncutation lol

[brittle beacon]

If you let $x = \sin^2(\theta)$ note that $4x^3 - 2x^2 - 3x + 1 = -\sin(3\theta) + \cos(2\theta)$ maybe

[clever fjord]

@brittle beacon

[ruby nacelle]

Hello chartbit

do you remember me before xd

[brittle beacon]

Yea I do

[ruby nacelle]

nice

[ruby nacelle]

let me try to comprehend that

isnt it -sin3θ-cos2θ?

oops my bad

i still dont know how can i use a lol

[brittle beacon]

That’s what I’m thinking not too sure whether that would be useful but it is interesting

[ruby nacelle]

yeah kinda

[brittle beacon]

Of course taking note of the roots of 4x^3 - 2x^2 - 3x + 1 I guess

[brittle beacon]

Oh also typo here, should be x = sin(theta)

[ruby nacelle]

yeah

[ruby nacelle]

(x-1)(4x^2+2x-1)

i found that sin(pi/10) is a root to 4x^2 + 2x - 1 = 0

[brittle beacon]

How did you find that?

Also you know the roots of 4x^2 + 2x - 1 = 0 as a "normal" polynomial
Guessing the argument is that letting x = sin(pi/10), you should get that x is a root of that cubic, so then either x=1 (which it isn't - pi/10 isn't e.g. pi/2), or x is one of those roots (but of course noting that sin(pi/10) has to be positive...)

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[thick pine]

(10 c r) (1)^10-r (-x)^r
as 10-7 = 3, r = 3
(10 c 3) (1)^7 (-x)^3
120 (1) (-x^3)
-120x^3

where did I go wrong?

<@&286206848099549185>

[hardy panther]

It's 7, not 3

Your first line is correct

@thick pine

Your second line is not correct after the comma

[crisp wadi]

You are looking for the coefficient of x^7, so you better pick r = 7

[hardy panther]

They did, they just got mixed up

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[wary cypress]

Learning trigonomety addition and subtraction at school and got this one math problem

[wary cypress]

There are a, b, and y as the triangle angles
Sin a = 3/5 and cos b = -12/13
Look for sin y

Sin y = ?

[sacred whale]

Have u tried drawing a triangle?

[wary cypress]

Yes

I've drawn the triangles for a and b

I have no idea on how can I find y

[sacred whale]

I dont think you should assume it is a right angle triangle unless it says so.

[wary cypress]

Hmmmm

I don't even know what is y in this problem

[sacred whale]

Okay lets start from scratch.
You know a,b, and y are angles in presumably the same triangle, so what should their sum equal to? (a+b+y)

@wary cypress
The sum of angles in a triangle equal 180.
This means u have a+b+y=180

[wary cypress]

Ohhhh

I got it

Sin a + sin b + sin y = 180
3/5 + 5/13 + sin y = 180

Like this?

[sacred whale]

No this is not it

You have to isolate a and b

then plug it into a+b+y=180

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[odd edge]

@stiff dawn Has your question been resolved?

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[molten quail]

hello i need to know if i got a right answer from this equation 5+(x-√7)(x+√7)≤(x-√3)^2+2√3

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@molten quail Has your question been resolved?

[fluid onyx]

What is your answer

[mystic saffron]

that's an inequation

[molten quail]

(-∞;√3+1) or just x≤√3+1

[molten quail]

sorry im not fluent in english 😬

[mystic saffron]

it's okk

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[hazy furnace]

Why is the one that i selected is wrong?

[slate radish]

For example A = {1}, B={1,2}

[hazy furnace]

yes

I thought that since A is a subset of B, then B would still have its elements

.close

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[wheat apex]

what is this saying?

[odd edge]

Please don't occupy multiple help channels.

[wheat apex]

fixed

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[mystic saffron]

how to get the 86000 weighted average outstanding shares?

i found a solutions manual for this question.

where are the numbers 1.2, 1.2, 1, and 1 coming from?

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[mystic saffron]

<@&286206848099549185>

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[outer wadi]

I'm having trouble representing this as an equation, its makes sense logically but how could I solve for N where 6 = ceil(N, 5)

[crisp wadi]

Start by recalling the definition of ceil

[outer wadi]

ceil gets the value of the next highest integer

hmm so that would be the qoutient of N/5 + 1

[crisp wadi]

So ceil(N/5) = 6 if and only if 5 < N/5 <= 6, i.e., 25 < N <= 30

It looks like you are looking for the smallest integer N, which is 26 in this case

[outer wadi]

ok that makes sense
So N/5 is always less than or equal to ceil(N/5) which gives N/5 <= 6. And N/5 must be greater than 5 or else the ceiled value wouldn't be 6. Which gives us
5 < N/5 <= 6. And so we just multiply by everything by 5 and choose the smallest value N can be.

So is the formal defintion of ceil(N/k) this -> k < N/k <= ceil(N/k) where N != k

[odd edge]

@outer wadi Has your question been resolved?

[outer wadi]

my question though is why must N/5 be greater than 5, it makes logically yeah but how can we conclude that

[outer wadi]

I'm just kinda confused on the 5 < N/5 part

[crisp wadi]

It looked like 5 < N/5 made sense to you earlier

[outer wadi]

yeah it makes sense sort of but I dont think im understanding it fully

[crisp wadi]

And no that's not the definition of ceil

ceil(x) = k if and only if k - 1 < x <= k

This is the definition

[outer wadi]

oh I see

that makes sense

ok I think I'm seeing it now

yeah I think my flawed defintion was just tripping me up that defintion makes more sense

so ceil(x) = k if and only if the value of x is between k - 1 exclusive and k inclusive and we have that left bound because the k would be a different value if our ceiled value x was more than one integer away from k since it takes the next highest integer

ok

[crisp wadi]

Right

[outer wadi]

so the value k gives us our domain to work with. If the ceil function is a relation of two sets R -> N, and if im not tripping would R also contain subsets where the real values can be represented as one integer {{0.01, 0.02..., , 1}, {1.01, 1.02..., , 2}} and each of those sets map the the last value of the subset which can be said to be k to k in the codomain N. And in this specific case we are looking the subset of R {5.01, 5.02.., , 6} which maps to 6 the final value. So since we are restricted to this set we know that N/k must be greater than k-1, 5. So in this problem they gave us a domain R and a domain within that domain 5.0001 - 6, idk if that is stated correctly. Then they also gave us 5 and now that I think about it could the information 5 tell us that we are relating a function x/5 to the domain of the ceiled function, to the range of the ceil function. So a composite function?

[crisp wadi]

I don't think I'm following

Those subsets have the same image, sure, but what's your point?

[outer wadi]

I'm just trying understand how all the information given relate to the ceil function and I think what I said for k=6 might be right, but I was wondering why the information 5 was needed, and its needed to find the value of x, N/5. Since N/5 is also a function I was just thinking that this can be represented as a composite function, and the question is giving us the value 5 the domain of the first function(N/5) which we can relate to the domain of the ceiled function
N/5 -> to the subsets, which we then can map to the range of the ceiled function. I was just trying understand and relate all the topics to see if I'm understanding all of this right.

[crisp wadi]

Basically, why is the part "k - 1 < x" necessary in the definition?

[outer wadi]

Well I think understand why k-1 < x is necceary because its the defintion of the ceiled function and is bounding all the values x k - 1 < x <= k to a set that we can map to the value of x or more specifc k in the codmain. I think my question was more of what information was 5 giving us in order to solve this problem and I think the answer to that is the domain of the function N/5

thanks for the help

.close

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[fringe lintel]

how do u use gamma with imaginary numbers?

[fringe lintel]

<@&286206848099549185>

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[wary berry]

does this domain look right

this domain format was wrong though

but this one was right

[ocean bramble]

no, that domain is incorrect

you're including -1 and 1 in the domain, but if x=1, or x=-1, you're dividing by 0

[wary berry]

ahh

so it's be like (-1, 1)

[ocean bramble]

yes

while arcsin(x) has that domain, you dont have tan(-90º) or tan(90º)

so those two points get excluded

[wary berry]

hmm that sounds super complicated

[ocean bramble]

not really, since you can look at the algebraic expression without the trig functions

you look where the root exists, and where the denominator is non-zero

[wary berry]

so, we're looking at the domain of sin(arccos(x)), right

the domain of arccos is umm -1 and 1

so it's ALSO (-1, 1)?

[wary berry]

as seen here

fuck it was NOT right

whatever math sucks anyways

.close

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[polar patio]

This question is about group theory

[polar patio]

I have a question about (b), this is how I attempted it but I’m not sure how I’m supposed to do it using the hint

[desert vector]

let it be true for i

[desert vector]

then (kgk-1)^i=kg^ik^-1

multiply both sides by kgk-1

[lucid solstice]

can soemone help me whit math tank

hey gazou

[polar patio]

This channel is taken bro, pls find an empty channel

[polar patio]

I’ll try that but is my method wrong ?

[lucid solstice]

wich channell

[polar patio]

Also do you mean to multiply on the left or right?

[lucid solstice]

guy what a channel can soemone can help me

[desert vector]

it shouldn't matter

[lucid solstice]

and for you ingnor me

plz

[polar patio]

There’s a bunch of help chats on the side, select one that says ‘open’

[mint mirage]

You already have a channel #help-11, just be patient

[lucid solstice]

ok thank you

[polar patio]

The only issue with that is

This is not an abelian group

[lucid solstice]

oh ok

[polar patio]

So how would you even raise kgk^-1 to the power i

[desert vector]

exponentiation is still well-defined for non-abelian groups though

[lucid solstice]

i ask but you guy can you help me guy ç

[polar patio]

How…?

[odd edge]

Please wait patiently, and do not interrupt other channels with your question. Helpers in this server are volunteers, and the server cannot guarantee that someone will be able to help you. By being impatient or begging, you will only turn potential helpers away.

In the meantime, please make sure your channel contains the original question, clearly describes what you have already tried, and states exactly what you are having trouble with. This increases your chances of getting a good response.

[desert vector]

for example a^n=a*a*...*a

there's only one way that you can do that

[lucid solstice]

i am sorry

sorrrrrrrrrrrrry guy

[polar patio]

No but like

Okay so i asked my professor this question the other day

When i was trying to do this question, here was his response

[desert vector]

ok I see the confusion I think

when you have (ab)^2 just think of it as abab

instead of a^2b^2 which isn't always true

[polar patio]

Yea but how would you apply that to a power that isn’t a fixed number

[desert vector]

(kgk-1)^i+1=(kgk-1)^i(kgk-1)

does that make sense

[polar patio]

Yes

Ohhhh

[desert vector]

yeah

[polar patio]

You’re multiplying by kgk^-1 to get a k+1 condition

[south plume]

$(kgk^{-1})^2 = kg\cancel{k^{-1}}\cancel{k}gk^{-1}$

[clever fjord]

artemetra

[polar patio]

Ahhh and so you have just the outer k’s

[south plume]

mhm

[polar patio]

And g^i

In the middle

[desert vector]

yeah it all cancels

[polar patio]

Okay okay makes sense

I think I’m just struggling with formalising this

Let me try and write this down

Is the induction fine up to here ..?

[desert vector]

yeah

[polar patio]

So like that

[desert vector]

yeah

[polar patio]

Okay nice, thank you!

Also

There’s a part (d) which is very similar to this but instead of the middle thing being raised to the power now it’s the outer ones

So like

Part d here

If we apply the same trick here, it doesn’t work

I’ve put the new group presentation at the base for a reference

All that tells us is that a and d commute, b and d commute and b^3 is conjugate to b by a^2

Also tried this but… hmm

I’m lost on this one

<@&286206848099549185>

[odd edge]

@polar patio Has your question been resolved?

[odd edge]

@polar patio Has your question been resolved?

[leaden quiver]

@polar patio What part would you like help with?

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[polar patio]

Hey @leaden quiver sorry i missed your message

[polar patio]

I hope you can see the older messages, i need help with part d

I’ve shown where i got up to with my solution but lost on where to go from there

I think this person may be asleep, I’ll try asking the question again tomorrow

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[small osprey]

guys i need help with this

[small osprey]

(5x+3)(x+1)

5x^2+8x+3

does that look right guys?

<@&286206848099549185>

[open onyx]

seems good

[small osprey]

ok ty

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[lime oyster]

can someone help me solve this using the shown method?

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[lime oyster]

.close

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[plain fern]

I need help with proofs. Im utterly clueless so I need feedback lol

Problem:
Determine for which integers n ≥ 1 there is a graph with n vertices where each vertex has degree n − 2.
Make sure to prove your result is correct.
That is, for a particular n, if no such graph exists, you must prove this, and if such a graph does exist you must give an example.

My "Proof":
Observations:
-All vertices have the same degree (n-2).
-Handshaking Lemma: The sum of degrees of all vertices in a graph, is twice the number of edges (2E).
-The sum of K odd numbers will always be odd, when K is odd.
-The sum of K odd numbers will always be even, when K is even.

Argument:
Since the sum of degrees equals 2E, this means the sum of degrees must always be even.
Each vertex's degree is n-2. If n is odd, then n-2 is also odd, meaning the degree of each vertex will also be odd.
The sum of an odd amount, of odd degrees, is odd. This contradicts the requirement that the sum of all degrees must be 2E, which is an even number.
Therefore, n cannot be odd.

If n is even, then n-2 is even, which satisfies the requirement that the sum of all degrees must be even.
This suggests that a graph may exist for even values of n.

Constructing for Even n:
For every even n, a graph can be constructed where each vertex is connected to all other vertices except 2, ensuring each vertex has a degree (n-2).
Basically equivalent of creating a complete graph of n vertices, and then removing an edge incident to each vertex, such that no two removed edges are incident to the same vertex.

Conclusion:
A graph with n>=1 vertices where each vertex of degree (n-2), exist if and only if, n is an even integer.

[open onyx]

assume simple graphs, then yes
however your construction is not really all that constructive

try finding an inductive construction

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[weary tusk]

How did we know that sin(30)=1/2?

[cold sage]

create an equilateral triangle of side lengths 2

cut it in half down an axis of symmetry

sin=opp/hyp

1/2

[weary tusk]

yes but how did we know that this angle is 60

[cold sage]

know what angle is 60

[weary tusk]

oh wait

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[maiden crow]

can someone please help me understand this

[dusk spire]

Ok, so the refracted ray is the ray that is going into the purple

The reflected ray is the one that just bounces off the purple part

[dusk spire]

When the ray is refracted into the purple area, the speed changes

It is now 0.66c

We know that light does not change colour with reflection because it doesnt change speed

The colour of something basically depends on the frequency of the wave,

and because v = f x wavelength, when the velocity changes, the frequency and wavelength change too and the wave changes colour because it has a different frequency

[maiden crow]

so

that means the refracted light is the one that changes color!

[dusk spire]

yep

but i dont just want to give you the answer

do you understand why it changes colour?

[maiden crow]

that makes sense

cause youre right the speed

[dusk spire]

alright cool

[maiden crow]

is the same for the other angle that makes alot of sense

[dusk spire]

i literally had an exam on this today lmao

[maiden crow]

roflmao

[dusk spire]

cool

[maiden crow]

physics 2?

or physics 1

[dusk spire]

im in australia so it might be different

its just physics

[maiden crow]

oh yeah!

that part kinda slipped to me so that was super helpful!

it makes sense though

[dusk spire]

nice if thats the only question you have you can .close the channel

[maiden crow]

.close

[odd edge]

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[desert marlin]

Let $f:\mathbb{R}^{2} \to \mathbb{R}$ be defined as
$$f(x,y)=\begin{cases} xy\sin{(\frac{1}{x})} \quad x\neq 0 \ 0 \quad \quad x=0 \end{cases}$$
Prove or disprove $f$ is differntiable at the origin.

[clever fjord]

Austin

[desert marlin]

Okay

I am wondering about my solution to this, I'm not sure if it is correct or not

So, I just tried using the limit definition of differentiability

[clever fjord]

Austin

[desert marlin]

and at a=(0,0) this becomes

[clever fjord]

Austin

Austin

Austin

Austin

[desert marlin]

accidentally deleted a message

Wondering if this is valid?

[quasi sparrow]

Your argument shows f is continuous at zero.

[desert marlin]

:/ I was thinking it looked way too similar to that

but I can't catch what is going wrong

Where did I lose the strength of the limit that I started with?

[quasi sparrow]

Wait this was wrong

[desert marlin]

wrong as in, I didn't even show continuity? Or wrong as in my work is fine?

[quasi sparrow]

Your squeeze theorem isn't convincing

[desert marlin]

well

xsin(1/x) -> 0

by squeeze

and if we just add an additional squeezing y

xysin(1/x)

it just squeezes it harder

[quasi sparrow]

that's f(h), not f(h)/|h|

[quasi sparrow]

Which is why I thought this

[desert marlin]

oh right

[toxic monolith]

$\text{I use such a definition of differentiability for the function of the 2 real variables: }\\\lim_{\left( h,k \right) \to \left( 0,0 \right)}\frac{f\left( x_{0}+h,y_{0}+k \right)-f\left( x_{0}.y_{o} \right)-\frac{\partial
f}{\partial
x}\left( x_{0},y_{0} \right)\cdot h-\frac{\partial
f}{\partial
y}\left( x_{0},y_{y} \right)\cdot k}{\sqrt{h^{2}+k^{2}}}=0$

[clever fjord]

Joanna Angel

[desert marlin]

Do you know how I might go about this then?

Because

this function isn't C1

so I can't use that the partials exist and are continuous at the origin

because they aren't continuous at the origin

So I'm not sure how else to prove it

[quasi sparrow]

Find a counterexample path

[desert marlin]

it is differentiable

at the origin though

it just isn't continuously differentiable

[quasi sparrow]

How did you prove it's differentiable

[desert marlin]

well my professor just said it was, I clearly don't have a proof yet

I can show what he posted, I don't understand it though, maybe that would be something I could get help with? Since my proof ended up being incorrect

[toxic monolith]

so he showed partial derivatives are ok

[desert marlin]

the partials are fine yeah but they don't give us differentiable yet

[quasi sparrow]

What's E

[desert marlin]

I wish I knew

I think he let h be (x,y)

the rest I'm not sure

normally E is like the error term

[quasi sparrow]

Probably quadratic term in Taylor expansion

[desert marlin]

likely

[toxic monolith]

$\lim_{\left( h,k \right) \to \left( 0,0 \right)} \frac{hksin\left( \frac{1}{h} \right)}{\sqrt{h^{2}+k^{2}}}=0$

[clever fjord]

Joanna Angel

[toxic monolith]

that you get when you use formual

using trig of your instructor

$|\frac{hksin\left( \frac{1}{h} \right)}{\sqrt{h^{2}+k^{2}}}|\leqslant \frac{|hk|}{\sqrt{h^{2}+k^{2}}}=\frac{r^{2}sintcost}{r}\to 0$

[clever fjord]

Joanna Angel

[odd edge]

@desert marlin Has your question been resolved?

[odd edge]

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[round sluice]

Because this module is on implicit differentiation, I assume I need to find that here, but I don’t really know where I went wrong? After I find that, I assume I just need to plug in the numbers but I don’t know (calculus 1 btw)

[upbeat drum]

applied chain rule on cos(xy) incorrectly

how did you get xy dy/dx in the LHS

[round sluice]

((x)(dy/dx))((y)(1))

Is that wrong?

@upbeat drum

[upbeat drum]

$$(f(g(x)))'=f'(g(x))g'(x)$$ so you need to diff the inner function, which would be xy

[clever fjord]

chlamydia

[round sluice]

Omg did I just forget how product rule works? Is it that you add the two products?

Instead of multiply

Or wait

Wait no

[upbeat drum]

product rule: $(fg)'=f'g+fg'$?

[clever fjord]

chlamydia

[round sluice]

yea you add them… I’m a bot lol

[upbeat drum]

lol

[round sluice]

Correct?

[odd edge]

@round sluice Has your question been resolved?

[round sluice]

Ok, surprisingly that was the only question I got wrong on that mcq :p

idk what I did wrong

tho

[round sluice]

Ok I’m looking over my very rushed math work, and it made sense to me when I did it but idk wtf is going on looking at it now lol

[round sluice]

Yea, so I distributed -sin(xy), and then I tried to divide just two terms by it in the equation, but you can’t do that so idk lol

Ok I literally need to go to sleep, so I’m going to do that now. I’m stumped on that problem, so hopefully I’ll wake up to an answer?

[odd edge]

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[chilly epoch]

From a solid sphere of mass M and radius r a maximum possible volume of cube is cut find the volume of cube

[chilly epoch]

When the volume of the cube is maximum, the longest diagonal of cube will be equal to diameter of the sphere. This is the soln

But i can't figure out how?

[odd edge]

@chilly epoch Has your question been resolved?

[odd edge]

@chilly epoch Has your question been resolved?

[odd edge]

@chilly epoch Has your question been resolved?

[odd edge]

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[mystic saffron]

does anyone know how to do monic quadratics?

[odd edge]

Please don't occupy multiple help channels.

[mystic saffron]

does anyone know how to do monic quadratics?

[honest linden]

^ Don’t occupy multiple help channels

@mystic saffron Either close this help or the other one, don’t occupy multiple help channels. Especially when you’re asking the same question.

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

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[weak cosmos]

hi

[weak cosmos]

when i do 70.4m * 5.5 m/s does it equal 387.2 m/s or 387 m^ 2/s

[zenith tartan]

second one

[clever fjord]

Aditya gamer

[cold urchin]

,calc 7.4 * 5.5

[clever fjord]

Result:

40.7

[zenith tartan]

boss is here

[gleaming cove]

Can i have help

[cold urchin]

#❓how-to-get-help

[gleaming cove]

thnks

oh i see it noe

[mystic saffron]

3

[weak cosmos]

then if i devide that by 3.2 so that will be 387.2 m^ 2 / s dividing 3,2 s the seconds will be ^ 2 as well ?

[mystic saffron]

@gleaming cove 3 will be answer

[cold urchin]

Where does 387 come from

Oh u fixed it

[weak cosmos]

i corrected i meant 70.4m not 7.4 sorry

[cold urchin]

Is this 3.2 figure dimensionless?

[zenith tartan]

telling answers like this will end up ur account being permanently terminated

[cold urchin]

ie. It has no unit

[weak cosmos]

it's for convertions only

[cold urchin]

Can you give me context. Perchance.

[mystic saffron]

Bro he literally asked from a quick help for his test. I could not answer him in full detail as he asked in an already occupied channel

[cold urchin]

dude

1. you don’t help with tests 2)stop messaging in this help channel.

[weak cosmos]

my exercice is simply do (70.4 m/s * 5.5m ) / 3.2s

[cold urchin]

Alright then 3.2 has a unit

It’s 3.2 seconds

[weak cosmos]

i understand how to do the first part but not the division

[cold urchin]

the first part gives u something in terms of m^2/s

Now you’re dividing by second

$\frac{\frac{m^2}{s}}{s}$

[clever fjord]

Pure

[cold urchin]

This will give you xxx m^2/s^2

You’d simplify it like normal fractions

[weak cosmos]

can i just understand how can we get s^2 if we are dividing

i didn't get it

[clever fjord]

Pure

Pure

[weak cosmos]

i understood thank you

have a nice day

. close

.close

[odd edge]

Channel closed

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[cold urchin]

[odd edge]

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[odd edge]

Please don't occupy multiple help channels.

[zenith tartan]

close ur previous channel pls?

[glad summit]

How do I find the equation of the line?

Yeah okay

Is it closed now?

[zenith tartan]

you can use 2 point form of a straight line to form an equation

[sacred whale]

depends on if ur willing to factor this

[glad summit]

No, no thank you lol

[glad summit]

Would the gradient be -1 then?

[sacred whale]

yeah, theres probably an easier way 🤷

[zenith tartan]

dont do such msgs, open ur new help channel and ask help in it

[sacred whale]

It related to their question - the same question they asked in their previous channel

[zenith tartan]

yes

hm

ok

but i think we should help with this one now first

[sacred whale]

also the slope isnt -1

[glad summit]

Hi again

[sacred whale]

Hello

[glad summit]

What is it, and how do I get it? I think you might've lost the channel when you were helping me last time btw

[sacred whale]

You sure you want to go with my approach? its probably not the most elegant

[glad summit]

I don't mind anything tbh

[sacred whale]

Okay so we are looking to get the value of x1 in terms of b

[glad summit]

Okay

[zenith tartan]

i concluded it from the 2 points written on the graph

[sacred whale]

Do you know how or shall i guide u

[glad summit]

Can you guide me please

[sacred whale]

They were written by Not_Abbas 🤷

ok so

We first need to get the slope in terms of b

[zenith tartan]

wow, it looked like it was printed

[sacred whale]

You can use the point at the y-intercept (0,b)
and the point (3,3)

[glad summit]

Okay

[sacred whale]

Do you know how

[glad summit]

No, I was planning on hearing you out then working backwards

[sacred whale]

Ohh okay fine i guess 🤷

Okay so the slope is going to be (3-b)/3

(derived from the two points 0,b and 3,3)

[glad summit]

Oh, from ∆y/∆x

[sacred whale]

Now we can write m in the equation, y=mx+b in terms of b

It will look like this

And of course, at x1, the y value is going to be 0

[glad summit]

Yep

[sacred whale]

So solving for x1, it will be

Now we can just use pythagoras with the two lengths, x1 and b
and the hypotenuse of 4*sqrt(10)

[glad summit]

Okay

[sacred whale]

and rthen you get to factor that thing 👀

[glad summit]

...

Should be fun

[sacred whale]

are you allowed to use calculators or?

[glad summit]

I don't think so

The teacher didn't specify so if the question requires one, I'd just use one. The question was just given as a problem question in maths class, not in a test or anything like that

@sacred whale?

[sacred whale]

yea

[glad summit]

Can you carry on please?

[sacred whale]

So do you have b now?

[glad summit]

I understand everything you've done so far

[sacred whale]

Yes, so you have set up the a^2+b^2=c^2
where a = x1, b= b, and c=4*sqrt(10)
?

[glad summit]

I can't write it out rn because I'm outside but I would know how to set that up

[sacred whale]

Thats what we need to solve, then you have b, and you can plug it in the equation for the slope in terms of b

and there u go

[glad summit]

Okay, I'll try it, thanks

[odd edge]

@glad summit Has your question been resolved?

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[fast coral]

if i graph the circle with the r 67.5 and centre at 0,67.5 i got expression below with a being angle and x being x coord

so

but that isnt right

so i need an explanation on how im supposed to do it

[stone shore]

What’s London eye

[odd edge]

@fast coral Has your question been resolved?

[odd edge]

@fast coral Has your question been resolved?

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[normal sundial]

which steps did they take to rewrite this equation?

[shy smelt]

They distributed the 2 into the (2x + 2y dy/dx)

and then distributed (leaving x^2 + y^2 alone though)

[normal sundial]

What do you mean with this?

[shy smelt]

$$LHS = (x^2 + y^2)\cdot 4x + (x^2 + y^2)\cdot 4y\frac{dy}{dx}$$

[clever fjord]

Jelle

[shy smelt]

and then they moved around some terms and factored again

[normal sundial]

Why do they do so difficult

[shy smelt]

Presumably so it can be solved with seperation of variables

[normal sundial]

Wait but where does the 4x come from

Is it the 2 * 2x

[shy smelt]

yeah

same with 4y

[normal sundial]

Why didnt they distribute the 2 to (x^2 + y^2)? Those are the rules right?

[shy smelt]

You can distribute the 2 either to (x^2 + y^2) or (2x + 2y dy/dx), doesn't matter which

2 * a * b = a * (2 * b)

[normal sundial]

Oh okay

Ty

.close

[odd edge]

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[supple rose]

How do you do this?

[supple rose]

can someone help?

...

<@&286206848099549185>

<@&286206848099549185>

[mystic saffron]

hii

[supple rose]

hello

Can u help?

[mystic saffron]

first you want to find the radius in the circle

im not sure if thats how you do it properly

[supple rose]

Wdym

like which circle

[mystic saffron]

because my math tutor told me alwaays try to find the radius in the whole big circle

[supple rose]

yea how do you do that?

1 + x?

[mystic saffron]

wait which outer circle?

[supple rose]

the bigger one?

[mystic saffron]

kk

r = √(A / π)

[supple rose]

<@&286206848099549185>

[mystic saffron]

here try using this

formula

im not sure if its the correct one or not 😭

[supple rose]

But that doesnt help + I already know that

[mystic saffron]

The outer radius of an annulus is the radius of the larger of the 2 concentric circles that form its boundary.

[supple rose]

what an 'annulus'

[slate radish]

I think 1+r is a good starting point, connect the center of the inner circle to the center of all small circles

and connect all the centers of the small circles

you get 10 triangles

[supple rose]

basically, all I know it that there is a 36 degree gap between each of the small circles

[slate radish]

Yep

[mystic saffron]

it means region between two circles

[supple rose]

but idk how to continue on from there

[slate radish]

focus on one triangle

[supple rose]

I only have 2 values so trig wont help

[slate radish]

The lengths are 1+r, 1+r, and 2r

and the angle is 36

[supple rose]

2r for what?

[slate radish]

and 72 and 72

excuse me not 2r

my bad.

you have all angles.

[supple rose]

yea

[slate radish]

Isosceles triangle

[supple rose]

but what then

oh wait

cant i just half the triangle?

to get a right angle one

?

[slate radish]

ye

[supple rose]

so I have 18 degrees , 1 + x and 72 degrees

[slate radish]

go for it

[supple rose]

idk what now

1 + x / sin ( 72 )?

oh wait no

sin(72) x 1 + x?

@slate radish

[slate radish]

?

[supple rose]

What do I do now?

[supple rose]

After this

[slate radish]

I don't think it's the right way to go

[supple rose]

so what do I do

<@&286206848099549185>

<@&286206848099549185>

[slate radish]

You should call for helpers only once

[supple rose]

K

[slate radish]

You could draw a line from the center of the inner circle to the touching point of each of the smaller circles

To have tangent to the circles

[supple rose]

What do you mean?

How is a tangent going to help?

[slate radish]

It's going to help you find a right angle

when connected to the center

[supple rose]

but you are finding the radius of the bigger circle

[slate radish]

We are going to go through the smaller circles

The radius of the smaller circles, let's call it r

[supple rose]

yea

[slate radish]

the bigger circle has radius 1+2r right?

[supple rose]

no

1 + r

[slate radish]

why?

no, 2r.

[supple rose]

oh u said radius

right

then its 2r

[slate radish]

Ok

Now look at one of the small circles

[supple rose]

mhm

[slate radish]

Let's call the center of the inner circle O

And the center of the small circle P

And draw another line to the touching point of the small circles

Let's call it T

So you have a triangle

OPT

the length of OP = 1+r

[supple rose]

wait what

[slate radish]

where two small circles touch each other

[supple rose]

like that?

[slate radish]

no

[supple rose]

ok

[slate radish]

Now you have a triangle OPT

OP = 1+r

PT = r

(the lengths)

[supple rose]

yea

[slate radish]

right?

Now the angle POT

Can you say what it is?

and its a right angle triangle right?

[supple rose]

18 degrees

[slate radish]

true

[supple rose]

is it a right angle triangle?

yea I guess

OPT is 90

[slate radish]

yes, because the tangent is perpendicular to the radius

[supple rose]

sorry

ok

[slate radish]

Now you can use trig

[supple rose]

but the trig will have unknowns

[slate radish]

True

let's try

[supple rose]

Sin ( 18 ) = r/1+r

[slate radish]

Draw the triangle

why 30?

[supple rose]

sorry 18

[slate radish]

alright, solve

[supple rose]

Sin ( 18 ) + Sin(18) r = r

[slate radish]

,w sin(18) = r/(1+r) solve for r

there you go

[supple rose]

r(1 + r) ?

[slate radish]

oops

[supple rose]

isnt it r / 1+ r

[clever fjord]

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[slate radish]

here

Typo

[supple rose]

2/ sqr root 5 + 1

?

[slate radish]

yes

[supple rose]

it says here the answer is 1+ Sin(18) / 1 - Sin(18)

[slate radish]

,w 2/Sqrt[5] + 1

[clever fjord]

Results provided by WolframAlpha

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Requested by cain0196

[slate radish]

,w 1+ Sin(18) / (1 - Sin(18))

[clever fjord]

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[slate radish]

Are you sure?

[supple rose]

yes

[slate radish]

Ah, I see where I got it wrong

wait

,w 1+1/sqrt{5}

[clever fjord]

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Requested by cain0196

[slate radish]

It looks like they did 1+r instead of 1+2r

[supple rose]

so they made r = to the diameter?

[slate radish]

,w 1+1/sqrt{5} == 1+sin(18)/(1-sin(18))

[clever fjord]

Results provided by WolframAlpha

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[slate radish]

I don't understand why

[supple rose]

same

[slate radish]

They probably had a different way of solving it

Their answer is also different as an expression

But I don't see where we've got it wrong

[supple rose]

Alright then Ill have to go now, Thanks for your time

.close

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[mystic saffron]

Is anyone good at algebra 1 in this chat? I’m in 9th grade and need help with my math homework. Idk what to do.

[summer cradle]

[mystic saffron]

There is a lot of questions on this homework tho

[odd magnet]

we should be able to help you with the idea behind them and you can work out the rest

[odd magnet]

so with a graph this is a lot easier - try to find each of the x points and see if the y point makes sense to be on the line. in some cases it will quite obviously not go through the point

and for "maybe?" you can do substitute the numbers in for y and x in the equation

for instance, look at the point (x=-5, y=2). does that look like it'll be on the line?

[mystic saffron]

No

[odd magnet]

so that one is not an answer, got it

can you apply the same idea to the others?

[mystic saffron]

Yeah

[odd magnet]

[odd edge]

@mystic saffron Has your question been resolved?

[odd magnet]

@mystic saffron if you need help, we can continue in here.

[mystic saffron]

Tbh your help didn’t rly help me

[odd magnet]

what didn't you get?

[mystic saffron]

All of it

[odd magnet]

okay, so you are trying to find which of these points is on the line

you can look at the values (x, y) and check the graph of "find this x value, then find the y value, is this on the line"

[wary berry]

I think I can help

if you're still there @mystic saffron

if not hopefully read this later

so a really basic way to find an ordered pairs, is to count from the origin

so we can see that (1, -3) is a point on the graph because the line of the equation crosses through that point on the graph

but since the equation is y = -4x+1, we can also start at +1, and subtract 4 from Y and follow that pattern along the X axis

using that, we find that 1 - 4 = -3, and 1 - 8 = -7, right

so we know that (1, -3) and (2, -7) are both points on the graph

you can do this in the opposite direction as well, through as you can see the y value increases and the x value decreases

so lets take (-1, y) for example, we know that the slope is -4, so we can add 4 when going to the left

so there will be a point at (-1, 5) because 1+4 (4 is the slope here, 1 is the y intercept) is 5

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[desert marlin]

Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function with $f(0)=0$ and $|f'(x)|\leq |f(x)|\quad \forall x\in \mathbb{R}$. Prove that $f$ is constant.

[clever fjord]

Austin

[desert marlin]

So I was working on this one for a little bit with MVT before I surrendered to Math stack exchange where I've been trying to follow along with this proof

But I am not understanding this step

I've been trying to see if I can like work backwards to derive that step somehow but I just don't know why it is true / it follows that

could anyone explain?

[odd edge]

@desert marlin Has your question been resolved?

[desert marlin]

<@&286206848099549185>

[open onyx]

$f’(c)=\frac{f(m)}{m}$ (mvt)

then $f(m)= mf’(c)\le mf(c)$ just by rearranging and applying restriction on derivative given

[clever fjord]

IV

[open onyx]

abs signs missing bc lazy

i will say until i read the phrase mvt i had 0 fucking clue what it’s talking about

[desert marlin]

Ohhh right the given that f(c) <= f’(c)

Crap I missed that

TY IV

I get it

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[urban plover]

Can anyone help me on this

[urban plover]

.close

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[proven pumice]

i dont get how these semi log plots work

and how my teacher got the answer that she did for part a

[cold sage]

do you know how to read them at all?

[proven pumice]

i know how to read them

i just dont understand how she did the AROC thing

and got 1/10 log3

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[safe jetty]

Why is

[safe jetty]

equal to

[undone mauve]

a*logb= log(b^a)

log(a) - log(b) = log(a/b)

now you can change the base to a since it's a common base in the whole thing and then simplify it, you'll get your answer

[safe jetty]

log(a)^(a/2) / log(1/a) how can i change base

i thought base change creating more fractions

[brittle beacon]

simpler: log(1) = 0

[undone mauve]

yeah actually that's better I don't know why I didn't point that out

[undone mauve]

If you have the same base across an entire equation you can change it

Since log(a)b is basically logb/loga

[odd edge]

@safe jetty Has your question been resolved?

[safe jetty]

yeah i think just knowing my logs better would have helped

log(1) = 0 is pretty important

thanks

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[molten panther]

Hello, can someone help me with this masterpiece

[molten panther]

I have to find the shaded area knowing this is a square with l = 1 and it's a right isosceles triangle

[nocturne belfry]

doesnt look right

oh, the red triangle is right

[molten panther]

um, this is a problem

I don't really know which is which

I just translated the problem in eng

[nocturne belfry]

why not translate the solution

just kidding

[molten panther]

cuz there is not

I found the problem on google

[nocturne belfry]

lets see

its trivial to get a side and an angle

[molten panther]

I am eyes and ears

[nocturne belfry]

can you link the problem?

[molten panther]

://www.facebook.com/profile.php?id=100078632036274

https in front of :

it's from a fb page with math problems

unfortunatelly the page is in romanian

[molten panther]

.

[fair prism]

You can find the coordinates of all the points of the red triangle, create equations for some of the lines to find intersection of grey line w/ red triangle, then find all the lengths of the triangle (which you can then plug into formula)

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[fresh sigil]

Hi can canyone check this

[fresh sigil]

Pls

[odd edge]

@fresh sigil Has your question been resolved?

[fresh sigil]

<@&286206848099549185>

[silver talon]

i can help you

[odd edge]

@fresh sigil Has your question been resolved?

[fresh sigil]

<@&286206848099549185>

[fresh sigil]

<@&286206848099549185>

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@fresh sigil Has your question been resolved?

[stiff briar]

what was the original question?

[fresh sigil]

Thanks @sullen ferry

.close

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[sly cliff]

How to solve vertex and x & y and graph for f(x)=2x²+4x-2

[sly cliff]

I'm stuck at solving for the x and y part idk how to get it

[prisma snow]

It's messed up that I know how to do this with calculus but forgot how to do it the way they teach you

I believe you use the formula for the x value of the vertex and then put that value into the function to get the y value

[sly cliff]

What is the formula and how do I get the x values too?

[prisma snow]

Let me derive the formula real quick on paper

x = -B/2A

Then, you calculate f(x) using that x value, to get both vertex points

[sly cliff]

That's the place I'm stuck at like finding the values

[prisma snow]

What do you mean the values

A and B?

[sly cliff]

What numbers should be put?

No finding x and y

[prisma snow]

Ok, I'll try to explain

Given a quadratic equation in standard form, f(x) = Ax^2 + Bx + C, we can find the vertex values, (x_v, y_v) by doing the following:

Calculate x_v = -B/2A
Calculate y_v = f(x_v)

Which part of these steps do you not get

[sly cliff]

The first one

[prisma snow]

Ok, so the step with x_v = -B/2A

Quadratic functions can be put in a standard form, Ax^2 + Bx + C, where A,B, and C are all constants.
For example, 6x^2 + 3x - 5 is a quadratic function in standard form, with A = 6, B = 3, C = -5

To calculate x_v for this example, we would calculate -(3)/(2*6) = -1/4

[sly cliff]

Oh

I get it now

Thanks

[prisma snow]

No problem

[sly cliff]

.close

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[wintry viper]

So that it is messy

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[sacred jewel]

whats is

[sacred jewel]

i understand this

[wheat raptor]

are you trying to simplify this or just arrive at an answer

[sacred jewel]

new to math so i want to know how do you multiply this or is it even possible

i dont think i have to simplify it

[wheat raptor]

you cannot simply it in the same way if this is what you mean, as the bases (large numbers) are different, 2 and 3

to do this you would evaluate each individual exponent;
what does 2^4 equal, and what does 3^2 equal?

[sacred jewel]

3^2 would be 6 and 2^4 would be 16

[wheat raptor]

i think you've made a mistake with 3^2, take a closer look

[sacred jewel]

opps 9

[wheat raptor]

Yes

And the question is 3^2 multiplied by 2^4

So 9 multiplied by 16 is the final answer, which you can work out

[sacred jewel]

ok so 3^2.2^4=9x16=144

[wheat raptor]

That's correct

[sacred jewel]

ok thx your the goat

[wheat raptor]

👍

[sacred jewel]

.close

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[worldly linden]

If lcm (a,b)=150 and if 2a=5b
How to find gcd (a,b)

[warped glacier]

150 = 2 * 3 * 5^2

so a = 3 * 5^2, b = 2 * 3 *5 will work

[warped glacier]

not sure if there's a better way but here you go

a, b can't have a higher power than 2, 3, or 5^2 or otherwise the lcm condition would be broken

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[sacred jewel]

would i write down the answer as 5.8^2

or can i not calculate this

[late dust]

You can't get an exact answer if that's what you mean

[sacred jewel]

ok

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[primal kestrel]

What is four hundred and fifty six written as a numerical expresion

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[lilac jacinth]

i have a relatively easier question. when we say for example 100 km per hour, why do we write it as 100km/hour. why divide km by hour

[forest sky]

that's the meaning of the unit "kilometer per hour". to find the average speed you can divide the total distance traveled (in kilometers) by the time it took to go that distance (in hours). the units are kilometers/hours. if you simplify that you get the amount of distance you travel in one hour at that speed, which is (for example) 150 kilometers / 3 hours = 50 kilometers / 1 hour = 50 kilometers in one hour = 50 kilometers traveled for each hour spent traveling = 50 kilometers per hour

[lilac jacinth]

i see

thanks

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[ember shell]

B and c for 16.27

And b for 16.28

I'm like kinda confusion

:D

[signal oar]

Well, for the domain, which values for x could you put in?

[zenith tartan]

what according to you is the definition of domain and range i wanna ask

well

kepe is here for help

[ember shell]

Isn't domain like the input kinda

Like the numbers you can put into the function

And then the range is like the result?

Wut

[ember shell]

Uhhh any number

Right?

[signal oar]

0 too?

[ember shell]

Like all real ig

[signal oar]

-1 too?

-2 too?

[ember shell]

Positive only?

😭😭😭