help-10

[hybrid glade]

I have a feeling this is wrong

I'm p sure it is

pls help me correct it

[meager glade]

!original

[obtuse pebble]

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.

[meager glade]

ah nvm i see it now

idk looks correct to me

do you know what the answer should be?

[hybrid glade]

no its on an assignment

if it looks correct il use it

[obtuse pebble]

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[mental yew]

I have a question

[mental yew]

Why did they use the discriminat on the quadratic.

I factoried it to get this graph

I thought it factorised to (x-2)(x+7)

Hence has roots at 2,-7?

[wary badger]

because they were looking for the zeros

to find the zeros set each factor equal to zero

but since the discriminant for the quadratic was negative it had no roots

or real roots atleast

two numbers that multiply to 14 but add to 5

[mental yew]

Okay so if I get a question similar to that and it has a quadratic do I have to use the discriminant to see if it had real roots?

[wary badger]

(x-2)(x+7) gives —14 not +14

[mental yew]

Ahhh yes!

Okay that makes sense

Thank you!!!

[wary badger]

yea unless u can just factor it right away

if u can’t factor it

before u do quadratic

[mental yew]

Okay that’s prefect thank you!!

[wary badger]

do the discriminant

yea no problem

[mental yew]

Yep!!

[obtuse pebble]

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[violet timber]

whats the formula i need for this type of quesiton i forgot

[uneven palm]

have you learned of sohcahtoa?

[violet timber]

well

not rlly

i know what it is but idk how to use it

fully

i just wanna know the sohcahtoa formula for this one

[uneven palm]

so it’s
sin(theta) = opposite / hypotenuse
cos(theta) = adjacent / hypotenuse
tan(theta) = opposite / adjacent

[violet timber]

so what do i do when i have an angle

[uneven palm]

so the sides in question are x and 14

[violet timber]

OHH

so

[uneven palm]

x is adjacent to the angle and 14 is the hypotenuse

[violet timber]

so cos?

[uneven palm]

yep

[violet timber]

so

what do i do in that case

how do i use cos

what should i wwrite in my calc

[uneven palm]

so the relationship is cos(30) = x / 14

and then solve for x

[violet timber]

so 14x30?

12.12

so a

[uneven palm]

,calc 14 * cos(30 deg)

[warm shale]

Result:

12.124355652982

[uneven palm]

yup

[violet timber]

alr

so for this one

sin(theta) = opposite / hypotenuse

right?

sin(53) = 14.5 / x

[uneven palm]

well 14.5 is adjacent to the angle

so not quite sin

[violet timber]

ohhh wait how do i nkow if it is

[uneven palm]

adjacent means next to

while opposite would be the side on the left

so the bottom side is “next to” the angle

[violet timber]

OHHH

the angle

that mkaes sense

cos(53) = 14.5 / hypotenuse

[uneven palm]

yep

[violet timber]

so 53x14.5?

wait nah

im stupiod

14.5 div 53

24

[uneven palm]

yeah

[violet timber]

alr

thank

you

[uneven palm]

np

[violet timber]

it makes sense now

anyways you can close this ticket idk how to

[uneven palm]

yeah it’s .close

.close

[obtuse pebble]

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[obtuse pebble]

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[distant ravine]

My notes say they got from one equation to the other by deriving, but I'm not exactly sure how it was done.

[distant ravine]

So I'm wondering, how was the deriving done in this case? Like if you had to write it step by step or something

[wary yacht]

its implicit differentiation

[final thunder]

y is a function of x

[timid silo]

^

important distinction they should have made

[distant ravine]

That's what I was assuming, that they "divided" everything by dx or something

[timid silo]

if y was independent, you would have needed to take the partial derivative. Which it isn't clear if it is or not

[timid silo]

no they used the chain rule that's what

[distant ravine]

Yeah, you get what I mean

hopefully lol

[wary yacht]

so when u do dz/dx = u cant just do db/dx so u need to add dx/dy

db/dx * dx/dy = db/dx * dx/dy

= b * dy/dx

is that clear enough?

its basically adding something to make it differentiable

[timid silo]

$y = \m y x$, so by the chain rule which states: [
\dv[y]x = \dv[y]u \cd \dv[u]x
]
you would get: [
\dv[y]x = 1 \cd \dv[y]x
]
which is kind of tautological

[distant ravine]

It wasn't differentiable before?

[wary yacht]

ur trying to differentiate y with x

its like talking german to japanese people

[distant ravine]

But y is a function of x, no?

[wary yacht]

oh in this case u see

[distant ravine]

at least I think it is

[wary yacht]

dz/dx that means u want to make the x go away

[distant ravine]

Yeah

(I haven't had a math course in a year now so I'm a little bit rusty, my apologies)

[wary yacht]

but y wont disappear that easily

u need to do something

[distant ravine]

Right

[wary yacht]

if u need to remember it for an exam

i do it that way

dz/dx = db/dy (do the normal differentiation) MULTIPLY by dy/dx or u can just write y' at the end

[distant ravine]

wait so what does that have to do with the second equation

I kinda lost you

You derive using x to get rid of the x

but you're left with dy/dx

[wary yacht]

yes

[distant ravine]

like you multiplied it by d/dx

[wary yacht]

like this right?

(sorry im trying to write with my mouse

oh and it should be b *dy/dx

[distant ravine]

so the (d/dx) * (x) becomes 1

and you can multiply by d/dx on both sides cause it doesn't change anything

[wary yacht]

yea or u can do it like

lets say now it is bx^2 then it is b(2x) * y'

it works better with actual numbers

x^3 = 2x + 3y + 10

d (x^3 )/dx = 2* d(x) /dx + 3* d(y)/dx

and u just cant solve dy/dx so u leave it there

exam question typically will ask u to find y'

so u just change the subject

and tada

[distant ravine]

Change the subject like:
"Find y' "
"So, uh, nice weather we're having"

thanks lol

[wary yacht]

np!!

[obtuse pebble]

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[meager glade]

Whats your question

[lyric crane]

get both in slope intercept, slope of the lines should be negative reciprocals

[boreal jay]

if two slopes are perpendicular then
slope1 * slope2 = -1

just convert them into the y = mx + c or slope intercept form and then plug it in like the other person said

[obtuse pebble]

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[chrome basin]

An yone can help

[low patio]

What are the sides multiplied by in terms of length?

[chrome basin]

Uh idk

[tidal gazelle]

Imagine an hypothetical 2x2 matrix with one column being (p q) and the second column being (r s). The goal is to find p, q, r and s

[low patio]

What is the length of D-C and D'-C'?

[tidal gazelle]

You can get the values for p,q, r and s by getting four equations using the transformations for any four of the available points

take point A for example which has coordinates (-1, 2). When the matrix is applied on this it must give A' which is (3, -6)

and like this do the same for B, C and D and you will have four equations

solve the equations to get the p, q, r and s

and you are done

[low patio]

he probably doesn't know what a matrix is

[obtuse pebble]

@chrome basin Has your question been resolved?

[obtuse pebble]

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[oak aurora]

There exists $g \in G$ st $|g|=n$, the maximal order. Let $h \in G$ and $|h|=m$. FSOC, assume $m$ does not divide $n$. Then there exists a prime $p$ whose multiplicity as a factor of $m$ exceeds that of $n$. Thus let $p^a$ be the highest power of $p$ in $m$ and $p^b$ be the highest power of $p$ in $n$, the earlier statement will imply that $a>b$. Consider the elements ${g^p}^a$ and $h^{m/p^a}$

Im not sure how to proceed

[warm shale]

Iusgnol

[oak aurora]

Oh okay just more thoughts, i also have that c cannot be coprime with with n since othewise <h>=<g^c>=<g>

.close

[obtuse pebble]

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[random berry]

if f'(x) = 0 and f''(x) = 0 how do i determine whether its a maximum or minimum

[timid silo]

you have to continue taking higher derivatives I believe

[dark stirrup]

There's not enough information to say for sure if it will be either or any

You could have f(x)=0, in which case, you'll get zero for every derivative

[random berry]

well i have to maximum points which are x = 1 and x = -1

[dark stirrup]

!original

[obtuse pebble]

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.

[random berry]

heres the function

[dark stirrup]

Is this a test?

[random berry]

no

[random berry]

what makes you ask that?

[dark stirrup]

Just the formatting of the problem

[random berry]

no its just homework problems

[dark stirrup]

If second derivative test fails, my only strat to test if it is a local min or max is to inspect points between the local extremum, and the adjacent extrema

There may be a more exact way, but that's all I've got for now

[timid silo]

<enumitem>
Let $f$ be a sufficiently differentiable function along an interval $I \subset \R$, let $c \in I$ and let all the derivatives of $f$ up to some positive integer $n$ be 0 but with the $(n+1)$ derivative being Nonzero. Then:
\env{enumerate}{
\ii if $n$ is odd and $f^{(n+1)}(c) < 0$ then $c$ is a local maximum
\ii if $n$ is odd and $f^{(n+1)}(c) > 0$ then $c$ is a local minimum
}

[timid silo]

if otherwise, the point should be a saddle point

[random berry]

oh so its a minimum okay thanks

[obtuse pebble]

@random berry Has your question been resolved?

[random berry]

so would all the critical points for this question be

part (a)

be

+1, -1, 0

[tame narwhal]

seems like it

[obtuse pebble]

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[solemn anvil]

Let x,y,z be positive real numbers such that x+y+z=3. What’s the minimum of $$x^4 + 2y^4 + 3z^4$$

[warm shale]

Faq

[solemn anvil]

This is from number theory class

[timid silo]

sounds like some AM-GM-HM shenanigans

[solemn anvil]

That’s what I thought at first but don’t see how to do it

[tidal gazelle]

use lagrange multipliers

[low patio]

Lagrange multipliers

[solemn anvil]

Guys it’s number theory

[timid silo]

maybe Cauchy Schwarz as well

[smoky vigil]

that is another way, but i'm pretty sure Am-Gm-HM is the way to go here

[tidal gazelle]

think of two functions f(x, y, z) and g(x, y, z). And thier gradients will be proportional

[solemn anvil]

It’s number theory

[timid silo]

I am pretty sure it's either Cauchy Schwarz or AM GM so try both

[solemn anvil]

Can you give me a hint?

Well I’m not sure what to make of the AM-GM

[tidal gazelle]

think of two vectors <x^4, y^4, z^4> and <1, 2, 3>. Thier dot product is x^4 + 2y^4 + 3z^4. This dot product must be lesser than the product of the lengths of the vectors

thats what i could come up with

after that i am quite stuck

[timid silo]

they have reiterated multiple times that this isn't a calculus class.

[obtuse pebble]

@solemn anvil Has your question been resolved?

[limber quartz]

What chapter of your number theory textbook is this from?

[solemn anvil]

It’s under a big inequality chapter

With the famous inequalities like Am gm Cauchy Schwartz holders etc

I’m just not sure how to do this one

[limber quartz]

What is the chapter called

[solemn anvil]

Balanced coefficients

[fluid snow]

See https://artofproblemsolving.com/community/c6h1619297p10130647

[solemn anvil]

Ah okay thanks

That’s so random omg

[tidal gazelle]

is the answer sqrt(42) by any chance?

[solemn anvil]

I don’t really get how they got to that tbh

[limber quartz]

I wonder if the solution to this problem has something to do with a "balanced coefficients" procedure described in the textbook 🤔

[limber quartz]

Ditto

Some theorem by a mathematician named Holder?

[solemn anvil]

Ah, yeah holder

Still a fairly weird choice

[limber quartz]

https://en.m.wikipedia.org/wiki/Hölder's_inequality

[solemn anvil]

How to know to use that weird expression^3

Is beyond me

[limber quartz]

Sounds calculus-y

[tidal gazelle]

Did I make any mistake?

[meager glade]

,rotate

[tardy epoch]

I thought this was faq's channel

[timid silo]

for the 1 millionth time THIS IS NOT A CALCULUS QUESTION!!!

[meager glade]

what about [\int x+y+z = 3] ?

[tardy epoch]

[smoky vigil]

sully

[warm shale]

Flappie

[tidal gazelle]

the only doubt i have is that P might be greater than sqrt(42) but sqrt(42) isnt necessarily a minima

[lyric crane]

please use your own help channel

[tidal gazelle]

the question is related to the one being discussed here

[lyric crane]

ah true mb

[fluid snow]

WolframAlpha agrees with what was obtained by that guy on AoPS.

[final thunder]

Admittedly that was hard to notice, but basically we look for real positve (a,b,c) with (a+b+c=3) and so
[
\wrb{(x^4+2y^4+3z^4)(a^4+2b^4+3c^4) >= (a^3x+2b^3y+3c^3z)^4}
]
Choose (a,b,c) such that (a^3=2b^3=3c^3=k^3) and we get
[\wrb{
x^4+2y^4+3z^4 >= \4{(3k^3)^4}{(a^4+2b^4+3c^4) }
}]

[warm shale]

Pure

[solemn anvil]

Damn

[tidal gazelle]

how are you supposed to come up with that though? Like what's the motivation or intuition to think of that

[final thunder]

Not sure, I only realised it after seeing the solution.

[solemn anvil]

I don’t totally get the choosing a,b,c part

[ebon yoke]

it's a special case of Holder's inequality for p=4 and q = 4/3

[obtuse pebble]

@solemn anvil Has your question been resolved?

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[solemn anvil]

Yeah I see it now thanks

[obtuse pebble]

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[compact tulip]

I'm a little stuck here

[forest sinew]

sorry, which task?

[compact tulip]

315

[compact tulip]

It's uh

The one with the blank spacr

Hopefully it's clear

[forest sinew]

yea, im trying to remember how to do these

[compact tulip]

😭

[forest sinew]

i believe you want to define like a unit-rate of work

[compact tulip]

That doesn't sound too go9f

Yes

I think so

[forest sinew]

no they are contrived problems

you only see them in algebra then they go away

[compact tulip]

😭😭

Ok

[forest sinew]

lets see so

Tom and huck would take 9 hours

[compact tulip]

Ye

I have thst in my work I think

But the thing is

Idk if my second rate is correct

[forest sinew]

1/x?

[compact tulip]

Yeah

[forest sinew]

i mean yea its unknown

[compact tulip]

Ok

So my equation was like

Idk what to write there though

[forest sinew]

hello kappa

[compact tulip]

Wut

😭

[forest sinew]

kappa was typing

[compact tulip]

Oh

[forest sinew]

this is gonna sound dumb but

so you got tom and huck working

they go 1/9th of the fence per hour

so that means when jim joins theres 5/9 of the fence left to paint right

[compact tulip]

Right

[forest sinew]

so now you should be able to get an equation, i think?

[compact tulip]

Ehm

You need like

[forest sinew]

because in two hours, their combined rate has to equal 5/9

[compact tulip]

So they use uhm 5/9/2?

Or something...

[forest sinew]

in two hours, how much of the remaining job can tom and huck finish?

how fast do they work?

[compact tulip]

Oh uhm

[forest sinew]

not a trick question, you already said it

whats their rate of work in hours/portion of fence

its 1/9, yea?

[compact tulip]

Ye

[forest sinew]

okay

[compact tulip]

Well uhhh

They use like

[forest sinew]

so we got all 3 of them

tom and huck do a part of the work in that 2 hours right

tom huck and jim take 2 hours to paint 5/9 of the fence

how much of that 5/9 do tom and huck paint?

[compact tulip]

Yes

[forest sinew]

use this

[compact tulip]

Uhmmm

[forest sinew]

they dont speed up or slow down painting, they still have the same rate

[compact tulip]

So that's like

1/9 plus 1/x equals to 2?

Or something ??

[forest sinew]

dont worry about setting up an equation

you can think through the problem from here

[forest sinew]

how much of the fence can tom and huck paint in 2 hours?

how much do they paint in a single hour?

[compact tulip]

Uhm

He paints like

2/9

Of the fence

Or they pain 2/9

Sorry

Uhh

[forest sinew]

they do, yea

in two hours

[compact tulip]

Right

[forest sinew]

tom and huck can paint 2/9 of the fence

you agree?

[compact tulip]

Yeah

[forest sinew]

okay

so heres the scenario

jim joins up

theres 5/9 of the fence left to paint

2 hours later, its all painted

[compact tulip]

Rifht

[forest sinew]

so of that 5/9

[compact tulip]

So he paints 3/9 of the fence?

[forest sinew]

tom and huck must have painted 2/9

[compact tulip]

Yeah

[forest sinew]

yea, i think so

so whats his hourly rate?

jim, i mean

the new guy

he can do 3/9 in two hours

[compact tulip]

Roght

Right

[compact tulip]

Wo he does 6

[forest sinew]

i mean really

[compact tulip]

Then right?

[forest sinew]

you dont even need his hourly rate to finish the problem

[compact tulip]

Oh

[forest sinew]

he paints a third of the fence in two hours

this is enough to finish the problem

what is it youre originally asked to find?

[compact tulip]

Yeah but it asks for hours

Not rate

[forest sinew]

it does

[compact tulip]

Lol

Ye

So 6

[forest sinew]

so he paints a third of the fence in two hours

yea

6 hours

told you that you dont need an equation

[compact tulip]

Oh

Ic

[forest sinew]

if you wanted to set up something i mean

you could have done like

$2 \cdot \frac 19 + 2 \cdot \frac 1x = 1 - \frac 49$

i think

[warm shale]

jan Niku

[compact tulip]

I don't understand thst ag all

At all

[forest sinew]

jim joins, then 2 hours pass

then theyre done

so tom and huck contribute 1/9 for each of those two hours

and jim contributes two of whatever his hourly rate is

call it 1/x

and in that two hours they finish whats left of the job, which was one whole fence, but minus the 4/9 that tom and huck have already finished

make sense?

[compact tulip]

Oh

Lol

Yeah I think so

So uhm

Like

When writing the equation

For example in the next problem would it be like

Or is it like bad idk

I'm rlly bad at this uhit

😅

[forest sinew]

i think i would do it like

[forest sinew]

you can turn this into the same problem as before

[compact tulip]

Alright

[forest sinew]

A knight is painting a fence for 15 hours. His partner joins him to help, and they finish in 10 more hours. If his partner had not helped him, it would have taken him 15 more hours to finish the fence. How long would it take his partner if he worked alone?

do you think you can solve this problem?

[compact tulip]

Uhm

So sure I can try

[forest sinew]

can you see how this problem I wrote is the same as the one in your paper?

[compact tulip]

yeah kind of

[forest sinew]

i mean actually in the end your problem will be off by a factor of two

but thats not important

[compact tulip]

😭

[forest sinew]

I mean more why you can apply the exact same process to this problem as before

[compact tulip]

ok

i c

[forest sinew]

try solving this

[compact tulip]

ok

[forest sinew]

you even have the framework of an equation here

youll need to fill in some pieces

[compact tulip]

alright

[compact tulip]

You take 30 in the end right

?

[forest sinew]

lets see

[forest sinew]

so the knight can paint one fence in 30 hours

thats a rate of 1/30

[compact tulip]

Ye

[forest sinew]

so,

he completes half of the fence on his own

[compact tulip]

ye

and then he finishes 10/30 in the other 10 hours i think

[forest sinew]

then, he finishes 10/30 of the fence while his friend helps

yup

[compact tulip]

so he like gets to 5/30

and then u like just multiply by 6

meaning that its 6

lol

[forest sinew]

so his partner must finish 20/30 in the 10 hours

[compact tulip]

wut

[forest sinew]

errr

sorry

youre right

[compact tulip]

😭

i was about to have a heart attack

[forest sinew]

theres 15/30 of the fence left

the knight does 10/30 of it

so the partner does 5/30 in 10 hours

that means per hour, they do 5/300

,calc 5/300

[warm shale]

Result:

0.016666666666667

[forest sinew]

its 1/60

[compact tulip]

uh ye

wait

uhmmm

cuz ur trying to find like the full fence

like

dont u try to like uhm

multiply te two numbers

cuz like

[forest sinew]

hmm?

[compact tulip]

errr

the thing is like

5/30 = 1/x

so u just flip the reciprocal or something

or no wait

[forest sinew]

well its not his rate

5/30 i mean

[compact tulip]

i mean like

ik but like

[forest sinew]

its 10 times his rate

[compact tulip]

errr wut

[forest sinew]

he can paint 5/30 of a fence in 10 hours

[compact tulip]

5/30 is 6 times more

or wait lol uhmm

ur trying to get to 1

i think

right?

[forest sinew]

or i mean think of it how we did last time

you remember how we got like

what was it

a third in two hours?

[compact tulip]

uhh i did it the same way

[forest sinew]

so we figure, third and a third and a third

[compact tulip]

uhmmm

wut loll

[forest sinew]

that means here, where we have 5/30 in 10 hours

we have a 5/30 plus a 5/30 plus a ...

for 60 hours

[compact tulip]

errr

its 30

[forest sinew]

the answer is 30, yea

because the way i rewrote the problem 😭

[compact tulip]

😭

pulling hairs out rn

[forest sinew]

naaaah i want you to believe the answer lol

okay knight one

hes alone

15 hours

[compact tulip]

ye

[forest sinew]

he paints three thirds in 15 hours

now, new fence

knight and his partner

they paint the same three thirds

[compact tulip]

ye

ye

[forest sinew]

but in 10 hours

[compact tulip]

right

[forest sinew]

well the knight painteds what, a third in 5 hours

so he painted 2/3 of the fence

[compact tulip]

right

[forest sinew]

he must work twice as fast as his partner, then

[compact tulip]

oh

[forest sinew]

since in the same time, his partner painted only 1 third

in theory you could solve this problem by inspection, if youre comfortable enough

[compact tulip]

no lol

[forest sinew]

but anyways

the process is the same

get the original person rate

[compact tulip]

hmm

[forest sinew]

how much of the new job did they do?

[compact tulip]

not for the last problem it seems

[forest sinew]

that leaves the rest to the unknown person

[compact tulip]

ah

i believe you have to subtract in the last one

right?

[forest sinew]

no

[compact tulip]

Pipe A leaks the full pool water out in 5 hours. How long will
it take for the two pipes to fill the pool if both are turned on at the same time?

[forest sinew]

its related to the one bofer it

[compact tulip]

yes

ik

i got 2 2/9 on that one

[forest sinew]

,w a/4 + a/9 = 1

[warm shale]

Results provided by WolframAlpha

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Requested by janniku

[forest sinew]

thats

2 + 10/13?

oh wrong numbers lmao

[compact tulip]

lol

[forest sinew]

,w a/4 + a/5 = 1

[warm shale]

Results provided by WolframAlpha

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

Requested by janniku

[forest sinew]

cool

[compact tulip]

ye

[forest sinew]

i dont understand the set up of the last problem

[compact tulip]

like

um

pipe a instead of filling a pool leaks?

ig

so that it drains the pool water out

so while pipe a is draining the pool water

pipe b is adding water

and since pipe b goes at a faster rate

it goes like

it fills up at one point

[forest sinew]

so wait

but it says both pipes turned on

[compact tulip]

ok uhmm

[forest sinew]

are there two pipe A's

[compact tulip]

let me make like a clearer example

lol

if someone is writing on a board

while someone else is erasing

assuming that the person writing on the board is faster

how much time will it take for the person writing on the board to fill up the entire board

since like

the rate of the person erasing is slower

they would eventually fill up the board completely

[forest sinew]

okay, i guess

if that is the problem, then you can solve it im guessing?

you said its something subtraction?

[compact tulip]

i have no clue

tbh

[forest sinew]

can you guess how to set up the problem?

you have a clue you said subtraction

[compact tulip]

true

[forest sinew]

the previous problem was solved by $$\frac x4 +\frac x5 = 1$$

how ca nyou solve this one?

[warm shale]

jan Niku

[compact tulip]

im assuming its something along the lines of

x/4 - x/5 = 1

right?

[forest sinew]

this seems fine to me

[compact tulip]

so it should be 20 horus right?

[forest sinew]

,w a/4 - a/5 = 1

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by janniku

[forest sinew]

seems good to me

does it sound reasonable

[compact tulip]

wooo

[forest sinew]

im having a hard time thinking what would be a reasonable time

[compact tulip]

lol

[forest sinew]

but if we got like

100

id think that was too high i think

so 20 sure

[compact tulip]

:D

thx lol

.close

[obtuse pebble]

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[jade mountain]

what is f’(x) when f(x)=2/x^2

[final thunder]

power rule

rewrite it as (2x^{-2})

[warm shale]

Pure

[robust bloom]

do dy/dx

[jade mountain]

so is n/x^y=nx^-y?

[final thunder]

,,\expolaws

[warm shale]

Pure

[final thunder]

[\42{x^2} = 2x^{-2}]

[warm shale]

Pure

[robust bloom]

the is a derivative q and you forogt exponent

💀

[jade mountain]

oh

i have some other random math questions, could u answer them?

[robust bloom]

first solve this

[jade mountain]

is it not just [/42{x^2} = 2x^{-2}]

[robust bloom]

Huh?

[jade mountain]

isnt f’(x)=2x^-2 for f(x)=2/x^2

[final thunder]

No

[f(x) = \42{x^2} = 2x^{-2}]

[warm shale]

Pure

[final thunder]

so the derivative of f is the derivative of 2x^{-2}

[jade mountain]

so its -4x^-3

[final thunder]

Yeah

[jade mountain]

ok

[robust bloom]

yeah

simpel

[jade mountain]

some other questions, how is e^i(pi)=-1

and how is 0.999 repeating = 1

[robust bloom]

x = 0.99...... eq2
10x = 9.999999999999999999999999999999999999999999999999999999.... eq1
eq1 - eq2
9x = 9
x= 1

@jade mountain

get it?

[jade mountain]

yes but it doesn’t make logical sense

[final thunder]

Are you trying to prove this at undergrad level

[jade mountain]

no im a middle schooler

[final thunder]

Oh well

[robust bloom]

that explains my explanation

i wont do calculus to explain it to him lol

[final thunder]

analysis more like

[jade mountain]

i know a good bit of calc tho

[robust bloom]

i see

where are you from?

[jade mountain]

im taking ap calc bc rn

california

[robust bloom]

Nice

MI

[final thunder]

Do you know about debekind cuts, for example.

[jade mountain]

u mean dedekind cuts?

[robust bloom]

for a middle schooler it does lol

[final thunder]

yeah

[jade mountain]

then yes i do

[final thunder]

Then it should follow nicely that 0.99... = 1

[jade mountain]

ye it does

its just hard to wrap your mind around

[robust bloom]

What grade are you in?

[jade mountain]

8

[final thunder]

The common thing I see people say ask is 'what is the number between 0.9999... and 1'?

[jade mountain]

there is no number between the two

but those numbers have different properties

[final thunder]

How so?

[jade mountain]

sqrt 1 is 1

sqrt 0.999… isnt

[final thunder]

what is it then?

[jade mountain]

idk

[robust bloom]

its 1

[jade mountain]

actually ye it would be 1

mb

[robust bloom]

both numbers are same

they cant have different sq roots

[jade mountain]

ye i guess

even weirder tho is that e^i(pi)=-1

[robust bloom]

e^i(pi) = i^2

[zenith raft]

.999999… is, by definition, the limit of the sequence
.9, .99, .999, .9999, …

which is 1

[robust bloom]

it tends to 1

[jade mountain]

i^2 is -1

[robust bloom]

thank you

[zenith raft]

the limit is 1

[minor furnace]

you use eulers id for this which can be derived multiple ways, notably taylor series and separable diff. equations

[final thunder]

This is getting chaotic. I'm out.

[zenith raft]

i thought you wanted chaos to take the world

[jade mountain]

how

[minor furnace]

unless you already know about either of these topics I wouldn't worry about letter soup making no sense

[jade mountain]

ok ill just quit it

bye and thx for the help

.close

[obtuse pebble]

Channel closed

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[hardy osprey]

ok crash course for you

first of all theres a thing called complex numbers, in the form ( a + bi ), where ( i ) is the imaginary unit with the property that ( i^2 = -1 ). euler's formula states that for any real number ( x ):
[ e^{ix} = \cos(x) + i\sin(x) ]

the proof for this formula is from taylor series, which is a infinite sum to make any function into a polynomial.

the Taylor series expansions of ( e^x ), ( \sin(x) ), and ( \cos(x) ):

• ( e^x ) can be expanded as:
  [ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots ]

• ( \sin(x) ) can be expanded as:
  [ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots ]

• ( \cos(x) ) can be expanded as:
  [ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots ]

if we substitute ( ix ) into the Taylor series for ( e^x ), we get:
[ e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \cdots ]
this simplifies to:
[ e^{ix} = 1 + ix - \frac{x^2}{2!} - i\frac{x^3}{3!} + \cdots ]

• real part: ( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots ) which is the series for ( \cos(x) )
• imaginary part: ( ix - i\frac{x^3}{3!} + i\frac{x^5}{5!} - \cdots ) which is ( i ) times the series for ( \sin(x) )

therefore:
[ e^{ix} = \cos(x) + i\sin(x) ]

so now, apply Euler's formula for ( x = \pi ):
[ e^{i\pi} = \cos(\pi) + i\sin(\pi) ].

knowing that ( \cos(\pi) = -1 ) and ( \sin(\pi) = 0 ), we get:
[ e^{i\pi} = -1 + 0i ]

which simplifies to:
[ e^{i\pi} = -1 ]

[final thunder]

Lol tim

[warm shale]

timliniscool

[hardy osprey]

haha

[obtuse pebble]

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[left lotus]

Is it possible to divide sound wave (e.g. voices) into pure frequency waves?

[tardy epoch]

What's the given equation

[left lotus]

Oh sorry I was asking for more general algorithm kinds of things -

[tardy epoch]

Not enough information to answer you then.

[left lotus]

What information do you need?

[obtuse pebble]

@left lotus Has your question been resolved?

[worn yoke]

sounds like you're thinking of fourier analysis, by which complex waves (e.g. real-world sound waves) may be modeled as a sum of simpler sine waves

[left lotus]

This looks cool, thank you, if there are no alternate options to consider, I'll look into this in more details, thanks

[obtuse pebble]

@left lotus Has your question been resolved?

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[cyan copper]

Why does the answer sheet say number 1 angle is 30? The number 1 angle belongs to diameter CE and it's bigger so it's logical to say it's 120 ÷ 2 = 60°, or does the right triangle rule gets applied here and this circle is actually collary 1 of the inscribed angle

[cyan copper]

You cannot say the number 1 angle belongs to diameter LA aswell because it will just result to 60° again

[low patio]

if mLE = 120, then what is mAE?

[cyan copper]

60°

Now we are inside the diameter of CE

Then AC would be 120

120÷2

So 1 is 60°

So number 2 would actually be 30

Same as 3, 4

Right triangle rules won't apply here right

[low patio]

wdym AC? angle ACE?

[cyan copper]

Ye

1 belongs to semi circle ACE

[low patio]

anyway, since mAE = 60, you divide by 2 to get the inscribed angle

that's it for 1

so it's 30

[cyan copper]

wait what

[cyan copper]

I thought 1 belongs to angle AC

[low patio]

in the picture i just sent, we can imagine the inscribed angle (1) as being related to the arc AE

[cyan copper]

💀

[royal basin]

angle 1 subtends arc AE though, not AC.

[cyan copper]

damn

[low patio]

just look at it

[royal basin]

also it doesn't make much sense in general to speak of angles as "belonging to a diameter"

diameter-less inscribed angles exist, even though in this picture there aren't any

[cyan copper]

So, number 2 subtends to arc AC?

CA*

[low patio]

No, that's 5

[cyan copper]

where does #2 subtend

[low patio]

it's simple: just look at the two line segments associated with the angle, and those are the two points for the arc it subtends

[royal basin]

without the word "to".

this is angle 2 highlighted

it subtends arc LE

[cyan copper]

Ohh

Okay

This is a big help

Thanks man

ohh that is why number 2 is 30°

60°*

So number 3 subtends arc CA?

[royal basin]

yes

[cyan copper]

wait, where does #4 subtend then

If it's 60° aswell

CA aswell?

[ashen dove]

No

Point N is not on the circle

[royal basin]

angle 4 subtends arc CL, but it's also a central angle and not an inscribed one.

which is important.

[cyan copper]

oh yeah central angle is congruent to the arc CL

Okay I get it now

[cyan copper]

It sucks cuz the problem didn't state that 4, 5, 6 would be central angle

[royal basin]

it called the circle by its center, N

there was no need to state that N is the center a second time

[cyan copper]

Thanks

It actually makes sense now

I solved the entire problem

Without looking at the sheet

.close

[obtuse pebble]

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[barren nest]

A and B are normal subgroups of a group G with |G| = 18, |A| = 2 and |B| = 3. Is the group AB cyclic?

[barren nest]

I had earlier proved that AB is maximal in G, in where i had shown that o(AB) = 6.

I can't figure out if AB is cyclic or not, or am i just overthinking

[viral blade]

Well it certainly can be, consider G = Z_18

[barren nest]

Ohh so it may be or may not be?

[viral blade]

Idk if it's possible for it not to be

If it isn't, then ofc it's D6

or D3 or whatever you call the dihedral group with 6 elements

but in that case, I don't think A is normal in AB, which means it's definitely not normal in G

[barren nest]

Can we prove it generally that we can't

[viral blade]

Can't what?

[barren nest]

I mean you're just giving me examples. I am asking if there is a more general proof to say AB can't be non cyclic

[viral blade]

that is a proof

If AB isnt cyclic, then it has to be isomorphic to D6

That's because Z6 and D6 are the only two groups of order 6 up to isomorphism

[barren nest]

Buuut we haven't studied isomorphism yet

[viral blade]

How did you show o(AB) was 6?

[viral blade]

Cause that takes away almost every group theory tool I can think of

[barren nest]

Since o(AB) = o(A)o(B)/o(A∩B) = 6/o(A∩B), i investigated what could o(A∩B) be. it can't be 2 or 3, therefore 1.

[viral blade]

That looks like the second isomorphism theorem

So that much is allowed

[barren nest]

Did u start group theory with morphisms 🐷

[viral blade]

pretty much

[barren nest]

Morphisms are like the uhhh last chapters in my syllabus

I think i have to like let AB isn't cyclic then contradict something

[viral blade]

If it's not cyclic, turns out there's no normal subgroup of order 2

Whenever A and B are normal and have an intersection of size 1, AB should be isomorphic to A×B

Ah

Take a in A and b in B, both not the identity. Show a*b has order 6

[barren nest]

Ahh

Right that is pretty eaasy

Dude thanks, i knew i was overthinking

[viral blade]

np

[barren nest]

.close

[obtuse pebble]

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[obtuse pebble]

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[fringe mango]

could i get some help with an asymptote question

[lost tree]

$f(x)=(-5)\cdot 6^x+2$ right

[warm shale]

SilverSoldier

[lost tree]

what have you tried

its also given u a hint

[fringe mango]

like this

[lost tree]

oh

right

so u can use the hint

[fringe mango]

so ive tried just using the traditional method

dont work

[lost tree]

what is the traditional method

[fringe mango]

putting it into desmos and going from there

thats how ive done it lol

[lost tree]

well why did it not help

[fringe mango]

because i still dont get how to get the x and y values of an asymptote

[lost tree]

do u know what an asymptote is

[fringe mango]

yeah

function that approaches a certain value

whether it be positive or negative inf

thats about it

[lost tree]

well an asymptote is a line

[fringe mango]

yes

[lost tree]

its a line that a graph gets closer and closer to

cant u see such a line

when u draw the graph?

[fringe mango]

it dips down

[lost tree]

thats when u go towards +ve infinity

what happens when u go to -ve infinity

[fringe mango]

still no idea

i haven't got a clue on what the hell -ve infinity is since ive just started asymptotes

[timid silo]

negative

[fringe mango]

oh

[lost tree]

to go towrads +ve infinity - keep going to the right

to go towards -ve infinity - keep going to the left

[timid silo]

-ve is negative +ve is positive

[lost tree]

oh

[timid silo]

thats what he didn’t understand

[lost tree]

oh lol 😅 👍

[fringe mango]

i also dont understand what they're asking me for

[timid silo]

The asymptotes

[fringe mango]

its a single line dipping down

[lost tree]

can u find a line that the graph is approaching as you go towards -ve infinity

is it getting closer and closer to a certain line, the more u go to the left?

[timid silo]

Basically the point on the y axis/x axis that the graph is infinite

[fringe mango]

the more you go to the right it dips down and as you increase the x value the y decreases

[lost tree]

yes

what about the left

[fringe mango]

its just a single one

also the left

[lost tree]

well zoom in

[fringe mango]

y remains 0

[lost tree]

zoom in on the graph

towards the left

[fringe mango]

i hope you know that its just a horizontal asymptote

it approaches -ve inf

[lost tree]

well yeah theres a horizontal asymptote to the graph

and the graph approaches it as u keep going to the left

[fringe mango]

im still confused

so like y=0?

as you keep going to the left

[lost tree]

yes y=0 is the horizontal asymptote

[fringe mango]

OHH

what about the vertical asymptote?

is it just +ve infinity

[lost tree]

can u see a vertical asymptote?

[fringe mango]

no

[lost tree]

yeah

well so do u know what a vertical asymptote is?

[fringe mango]

the positive thing on the black 1/x graph?

[lost tree]

u seem to not quite know exactly what an asymptote is

an asymptote is a line

[fringe mango]

auugh

1/x line

[lost tree]

1/x is not a straight line

[lost tree]

a straight line

it is a straight line that the graph keeps getting closer and closer to, but never actually meets

[fringe mango]

hmmm

so it approaches a value but it doesn't actually meet it?

[lost tree]

yes

so like the graph never crosses that line

[lost tree]

so here
the black curvy lines are the "graph lines"

the y axis (or the line x = 0) is a straight line, and u can see that the black lines keep coming closer and closer to the y axis, but never actually cross it

[fringe mango]

OHH

[lost tree]

and because the y axis is vertical, its a vertical asymptote

[fringe mango]

so it never actually touches the x or y axis

but it always gets closer

[lost tree]

yes

[lost tree]

yes

[kind hawk]

(in general a function can cross the asymptote, imagine something like sin(x)/x and the x axis)

[lost tree]

yes though...but this might confuse the OP wudnt it?

[kind hawk]

well not like it would be better to say wrong stuff

and then later they might ignore an asymptote because it actually does cross it

[lost tree]

well

[fringe mango]

actually i reset the problem

its this one now

ive mapped the line in desmos

[kind hawk]

that graph seems wrong

[lost tree]

so an asymptote is a line that the graph gets closer and closer to, but never actually meets, as you keep going towards +ve/-ve infinity
is this better?

[kind hawk]

where exactly did you place the +6?

well the graph can meet the line. thats the point

[kind hawk]

thats wrong

that would be (8)3^(x+6)

[lost tree]

oh right then just "keeps getting closer to"

[kind hawk]

yes

[fringe mango]

isnt that the same thing

[kind hawk]

not at all

$8 \cdot 3^x +6$

[warm shale]

Denascite

[fringe mango]

it is

[kind hawk]

enter this into desmos

[fringe mango]

ok

...

am i stupid

or am i dumb

[kind hawk]

you just made a mistake

[fringe mango]

oh my god

it all makes sense now

horizontal asymptote is 6

[kind hawk]

yes

[fringe mango]

THANKS

it was correct

.close

[obtuse pebble]

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[boreal compass]

is my proof correct

[boreal compass]

statement: square matrix nxn, if A invertible then r(A) = n
is my proof valid?:
A invertible so exist B s.t AB = I, thus we know col(AB) subset of col(A), thus dimcol(AB) <= dim(col(A)) which means rank(AB) <= rank(A) but AB = I so rank(AB) = n.
so n<=r(A) but also r(A)<=n so therefore r(A) = n

i assumed already dim(row(A)) = dim(col(B)) and dim(row(A)) = rank(A)

[obtuse pebble]

@boreal compass Has your question been resolved?

[boreal compass]

bumping

idk how it works

[obtuse pebble]

@boreal compass Has your question been resolved?

[obtuse pebble]

@boreal compass Has your question been resolved?

[obtuse pebble]

@boreal compass Has your question been resolved?

[meager swift]

Looks pretty good to me. Only thing I wish you elaborated further on is why r(A) <= n, but this is just nitpicking

[obtuse pebble]

@boreal compass Has your question been resolved?

[obtuse pebble]

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[timid silo]

Determine whether the set S is a subspace of $R^3$:
$S = {(x,y,z) \in R^3 | x=2y}$

[warm shale]

Derivative

[timid silo]

here is what I do

so first we check if it is not empty

if x=0, y=0, z=0 then 0-2(0) = 0 so this is satisfied

If I check if it's closed under addition: $(x_1, y_1, z_1) + (x_2, y_2, z_2) = (x_1+x_2, y_1+y_2, z_1+z_2)$

[warm shale]

Derivative

[timid silo]

and let x = x_1+x_2, so 2y = x_1+x_2, so this is satisified

for the multiplication closure: $k(x,y,z) = (kx, ky, kz) = (2(ky), ky, kz)$