help-42

[plain bone]

You have to go in "order"

For example, in ASA, you have to start with an angle, immediately go to a side, and immediately go to another angle in the same direction

These triangles are congruent by AAS because you start at angle U, go to angle T, and go to side TR

[normal gull]

Thank you !! FOR REAL FOR REAL FOR REAL

THANK YOU MAN

Ur the best

Can i give you any vouch or smth in this server ?

[calm coral]

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[calm coral]

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[midnight dove]

Why is the span of the input what determines the column length and output span determines row length. Am I plotting the second question correctly or should I be connecting the dots before and after transformation

[main oak]

For the first question, you need to be able to multiply a vector from the domain by the mapping matrix, so the number of columns has to be equal to the dimension of the domain (4). The number of rows will determine the amount of rows your output vector will have (5).

[midnight dove]

ohh like how a 2x2 matrix multiplies with a 2x1 and the outter dimensions of the equation determines the matrix?

2x2 2x1

[main oak]

yeah, basically
the inner dimensions must match and the outer dimensions determine the resulting dimensions

You could make up an example and see what happens when you change the number of rows/columns if you are not completely convinced.

[midnight dove]

im lowkey confused conceptually what is going on with this mapping thingy and stuff

[main oak]

Regarding the second question, I am not sure if you are supposed to connect the dots or not and I'm assuming you see what the transformation does to each vector.

[midnight dove]

my professor did this but he was given more coordinates

it makes sense to not connect the dots because i can simply say each dot maintained their y coordinate but went to x=0

[main oak]

Yeah, it's easy to see what the transformation does without connecting the points. You could still connect the points before and after the transformation if your professor did so.

[midnight dove]

either connect or no connect my answer should be correct right

and my professor seems generous

I pray 🙏

.close

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[main oak]

good luck then haha, my linear algebra professor was quite generous as well

[midnight dove]

😎

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[mossy geode]

why wont it represent the min arc

[mossy geode]

since angle is acute

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[vital scaffold]

okease helpp

please*

[remote mural]

i think organic chemistry tutor

or chatgpt

[jolly pilot]

this is basically just vectors. Khanacademy has it

[remote mural]

wait is that ap calc bc?

calc 2?

[jolly pilot]

https://www.khanacademy.org/math/algebra-home/alg-vectors

this is just the prerequisites to the actual course

[remote mural]

khan academy has so many calc 2 things

[jolly pilot]

multi var calculus has actual calculus. that pic has only basic concepts in vectors

which is basically just algebra

[remote mural]

its called AP Calculus BC though cause thats the class in america

[jolly pilot]

on khan academy

[jolly pilot]

thats fine. but the contents of the topics mentioned in the picture form just the prerequisites of the course. They are not the actual contents of the course itself. These topics should be revised from when they were taught in previous classes (whatever they may be named)

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[magic otter]

Could someone help me find where I went wrong?

[magic otter]

Looking for F=c, looking for c

[swift dragon]

f(y) cant be in terms of x

$f'(y)=\dv{y} f(y)$

[potent lotus]

Bonk

[swift dragon]

you write f'(y)=6y^2-6xy

but clearly there is an x

so its f'(x,y)=6y^2-6xy

which this shouldnt bethe case

,w (6x^2+5y^2)dx+(10xy+6y^2)dy=0

[potent lotus]

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[swift dragon]

uhm.....

[magic otter]

Sorry one sec I'll get on it I was grabbing coffee

[magic otter]

Rough

And this is to solve for c?

That's crazy

[swift dragon]

ye im not sure lol

[magic otter]

I found a previous mistake but I still have an error somewhere

[swift dragon]

no

you have f(y)=6y^2+C

so F=2x^3+5y^2+2y^3+C

[magic otter]

No it's not, you have to set F = c

I know it's weird, it doesn't make sense to me

There's a proof for it using a rectangle

[swift dragon]

F(x,y(x))=2x^3+5xy(x)^2+2y(x)^3=C

[magic otter]

Yeah

The C's do not add to 0

Oh nevermind! I just put it in wrong lol

These are sooo difficult

[swift dragon]

$\dv{x} F(x,y(x))=6x^2+5y(x)^2+\left(10xy(x)+6y(x)^2\right)\dv{y}{x} = 0$

[potent lotus]

Bonk

[swift dragon]

$F(x,y(x))=2x^3+5xy(x)^2+2y(x)^3+C$

[potent lotus]

Bonk

[magic otter]

Yeah, like all that plus C_1

[swift dragon]

yeah yeah

[magic otter]

Yep

[swift dragon]

its just some constant

[swift dragon]

thats why you practice 😄

[magic otter]

Duuudddeee these are so hard D= :

Tysm for helping me out, bestie

I appreciate you ❤️

[swift dragon]

❤️

[magic otter]

This is what I have for my next problem

Wish me luck

Place your bets xD

[swift dragon]

youve got this

100%

[magic otter]

I bet it's wrong. The numbers are weird

IT WAS FRIGGIN RIGHT!!!

lesgooo boys!

[swift dragon]

lets gooo

!done

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[magic otter]

I've messed up

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[blazing coyote]

I would start off with x+2y=v

and y-x=u

so y=v+u/3

and x=v+2u/3

hmm

This is sus

[swift laurel]

i would double check how you solved for x,y

[blazing coyote]

oops

x=(v-2u)/3

so y=x

or u+v=v-2u

or u=0

The other line is v=2

then we have v-u/3=0

so v=u

and v=u+2/3

It's essentially this region( the upper triangle)

so $\int_{0}^{2} \int_{u}^{u+2/3} Jve^u dv du$

[potent lotus]

What a wonderful world it is !

[blazing coyote]

is this fine

I cna find the jacobian

[swift laurel]

this isn't correct

[blazing coyote]

huh?

x+2y=v

y-x=u

adding them gives 3y=v+u

so y=(v+u)/3

forgot the brackets

oops

[swift laurel]

so i would double check your bounds for v as a result

[blazing coyote]

It's wrong?

I kept track of it on my rough work

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[spice wasp]

i got 123/99 (simplified to 41/33) but apparently thats wrong so now im lost

[velvet osprey]

123/999, not 123/99

[spice wasp]

I MADE A FUCKING ERROR IN MY WRITING

I WROTE times100 INSTEAD OF times1000

thank you ann in a teacup

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[exotic oracle]

I have a cloud of water particles of various sizes, and I want to calculate their size distribution. These particles falling speed is proportional to their radius squared. I can measure transmittance at any point in the cloud too, which is proportional to both particle size and particle density

[exotic oracle]

Assuming ideal conditions (perfect distribution to start, no air movement no motion to start they instantly hit their terminal velocity etc) is it possible to extract the distribution based on how transmittance changes over time?

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[upper lance]

no clue how to solve this

[upper lance]

part b

how do i write it in the form?

[velvet osprey]

you need to separate out the "integer part" from the fraction

there's also a 'dumb'/bruteforce way to do it if you'd like to hear that

[deep violet]

Isn't part (b) easy?

[velvet osprey]

"easy" is relative

very relative

[upper lance]

probably im just dumb

[deep violet]

Any logical person would figure it out in 10 seconds.

[velvet osprey]

neither a nor b are particularly hard for me or you but there's no way we would expect OP to find this easy given he's shown up here with this problem

[upper lance]

do i need to perform synthetic dibision

[velvet osprey]

do not do this.

[velvet osprey]

thatll get you there yeah

well you don't like need it

[upper lance]

no im sure you can do it mentally

i mean i can do it mentally

but to be fair

im not gonna get marks for that in an exam

A = constant and B = remainder then?

[velvet osprey]

yes

[upper lance]

through synthetic division

so the answer would be 1 and 2

[velvet osprey]

well A is the quotient, which in this case is a constant

[upper lance]

yeah that makes sense

got it ty

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[simple musk]

Let $L_1:\lambda(1,2,-1)+(a,-3,5)$ and $L_2:\lambda(0,-1,2)+(-1,2,a)$. The value of $a\in\mathbb{R}$ for which there exists a plane that contains both $L_1$ and $L_2$ simultaneously is

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[simple musk]

I actually have no idea

I get that the normal of the plane is (1,2,-1) x (0,-1,2)

,w (1,2,-1)x(0,-1,2)

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[simple musk]

3x -2y -z + D = 0

((a,-3,5)-(-1,2,a)) = (a+1,-3-2, 5-a)

,w (a+1,-5,5-a).(3,-2,-1)=0

[potent lotus]

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[calm coral]

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[brittle valve]

try expanding first the given expression

[tall moon]

sub x=a-b y=a+b if your feeling frisky

[unkempt drift]

this is the best approach

you should show what you've tried so far first

we can't help you if you don't show work

[brittle valve]

thats what u get after u expand, u can factor that

[unkempt drift]

,w expand (a-b)^2-7(a^2-b^2)+12(a+b)^2

[potent lotus]

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[unkempt drift]

ah okay

[brittle valve]

then it can still be factored rightt

[unkempt drift]

try (3a .... )(a ..... ) for the quadratic

weird, one sec

[brittle valve]

yea u can still factor it

[unkempt drift]

ah there's a special way you can break 10 into two factors

[brittle valve]

yes 😭

u can use ac method or whatever u r comfy w using in factoring

[unkempt drift]

did you try 5b and 2b?

try the same order, 5b and 2b

see what you get

try expanding (3a + 5b)(a + 2b)

[exotic cosmos]

are we not going the x = a - b and y = a + b route anymore?

😭

[unkempt drift]

yeah I was going to say, that would have been much easier to factor

[exotic cosmos]

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[waxen swan]

helloo would appreciate if someone could check on this….

[brittle valve]

sec^2 -1

is

tan^2 rightt

[unkempt drift]

yes, that's all correct

[brittle valve]

or am i trippin

[unkempt drift]

yeah, the way how I remember it is sec^2 (0) = 1/cos^2(0) = 1

[brittle valve]

oh wait

[unkempt drift]

and tan^2 (0) = 0, so you subtract 1 from sec^2

[brittle valve]

yea

MB

😭

[unkempt drift]

,w d/dx sqrt(2)/2 ln(abs((x + sqrt(x^2 - 2))/sqrt(2))

[potent lotus]

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[unkempt drift]

just checking, hooray

[unkempt drift]

nice work sis!

[waxen swan]

alrr thank you all!!

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[foggy crypt]

Hi chaps. Any chance of help with this question? (actually, I've read the solution - triangle inequality, bound the two guys, etc. - but I don't get the `idea' behind it or how I would think of that, so any insight into that would be appreciated)

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[potent smelt]

.reopen

[calm coral]

✅

[potent smelt]

@foggy crypt you still around?

I'm not entirely convinced of the second part of this prompt.

If we take F to be all rational numbers, for instance, then because the rationals are dense g(x) = 0 for all x, not just the rationals.

[velvet osprey]

Q is not closed

it is stated that F is closed

[potent smelt]

Ah!

I did miss that, thanks for the correction

Critical part that

Anyway, if you're still around, OP, I could use a little bit more information about your understanding, and where the gap in intuition is. Otherwise, I'll just let this time out again.

[foggy crypt]

Yep. Sorry

I just left to get lunch and I didn’t want to… not respond while eating. 😛

So I was going to re-ask later

[velvet osprey]

wait so which part are you interested in

[foggy crypt]

Essentially, I’m understand what the question means (it’s the smallest distance between x and a member of the set) but I’m not sure how one can ‘see’ what to do from there.

[velvet osprey]

continuity, or nonzeroness outside of F?

[foggy crypt]

Ah, continuity

I think nonzeroness is quite easy

I understand my question is quite …abstract.

Initially I tried to use the sequence characterisation

[calm coral]

@foggy crypt Has your question been resolved?

[velvet osprey]

that it is

wait are you allowed to use basic continuity results like

|x| is continuous

[foggy crypt]

Yes, that should not be a problem

[velvet osprey]

oh then it's not too hard is it

[foggy crypt]

I think things like that are fine to assume. Proving them would just market things tedious

[velvet osprey]

do you also know any open set in R decomposes into a union of disjoint open intervals

[foggy crypt]

Hmm… I know that the union of open sets is open. Probably this follows from that

[paper mason]

I don't quite know where I should ask for math questions, but could someone help me with this? "A recipe for oat balls made 32 pieces. Next time, every other oat ball was made with double the diameter, and every other with half the diameter. When the dough was no longer enough to make oat balls with double the diameter, oat balls with half the diameter were made from the rest. How many oat balls did the recipe make then?"
The answer i found on the book was 67 oat balls, but I don't know how it came to the answer

[foggy crypt]

You've got to go to an available channel

In the math help (available) section

[paper mason]

How do i find available channel?

[foggy crypt]

#help-25

that's open

[velvet osprey]

not quite

[foggy crypt]

Oh, maybe the fact they have to be 'intervals' complicates matters

I guess you could make an open set out of open intervals but maybe turning an arbitrary open set into the constituent intervals is harder.

But what relation does that have with the task anyway?

[foggy crypt]

.close

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[novel dock]

I have function
f : [0, 1] → [0, ∞)
integral from 0 to 1 of this function > 0 - true/false
my book says its true but what if f(x) = 0

[plain bone]

One could argue that if f(x) = 0, then the range does not extend to infinity.

I'm not 100% sure about this though

[novel dock]

the function is continuous

chat gpt told me the [0; ∞) means the possible outputs of the funtion are between 0 and ∞

[deft tree]

chatgpt is unreliable

[sweet stag]

can you send a picture of the problem? @novel dock

[novel dock]

you can see integral and function

the problem is in my langueage

[novel dock]

.

[sweet stag]

the whole line

ur cutting it in the middle

[sweet stag]

okey better now

[sweet stag]

The phrase "the range extends to infinity" is not a formal mathematical requirement. Instead, it is often used informally to describe functions whose values can grow arbitrarily large

[sweet stag]

and no it isnt true, consider f(x) = 0 forall x in [0,1]

[novel dock]

Ye i needed to ask because it's in our exam sheets and I guess someone made mistake and put true in answers

Thanks

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[mortal orbit]

if you suppose f is continuous and not the 0 function, THEN the statement will become true

[novel dock]

.reopen

[calm coral]

✅

[novel dock]

yes

but we dont know if f = 0

so it's not always true

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[uneven rose]

I tried to do it for some small numbers but it doesn t take my anywhere

And I need a general term formula

[remote mural]

i like your writing

[uneven rose]

It s from romanian math gazette so I translated it on sheet

Can you help me with it?

[remote mural]

are you from romania?

[uneven rose]

Yes

[remote mural]

dm me

ill help you

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@uneven rose Has your question been resolved?

[uneven rose]

<@&286206848099549185>

[ornate meadow]

umm, could you maybe type that out?

its a bit difficult to read

[uneven rose]

There (-1)^n is a power

Not product

[ornate meadow]

cant you just set a1 equal to 1?

then a1 and a2 are consecutive integers

[uneven rose]

Well yes that is a solution

But how I demonstrate it s the only one

[ornate meadow]

hmm from the wording the question doesnt require you to do that

but you can combine all the terms

to the previous number

$a_{n+1} = \frac{1}{a_1}+ \frac{a_2}{2}+...+\left(\frac{a_n}{n}\right)^{(-1)^n}=a_{n}+\left(\frac{a_n}{n}\right)^{(-1)^n}$

[potent lotus]

DeadTomato

[ornate meadow]

and then look at the two cases, where n is even or odd

[uneven rose]

If it s even than (a)n+1 is bigger than (a)n

If it s odd it s the opposite

[ornate meadow]

also i misread the question a bit 😅, yeah there are more solutions

[ornate meadow]

the sequence is always increasing, as its a sum of positive real numbers

[uneven rose]

Ah yeah sorry

I was thinking about something else

Than what should I observe?

[ornate meadow]

try to prove first, that if n is even and an an integer, the next one cant be

[uneven rose]

Can t be what?

An integer?

[ornate meadow]

yeah

[uneven rose]

Well yeah because n is smaller than (a)n

And it will be something under 1 plus something integer

[ornate meadow]

you dont know that

[uneven rose]

They are positive

[ornate meadow]

take a1 = 1, a2 = 1, a3 = 3/2

none of the a_ns are bigger than n

[uneven rose]

Well they are small

So they are out of the question

I think the bigger ones work

[ornate meadow]

why are they out of the question

isnt that sequence even an example for ones that work?

$a_{n+1} = \frac{1}{a_1}+ \frac{a_2}{2}+...+\left(\frac{a_n}{n}\right)^{(-1)^n}=a_{n}+\left(\frac{a_n}{n}\right)^{(-1)^n}$

[uneven rose]

So I think that an is bigger than n for n>=4

[potent lotus]

DeadTomato

[ornate meadow]

look at the term on the right

how can you manipulate it when n is even

[uneven rose]

The fact that n is even shows only the fact that n is the numitor

[ornate meadow]

you mean denominator

[uneven rose]

Yeah sorry

English is not my first

[ornate meadow]

yeah so if n is even and with some manipulation $a_{n+1}=a_n+\frac{a_n}{n} = a_n(1+\frac{1}{n})$

[uneven rose]

Yes

[potent lotus]

DeadTomato

[uneven rose]

Yes

[ornate meadow]

now

can a_n+1 be an integer?

[uneven rose]

No

If n is not 1

În a1 case

[ornate meadow]

yeah but we said n is even

[uneven rose]

Ah yeah sorry

Than no

[ornate meadow]

yeah

now try to see if you can make some claims about when n is odd too

[uneven rose]

Still the next isn t integer

Cause it s 1/an ×(an+n)

=(a)n+1

Dang

Both cases are not good

[ornate meadow]

not quite

[uneven rose]

Sorry

I just realized

Its an squared

Right?

It s the same tho

[ornate meadow]

yeah but you dont need to simplify it to see it

just look at $a_n+\frac{n}{a_n}$

[potent lotus]

DeadTomato

[uneven rose]

Anyways that means the only case is a1 a2?

[ornate meadow]

you suppose an is already an integer, when is that term also an integer

[uneven rose]

Yes but it can t

So a1 a2 is the only right?

[ornate meadow]

hmm but we found an example so somehow it can

[ornate meadow]

what condition can you extract from that term for it to be an integer

[uneven rose]

an being 1?

[ornate meadow]

no but an has to divide n

[uneven rose]

Ah yeah true

So there are more strings?

Bro tell me

Wouldn t that mean in the other case n has to divide an

?

[ornate meadow]

yes but we showed that thats not possible

[uneven rose]

So the only problem is in this case?

[ornate meadow]

yeah

[uneven rose]

And how i finish the problem?

[ornate meadow]

wel now the question is, when can we construct back a sequence, from a chosen n and an, where we let an divide n

well also i just realized i made an oopsie xd

in fact even and odd dont quite differ

[uneven rose]

Anyways the problem is solved in my opinion

[ornate meadow]

it is indeed possible for even to have the next thing be an integer

like you said you just have to have n dividing an for that

[uneven rose]

So the case are a1 a2

And the ones in odd case where an has the form k×n

[ornate meadow]

if an is even, it has the form k*n, if its odd, n has the form k*an

[uneven rose]

Good thanks man

How do i end the conversation?

I mean the channel

@ornate meadow

@calm coral

[wintry geode]

.close

[swift dragon]

!done

[calm coral]

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

[swift dragon]

.close

[calm coral]

Channel closed

Closed by @swift dragon

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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[deft tree]

Find functions f(x) and g(x) so the given function can be expressed as f(g(x)). p(x) = (2x3 + 1) / (x3– 1)

[deft tree]

g(x) equals x^3

i can solve for f(x)

but is there an easier way

than just thinking abouit for like 3 minutes straight

[swift dragon]

$f(g(x))\cdot p(x)=\frac{2x^3+1}{x^3-1}$?

[potent lotus]

Bonk

[deft tree]

ok

so now how do i solve for f(x)

fast

instead of just

u know

[mystic burrow]

How are you doing it right now?

[deft tree]

just thinking about it

[mystic burrow]

Okay, so what's your method?

[swift dragon]

if you have g(x)=x^3, then what could f(x) be?

[deft tree]

2x+1/x-1

[swift dragon]

BRACKETS

[deft tree]

bruh

[mystic burrow]

Try plugging in g(x)=x^3 into the f(x).

[deft tree]

thinking about it

[mystic burrow]

What do you have now?

[deft tree]

what why

its gonna make

[deft tree]

tthis

[mystic burrow]

Because youre trying to find fx??

[deft tree]

but i already did?

[mystic burrow]

$f(x^3)\cdot p(x)=\frac{2x^3+1}{x^3-1}$

[potent lotus]

pratikitiki

[deft tree]

(2x^3 + 1) / (x^3– 1)

[mystic burrow]

Now, from this equation, f(x) could either be 1/(x-1) or 2x+1.

Just by thinking about it.

[deft tree]

huh

no

wait what

(2x+1)/(x-1)

thats f(x)

g(x) is x^3

[swift dragon]

now what is p(x)?

||1||

[deft tree]

(2x^3 + 1) / (x^3– 1)

.

[swift dragon]

thats p(x)?

[deft tree]

yeash

is it mnt?

not?

[swift dragon]

i thought that was f(g(x))

[deft tree]

well yeah

it is

fgx is px

we have to find gx and fx so that when we plug it into fgx it makes px

[swift dragon]

i thought that this was the question

ahhh

okay

[deft tree]

no

so yeah i was asking

[swift dragon]

so yeah then youre done

[deft tree]

what

jpw

no i know

but

[swift dragon]

$f(x)=\frac{2x+1}{x-1}$ and $g(x)=x^3$

[potent lotus]

Bonk

[swift dragon]

then whats your question now?

[deft tree]

i was asking how to find f(x)

the official way

because my mehtod

is too slow

[swift dragon]

well the issue is

there are infinitely many solutions

[deft tree]

hm

ok

[swift dragon]

like, if oyu take g(x)=x

then f(x)=(2x^3+1)/(x^3-1)

also works

[deft tree]

oh what

[deft tree]

wait

how

because in the denom it is x^3-1

[swift dragon]

apologies

ive corrected it now

[deft tree]

kk

[swift dragon]

and similiarly you can take g(x)=x^2 and f(x)=(2x^(3/2)+1)/(x^(3/2)-1)

etc etc

[deft tree]

hm ok

so

what ur saying is

there is no fast way?

[swift dragon]

i think the fastest way would be to look at repetitions

you noticed x^3 appearing twice

so that could be a possible g(x)

[deft tree]

so i put gx as x^3

oh oka

well in this: h(x) = √(2x – 1) /√(3x + 4) i see √ comes twice

so how would i do that

√x?

as gx

?

[swift dragon]

like if i give you $h(x)=\frac{(4x^2-3)^2}{3+(4x^2-3)}$

[potent lotus]

Bonk

[swift dragon]

you can see the 4x^2-3 is repeated twice, so thats a good start for g(x)

[deft tree]

ohhhh

[swift dragon]

this is a bit trickier

and i dont think there is a nice and clean f(g(x)) for that

$\sqrt{\frac{2x-1}{3x+4}}$

[potent lotus]

Bonk

[swift dragon]

there isnt rly a good f(g(x))

[swift dragon]

but here there clearly is

[deft tree]

yeah for that i just put √2x-1 anf for fx i put x/√3x+4

is that corrct

[swift dragon]

no

[deft tree]

bruh

[swift dragon]

then you would get $f(g(x))=f(\sqrt{2x-1})=\frac{\sqrt{2x-1}}{\sqrt{3\sqrt{2x-1}}+4}$

[potent lotus]

Bonk

[swift dragon]

see the issue?

[deft tree]

ohhh

yeah

what grade are u in?

[swift dragon]

can you determine g(x) and f(x) here?

[swift dragon]

im in uni

[deft tree]

yeah

[deft tree]

ohh

[deft tree]

should i

or

like

idk what ur asking

like ar eu asking am i able to

or are u telling me to do it

[swift dragon]

im giving you a question

[deft tree]

oh ok

[swift dragon]

thats similar to what youre doing

[deft tree]

yes

i can

[swift dragon]

(just extra practice i suppose)

[deft tree]

gx =

what

[swift dragon]

hm?

[deft tree]

gx= 4x^2-3 and fx =x^2/3+x

[swift dragon]

BRACKETS

[deft tree]

lol

[swift dragon]

(can you add brackets pls? <3)

[deft tree]

fine 😭

g(x) = (4x^2 - 3) and f(x) = (x)^2/(3 + x)

holy

that took so long

[swift dragon]

you dont have to be that excessive lol

the only issue was f(x)

[deft tree]

what

aw

[swift dragon]

you wrote: $x^2/3+x$ which is interpreted as $\frac{x^2}{3}+x$

[deft tree]

uh

[swift dragon]

texit is dying idk

[potent lotus]

Bonk

[deft tree]

ok

[swift dragon]

but what you meant was: $x^2/(3+x)$ which is equal to $\frac{x^2}{3+x}$

[deft tree]

wait

[potent lotus]

Bonk

[swift dragon]

see the problem?

[deft tree]

ohhh

so i got it correct right

in the second one

?

[swift dragon]

yeah

but you did it excessively

[deft tree]

with the ()?

[swift dragon]

(4x^2-3)=4x^2-3

[swift dragon]

yeah

[deft tree]

oh lol

[swift dragon]

anyway

im gonna give you a very similar one

[deft tree]

okie

[swift dragon]

i wanna see if you can spot it

$h(x)=\frac{(4x^2-3)^2}{4x^2}$

[potent lotus]

Bonk

[swift dragon]

what is g(x) and f(x)?

[deft tree]

hm

g(x) = 4x^2 and f(x) = ((x-3)^2)/x

pls be correct

[swift dragon]

i suppose you can do that aswell

but you can take the exact same g(x) and f(x) as you did before

g(x)=4x^2-3 and f(x)=x^2/(3+x)

[deft tree]

what.

[swift dragon]

$h(x)=\frac{(4x^2-3)^2}{4x^2}=\frac{(4x^2-3)^2}{4x^2-3+3}=\frac{(4x^2-3)^2}{(4x^2-3)+3}$

[potent lotus]

Bonk

[swift dragon]

see it now?

[deft tree]

oh

oh wow

thats kinda cool

okay i think im good now

thank you so much bonk

for taking the time

to help my dumb ahh

.close

[calm coral]

Channel closed

Closed by @deft tree

Use .reopen if this was a mistake.

[swift dragon]

neat trick huh

being able to spot stuff like this is a really useful ability

[deft tree]

yeah

.reopen

[calm coral]

✅

[deft tree]

ahem.

how would i do this one

h(x) = √(2x – 1) /√(3x + 4)

[swift dragon]

i dont think there is a nice way

!original

[calm coral]

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.

[swift dragon]

i think they want f(x)=sqrt(x)

then what could g(x) be

[deft tree]

um

lemme cook one sec

[swift dragon]

thats g(x)?

[deft tree]

waity what

no

it cant be

[swift dragon]

and then whats f(x)?

[deft tree]

ye thats gx

[swift dragon]

you sure?

[deft tree]

sqroot x

can be fx

[swift dragon]

no

you had it right the first time

be confident in your answer

[deft tree]

k

so its done?

wait but

[swift dragon]

g(x)=(2x-1)/(3x+4)

f(x)=sqrt(x)

[deft tree]

would it not just all cancel out

its fine

[swift dragon]

wdym?

no, explain

[deft tree]

well

if that was fx

i mean gx

and if u substitutes that into sqroot x

OHHH

What am i doing

yeah

i see

okay

LAST THING I SWEAR

Find f(12) for the given piecewise function:
f(x) = {-18x + 20, if x < 19}
{-16x2
, if x ≥ 19}

for this

do u just substitite

[swift dragon]

!original pls

[calm coral]

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.

[deft tree]

that is orignianal

[swift dragon]

can you send a photo

is what im asking for

[deft tree]

The middle one

is it just [-18(12) +20]?

what about the stuff with x<19

i gtg soon

@swift dragon sorry

[swift dragon]

yup

because you are in the case x<19

f(12)=-18*12+20

[deft tree]

oh and if ur not

then no answer?

[swift dragon]

with this function youd be in the other case

if x=25, then you are in the -16x^2 part

[deft tree]

wdym im in that [part

[swift dragon]

look at the case

x=25

25>=19

[deft tree]

then its in the second one

[swift dragon]

thus you look at -16x^2

f(25)=-16*(25)^2

[deft tree]

hm sop in this case x=12 so its in the first case

[swift dragon]

yup

now

slightly tricky question

what about f(19)?

[deft tree]

second

cmon now

im not that dumb

so u leave the otther one aslone?

[swift dragon]

[deft tree]

💀

[swift dragon]

yeah you can only be in one case

[deft tree]

i see

[swift dragon]

since function cannot have two values at one x

[deft tree]

oh

ok

ok

im finally done

with ALL my questions

[swift dragon]

nice

[deft tree]

god i just realized i have been talking to a stranger about math for one hour

[swift dragon]

😄

[deft tree]

can i add u

.

wow that felt awkward

[swift dragon]

read the bio

[deft tree]

oh lol

okay thank you so much bonk

for helping mew

[swift dragon]

🙂

[deft tree]

me

okay cya

[swift dragon]

np

[deft tree]

.close

[calm coral]

Channel closed

Closed by @deft tree

Use .reopen if this was a mistake.

[swift dragon]

open a channel anytime 🙂

[calm coral]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[gusty lance]

bit confused on whats going on after the teacher differentiates with respect to time

[gusty lance]

can someone explain this to me like im 5

[calm coral]

@gusty lance Has your question been resolved?

[keen flare]

after $\frac{dy}{dt} = 8\cos{\theta} \cdot \frac{d\theta}{dt}$?

[potent lotus]

BuilderDolphin

[gusty lance]

no like getting to that step

[keen flare]

oh from the y =

[gusty lance]

when she says differentiating y = 8sinx

yeah

[keen flare]

do you know implicit differentiation

[gusty lance]

yep

but

im kinda shaky on the working of it here

usually, i would only be dealing with one variable

like

with the implicit differentiation of for ex sin(x+y) = 1

i would be familiar with dy/dx

and so that way it made sense since for x we would js take the deriv normally

but this time its like 2 variables? idk how to explain

[keen flare]

this is just a more universal application of implicit differentiation

[swift dragon]

$\dv{y}{t}$

[potent lotus]

Bonk

[keen flare]

oh i did not know that

thanks

[keen flare]

well you're trying to find the derivative of the variables with respect to time

which basically means that you perform the chain rule on the variables that you differentiate that are not t (im sort of bad at explaining that)

in this case y and theta

y and theta are treated as implicit functions of t

[keen flare]

eg in this case $y = 8\sin{\theta}$
treat $y$ and $\theta$ as functions of t:
$y(t) = 8\sin{({\theta}(t))}$
differentiate:
$y'(t) = 8\cos{({\theta}(t))} \cdot {\theta}'(t)$
rewrite:
$\dv{y}{t} = 8\cos{\theta} \cdot \dv{\theta}{t}$

[potent lotus]

BuilderDolphin

[gusty lance]

so we treat them as if they are t?

[keen flare]

functions of t

because y and theta change with respect to t

[gusty lance]

can you give me a simpler exmaple 😭

[keen flare]

uh

ok

[gusty lance]

cuz from what im understanding, when its like a rate we have to differntiate every variable with respect to time, so every variable is being implicitly differentiated?

[keen flare]

let's say we have $1 - 2y^2 = x^2$ and are trying to differentiate with respect to x. we know that y changes based on x so we can treat it as a function of x, let's call if $f(x)$: $1 - 2(f(x))^2 = x^2$. Differentiating this, we get: $-4f(x) \cdot f'(x) = 2x$, which is the same as $-4y \cdot \dv{y}{x} = 2x$, which is how it would look like if you did implicit differentiation normally

[potent lotus]

BuilderDolphin

[keen flare]

well when we're differentiating with respect to a variable (t) we are implying that the other variables change based on the variable (t)

so we can just rewrite it as a function of t

[gusty lance]

okok i see

[gusty lance]

right okk

that acc makes much more sense now

thank you

.close

[calm coral]

Channel closed

Closed by @gusty lance

Use .reopen if this was a mistake.

[calm coral]

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[vestal tree]

H

[vestal tree]

Does this

Server help with business economy Matt

Matt

MATH

[pallid halo]

if it's a math question then probably

[vestal tree]

It is math but like it had some economic terms in it

[pallid halo]

yea people ask that sort of question here

[vestal tree]

Okay so I’m having a hard time using the formula fixed costs/(cost per product - variable costs)=the break even point

Like I have the formula but everytime I’m supposed to use it my brain goes kappor

Wait I’ll find a question with it in it

Okay I found one

How do I use that formula in this question

<@&286206848099549185> anyone?😭

[sacred trellis]

I'll help

[vestal tree]

THANK YOU

If I’m gonna be honest here the first question confuses me too cause all the costs are fixed…

The income is only variable

[sacred trellis]

Variable include ad & transaction fee

Everything else is fixed

50 per ad

60k

= 60000×50

=3000000 sek

[sacred trellis]

Minus total variable cost & you get profit

Break even= (fixed cost/ price per ad)- variable cost per ad

[calm coral]

@vestal tree Has your question been resolved?

[vestal tree]

Ohhhhhh

Thanks so much 😭

[calm coral]

Channel closed

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[mental tendon]

what happened to the b

[main oak]

lhopital

[mental tendon]

How

[eternal shard]

e^b derivative is e^b
b derivative is 1

-1 vanishes

[mental tendon]

O wait

I thought this was before lhopital

Thanks

.close

[calm coral]

Channel closed

Closed by @mental tendon

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as this action is irreversible, and this channel may abruptly lock.

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[calm coral]

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[ripe coral]

i forgot what to do with the amplitude

[ripe coral]

do i just move it 3 units up

oh yea i do

.close

[calm coral]

Channel closed

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[marble holly]

😂

[tranquil wasp]

aw you gotta give em that sohcahtuah

[calm coral]

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[covert orchid]

Assume f(x) is a continuous function on $[x_0, \infty)$ and differentiable on $(x_0, \infty)$ such that $f'(x) \geq a > 0$ for every $x > x_0$, prove that $\lim_{x \to \infty}f(x) = \infty$ \
My proof:
Let's take the segment $[x_0, x]$ the function is continuous and differentiable on $[x_0, x]$ therefore using Lagrange mean value theorem there exists a point $c \in (x_0, x)$ such that $f'(c) = \frac{f(x)-f(x_0)}{x-x_0}$ from the given that $f'(x) \geq a > 0$ for every $x > x_0$ we get $f'(c) = \frac{f(x)- f(x_0)}{x-x_0} \geq a$ therefore $f(x) \geq a(x-x_0)+f(x_0)$ we got a linear function with a positive gradienr and therefore it diverges to infinity, using pizza theorem we get $\lim_{x \to \infty}f(x)= \infty$ \
The answer sheet:
Let $x_0 < a_n$ be an arbitrary sequence that diverges to infinity. Therefore for every natural b in $[x_0, a_n]$ we can use LMVT, therefore there exists $c_n \in (x_0,a_n)$ such that $\frac{f(a_n)-f(x_0)}{a_n-x_0} = f'(c) \geq a > 0$. Therefore $f(a_n) \geq a(a_n-x_0) + f(x_0)$ using "pizza" and infinite limits arithemtics we get $\lim_{n \to \infty}f(a_n)= \infty$. We showed that for every sequence $a_n > x_0$ that diverges to infinity we get $\lim_{n \to \infty}f(a_n)= \infty$. Using heine theorem we get $\lim_{x \to \infty}f(x)= \infty$

[potent lotus]

prograce

[covert orchid]

My proof is different from the answer sheet, is it wrong? If so then why? Also I don't understand why they had to use sequences in the answer, also note that the answer sheet is not formal writing

[ornate flint]

you cant disentiable at x0

see the range

Other than that, in the open section you can use MVT

[quiet swan]

as noted already you need MVT only need differentiability at (x_0,x), it doesn't matter for your proof but you need to fix the part where you said it is differentiable on [x_0,x].

what is pizza's theorem?

[ornate flint]

ps. think Proof by contradiction

[quiet swan]

the definition of limit should just give lim_{x\to \infty}f(x)=\infty, I sort of don't get why the official solution uses the "every sequence thing"

Is your definition the following: We say $\lim_{x\to\infty} f(x)=\infty $ if for all $C\in \mathbb R$ there exists some $N\in \mathbb R$ such that for all $x>N$ we have $f(x)>C$.

[potent lotus]

qwertytrewq

[covert orchid]

I think It's a typo It's supposed to be differentiable on (x0, x)
Pizza theorem is when you havetwo function f(x)<=g(x) and you know that f(x) diverges to infinity then you conclude immediately that g(x) also diverges to infinity

[covert orchid]

I don't understand what is disentiable ?

[quiet swan]

differentiate is what they meant i think

[covert orchid]

Ohh, then yea.

[covert orchid]

That's what I'm asking about and I really reallly wanna know😭

[quiet swan]

yeah yours looks fine

theirs looks weird lol

your professor might be speedrunning solution write up that they missed it

[covert orchid]

They use this method a lott and I really wanna know whyy why can't I just not use sequences and do what I did

[covert orchid]

Lmaoo

[quiet swan]

well, is pizza thing stated for sequences or functions?

if it is for sequences then theirs makes more sense

[covert orchid]

I used it for functions but let me recheck

That might be why

[quiet swan]

honestly the proof is pretty short anyways... you can just write your own pizza function lemma

[covert orchid]

It is indeed for functions

[quiet swan]

i see math and symbols 😵‍💫

[covert orchid]

Hahaha

Focus on f(x)<=g(x) and limf(x)=inf then limg(x)=inf

[quiet swan]

yeah you should be fine then you should ask your prof if they allow one to use the function version

who knows, maybe they are picky

[covert orchid]

Okay then I will tomorrow

Ty for checking it!

.close

[calm coral]

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[quiet swan]

nw

[quiet swan]

oh what that is limit to a not to infinity

you might also want to confirm the infinity case

[covert orchid]

Ohhh

[quiet swan]

mb i missed that lol

[covert orchid]

It says "if f,g are functuions that are defined on a segemnt of point a, given f(x)<=g(x), if limx->af(x)=inf then limx->ag(x)=inf"
I guess if the functions are defined on R then it also works for limit to infinity no ? I'll try to prove it anyways

[quiet swan]

it does work for infinity... you need f to be defined on some [c, infty)

but the proof is the same lol

[covert orchid]

Yeahh in my questions the funcrions are defined

[quiet swan]

yeah so it is applicable but ig they probably didn't do the infinity version

[covert orchid]

They did tho

What the they used pizza theorem on sequences in their answer

I mean it should works on sequences too

Ah it exists in sequences too

[quiet swan]

my guess is that they did pizza on sequences, and proved it for functions converging to a, but not functions going to infinity

idk tho you should check

[covert orchid]

Alr will do🫡

[calm coral]

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[hot olive]

why is the lines not opposite

directioms

like

/c

c/

[velvet osprey]

solve that equation for y

and you will see y = -4/(x-1) eventually, or an equivalent form

+4/(x-1) would give you curves NE and SW of the asymptote intersection point like you're describing

[hot olive]

i divide y-xy yby y and got y(x-1)=4 so divide both sides by x-1

wait

itw ould be -x+ 1

so jsut gotta make bottom postive and top ends up becoming negacive?

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[calm coral]

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[calm coral]

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[knotty lake]

I'm certainly making this harder than it has to be

Where did l go wrong

I'm trying to find X y and z

[versed trout]

this is the system?

[knotty lake]

Yes

[versed trout]

have u learnt gaussian elimination

[knotty lake]

NO clue what that is

[versed trout]

okay u can do this then

isolate x from the first equation

so express it in terms of y and z

then sub it into the x of the second equation

then isolate y (expressing only in terms of z)

[knotty lake]

thats what i did

thats that entire process

[versed trout]

yeah thats the next simplest method

[knotty lake]

everything there is that

[versed trout]

gaussian elimination is the easiest method to do this

[knotty lake]

just trying to get it in z

im sure of it and i think i can look that up by myself but like

i feel like there should not be this much work needing to be done

i feel like im making extra steps where i dont need to

[versed trout]

u could proceed with standard elimination as well

eliminating x first

then eliminating y

[knotty lake]

elimination i dont know too well

and it eventually succumbs to being substitution as well right

[versed trout]

can u use calculators then

[knotty lake]

not like computer algebra calculators

just like ur typical ti 84

[versed trout]

yeah a normal calculator

[knotty lake]

which i dont have

[versed trout]

yeah ur ti can do it

oh

[knotty lake]

its about 3 hours northeast of me atp

lucky me

[versed trout]

well then substitution is the best method here

[knotty lake]

thats what ive been doing though

should i just like

bite the bullet and keep going where im going

[versed trout]

yes

[knotty lake]

if so i dont know how to comfortably solve that -45x = 35 + 6/5

[versed trout]

https://www.youtube.com/watch?v=eDb6iugi6Uk&t=52s

do learn this though. it will be useful later and useful for this as well

its very simple

[knotty lake]

i believe it

simpler than this

bravo ill figure it out eventually

.close

[calm coral]

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