help-0

[vivid badge]

i lost the + c

haha

[coral flower]

Let me see

Ok the problem is

U dont multiply 1/2 with the whole thing

[vivid badge]

ohhhhh

that means that

[coral flower]

Yes

[vivid badge]

ohhh

thats new for me (im begginer) thanks for teaching me this haha

[coral flower]

Just think of it like this

[vivid badge]

with that i should cancel 2/-4 and getting the -2 im looking for

great

[vivid badge]

ohhh

[coral flower]

Those are brackets btw

[vivid badge]

makes sense

niceeeeeeeeeeee

thanks jtp

.close

[lone heart]

Channel closed

Closed by @vivid badge

Use .reopen if this was a mistake.

[coral flower]

Np

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rugged cosmos]

why is k = -2 and not 2? where is the negation coming into play?

[north heart]

the general eqn is (x-h)^2+(y-k)^2=r^2

so here -k=2

so k=-2

[rugged cosmos]

-k=2?

i dont really understand that phrasing

a negative before the variable name

i understand k = -2

you mean --2 = -2?

[median oar]

What’s wrong with -k = 2

[rugged cosmos]

-k(where k is -2) = --2

[median oar]

That’s not right

[north heart]

i am comparing the standard eqn with your eqn

like (y-k)^2=(y+2)^2

[rugged cosmos]

is there implicit simplicification happening here or something

why do you go from (y-k)^2 to y+2

makes sense if k is a negative already. then double negative is positive

but is k always guaranteed to be negative in this equation

[north heart]

nope if the eqn had y-2 then k would would have been postive

[rugged cosmos]

can you say why -k goes to 2 exactly

i feel like an explanatory step is being skipped

like some translation is happening

ok i think i get it. You are obvserving that the standard form is x - something, y - something. So the fact that the equation they show us is y + something, means that the original something must have been a negative, making y - something a double negative thus y + more simple

[north heart]

exactly

i am sorry but i could not explain into words as i did not understand what you meant at first

[rugged cosmos]

no worries you helped me get there

[north heart]

aight

[rugged cosmos]

ty

.close

[lone heart]

Channel closed

Closed by @rugged cosmos

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[magic arch]

Hi
I know the differential equation is separable, but we were asked to solve it as if it was a linear non separable one

But we have no idea where it went wrong

(the integral at the end is incorrect)

[north heart]

i am dumb i differentiate it

mb

[alpine sable]

can you write out the ode your handwriting isnt clear

like the first line

[magic arch]

y' - 3x²y = 6x²

[alpine sable]

ok

[magic arch]

I know the solution, just have no clue why solving it this way leads to nowhere

y = -2 + e^(x³)

[lone heart]

@magic arch Has your question been resolved?

[subtle light]

this differential equation is not separable, you have to use the integration factor

[magic arch]

it is tho

[ocean seal]

y0shi

[subtle light]

if you move the y term to the other side

[magic arch]

yea

[subtle light]

there is no way to factor out a y

[magic arch]

so it is

[subtle light]

nope

look $y’=6x^2+3x^2y$

[ocean seal]

y0shi

[magic arch]

y' = 3x^2 (1+y) ?

[subtle light]

well it’ll be 2+y instead

[magic arch]

y mb

[subtle light]

yeah i see now then

this isn’t what you did above tho

[magic arch]

yes

as ive said

i dont care about the end result

my question is where did it go wrong

it should be solveable the way we started

[subtle light]

your C’(x) is wrong

and honestly i’m not sure if you can solve it using homogeneous and particular solution, but you probably can

[magic arch]

is it?

[ocean seal]

y0shi

[subtle light]

they are being multiplied together

not added together

[magic arch]

bruh

my friend di it btw

but w/e

[subtle light]

oh my

no way they integrated e^x^3

[magic arch]

yea

they tried

[subtle light]

yeah that’s not how that works

maybe it’ll work if you do it properly idk

[magic arch]

oohh

yes

cuz

f(g) * g' has a primitive

so

6x^2 / e ^(x^3)

where

x^3 = g

and we get
-2 * e ^(-(x^3))

[lone heart]

@magic arch Has your question been resolved?

Channel closed

Closed by @magic arch

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[true dagger]

Hello i dont get this lines

[kindred nexus]

just put it in chat gpt or sum

its easy

[long axle]

the solution to the box will be the answer on the line u need to go to

for example if the answer to the first box of 6, you would go down

if it was 5, you'd go right

[true dagger]

the answer is 2 thats why i dont get it

[long axle]

u need to put the numbers from least to greatest

only then can you find median

[true dagger]

so its 5?

[long axle]

no

what did u get upon putting them from least to greatest

[true dagger]

oh its 6

lol

[true dagger]

why does it has to be least to greatest?

[long axle]

these metrics we use, mean, median mode, are all about finding the "center" or "average" values

if we had a sequence

9, 9, 2, 9, 9

the median is 2

however, thats not helpful, because everything else is a 9, thus the 2 is a bad representation of the actual sample

or if we have

10000, 3242, 12342, 9, 3282, 8779, 87721

the median is 9

"is 9"

but we havent ordered it from least to greatest

the median is supposed to give you a good "center" read on the sample

9 doesnt do that at all

[true dagger]

in range does it has to be least to greater also?

[long axle]

range is just greatest number - least number

[long axle]

for this, range would be 9-2

[long axle]

for this itd be 87721 - 9

[true dagger]

alr i get it now

thank you

.close

[lone heart]

Channel closed

Closed by @true dagger

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lime garden]

hey guys, i'm currently learning proof by induction and the video my uni gave me confused me. i've watched a couple of other videos now and i think i understand it a lot better, however the original video confused me. when we're aiming to prove that LHS < RHS for k+1, what's with the additional steps, is it not just enough to say that because k >= 5, that for all cases (k+1)^2 < 2^(k+1)

[lime garden]

this is the video if you wish to take a proper look, i've timestamped the start of the question https://youtu.be/So-Sd5K7PgE?si=SE7tEsctEtOSX7Ob&t=367

[royal grail]

it's a terrible wording, instead of saying suppose k>=5 and P(k) true, they should suppose P(k) true for some k>=5 (but not necessarily all)

[odd dome]

For the case of considering k, you actually consider it to be true, assuming it to be true then gives you the motivation (normally using algebra) to prove that "Well if the number before it is infact true, then this case has to be true" That is the motivation for everything after the green text

[lime garden]

ok that explains it

[lime garden]

i believe they're more specific in the video

[lime garden]

why does it suddenly go from k^2 +2k +1 to k^2+2k+k

i know it's obviously greater

but why go in that direction

[odd dome]

Effectively you are supposing that if you can prove that something is true if it's previous step is true this means that everything up "k" is true

[lime garden]

yea i remember that about induction

[odd dome]

It's an inequality for the convenience of proof since we know that k^2 +2k + 1 < k^2+2k+k and that k^2+2k+k = 2^(k+1) which is of the same form of the required proof

Effectively this inequality just proves your statement

[lime garden]

okay so it's shifting it into the same form as the inequality in order to show that it's greater

[odd dome]

This statement is trivially true and it allows us to prove the more general statment that:
(k+1)^2 < 2^(k+1)

[lime garden]

okay gotcha

[odd dome]

Which is what you are required to prove

[lime garden]

is this particularly just for inequalities

i've practiced a lot of others and the ones making both sides equal seems far esaier

[odd dome]

This property is most useful with inequalities

[lime garden]

okay gotcha

thank you :)

is there any chance you could quickly look over one of my proofs

[odd dome]

Sure

[lime garden]

seperate proof by contradiction just learning proofs and not too confident atm

[odd dome]

Send it through

[lime garden]

most other people i spoke to chose to expand the quadratic out and then determine the parity of each part of it so you end up with like even + odd = even contradiction

not sure how much my proof relates to z being irrational in the original statement tho

[odd dome]

For this contradiction you really just want to assume that z is rational for odd a, b and c and then prove that this cannot work

[lime garden]

okay so that would be by expanding further before rearranging

then determining parity of each part and concluding that for rational z, a, b and/or c must be even

?

[odd dome]

You can certainly do that

It's normally that a and b are of the same parity and that c is odd

Just as a hint

[lime garden]

yea ok that was my original plan but i found the implication thing messing around

ahh okay cool

[lime garden]

from a marking perspective do you think this proof is inherently wrong?

[odd dome]

And just remember, if x is even x^2 is even and if x is odd x^2 is odd

[lime garden]

yea cheers

[odd dome]

Certainly not, rather it may miss what the question expects you to do, which will see you lose marks irrespective of wether or not your proof is true

[lime garden]

ok gotcha

[odd dome]

I ran into the same issue in 1st year with vector space proofs, I would prove it say using properties of triangles, rather than properties of vectors because the question constructed a triangle, and this would see me be decucted even if my mathematics and wording were perfectly adequite for a perfect score

[lime garden]

alright will keep that in mind

will redo it if i get time but probs not

thank you so much for your time

i'll probs be back in like 5 with another induction question

[odd dome]

Feel free to @' me

[lime garden]

thanks man

would DM work easier?

[odd dome]

sure

[lone heart]

@lime garden Has your question been resolved?

Channel closed

Closed by @lime garden

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[near hollow]

what have u tried

[fleet grove]

i have integrated it to x^4+ k/2 x^2 +c

[near hollow]

not quite

$\int 4/x^3+kx :dx=\int 4x^{-3}+kx: dx=\frac{4x^{-2}}{-2}+\frac{kx^2}{2}+c$

[ocean seal]

Max

[fleet grove]

ah ok

@near hollow i am not sure how i'm ment to get the vale of k tho

oh weight its not asking for me to

sorry silly

thanks

.close

[lone heart]

Channel closed

Closed by @fleet grove

Use .reopen if this was a mistake.

[near hollow]

@fleet grove

.reopen

[lone heart]

✅

[fleet grove]

yes?

[near hollow]

there is no information

so you cannot find the value of c or k

[near hollow]

okay

just add the bounds to the integral

$\int_0^2 4/x^2+kx:dx = \left[\frac{4x^{-2}}{-2}+\frac{kx^2}{2}\right]_{0.5}^2=8$

[ocean seal]

Max

[fleet grove]

then you sub in and take away thing

[near hollow]

yea

[fleet grove]

is it the top number - bottem?

so like the 2 subbed in squar brakets - 0.5 subbed in squar brakets

[near hollow]

yea

[fleet grove]

i subed it in and now im confused a bit sorry

[royal grail]

what did you get after you made the substitution

[fleet grove]

I did the square bracket thing with subbing in the 2 and 0.5

[lone heart]

@fleet grove Has your question been resolved?

[royal grail]

yeah what did you get

preferably show us your working

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[magic falcon]

hello

my current lesson is about using derivatives for related rates and optimization

im having a hard time with it esp on this problem, could someone guide me through the solution

[lone heart]

@magic falcon Has your question been resolved?

[sly ocean]

hey, im new here so i dont really know how to proceed, but imma try my best to help

I think u are supposed to develop a function related to the volume of the cone and derivate it to obtain the speed and acceleration at which the water rises, and the same for the radius of the balloon

[magic falcon]

this is where i ended and i dont really know how to proceed from here

i think i should equate negative dv1/dt to dv2/dt

[sly ocean]

Ok, so the problem is asking you to derivate w/ respect of t, which u dont have in the equation of volume

But you can put both the radius and height as a function of time using the information the problem gives

Bc in this case height and radius are not constant

I dont know if im properly explaining myself 😅

[crisp nacelle]

Is it okay to explain Calculus to a 5th grader?

[sly ocean]

I mean, this is what this server is for isnt it?

[magic falcon]

r1 would be 2
h would be 6

I don’t have an idea on what to do regarding the other variables

[crisp nacelle]

I am in a situation where I have to teach my friend's son, for a math competition, now I am just asking if a 5th grader can pick up Calculus and understand it deeply.

[nocturne whale]

Anything is hypothetically possible - it's question of many variables, though.

[sly ocean]

oh ok, i didnt understand the question srry 😅

[crisp nacelle]

Ah okay, I am very new to teaching and I don't really know a 5th grader's limits.

[nocturne whale]

My prognosis is not optimistic, but it does depend on the amount of time and patience allowed.

[crisp nacelle]

The competition is in 2 weeks.

[nocturne whale]

What exactly would you plan to teach in calculus?

[crisp nacelle]

I'm not sure, I haven't seen the questions yet in the warm-up tests.

[sly ocean]

gimme a second. im grabbing my notebook so i can explain it visually to u

[magic falcon]

thank you

[sly ocean]

Check this out while I write the next explanation

As u see I borrow the equation of motion to get height as a function of time

Also at the very end whats cut off is h = 0.2 t

[magic falcon]

but 0.2cm/s isnt like a constant throughout the whole process, only when the water level is 5cm deep

[sly ocean]

Oh shi

True

then we'll need to get the acceleration

[magic falcon]

i was thinking i need to somehow get dh/dt from that information but i have no idea how

r1 is 2
h is 6
r2 is 1

the unknown is dh/dt and dr1/dt to solve for dr2/dt

im not quite sure what dr1/dt represents

i know dh/dt represents rate of water level increase within the cone, for dr1/dt i dont get because r1 is already a constant?

[sly ocean]

dr1 is the derivative of the radius w/ respect to t, but to do that you need r as a function of t

[magic falcon]

wait wouldnt it just be 0? the radius of the cone will never change

therefore rate of change is 0

hmm

[sly ocean]

dr1 is not a constant. they just give u the height and radius of the cone so you can know the angle of the generatrixes

[magic falcon]

i meant r1 since r1 refers to the radius of the cone, 2, which doesnt change

[sly ocean]

Ok

[magic falcon]

so i was thinking it cant have a derivative if its a constant

[sly ocean]

so

[magic falcon]

sorry is that stupid

im honestly unsure of everything im doing

[sly ocean]

Nono

dont worry

This is not an easy problem

Let me make a sketch to simplify the problem

[magic falcon]

with this logic that must mean dh/dt is also 0 because the height of the cone also does not change

which makes the right hand of the equation equal to 0

so im probably not right at al

i realized i havent taken this into consideration yet

trying to figure out where this info fits in

[sly ocean]

Ok, so we maybe shoud have started from there

[magic falcon]

do i need a new variable for water level?

[magic falcon]

no variable representing it here

ill name it dw/dt then

[sly ocean]

W is the rate of water right?

[magic falcon]

dw/dt is the rate of which the water level in the cone is rising

whilst w itself represents the water level height

i think(?)

so rises .2 per second when 5cm deep

dw/dt = .2 when w = 5

wait then ill need another variable for the water leaking onto the balloon or rather the rate of water entering the sphere

i am so lost

[sly ocean]

what i dont really get is why -dV/dt=dW/dt

bc -dV=dW

so -dV=dV(balloon)

[magic falcon]

i just took it from the hint

oh man i dont get anything anymore, i missed that first negative sign at the top

[sly ocean]

that might be it

anyway

i think ive never solved a problem the way your doing it

i guess bc ive been taught differently

[magic falcon]

ive been taught to just equate the derivatives

and plug the variables in

could u solve it the way u would solve it?

[sly ocean]

ok, lemme try

[lone heart]

@magic falcon Has your question been resolved?

[sly ocean]

sorry for the wait

ill send u what i have so far

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stoic turtle]

i dont even know where to start

[lone heart]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[gilded galleon]

so how would you figure this out without actually solving the integrals. I assume the answer is false since if you switch the intervals you would have to make the function negative?

[lone heart]

@gilded galleon Has your question been resolved?

[lone heart]

@gilded galleon Has your question been resolved?

[near hollow]

Draw the region

Is it a simple region?

If so then you can swap the order of the bounds hence they are the same

Otherwise they are not

[gilded galleon]

ok so its true since the region is a box and it doesnt matter the order you integraet it in since they are just constant bounds

for some reason i thought the intervals were switched too

.close

[lone heart]

Channel closed

Closed by @gilded galleon

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stark flicker]

How can I find the angle θ?

[jagged cobalt]

only info youre missing is the radius, then you should be fine using soh cah toa

[stark flicker]

The radius is sqrt(5)-1

[jagged cobalt]

then thats your hypotenuse

[stark flicker]

I don't know the other angle though, or the third side so I'm missing something from soh cah toa

[jagged cobalt]

?

to find the angle you only need to know the lengths of two sides of that triangle, which you do

[stark flicker]

I need the length of the long side to do that though

I only know the length of the short side

[jagged cobalt]

i just told you the long side is the radius

[jagged cobalt]

^

[lone heart]

@stark flicker Has your question been resolved?

Channel closed

Closed by @stark flicker

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[whole scaffold]

what do i do here?

[short sonnet]

Ive been stuck on this for about 10 minutes

[whole scaffold]

hey using this channel already :))

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[lone heart]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[whole scaffold]

I'm not really sure what to really do here. I've been brute forcing to find f(x,y) and not really sure what bounds to integrate by

[lone heart]

@whole scaffold Has your question been resolved?

[whole scaffold]

<@&286206848099549185>

[charred flint]

path independent means you only care about the start and end, it's f(5,1)-f(1,0)

[whole scaffold]

so i just integrate one from 5 to 1 and another from 1 to 0?

but which variable od i integrate by?

[charred flint]

you did the integral already

gradf = F is basically derivative of f = F, so little f is the integral equation already

[whole scaffold]

like

[whole scaffold]

my integral is y^2+(x^2)(e^-y)

what variable do I integrate by, cuz i have x and y

[charred flint]

that's it already

just plug in the points and subtract

[whole scaffold]

oh like

i just evaluate?

oh

so i know i should probably make a new channel but what about this one

is it just the same thing?

i just repalce x and y with the parametric formula and just plug in 1 and 0 for t?

instead of integrating from 1 to 0?

[charred flint]

not sure what you mean, but you find the start and end and do f(end)-f(start) again yes

just from the parametric part with t

[whole scaffold]

oh

oh

i got it

tysm 🙏

.close

[lone heart]

Channel closed

Closed by @whole scaffold

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[primal thicket]

have i done this question right?

[past musk]

1+2x is not 1+2

should be just 1

[primal thicket]

ooohhh

truee

[past musk]

other than that looks good

[primal thicket]

awesome!

thanks!

.close

[lone heart]

Channel closed

Closed by @primal thicket

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[karmic saddle]

Why am I unable to graph this onto my calculator?
I'm able to graph tan(pi/x) but not xtan(pi/x)

[karmic saddle]

also how would I have found the asymptote?

[past musk]

compute the limit of the function as x->+inf

[karmic saddle]

wth is that 😭

[past musk]

uhh

ok i think you were just supposed to visually see that from the calculator

[karmic saddle]

I can graph tan(pi/x) just fine it shows what the answer shows but if I graph xtan(pi/x) it doesnt show anything

[muted hornet]

https://www.desmos.com/
use this website

[past musk]

sounds like a calculator issue

yeah desmos

[karmic saddle]

i wont have that during the exam bro what 😭

its a graphical calculator

i have to use

[past musk]

calculator issue

[muted hornet]

hmmmm. post input and output

[past musk]

are you waiting long enough

i think some calculators can be slow

[karmic saddle]

nah mine isnt

its the ti nspire one

[past musk]

then idk

[karmic saddle]

okay forget about that then

but like

how would I have found the asymptote

to equal pi

[past musk]

just look at what it is for large values of x

[karmic saddle]

oh yea ur right

sub in a large value and i get 3.14 so its pi

my calculator graphs tan(pi/x) just fine and it shows what the answer shows so i'll just simple shift the graph up and add an asymptote at 3.14... simple work around

thanks for the help

appreciate it

.close

[lone heart]

Channel closed

Closed by @karmic saddle

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wintry sluice]

can someone tell me how to proceed

[wintry sluice]

im lost here

[subtle birch]

what is y_2 and y_1?

[past musk]

maybe the subscript is the p

[wintry sluice]

thats my question too

[past musk]

just a guess tho

[wintry sluice]

this is the solution

i understand the first 2 steps

what is y1

[worn fox]

the first derivative of y wrt x (bad notation)

[wintry sluice]

how did sin(pt)

turn into sqrt(1 - y^2)

[worn fox]

well plug y = cos(pt) into this

[wintry sluice]

ohh

oh

[worn fox]

its using the pythag trig identity

[wintry sluice]

oh shi

[worn fox]

to write sin in terms of cos

[wintry sluice]

alr

thankyou

.clos

.close

[lone heart]

Channel closed

Closed by @wintry sluice

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cobalt forge]

.open

Need to find out if xsin(1/x) has infinite number of maxima and minima on the interval (0,1)

[wraith stratus]

do you mean local min/max

or global

[cobalt forge]

local actually

let me show my progress

[wanton peak]

You can prove by limit as x approaches zero, the frequency of the oscillations of sin(1/x)) increases to infinity, leading to an infinite number of oscillations within the interval (0,1)

Since for every period of oscillations return 1 local max/min

[cobalt forge]

yes

but what about the x part

[wanton peak]

then u also need to prove that these local max min doesnt = to each other

[cobalt forge]

near 0, x=0

[wraith stratus]

take the infs and sups

[wanton peak]

x*sin so x acts as altitude

[wraith stratus]

and show the sequence of infs is decreasing

and sequence of sups is increasing

[wanton peak]

Thus each local max/min doesnt = to each other

So infinity local max min still hold

[cobalt forge]

that is true

[wraith stratus]

and start by considering the subintervals x in [1/(4pi n), 1/(2pi n))

[wanton peak]

Basically, y = sin(1/x) doesnt have infinite local max min

Only y = x sin(1/x) does

[cobalt forge]

so only way to prove it by showing that the given function oscillates too much, and hence there are infinite number of extremums ?

[wanton peak]

Not too much

as x approach 0, the period approach 0

[cobalt forge]

I mean infinitely

[wanton peak]

Thus there's an infinite amount of oscillates

Yes!

[wraith stratus]

it does

[wanton peak]

Oh not local

[cobalt forge]

but using the derivative tests, can't it be shown ?

[wanton peak]

Global

[wanton peak]

Hmm

[wraith stratus]

what derivative tests

[cobalt forge]

first or second

[wraith stratus]

which are

you mean

f has min/ max at x0 implies f'(x0) =0 ?

[cobalt forge]

[ \frac {d} {dx} xsin(\frac{1}{x} = 0 \ ]

[wanton peak]

We can use it, then it repeat the very arguement that we made from the start

[cobalt forge]

yes I was trying to do so

[wraith stratus]

no

the assumption is f has a local min/max

not that f' = 0

the converse isn't true

[cobalt forge]

[ \frac {d} {dx} xsin(\frac{1}{x}) = 0 ]

[ocean seal]

shanks44

[wraith stratus]

you can't just say f' = 0 for inf many x in (0,1) implies f has inf many min max

[cobalt forge]

no but solving the equation f'(x) = 0 ?

[wraith stratus]

i mean you can if you want to get an idea of what's going on

but if you're asked to prove it

then you need to prove it

[cobalt forge]

[ xcos(\frac {1} {x}) \frac{-1}{x^2} + sin(\frac{1}{x}) = 0]

[ocean seal]

shanks44

[wraith stratus]

please don't solve for x

[cobalt forge]

oh

then ?

[wraith stratus]

use ivt on the intervals x in (1/(1 + 4pin), 1/(1 + 2pin))

but again

[wraith stratus]

what you're doing is not a valid proof

[cobalt forge]

oh, I don't get it, may you explain ?

[wraith stratus]

which part

why it's not valid

or using ivt

[cobalt forge]

yes

what is ivt ?

[wraith stratus]

intermediate value theorem

[cobalt forge]

ok

[cobalt forge]

and this part also

[wraith stratus]

do you know what the statement of the theorem is

[cobalt forge]

oh, let me quickly check

[wraith stratus]

the assumption is f already has a min/max

you're assuming f' = 0

you can't apply this theorem

[cobalt forge]

but I was thinking about finding out all the critical point

[wraith stratus]

and then what

[cobalt forge]

so at critical points the given function will be horizontal

[wraith stratus]

why

[cobalt forge]

that is some local extremum is present

[wraith stratus]

you're assuming the converse of this statement is true

which it is not

[cobalt forge]

ok

[cobalt forge]

but what about this fact ? I know I am missing something very basic

slope of the function will be 0 right

[wraith stratus]

the statement only works one way

IF f has a min/max THEN f' = 0

it is not true that
IF f' = 0, THEN f has a min/max

what is the derivative of x^3 at 0

[cobalt forge]

0

[wraith stratus]

is there a minimum/maximum of x^3 at 0?

[cobalt forge]

actually neither

[wraith stratus]

exactly

the theorem doesn't work in the other direction

[cobalt forge]

ok, I am getting some of it

so the only way to show that the function has inf. number maxima and minima, is because the function oscillates infinitely then

[wraith stratus]

i mean

im sure there's another way to prove it

but taking all the subintervals [1/(4pin), 1/(2pin)) is the most obvious to me

[cobalt forge]

what is 1/4pin ?

[wraith stratus]

n is natural number

$\frac{1}{4\pi * n}$

[ocean seal]

chebyshev's infinite pee norm

[cobalt forge]

ok

so you are saying, finding inf and sup in all the sub intervals ?

[wraith stratus]

yes

[cobalt forge]

and why choosing the sub intervals in that specific way ?

[wraith stratus]

try it

let n = 1

[cobalt forge]

anyway thank you both @wraith stratus , @wanton peak

[wanton peak]

You haven't finished it yet?

[cobalt forge]

yes as you suggested the first way, by showing the inf. no. of oscillations

I found out that f' = 0 cannot be used here

[cobalt forge]

also why semi open intervals ?

open intervals also serve the same purpose right ?

[wraith stratus]

yh

[cobalt forge]

ok, so the entire (0,1) can be split in (1/4pi,1/2pi), (1/8pi,1/4pi) and so on

and we need to show that each such sub interval has an infimum and supremum

so there are infinite numbers of extremums

right ?

@wraith stratus ?

[wraith stratus]

yh

[cobalt forge]

ok thanks

.closed

[wraith stratus]

well they should be min /max

not inf /sup actually

[cobalt forge]

yes yes

[wraith stratus]

but ye

[cobalt forge]

.close

[lone heart]

Channel closed

Closed by @cobalt forge

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[frail grove]

can you give an example of 2 functions where :
$\ \lim_{x\to a} f(x) = \lim_{x\to a} g(x) = \infty$ and $\lim_{x\to a} \frac{f(x)}{g(x)} = 1$

[ocean seal]

Adam Chebil

[wind cloak]

x^2 + x
x^2

[frigid mirage]

do you want a to be a finite number

[frail grove]

no

[frail grove]

can you give an example wher lim f = lim g = 0

[wind cloak]

1/x^2
1/(x^2 + x)

[frail grove]

is this possible :
$\ \lim_{x\to a} \frac{f(x)}{g(x)} = 1$ and $\lim_{x\to a} f(x) = 0$ but $\lim_{x\to a} f(x) \neq \lim_{x\to a} g(x)$

[ocean seal]

Adam Chebil

[wind cloak]

can't think of an example off the top of my head

im guessing no because if the limit of g is any other real number it would be 0 and not 1

[wraith stratus]

i was thinking about this yesterday

backwards lhopital gets you lim f = lim g

but obviously can't assume the converse of lhopital

but i couldn't come up with any counter example as well

[frail grove]

I'm trying to verify if this statement is true for all functions f(x) and g(x) such that lim f and lim g exist

but i'm guessing it's not true if lim f = 0 or lim g = 0

[lone heart]

@frail grove Has your question been resolved?

[lone heart]

@frail grove Has your question been resolved?

Channel closed

Closed by @frail grove

Use .reopen if this was a mistake.

✅

Channel closed

Closed by @frail grove

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dusk cave]

hello! i'll try to translate this the best I can, but I'm supposed to calculate the coördinates of point A and B. while trying to coördinate point A, I bumped into a small problem on where I didn't understand why t = 5/6pi. Why isn't it 1/6pi or 1 1/2 pi instead? I was just wondering, since they're within the given domain(?is that how you call it in english) too

[dusk cave]

i basically understood everything up until the last step, so if anyone can explain it to me i'd appreciate it alot!!

[lone heart]

@dusk cave Has your question been resolved?

[dusk cave]

.close

[lone heart]

Channel closed

Closed by @dusk cave

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fluid moon]

hopefully these questions too much, but I'm confused on how a converges. When I do the ratio test, I obtain $\abs{-5}$ which is 5 and thus does not satisfy the ratio test

[ocean seal]

parakeeti

[fluid moon]

my work is as follow

$(cn(-5)^(n+1))/(cn(-5)^n)$

[ocean seal]

parakeeti

[fluid moon]

I think I messed up the latex thing, but it cancels out to just (-5)^1

[vale crag]

should be c_(n+1) in the numerator

the problem is you don't know what c_n is

what if it's 1/42^n

[fluid moon]

that makes sense, but if we don't know what Cn is, how can we conclude anything from it

[vale crag]

well you're given that sum c_n 7^n converges

the ratio test should tell you something about the limit of c_(n+1)/c_n

[fluid moon]

do we have to find r

[vale crag]

you can't know r for sure

what I'm suggesting is using the ratio test in reverse essentially

[fluid moon]

comparison test?

[vale crag]

you know the series converges

[fluid moon]

oh lol xD

[vale crag]

so |r| is less than 1

[fluid moon]

could we multiply c_n on both sides

wait

nvm

doing so would get c_n+1 < cn

[vale crag]

$r = \lim_{n\to\infty} \frac{c_{n+1}7^{n+1}}{c_n7^n}$ 

[ocean seal]

aPlatypus

[vale crag]

that thing has to have absolute value less than 1

[ocean seal]

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

[vale crag]

just use paint if latex screams at you

[fluid moon]

I think I understand it now, there was a video with the web-assign

I'm a little confused on where -4 < x < 4 comes from and why it's not just {4}, I'm assuming it's because when ur doing the ratio test x is abs value right

[vale crag]

it's the domain where you know the power series converges

[vale crag]

so pretty much yes

[fluid moon]

okay, that makes sense

my brain is so fried

so much calculus today

but thank you

.close

[lone heart]

Channel closed

Closed by @fluid moon

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sour sphinx]

How do I know when to use p equals p primed and ek equals ek primed

I don’t know how I would know that Batman’s final velocity is solved with p equals p primed

[wanton granite]

your starting point should always be conservation of momentum as it is always valid and a safe thing to do

Conservation of kinetic energy is only true if we assume no energy is lost in any other form (heat, sound etc)

In this case, you need both because you have 2 unknowns (your number of unknowns has to be equal to the number of equations)

[sour sphinx]

Whoops

[wanton granite]

So if you are given no idea of the final velocities/momentums/coefficients pertaining to how the collision took place (which I imagine you haven't done yet so dw about it), then the second thing to do is to consider Ek

[sour sphinx]

I just don't know which formula I employ for which collision object

Does it have to do with the magnitude of V or the mass?

[wanton granite]

It shouldn't matter, provided you're consistent with which way is positive in the case of momentum

I think the red arrows are probably confusing you

Ignore them

[sour sphinx]

The red arrows I put indicate which formula goes for what object

Wait

[wanton granite]

It doesnt matter though

[sour sphinx]

But I don't think it matters though

[wanton granite]

You use both for both objects

[sour sphinx]

Dang thats odd

So it's up to me to match the final value to the object

[wanton granite]

m1 and v1 relate to car 1, m2 and v2 relate to car 2, put 's on the vs and it relates to the final velocity of 1 and 2

Just be consistent with your notation

We know that v1 and v1' are the initial and final velocity respectively of car 1

[sour sphinx]

yeah

[wanton granite]

So the value you get for v1' refers to car 1

No matching required

[sour sphinx]

ah

thanks man

.close

[lone heart]

Channel closed

Closed by @sour sphinx

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[humble totem]

pls help

[paper mango]

with?

[humble totem]

why can't the mean be: 25/5

[jagged cobalt]

of?

[humble totem]

look

here

[jagged cobalt]

👀

[humble totem]

yes?

Why can't the mean be: 25/5

I don't understand! Just why not!?

Why do we need to multiply the capacity with the capacitors to find the sum then count the total of the capacity!? Why not just take the sum of the capacitors over the number of the capacity?!!!!

JUST WHY NOT.

[jagged cobalt]

[6(21)+5(24)+3(27)+5(30)+6(32)]/25

we want the mean of the capacity, so we have to add all said capacities

[humble totem]

Not what I was asking for 😡

[jagged cobalt]

before we divide by how many capacitors

i dont really understand what you mean

[humble totem]

I wanna make it 25/5

Pls

[jagged cobalt]

on what basis

[humble totem]

I dont want [6(21)+5(24)+3(27)+5(30)+6(32)]/25

[humble totem]

Logic

Just count the number of capacitors= 25

over the number of capacity (the x axis)= 5

[jagged cobalt]

why on earth would that work, that would just give you the mean number of capacitors for each capacity

[humble totem]

yes thats how it's supposed to go like

Like why not?

Why not my solving?

[jagged cobalt]

i dont understand why you believe that would work?

at least i dont see the logic behind the thinking

[minor bone]

sum/25 is capacity/#each capacity

[humble totem]

What about the logic behind your thinking

[minor bone]

sum/5 is capacity/#each type of capacity

[humble totem]

proof?

[jagged cobalt]

the logic of mine is i summed all the capacities and divided it by how many capacitors there are

thats the definition of mean value

[humble totem]

what-

[humble totem]

prove mines didnt sum all the capacities and divided it by how many capacitors are there

[jagged cobalt]

you took the number of capacitors and divided it by how many values of capacity are available

[humble totem]

Prove that

[jagged cobalt]

whats the basis of that giving you a mean capacity if you didnt even use capacity

[minor bone]

mean capacity with respect to capacitors is amount of capacity/number of capacitors

mean capacity with respect how many types of capacitors is amount of capacity/number of different types of capacitors?

[humble totem]

the capacity is the many values of capacity that were being used at the denom

[humble totem]

HEADACHE

[minor bone]

bruh

[jagged cobalt]

5 isnt even a capacity

[humble totem]

Why?

its the sum

[minor bone]

that's why i put a question mark, maybe you seeing it that way?

[jagged cobalt]

youre saying 5 as in there are 5 capacities

[humble totem]

what way?

[jagged cobalt]

but thats not the capacities themselves, just how many unique values of them you have

[humble totem]

total no./sum of capacity: 5

[jagged cobalt]

25/5 would be the mean number of capacitors for each capacity

[humble totem]

thats completely correct?

[jagged cobalt]

thats not what it asks for though

it asks for the mean capacity

of all the capacitors

[humble totem]

thats what its asking bruh

[humble totem]

mean capacity? so two equations?

[minor bone]

well it says "capacities w.r.t capacitors", not "capacities w.r.t types of capacitors"

[jagged cobalt]

im just going to step away
all im saying is it wants the mean capacitance thats it

good luck

[humble totem]

capacities = 25

[humble totem]

ok? i gave him the capacitance

good luck for you too

[minor bone]

ok im out too

[humble totem]

alr bye

[minor bone]

just "disassemble" the diagram

it never says you need to organize them by 4uF, 7uF, 10uF, 13uF, 16uF

in total there are 25 capacitors

and each capacitor has a unique capacity

[humble totem]

disamble?

[humble totem]

thatswhat i said

[minor bone]

ok? and how you define mean?

[humble totem]

sum over total

[minor bone]

yes

what is the total capacity?

[humble totem]

5

[minor bone]

that's the number of TYPES of capacitors

[humble totem]

no?

How u knew?

[minor bone]

know what the answer is 59159190610369

[humble totem]

?

[minor bone]

bye

[humble totem]

😭

Go sleep

[summer dirge]

err

could you post a better screenshot

I can't really read the one you posted

[humble totem]

It's the best one I could screnshot

[summer dirge]

really?

zz

okay

[humble totem]

Yes.

[summer dirge]

I seriously can't read this properly lmao

[humble totem]

Then don't

[summer dirge]

what's being said here?

[humble totem]

microfarads greater than the capacity of each

[summer dirge]

I see

[humble totem]

But you don't

[summer dirge]

right.

so what's the issue?

[humble totem]

Here.
Mean value : 25/5 ❌

[summer dirge]

what does that mean

(pun not intended)

[humble totem]

25/5 not correct answer

[summer dirge]

yeah it isn't

[humble totem]

It is?

[summer dirge]

why would it be?

[humble totem]

Because

25(capacitors)/5(capacity)

[minor bone]

that's mean for amount of capacitors

[summer dirge]

^

[minor bone]

and you need mean for capacity

[humble totem]

they want that

[minor bone]

it says mean for capacity

[humble totem]

same thing

[summer dirge]

they are not the same thing

[humble totem]

can u prove it

[jagged cobalt]

capacitors have capacitance/capacity

[humble totem]

then?

[summer dirge]

I have a bunch of capacitors.

[jagged cobalt]

that was my last contribution apart from reactions, salut

[summer dirge]

I want to find out how good they are on average

[humble totem]

ok

[humble totem]

ok

[minor bone]

ok suppose you have power, engines and different types of engines

what u doing is mean for engine/different types of engines

not power/engines

[humble totem]

what's the difference?

[minor bone]

$amongusamongusamoagnawuigigh69834 ihu609ytwi 094y$

[ocean seal]

svc

[humble totem]

Clearly understood that.

[summer dirge]

the mean is the sum of all the capacitors' capacities, divided by the number of capacitors

you have 5 types of capacitors

each type has a different capacity

[humble totem]

Oh okay

[summer dirge]

you want to sum up all of their capacities to get a total capacity, and then divide that by the number of capacitors to find the average capacity

[humble totem]

Thank you for your clarification 🥰.

[summer dirge]

...does that solve your problem?

[humble totem]

Yes.

.close

[lone heart]

Channel closed

Closed by @humble totem

Use .reopen if this was a mistake.

[minor bone]

i dont know what to say

[summer dirge]

me neither

I don't even know where the confusion lay lmao

I guess in the definition of mean

[humble totem]

No, it's just that the other two prolly never once presented for class

.close

@paper mango go away

(jk)

[paper mango]

Keep in mind that independent of other's backgrounds, everyone helping here is sacrificing their free time voluntarily to aid others with problems they're likely too good for. Mutual respect is the least I'd ask for.

[humble totem]

? 😭

[paper mango]

you're aware what I refer to

[humble totem]

Not everything has to be taken serious btw.

[summer dirge]

what does this even mean

[humble totem]

Class presentation

[summer dirge]

the other two have almost certainly done presentations before, but I don't see why that's relevant

[humble totem]

Dw I've got you

[summer dirge]

not sure what you mean

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[gilded nimbus]

hi! i'd like help on this problem please!

[gilded nimbus]

I have no idea where to start, what formula to use, or just how to go about this problem 😦

[outer pollen]

one second I got you

[gilded nimbus]

tysmm

[outer pollen]

Here we will likely use the intersecting secents theorem

(in case you need it https://www.varsitytutors.com/hotmath/hotmath_help/topics/intersecting-secants-theorem)

Let me know if you need any more help

[gilded nimbus]

TY!

[lone heart]

@gilded nimbus Has your question been resolved?

Channel closed

Closed by @gilded nimbus

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vapid shuttle]

Hi, I was hoping someone could check over my reasoning in my proof for part (b) of this

Just the forward direction of the proof actually

Let $a\in H$ and $b\in K$, then $ab\in H \cup K$ and assuming $H\cup K$ is a subgroup this implies that $(ab)^{-1}\in H\cup K$. So either $b^{-1}a^{-1}\in H$ or $b^{-1}a^{-1}\in K$. If it is in $H$ then by closure $b^{-1}a^{-1}a\in H$ and $b^{-1}\in H$ which implies since $H$ is a subgroup $b\in H$ so $K\subseteq H$. And the argument is the same if $b^{-1}a^{-1}$ was instead in $K$ we'd get the opposite inclusion and hence prove the forward direction.

[ocean seal]

Austin

[vapid shuttle]

Does that work?

[marsh rapids]

no

because you could have some (a, b) pairs for which you show b in H
and some (a, b) pairs for which you show a in K

so you don't get inclusion

[vapid shuttle]

What do you mean?

[marsh rapids]

for all (a, b), (a in K) or (b in H)
That doesn't let you claim K in H or H in K

[vapid shuttle]

There's no ordered pairs here

I don't understand what notation you're using

let a be an element of H and b be an element of K. I have to show either a in K, or b in H.

[marsh rapids]

you start with "let a in H, b in K"
That specifies a pair (a, b) on which you reason
You do that reasoning for all pairs (a, b)

you have shown that

I claim it doesn't trivially imply inclusion of the whole groups

[vapid shuttle]

I don't understand

this might be easier to read

same reasoning

I just typeset

[marsh rapids]

ok then
explain how "b in H" implies "K subset H"

[vapid shuttle]

b was arbitrary in K

[marsh rapids]

but for a given pair, you've got one of these
it need not be always the same

you could sometimes show a in K and sometimes show b in H

in which case it could be that you failed to show either total inclusion

[vapid shuttle]

I think I follow what you're saying

What should I do instead, any hint?

[marsh rapids]

well how could that fail ?

[vapid shuttle]

I'm not sure

[marsh rapids]

then think more

[vapid shuttle]

Your question isn't even clear what I should be thinking about

[left wharf]

I don't see why ab\in H U K

[marsh rapids]

if you can formulate how it can fail
then you might see how it can fail to happen
i.e. the failure case is actually unachievable

[vapid shuttle]

ab in H by closure of the group operation

[marsh rapids]

assuming H U K is a group, then it is closed under multiplication

[left wharf]

oh yeah true thanks :D

forgot about the assumption

[marsh rapids]

assuming neither K subset H nor H subset K
Then you have 2 counter-example elements. What can you say about them ?

[vapid shuttle]

there exist g1 in H\K and g2 in K\H

then apply the above reasoning to these g1, g2

and we will achieve a contradiction?

Suppose H U K is a subgroup but for the sake of contradiction that there exists $g_1\in H\setminus K$ and $g_2 \in K\setminus H$. Then $g_1g_2 \in H \cup K$ which under the assumption that it is a subgroup implies $g_{2}^{-1}g_{1}^{-1} \in H$ or $g_{2}^{-1}g_{1}^{-1} \in K$. If it is in $H$ then $g_{2}^{-1}g_{1}^{-1}g_{1}\in H \implies g_{2}^{-1} \in H \implies g_2 \in H$ which is a contradiction, and the same for the other case.

[ocean seal]

Austin

[vapid shuttle]

I assume the thumbs up means yes, but is that now correct Bezier?

[marsh rapids]

ye

it's the same thing, except we showed there's an instance in which this leads to a contradiction

[vapid shuttle]

Thank you, I wouldn't have noticed the edge case myself

.close

[lone heart]

Channel closed

Closed by @vapid shuttle

Use .reopen if this was a mistake.

[lone heart]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[full nimbus]

How would I solve for a?

[sharp gate]

Did you differentiate the curved graph?

[full nimbus]

huh?

[lone heart]

@full nimbus Has your question been resolved?

[full nimbus]

<@&286206848099549185>

[civic swallow]

start by differentiating the curved graph, which is the graph of y = ax^2 + ....

[full nimbus]

what does that mean?

[civic swallow]

it means to calculate y', the derivative if you prefer

[lone heart]

@full nimbus Has your question been resolved?

[full nimbus]

idk how to do that

I think I need to solve it by doing something with the discriminant

[full nimbus]

.

[civic swallow]

you dont need the discriminant to differentiate a function, so you dont need it here

you should probably have a table or a spreadsheet with the common derivatives

[full nimbus]

I'm in pre calc, idk what that is

[full nimbus]

.

[civic swallow]

oh, then im not sure what to tell you

i thought that was how you were supposed to solve that

cant help you then sorry ;/

[lone heart]

@full nimbus Has your question been resolved?

[karmic saddle]

just derive it with inspection

[tribal heart]

if two graphs intersect

they have the same point (x,y)

try plugging in y from the first

into the second

-4x+5=ax^2+(a-4)x+3

and then find a such that it has one solution

(discriminant 0)

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vestal hound]

Anyone here???

[lone heart]

Please don't occupy multiple help channels.

[vestal hound]

Bruh

Ok

[twin nimbus]

.close

[lone heart]

Channel closed

Closed by @twin nimbus

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[daring echo]

How am I supposed to apply those fractions to real life problems? I don't understand how to interpret that into a real math problem

[granite iron]

Anyone got time for some pre algebra /algebra problems

[cloud nacelle]

just ask

[cloud nacelle]

its used to make various objects