help-49

[autumn canopy]

And the former is 1, the latter -1

[sand flume]

isn't all we want to do is find a case where |sin(1/x) - A | < epsilon not true

I want to set sin(1/x) = 1 and have |1-A| and then I would like to break it up in such a way that I could subtract 1 from epsilon. But then I run into what I did earlier

[autumn canopy]

\li \ Now, this is the argument: Let $\varepsilon > 0$ be arbitrary. Assume that $\sin(1/x) \xrightarrow{x \to 0} A$. Then, since $x_n \to 0$ and $y_n \to 0$, we get [|\sin (1/x_{n})-A|<\varepsilon ] and also [|\sin (1/y_{n})-A|<\varepsilon] for large enough $n$ (because then $|x_n - 0| < \delta$ and $|y_n - 0| < \delta$ hold). But as we just calculated, that's [|1-A|<\varepsilon \quad \text{and}\quad |-1-A|<\varepsilon.] But also [|1-(-1)|=|1-A+A+1|\le |1-A|+|A+1|<\varepsilon +\varepsilon =2\varepsilon ] and so $2 < 2\varepsilon$, i.e. $1 < \varepsilon$. Contradiction!

[grand pond]

Kepe

[autumn canopy]

So yeah, with these two sequences I suggested we got the contradiction

[sand flume]

give me a moment to digest this

[autumn canopy]

Basically, the moment we found these two sequences, we "had" the contradiction already from what we calculated, we just had to mess around until it became rigorously visible

[sand flume]

is it really sequences? because all you did was find to concrete values for sin (1,-1) and then you rewrite 2 as the two conditions for them to be continuous and show that 2 < 2epsilon meaning 1 < epsilon and then we can't have it true for all epsilon > 0

[autumn canopy]

Yes, we needed sequences to get to this part in the first place

The < epsilon inequalities only hold when for what is inside of sin, 1/x, x goes to 0

So we had to let x go to 0, but in a nice way that gives us the contradiction

And we did that with sequences

[sand flume]

isn't that the defintion of continuoity tho?

[autumn canopy]

Continuity tells us that |sin(1/x) - A| < epsilon when |x - 0| < delta

We ensure |x - 0| < delta by making x -> 0

The continuity condition reads roughly as "sin(1/x) is arbitrarily close to A when x gets arbitrarily close to 0"

[sand flume]

so your saying that the x-> 0 is part of sequences? but then how have I been doing limits if that is the whole concept of a limit? To make x -> a

[autumn canopy]

Ok maybe I should phrase it differently: Contuinity tells us: [\forall_{\varepsilon > 0} \exists_{\delta > 0} \forall_{|x - 0| < \delta}: |\sin(1/x) - A| < \varepsilon.] Now, pick an $\varepsilon$, then we get a $\delta$. If we now pick a sequence $x_n$ that goes to $0$, then we can surely find alarge enough $n$, so that $x_n$ is closer to $0$ than $\delta$, right?

[grand pond]

Kepe

[autumn canopy]

And so beginning from those large enough n, the condition is fulfilled

And so |sin(1/x_n) - A| < epsilon holds, beginning from that n

[sand flume]

ok so this is because x_n < delta?

[autumn canopy]

Yes

[sand flume]

but we have |x| < delta can't we find any delta that will satisfy this condition

if our |x| is bigger just find a bigger delta

[autumn canopy]

Here, continuity is given. So we give an epsilon, and they give us back a delta that we know will satisfy "for all |x| < delta it holds |sin(1/x) - A| < epsilon"

If we had to prove continuity, they give us an epsilon and we give back a delta

[sand flume]

ok so then you want to make sure that the values of x are less than the deltla they give us

[autumn canopy]

Yes

[sand flume]

ok one moment let me look through everything again

these are the x_n and y_n that we want to use right?

[autumn canopy]

Yes

both x_n -> 0 and y_n -> 0 (so the argument we just talked about works with them)

But also sin(1/x_n) and sin(1/y_n) give different values (1 and -1)

[sand flume]

if x_n -> 0 and likewise for y_n how can we get values for sin?

[autumn canopy]

By plugging in

[sand flume]

sin is 1 at pi/2 and -1 at 3pi/2

oh it is just adding by a factor of 2pi

[autumn canopy]

$\sin(1/x_n) = \sin\l(\frac{1}{\frac{1}{2n \pi + \pi/2}}\r) = \sin(2n \pi + \pi/2) = \sin(\pi/2)$

[grand pond]

Kepe

[sand flume]

Ok I think I see it. Do you have any advice for solving problems in this area?

[autumn canopy]

This here is a pretty standard trick, next time you see a similar problem you know you will want to construct sequences like this

The main step was finding a sequence that goes to 0 but also sin(1/x_n) = 1 for example

And then you just reverse-engineer

We could have also found one that made sin(1/x_n) = 1/2

And another that made it -1 or whatever

[sand flume]

so find sequences that approach the same value as the point we are interested in

[autumn canopy]

And then that would also give us a contradiction

[autumn canopy]

Yeah, and the function evaluated on them should give a value independant of n and different from the other sequence

[sand flume]

ok and then we show that epsilon cannot be greater than 0 but greater than something else positive

[autumn canopy]

The contradiction comes from the assumed continuity saying it holds for all epsilon > 0

But in the end we conclude we must have epsilon > 1

[sand flume]

yes

ok thx

[autumn canopy]

np

[sand flume]

.solved

[midnight plank]

Channel closed

Closed by @sand flume

Use .reopen if this was a mistake.

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[leaden seal]

Can someone explain why my working for Q4 is wrong and I don't get 9k^2

[small jasper]

,w \int^{2k}_{k} (3x-2) dx=10

[grand pond]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[small jasper]

are you sure that you copied the question correctly?

because this answer is correct if you simplify it

@leaden seal

[leaden seal]

hmm

[safe oriole]

no

it is correct as civil said

[safe oriole]

here is the verification of wolfram too

[leaden seal]

wait

[safe oriole]

just like how you did it

$\int_k^{2k}3x-2\ \dd x=\frac{3x^2}2-2x\big\vert_k^{2k}=(\frac{3(2k)^2}2-2(2k))(\frac{3k^2}2-2k)=\frac{9k^2}2-2k$

[grand pond]

ali yassine

[midnight plank]

@leaden seal Has your question been resolved?

[leaden seal]

hmm

[safe oriole]

is there still a problem

[leaden seal]

yeah i will show u

[safe oriole]

sure go ahead

[leaden seal]

here

and this doesnt give me 9k^/2 -2k

[safe oriole]

6k^2-1.5k^2=4.5k^2=9k^2/2 and -4k+2k=-2k

i dont see a problem here

[safe oriole]

(also note that the RHS in the 2nd and 3rd lines is 10 and not 0)

[leaden seal]

oh yeah i see

thanks

.close

[midnight plank]

Channel closed

Closed by @leaden seal

Use .reopen if this was a mistake.

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[sacred bramble]

give me any question

[sacred bramble]

I will solve it in seconds

[soft stone]

this is a help channel

[frozen rain]

uhh

wrong channel

[soft stone]

you're supposed to ask the question

[sacred bramble]

my bad

then...
what is 233333499+3983938938938

[frozen rain]

plug it into the calculator...

[soft stone]

didn't you say you can solve any question in seconds

[sacred bramble]

...

I..

I am just testing you

[rich yew]

OP do you have a question? Otherwise, this is not what help channels are intended for.

[hard shard]

its an expression
hope this helps!

[sacred bramble]

I don't have a plug

[stiff bison]

<@&268886789983436800> potential troll

[gaunt imp]

A natural number

[formal blade]

Don't troll here

[rich yew]

.close

[midnight plank]

Channel closed

Closed by @rich yew

Use .reopen if this was a mistake.

[soft stone]

testing someone with a calculator question after claiming any question is a bit self defeating don’t you think?

[dusk remnant]

please read #❓how-to-get-help . this is for asking your own questions, not asking for other people's questions.

[hard shard]

hi can you solve this one please

[safe oriole]

lol this was the best move

[rich yew]

he unfortunately didn’t specify how many seconds

also ai pfp and “tuff” bio? 😭

probably going to ask chatgpt

[stiff bison]

yeah bro literally got left speechless

[gaunt imp]

Most likely, also

We're probably giving him too much engagement

[rich yew]

[gaunt imp]

Most likely what he came here for

[last slate]

Hello

[hard shard]

the correct answer with exllanation is ||F. A line segment because thats how its defined||

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[uneven sandal]

Does it follow from the definition of time-homogeneous processes that any time-homogeneous process

X(t_1) - X(t_0) (equals in distribution) to X(t_1-t_0) - X(0).

No right it just Tells me that X(t_1) - X(t_0) = Y(t_1-t_0) For some r.v. Y(t_1-t_0) but this feels so useless.

[soft stone]

yes

by definition of time homogeneity the law of an increment depends only on its length

hence for any $t_1>t_0$ we can say $$X(t_1) - X(t_0) \stackrel{d}{=} X(t_1 - t_0) - X(0)$$

[grand pond]

anflo

[soft stone]

but remember that this is purely a distributional statement (not pathwise equality)

[midnight plank]

@uneven sandal Has your question been resolved?

[uneven sandal]

ehm why should this be true?

[soft stone]

writing $X(t_1)-X(t_0)=Y(h)$ is not weaker or useless, its exactly what equal in distribution means

[grand pond]

anflo

[soft stone]

the Y(h) notation just packages this family of increment laws

[uneven sandal]

But Why can we write
$$X(t_1) - X(t_0) \stackrel{d}{=} X(t_1 - t_0) - X(0)$$

[grand pond]

tobi

[uneven sandal]

since by definition it only tells us that there exists a random variable Y(h)

[soft stone]

because time homogeneity says the distribution of an increment depends only on its length

both increments $X(t_1)-X(t_0)$ and $X(t_1-t_0)-X(0)$ have the same length $t_1-t_0$ right?

[grand pond]

anflo

[soft stone]

so we can say their distributions are equal

[uneven sandal]

but why subtract the constant X(0) ?

Oh

since we are only allowed to consider differences?

[soft stone]

exactlyy

[soft stone]

because $X(t_1)-X(t_0)$ starts at $t_0$ and $X(t_1-t_0)-X(0)$ starts at $0$

[grand pond]

anflo

[uneven sandal]

How can we say that the difference starts anywhere?

[soft stone]

$X(t_1) - X(t_0)$ is the displacement of the car during the interval $[t_0, t_1]$

[grand pond]

anflo

[soft stone]

basically we "start" our stopwatch at t_0 and see how far it goes

a more general way to write that would be X(t+h)−X(t) is the change of the process over the interval [t,t+h]

[uneven sandal]

Thank you, this really helped

:)

.close

[midnight plank]

Channel closed

Closed by @uneven sandal

Use .reopen if this was a mistake.

[soft stone]

np!

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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 geyser]

i am thinking D but given answer is a WHY?

[soft stone]

given answer is WHY?

[near geyser]

A

[lyric charm]

ANY atmosphere is a "mixture of gasses"

[frozen rain]

theres too many words

[near geyser]

scary words

[frozen rain]

lost me at statement

[pure wraith]

The reason needs to explain the assertion, which it doesn't (as Ann pointed out)

Even Venus's atmosphere is a mixture of gasses but you can't live there

[lilac finch]

Even A needs to specify solar system

[lyric charm]

no exoplanets are known to harbor life anyway

[frozen rain]

D is false causality because of other planets having an atmosphere with a mixture of gases

[lyric charm]

some are potentially habitable but we don't know if anyone's there

[lilac finch]

Well nvm

[near geyser]

tq guyssssssssss

my science is weak now lol

[frozen rain]

lowk dont want to take act science 😭

[soft stone]

it should say “in the solar system”

[frozen rain]

!done

[midnight plank]

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

[near geyser]

so we should ask question to reason?

[soft stone]

yeah you should ask smth like does R actually explain A?

[near geyser]

reason should clarify the assertion properly?

[soft stone]

yeah a good trick is to join them with because

if it doesn’t genuinely explain it (or sounds half true) then you know that option’s out

[midnight plank]

@near geyser Has your question been resolved?

[lyric charm]

@near geyser what remains in doubt?

[near geyser]

.close

[midnight plank]

Channel closed

Closed by @near geyser

Use .reopen if this was a mistake.

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[pure wraith]

Do you know how to begin?

[robust field]

Nope

[grand pond]

jewels!

[pure wraith]

Can you identify the three roots from the diagram

[lyric charm]

if you were given a polynomial graph then would you know how to write down its eq

[midnight plank]

@robust field Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[viral dagger]

lowkey im hard stuck 😭

[viral dagger]

the closest to an idea i could form was trying to provr AIX=AIY, as AIX=270-ABX, but that just stops there (?)

any other idea i feel like is stuck at _I_ where you cant really go anywhere from the I in the center, and I__ but i dont know what to use

source is IGMO Christmas Mock 2025 P2 btw, i cant find a sol online tho :<

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[raw vector]

What do you know about mixtilinear circles?

[raw vector]

I gotta go for a few hours rn, but i'll help if this channel is still open when i come back

[viral dagger]

i think ill sleep first :<

[raw vector]

Hopefully someone will help before that

[viral dagger]

🙏

[tawdry laurel]

ask in https://discord.com/channels/533153217119387658/1403434610158731315

im outside rn and having an mo crisis so might not respond

[midnight plank]

@viral dagger Has your question been resolved?

Channel closed

Closed by @viral dagger

Use .reopen if this was a mistake.

[viral dagger]

sure

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[fringe yarrow]

Hi ok so
I’m doing a math question, I think its called a geometric progression?
The question isnt in english so I cant really provide it, but ill try to explain it
The common ratio is just ‘q’
I was given a1,a2,a3, etc
And also that they made a new sequence/series,
b1 = (a1/3), b2 = (a2/3), etc
According to google translate, they asked this
“Express the quotient of the new geometric series using q.”
How do I do that ?

[midnight plank]

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.

[placid ridge]

Do you know what the quotient is?

[fringe yarrow]

i do not

[fringe yarrow]

I think theyre asking what the common ratio for the new sequence/geometric series is

[placid ridge]

Yes thats the quotient

[fringe yarrow]

oh

[placid ridge]

To find it, you divide the nth term by the n-1th term

[fringe yarrow]

im not sure i understand

[placid ridge]

Well what does the common ratio mean

[fringe yarrow]

its the difference between the numbers

[placid ridge]

Between 2 consecutive terms to be exact

[fringe yarrow]

yes

[placid ridge]

So a(2)=a(1)×q

This would mean a(2)/a(1)=q right

[fringe yarrow]

yeah

[placid ridge]

We can do the same to find the ratio of the sequence b(n)

Find b(2)/b(1)

[fringe yarrow]

?

[placid ridge]

Yes

Now simplify it

[fringe yarrow]

um i forgot how to

[placid ridge]

Both the numerator and denominator has a factor of 1/3

[fringe yarrow]

Oh
So i can get rid of it
Which would leave me with a2/a1
Which i already know equals q
So the common ratio would be the same?

ohhhhh
i see

thank you :)

[placid ridge]

Np

[fringe yarrow]

how do i close it

[blazing nacelle]

" .close "

[fringe yarrow]

thanks

.close

[midnight plank]

Channel closed

Closed by @fringe yarrow

Use .reopen if this was a mistake.

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[twilit field]

Here it is enought to say cos(n) doesn't converge to 0, right

[rich yew]

I believe so yes.

[twilit field]

had a few more answers I wanted checked. Here I can use the comparison test with 1/n^2

[silent dock]

yes

[dawn dagger]

no

[twilit field]

( I know I was doing series of functions in #real-complex-analysis , but I just realised I haven't practiced many problems on series, sorry)

[dawn dagger]

u can have negative terms

[twilit field]

oh right

[dawn dagger]

but you can show it converges absolutely i think

[silent dock]

hmhmhmhmhmhmhmhmhmhmhm

[twilit field]

mhm, yea

fair enough

[silent dock]

I thought that's what you meant anyhow

[twilit field]

I forgot that cos can have negative terms

😔

[silent dock]

tsk tsk

[twilit field]

Last two questions for now
for 8, follows from the product rule for series
for 9 a counter example is the series a_n=1/n^2

[pure wraith]

Works

Although I don't know what exactly you mean by "product rule for series"

[twilit field]

The product of two convergent series is convergent and is the product of the individual limits

[safe oriole]

i am assuming you mean something like this?

[pure wraith]

That's not what's happening here though

[safe oriole]

a convergent series with all terms >0 is also absolutely convergent

but ah yea the product here isnt exactly the same lol

[twilit field]

hmm? $\sum a_n \cdot \sum a_n = \sum a_n^2$

[pure wraith]

(x + y)(x + y) = x^2 + y^2?

[twilit field]

no

[pure wraith]

That's basically what you've written here

[safe oriole]

thats not quite right

[twilit field]

isn't that what's going on here though....

[safe oriole]

at least not with the cauchy product definition of product of series

[pure wraith]

$\left ( \sum a_n \right ) \left ( \sum b_n \right ) \neq \sum a_n b_n$ in general

[twilit field]

yes

[grand pond]

jewels!

[pure wraith]

So...

You're gonna have to come up with an actual justification

[twilit field]

oops, yea, got confused with the summation law

😔

sorry and thanks

[pure wraith]

As a hint, you know that if $\sum a_n$ converges, then $\lim a_n = 0$

[grand pond]

jewels!

[twilit field]

I'm trying to think of a cauchy argument

okay, that works, I guess

[pure wraith]

That's not really necessary, but sure

[subtle blaze]

Do you think a^2 will converge

[safe oriole]

hint: ||\sum a_n^2\leq (\sum a_n)^2||

[twilit field]

$\abs{ a_{n+1}^2+a_{n+2}^2+ \dots+ a_{m}^2}< \sum_{i=n+1}^{m} \abs{a_i}^2$

[twilit field]

my idea now , yea

[twilit field]

yes

[subtle blaze]

Well I was thinking more along the lines that a_n is less than 1 from some point onwards

[grand pond]

wai

[subtle blaze]

Now if you square the terms get smaller

Or rather, closer to 0

[twilit field]

mhm

[subtle blaze]

But now you have a problem

What if the terms were negative

[zealous schooner]

what

[subtle blaze]

How do you guarantee that without the minus terms we can still converge?

[twilit field]

triangle inequality

[twilit field]

the problem assumes a_n is positive

[subtle blaze]

Oh

[twilit field]

I mean I get its not needed

[subtle blaze]

Then yeah term by term it’s smaller so it’ll converge

[twilit field]

I think I get it from here, we say this is less than m eps as (a_n^2) converges

and thus the sum converges

[subtle blaze]

This isn’t true you need equality

But also both sides just say the same thing

[twilit field]

hmm, yea

[subtle blaze]

They are in fact equal when a_n > 0

[twilit field]

and then this works?

[subtle blaze]

Can you be more clear

[pure wraith]

You can't have the epsilon-bound depend on m or n

[twilit field]

wait, why not

[subtle blaze]

Well

Be more clear

Real analysis is about being precisely anal about every detail

[twilit field]

oh right, m depends on N

yea, noted

[subtle blaze]

So to sweep everything under the rug like you did here goes directly against the spirit of RA

[subtle blaze]

And then again

Yes it’s a hassle to write out all the details but that’s the point of RA

[twilit field]

I have an idea, I'll work on it and post it in a bit

[twilit field]

Yea, got it

Thanks so much everyone!

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[twilit field]

OH GOD

this was just the comparison test?!

🤦

[pure wraith]

yes

I was hoping you'd pick up on that

[twilit field]

yea, idk why I went into cauchyness and all that

I"M AN IDIOT MY GOD

thanks so much

[safe oriole]

yea that was why i wrote my hint

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[woeful turret]

i dont know how to start

i thought of using that angles bisected by B are equal

so we can equate that tan theta = (m2-m1)/(1+m1m2 formula)

but i dont have enough info for that

[earnest quest]

need a sec to see if I can do it first, but have you tried drawing it?

[woeful turret]

yea i drew it..cant send the diagram rn

[lusty python]

just send it

[woeful turret]

bro i dont have my phone

wdym just send it

[hard shard]

have you found the coords of B?

[woeful turret]

i mean i know that B is on the angle bisector so B is (alpha,alpha)

[hard shard]

not alpha

[woeful turret]

bruh the whole question is find coords of B

[hard shard]

but yes, (t,t)

[woeful turret]

ye

and i have no idea how to use that 2AB = BC info

[hard shard]

plug x=t, y=t to 2x-y=2

[slow thorn]

its quite easy to find A and C using diagram and this property

[hard shard]

i am going to scream

[woeful turret]

ohh bruhh

[woeful turret]

same 😭

[hard shard]

that theorem isnt that helpful imo

[woeful turret]

i remember learning this but i have never used it once

[hard shard]

in this problem at least, its a pretty good theorem

[woeful turret]

yeah..anyway could u tell me how to proceed without that if u know?

[hard shard]

oh wait

i misread the problem

[woeful turret]

bro

😔

[slow thorn]

you know the coordinates where the angle bisector meets the opposite side AC

[hard shard]

my bad

[woeful turret]

yes (2,2) thats the first thing i found

[hard shard]

so i dont feel completely useless, heres a diagram to find the point given the angle bisector and the two other points

i have to go though

[woeful turret]

ok i got C= (-2,-6)

now what to do?

[woeful turret]

oh alr alr nw 👍 thanks for helping

[worldly geyser]

You could find C using the triangle angle bisector theorem sent above

[woeful turret]

^^

[worldly geyser]

Is this for jee?

[woeful turret]

ye

[worldly geyser]

I see

[woeful turret]

@slow thorn now distance formula with condition 2AB = BC?

[worldly geyser]

Just a few mins ill brb

[slow thorn]

yyup

[slow thorn]

no uh didnt you already use that like right now

[slow thorn]

you need B using this

[woeful turret]

no i used that theorem u sent..now im asking if i should use it with the points

[worldly geyser]

I'm a jee aspirant too, i tried to do it but I got C as 1,0 🤷🏽‍♀️

I took the ratio wrong 🙏

Section formula right?

[woeful turret]

@slow thorn i got it ty

.close

[midnight plank]

Channel closed

Closed by @woeful turret

Use .reopen if this was a mistake.

[slow thorn]

ah alr

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[inland patio]

Let X ~ N(m, A) be a normal random vector. My book says that if det A = 0, there is no density, and if det A > 0, there exists a (unique) density. Why is this true?

Obviously A is (not) invertible in the latter (former) case, but I don't understand how the covariance matrix relates to density.

[wary thorn]

well

you can derive

the density

dependent on det(A)

being non-zero (and positive)

that's the short answer

does that help

[inland patio]

.close

[midnight plank]

Channel closed

Closed by @inland patio

Use .reopen if this was a mistake.

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[balmy cypress]

$2x^2-3x+2=0$

[grand pond]

Professor Edward C. Hawthorne

[balmy cypress]

$\frac{3+/-\sqrt{-7}}{4}$

[gaunt jetty]

yeah, that's not going to work

[grand pond]

Professor Edward C. Hawthorne

[gaunt jetty]

you can't apply partial fraction decomposition if that's your goal

[balmy cypress]

I think i can apply the formula instantly

[gaunt jetty]

what formula

[balmy cypress]

The one for when the determinant is less than 0

The first One

[snow niche]

you have to get this form

[balmy cypress]

?

I have the shortcut

[keen crag]

This pretty simple bro

Do u still want the solution

[balmy cypress]

Nope

[keen crag]

This can do too

[balmy cypress]

I was working on this anyway

1+sinx=t

cosx dx = dt

Mmmm

[keen crag]

Nah

Take

Wait

Rest u can do by ur self

[balmy cypress]

Maybe

$\frac{1-sin(x)}{1-sin^2(x)}$

[keen crag]

And is -1/4cotx bruh+c

[grand pond]

Professor Edward C. Hawthorne

[balmy cypress]

$\frac{1-\sin(x)}{\cos^2(x)}$

[grand pond]

Professor Edward C. Hawthorne

[balmy cypress]

Yes

[keen crag]

Wait what’s the answer

[balmy cypress]

1-sin(x)=t
-cos(x)dx=dt
dx=-1/cos(x)dt

[keen crag]

Udh mb

[balmy cypress]

Mm

[keen crag]

I messed the formula up

Yes ur supposed to rationalise it bro

U hv another question

I like integrating

[balmy cypress]

Ok

$\int sin^6(x) cos(x)dx$

[grand pond]

Professor Edward C. Hawthorne

[keen crag]

Ye got it

[balmy cypress]

How?

[keen crag]

I’m solving it but I have a good idea

So I might get it

I will send u

[balmy cypress]

Do you know the chain rule for integrations?

[keen crag]

There is no chain rule for integration

Only substitution

Or byparts

[balmy cypress]

$\int f'(x)[f(x)]^n dx = \frac{[f(x)]^{n+1}}{n+1}$

[grand pond]

Professor Edward C. Hawthorne

[balmy cypress]

n≠-1

[keen crag]

I think I got it one min

Broski I got this as the simplification

It’s long

But rest can be solved using cosXcosy formula

[balmy cypress]

Thanks

.close

[midnight plank]

Channel closed

Closed by @balmy cypress

Use .reopen if this was a mistake.

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[brave berry]

i just need help on where to start

for part a i ahvent learnt dot product so i cant use it

bro i ahte vectors so much man

@flat veldt idk if u wanted to help but u did !volunteer so ehre

[flat veldt]

That's just a command for the bot

[brave berry]

oh

mb

[dawn dagger]

How have you defined perpendicularity

[brave berry]

we havent

thats what i tried to do in my first step lol

its meant to be like a challenging q

[dawn dagger]

?

[atomic mist]

they heavent taught you product of 2 vectors right?

[brave berry]

no

idek if u can do dot product for that q

[atomic mist]

but in the question they are doing ab-ba and dont state which product is it

[atomic mist]

you can, dot product is the only way i can think of rn

[brave berry]

wdym

i think b and a are just some scalar quantity not vectors

theyd be bold if they were vectors

[atomic mist]

like they heavent taught you the product of 2 vectors yet in the question itself they are using product of 2 vectors

[brave berry]

.

[atomic mist]

read the first line

[dawn dagger]

you could use trig i guess

[brave berry]

oh good point

[brave berry]

how

[dawn dagger]

the question is not stating from what set a and b are so you could make a lil sketch

in a 2d plane

[brave berry]

i did that

[dawn dagger]

can u show?

[brave berry]

ye 1 sec

I think dot product might be the inly way to do it tbf

[dawn dagger]

ok let him cook

[brave berry]

Bcus the psheet that i got set had some second year stuff in it i ahd to self teach

the ans/mark scheme doesnt use dot product

am i cooked

[midnight plank]

@brave berry Has your question been resolved?

[raw hamlet]

Your diagram is not correct

Draw it with a and b having different magnitude to avoid getting mislead

I do not think you need the dot product because you can reason about it purely geometrically

Note that the notation is horrid, it's asking you to do (length of b)(a as a vector) - (length of a)(b as a vector)

[brave berry]

Ohhh

So

|b|a

[raw hamlet]

Yes. It's hidden above where it tells you OA = a

[brave berry]

Okay

How would I stat the proof tho

[raw hamlet]

And they're using bold a to mean the vector and non bold a to mean OA which is just the |a|

[brave berry]

Like how do u define perpendicularity geometrically

[raw hamlet]

Well you should be able to say something concrete about |a|b and b|a|

There's something very clearly special about those two vectors

That they have in common

[brave berry]

well

Okay so

Ill do A, B as the lengths and

a b as vectors

Bcus i cant rlly do it any better as text

[raw hamlet]

It might help to draw it out with like trying to make a 1/3 the length of b

Or vice versa

That way your diagram will actually be less misleading about what's going on

[brave berry]

Ok pnr sec

[brave berry]

Ill be completely honest i havent got a clue

[atomic mist]

isnt it |b|a and b|a|?

[brave berry]

Other than theyre relative to O

[raw hamlet]

Yeah I'm getting lost in the sauce

On my phone

[raw hamlet]

What's the magnitude of |a|b?

[brave berry]

|a||b|

oh they have the same magnitude?

[raw hamlet]

Yes

[brave berry]

i was thinking that but i wasnt too sure

[raw hamlet]

So in your diagram you drew b and a have the same magnitude but it's actually the new vectors that do

[atomic mist]

wait in the diagram OC is perpendicular to AB right?

[brave berry]

so theres another similar triangle with sides ab and ba

[raw hamlet]

In his diagram it is but that is not given by the problem statement.

[atomic mist]

no i think its part of the problem statement

cause then you can use triangle law

[raw hamlet]

The problem statement says it bisects the angle

[noble forge]

Wait what? ba-ab is just a null vector

[raw hamlet]

It is not

[noble forge]

It is, bruv

[atomic mist]

nah notation is weird its |b|a - b|a|

[raw hamlet]

It represents something concrete relative to |a|b - |b|a

[noble forge]

Oh okay

[raw hamlet]

a and b are not the same magnitude so the picture is wrong but also "correct" once you view it as the vectors a|b| and b|a|

Basically that's the triangle we need to make not the one we are given

[brave berry]

is this correct or am i just losing my marbles

[raw hamlet]

Depends what you mean by similar. It's not similar to the original triangle, but it's similar to your diagrams triange

[brave berry]

oh so its not mathematically similiar?

[raw hamlet]

And the vector magnitudes being equal tells you it's isosceles

The original triangle was not guaranteed to be isosceles but the new one is

[brave berry]

how does this help us

[raw hamlet]

What do you know about the bisector of the angle (between the two equal sides) of an isosceles triangle?

Draw it out and do some basic algebra on the angles if you have to

[brave berry]

it splits it into two triangles with equal hypotenuse?

[brave berry]

alr one sec

[raw hamlet]

You used the word hypotenuse there which is only relevant when?

[brave berry]

right angle

so it makes two right angled triangles

[raw hamlet]

And you are asked to prove?

[brave berry]

that ba - ab is perpendicular to c

is that the proof?

[raw hamlet]

Can you see how that's the same thing as those two right triangles?

[brave berry]

kind of

im a bit

ok so

the fact that we have two right triangles

how does that prove that the vector c is perpendicular to ba - ab?

i get that the bisecting line would be the

like

the straight edge of the two triangles

[raw hamlet]

Do you get what ba - ab is geometrically?

[brave berry]

is it just the vector |b|a - |a|b

[raw hamlet]

Yeha I mean that

Is the 3rd side of the triangle

[brave berry]

yea

i get that

[raw hamlet]

And c is the angle bisecotr

[brave berry]

yes

[raw hamlet]

As given in the problem

[brave berry]

yes

[raw hamlet]

So when u extend c

It hits that side

And gives you two right triangles

[brave berry]

but how do u know that the bisecting line creates two right angled triangles with the same length of base?

like

do u get what I mean

[raw hamlet]

That's where the triangle being isosceles is important

[brave berry]

ohhhhhhhh

ohh

bcus a right angled triangle with two equal sides has to be like

the same?

[noble forge]

What if it's not

[raw hamlet]

It's easily verified with some basic algebra on the angles if you draw out the diagram and use angles you know are the same

[raw hamlet]

It's already proven to be isosceles

[brave berry]

but if b > a in our original triangle it wont make an isoceles triangle?

[raw hamlet]

That's why we are working with the new triangle

The |b|a and a|b| one

Not the original one

[brave berry]

ohhhhhhhhhhhhhhhhhh

because |b|a = |a|b

and thats the sides of our triangle

[raw hamlet]

Their magnitudes are equal yea and yeah they are the sides

[brave berry]

ok i get that bit now

but so ur saying

lets say the bisected angle = theta

ur saying that this is true no matter what theta is?

like within the ranges of our question

bcus what if theta is some weird value where the bisecting line doesnt create two equal right angled triangles

[raw hamlet]

Yes. You can prove it geometrically (putting aside the question for now) that the angle bisector of an isosceles triangle is perpindicular to the third side

I recommend working thru that first if it confuses you. Draw out a diagram, label which angles are equal, and consider which ones sum to 180

[brave berry]

wait

nvm

im a bit stupid

i just like

visualised it for a sec

ur righ

like

wait

it makes sense now

how do u spot stuff like that

[raw hamlet]

Experience

[brave berry]

aura

[raw hamlet]

A series like art of problems solving will have more focus on that kind of stuff than something in your normal curriculum

[brave berry]

art of problem solving?

is it a book

[raw hamlet]

Ye

[brave berry]

idk where ur from but

im currently in y12

doing a levels

is it of that level?

[raw hamlet]

Aops is sort of an alternative mostly highschool level math textbook series that goes into the same topics but in much heavier detail with much harder problems

[brave berry]

whats like the furthest extent subjectwise that they cover

[noble forge]

You don't really need this isoceles logic. You can just show that cos ac = cos bc because angles are the same
Meaning
$\frac{\mathbf{c} \cdot \mathbf{a}}{|\mathbf{c}| , |\mathbf{a}|} = \frac{\mathbf{c} \cdot \mathbf{b}}{|\mathbf{c}| , |\mathbf{b}|}$

[grand pond]

a handsome russian dude

[raw hamlet]

He said at the start he didn't want to use the dot product

[noble forge]

Okay

[raw hamlet]

Cause not covered yet

[brave berry]

wait

[raw hamlet]

Calculus but the non calculus parts cover way way way more than a standard curriculum

[brave berry]

holy

u lit got it

word for word

fairs

[brave berry]

wdym non calculus partts

like methods within calculus e.g integration via partial fractions?

[raw hamlet]

Like their algebra book will teach you way more techniques than a normal high school course

[brave berry]

this is the cover right?

[raw hamlet]

That one is geared towards math Olympiad students

But yeah

[brave berry]

perfect

wait

can i dm u rq?

[raw hamlet]

Sure

[midnight plank]

@brave berry Has your question been resolved?

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[twilit field]

working on part(c) rn, for now need to verify my pointwise limit

[twilit field]

$h(x) = \begin{cases} x& x = \frac{1}{n} \ 0 & \text{ otherwise } \end{cases}$

[grand pond]

wai

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

<@&286206848099549185>

[frozen rain]

yeah this is correct

[twilit field]

Awesome

thanks!

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[radiant roost]

can h really depend on n?

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[frozen rain]

explain

[radiant roost]

oh i think they meant h(x) = x if x = 1/n for some natural n
i was confused since the limit variable was also n
but i think these are different variables

[frozen rain]

yeah

.close

[midnight plank]

Channel closed

Closed by @frozen rain

Use .reopen if this was a mistake.

[frozen rain]

@twilit field add this to your pointwise limit

[twilit field]

.reopen

[midnight plank]

✅ Original question: #help-49 message

[twilit field]

add what?

I'm confused

[frozen rain]

x if x = 1/n for some natural n

[twilit field]

yea, that's what I meant when I used n

[frozen rain]

oh

alr

[twilit field]

thanks agin!

[frozen rain]

.close

[midnight plank]

Channel closed

Closed by @frozen rain

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[scenic wyvern]

<@&268886789983436800>

[midnight plank]

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.

[lilac finch]

Lessgo 69

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[lilac finch]

Claim

[pure wraith]

This just seems bashy

[lilac finch]

Lol

[wet locust]

this feels unbelievably contrived

[twilit field]

Average JEE problem ( I think this is JEE?)

the "July 29, 2022(II)" suggests it's a past JEE mains question

[lilac finch]

It is

Maybe

brb, don't let dis close

[wet locust]

do you like

rotate the diagonal 45 degrees

or something

ive been staring at this for too long

like surely there are not this many trig functions if not for some contrived rotation

[junior flower]

average jee problem

[pure wraith]

Actually sounds good lol

Rotate by α

[modern sapphire]

jee in july

[lilac finch]

M back

[modern sapphire]

oh, i forgot about covid entirely oops

[lilac finch]

Then

[late rover]

I feel like we can find a

[marble hedge]

We're stuck with slopes m1 and m2

[late rover]

chug that point into the equation

if it does not lie on that line then

Its distance would be a*sqrt{2}/2

[marble hedge]

Ohh yep

We get a

[late rover]

then m

m1*m2=-1

and that line equation also

[marble hedge]

No we can't

[late rover]

why not

[marble hedge]

That point isn't satisfying on that line

Because that point is not on that line

[late rover]

hm

so we can't find that line equation huh

[pure wraith]

It'll make the equations and the point independent of α so you can actually plot it and work from there

[late rover]

I bet we can still find the slope of the diagonal lines

[marble hedge]

Got it

[marble hedge]

Yeah

So that line isnt satisfying that point right?

[late rover]

yeh yeh

[marble hedge]

But we know which line will satisfy it

[tawdry bramble]

Who wants to place chess with me

[marble hedge]

The line just perpendicular to that line

[late rover]

#chess-go-shogi

[late rover]

it sure is

[lilac finch]

Hmmm.

[marble hedge]

We can find its equation

[tawdry bramble]

Yes

[pure wraith]

After undoing the rotation, you just get one vertex at (10, 10), and the diagonal is x + y = 10

[marble hedge]

So its slope will be
(c+s)/(c-s)
c=cosa
s=sina

We get value of a

[pure wraith]

This makes the other diagonal x = y

[late rover]

igtg

[lilac finch]

How

[pure wraith]

Do you know the rotation matrix?

[late rover]

It seems a bit chaotic here, one helper at a time should be more helpful, Im out

[marble hedge]

I don't think the question is telling us to use rotation

[lilac finch]

No

[pure wraith]

Hmm alright

[marble hedge]

Listen

[pure wraith]

Makes it a lot easier though

[lilac finch]

Yes

[marble hedge]

.

[verbal pumice]

Shifting and rotating the coordinate axes is in the JEE syllabus though

[lilac finch]

It is

[marble hedge]

Yep

[verbal pumice]

Then why not just get rid of α like Jewels said

[lilac finch]

But they teach matrix at the end

[verbal pumice]

Right but you have SOME formula for rotating the axes I'm sure

It's just better expressed as a matrix

[lilac finch]

Lemme see

[lilac finch]

Hmmm. Right ts makes sense

Makes a lot easier

[lilac finch]

This also works

Tysm guys

.close

[midnight plank]

Channel closed

Closed by @lilac finch

Use .reopen if this was a mistake.

[marble hedge]

Whats the answer

128?

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[lilac finch]

.Yes

[marble hedge]

👍🏻

[marble hedge]

.close

[midnight plank]

Channel closed

Closed by @marble hedge

Use .reopen if this was a mistake.

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[twilit field]

So I want to show $\sum_{i=1}^{ \infty} \frac{1}{n(n+1)}$ converges.

My idea simply is to look at the partial sum $\sum_{i=1}^{n} \frac{1}{i(i+1)} = \sum_{i=1}^{n}\frac{1}{i} -\frac{1}{i+1}= 1- \frac{1}{n+1}$.

The definition of the sum of a series IS the limit of this partial sum. So we look at $\lim_{n \to \infty} 1 - \frac{1}{n}=1$. As the limit of the partial sum exists and is finite, the series converges.

[grand pond]

wai

[upper coral]

telescoping?

[twilit field]

Seems to be, I think that's essentially what I've done.

[sage olive]

that works, could also do comparison with a p-series

[twilit field]

Oh before I forget , no comparsion test

[sage olive]

oh nvm

[twilit field]

Awsome

Thanks!

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[robust field]

I don’t know how to do

[sudden yacht]

Start by finding b

You should have been taught how it is defined. There's a special formula for it

[robust field]

3/8

[robust field]

Found it

[sudden yacht]

Good

Now use the info that f(320°) = 0

[robust field]

Ok

[sudden yacht]

Together with the already found a+c=6

[robust field]

Aight got it tysm

I thought this mean 15C3 (1) (kx)^3

[pearl hull]

,rccw

[lilac finch]

you are right

,calc -29120/455

[grand pond]

Result:

-64

[midnight plank]

@robust field Has your question been resolved?

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[solemn coral]

hello

[solemn coral]

(btw the problem is only translated with chatgpt)

we note AD=2a AB=2b DC=2c BC=2d

AQ1=DQ1=a*sqrt5

and analogs

(⇒)

if M1 N1 P1 Q1 are coplanar then note M1P1 intersect N1Q1 in O ?

and im pretty stuck

<@&286206848099549185>

[runic pebble]

Hey

[solemn coral]

hi

[lusty python]

!15min

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[late rover]

It sounds weird

Can you send the original

[solemn coral]

its in romanian

[midnight plank]

@solemn coral Has your question been resolved?

[solemn coral]

<@&286206848099549185>

[livid merlin]

How about drawing it and visualizing it? Then, trying to come up with an answer.

[solemn coral]

i tried

but its kinda messy

[midnight plank]

@solemn coral Has your question been resolved?

[midnight plank]

@solemn coral Has your question been resolved?

[quiet finch]

we antonio ciao

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[twilit field]

I know I've done this before, but I don't think I really understood it then , so I'd like to try again

[twilit field]

Showing $(I \cap J)$ is an ideal of $I$ is pretty easy so I won't do it again

[grand pond]

wai

[twilit field]

Just want to be sure as my proof depends on this, I+J need not be ring, does it

[grim vector]

well, it is

$I + J = {i+j, i \in I, j \in J}$

[grand pond]

υἀκυβρος

[twilit field]

oops

right

I was thinking we start off by considering an injective homomorphism $\psi: I \to I+J$.

Further define $\phi(r(I \cap J)) = \psi(r) J$

I then believe this map is an isomorphism b/w the two rings

[grand pond]

wai

[grim vector]

I was more thinking of
$\phi : I \to I+J/J$
with $\phi(i) = i + J \quad i \in I$
and then said it is a ring homomorphism and find the ker to apply first theorem

[grand pond]

υἀκυβρος

[twilit field]

huh

I see

[grim vector]

*say

[twilit field]

yea, that works

I'll work with this hint

Thanks so much!

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[twilit field]

.reopen

[midnight plank]

✅ Original question: #help-49 message

[twilit field]

You mean $(I /(I \cap J) \to (I+J) / J$ right

[grand pond]

wai

[twilit field]

The kernel is {0} is it not

[grim vector]

well i start with I -> I+J/J and then i find that the ker of it is I \cap J and this result to this being an isomorphism by first theorem

[twilit field]

How's the ker (I \cap J)

let x in (I \cap J)

[grim vector]

since phi : I -> I+J/J is a surjective homomorphism with ker I cap J

[twilit field]

then phi(x) = (x+0)/J=x/J

oh

[grim vector]

phi (i) = 0

for i in ker

[twilit field]

oh right

[grim vector]

you have i in I and so by this you have phi(i) = i + J = J (0 of I+J/J)

[twilit field]

noted

[grim vector]

i + J = J implies i in J

so i in I cap J

ker = I cap J

[twilit field]

yea, okay

[grim vector]

then 1st theorem conclude

[twilit field]

so I/(I \cap J) iso to im(I+J)/J

here, I'm just trying to figure out why it is (I+J)/J and not I/J

[grim vector]

well thats the theorem

[twilit field]

Okay, got it

tysm

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[safe oriole]

the quotient I/J doesnt make sense since J is not necessarily a subset of I let alone an ideal of I

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[gentle crown]

What to do
Question
C of 9

[gentle crown]

@keen saddle

Aadil bro check this out

[lyric charm]

!noping

[midnight plank]

Please do not ping individual helpers unprompted.

[lyric charm]

or do you want help from him and him only rn

@gentle crown

[gentle crown]

He has said that ping me that's why I had pinged him

[lyric charm]

in case you're open to help from others: take a look at 9a. and also consider parallelograms ABCD, AGED and DEFH.

[gentle crown]

I know those theorems but from where do I start?
I don't see anything to start

[lyric charm]

9a says: parallelograms standing on the same base and between two parallel lines are equal in area.

like this is simply reading off the given answer of 9a in square brackets

or it would come from your knowledge of the theorem

you wish to prove that some two other parallelograms are equal in area

they dont share a base nor a pair of parallel lines as sides, so you can't do it directly

but i am giving you here a link between them

what could i be suggesting here?

think about it.

if i say anything more i will spoil the question for you.

[gentle crown]

Now I'm stuck to equal ADH and EFG

[lyric charm]

there is no need

[keen saddle]

hi bro

[gentle crown]

Hii

[lyric charm]

aight well i will leave yall to it then

[gentle crown]

Lemme send you my progress in this question

[tribal ice]

yo bro

[gentle crown]

Done

Tysm Aadil bhai

.close

[midnight plank]

Channel closed

Closed by @gentle crown

Use .reopen if this was a mistake.

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[paper dagger]

hi

[paper dagger]

i have a doubt

[late rover]

What? Why the equal sign is inside abs value

[keen saddle]

why would you use I for angle

[shell wigeon]

That's not even I, that's |

[midnight plank]

@paper dagger Has your question been resolved?

[midnight plank]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[vague wigeon]

what do i do in this

[vague wigeon]

we legit have nothing to start with

[spiral rock]

You have some things from calculus

IVT for example

Other then that maybe some trig identities can help

[vague wigeon]

idk what that is

[spiral rock]

Intermediate value theorem