help-0

[hard patio]

in such cases you find the class marks for all of those ranges

[tired sparrow]

hm

waitim tryna let it sink

[hard patio]

alr

[tired sparrow]

okay I think I kinda get it

[hard patio]

okay so do you know what a class mark is?

[tired sparrow]

im sorry man i really aint getting anything, sorry for wasting yalls time

[crystal berry]

a proper explanation would be that since most natural phenomena follow a normal distribution (bell shaped curve) that's why its more accurate to use the midpoint of a group. Since the majority of the frequency will most likely be in the middle.

[tired sparrow]

.close

[lone heart]

Channel closed

Closed by @tired sparrow

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.

[vocal tapir]

was there a name for this type of a determinant?

[random tinsel]

@vocal tapir

name：
Tridiagonal determinant

[vocal tapir]

tysm

.close

[lone heart]

Channel closed

Closed by @vocal tapir

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.

[old gust]

I understand everything I did until f

I'm not sure if e is right either

[lone heart]

@old gust Has your question been resolved?

[lone heart]

@old gust Has your question been resolved?

[old gust]

<@&286206848099549185>

[lone heart]

@old gust Has your question been resolved?

[bright elk]

anyone please help me with this question-Consider that a, b, c, d are positive real numbers satisfying (a + c)(b + d) = ac + bd.
Find the smallest possible value of S=a/b+b/c+c/d+d/a

[lone heart]

@old gust Has your question been resolved?

[lone heart]

@old gust Has your question been resolved?

[balmy grail]

You can demonstrate e a little clearer by saying inflation is the percentage growth in the price level over time. $$P_{t+1}=P_t*(1+i)$$ therefore as the number of years increases we can increment t and define the price level recursively so that $$P_{t+2}=P_{t+1}(1+i)=P_t(1+i)^2$$ and so on as t gets larger. We can rewrite this as an exponential growth equation $$P_T=P_0*(1+i)^T$$ for a constant inflation rate. Then since the price level is growing exponentially we can say the purchasing power decreases exponentially since if $$A=M/P$$ if M does not grow proportionately, A will need to compensate.

It was at this point I saw the rest of the question. Yeah for f you just multiply by 100/100

[ocean seal]

Pixelius

[lone heart]

Channel closed

Closed by @old gust

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.

[peak wagon]

ohh

[peak wagon]

is easy

but.........

I'm busy with my things right now so I'm sorry but I can't help you this time

[lone heart]

@peak wagon Has your question been resolved?

[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.

[fringe swallow]

Why is 9÷1.5 = 6

[strong compass]

What do you mean by why? Do you mean how do you calculate it?

[fringe swallow]

I just don't understand why 9 ÷ 1.5 = 6

[rose sigil]

because 6*1.5 = 9

[fringe swallow]

What if you were given
9÷1.5 on its own how would you find the answer to that

[hollow vapor]

Long division???

[lone heart]

@fringe swallow Has your question been resolved?

[fringe swallow]

Yes

[lone heart]

Channel closed

Closed by @fringe swallow

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.

[rancid jay]

how do i maximize this?

[lone heart]

@rancid jay Has your question been resolved?

[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.

[ruby elbow]

help

i got it wrong because I get every single question wrong about the subject im in so I come here every single question and every single question takes 2-3 hours because nobody wants to do geometry

i've been DOING THIs FOR 8 HOURs

khan academy is a terrible teacher

i've been punching walls and banging my fists on my desk for 8 hours

cause nothing what khan academy teaches makes sense

[oak chasm]

Does it say that that proof is wrong?

[ruby elbow]

no

[ruby elbow]

the proof has to be right for the question

its not the type of question where you prove someone elses proof wrong and find their mistake

its you making a proof

about

this

[oak chasm]

OK, is that your proof on the left?

[ruby elbow]

yes

and I got it wrong

because I get everything wrong

[oak chasm]

OK, look at line 4.

It uses substitution.

[ruby elbow]

of 2 and 3

[oak chasm]

It uses something from line 2 and something from line 3.

But in line 4, it talks about the measure of angle D.

[ruby elbow]

ok

[oak chasm]

But there's no D in lines 2 or 3.

[ruby elbow]

yes

[oak chasm]

You can't change line 2, but you can change line 3 to something with D in it.

[ruby elbow]

angle A + angle D = 180?

that is one of the options

[oak chasm]

OK, so let's try that.

Then, it substitutes the measure of A with 90 degrees, because line 2 allows that.

[ruby elbow]

ok

what would be the reason

[oak chasm]

So, line 4 now makes sense.

The reason for line 4?

[ruby elbow]

ye

I clicked check

but it still says its wrong

[oak chasm]

OK, so we've only dealt with line 4 so far.

Do you see why it now makes sense that line 4 is true?

[oak chasm]

You get m<a = 90 in line 2, so you can replace m<a with 90 anywhere you want.

Then you have m<a + m<D = 180 in line 3.

Then, in line 4, you do that replacement of m<a with 90.

Does that part make sense?

[ruby elbow]

yes

[oak chasm]

OK, now if you look at m<A + m<D = 180, is it saying two angles are congruent?

[ruby elbow]

no

[oak chasm]

OK, so the reason for line 3 can't be the top two options.

The third option says that two angles sum to 180.

That matches what line 3 says.

Are there more than 3 options for the reason?

[ruby elbow]

no

what is a same-side interior angle

[oak chasm]

OK, so since step 3 is summing two angles to get 180, the only reason that fits is the third one.

[ruby elbow]

khan academy never taught me that

[oak chasm]

Well, you have a quadrilateral.

It's like four lines put together.

So, all of the lines each have two endpoints.

[ivory pivot]

[oak chasm]

AD has A and D on the ends of it, right?

So, A and D are on the same side (AD).

Does that make sense?

[ruby elbow]

hm

yes

[oak chasm]

So, what you're doing here is you're making things make sense.

[ruby elbow]

in step 3

[oak chasm]

You see in line 4 that you need to be able to use a substitution from line 2 into something in line 3 to get line 4.

[ivory pivot]

its like stonks line

[ruby elbow]

it says "when a transversal crosses parallel lines"

[ruby elbow]

so like this

so

with that

what would a same side interior be

[oak chasm]

AD and BC are parallel lines.

[ruby elbow]

yes

so are ab and dc

[oak chasm]

Sorry, AB and DC are parallel lines.

And line AD crosses both of them.

When that happens, the angle at A + the angle at D = 180.

[ruby elbow]

would line ad be the transversal

[oak chasm]

Yes, that's right.

[ruby elbow]

ok

so

angle a

and angle d

are same side interior like this?

[oak chasm]

Yes, that's right.

[ruby elbow]

ok

[oak chasm]

They're inside (in the interior of) the quadrilateral.

And they're on the same side (AD).

[ruby elbow]

ok

[oak chasm]

But the main thing is that you have to look at the lines and figure out what fits.

Like line 4 needs you to use line 2 for a substitution. Then you use that substitution on something in line 3.

[ruby elbow]

ok

[oak chasm]

So, you need line 3 to be exactly like line 4, except without a substitution done just yet.

[ruby elbow]

yes

what is the difference between a rectangle and a rhombus

rectangle requires 4 right angles

what does a rhombus require

[oak chasm]

A rhombus is like a square but without needing right angles.

The sides all have to be the same length.

A rectangle is like a square but without needing the sides to all be the same.

The angles all have to be right angles.

So, there are two differences.

Rectangles don't need all sides the same length. Rhombuses do need all sides the same length.

Rectangles need all angles to be right angles. Rhombuses don't need all angles to be right angles.

[ruby elbow]

ok

thank you

[oak chasm]

You're welcome.

[ruby elbow]

.close

[lone heart]

Channel closed

Closed by @ruby elbow

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.

[ruby elbow]

i know its ABE and CDE

but is it angle side angle

or side angle sde

side

would DE and CE be the two sides

and vertical angle E

be the angle

?

nope

of course

as always im wrong

im never correct

oh wow

guess what

[ivory pivot]

you are correct to say you are never correct

[ruby elbow]

i wasn't even correct about this

sometimes i want to kms because of this stuff

im never right with geometry congruence

[oak chasm]

I'm not sure how Alessandra would prove it from that, but you know that AB and CD are congruent, so those have to be in there.

[ivory pivot]

they are not so easy actually

Alessandra is very special ...she can do really great

[oak chasm]

Yeah, geometry proofs are hard for a lot of people.

[ivory pivot]

@Ann is very good in geometry indeed

[oak chasm]

Since you know they're using the sides AB and CD, it's either ASA or SAS.

[ruby elbow]

why tho

[oak chasm]

But there's no easy way to get the half-diagonals.

Well, that's given to you.

[ruby elbow]

i thought the answer was ABE and CDE because

[oak chasm]

It says that AB and CD are congruent and parallel, so AB and CD are going to be in there.

[ruby elbow]

there was nothing

on the other sides

and it looked like we were only wokring with ab and dc

and not bc and ad

[oak chasm]

Right.

[ruby elbow]

so why is it ABC and CDA

[oak chasm]

Because you can use DC and AB congruent and AC congruent in both.

Do you see how those pairs of sides match in each triangle?

[ruby elbow]

?

[oak chasm]

Well, you have ADC and ABC, right?

[ruby elbow]

yes

[oak chasm]

OK, and AC is on both of them, right?

[ruby elbow]

yes

[oak chasm]

And CD in ADC and AB in ABC are congruent, right?

[ruby elbow]

?

[oak chasm]

Do you see the | marking on AB?

You circled it in purple.

Do you see it?

[ruby elbow]

yes

[oak chasm]

Do you see the | marking on CD?

[ruby elbow]

yes

they are parallel

[oak chasm]

That means they're congruent.

No, the > is for parallel.

[ruby elbow]

ok

[oak chasm]

So, ABC and CDA have AC and AC congruent.

And ABC and CDA have AB and CD congruent.

Does that make sense?

[ruby elbow]

hold on

im drawing it out

[oak chasm]

Right.

[ruby elbow]

ok

[oak chasm]

So, AC is the same in both of them.

[ruby elbow]

yes

[oak chasm]

And CD and AB are the same in both of them.

[ruby elbow]

yes

[oak chasm]

OK, if we use ASA or SAS, and those three things (either ASA or SAS) are the same, the triangles are the same.

So, we have two pairs of sides that are the same length in both triangles.

If we can prove that the angle in between those sides is the same in both triangles, we know the triangles are the same.

So, angle DCA and angle BAC.

Does that make sense?

[ruby elbow]

ok

[oak chasm]

The main idea with SAS or ASA is that if you know SAS, you know all the sides and all the angles. You know the whole triangle just from two sides and an angle.

Same with ASA.

The idea is to use a little information to get all the information.

[ruby elbow]

yes

[oak chasm]

So, if the little information is the same in both, all the information is the same in both.

[ruby elbow]

ok

but why is it

[oak chasm]

Why is what?

Oh.

[ruby elbow]

why is it abc and cda

and not abe and cde

[oak chasm]

Because we only know one side and no angles of ABE and CDE.

With ABC and CDA, we know two sides and no angles.

So, ABC and CDA are better to use.

[ruby elbow]

ok

[oak chasm]

Plus, you can see that transversal crosses parallel lines thing from the previous problem to get that angles CAB and DCA are the same.

CA crosses both AB and DC, which are parallel lines.

[ruby elbow]

yeah

[oak chasm]

So, it's not just that we know more sides, it's that the angle in between can be gotten so that we have SAS.

[ruby elbow]

thank you again

[oak chasm]

You're welcome.

[ruby elbow]

.close

[lone heart]

Channel closed

Closed by @ruby elbow

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.

[wooden jewel]

My finals are in 2 days please help

[wooden jewel]

How do i do b)

I need to prove that the subspace is closed under addition and multiplication

but idk how to set it up

[brittle ember]

I think statement b is false

for ex (0,1,0) + (1,0,1) =(1,1,1) is not in W

where $ (0,1,0),(1,0,1) \in W $

[west cipher]

what does the | mean after R^3

does it mean such that

[wooden jewel]

uh i guess yeah

[wooden jewel]

ohhh u right

alr TY

how about a) tho

[west cipher]

that shows W is not a vector space

his example

[wooden jewel]

how do I know if a point is in between two planes

[wooden jewel]

yeah

[west cipher]

oh

lol

[wooden jewel]

i mean a)

[west cipher]

Well if its between the planes then obviously it wont be on them

whic hseems to be the case

[brittle ember]

I think you can get two point on each plane that has same x,y coord with (1,2,1)

[west cipher]

^

[brittle ember]

Then you might be able to compare z values

[wooden jewel]

like for this

[brittle ember]

im not sure how "between" is defined

[wooden jewel]

i mean its like sandwiched in between

i guess thats what it means

alr ty i think i got it

.close

[lone heart]

Channel closed

Closed by @wooden jewel

Use .reopen if this was a mistake.

✅

[wooden jewel]

I also need help with this pls

how do i do a)

[brittle ember]

It's because $(B^TAB)^T=B^TA^T(B^T)^T$

[ocean seal]

Dri111

[wooden jewel]

OHHHHH TYYYY

i forgot about that

omg

.close

[lone heart]

Channel closed

Closed by @wooden jewel

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.

[wary jacinth]

a is easy, 2300

b i can brute force but looking for a more elegant solution

c can be obtained after b is obtained

so mainly find an elegant solution for B

[surreal meadow]

to an extent there's a need for brute force, i would think

you can count for one horizontal line and multiply by 10, 5 for all the horizontal and 5 for the vertical

then consider the diagonal cases

[wary jacinth]

yeah ik, you can do horizontal, vertical, diagonal w/ gradient +-1 then w/ gradient +-2 (multiply by 4 to account for +-1/2)

but i wanna know if there is a better solution, faster, elegant and more beautiful

[lone heart]

@wary jacinth Has your question been resolved?

[lone heart]

@wary jacinth Has your question been resolved?

[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.

[wise summit]

Can someone explain how the Horizontal asymptote is y=x+2

[wise summit]

Please help me

<@&286206848099549185>

[remote heron]

well u can think of what will happen as x gets very positive, or very negative

as $x \to -\infty$, then $\sqrt{x^2+1} \approx \sqrt{x^2} = x$

[ocean seal]

jan Niku

[shadow quail]

So it will be x-2?

[remote heron]

it depends on if you are talking about positive x or negative x

https://www.desmos.com/calculator/4ocs4sf1hy

look here

x-2 describes the asymptote for positive x

x+2 describes the asymptote for negative x

[wise summit]

I don't know to explain English but did you understand me?

[remote heron]

it is because $\frac{-2x}{\sqrt{x^2+1}} \approx -2$ for really big $x$

[ocean seal]

jan Niku

[remote heron]

im sorry i dont know arabic

oh it says -infty

[wise summit]

Yea

[shadow quail]

I understand now

[remote heron]

for $x \to -\infty$ then $\frac{-2x}{\sqrt{x^2+1}} \approx 2$

[ocean seal]

jan Niku

[remote heron]

because $x^2+1 \approx x^2$

[ocean seal]

jan Niku

[remote heron]

so $\sqrt{x^2+1} \approx \sqrt{x^2} = |x|$

[ocean seal]

jan Niku

[remote heron]

but x is negative

so $-2x$ is positive

[ocean seal]

jan Niku

[remote heron]

does this help?

[shadow quail]

Can I ask one doubt?

[remote heron]

yea

[shadow quail]

A method says horizontal and vertical asymtotes will be coefficient if the highest power of x should be 0

[wise summit]

Thanks for help

[remote heron]

the way of looking at coefficients works well when you have polynomials

[wise summit]

Yes thanks

[remote heron]

but here you have a square root too

its more complicated

I'm not sure i understand your question completely

if you have something like $\frac{2x}{5x+1}$, then yes this will have asymptote $\frac 25$

[ocean seal]

jan Niku

[remote heron]

for example

[wise summit]

Yea it's eazy

[shadow quail]

I was wrong

Sory

[wise summit]

But my example it's very hard

[shadow quail]

If we need to find asymtotes parellel to x axis or y axis then we look at cofficent of highest powers

[wise summit]

Can you write the solution in paper and take a picture of paper

[remote heron]

hmm I typed a lot of stuff, is there something you dont understand?

[remote heron]

asymptotes parallel to the y axis always come from the denominator, unless you have special functions like logs

[wise summit]

I am sorry my english is not good a understand something

[shadow quail]

Same

[remote heron]

[shadow quail]

Suppose we have 1/x

So parallel to y axis will be?

[remote heron]

1/x has a vertical asymptote, yea

the asymptote is parallel to the y axis

i guess you could say

it also has horizontal asymptotes

,w plot 1/x

[ocean seal]

Results provided by WolframAlpha

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

Requested by janniku

[shadow quail]

xy=1

[remote heron]

but since this is $\frac{x^0}{x^1}$ the horizontal asymptote is at 0

[ocean seal]

jan Niku

[shadow quail]

So here we can see highest powwr of y is x and vice versa

[remote heron]

sure, yea

[shadow quail]

How many asymtotes depends on highest powers?

[remote heron]

it depends on many things

roots of the denominator can give you vertical asymptotes

you could look at something like

[shadow quail]

Or highest coefficient y

[remote heron]

,w plot (x^3+x^2+x+1)/(x^2-5x+6)

[ocean seal]

Results provided by WolframAlpha

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

Requested by janniku

[remote heron]

this is really complicated and has a lot of asymptotes

i dont know if there is an easy rule to count them all

[shadow quail]

X^2-5x+6 roots

[wise summit]

I wrote example about what i mean

[remote heron]

,rotate -90

[remote heron]

whot do you mean? what is this an example of?

[wise summit]

About what i sent

Before

It's not the same

That different example

[remote heron]

about the coefficients of the leading power?

[wise summit]

D : y = x+2

Did you know how i got this answer of the last example

?

[remote heron]

im assuming you just looked at limits?

[wise summit]

Yes

First thing i did

[wise summit]

Thanks you for helping i bother you a lot im sorry but my english is weak

[remote heron]

you're fine I just think i do not understand what you are asking

I asked if any other helpers can take a look but i'm not sure anyone else will understand

[wise summit]

There is something called a diagonal asymptotic. I just want to know how this answer turned out

In the first expmle

[remote heron]

oh, in the original question?

[wise summit]

Yes

[remote heron]

i think your asymptote is right

[wise summit]

I know that

But how

[remote heron]

is it the sign that is confusin gyou?

or what part

[wise summit]

Yes

[remote heron]

is is just because $\frac{-2x}{\sqrt{x^2+1}} \approx -2 \frac{x}{|x|}$

[ocean seal]

jan Niku

[remote heron]

for large x

do you feel okay with this part?

[wise summit]

Yes

[remote heron]

okay

$\frac{x}{|x|}$ actually has a name

[ocean seal]

jan Niku

[remote heron]

its called sign (x)

,w graph x/|x|

[ocean seal]

Results provided by WolframAlpha

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

Requested by janniku

[remote heron]

its either 1, or -1

its 1 where x is positive

[wise summit]

1

[remote heron]

and -1 where x is negative

[wise summit]

Yes

[remote heron]

so $\frac{-2x}{\sqrt{x^2+1}} \approx -2 \frac{x}{|x|} = -2\text{sgn} \qty(x)$

[ocean seal]

jan Niku

[remote heron]

yea?

[wise summit]

Yea

Thanks for help

[remote heron]

im sorry but i gotta go to bed

i hope you are able to sort it out

[wise summit]

Ok thanks

[wise summit]

I got it but in our country we use different way to solve this

When i have our solution of solve i will show you

[lone heart]

@wise summit Has your question been resolved?

Channel closed

Closed by @wise summit

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.

[alpine sable]

hello

[ruby sinew]

Btw when asking for help rather than greeting ask the question first so we can understand it quicker

[alpine sable]

oh okay

any easy way to finish this chapter?

..

.close

[lone heart]

Channel closed

Closed by @fathom dagger

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.

[shy fractal]

Can someone help me with understanding this

[nimble fern]

I think it's easier to understand using a unit circle representation

if you are ok with reading, you can try this
https://math.libretexts.org/Bookshelves/Precalculus/Precalculus_1e_(OpenStax)/05%3A_Trigonometric_Functions/5.02%3A_Unit_Circle_-_Sine_and_Cosine_Functions

if you are better with videos, use this
https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:trig/x2ec2f6f830c9fb89:unit-circle/v/unit-circle-definition-of-trig-functions-1

but either way cannot be explained using a simple single sentence

Hope that helps!

[shy fractal]

ye it did help thanks

originally I was thinking of using an unit circle but I didnt get how to solve the other value for c n d

[nimble fern]

oh i see

it's pretty tricky for x's after pi

in case you have a clear representation like this question you've post, you can try plug in some suitable numbers and use the "rules" to solve it

sin(x)=sin(x)
sin(pi-x)=sin(x)
sin(2npi+x)=sin(x)

[shy fractal]

so do I still get the same answer??

[nimble fern]

well, after you plug in a number, you can read the pattern, and then the final step would be replace that number back to an unknown

this however

is kinda dangerous

since you might miss a step or 2, but

[shy fractal]

I think I’ll just remember the formula given n if I come across a question like that I’ll just write it like that

[nimble fern]

it's easier to visualise

another method would be drawing on the graph

[shy fractal]

yes but I don’t trust my graphing skills

[shy fractal]

anyways it was just helpful that even undergrads find this hard

[nimble fern]

Cheers mate!

[shy fractal]

cheers

.close

[lone heart]

Channel closed

Closed by @shy fractal

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.

[waxen hamlet]

How could I start this question?

Wow, help 0

[alpine sable]

do you know the fundamental theorem of calculus

[waxen hamlet]

Ya

[alpine sable]

[
\m{F'}x = \dv x \int_1^x \int_1^{t^2} \f{\s{5+u^2}}u \dd u \dd t
]

[waxen hamlet]

[alpine sable]

do you have an idea on how to do this

[waxen hamlet]

No

We went over the fund. theorem

havent touched this far yet

If you could guide me on how to put the parts together, I can do the math

[alpine sable]

[
\m{F'}x = \dv x \int_1^x \int_1^{t^2} \f{\s{5+u^2}}u \dd u \dd t= \dv x \int_1^x \m ft \dd t
]
Try using the fundamental theorem of calculus here that states [
\dv x \int_a^x \m ft \dd t = \m fx
]

[alpine sable]

does this help

[waxen hamlet]

YEa

Lemme do the math and I'll post it here

[alpine sable]

ok

[waxen hamlet]

[alpine sable]

dont forget to take the second derivative

[waxen hamlet]

I got sqrt(21)/2

Although it could be a flunk cause I based it off of a sample problem on google

[alpine sable]

show your work

[waxen hamlet]

taking photo

one sec

Actually got sqrt(21)

ignore the first thing on top

[lone heart]

@waxen hamlet Has your question been resolved?

[nimble fern]

i think it's correct

,w d(integrate_1^(t^2) sqrt(5+u^2)/u du)/dt

[ocean seal]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by biscuityxd

[nimble fern]

hmmm

,w d(integrate_1^(t^2) g(u) du)/dt

[ocean seal]

Results provided by WolframAlpha

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

Requested by biscuityxd

[nimble fern]

yea, looks good

[lone heart]

@waxen hamlet Has your question been resolved?

[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.

[vagrant agate]

if (2x-9) (4x+3) and (2x-3)(4x-9) both = 8x^2 - 30x + 27 what would the roots of the equationbe?

[echo socket]

Why do you think they are both equal to 8x^2 - 30x + 27?

[vagrant agate]

Well when i expanded both thats what it became

[vapid shuttle]

You expanded one wrong

They don’t both equal that

[vagrant agate]

Oh wait one is -27 ok sorry for wasting your times

.close

[lone heart]

Channel closed

Closed by @vagrant agate

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.

[alpine sable]

!show

[lone heart]

Show your work, and if possible, explain where you are stuck.

[hard inlet]

I don't know how to start

[rose sigil]

[vale wigeon]

@hard inlet given a pen costs 3 times as much as a file, how many pens would match the total price of 5 pens and 12 files?

[abstract furnace]

[hard inlet]

41 pens

[tall topaz]

i thought this was kanna

[vale wigeon]

are you sure about this?

how did you get 41 pens

[rose sigil]

[lone heart]

@hard inlet Has your question been resolved?

[vale wigeon]

👻 ❓

[hard inlet]

<@&286206848099549185>

[vale wigeon]

??

do you still want help with this or not? @hard inlet

[hard inlet]

No

.close

[lone heart]

Channel closed

Closed by @hard inlet

Use .reopen if this was a mistake.

[vale wigeon]

as you say

[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.

[tiny bay]

(3i) / (1+3^(1/2))

i want to find the absolute value and the argument

first thought i was thinking of marking them on a x, y grid where y = imagary part and x = real part

and then then to find absolute value, i do sqrt( (3i)^2 + (1+3^(1/2))^2 )

but i am not sure

[nimble fern]

Do you mean $\frac{3i}{1+\sqrt3}$

[ocean seal]

Biscuity

[tiny bay]

yea

[nimble fern]

oh in that case

[nimble fern]

this number is purely imaginary

oh ..

you missed an i

in your original post

i see

[tiny bay]

yeah obs i meant to have i like in the pic

sorry

[nimble fern]

it's okay

do you know if we have complex denominator

we can expand the fraction by the complex conjugate of the denominator?

[tiny bay]

yeah wait

[nimble fern]

$\frac{3i}{1+i\sqrt3}\cdot\frac{1-i\sqrt3}{1-i\sqrt3}$

[ocean seal]

Biscuity

[tiny bay]

Could you say that if you have a complex denominator you should always expand it with the conjugate of the denominator?

[nimble fern]

not always, but in many situations, yes

and specifically in the type of question you are doing, mostly yea

[wraith chasm]

Is the part where I halved du legal or do I have to use integration by parts?

[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).

[tiny bay]

first attempt

[nimble fern]

lets see

looks good

although my usual practice would be make it into the form a+bi, but it's only my habit

so what i wrote is
$\frac{3\sqrt3}4+\frac34i$

[ocean seal]

Biscuity

[nimble fern]

which is the same as your final expression

[tiny bay]

yeah

[nimble fern]

now we have to find modulus and argument of the complex number

[tiny bay]

ok so now i know the coordiantes

[nimble fern]

modulus is easy, you just do the Pythagorean thingy XD

[tiny bay]

(a + ib)

sqrt (a^2 + ib^2)

[nimble fern]

and the argument, did you teacher/prof tell you to write the Principle Argument or just argument?

[nimble fern]

nah

for any complex number a+bi
we have modulus=sqrt(a²+b²)

no "i" inside the sqrt

[tiny bay]

ah thanks

i am not sure what principle argument is

wait

when they say argument i understand it as the angle between the modulus and the real

[nimble fern]

oh ok

if that's the case

you can use the method they teach to find the arg

since from the complex number, we know it's in first quadrant.

so the argument will be between 0 to pi/2

[lone heart]

@tiny bay Has your question been resolved?

[tiny bay]

i calculated the value of modulus as

(sqrt 6 + sqrt 3)/2

i am not sure how to find the argument

the method they teach i looked at my notes and i have this

i remember something about the teacher said, it can only be two different ones, does this apply here? like it can either be small or big

i have some more notes i am checking but i am having lunch now

thanks so far

[lone heart]

@tiny bay Has your question been resolved?

[tiny bay]

z should be 3/2 but i dont really understand the double sqrt part

<@&286206848099549185>

[charred jewel]

dont you have to square the real and imaginary part

and second thing you cant just break square roots up like that

if its a multiply in the middle, sure, but if its addition you cant do that

[tiny bay]

thanks

now i just need to find the argument

how do you find it for the denominator

[charred jewel]

i mean you can just find the argument by finding the arctan of im/re

it simplifies pretty nicely

[tiny bay]

and how do you do that

i have been prioritizing other courses so my math is rusty af

[tiny bay]

sqrt(3) / 1
which is sqrt(3)
and then you should know by heart that sqrt(3) is 60 deg which is pi/3

also a little bit unrelated, i am having a test in 1 week and i had to prioritize other courses to be able to pass, now i passed those and i am going to do my best at this math test but i might have to do it the 2nd time

but still want to do my best

.close

[lone heart]

Channel closed

Closed by @tiny bay

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.

[upbeat eagle]

Graph 3

[alpine sable]

what about it

[upbeat eagle]

increasing acc, decreasing acc, and then the star is 0 acc?

[alpine sable]

it fluctuates

[upbeat eagle]

wym

a bit confused

[echo socket]

The star is not 0 acc

[upbeat eagle]

no?

[alpine sable]

no

look

[upbeat eagle]

[alpine sable]

the graph might or might not continue its fluctuacity

by the looks it will just fluctuate like that

so it never touches 0

[upbeat eagle]

Wont this line be areas of 0 accelration tho?

Oh so it has to touch 0 velocity

for 0 acceleration?

[restive cloak]

the star graph is increasing to a certain point and then it start's going down [decreasing] again

[upbeat eagle]

Yeah, ive understood that

[alpine sable]

no

that wont be 0 acc

[upbeat eagle]

I just remember my teacher saying something that there are 2 stationary points of points where acceleration is monetaryily zero

Thats why im a bit confused

Maybe i heard wrong

[restive cloak]

it means that the derivative of this graph touches 0 at a certain point

[upbeat eagle]

oh

So this is what x^2 graph so dy/dx = a straight line

and it goes through 0?

So 0 acceleration there ?

[restive cloak]

if f(x)= x^2 then the derivative of f(x)=2x

[upbeat eagle]

Yes?

[restive cloak]

so you know that 2x=0 for x=0 and that 2x>0 for x>0

it means that the graph oh x^2 is increasing for x>0

[upbeat eagle]

nvm understood

[upbeat eagle]

ah ok got that

[restive cloak]

try opening geogebra and graph it

graph some functions and their derivatives and observe them

[split mantle]

Increasing: $$f(a)<f(b), a<b$$ Decreasing $$f(a)>f(b), a<b$$ Zero: $$f(a)=f(b), a<b$$

[ocean seal]

Good

[upbeat eagle]

Would i find the 2 stationary points of acceleration there?

[upbeat eagle]

yes yes

[restive cloak]

in the graph of x^2 there is only 1 stationary point

send me graph of which you need to find the stationary points

[restive cloak]

do you know the concept of tangent line?

[upbeat eagle]

Yes

dy/dx = 0 is max /minima

[restive cloak]

yes

[upbeat eagle]

so here there are 2 max/min so 2 stationary acceleration points

?

[restive cloak]

yes right

[upbeat eagle]

bc i think velocity time dy dx = acceration

OHhh

Ok thank you haha

d2y/d2x= >0 <0

do you think you can remember Which one is a maxima and a minumum

Do you know, i forgot and cant find it

sry lol

[restive cloak]

you see them in the graph

the maximum is higher than the minimum

[upbeat eagle]

So its >0?

and minimum is <0 ?

Ill find it somewhere in my book again tehe

thank you JO

and everyone esle

:3

.clsoe

.close

[lone heart]

Channel closed

Closed by @upbeat eagle

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.

[rapid trellis]

it's asking for the sum of the roots within the given interval and I'm asking what I did wrong

[rapid trellis]

I will try to explain what i did

[lone heart]

@rapid trellis Has your question been resolved?

[robust bear]

can you explain the last part

from where you wrote the two eqns

[rapid trellis]

i did sin2x = m, to not have to write sin2x everytime

[robust bear]

not that

[rapid trellis]

oh

[robust bear]

the general solution

for

sin

[rapid trellis]

if sinx = sina

x = a + 2πk
x = 180 - a + 2πk
no?

[robust bear]

ah alr, you separated the solns

okok

i remember them like this

[rapid trellis]

so......

[robust bear]

one sec

[robust bear]

whats the ans

[rapid trellis]

E

[robust bear]

okay

ye

[robust bear]

try using the general sol from this form

the ans will match

remember

[lone heart]

@rapid trellis Has your question been resolved?

[rapid trellis]

I've never seen this

we're not taught that

[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.

[fallen raptor]

How do I get this area using double integration

[lone heart]

@fallen raptor Has your question been resolved?

[lone heart]

@fallen raptor Has your question been resolved?

[lone heart]

Channel closed

Closed by @fallen raptor

Use .reopen if this was a mistake.

[fallen raptor]

.reopen

[lone heart]

✅

[lone heart]

@fallen raptor Has your question been resolved?

[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.

[cobalt needle]

I have done part a but part B has me stumped

[cobalt needle]

wow help 0 what an honour

for part A I got $512-114x+18x^2$

[ocean seal]

sealpup321

[cobalt needle]

which is correct

so for part B I tried putting it so $512a + 512bx -128 = 0$

[ocean seal]

sealpup321

[opal ingot]

Wait, why?

[cobalt needle]

I wasn't sure what to do

?

[opal ingot]

Well, perhaps you could multiply the a+bx through what you have?

[cobalt needle]

sorry I am still confused

[opal ingot]

(a+bx)*(512-114x+18x^2 + (higher order terms in x))

[cobalt needle]

ah

so i did this for the first and second terms

[opal ingot]

You already know how to expand (2-x/16)^9, or at least the first part of it

OK

[cobalt needle]

but was confused after that

[opal ingot]

OK, so what did you get?

[cobalt needle]

0=0

...

i separated bx on one side of the equation then set them equal to eachother

which just simplifies to 0 = 0

[opal ingot]

Well, what if you just multiply (a+bx)*(512-114x+18x^2 + ...) -- what does that give you?

[cobalt needle]

huh

i... didn’t think about that

512a = 128

right?

[opal ingot]

This isn't really an equation (no equal signs), at least not at this point. It's more of a "simplicatio..." I think you got it

[cobalt needle]

yup got it thanks I just completely overlooked that

its a bit like how you solve partial fractions right?

[opal ingot]

Yes, that's one way to look at it

[cobalt needle]

great thanks for the help 😄

.close

[lone heart]

Channel closed

Closed by @cobalt needle

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.

[iron tapir]

A hen knows how to count. We put it in front of a packet of 2024 eggs. She counts the eggs from the first packet and places them in a second packet. Each time she has counted 4 eggs, she lays an egg which she places in the packet of those she still has to count.

How many eggs will there be in total to count for the hen?

[iron tapir]

my answer is 2698

<@&286206848099549185>

[frail jewel]

whats the q?

[frail jewel]

this?

[iron tapir]

yes

[fallen verge]

Show work

[iron tapir]

if she counts the first batch of 2024 eggs, she will lay 506, then she will count 506, and lay 126, then she will count 126 and lay 31, then she will count the 31 and lay 7, then count the 7 and lay 1, and she will count the 1; 2024+506+126+31+7+1

[fallen verge]

You forget that there is still some left over from other counts

[lone heart]

@iron tapir Has your question been resolved?

[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.

[gray oxide]

Hello

[gray oxide]

Rralosed this is more of an circuits question .close

.close

[lone heart]

Channel closed

Closed by @gray oxide

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.

[alpine sable]

How can I check the right totality of a function?

[alpine sable]

Can I set it to a number c and see if there’s any possible solution?

And if the solution is unique, then it’s injective too right?

post your question

do you know the definition to injective functions

[lone heart]

@alpine sable Has your question been resolved?

[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.

[royal meadow]

i'm going insane. why is sample variance chi-squared distributed again? it makes sense but i can't prove it

[balmy grail]

The the observations are normally distributed so the sum of the obs in sum((x-xbar)^2) is the sum of squared normal variables which has a chi2 distribution https://stats.stackexchange.com/questions/121662/why-is-the-sampling-distribution-of-variance-a-chi-squared-distribution

[solemn juniper]

X-Xbar is not standard normal, be careful

[royal meadow]

yeah

that's what's tripping me up

because X and X-bar aren't independent, right

[balmy grail]

Sorry you also divide by sigma

[royal meadow]

still not independent

[balmy grail]

The draws are independent

[royal meadow]

right

but X_i is not independent of X-bar

so X_i - X-bar is not necessarily normal

right?

[balmy grail]

The X itself is normal, the (x-xbar)/sigma transforms it into the standard normal then the chi2 is the sum of squared standard normal distributions

[royal meadow]

X_i is normal and X-bar is normal

wait

no ok so

lemme write it out

$\sum_i (X_i - \bar{X})^2 = (X_1 - \bar{X})^2 + (X_ 2- \bar{X})^2 + ... $

pls bot

[balmy grail]

Use 2 $

[royal meadow]

$$\sum_i (X_i - \bar{X})^2 = (X_1 - \bar{X})^2 + (X_ 2- \bar{X})^2 + ... $$

[ocean seal]

Kaisheng21

[royal meadow]

ok ok

right so

why should those first two terms on the RHS be independent

they both involve X-bar

[balmy grail]

So the handwavy explanation would be xbar is constant.

[royal meadow]

it's noooot

what

i also considered $$\sum_i (X_i)^2 - \bar{X}^2$$ but these aren't independent either

[ocean seal]

Kaisheng21

[balmy grail]

Look at cochranes theorem

[royal meadow]

you what

jesus christ wtf is this

is it actually this hard

well ok this works, i didn't think i'd have to bring in like industrial machinery for it

[balmy grail]

Haha yeah if you’re not satisfied with mu and sigma are parameters, these give us std normals Z, the Z are squared and summed into a chi2. Then you have to use theorems to be more rigorous

[royal meadow]

yeah if they were actually independent this would be so easy

[solemn juniper]

welcome to stats at large

[iron tapir]

wdym

[lone heart]

@royal meadow Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[royal meadow]

.reopen

[lone heart]

✅

[royal meadow]

.close

[lone heart]

Channel closed

Closed by @royal meadow

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.

[tacit arch]

What part don't you understand

What part exactly

Read it one line at a time and understand one at a time

????

What question are you even asking

You should understand that problem first

[kind sparrow]

If you've never been exposed to how to solve a particular problem, maybe you should consider that asking an AI how to do it, then asking a human to interpret the output for you is a bad strategy.

[tacit arch]

????

Then why are you competing

[kind sparrow]

You do realize that most communities find cheating in competitions to be far worse than cheating on your homework. It is a much more direct causative link that you are not just cheating yourself, but directly cheating your competitors.

Asking people to correct an AI's work instead of your own is the easiest way to annoy people. Not only are you cheating in a competition, but you are cheating via two layers of stupidity.

[tacit arch]

Too late

[lone heart]

@royal wharf Has your question been resolved?

[worn fox]

.close

[lone heart]

Channel closed

Closed by @worn fox

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.

[nocturne iris]

HELLO AGAIN

[nocturne iris]

Given $A \subseteq \mathbb{R}^m$, $x \in \mathbb{R}^m$; how do I show that $d(x, A) = 0$ if and only if there exists a sequence $(a_n)_n$ of elements in $A$ converging to $x$?

[ocean seal]

lilisworld.

[nocturne iris]

idk how to do the first implication

<@&286206848099549185>

[rose sigil]

which one is that?

d(x,a) = 0 implies ... ?

[nocturne iris]

yes

[rose sigil]

you should try to translate d(x,A) = 0 to something like "there are points in A arbitrary close to x"

[nocturne iris]

d(x, A) = inf(d(x, a), a in A) = 0 ?

[rose sigil]

yea

what's another way to say inf(d(x, a), a in A) = 0?

with an epsilon style statement

[nocturne iris]

i dont know

[rose sigil]

what i was thinking of was: for every epsilon > 0, there is an a in A with d(x,a) < epsilon

does it make sense why that's true?

[nocturne iris]

yes

[rose sigil]

and do you see why it helps?

[nocturne iris]

i need to think but i see that it looks like the definition of the limit

[rose sigil]

yea sure

looks kinda like convergence to x

which is what we want

[nocturne iris]

but i still dont know how to make it look like the actual convergence definition

[rose sigil]

ok so we want to make a sequence

one that converges to x

[nocturne iris]

yes

[rose sigil]

this is true for every epsilon

[nocturne iris]

ok

[surreal meadow]

maybe it's worth asking if you see why it is true from that alone

before making it rigorous

[rose sigil]

you can take over if you want i wanna do laundry

[surreal meadow]

lili, do you see how this condition translates to

there are points in A arbitrary close to x

[nocturne iris]

i see why it's true yes

[surreal meadow]

then can you kind of see why there would be a sequence converging to x

for a concrete example, consider A = (0,1), and x = 1

[nocturne iris]

do you want me to give a specific sequence for a in (0, 1) that converges to 1?

like for ex a/n + 1?

@surreal meadow

[surreal meadow]

no, its more like

to visualize this

our goal is to construct a sequence converging to x

and what you should imagine is epsilon slowly getting smaller

meaning we need to get arbitrarily close to x

[nocturne iris]

how do i do that

[surreal meadow]

well there is no explicit construction here since A and x are arbitrary

at least not in terms of explicit elements of A

what you should notice is that

for every epsilon > 0, there is an a in A with d(x,a) < epsilon
is an existence statement

and our goal is to produce a sequence, for which we need the existence of some elements

does that make sense

do you see how they are related in that way

[nocturne iris]

im lost, do we explicitely produce the sequence?

[surreal meadow]

we can't give explicit elements of A because we do not know what they are

but the condition is

for every epsilon > 0, there exists an a in A with d(x,a) < epsilon

so if we are trying to construct a sequence

we probably want to use those elements we know exist

[nocturne iris]

yes

[surreal meadow]

ok so that's the abstract idea we are getting at

[surreal meadow]

what i meant is you cant do something like this

we dont know what A looks like

but anyways

do you know the condition for a sequence to converge?

[nocturne iris]

is it the definition i gave earlier? xn converges to L?

[nocturne iris]

this

[surreal meadow]

yes

im trying to see how i can hint at the following

but we could construct a sequence of a's from a sequence of epsilons

so if you consider some sequence of epsilons that goes to 0, we would want a corresponding sequence of as that converges to something (ideally x)

so for example

say i gave you the sequence eps_n = 1/n

for every epsilon > 0, there exists an a in A with d(x,a) < epsilon
is 1/n > 0?

[nocturne iris]

yes

[surreal meadow]

so what does that condition tell us

[nocturne iris]

i dont really know

[surreal meadow]

for every epsilon > 0, there exists an a in A with d(x,a) < epsilon

we have an epsilon > 0

so what do we know exists

[nocturne iris]

a

[surreal meadow]

such that what

[nocturne iris]

there exists a in A such that d(x, a) < 1/n?

[surreal meadow]

yes

and this is true for any arbitrary n

since 1/n > 0 for any n

[nocturne iris]

ok

[surreal meadow]

so you can take each of these

and call them a_n

so for every n in N, there is an a_n in A such that d(x,a_n) < 1/n