help-39

[fathom sequoia]

thanks for this

[crystal dew]

tbf you kinda stepped in while Nicole is helping but she yielded so all good ig

[wary tusk]

Still a decimal

[frank dawn]

Ok

[fathom sequoia]

you forgot to carry over the uh

actually hold on

yeah i see the problem.

[wary tusk]

What is it

[fathom sequoia]

you forgot to cross over the constant term to RHS and for some reason didnt subtract 1/4 from 2

you had the right idea in mind

just didn't execute it right

[wary tusk]

Omd

It worked

Thank you so so so much

[fathom sequoia]

you're welcome

[west sail]

well done!

[wary tusk]

Can I add u 🙏

[fathom sequoia]

go ahead

im helping out cause the server carried me on math induction 🙏

[wary tusk]

When im done im gonna do the same

[fathom sequoia]

yeah always remember that b's value is always the value of the linear term (x)

[wary tusk]

I will, im gonna do another similar one now

[fathom sequoia]

so when you have a quadratic equation like lets say 3x^2 + 7x + 9; b = 7

👍

[wary tusk]

Ohhh thanks

I GOT THE NEXT ONE COREECR

[fathom sequoia]

nice

@wary tusk if you don't need further help i'll close the channel now

.close

huh

[wary tusk]

Yes

[fathom sequoia]

can you close the channel

using .close

so someone else can use it

[crystal dew]

.close by OP agreement

[pearl pond]

Channel closed

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[wary tusk]

I need some help to check something

[toxic lichen]

i can tell you the last step is bogus for sure

[wary tusk]

😭 😂

🙏 🙏 🙏 🙏

[toxic lichen]

x(x+4) = 4 does NOT imply either of x or x+4 must be 4.

this sort of trick ONLY WORKS WHEN THE RIGHT SIDE IS ZERO

[wary tusk]

Mb mistake

[supple cipher]

Is this linear function?

[wary tusk]

It said to solve simultaneously

[fathom sequoia]

it's a circle

yes?

oh god

this solution is ugly

so uh

hm

we can begin with the fact you totally disregarded y when you did the CTS

[wary tusk]

I substituted

[fathom sequoia]

wait can you show the original question

idk what im dealing with here

[wary tusk]

Solve for x

[fathom sequoia]

is this the original system?

[wary tusk]

Yes

[fathom sequoia]

then how did we arrive to the equation for a circle?

[fathom sequoia]

as shown here

[wary tusk]

No wait wIt

[graceful nacelle]

do you know linear algebra

[wary tusk]

Wait

[graceful nacelle]

sm12986

[wary tusk]

Wrong number

[wary tusk]

Yes

[graceful nacelle]

thats great, we can solve using that instead of substitution and allat

[fathom sequoia]

ooh are we using matrices

[wary tusk]

I wasn't taught that

[fathom sequoia]

oh

[graceful nacelle]

im jk

[fathom sequoia]

so we can't use cramer's rule

meh its fine

[tawdry sage]

what abt simple substitution ?

[fathom sequoia]

just use sub

[wary tusk]

I just tried that

[tawdry sage]

y is already isolated

[wary tusk]

I did in my pic

[wary tusk]

Here

[fathom sequoia]

one moment

solving

[wary tusk]

Oki

Thanks

[fathom sequoia]

fairly tedious

taking me a bit

one moment

found it

you forgot to multiply 7 by 9 when you eliminated the denominator

@wary tusk

[wary tusk]

Where

[wary tusk]

Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

I got x =4 and -8

Correct?

[fathom sequoia]

hm

you somehow got the second root right'

can ypu show me the revised solution

[fathom sequoia]

oh wait

yeah this checks out

nice job

its correct

[wary tusk]

Thnkksss

[fathom sequoia]

just use .close if no further help is needed

[wary tusk]

.close

[pearl pond]

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[pearl pond]

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[earnest sierra]

Can someone help me understand my mistake with probability?

I thought that for the second card you draw, your goal is to make a pair, so you have 3 cards to pick from to make that pair. The rest of the cards do not matter.

[west sail]

I assume the question wants you to draw strictly a pair.
your third card chooses from the remaining 50. but amongst the remaining 50, 2 of them are the same rank as the drawn pair. as such there is a chance you are drawing either card of the same rank as the pair, turning it into a 3-of-a-kind.

the same issue happens with the fourth card.

[earnest sierra]

suppose that you dont have to draw strictly a pair. you can draw two pairs, three of a kind, a full house, four of a kind.

then is there a mistake?

[west sail]

so basically the hand must contain at least one pair.

[earnest sierra]

yes

[west sail]

if so I cannot immediately pinpoint a mistake, sorry.

I'll leave the question to other helpers who may be able to assist.

[earnest sierra]

no worries, ill show more more work to show why there might be a mistake

[toxic fractal]

you seem to be assuming that you match the first card, but you can match any card in the hand...?

[pearl pond]

@earnest sierra Has your question been resolved?

[earnest sierra]

So it could be calculated like this?

Wait I think Im starting to get it

[buoyant stump]

That way, in which draw you form a pair, does that change the probability though? I think it would still be the same as long as you cared about drawing that one particular pair.

[earnest sierra]

Maybe? I can show my work in a second

yeah it's still wrong

I know this is the right answer:

[earnest sierra]

and this is what I tried to do for this

[buoyant stump]

This means they are asking for at least one pair.

[earnest sierra]

yes

[buoyant stump]

So this won't cover all the cases.

[pearl pond]

Channel closed

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[earnest sierra]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[buoyant stump]

There would be too many cases to consider unless you go other way around.

[earnest sierra]

the complement?

[buoyant stump]

Yeah. Basically, what the answer shows.

[earnest sierra]

okay I see, thank you

just wanted to see if you could compute it without the complement.

[buoyant stump]

You could try. There are at max four different kind of pairs we can draw while drawing 8 cards. We don't care about three-of-a-kind and further on.

So you would have to make cases for exactly one pair, then exactly two, then exactly three, and exactly four, I think.

[earnest sierra]

ohhhhhhhhhhhhhh

I see, I think so far I have only computed exactly one pair

actually i havent

[buoyant stump]

But there are many kind of pairs also possible so this goes crazy.

[earnest sierra]

okay

yeah okay we're just gonna do the complement

.close

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[pearl pond]

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[sharp galleon]

yo guys am i stupid or smth

[sharp galleon]

im doin this question where i needa differentiate -lnxcosx

and the mark scheme says its tanx

[toxic lichen]

-ln(cos(x)), you mean.

[sharp galleon]

how is it tanx tho

Ye i sent it

[toxic lichen]

do you know the chain rule?

[sharp galleon]

OHH

WAIt

THE COSx

IS INSIDE

LOL MY BAD

[toxic lichen]

-lnxcosx has an extra x that shouldnt be there and two missing sets of brackets btw

[sharp galleon]

i thought it was

like

-lnx * cosx

im trippin bro

my bad thank you

.close

[pearl pond]

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[toxic lichen]

<@&268886789983436800> another misgendering...

[sharp galleon]

thats just

how i talk

[radiant terrace]

The little diamonds next to peoples names tells you what their gender is.

So you do have that cue

[toxic lichen]

we've been over this. it's a habit worth breaking. some people dont like being called "bro".

[sharp galleon]

the bro wasnt even meant for her i just say it to myself

[toxic lichen]

"bruh" exists

[sharp galleon]

bruh and bro is the same shit what

[radiant terrace]

It is pretty plausible that they didn't mean anything malicious by this

[sharp galleon]

yeh

[radiant terrace]

Just trying to be more cognizant of it going forward is probably the best course of action here.

[sharp galleon]

bruh why would i try to insult someone helping me with math

a bit cringe no?

[radiant terrace]

Some people can be unnecessarily rude to people helping

[sharp galleon]

Lol fair

.close

[warped violet]

can't close something already closed 🙂

[pearl pond]

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[quick venture]

can someone check a please

[buoyant panther]

you only change the sign of the imaginary part

then which one do you think is true?

[quick venture]

the second one

[verbal whale]

Yup

[quick venture]

is this how we find the complex conjugate? (only change the sign of the Im?)

[lilac jackal]

yes

[quick venture]

i need help with part b

[verbal whale]

z² = 1²/(2 + j)², isn't it?

So you just need to expand the denominator 😉

[verbal whale]

Yeah that's another way

You'd get the same result (of course)

[quick venture]

ok so for b

z=2-j/5

and 1/z=?

[verbal whale]

It will be 2+j

Since z = 1/(2 + j)

[quick venture]

i dont understand what they've done for the working

they've squared Z

but if 1/z=2+j

[verbal whale]

It's a typo

[verbal whale]

The ² on the (2 + j) is wrong

[quick venture]

so

thanks

.close

[pearl pond]

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[amber compass]

how do i find the critical points

[amber compass]

is it like y = 0?

[amber compass]

i thought u were a bot for a second

[amber compass]

i don't really understand that

[buoyant panther]

Think about what the derivative of this graph might look like

[amber compass]

i just guessed

its when derivative equals 0

or turning points righrt

imma close this channel now

thanks modus for helping

.close

[pearl pond]

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[scarlet glade]

This is all I know. I don't know how to convert this

[ruby cargo]

,rccw

[pearl pond]

@scarlet glade Has your question been resolved?

[pearl pond]

@scarlet glade Has your question been resolved?

[pearl pond]

@scarlet glade Has your question been resolved?

[sudden sigil]

You're missing an important one: the sum of the squares of x y z is p squared.
First multiply both sides by p.
Then the right hand side is 4x, and the left hand side is x^2+y^2+z^2

@scarlet glade

[pearl pond]

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[swift spindle]

can somenoe show how they did this

im honestly so confused

sqrt (-3 + 4i) to (1 + 2i)

[tardy reef]

let x+iy be the sqrt, so if you square it you should get -3 + 4i. So actually square it and youd get two equations for the real and complex parts

OR write -3 + 4i as e^ix and then square root would be at half the argument

and simplify it that way

[swift spindle]

lowkey i still dont get it

[fierce sky]

$-3+4i=\left(1+2i\right)^{2}$

[jolly parrot]

Roy

[tardy reef]

Lets go with this method

you have some (x+iy)^2 = -3+4i right?

so just expand the left side

[swift spindle]

okay

x^2 + 2xyi - y^2 = -3 + 4i

so u get 2 eq? is that the idea

[tardy reef]

indeed, and since x,y are real numbers, you can treat the real and complex part as two separate equations

[swift spindle]

x^2 - y^2 = -3
2xy = 4

[tardy reef]

mhm

[tardy reef]

quite easy to solve these two equations and youd find the two roots

[swift spindle]

am i doing this wrong

[tardy reef]

nahh, this is correct

let x^2 be some u, and you have a quadratic in u

[swift spindle]

so x = 2i, 1

[swift spindle]

but shouldnt there only be 1 value for x

which one do i reject

or know to reject

[tardy reef]

technically only 1, not 2i

[swift spindle]

ohh

[tardy reef]

coz the representation x+iy has x and y in reals

[swift spindle]

so i write

reject as x belongs to R but R with the double line

[tardy reef]

yea

[swift spindle]

i see

tysm

[tardy reef]

also, these should both be +/- if im right

coz you got x^2 = smthn, and that has two roots

[swift spindle]

oh

true

bruh

wait so then i get x = +- 1

but then how do i know which one to reject

[tardy reef]

mhm

[tardy reef]

-3+4i also has two squareroots

each of x give you one sqrt value

[swift spindle]

doesnt that mean i get like 4 values

[tardy reef]

no...

x = -1 would give you y = -2

x = 1 gives you y = 2

[swift spindle]

oh wait

so basically

[tardy reef]

so this

[swift spindle]

i get x + yi = 1 + 2i and -1 - 2i

but because it is already +-

then i just write -1 +- (1+2i) / 2?

[tardy reef]

where did this come from?

[frosty sentinel]

yea

[swift spindle]

here

[frosty sentinel]

you only need the ± once

[swift spindle]

i mean when i put it back in

okay th

thx

[tardy reef]

ahh, I see, you missed that extra -2i there then

yea, then it gives you the exact same stuff written in there

that sqrt usually indicates the principal root

so +- sqrt covers both the values

[swift spindle]

ohh okok

yeah thx

.close

[pearl pond]

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[sharp smelt]

how do I do this

[sharp smelt]

part(b)

idt there's an inequality I can use is there

maybe CLT

[bitter herald]

\lm it comes down to showing [
\lim_{n\to\infty}\vb P(Y_n\ge\epsilon) = 0
]
does it not

[sharp smelt]

yeah

[bitter herald]

fix an arbitrary epsilon > 0 and wlog assume n > epsilon

[sharp smelt]

oh right

so then its sufficent to show P(X_n>esp)=0

[bitter herald]

not quite

if X_n = 0 then you have Y_n = 0 which is lesser than epsilon

what happens if X_n >= 1?

[sharp smelt]

wdym

[bitter herald]

well like if X_n >= 1 then Y_n >= n

[sharp smelt]

yeah

[bitter herald]

since n > epsilon this means that Y_n > epsilon

so Y_n >= epsilon is exactly the same as X_n >= 1

[sharp smelt]

ooh

okay

got it

tq

[pearl pond]

@sharp smelt Has your question been resolved?

[pearl pond]

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[fading ledge]

What's the difference between 3 and 2?

[olive fractal]

zeros vs poles right

[fading ledge]

Yes

[olive fractal]

yeah that's the difference

i don't understand

[fading ledge]

Yes explain please

Why one thing is essential singualrity

And other is non isolated

[olive fractal]

im not sure actually im not well versed in complex analysis

i think how it is

is that for case 1 as n approaches infinity

it forms a singularity at z = a

for the second case there are infinitely many poles that accumulate at z=0

so it's crowded

[pearl pond]

@fading ledge Has your question been resolved?

[pearl pond]

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[tawdry solar]

!status

[pearl pond]

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

[tawdry solar]

1

[ivory basin]

Are you familiar with the unit circle

[tawdry solar]

yes

[olive fractal]

you can draw an acute theta

[ivory basin]

Draw one

[olive fractal]

and wtv they specify

[tawdry solar]

yeah

okay i drew one

[olive fractal]

yk how sin theta and cos theta are the y and x values respectively?

[tawdry solar]

ye

[olive fractal]

you can show those satistfy the relation asked in the question

[tawdry solar]

huh

how

[olive fractal]

for example

in a bit

they ask you to show that

cos(180-x) = -cos x

i replaced theta with x for simplicity

so you can plot x and 180-x as angles on the unit circle

and show that their x values (cos(180-x) and cosx) are related as showin in a bit

[tawdry solar]

ohh

cuz the chapter was about like drawinf the graph of sinx

i forgot that i can use unit circle

i thought we had to use the graph

thanks

.close

[pearl pond]

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[royal galleon]

If you have F proportional to m and F proportional to M. How can you combine them to get F proportional to mM. What is the intuition and proof behind this?

[unborn abyss]

well, what does F proportional to m mean?

algebraically

[pearl pond]

@royal galleon Has your question been resolved?

[royal galleon]

it means that it only depends on m

I get that the force depends on both but connection that then force = mM alludes me

[unborn abyss]

no it means it depends on m, not only

it means $F = \bsq m$ and we also have $F = M \psq$

[jolly parrot]

hayliänus austrǎlis

[unborn abyss]

and like it's kinda clear that one solution of that would be $F = \gsq Mm$

[jolly parrot]

hayliänus austrǎlis

[unborn abyss]

https://math.stackexchange.com/questions/174948/proportional-to-2-separate-variables-vs-proportional-to-product-of-2-variables

this one's pretty straightforward

[west sail]

consider the area of a quadrilateral.
its area is proportional to the length of the quadrilateral and the width of the quadrilateral.
and unsurprisingly, the area of a quadrilateral is A = lw (length times width).
the proportionality constant here is 1, so in this case A = lw and A \propto lw mean the same thing, but this could be a good example too.

[royal galleon]

not seeing it sorry

[west sail]

if you want a non-trivial proportionality constant, consider the volume of a rectangular prism with a fixed height.
this time, the volume of the prism is again proportional to its length and its width, but similarly, V = lwh.
since the height is fixed, we can treat it as a proportionality constant and say that V \propto lw as well.

[royal galleon]

ok I am starting to see it better

[royal galleon]

oh since the green constant is equal to say the blue constant * m or the red constant * M is equal to the blue constant

.solved

[pearl pond]

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[still hamlet]

Let H be a subgroup of G. Prove H is normal if and only if $\forall a,b \in G$, $ab \in H \Leftrightarrow ba \in H$.

[jolly parrot]

Green

[still hamlet]

for $\Leftarrow$ I'm thinking it has something to do with $abxb^{-1}a^{-1}$ for $x,ab \in H$ I just can't figure it out

[calm wing]

is x in H or ab in H?

or both

[still hamlet]

i suppose both

[jolly parrot]

Green

[calm wing]

[still hamlet]

this may be entirely wrong i just feel I have to work with this or something similar

actually if ab is in H this is automatically in H

yeah I'm pretty much clueless

[calm wing]

oh this isn't hard actually

unless i misunderstand something

the condition basically says that H is abelian

so if you have any $x \in H$, then $\forall g\in G, gxg^{-1} = gg^{-1} x = x \in H$. since $x$ and $g$ are arbitrary this shows that the subgroup is normal

[still hamlet]

I don't understand how H is abelian 😭

[calm wing]

wait

[still hamlet]

cuz it only says ab and ba are in H but not that they are equal

[calm wing]

yes damn

you are right

[still hamlet]

also if H is abelian and if g is not in H, does any x in H necessarily commute with g

[calm wing]

yeah that doesn't work

i think i have an idea tho

pick $g \in G$ and $h\in H$ and pick $a=g$ and $b=h g^{-1}$

[jolly parrot]

artemetra

[calm wing]

then you get $ab\in H \iff ba \in H$ simplifies to $ghg^{-1} \in H \iff hg^{-1} g = h\in H$

[jolly parrot]

artemetra

[calm wing]

and h in H is obv true

[still hamlet]

ohh nice

okay how would I prove =>

[calm wing]

oh i thought you did that one

xd

[still hamlet]

oh 😭 I couldn't do either throught of asking <= first

[calm wing]

if ab in H, then conjugate by a^-1 in G

a^-1 a b a = ba in H

[still hamlet]

oh 😭😭

[calm wing]

if ba in H, then conjugate by a

[still hamlet]

thanks a lot ❤️

[calm wing]

of course, i need to get my group theory back together lol

so thanks for the fun problem

[still hamlet]

oh nw

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[exotic gale]

is the answer just half of 5!

[exotic gale]

so 60?

thanks

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[prisma kernel]

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[tacit bough]

Does anyone see where I'm going wrong here? I suspect its related to a distribution error

[indigo pewter]

when going to line 4 to line 5, you say that 4 * 2 * 3/4 * (x +4) = 2x + 8

is that correct?

[tacit bough]

ohh

2 3/4 should become 6/4?

[indigo pewter]

yes

and then when you multiply that by 4, what do you get

[tacit bough]

24

[indigo pewter]

uh

$\frac{6}{4} \times 4$

[jolly parrot]

dork9399

[marble finch]

das ez

[tacit bough]

6

[marble finch]

24/16

[tacit bough]

isn't it 4/1?

[marble finch]

wait is the denominator 1

oh yeah forgot

answer is 4

[indigo pewter]

yeah its 6

so then instead of 2x + 8, it should be 6 * (x + 4)

[marble finch]

hmm

6x + 24

[indigo pewter]

yes

[tacit bough]

thank you!

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[grizzled matrix]

Hi, I got problems with solving question a and b pls help me. I translated it from chat gpt so i hope there are no translation mistakes :
Exercise 3: Favorite Music

60% of all teenagers like listening to rock music.
30% of the teenagers are female and do not like listening to rock music.
40% are male and like listening to rock music.

a) Create a tree diagram and determine the probabilities of the following events:
A: A randomly surveyed teenager is female.
B: A randomly surveyed girl likes listening to rock music.
C: A randomly surveyed person is female and likes listening to rock music.

b) Create a contingency table (four-field table) based on 100 teenagers.

[pearl pond]

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[quick venture]

i dont understand why its negative x

[azure ether]

Because it will be -In(7-y) on the left side

[quick venture]

it is becus i have to use substitution?

[azure ether]

Yep

-In(7-y)=x+c

Both side divide by - gives u =-x-c

[quick venture]

hmm ill use substitution then

[azure ether]

Yep, du=-dy

[quick venture]

thats wrong

[azure ether]

It became -1/u du

Cuz dy=-du

U substitute the dy in the integral with -du, that’s how u get a negative

[azure ether]

!occupied

[pearl pond]

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).

[azure ether]

Post in discussion

@quick venture everything clear?

[quick venture]

yeah im just checking it

[azure ether]

Alr

[quick venture]

so we multiply it by -1?

[azure ether]

Yeah to both side

[quick venture]

okay got it thanks ❤️

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[royal galleon]

when he says r zeros for part a does he mean the multiplicity or distinct zeros?

[cursive wraith]

"The zeros are to be counted as often as their multiplicity indicates"

[royal galleon]

🤦‍♂️. Can't read

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[carmine portal]

why is it that for example when we try to find out the integral of a function say sin(3x) its -cos(3x) divided by 3 which is the derivitive of the inside function. But, if we make the function more complex to the point where we need to use the u-sub method suddenly we don't have to divide by the derivative of the inside function anymore. Why is that?

[hazy pilot]

could you give another example of what you mean exactly

[summer imp]

You do divvide by the inside derivative implicitly.
If you use u = f(x), then du = f'(x) dx so you would have to rewrite dx = du/f'(x).

It's just depending on the shape of the integral you might just spot f'(x)dx directly in there and replace it with du without going through the whole process.
You can always just "solve" for dx and replace that in the integral which involves dividing by the f'(x), which is the "inside" derivative.

[carmine portal]

Say we have a function sin(3x) when we integrate it the result would be -cos(3x) divded by 3 which is the derivative of the inside function
But, lets say the function was sin(3x^2) * 3x, now the result is only -cos(3x^2) without division

[hazy pilot]

well lets just look at it like this

we have integral of g(f(x)) 1/2 * f'(x) dx

true?

where g(x) = sin(x) and f(x) = 3x^2

[carmine portal]

ok

[hazy pilot]

now we substitue u = f(x)

to get our du in our integral, we differentiate both sides, which gives f'(x) dx

[verbal whale]

<@&268886789983436800>

[radiant terrace]

Stop trolling here

[hazy pilot]

which then gives integral g(u) 1/2 du,

[delicate thicket]

gru from despicable me?

aight Ill stop now seems you got it haha

[hazy pilot]

think of your first thing: integral sin(3x) dx

take u = 3x

du/dx = 3 => du = 3 dx

=> dx = 1/3 du

=> integral sin(u) 1/3 du

=> 1/3 integral sin(u) du

=> 1/3 (cos(u) + C)

notice how the 1/3 came to be?

it's the same with the other integral

[carmine portal]

Oh, so we always divide by the inside function but sometimes it gets cancelled doing the integration so it doesn't make it to the result

[hazy pilot]

that's one way to look at it

and also a good step up to learn how partial integration works

[carmine portal]

is that the same as integration by parts ?

[hazy pilot]

have you had integration by parts?

[carmine portal]

I'm not sure, I think I took it briefly in calc 1 but not in calc 2 yet

[hazy pilot]

well lets just stay on this then

always use u substitution if available

it reduces mistakes

also look for chain rule results. like cos(3x^2) 6x, which looks very much like the derivative of sin(3x^2).

it's a bit like the reverse chain rule

like how integration is the reverse of differentiation :)

[carmine portal]

yeah haha

I get it now

Thanks

[hazy pilot]

np

[pearl pond]

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[glossy blade]

hello, how can i solve for the limit ?

[hazy pilot]

what have you tried?

[glossy blade]

i was thinking to cancel out the sin2x to equal it to 1

and go from there

i don’t know how though

[hazy pilot]

what kind of stuff / tricks do you know which have to do with limits

[glossy blade]

well it’s just the asymptotes right

or whatever value it is when x is approaching 0

[autumn trellis]

You have done the correct first step

What happens to cos(5x) and cos(2x) as x->0?

[glossy blade]

would it equal 5 and 2?

[autumn trellis]

Um how?

lim x->0 cos(5x)=5?

They are both 1

Simply put x=0, cos(5*0)=cos(2*0)=1

So you are just left with lim x->0 1/(8x*1/sin(2x))

Now it should be easy with the information they gave you about the limit of sin 2x/(2x)

[sly zealot]

idk if this helps but another way to do it is to use the fact that 1/cot(x) = tan(x) and the fact that lim f(x)g(x) = lim f(x) * lim g(x)

note that the second fact only holds if both limits exist

[glossy blade]

got it ty

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[swift spindle]

ok so

if r^2 = 2

then r = +- root 2

so where did his negative value go

oh wait is r just always > 0

[dense jasper]

yeah usually that’s taken as convention

Assuming you don’t have the complex number 0

[rancid depot]

-sqrt(2)cis(pi/6) = sqrt(2)cis(pi/6+pi)

the 2theta takes care of the sign issues

[swift spindle]

ohhh

i think i kinda get it

[swift spindle]

alr

thnaks guys

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[tawdry solar]

Evaluate [\log_8 \sqrt2]

[jolly parrot]

1.732050807568877293527446341505

[tawdry solar]

idk how to do this

[rancid depot]

are you familiar with change of base

this one

[fair creek]

you should know what $\log_{a}{b^n}$ equals to as well

[jolly parrot]

parabolicinsanity

[fair creek]

don't think that's necessary

[rancid depot]

sure its not necessary, but its much easier to understand and explain

plus it can be useful later on

[tawdry solar]

not yet

the hats next chapter

[rancid depot]

alright

thats fine

[tawdry solar]

nlog_a(b)

[rancid depot]

nice

do you know how to rewrite sqrt(2)?

[tawdry solar]

2^1/2

[rancid depot]

ok nice

so we can write $\log_8(\sqrt 2)$ how?

[jolly parrot]

LocalLunatic

[tawdry solar]

,,\frac12 \log_8 2

[jolly parrot]

1.732050807568877293527446341505

[rancid depot]

do you know how 8 and 2 are related?

[tawdry solar]

8 is a power of 2

[rancid depot]

yup

8 = 2^?

[tawdry solar]

3

8=2^3

so 2 = 8^{1/3}

[rancid depot]

close

2=8^(1/3)

[tawdry solar]

mb

so now what

so its jusr 1/3

[rancid depot]

so we can write 2 as 8^1/3

[tawdry solar]

so the answer is 1/6

[rancid depot]

indeed

[jolly parrot]

1.732050807568877293527446341505

[tawdry solar]

= 1/6

okay thank you

[rancid depot]

youre welcome

[tawdry solar]

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[tawdry solar]

\textbf{13} $V = \frac43\pi r^3$ is the volume of a sphere of radius $r$. Express $\log_2 V$ in terms of $\log_2 r$.

[jolly parrot]

1.732050807568877293527446341505

[plush bramble]

,tex .log rules

[jolly parrot]

riemann

[plush bramble]

First and third

[tawdry solar]

idk

can you help please

[jolly parrot]

1.732050807568877293527446341505

[tawdry solar]

then i can multiply?

[fiery willow]

You can just cancel log_2

[tawdry solar]

why would i do that

[fiery willow]

Same logs bases can cancel

[plush bramble]

Write the right side as 4/3 pi * r^3 first and use product rule from the table

[tawdry solar]

,,\log_2 V = \log_2\frac43 \pi + 3\log_2 r

[tawdry solar]

read the question?

[plush bramble]

Missing log_2 on first term

[jolly parrot]

1.732050807568877293527446341505

[plush bramble]

Yea you're done

[tawdry solar]

is that it

bruh

nah i was getting confused cuz there was a variable

thanks riemann

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[fiery willow]

Wait

Since it has a fraction

Make it subtract

[tawdry solar]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[tawdry solar]

\textbf{1 , a} , By changing to base 3, calculate $\log_9 243$

[jolly parrot]

1.732050807568877293527446341505

[tawdry solar]

im not sure how to do it

ik the rule is

[jolly parrot]

1.732050807568877293527446341505

[plush bramble]

Yea so the left side is what you're starting with and the right side is what you're asked to find

[tawdry solar]

is a =3

[plush bramble]

Yes

[tawdry solar]

okay

thank you

is it

,,\log_9 243 = \frac{\log_3 243}{\log_3 9}

[jolly parrot]

1.732050807568877293527446341505

[plush bramble]

Yes

[tawdry solar]

thank you

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[fading ledge]

Cortical point is that
Where the derivative of function is either zero or not defined

[fading ledge]

Someone told me this definition

But derivative did not mention?

[plain hollow]

what do you mean

[fading ledge]

Function derivative should be 0 and undefined

[worldly jewel]

what do you mean "0 and undefined"

[toxic lichen]

||sounds like op didn't think before he spoke||

[worldly jewel]

[plain hollow]

0 or undefined

[fading ledge]

I see

Thanks

[fading ledge]

I meant first derivative

[fading ledge]

What should I say then?

[plain hollow]

you just made an error saying "0 and undefined" rather than "0 OR undefined" since a function can't be 0 and undefined at the same time

[fading ledge]

Thanks

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[fleet stump]

hii. so this is a graph and i just need a confirmation whether or not this is correct. if you need me to translate something lmk. va is vertical asymptote, ha is horizontal and ka is oblique as.

'ekstremi' means extremes and 'nema rj' means no result

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[autumn fossil]

Prove that the red tangents are parallel to the sides of the yellow triangle

[autumn fossil]

What I know is that the centers of orange circle lie on the the circmucircle of yellow triangle and they lie exact in the middle of the relevant arcs

That implies that they also lie on the angle bisectors of the yellow triangle

but then idk how to proceed

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[autumn fossil]

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[autumn fossil]

(gtg for now)

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[pearl pond]

Please don't occupy multiple help channels.

[fading ledge]

.close

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[robust dock]

I want to prove that D is dense in R,is this rightN

[robust dock]

<@&286206848099549185>

[warped violet]

!15mins

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.reopen

[pearl pond]

✅ Original question: #help-39 message

[robust dock]

<@&286206848099549185>

[ivory basin]

What the heck is D

[robust dock]

Decimal numbers

[ivory basin]

What is a decimal number

[robust dock]

D = {n/10^p/(n,p) in ZxN}

[pearl pond]

@robust dock Has your question been resolved?

[pearl pond]

@robust dock Has your question been resolved?

[tawdry stag]

@robust dock which grade you in?

[robust dock]

idk we don't count like y'all

lets say an equivalent to second year at uni

[fluid rose]

.

[rapid token]

W name gng

[pearl pond]

@robust dock Has your question been resolved?

[wary vault]

what is E(10^n x) ?

[naive zinc]

Any open interval (a,b), there will be p such that 1/10^p < b-a, meaning some n will exists such that n/10^p is in (a,b)

[lapis trout]

how would you prove that? " there will be p such that 1/10^p < b-a"

we have to use the inequality satisfied by the floor function

[wary vault]

this is above my pay grade im sorry

[calm wing]

what is E?

[pearl pond]

@robust dock Has your question been resolved?

[robust dock]

the floor function

or whatever yall call it

E(x) = [x]

its the same thing

[tawdry stag]

prob still not solved?

@robust dock

[robust dock]

Nah I'm just asking about if its correct

I may have made an error

[tawdry stag]

k gl may Allah make it easy for u brother

[robust dock]

for you too akhi

[pearl pond]

@robust dock Has your question been resolved?

[pearl pond]

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[pearl pond]

@flat falcon Has your question been resolved?

[still hamlet]

what have you tried

[pearl pond]

@flat falcon Has your question been resolved?

[crude quarry]

whats the problem @flat falcon

[pearl pond]

@flat falcon Has your question been resolved?

[pearl pond]

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[quick venture]

can someone explain why we subtract

the areas

[ivory basin]

Cuz you want the area between 0.45 and 2

You have the area between 0 and 0.45, and then you have the area between 0 and 2

Subtracting gives you the area between 0.45 and 2

[pearl pond]

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[fading ledge]

i did not understand the last line

where does this coming from?

[tropic saddle]

which part

the line above shows that (a,b,c) in L(S)

L(S) subset V is obvious

[fading ledge]

linear span of S means?

S is smallest subspace of V am I right?

[tropic saddle]

<@&268886789983436800>

[fading ledge]

???

[tropic saddle]

just a spambot. ignore

L(S) is the smallest subspace of V which contains S

hence L(S) is a subset of V

[fading ledge]

how did this come?

[tropic saddle]

they took an arbitrary element (a,b,c) in V

and showed that it is in L(S)

that is how you show that something is a subset of something else

[fading ledge]

i am feeling but how

suppose V is vector space

and we have subset S

and i will make lots of subspaces having S

so l(s) subset of V

which seems good

[tropic saddle]

a subspace that contains S contains the elements of S

and is closed under addition and scalar mult

so if v and w are in S, then also v+w is in L(S)

and 17v-32w

and so on

here, (1,0,0), (0,1,0) and (0,0,1) are in S and therefore a(1,0,0)+b(0,1,0)+c(0,0,1) is in L(S)

[pearl pond]

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[calm wing]

what's your question

[fading ledge]

i did not understand their solution

[cursive wraith]

"f has a pole of order n at z = 0" means that around 0, $f(z) = \frac{c_{-n}}{z^n} + \frac{c_{1-n}}{z^{n-1}} + ...$, with $c_{-n}\neq 0$

[jolly parrot]

Rafilouyear2026

[cursive wraith]

So $z^nf(z) = c_{-n} + c_{1-n}z + ...$

[jolly parrot]

Rafilouyear2026

[cursive wraith]

So $\lim_{z\to 0}z^nf(z) = c_{-n}\neq 0$

[jolly parrot]

Rafilouyear2026

[fading ledge]

(Sinz-zcosz)/z^2

[cursive wraith]

to check if this function has a simple pole at z = 0

look at limit of z * (Sinz-zcosz)/z^2

[fading ledge]

If we look at laurent series would be

z-z^3/6+......-z+z^3/2-

[cursive wraith]

Right I see how the solution is wrong too

f has a removable singularity

[fading ledge]

So you agree that solution is wrong?

[fading ledge]

Pole of order 2 no?

[cursive wraith]

there's no pole

As you pointed out, when you develop sin(z)/z^2 - cos(z)/z into laurent series

there is no term in 1/z^... of any kind after simplifying

so no pole

[fading ledge]

Yes there is no pole

[cursive wraith]

removable singularity

[fading ledge]

We don't have any problamatic term

[cursive wraith]

Correct answer is 3)

[fading ledge]

I don't know complex integration actually

I meant properly

|z|=1 circle around 0 radius 1

[cursive wraith]

you have a meromorphic function on the disk |z| <= 1

so you can use residue theorem

or you can just say that the function is holomorphic on the entire disk which is simply connected

so integral on closed loop is 0

[fading ledge]

Meromorfic means we remove poles

[cursive wraith]

meromorphic = Holomorphic on the entire domain except an isolated set of points which are poles

[fading ledge]

And holomorpic means analytic at all points of domain right

[fading ledge]

But we have some points where it is not analytic

Inside the circle

[cursive wraith]

which ones?

[fading ledge]

=0 point

[cursive wraith]

If you say "0" then you forgot that it's a removable singularity

meaning, if you define f(0) as the limit it's supposed to be, then it becomes analytic even at 0, so everywhere

[pearl pond]

@fading ledge Has your question been resolved?

[pearl pond]

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[gloomy matrix]

you need to find a 4 numbers pin: 1258, 2259, 3269, 4369, 5360, 6470. in each of this combination only 1 number is right. Can someone tell me if theres a easy way to solve it?

[unborn abyss]

i'd put it in a SAT solver lol

[glass meadow]

1258
2259
3269
4369
5360
6470

The first digit is different for all, so check which line has the correct first digit

[spiral pivot]

@gloomy matrix does the "1 number is right" apply to position and value or just value?

[thorn agate]

wordle type shi

[gloomy matrix]

the position is right and 1 being good means the others are not

[glass meadow]

If you test the second number for example, and assume 2 is the correct first digit, then 2 cannot be the correct second digit (numbers 1 and 3 are out), and 9 cannot be the correct fourth digit (numbers 3 and 4 are out). That leaves only numbers 5 and 6 to give you 3 more digits, which is not possible

So the first digit is not 2

[gloomy matrix]

but the first digit is 2 🤔

[eager jewel]

💀

[iron basin]

wait wdym "numbers 1 and 3 are out"

[glass meadow]

First and third numbers have 2 as their second digit

[iron basin]

yeah then why are they out if 2 is the correct first digit in the second number in your list
why cant all three have 2 as the incorrect second digit

oh uh wait im confused

ok yeah

[autumn fossil]

||4200
4220
4210
4240
4280
4230
4290
2468||

these should be all the sols

I feel like the last one of these was the intended one, but the authors of the Q messed up

[glass meadow]

No you're right, I was wrong

[iron basin]

oh

[glass meadow]

None of the numbers have 0,1,2,3,4,8,9 as a third digit

Oh wait that's the point

Yeah never mind

[autumn fossil]

Wait wdym?

[glass meadow]

My thinking was backward in some places here

[iron basin]

okay so one confusing fact is that the same digit can be in the correct place in a set of numbers

[glass meadow]

Yea, I feel like the question was not correctly written by OP but whatever

[pearl pond]

@gloomy matrix Has your question been resolved?

[iron basin]

wait true !!!

[autumn fossil]

indeed true, it was done by a SAT solver lol

[iron basin]

no idea what that is

[pearl pond]

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[green aurora]

From a deck of French cards (without jokers), a card is subtracted without looking at it. Then,
another card is turned over: what is the probability that this is a club?

[zenith bramble]

Oh god

I m sorry

[green aurora]

i did P(AUB)=12/51+13/51=25/51 but its wrong

I distinguished between the two cases, whether I drew the club card or not.

[west sail]

you forgot to consider the odds of the first card being a club or not in the first place.

[green aurora]

why i need to use P(A|B)?

[west sail]

because of what I said.

[green aurora]

but i remove the card

52-1

[west sail]

did you stop to consider that the card removed could be a club or not be a club, but they have different probabilities of happening?

[green aurora]

why can't it be done?

[thorn agate]

I think result will be similar to the balls one done few days ago

[west sail]

so what you actually want to find is the 2nd card being a club, conditional on the first removed card either being a club or not.

[green aurora]

ok

i try

[green aurora]

yes

[bitter herald]

another perspective: do you know of the concept of the 'Law of Total Probability'? @green aurora

[green aurora]

yes

but here the events are dependent

i think

I can't calculate P of the intersection

[bitter herald]

which intersection?

[green aurora]

of A=draw club cards B=a club card has been removed

because A and B are dependent

[west sail]

I don't think you should consider intersections, but rather two separate conditional probabilities.

[green aurora]

so P(A)+P(B)+P(AUB)

[west sail]

I'll let Lex explain better. I don't think I can come up with a satisfactory answer.

[green aurora]

all divided by P(B)

[bitter herald]

ok but remember that $\vb P(A\cap B) = \vb P(A\where B)\2\vb P(B)$

[green aurora]

I got confused with P(AUB)

[green aurora]

yes

but then if A and B are dependent it is very difficult to find intersecting P(A and B)?

[bitter herald]

well no you already know what P(A|B) and P(B) are in this case

ok wait

ill give you a proper mathematical justification

[green aurora]

but what's the point of that formula, then everything simplifies

i get P(A|B)=P(A|B)

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

[jolly parrot]

Goofy Joe

[green aurora]

this?

[bitter herald]

Define clear symbols for the events occurring at each step, for the sake of clarity:
\e{itemize}{
\ii Let $C_1$ be the event that the first card removed is a club.
\ii Let $\comp{C_1}$ be the event that the first card removed is not a club.
\ii Let $C_2$ be the event that the second card turned over is a club.
}
Hence, by the Law of Total Probability, you must have [
\vb P(C_2) = \vb P(C_2\where C_1)\vb P(C_1) +\vb P(C_2\where \comp{C_1})\vb P(\comp{C_1})
]