help-27

[restive river]

Divide 2pi?

[final drum]

there we go

now with radius

use it to find area

remember

so area of circle is pi*(radius^2)

so pi * (p/2pi)^2

so we simplifiy to p^2/4pi

so ratio square:cirlce

is (p^2/16):(p^2/4pi)

[restive river]

You lost me

[final drum]

so area of circle is pi*r^2

[final drum]

right?

[restive river]

Uh

[restive river]

Yea

[final drum]

so already know the radius is p/2pi

now we plug it in

p*(p/2pi)^2

[final drum]

you got it?

[restive river]

Uhhhh

Bro I’m so lost

[final drum]

y

we know the radius is p/2pi

[restive river]

Can we start over 😭

[final drum]

r u trolling

i feel like ur

[restive river]

Bro

How do I prove I’m not trolling

[final drum]

take picture of paper work

[restive river]

This is literally the only question I’ve not done

[final drum]

start writing stuff down

[restive river]

Ok but can we start over

[final drum]

yes

[restive river]

I will start

[final drum]

but pls take picture

step by step

i tell u

[restive river]

Ok but wait

So the perimeter of a square is 4(1/4p)
And the perimeter of a circle is r^2?

[final drum]

draw it

draw out a square

and a circle

[final drum]

no

[final drum]

good

what is the area

of the square now

[restive river]

P/4^2?

[final drum]

yes

btw ^2 applies to both p and 4

[restive river]

Okay so do I need to write that?

[final drum]

yep

thats ur final solution for the area of the square

[restive river]

So it should be like this?

[final drum]

yes

right

[restive river]

Circle is r^2?

[final drum]

no

okay

lets write these 2 things down

Area of circle = πr^2
Perimeter of circle = 2πr

r is the radius of the circle

[restive river]

Question- what is the perimeter of the square? P/4 times 4?

@final drum

[final drum]

YES

ur right

[final drum]

there

now we know the perimeter of the circle is p

[restive river]

Wait what now

[final drum]

2πr = p

[restive river]

Oh ok

[final drum]

its just a variable

[restive river]

Ok continue

[final drum]

isolate for r

then send pic

[restive river]

@final drum

[final drum]

yep

now we know r = p/2π

[restive river]

Yea

[final drum]

subsititude r into the area formula

[restive river]

@final drum

[final drum]

yep

now simplify

[restive river]

…

I will try my best but honestly I don’t think I can do that

[final drum]

you can

[restive river]

I can’t do it

😭

[final drum]

yep

now you can cancel the π

with one of the π in the demoniator

so its just (p^2)/(4π)

[restive river]

@final drum I don’t understand

[final drum]

take this

y(x/y)

what is that

[restive river]

X

[final drum]

there we go

do the same thing

[restive river]

Wait did I do it wrong

[final drum]

u

forgot the

^2

on the bottom of the demoniator

[final drum]

look

what you do (x/y)^2

the ^2 applies to the top and bottom

so x^2/y^2

[restive river]

Wait was I supposed to multiple top by tip bottom by bottom?

[final drum]

what?

just go back to π(p/2π)(p/2π)

[final drum]

yes

now

[restive river]

Wait

[final drum]

put the π

[restive river]

Pause

[final drum]

yes

[restive river]

So in this situation- top by top bottom by bottom?

[final drum]

what does that mean

which ever means πp^2/4π^2

[restive river]

Multiply like that?

[final drum]

yes

ur correct

:)

[restive river]

Ok

[final drum]

now

[restive river]

Now do the same thing again?

[final drum]

yes

π/1

[final drum]

yep

now keep simplifiy

[final drum]

YES

thats the area of the circle

[restive river]

Okay. And the area of the square is a^2 right.

I made some changes

Because I wasn’t understanding

That perim should be 4a

[final drum]

no

bro

[restive river]

Wait

[final drum]

the perimeter

is the outside lenght

[restive river]

It’s 4a

[final drum]

what is a

[restive river]

A side

[final drum]

yes

but u were right

[restive river]

That’s right then yea?

[final drum]

yes

area is p^2/16

[restive river]

This way is easier

Calling a side a

[final drum]

okay

[restive river]

Now what

[final drum]

now just write down the solution

[restive river]

…

[final drum]

area of circle : area of square

[final drum]

you cannot just add ur own variables

BRO

ur 100% torlling

[restive river]

Bro

Wdym

[final drum]

area of square p^2/16

we just talked about this 10 minutes ago

[restive river]

I thought I said area of square is a^2

[final drum]

yes

but u cannot leave that as the solution

they cannot find the ratio

if you given them a random variable

[restive river]

So what do I do

[final drum]

you ahve to include the perimeter

[restive river]

I don’t understand

[final drum]

if i ask you to find teh area of a square

and given you the length of a car

will that work?

[restive river]

Bro what

[final drum]

thats what u r saying

you have to have the perimeter

[restive river]

😭😭😭😭

[final drum]

to find the area

[restive river]

We never even found the perimeter of the circle tho

We just left it at 2pir

[final drum]

but ur given

[restive river]

😭😭😭

[final drum]

dude just write down are is p^2/16

for the area of the square

im busy

[restive river]

Well I’m trying to understand why I’m doing what I’m doing bruh

[final drum]

okay

given the perimeter

of a square

[restive river]

Which is 4-

4a

[final drum]

no

its not

you cannot just add random stuff

[restive river]

The formula for the perimeter of a square is 4-

4a

[final drum]

dude

listen

[restive river]

The formula for the area of the square is a^2

[final drum]

okay

okay

solve for a

given the perimeter

4a=p

what is a

[restive river]

P/4

[final drum]

yes

now use that to find the area

what is the area using the perimeter

[restive river]

P/4^2

[final drum]

^2 applies to the p too

[restive river]

How and why

[final drum]

(p/4)^2 = (p/4)(p/4)

[restive river]

Understood but I don’t need to write p^2 then

[final drum]

simpleist form

[final drum]

there we go

[restive river]

Okay so now we got ur

[final drum]

now area of circle : area of square

[final drum]

there we go

correct

[restive river]

Yeah why?

[final drum]

no wounder

[restive river]

How did you know?

[final drum]

i also go to school in canada

and teachers are stupid

and they cannot teach

okay, did you write that in solutions

[restive river]

What is the next step?

Yes

[final drum]

your done

[restive river]

But this is the not the final answer in the answer key.

[final drum]

what is the answer

[restive river]

4:pi

[final drum]

its the same

they just cancelled the p^2/4

form both side

its the same

its like how 1+1 and 2 is the same thing

[restive river]

How do i get their answer

[final drum]

you divide both side

by p^2/4

[final drum]

what r u doing

what

divide both sides by p^2/4

(p^2/4)/(p^2/4) = 1/pi

(p^2/16)/(p^2/4)=1/4

1/pi:1/4

[final drum]

there we go

cancel the 4 with 4 pi

and 4/16

[final drum]

1/pi

not pi

it was in the bottom of the fraction

[restive river]

What?

[final drum]

4/4pi is 1/pi

[restive river]

So now does the top cancel?

[final drum]

there we go

no

you multply both sides

by 4pi

[final drum]

both sides

4pi/pi : 4pi/4

[restive river]

Bru what

[final drum]

multiply both sides by 4pi

[restive river]

Why thi?

[final drum]

their gcd

lOWEST common demoniator

[restive river]

You mean lowest common denominator?

[final drum]

yes

[restive river]

Bro aren’t we back to the same point?

[final drum]

no

cancel out the pi with the pi

on teh left side

and with the right side cancel 4 with 4

[final drum]

THERE WE GO

good job

[restive river]

😭😭😭

How am I going to remember how to do this on the test 😭😭

[final drum]

don't

[restive river]

That took ages

[final drum]

highschool teachers

give free passes

in canada

[restive river]

What?

[final drum]

i ahve witness people in my class who pass with 40s

dw

ur fine

[restive river]

No I need a high mark

[final drum]

ur not going to get one

i am going to be honest

[restive river]

I need to know how to do this

[final drum]

you will not

you cannot even do basic arithmetic

i am not trying to be mean

and chances are

its not going to be on the test

what course is this?

functions?

[restive river]

Pre calculus 11

[final drum]

so functions

okay

[restive river]

I don’t even see how this question relates to the chapter

[final drum]

thats whats its called in ontario

[restive river]

This is a 1 off question

[final drum]

it doesn't

[final drum]

it is not

they won't test u on this

ur fine

[restive river]

It’s very odd I don’t see how/why it’s in the chapter review

[final drum]

its not

[restive river]

I think they might put 1 of these on the tedt

[final drum]

average garbage highschool textbook

[final drum]

its not

i alr took grade 11 math

its not

[restive river]

So you think I shouldn’t spend time on it?

[final drum]

they are not even slightly related

[final drum]

no

at least not for this stupid question

if u rly want to

just

understand it

[restive river]

Yeah this is the Linesr systems chapter

[final drum]

and do it a few more times

[final drum]

okay no way thats related

[restive river]

THATS WHAT IM SAYING

[final drum]

u done with the help?

[restive river]

Well unless you want to help me practice another one

[final drum]

no thanks

i have work due tmr

[restive river]

Okay

[final drum]

.close

[devout snow]

Channel closed

Closed by @final drum

Use .reopen if this was a mistake.

[restive river]

Bye! Thanks so much for your help!

[final drum]

np

:)

[devout snow]

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[agile narwhal]

In an undirected weighted graph, where all edge weights are distinct, the two cheapest edges belong to some minimum spanning tree.

[agile narwhal]

is it true?

i cant find any counterexample which would disprove that

[neat night]

if the 2 cheapest edges are the two that form the circle

it won’t be in the same mst

@agile narwhal

[agile narwhal]

That's true

[agile narwhal]

Thanks, I could not find any counterexample

[devout snow]

@agile narwhal Has your question been resolved?

Channel closed

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[devout snow]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[agile sentinel]

I struggle to make a formal proof out of number (ii). I made a table to find out if they are equal and they are, but how do formulate that correctly? Can i just make 7 cases and show each of them seperately? Or is there a better method?

[agile sentinel]

Or should i try to proof that the left side is a subset of the right and that the right side is a subset of the left and they are thus equal?

[devout snow]

@agile sentinel Has your question been resolved?

[agile sentinel]

<@&286206848099549185>

I dont know if you can read that but i tried it like this. Is there a faster method than doing each case?

[spiral mural]

You're tryna' to derive that the symmetric difference is associative?

[agile sentinel]

Yes

[spiral mural]

Aight

okay well i can't really go through it since it's been a while since i've touched up on set theory but you could just search up a formal proof

https://www.youtube.com/watch?v=0qrrIY-13co

i'm unsure if there's a quick way

but you have to consider all cases

[agile sentinel]

Okay then i think i will leave it at that. I need to do (iii) next, do you have any tips or should i just try it myself first?

[spiral mural]

hmm

i'm unsure i'm not thinkin abt it rn as i'm eating dinner but if you don't have a response by the time i'm done i'll check on you

[agile sentinel]

Ok i will try it myself first

Im not sure if the last step is allowed or if i should make it more clear why x is element of (a\b)u(b\a) (instead of \ my book uses -)

,rotate

[agile sentinel]

The Problem says: Let n be a natural number and let a_1,...,a_n be sets such that a_1 is superset (or equal) of a_2 and so on up to a_n. Then for even n (i) is true and for odd n (ii) is true. My first idea was induction but i dont think that works (also the book hasn't introduced it yet. Is it enough if i proof it for an arbitrary a_k such that k=2n for (i) and k=2n+1 for (ii) ?

[agile sentinel]

For n=4 the equation becomes a_1-a_2-a_3-a_4=a_1-(a_2-(a_3-a_4)) a_4 is not in (a_3-a_4) but it is in a_2 since a_4 is a subset of a_2 and thus it is in (a_2-(a_3-a_4) and thus is not in a_1-(a_2-(a_3-a_4)) since a_1 includes every element of a_2.By the same logic you can find that a_(4-1) is in a_1-(a_2-(a_3-a_4)) etc. and so you have (a_1-a_2)u(a_3-a_4)=a_1-a_2-a_3-a_4

Is that a valid approach?

[devout snow]

@agile sentinel Has your question been resolved?

[agile sentinel]

I skipped the last problem but i have a question for the next one: i got {0} here instead of {}, did i do something wrong/misunderstood something?

I understood it that the set of wich i have to take the Intersection of is {{0},{0,1},{0,1,2},...} Or did i misunderstand it?

[agile sentinel]

For any set a out of M, 0 has to always be part of a since m greater or equal to 0 is always true for any m out of the natural Numbers. Im supposed to arrive at the empty set though, so what did i miss?

[devout snow]

@agile sentinel Has your question been resolved?

[devout snow]

@agile sentinel Has your question been resolved?

[devout snow]

@agile sentinel Has your question been resolved?

[devout snow]

@agile sentinel Has your question been resolved?

[devout snow]

@agile sentinel Has your question been resolved?

[agile sentinel]

My question hasn't been resolved btw

[mild comet]

is this currently the question you are asking about?

[agile sentinel]

Yeah

[agile sentinel]

And the 2nd one on this screenshot

[agile sentinel]

Do you know why {0} isnt the right solution? I want to understand this

[mild comet]

what is (i)

[mild comet]

what does it say in English?

[agile sentinel]

The equation is the same in english as in german, it says "the following equation is true:" but i cant really proof why

[agile sentinel]

For even n the equation holds that a_1-...-a_n=(a_1-a_2)u...u(a_(n-1)-a_n) btw instead of - the notation \ is also used

[mild comet]

0 is defined as being the empty set

[agile sentinel]

Oh so there is no difference between {0} and {} ?

[mild comet]

No, because {0} is {{}} ≠ {} = 0

[agile sentinel]

Then i dont understand what you mean

[restive river]

wait no, both are different?

[agile sentinel]

Oh nvm i get it

[mild comet]

0 := {}

[restive river]

{} is a null set

or empty set

whereas {0} is when, the set has 0 as its element

[agile sentinel]

So when i have the intersection of a set i dont have to put the answer as a set, right?

Or atleast if it is only one answer

[mild comet]

the intersection of two sets is always a set

[agile sentinel]

No its not of 2 sets but of one set

[mild comet]

what is the intersection of one set?

[agile sentinel]

Let me find the name in english im not sure

This thing. Its not the intersection of one set but of one system of sets

[agile sentinel]

But i thought the equation holds for 0 so why do i write the empty set instead of {0}

[frosty cradle]

because {0} is the set containing a single element, 0

[agile sentinel]

i.e. 0 is part of every such set

[restive river]

no

ø is part of every set

[agile sentinel]

Can you explain why not?

[restive river]

give me an example of set which contains natural numbers

the natural number starts with 1,2,3

[frosty cradle]

yeah, consider e.g. {1, 2, 3}. 0 is not in that set

[restive river]

yes

if you intersect {1,2,3} with {5,6,7}
the intersection will be ø

Ø represents nothing

[agile sentinel]

Not with 0?

[restive river]

0 is part of whole numbers

[frosty cradle]

some authors will consider 0 a natural number, but I think most would not

[restive river]

but that's second the thing

[agile sentinel]

I think i get it. What i previously thought it meant was that you take some m out of the natural numbers and then every n less than or equal to m is part of a set, then you take the intersection of every set with m from 0 to infinity

Well my book clearly defined 0 as a natural number "The natural Numbers shall contain the Element 0" ( i dunno if thats the right translation)

[frosty cradle]

are you referring to a specific problem?

[restive river]

cool

[restive river]

uh

[agile sentinel]

Yes this one the blue thing

[frosty cradle]

that's fine, I've seen another book that includes 0 as a natural number. but it is irrelevant to your confusion about the empty set

[agile sentinel]

K

[restive river]

in exams, just go with your book (incase asked in exams regarding natural numbers or whole number)

[frosty cradle]

I see. the blue one can be tricky. but it looks like it is taking the intersection over all n

[agile sentinel]

This is the problem in question

I dont understand how {1,2,3} is in that set that you take the intersection of

[restive river]

it was just an example

to make you understand that 0 is an element and can/cannot be in every sets

[frosty cradle]

so it's taking the intersection of {0, 1, 2, ...} with {1, 2, 3, ...} with {2, 3, 4, ...} and so on. do you see how the intersection over all sets constructed in that way will be empty?

[agile sentinel]

I see. I thought this notation meant the sets {0}, {0,1}, {0,1,2} etc.

[frosty cradle]

no, it would be the set of all natural numbers greater than/equal to "n", for a given "n". I assume that it is "iterating" over all "n", even though it isn't clear to me in the notation

[agile sentinel]

Oh okay, i thought it was the other way around. Thanks for clarifying.

[agile sentinel]

Can you help me with this one?

[agile sentinel]

So i would just say, if there were a k out of the natural numbers such that k is in everyone of those sets, choose n=k+1 and you will reach a contradiction, right?

[frosty cradle]

yes I think that's right

[agile sentinel]

Ok i think i will try the other one myself and leave it out in case i cant do it. Thanks for the help @frosty cradle and others.

.close

[devout snow]

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[lofty monolith]

for 11 do u have to find an equation for x in terms of y

[lofty monolith]

or does it not matter

i found one for y in terms of x but markscheme didnt

[frosty cradle]

I would've assumed y in terms of x probably

[lofty monolith]

yeah

okay thanks

.close

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[steel escarp]

calculate the circulation

The thing I get lost over in this exercise is when I go to polar form

i get x=cos(t) and y=sin(t)

Because it's a unit circle, and therefor r=1, right?

the solution:

[devout snow]

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[keen solar]

Hello

[keen solar]

So

I have a doubt in simplifying this

Then cut

So u rationalise it

The cut it

But none of them are the options

[jovial mauve]

!show

[devout snow]

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

[keen solar]

Oh ok

Now how do take out stuff

[jovial mauve]

Right

[pastel pasture]

Just write √8 as 2√2 lol

[jovial mauve]

Bruh

[keen solar]

What

[pastel pasture]

There was no need to do this

√8 = 2√2 so the denominator is just 3√2

[jovial mauve]

It's ok

[keen solar]

?

[jovial mauve]

Your work is alright

[keen solar]

It's 2√8 -2√2

[pastel pasture]

I don't think you get what I'm saying

Nvm

[keen solar]

How do I make it 2√2

[jovial mauve]

Just write √8 = 2√2 and simplify 2(√8-√2)

[keen solar]

What I did not get it

[pastel pasture]

$\sqrt 8 = \sqrt{4 \times 2} = 2\sqrt 2$

[woven radish]

NEONPerseus

[keen solar]

8 is cibe root of 2

[keen solar]

I mean 8 is 2 cube

Ohhhh

So two root 2 is

1 root 8

[pastel pasture]

well yes

[keen solar]

Ty

Lemme complete n send

I did not get it cuz i was thinking in the cube way

Forgot it was root 🌚

Problem solved

I have a doubt in this

How do u do 8

[jovial mauve]

Have you tried anything?

[keen solar]

I am not getting the image of what to do

[jovial mauve]

You should simplify the powers of a

Use the exponentiation rules

[keen solar]

?

Second one u can fo

4*1/3

[hollow mantle]

simplify this one first

[keen solar]

It is in a root right

How do simplify it

[hollow mantle]

You don't have to worry about signs because a is to the cube.

[keen solar]

I know that but

A is in a root

[jovial mauve]

Yes

[keen solar]

2/3 is outside the root

[hollow mantle]

So put them in order

imagine a^3 = x

[keen solar]

Ok

Root of a 3 is

A^3 ^1/2

[hollow mantle]

no

[keen solar]

No?

I mean root of a^3 is

Can be also weitten

(a^3)^1/3

?

[hollow mantle]

yes

they are not the same

[keen solar]

I simply like this?

[keen solar]

They are not???

How do I simplify the powers uh

[hollow mantle]

you have to be careful with parenthesis

they are not the same

[keen solar]

Ohhhh

Mb

[hollow mantle]

so the first one is the correct

a^3/2

but that's also raised to other thing if u check the exercise

[keen solar]

Wait what

[keen solar]

I did not get the second one

[hollow mantle]

the second one is wrong, it's what you wrote in the first place

[keen solar]

Ohhh my bad

[keen solar]

So this is

I multiply 1/2 with 2/3?

[hollow mantle]

right now you have this

[keen solar]

Ye ye

I got till here

Ohhhhhhh

1/2*2/3

[hollow mantle]

if base > 0 you are free to multiply

[keen solar]

2/6

Then 3*2/6m

[hollow mantle]

ok so u got what

[keen solar]

6/6

[hollow mantle]

which is

[keen solar]

A?

[hollow mantle]

yes

so all that was only a

[keen solar]

Omg

[hollow mantle]

now this one

[keen solar]

So it was just to confuse u

Tricky

But i have a doubt

U can cut exponents?

[hollow mantle]

what do u mean?

[keen solar]

If it is b^2/4

Can I write it is b^1/2

Meaning cutting out the factors

That is how we got a

[hollow mantle]

you can sometimes

[keen solar]

Uh

[hollow mantle]

when the base is bigger than 0

[keen solar]

So i can't cut?

Oh

So with positive u can cut

[hollow mantle]

we assume that

[keen solar]

This ain't in my text

But it is in my practise book

So i may be a bit slow

So now we got first term as a

[hollow mantle]

yes, but always assuming a>0

[keen solar]

We simplify inside the root first?

[hollow mantle]

in the second one?

[keen solar]

a^4/3

Ye

Wait that is wrong

[hollow mantle]

you can do the other way

[keen solar]

I don't know any ways

My first time doing such a problem

[hollow mantle]

from outside to inside

[keen solar]

So

uh

Root 12 is

1/2

1/12

[hollow mantle]

yes

[keen solar]

So we put that after 1/3?

After a whole square

[hollow mantle]

you would have what raised to what raised to what

[keen solar]

A^4

Ok in brackets

^1/3

Then brackets

Raised to

^1/12

[hollow mantle]

((a^4)^(1/3))^(1/12) like this, right?

[keen solar]

Ye ye ye

Then just open it

Property?

[hollow mantle]

from outside to inside

1/3 and 1/12

[keen solar]

Ye ye

[hollow mantle]

what do u get

[keen solar]

1/36

[hollow mantle]

so now we have (a^4)^(1/36)

[keen solar]

So

4/36???

[hollow mantle]

now here we have 2 different approaches

if we assume a>0, than you already solved it

a^(1/9) which is what you said

[keen solar]

Well i am in school

I have no negative

Till now

[hollow mantle]

if a is a real number, then the solution would be (|a|)^(1/9)

Then, the first one

[keen solar]

How do I prove x

[keen solar]

Smh managed this bro

[keen solar]

What happens if it is not

The way u said now

[hollow mantle]

you had something before that

[keen solar]

More complex?

[keen solar]

Ye ye

Ohhhhhh

[hollow mantle]

you know now that the first one is a

the second one is a^(1/9)

[keen solar]

Lemme write steps now

[jovial mauve]

Appreciate your patience @hollow mantle!

🫡

[keen solar]

Omg still don't know how I came this far

I had no idea

[hollow mantle]

come on, we alsmot finished

[keen solar]

Almost done sorry

[hollow mantle]

Remember we always assume a>0, if not, some things are not complete

[keen solar]

Wait

Is there negatives in class 9

I never learned till jow

[hollow mantle]

i don't know about classes, how old are u?

[keen solar]

I can't reveal that

[hollow mantle]

ok

[keen solar]

💀

[hollow mantle]

keep finishing

[keen solar]

Ok so

I leaerned real no 1 month back

So based on that u can say right

I learned irrational like 3 weeks back

Lemme see my text rq

Yep no no has negative

Ok let's finish

Lemme post my steps

[hollow mantle]

ok

[keen solar]

?

Ok now division

a/a^1/9 right

[hollow mantle]

yes

[keen solar]

How do I do that now uhh

[hollow mantle]

do you know how much is x^2 * x^2?

[keen solar]

?

[hollow mantle]

just a quick question

[keen solar]

Can u write it in the maths way

[hollow mantle]

wait

do you know how much is this?

[keen solar]

Oh

That is

Hold up

Xxxxx

Ahhh

Basicly

Multipikt 5 x

Am i right

[hollow mantle]

you add the powers

2+3 = 5

so x^5

[keen solar]

Ye

[hollow mantle]

now this is because it's a product

[keen solar]

U said property i said the way

[hollow mantle]

but we have a division

so we don't add them up

we substract

[keen solar]

Ye

[hollow mantle]

so we have a/a^(1/9)

what would that be?

[keen solar]

Ohhhhh

Wait

(a/a)^1/9

[hollow mantle]

no

[keen solar]

I was dumb ahh

[hollow mantle]

careful with parenthesis

[keen solar]

Writing in chat is hard for me

[hollow mantle]

it's ok

[keen solar]

Ohhh

Wait

a^1-1/9

Omg is it correct

[hollow mantle]

correct

how much is that?

[keen solar]

Omg les goooo

U gotta take lcm

That is 8/9

[hollow mantle]

so we got

[keen solar]

omg

Omg

Omg

Omg

Omg

Omg

[hollow mantle]

a^(8/9) = x, but you didnt' finish yet

the question still isn't finished

[keen solar]

Lemme write steps

Man i feel so good

I got like hella problems rest to do but whateber

I could have done 5 simple ones

[hollow mantle]

You didn't finish yet

[keen solar]

Ye ok

Ik

[hollow mantle]

Can you finish it from here?

[keen solar]

Lemme see uh

a^8/9=x

Nvm i can't

[hollow mantle]

yes you can

it's 2 simple things

[keen solar]

Idk the step

[hollow mantle]

i teach you

with an example

[keen solar]

Okk

The thing is

I told u

First time

I don't know any method or steps

Hmm

How uhh

[keen solar]

My time is ticking

Hmmm

How did a become 9

[hollow mantle]

this are examples

not your exercise

so you can use this hints to solve your exercise

[keen solar]

Thinking

Uhh

Wait

I do what I do with this and lha

I mean

Rhs and lha

Lhs

[hollow mantle]

whatever u do in one side u do in the other

[keen solar]

I know this

[hollow mantle]

so you have to eliminate 8/9 in your a

[keen solar]

I did it to find

[hollow mantle]

and give that to the x

[keen solar]

How do I eliminate it rho

By 9/8?

[hollow mantle]

yes

[keen solar]

I did not get it agh

Ah

[hollow mantle]

because if you multiply 8/9 and 9/8 you get 1

[keen solar]

Yep i knew rhay

Sooo

I do that with the x but that is not our

Answer

It is in root

[hollow mantle]

remember denominator is root

[keen solar]

Ohhh

[hollow mantle]

a^(1/2) = sqrt(a)

[keen solar]

Ohhh

So doneM

Question done?

[hollow mantle]

yep

[keen solar]

Man so ty

Got like 20 more to do

Ima actually do the easy ones first now

[hollow mantle]

go for them

[keen solar]

Lemme send the steps

Wait

In the question it is not whole raised to

The 9 is also in the root

The property is diferent

[hollow mantle]

explain what do u mean

[keen solar]

Man i can't text ahhhh

The properlty

Is

The number on top is whole raised to

That is after the root

In the exercise it is inside the toot

Root

[hollow mantle]

you have this, right?

[keen solar]

Yes

[hollow mantle]

what would you do if instead of 9/8 is 1/2?

isnt' that sqrt(x)?

[keen solar]

?

Did not get it

[hollow mantle]

if it were 1/2

[keen solar]

Just write root

[hollow mantle]

right?

[keen solar]

Yes

[hollow mantle]

that's this

[keen solar]

Yes

[hollow mantle]

do you see the 1 over the x in the root?

[keen solar]

No i learned the property as

[hollow mantle]

we don't put it because it's not necessary, but it's there

[keen solar]

One outside

The root

Like this

[hollow mantle]

the root takes the number in the denominator

[keen solar]

(√x)1

[hollow mantle]

x takes the number in the numerator

[keen solar]

The no on top is whole raised ro

To

[keen solar]

No

I meant

The 9

Should be outside the root

According to the property

[hollow mantle]

it's the same when we assume it's possitive

[keen solar]

Nvm

[hollow mantle]

you can try with a number

[keen solar]

So it's only whole raised to

If it is negative

[keen solar]

It it is positive i can write inside root?

[hollow mantle]

yes

[keen solar]

Omg

[hollow mantle]

if it's negative it doesn't work

[keen solar]

They teach me complex stuff

Then make us learn basic stuff

My bad

[hollow mantle]

you'll reach to that later

[keen solar]

I thought both made a difference

[hollow mantle]

now don't worry about it

[keen solar]

Man when i come to this server

I feel like i have decades more to learn to learn maths

[hollow mantle]

we all do

[keen solar]

Now time to do 20 problems lol

[hollow mantle]

go for it

[keen solar]

Could have asked an excuse to my maths teachers saying this was too tough

And did 20 problems

But what is the use of that loo

We study to learn

[hollow mantle]

correct

[keen solar]

Not for showing others

[hollow mantle]

good mentality

now .close this one to free the channel

[keen solar]

Hold up

Lemme

Just clear this tint doubt

I just gotta rationalise it right

Then i get x and y

?

[hollow mantle]

the good thing about this is that you have to try what you think it's the steps to solve it

and if that's true

you will feel better than if I say to you the answer

[keen solar]

Oh

[hollow mantle]

you better get lost and ask for help later

than asking before even failing

[keen solar]

So do i close this?

Or i do it fast

[hollow mantle]

yes, try and if u get more questions open a new one

because if I leave nobody will see this channel

cause was opened long time ago

[keen solar]

Ohh

Idk to close

[hollow mantle]

type .close

[keen solar]

.close

[devout snow]

Channel closed

Closed by @keen solar

Use .reopen if this was a mistake.

[devout snow]

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[celest tide]

any hints/thoughts on this?
I'm unable to figure out a geometrical interpretation and a "x + iy" sub just messes things up

[restive river]

try using the triangle inequality

[celest tide]

on the LHS, right?

[restive river]

yeah

[celest tide]

|z|^2 - |z| - 1 ≥ 0

|z| ≥ (1 + √5)/2

[restive river]

yup

[celest tide]

this does give me the minimum value, thank you

I'm guessing a similar approach for maximum?

wait nvm sorry I don't why im asking too many ques

I'll try it

[restive river]

isnt this a multi choice question?

[celest tide]

yes

multiple ans

[restive river]

hm ok

[sly viper]

yes

[celest tide]

|z| + 2 ≥ | |z|^2 - 1|

the approach here is to square on both sides, right?

ah that does work

thank you

.close

[devout snow]

Channel closed

Closed by @celest tide

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[devout snow]

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[restive river]

I have biquadratic equation, where a belongs to set of real numbers..
ok.. i solved it.
but there's one detail:

x² = a², a is in R set of numbers
this equation is equivalent to
x = +-sqrt(a²)
x = +- abs( a ) , because a is in R set
x = +-a

Correct?

If a were just greater than zero, a>0 then we wouldn't have abs parenthesis.

[flat rain]

sqrt(a²) = a in this case

no need to worry about abs

[devout snow]

@restive river Has your question been resolved?

[restive river]

because I have +- in front of a?

[flat rain]

Yes

the +- considers both cases already

[restive river]

@flat rain thanks! (:

[flat rain]

👍

[devout snow]

Channel closed

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[native oyster]

Hey everyone. I could need some help solving this Quantum mechanics task. I figured that eventhough it is physics related, this question is more mathematical than of physical difficulty. Could somebody indicate to me how to approach this question?

[native oyster]

Translation of the text above :

"The adjunct of an operator Â, that operates on Hilbertraum H, is denoted as  dagger and can be defined as follows:
....
for arbitrary vectors phi-ket and theta-ket in H. The star * denotes the complex conjugate of a complex number z. Use the definition above to show that for the Operators Â,^B, that operate on H, holds:
...."

[flat rain]

Use the fact that (AB)^dagger = B^dagger A^dagger

[native oyster]

Im not sure this holds here, because neither phi-ket nor phi-bra (behaves like a linear functional) are Operators.

[flat rain]

but they are vectors

[flat rain]

A and B are in general, matrices

where A is m x n and B is n x p. Here it is valid for B to be a vector where p = 1

[native oyster]

I can see what you mean in a finite dimensional space. But this space in its general form is not finite dimensional. Also, I am a bit afraid to use operator rules for my bras and kets, as quite some emphasis is put onto the difference between operators, bras and kets 😄

But thank you anyways 🙂

[flat rain]

Fair enough but I believe it still holds

bras are just vectors in a Hilbert space

kets are vectors in the dual space

not sure about the operators but I believe they hold

I.e. not sure what "space" they belong to

[flat rain]

specifically covector

[native oyster]

hmm ok. Given that I can not find a solid reason why this should not work, I'll use it 😄 thanks again!

.close

[devout snow]

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• 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.

[unreal pollen]

Can someone help, I really don't want to get it wrong. I have one last attempt on this question

What I did is multiply both sides by cos-1

so I got

$$theta=cos-1(-0.7986)$$

[potent moon]

?

[woven radish]

dasa

[unreal pollen]

to get theta alone

[potent moon]

ohh, you mean you took the inverse

[unreal pollen]

yeah sorry

wrong terms

and I got 143 degrees

[potent moon]

I thought you literally multiplied by cos(-1) lol

[unreal pollen]

haha

no sorry!

then I know cos in Q2 is negatice

negative*

so that's -143. But the restrictions are between 0 and 360

so I did 360-143=217

but it's wrong