help-26

[supple gulch]

BD

Sorry

[wise ivy]

It's a little bit strange that trapezium is not equilateral (AD and CB is unequal) but diagonals is perpendicular
Are you sure your conditions right

[supple gulch]

Yes

I dont think it has to be iscosceles if they ar eperpendiculae

<@&286206848099549185>

[tidal moss]

Hi

Where is x?

Sorry I am blind as fuck

[supple gulch]

X is b

On the picture

[tidal moss]

Alright

Do we know anything about the triangle

[supple gulch]

What triangle

AQC, is right angled

[tidal moss]

The one on the right

[supple gulch]

Yes

[tidal moss]

Q is the point where the diagonals meet?

[supple gulch]

No

Uh

O is the point

[tidal moss]

I don't see any Q on it that's why I am asking

[supple gulch]

My bad

[tidal moss]

So AOC

is right angled

[supple gulch]

Also the prblem says that a circle can be inscribed in the trapezium,, which means that the sum of the 2 pairs of opposite sides is equal

Thats where i got 14- b and b from

[supple gulch]

Yes

[tidal moss]

So the triangle BOC is right handed right?

[modern rock]

aint no way, its duskblade

[tidal moss]

doesn't sound plausible tbh

[supple gulch]

Yes

[tidal moss]

And AO is 4✓2?

[supple gulch]

Because the angle between the diagonals is r90

[supple gulch]

No

It was some guessing by me dont mind it

[tidal moss]

Oc*

gotcha

My battery is dying btw it is at 4%

And I will get home in like 2-3h.

Try using the similar triangle theorem

@supple gulch

[supple gulch]

On which ones

[tidal moss]

Maybe DAC and CAB or try AOB and DOC

See ya in a few hours if you didn't manage to solve it send me a message I will be home by then I got 2%

[supple gulch]

@prisma mesa

[wise ivy]

Ohh, i have an idea

Hope you get it too

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[supple gulch]

i dotn get it

please tell it to me

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[rugged oracle]

Not specifically math related but it can be. Does anyone know the exact term for the building blocks in a curriculum? For example in order to learn calculus, you would need to know basics of mathematics first, and before mathematics, you would need to know numbers, reading, writing, etc.

In my case, my goal is a mechanical engineering degree. I can pick an advanced subject and break that down into pre-requisite knowledge bases, but I'm not sure how to "define" them or do research on it, I feel like I'm missing the exact word. Best I could find was Building Block

[keen matrix]

<@&268886789983436800>

[keen matrix]

idrk, maybe prerequisites or foundations?

[unreal crow]

mb

im so sorry bru

wrong server

[keen matrix]

no worries :)

[flint stump]

Mechanical engineering is still somewhat vague target

Are you focused enough to have decided what you wanna do in that? Like robotics? Structural mechanics? Fluids and Thermal systems?

Like each of these subtopics have their own prerequisites and future scopes

e.g. Robotics is interdisciplinary so in addition to general mechanical engineering youd need some programming as well

[rugged oracle]

Right, so if I look up the university I would like to get into, their mechanical engineering program would start like this

and it's structured in an order that you learn the building blocks needed for the next set of courses and concepts. But is there a better word to use instead of building blocks to describe that?

[flint stump]

Looks to me as if they have listed the prerequisites? Like isnt that what you want?

[rugged oracle]

yes and no, because I'm not sure if I have the prerequisites to their prerequisites. Sorry if its confusing but I just want to make sure I'm prepared before applying, but its hard to narrow down everything I need to know beforehand

[flint stump]

Like, lets say you want Solid Mechanics CIVE207 from term 3. Then that needs basic 1st yr course of MECH 210 thats Mechanics 1

and so on

I see

[rugged oracle]

their program makes sense, what baseline of knowledge do i need before all of that

[flint stump]

I also did a MechE bachelors before shifting to a Maths Dept Masters. So in my experience, I would say, knowing basic topics, generaly known as precalc are enough

And some basic physics like velocity and acceleration is enough

But, anyways most Unis have their first lectures for any course as refreshers

[rugged oracle]

Okay, that works for me. For context I'm 30 and just have a bachelors in Business Admin, so I just want to refresh myself on concepts I learned 15 years ago so I'm ready beforehand

Thanks

[strange kindle]

<@&268886789983436800>

[rugged oracle]

.close

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[grim swift]

Guy im having hard time understanding how these rulle were applied here

ln'(x) = 1/x i know

is the entire expression just quotient rule?

was chain rule even used?

or quotient rule IS chain rule in itself ? 😮

just a different version

for when there is a fraction

[coarse tusk]

no

[coarse tusk]

yes

$\left(\log(\frac{x}{1-x})\right)'=\frac1{\frac{x}{1-x}}\cdot\left(\frac{x}{1-x}\right)'$

[thorny flame]

kheerii

[coarse tusk]

that's the chain rule bit

[grim swift]

woah

[coarse tusk]

to calculate the derivative on the right hand side we use the quotient rule

[grim swift]

oh so the entire fraction is the inner function

in the original function form

I thought ln immediately multiplies with the numerator

and then numerator = u and denominator = v

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[candid lance]

compute the volume of the solid above the cone x^2 =y^2+z^2 and inside the sphere x^2 +y^2 +z^2 =4

[candid lance]

I need help with this HW proble,

I intergrated with respect p, phi, and theta

p I got 8/3, PHI I got 1 - root2/2, and theta I got 2pi

then I simply combined them and got 16pi/3 (1-root2/2)

<@&286206848099549185>

is this correct

[neon iron]

1/3 (theta)

2/phi - 16pi/3 (1-root2/3pi)

[candid lance]

my p and phi are correct though right

[neon iron]

Ye

[candid lance]

do we need the simplify that answer any more

[neon iron]

I'm not 100% sure tho

[neon iron]

I don't think you need to

But if you can then ye

[candid lance]

can you show me the work for theta pls

how did you get thatr

@ helpers

<@&286206848099549185>

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[tawdry wyvern]

I need help Creating a multi step equation where x is 6 and has the distributive property and combining like terms on the same side

[tawdry wyvern]

<@&286206848099549185>

[hallow depot]

i don t understand .. wdym with " has the distributive property and combining like terms on the same side", are we still talking about equation ?

[fresh copper]

Please elaborate

Also

!15m

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[tawdry wyvern]

I need help on B

The solution needs to be a number from 6-12

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[tawdry wyvern]

<@&286206848099549185>

[thick jay]

wuts a multistep equation

[tawdry wyvern]

An example is like 3x + 5 = 9-5

[thick jay]

so an unsimplified linear equation in 1 variable?

[tawdry wyvern]

Yes

[thick jay]

x(4+2)=3+12+21

is this good

[tawdry wyvern]

Let me check

[thick jay]

x=6

[tawdry wyvern]

Thank you

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[toxic stirrup]

how is this row reduced form?

[toxic stirrup]

for the

0 2
0 1

ah i think its a typo

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[hollow rose]

Say I have N bananas and take 1/3 of them, but leave one behind, how many are left?

[hollow rose]

Because I think I can say I took (N-1)/3 of them

So I think that N - ((N-1)/3) is the amount left

But that's not true

So what am I doing wrong?

[inner wren]

wdym leave

[hollow rose]

Ill show you the question

Im trying to figure out how many are left after Arthur took his third

[inner wren]

ok

so

by 1 banana being left we mean that the number of bananas is in the form of 3k+1

3k+1 = N

[hollow rose]

Okay

[inner wren]

what;s the question though

[hollow rose]

Each person does the exact same thing as Arthur, and by the end, there are 3m+1 bananas left

(excluding the orangutang who doesn't take a third)

And I want to find an expression relating N and m

[inner wren]

I think we have to use the GIF function

[hollow rose]

Never heard of that in my life

Ohhh

The floor function?

Yes

That's the topic behind the assignment

[inner wren]

mhm

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[mortal spear]

Could someone help me verify this is a metric? Where x and y are in R

[mortal spear]

I am struggling to prove the triangle inequality for metrics

[winter egret]

the good old triangle inequality

what have you tried?

[mortal spear]

the triangle inequality haha

[winter egret]

Have you written out d(x, y)+d(y, z) ?

[mortal spear]

but the fractions and its denominators are my problem

[mortal spear]

yes

[winter egret]

can you show me that so I don't have to do it?

[mortal spear]

$$\frac{\left|x-z\right|}{1+\left|x-z\right|}\le \frac{\left|x-y\right|}{1+\left|x-y\right|}+\frac{\left|y-z\right|}{1+\left|y-z\right|}$$

wait

[winter egret]

thanks

[thorny flame]

Halex

[mortal spear]

There

[winter egret]

have you tried combining the RHS into a single fraction?

one thing I am kinda thinking about is I think there's some like general result where you can do convex mappings to a metric or something like that and what you get is another metric

so idk if you're supposed to be somehow clever about this or if you should just bruteforce it

[mortal spear]

Yes, it turns into something horrible

[mortal spear]

This?

[winter egret]

well that would certainly help if we could assume that

lol

https://math.stackexchange.com/questions/309198/if-dx-y-is-a-metric-then-fracdx-y1-dx-y-is-also-a-metric

so basically

you're putting your metric through an increasing function

and then that helps you in the way outlined in the stack exchange post 😂

sorry if all I did was help you cheat on your homework

[mortal spear]

yep, but that is a general result and can't use it atm

since I just started to learn what metrics are

[winter egret]

well if you really wanted to you could write this but instead of saying d(x, y) you'd write |x-y|

and then that's it

I doubt there's any easier solution

[winter egret]

but if this is in your textbook then I think it would make a lot of sense to just refer to that result

[mortal spear]

I'm kinda lost in the second step, first inequality. Does it really holds?

[winter egret]

remember that everything that you're dealing with is nonnegative

[winter egret]

if you add a nonnegative number to the denominator, your expressions could get smaller or could stay the same but they will certainly never get bigger

that's a very common trick in university level estimation

if A, B, C are positive numbers then A/B >= A/(B+C)

[mortal spear]

well, everything makes sense now

[winter egret]

I'm glad ^^

[mortal spear]

Thank you!

[winter egret]

no problem!

well done

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[supple kayak]

i needed help with the question and there’s a tutor thing here to help you but i’m not understanding how it got 2,4,6 as integer answers

[supple kayak]

can someone also explain consecutive integers?

[alpine mist]

Consecutive:
if n is an integer, then n+1 is the next largest integer.

if n is an ever integer, then n+2 is the next largest even integer.

e.g.
2,3,4,5,6 are a list of consecutive integers
2,4,6,8,10 are a list of consecutive even integers

[dense lily]

Successive even numbers

[dense lily]

In this case

[steep nebula]

after that they translated the sentence into an equation and solved for n, almost, you just need to do the last 2 steps

@supple kayak do you need help with anything else?

(aaah pinged the wrong person save meeee)

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[mortal spear]

For ( x = (x_1, \ldots, x_n) ) and ( y = (y_1, \ldots, y_n) ) in ( \mathbb{R}^n ), define

[
d(x, y) = \min{|x_1 - y_1|, \ldots, |x_n - y_n|}.
]

[thorny flame]

Halex

[mortal spear]

Is d a metric?

I don't know much about the minimum function, so I'm not sure how to approach it

[surreal mural]

do you think it is?

I would just go through all the axioms and try to prove them, and if you struggle at some point it may be an indication that this axiom does not hold and you could try to construct a counterexample

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[mortal spear]

how do one make counterexamples when working with R^n?

[jade thunder]

Look at the axioms

What do they say

[surreal mural]

you can choose any n

like n=2

[mortal spear]

Which ones specifically?

[jade thunder]

My go to is to check if I can make simple counterexamples to the axioms

If that doesn’t seem doable by inspection maybe it is a metric

[mortal spear]

fixing an n and just make a counterexample

[jade thunder]

But it’s always easier to prove it’s not a metric than to prove it’s a metric

Yeah but look at the axioms

Find one that looks easy to “break” and test it first

If you don’t look at the axiom you don’t even know what sort of counterexample you want to construct (if it’s even possible to construct one)

[mortal spear]

I get it now, thank you

may you help me with this one as well?
For ( x = (x_1, \ldots, x_n) ) and ( y = (y_1, \ldots, y_n) ) in ( \mathbb{R}^n ), define

[
d(x, y) = \max{|x_1 - y_1|, \ldots, |x_n - y_n|}.
]
I know this is a metric but not sure how to justify the triangle inequality

[thorny flame]

Halex

[jade thunder]

I think just put things in to try

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[thorn talon]

https://i.imgur.com/zBmf76Y.png

[thorn talon]

i dont even know where to start

[plush hollow]

P = 16 = 2(L+W)
8 = L+W
8-L = W
so area, A = L(8-L)
also from the condition in the second part, L>8-L

[thorn talon]

https://i.imgur.com/MlD8kqs.png so i used chatgpt to help me get this answer before hand but i just dont understand the whole process of how you get to the answer. im sorry if this doesnt make sense. its been a while since ive taken any math classes

[plush hollow]

what is the formula for the perimeter of a rectangle?

[thorn talon]

P=2(l+w)
(should this be common sense. because i didnt know that off the top of my head?)

[plush hollow]

it is common sense, basically adding up all the sides of the rectangle

now we know P=16, yes?

[thorn talon]

so im cooked. got it

[plush hollow]

so 16=2(L+W)

now rearrange to solve for the width

[thorn talon]

so divide each side by 2

so itd be 8 = l+w

then 8-l = w

im not sure where to go from here

[plush hollow]

now whats the formula for the area of a rectangle

[thorn talon]

a =wl

[plush hollow]

now we have a formula for w

input that into a=wl

[thorn talon]

so 8=wl yes?

or no

[plush hollow]

no

A = wl

but w=8-l

sooo

A = (8-l)l

[thorn talon]

im so sorry. im not understanding how you are getting form this step to the next

[plush hollow]

if A=wl, what is the area if w=4 in terms of l?

[thorn talon]

A=4L?

[topaz sinew]

@thorn talon Has your question been resolved?

[night imp]

@thorn talon do you still have doubts?

[thorn talon]

yes https://i.imgur.com/7qIlYS9.png i tried a practice problem and got stuck again

[night imp]

Yes after A=WL
put the value of W in that and find A in terms of L ie
A= (12-L) * L

[thorn talon]

ok lemme try and work it out

[night imp]

Now about finding domain, did you understand that?

[thorn talon]

absolutely not

[night imp]

Did you get this , you will get a quadratic equation for A in terms of L

I will help in domain next

[thorn talon]

ngl im still confused. idk how yall restructured the whole thing. im so sorry.

oh wait

they distributed the l

yes?

[night imp]

Yesssss

[thorn talon]

oh

ok

that makes sense

[night imp]

Now give me this A value in terms of L

I will explain domain after that

[thorn talon]

12L -L^2

yes?

[night imp]

Yes

A = 12L-L²
Now to find domain
In question it's given length will be greater than width correct?

[thorn talon]

correct

[night imp]

So length is L width in terms of L in 12-L , how to write this in terms of L>W?

L>12-L correct?

[thorn talon]

yes that makese sense

[night imp]

What will be value of L if you solve this inequality?

[thorn talon]

7?

12-7=5

[night imp]

Noo😅
L>12-L
Add L both side

[night imp]

You don't know L value yet

[thorn talon]

2l >12

l>6

[night imp]

Bingo

That's your start limit of domain

Now to find the end limit

[thorn talon]

ok idk what that means tbh

but continue lol

[night imp]

Now what was the area in terms of L ? Look above and type

[night imp]

It means L will always be greater than 6 no matter what happens

[thorn talon]

ah ok ok

sorry im dumb

[night imp]

@thorn talon

[thorn talon]

anyway back to math

[night imp]

No problem at all ,I am here to help

[thorn talon]

was it this?

[night imp]

Yes
A=12L-L²
Now you know area cannot be negative, so can we write
12L-L²>0?

[thorn talon]

ok

[night imp]

Now can you solve this quadratic equation and give me values of L?

[thorn talon]

so im assuming i subtract 12

then l^2 >-12

yes?

[night imp]

-L² not L²

[thorn talon]

oh

[night imp]

There is a negative sign before L²

[thorn talon]

ah yes ok

so -l^2 >-12

would i square root?

[night imp]

One second there was one slight Mistake
It was
A=L*(12-L)
A= 12L-L²

So
12L-L²>0

[thorn talon]

where did the 22 come from?

[night imp]

It will be 12 mb

[night imp]

This ^^

[thorn talon]

im so lost

[night imp]

What is the formula of A , type it again . I will make sure I don't type any Mistake

[thorn talon]

12L-l^2

yes

[night imp]

Yes now area should always be greater than 0 so you can write
12L-L²>0

[thorn talon]

ok

makes sense

[night imp]

Now solve this quadratic and find 2 values of L

You know how to do that right?

[thorn talon]

why isnt it greater than 5 or sum since its 6

do you not use starting domain thing

[night imp]

Not yet

That will come once you find 2 values of L from above equation

[thorn talon]

no

[night imp]

12L-L²>0
Add L²-12L both side , you will get
0>L²-12L
Now
0>L*(L-12)

[night imp]

Well that is necessary when we have square terms 😅

[night imp]

Let me know if you understand this or not @thorn talon

[thorn talon]

why did you undistribute

[night imp]

Because then I will get two values of L
One L<0
Another L-12<0

[thorn talon]

uhm ok i think i get the idea but still a little fuzzy

[night imp]

No. Worries just deal with it
Now we have two L values
L<0
L<12

Right?

[thorn talon]

uh yes

confused about the second one

did you isolate

[night imp]

Yes
L-12<0
Adding 12 both side

[thorn talon]

oh ok

makes sense

[night imp]

Now what was the starting point of L that we had found out previously?

[thorn talon]

6

[night imp]

Yessss so we need to discard L<0 value from here
Only L< 12 holds true for that

So our final solution is
L will always be greater than 6
L will always be less than 12
So 6,12 is your domain

[thorn talon]

https://i.imgur.com/VTBvifY.png

uhm

did i do something wrong

[night imp]

Nooo, ig we found domain of L not A

[thorn talon]

._.

ok

so do we need to backtrack to somehwere

[night imp]

Can you put 0,36 in that domain and see if that is correct or not, just wanted to check something

[thorn talon]

https://i.imgur.com/ixezldk.png

womp womp

[night imp]

Ohhh I get it, can you try 0,12 last one for me

[thorn talon]

https://i.imgur.com/x6wyQQo.png

nope

dang

hold on

https://i.imgur.com/O1W17Jd.png

it said its 6, 12

but said it was wrong when i did it

what

[night imp]

You need to use interval notation also

(6,12)

Like this ^^

[thorn talon]

https://i.imgur.com/yfufpEL.png

but when i did it on the actual problem i didnt?

why?

[night imp]

[thorn talon]

real

omw to drop calc frogrun jkjk

[night imp]

[thorn talon]

ok well i guess ill just try another problem and see if i can do that one correc haha

thanks lol

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[neon iron]

Would like to check my answer

$$\int{\frac{\sqrt{4x^2-1}}{5x}} dx$$

[thorny flame]

Edmund Cloudsley

[neon iron]

I'll just type it out"

[pearl fog]

!nosols

[topaz sinew]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[pearl fog]

but dont just give it immediately

[neon iron]

$$\text{Let x = \frac{1}{2} \sec{\theta}$$
$$\frac{dx}{d\theta} = [\frac{1}{\sec\theta}]^\prime$$
$$\frac{dx}{d\theta} = \frac{\sec{\theta \tan\theta}}{2}$$
$$dx = \frac{\sec\theta \tan\theta}{2} d\theta$$

yeah gimme a sec

I'll fix this

is it okay if I ping you once I am done

[thorny flame]

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

[neon iron]

jesus

one sec

[pearl fog]

what does "^\prime" mean?

[neon iron]

$$\text{Let } x = \frac{1}{2} \sec{\theta}$$

$$\frac{dx}{d\theta} = \left[\frac{1}{2} \sec{\theta}\right]^\prime$$

$$\frac{dx}{d\theta} = \frac{\sec{\theta} \tan{\theta}}{2}$$

$$dx = \frac{\sec{\theta} \tan{\theta}}{2} d\theta$$

[thorny flame]

Edmund Cloudsley

[pearl fog]

oh im stupid ncm

[neon iron]

ok so this is the first part, the u-substitution part

now I have to substitute this into the main integral

yeah the question asks me to do this long sub

they specifically say use x = 1/2 sec theta

$\int \sqrt{{4(\frac{\sec{\theta} \tan{\theta}}{2})^2 - 1}} \times \frac{1}{5(\frac{\sec{\theta} \tan{\theta}}{2})} \frac{\sec{\theta} \tan{\theta}}{2} d\theta$

[thorny flame]

Edmund Cloudsley

[neon iron]

Holy moly it worked

first try

once sec

oh shit I mistyped it

$$\int \sqrt{4\left(\frac{\sec{\theta}}{2}\right)^2 - 1} \times \frac{1}{5\left(\frac{\sec{\theta}}{2}\right)} \cdot \frac{\sec{\theta} \tan{\theta}}{2} , d\theta
$$

[thorny flame]

Edmund Cloudsley

[neon iron]

yeah this is it

and then we

\int \sqrt{4\left(\frac{\sec{\theta}}{2}\right)^2 - 1} \times \frac{1}{5\left(\frac{\sec{\theta}}{2}\right)} \cdot \frac{\sec{\theta} \tan{\theta}}{2} , d\theta = \int \sqrt{\frac{4\sec^2{\theta}}{4} - 1} \times \frac{2}{5\sec{\theta}} \cdot \frac{\sec{\theta} \tan{\theta}}{2} , d\theta = \int \sqrt{\sec^2{\theta} - 1} \times \frac{\tan{\theta}}{5} , d\theta

god if this doesn't work

$$\int{\frac{\sqrt{4x^2-1}}{5x}} dx$$
$$\text{Let } x = \frac{1}{2} \sec{\theta}$$

$$\frac{dx}{d\theta} = \left[\frac{1}{2} \sec{\theta}\right]^\prime$$

$$\frac{dx}{d\theta} = \frac{\sec{\theta} \tan{\theta}}{2}$$

$$dx = \frac{\sec{\theta} \tan{\theta}}{2} d\theta$$
$$\int \sqrt{4\left(\frac{\sec{\theta}}{2}\right)^2 - 1} \times \frac{1}{5\left(\frac{\sec{\theta}}{2}\right)} \cdot \frac{\sec{\theta} \tan{\theta}}{2} , d\theta = $$
$$\int \sqrt{\frac{4\sec^2{\theta}}{4} - 1} \times \frac{2}{5\sec{\theta}} \cdot \frac{\sec{\theta} \tan{\theta}}{2} , d\theta =
$$

$$ = \int \sqrt{\sec^2{\theta} - 1} \times \frac{\tan{\theta}}{5} , d\theta$$
$$ = \int \tan {\theta} \times \frac{\tan{\theta}}{5} , d\theta$$
$$ = \int \frac{1}{5} {tan^2 \theta} d\theta$$
$$ = \frac{1}{5}\int {sec^2 \theta - 1} d\theta$$
$$ = \frac{1}{5} (\tan \theta - \theta) + C$$
$$ = \frac{1}{5} (\tan{\arcsec{2x}} - \arcsec{2x}) + C$$

Let's go

nope

just a moment

I'll compile everything together

Okay phew finally typed it out

yeah could you check if this is fine @charred dagger

[thorny flame]

Edmund Cloudsley

[neon iron]

really

cool

thanks

this was a long one

yup we need to sub back x right?

I get arcsec 2x = theta

yup

gosh this was a horrible integral

felt like I was doing MIT integration

bee

tho that must be harder

anyways

have great day

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[vivid stone]

how can i by part this

[vivid stone]

as n is constant

[wispy pier]

Cool.

Do you want to solve it by proving ?

Or formula ?

[vivid stone]

i got this

but it look like it go on to infinity term

[vivid stone]

formula

[wispy pier]

Let me say you

Notice this

[wispy pier]

If we consider the integration to be I

We get I again on LHS

We can take it to RHS and divide by coefficient

To get the derived result

Got me ?

[vivid stone]

i guess

ill try

[wispy pier]

Ofcourse

[vivid stone]

so i need to divide both side by n^2/4?

ohhh wait

i can +n^2/4 both side

thx so much bro

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[prisma mesa]

I have few questions about this:

1. why is the probability 1 / 2^(# of supersets)?
2. why is the # of superset = n - |S|, shouldn't it be 2^(n-|CS)?

[flint stump]

For 1 - coz even if a cookie is not directly chosen, it'd still be eaten if a cookie that has it in its subset is chosen

[prisma mesa]

Okay, but why is there 2^(# of supersets)?

Oh, I think i kinda get it

# of sets that are not supersets / 2^n should be the probability

okay that makes sense actually

But why is the number of supersets n - |S|

If |S| = 5, then there are 3 (n-5) elements not in S

and I can pick any (nonempty?) subset of those 3 elements and add it to S to get a superset

[prisma mesa]

Oh this is incorrect too

i think

[flint stump]

But you are at the same time taking away all the subsets too

[prisma mesa]

oh, how does it influence the probability?

[flint stump]

so you are choosing in the end the whole subset as a single cookie

that makes the Powerset of S a single cookie

can you see what im saying?

[prisma mesa]

not quite

[flint stump]

Sorry, I have the intuition, but cant put it in text

[prisma mesa]

okay, I'll try to think about it

thanks

i probably have too low intuition in probability for this

[flint stump]

just a question, are you going thru all olympiad problems?

[prisma mesa]

I'm just reading evan chen's OTIS book

I'd like to try national olympiad this year, so im practising a bit

[flint stump]

ahh I see

gl in that case

[prisma mesa]

thanks

Just found this solution which I can understand

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[untold matrix]

how to do this

[untold matrix]

<@&286206848099549185>

[late quiver]

Okay

so are you ready

[untold matrix]

hi

its a part of my school assignment but my teacher didnt teach us this

anyone knows how to do
second order partial differentiation that involves intermediate variables? I can solve till first order but I have no idea how to proceed

<@&286206848099549185>

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[opaque nymph]

hey i dont really know how to navigate discord well, but could someone direct me (if there is a page?) to where i can find formula sheets and useful links. I need to make a matrices and functions "cheat sheet" for a first year engmath test and would like to save some time by modifying an original one if possible. Thank you

[opaque nymph]

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[hoary fable]

Idk the first part, showing the volume

[crisp raptor]

you could integrate the Identity function for a regular tetrahedron in R^3, however that's probably overkill for such a simple geometric object

[topaz sinew]

@hoary fable Has your question been resolved?

[hoary fable]

i hvent learnt integration

[astral juniper]

something like this

[hoary fable]

oh alr tyvm

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[snow gyro]

Instructions:

1. Get the equation and diagram
2. Get diagram and it has a specific number -3 to 3
3. Get diagram

[pearl fog]

wdym "it has a specific number -3 to 3", like the graph from -3 to 3?

[snow gyro]

yes

but the graph must be curved

<@&286206848099549185>

@pearl fog

[pearl fog]

do you do all 3 of the instructions per question or just like for number 1 you do instruction 1, num 2 do ins 2 ect

[snow gyro]

wdymm

[pearl fog]

actually nvm

[snow gyro]

the instruction is per number

[fallow heart]

Wdym?

[pearl fog]

1. just graph it and estimate
2. notice its a parabola, find its zeroes and find its extreme point and try to find some points to try to graph it roughly
3. change it to the y=something and graph it

[snow gyro]

what latter?

[snow gyro]

WUTTT 1.1 why is there a point .1

[snow gyro]

oh so this is for no.1?

[pearl fog]

sorry

[snow gyro]

huh why sorry

@pearl fog

[pearl fog]

bruh just ignore it 😭

[snow gyro]

ok okkk

[snow gyro]

This is for all numbers?

[pearl fog]

yeah

[snow gyro]

What is parabola

And what are the kinds

[pearl fog]

these are parabolas

[snow gyro]

Ohhh

So uhm

can yall help me

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@snow gyro Has your question been resolved?

[snow gyro]

no

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[snow gyro]

<@&286206848099549185>

[snow gyro]

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[fallow igloo]

you can do that again

[timber plover]

wait

I need help pls

[vernal matrix]

.(yea make sure you use the reacts on the bot next time, rather than typing the response to it )

[timber plover]

it's urgent

pls

Idk how to solve this problem pls someone help

[fallow igloo]

claim your channel bro

[timber plover]

how

[fallow igloo]

consider checking #❓how-to-get-help

[topaz sinew]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[vernal matrix]

(underneath there, there are available channels you can take)

[snow gyro]

.close

[fallow igloo]

it's closed Xd

[prisma mesa]

It's already close, it will lock soon

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[fathom brook]

Solve the system of equations by elimination.
3x–y–3z= -2

x+3y–3z=6

2x+2y+3z= -14

[north lion]

are there any variables whose coefficients are the same?

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@fathom brook Has your question been resolved?

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[thorny flame]

IndyRishi

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[maiden root]

Consider the energy levels of a hydrogenic atom for which $n$=3. In terms of appropriate quantum number values, list the different eigenstates, first assuming no spin-orbit coupling and then with spin-orbit coupling. Compare the degeneracies for the two cases (including electron spin) and sketch the component levels that result from spin-orbit coupling.

[thorny flame]

IndyRishi

[maiden root]

How can I find the eigenstates for n=3? They showed a diagram for n=2

I figured it out, this is my answer:

For n=3, there are 3 states of L: 0, 1, and 2.
For L = 0, there's only one magnetic quantum number state: 0
Since there are 2 spins, + or - 1/2, that means there are 2 states for l = 0.
Repeating the processes for L = 1 and L = 2 yields 6 and 10 states.

With spin orbit coupling, j can be l +/- 1/2 (spin of electron), except for if l is 0, then j is just 1/2
So j can be 1/2 for l = 0, 3/2 & 1/2 for l = 1, and 5/2 & 3/2 for l = 2.
M_j can go from -j to +j in integer steps. For each value of j, 1/2, 3/2, and 5/2, there are 2, 4, and 6 values respectively.
To calculate the states with spin-orbit coupling, multiply the m_j values calculated before by 2, because s can be +/- 1/2 (2 values).
In lieu of sketch, text diagram below:
If l = 0, j = 1/2. There are 2 states for m_j that we calculated earlier. Multiply by 2: there are 4 states.
If l = 1, j can be 3/2 or 1/2. For 3/2, there are 4 states for m_j and 8 states in general with spin. For 1/2, there are 2 states for m_j and 4 states in general with spin.
If l = 2, j can be 5/2 or 3/2. For 5/2, there are 6 states for m_j and 12 states with spin. For 3/2, there are 4 states and 8 states with spin.

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[daring schooner]

Guys how do I do question number 2?

I need help with that

[celest bough]

I will try something

[sudden temple]

have you tried anything?

okay nvm i am blind

[daring schooner]

No not yet

[sudden temple]

Yeah I saw that thing down there

😶 .

[sudden temple]

try to separate the log terms from the others

I mean, keep the log terms on a single side

then it'll be fine to work on

[daring schooner]

I got this step

[sudden temple]

erm yeah seems alright

btw, it is better if you multiply both sides by -1

and then proceed by properties of logarithm

[daring schooner]

okok

[sudden temple]

the reason is that, we have 5^{2-x} thing in the logarithm so it is better to have 2-x on the other side too

rather than x-2

[daring schooner]

how do i have it on the other side

wait let me show you

[sudden temple]

logarithm doesn't distribute over addition/subtraction

[daring schooner]

ohh my bad

i will fix it

[torpid sparrow]

Hint (from the beginning), rewrite every single term like log_5(something)

[daring schooner]

okok

Is it correct

😭

[daring schooner]

.

anyone?

[sudden temple]

here

[daring schooner]

okok

sorry

[sudden temple]

what's all that

[sudden temple]

oh you tried it differently here

[daring schooner]

finding x

yeah

[sudden temple]

you know what

if you wanna do it that way

treat 2-x as a single term

it'll be quicker to solve it

[sudden temple]

by doing that everything will be simplified in just 2-3 steps

just treat 2-x as a single entity

[daring schooner]

Alright

[sudden temple]

bruh where did log go?

with 8

[sudden temple]

just combine the logarithms on the left hand side, after writing log(2^3) as log(8)

and you'll be done

[daring schooner]

Okok

how do I combine 🥲

I got log (8)

[sudden temple]

$\log{a}+\log{b}=\log{ab}$

[thorny flame]

77²

[sudden temple]

dumb latex

if bases are same

[daring schooner]

I combined it here

[sudden temple]

bro that multiplication symbol is crazy

[daring schooner]

You thought it was a plus ?

[sudden temple]

fr

it looked like plus

ngl

[daring schooner]

My handwriting was bad

[sudden temple]

last step

can't do 8*5

exponents aren't same

[daring schooner]

yhh

so what do I do

I’m so confused

[sudden temple]

assume 5^{2-x}=Q

then 8Q=3^x - Q

then?

[daring schooner]

Yeah I got

9Q=3^x

[sudden temple]

now sub back Q

and solve for x

[sudden temple]

no

[daring schooner]

NO?😭😭

[sudden temple]

I mean the way is not rright

[daring schooner]

What is wrong

[sudden temple]

5^2 times 3^2 is not 15^4

it is 15^2

also 3^x times 5^x is not 15^2x

it is just 15^x

exponents add when the bases are same

I think you got it right?

[daring schooner]

yhhh

Yep I got it now

thank you brother ‼️‼️

[sudden temple]

np

!done

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[buoyant rose]

hey, any idea how the answer is 1,8 ?

[buoyant rose]

y=2x+6
y=−4x+12

[buoyant rose]

I need to find the point in which they contact

I got 1,12

anyone has any idea? <@&286206848099549185>

[fringe maple]

Two affine function intersect when their values (y) are equal hence they intersect when 2x+6=-4x+12

[wispy frost]

The intersection

[fringe maple]

Which is trivially solvable

[buoyant rose]

for some reason I wrote 16 in my calculation 💀

hold on a second, let me re do it

[wispy frost]

Wait, what?

You can find the intersect here

[fringe maple]

Solution should be 1

[fringe maple]

You like solve it rigorously instead on just looking at the answer ?

[buoyant rose]

how did you get 1?

I'm now re-solving for x

[fringe maple]

You do see that by imputing x=1 in the two function you find the same y

[buoyant rose]

I keep getting different results eh

I have 2x+6=-4x+12

[fringe maple]

You put the X on one side and the whole number on the other

[buoyant rose]

ok and I get -2x = 6

[fringe maple]

You want to privilege removing any negative sign

[fringe maple]

You put the -4x on the other side so 6x and remove the 6 from the 12 on the other so 6x=6 is the correct solution

Do you get it or do you want me to explain again ?

[buoyant rose]

BRO I WROTE IT WITHOUT THE -

one second

[fringe maple]

No worries

[buoyant rose]

okay so we get x=1?

[fringe maple]

Yep

[buoyant rose]

what do we do with that

we need x and y

[fringe maple]

^

[buoyant rose]

I see, Lemme try

[fringe maple]

You imput X in one of the equation and you should get your y

[buoyant rose]

I got y=12

is that right?

[fringe maple]

No y=8 ?

[buoyant rose]

how?

I did y = 2*1 + 6

2 * 1 is 2

and I did 6*2

so 12

[wispy frost]

here

[fringe maple]

y = 21 + 6 = 8
Or
y = -41 + 12 = 8

[buoyant rose]

OH

I NEED TO ADD AND NOT MULTIPLY

ok so yeah it's 8

[fringe maple]

Yep so intersect point is (1;8)

You understood the method ?

[buoyant rose]

yeah, but what do I do if I need to find y intersect point

so for example, we have: y=4x+8

and I need to find the intersect point on the Y axis

[fringe maple]

Well what we did here is finding the intersect on the x axis and on the y axis at the same time. On the x axis it’s 1 and on the y axis it’s 8

[wispy frost]

y = mx + b

b is the y-intercept

so,

y = 4x + 8

8 is the y int

M is slope (4)

[buoyant rose]

b is just y ?

[wispy frost]

y-intercept

[buoyant rose]

lmao the function's wording is kind'a confusing

[fringe maple]

Ohhhhhhhh y axis intersect

[wispy frost]

it's slope form

or wtv

[buoyant rose]

wdym y-intercept? like, they height of it

[wispy frost]

where it crosses the Y

so for exaxmple

[buoyant rose]

so, the answer is literally in the formula

[fringe maple]

I actually have a whole note on that

[buoyant rose]

there's no need for any calculations 😂

[wispy frost]

Y-INT is 3.5

because on the Y axis, the line crosses it on 3.5

[buoyant rose]

Hold on a sec, I don't get it

[wispy frost]

if you're trying to find the y int in

y = 4x + 8

it's just... 8. No calculations needed.

[buoyant rose]

oh okay

[wispy frost]

now for X intercept, it's different

but you said they only asked for y, right?

[buoyant rose]

nope?

[wispy frost]

idk the language

what does it say

[buoyant rose]

It says I need to find the point

[wispy frost]

which point

[buoyant rose]

of B

[wispy frost]

you might have to put it in proper form

x, y

[buoyant rose]

ohh, x just is 0

[wispy frost]

0, 8

[buoyant rose]

I think

yeah okok

ayyee correct

[wispy frost]

[buoyant rose]

ok now we need to do the same thing with x

[buoyant rose]

so y is 0

[wispy frost]

same graph?

[buoyant rose]

yes

[fringe maple]

A note I made for a friend taking a test, might be helpful in your study

[wispy frost]

@buoyant rose to find the x intercept

set y = 0

[buoyant rose]

so x = -2 ?

yeah yeah

[wispy frost]

and solve with y = 0

[buoyant rose]

I did 0 = 4x + 8

and then badabim badabam

and x = -2

right?

[wispy frost]

should be

yeah

[buoyant rose]

no?

it should be correct? what

[wispy frost]

this looks like -2

[buoyant rose]

it does

[wispy frost]

0 = 4x + 8 is -2

[buoyant rose]

but it's not working here s

[buoyant rose]

did I input it correctly?

[wispy frost]

what does the question say

maybe you read it wrong

[buoyant rose]

"find the intersection point in the x axis"

yeah, I read it right

[wispy frost]

and it's the same equation?

[buoyant rose]

yes

[wispy frost]

calculator says it's right

[buoyant rose]

DA FUCK

[wispy frost]

yeah

[buoyant rose]

😂 WHA

[wispy frost]

it was just a software issue

you should tell your teacher or whoever

uh

made it

[buoyant rose]

it's a random free course to catch up on lots of studying..

It works

[wispy frost]

oh

then yeah it's fine, you got it right

[buoyant rose]

exam in a week

got an entire syllabus to catch up on

I'm here

EACH ONE IS A TOPIC

SO HARDD

[wispy frost]

i believe in you

[buoyant rose]

allright

so- next question (same formula, same line)

I need to find "for what X value, the line gets positive values"

whatever the heck that means

[wispy frost]

plug an X value into the equation

and see what you get

trial and error

[buoyant rose]

this is the answer

how do you get to it?

[wispy frost]

all values

not just one

bc there are multiple values

that would have a positive y

[buoyant rose]

a positive y..?

what

I don't get it.. how do you get to x > -2 ???

[wispy frost]

is there like

lessons

associated to the exams

or are you learning from exams

[buoyant rose]

I'm learning...

from a course that the teacher gave me

I'm essentially not supposed to be trying 4 units, but I want to rank up to it

so I need to study since the exam is soon

[wispy frost]

you should read the lessons associated to the exam not just take the exam itself

[buoyant rose]

There isn't anywhere to "read"them

you come to class, you do the stuff the teacher says and u try to learn

but I was put in a lower-tier math's class, and got 100 on the final exam. so now if I wanna rank up- I gotta catch up on all of this material

[wispy frost]

do you have to use this platform

[buoyant rose]

no, not really

it just sums it up really well

[wispy frost]

learning from just taking quizes isn't good

it should be a refresher

not how you learn

[buoyant rose]

Yeah, but there aren't any video classes

[wispy frost]

you can look up ones from different resources

like find a YT tutorial on the section that exam is

i think your resource isn't really to teach, but mainly to quiz from things you learned elsewhere

[buoyant rose]

yeah it's annoying

Do you know where I can find this kind of thing?

starting with analatics

[wispy frost]

analytics?

[buoyant rose]

I don't even know how to spell it bro 😭

It's a different name in my language

[wispy frost]

is this for algebra 1, what class is it

[buoyant rose]

yeah, I don't know what this subject that we're working on rn is caled in english

but in translation it's analytics

[buoyant rose]

uh..

we just have.. units here

I guess it's algebra, I don't know

there's 3, 4, and 5 units

[wispy frost]

I have no idea how your education system works so I couldn't give you any recommendations

[buoyant rose]

allright.. I guess I'll keep trying

thanks

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[twin walrus]

How would I solve the integral "(3x+1)/((x^2)+1) dx" ?

[twin walrus]

so far I've substituted ((x^2)+1) for u(x)

[agile harness]

then what

[twin walrus]

and gotten dx = 1/2x du

[crisp raptor]

use the linearity of Integration operator

[twin walrus]

what is that

[crisp raptor]

$\int (f(x) + g(x)) dx = \int f(x) dx + \int g(x) dx$

[twin walrus]

and ive also split the integral into two

yes ive done that

[thorny flame]

tobi

[agile harness]

so then what is the issue

[crisp raptor]

$\int \frac{1}{1+x^2} dx$

[thorny flame]

tobi

[crisp raptor]

this is a well known integral

[twin walrus]

and I have (3x+1)/((x^2)+1) on one side and the thing that turns into arc tan on the other

oh

[neon iron]

tan

[twin walrus]

yeah but I dont know what to do with the other integral

[agile harness]

use your u sub

[twin walrus]

Mhmm?

[neon iron]

you just split it

[agile harness]

if du=2xdx then dx=1/2x du

the x will cancel

[crisp raptor]

3 can be brought outside of the Integral using the linearity property

[neon iron]

{3x/x^2 +1 } + 1/x^2 +1

[agile harness]

and you’re left with 3/2u

[twin walrus]

wow the x does cancel

[crisp raptor]

yeah then 3/2 can be brought outside of the Integral and you are left with 1/u du

[twin walrus]

I only wrote 1/2 du

not 1/2x du

wow

[agile harness]

^

[twin walrus]

Thank you all so much

but I do have another problem

[crisp raptor]

whenever you 1/f`(x) this works!

[twin walrus]

that I didnt even know where to start with

[agile harness]

send it

[twin walrus]

integral from pi to -pi, sinxcosx dx

[neon iron]

well solve more questions..... you will notice the similarities

[agile harness]

what’s sin(2x)?

[crisp raptor]

this is the rule, you multiply with the derivative of your substitution

[neon iron]

sin2x=2sinxcosx

[agile harness]

apples

[twin walrus]

I dont know