help-23

[final loom]

Higher

more higher

[frigid geyser]

okay, i see your point

[final loom]

nope, not really

you don't see it, it's not even in the first 100 numbers range

I tried brute forcing but then succumbed to Pell and solution came in 5 mins

[frigid geyser]

if i hit 85 and i didn't get a perfect square i'd shut it down and look at a more algebraic solution lol

[final loom]

(ヘ･_･)ヘ┳━┳

moreover, I did calculations via a stupid calculator instead of using wolfram(!?) wow waste of time

[frigid geyser]

pell works, but there has to be an easier way

[final loom]

Yep there sure is

restrict the values

3n^2 = 9x^2 + 21x + 24 < (3x + ...)^2 ? or something along those lines

to start with:

(x + 2)^2 < n^2 < (2x + 3)^2

is visible

just need to make the bounds smaller

by multiplying(!?) ig

[frigid geyser]

that has a crap ton of test cases

[final loom]

Yes it does...

[frigid geyser]

what did pell get you anyway?

[final loom]

but trust me,

that's probably the only alternative I can think of,

that I've previously came across,

during a very familiar situation

[frigid geyser]

i'll admit pell is probably the best way

if we could factorise things nicely we could have used simul

but i haven

haven

[final loom]

haven

[frigid geyser]

haven't found a factorisation*

gaaaah

enter key

[final loom]

Did you not see it?

[frigid geyser]

i saw it

[sudden merlin]

when would u be expected to know about pell

[final loom]

Don't worry Savage

Shrikar

Shreaker

idk

anyways

There is some solution relating to bounding the function

, bounding the quadratic in x

$(3x+1)^2 < 3n^2 = 9x^2 + 21x + 24 < (3x + 5)^2$

[final loom]

OwO

@frigid geyser camieee does this work?

[frigid geyser]

oh?

[final loom]

well hell yeah this works

mwahahahaha

[frigid geyser]

yes, this is a much tighter bound

[final loom]

only 3 integers

wait -,-

not 3

[frigid geyser]

LOL

[final loom]

hmmmmm

okay time to go brrr

$$(6x+7)^2 < 12n^2 = 36x^2 + 84x + 96 < 36x^2 + 95x + x + 64$

ehhhhhhhh

damn that constant -,-

wait, but x > 25

[frigid geyser]

oui

[final loom]

I can use that fact somehow

[frigid geyser]

oui

[final loom]

lemme see

[sudden merlin]

well i think they said that since there is another solution less than 25

[final loom]

well

yha

n = 16 works

UwU

Hmpfff

[frigid geyser]

sad

[final loom]

25 + 64 > 96 right 🤔

oh well

nvm

√12 would cause an issue now

but now I'm pretty sure, the bounding game would get the soln.

$(3x+3)^2 < 3n^2 = 9x^2 + 21x + 24 < (3x + 5)^2$

Yep, thats the most tightest bound I could produce

lmao

$3x + 3 < \sqrt{3}n < 3x + 5$

[frigid geyser]

wooooo

[final loom]

[sudden merlin]

do you think someone else has some insight

maybe do some fancy math

[final loom]

welp,

If it's Alg1, I suppose they would probably give the perfect boundary condition in 1 go

or maybe they'd give the pell first who knows

Look for the pins in #competition-math

check the invite to oly server

and ask over there

but don't expect them to explain themselves

[frigid geyser]

the oly server's pretty dead too

compared to this one

[final loom]

okay wait

this is pretty hecked

[frigid geyser]

?

[final loom]

idek what kinda boundary I'm looking for that'd help me out

I mean, if I could bound a perfect square between two perfect squares

sure that helps

but what's with bounding a multiple of perfect square!?

[frigid geyser]

LOL

[final loom]

I mean, $$(x + 2)^2 < n^2 = 3x^2 + 7x + 8 < (2x + 3)^2$$ at least gives me $$(x + 2) < n < (2x + 3)$$

[final loom]

I should be looking to bounding perfect square multiples of n

smh

[frigid geyser]

(3x+4)^2 = 9x^2 + 24x + 16 = 9x^2 + 21x + 24 implies ???

[final loom]

implies?

[frigid geyser]

yeah, ain't workin'

sorry

[final loom]

mhmm, that's why

I found it stupid
Plus recalled what happened back then

I had something similar, a quadratic

and I bounded it between two consecutive perfect squares

[frigid geyser]

i am suddenly so happy my competition solving days are over and my competition setting days are in

[final loom]

Camie,

ahem

lmao

[frigid geyser]

[final loom]

I thought of sth and felt like I should stop or I'd get slapped or smth

$$4n^2 = 12x^2 + 28x + 32 < (4x)^2$$

[frigid geyser]

bounding game?

[final loom]

hmmm?

[frigid geyser]

go on?

[final loom]

I'll continue probs, it looks promising

n < 2x

✓✓

meh, need pen paper to think this through

[frigid geyser]

you probably want to compare with (4x-k)^2, but this will eventually blow up faster than 4n^2, so best bound has to come from (3x+m)^2

which is a bit of a problem, oui

maybe (4x - k(x))^2, where k is now a function of x hoho

[final loom]

*not being able to efficiently use x > 25 is truly troublesome

if only you can show that 2x - 50 < n < 2x

it'd be an easier computation

I mean, at this point I'd just spoil the answer: ||(x, n) = (127, 222)||

but, yha, apart from Pell, the only immediate solution I think comes from the bounding game, which really would require pen n paper, which is not accessible to me atm

[frigid geyser]

i'd work on it, but i'm busy today

i'm waiting for breakfast, but after that i have to do my prep job as a contest judge

[final loom]

waiting for breakfast

Also,

[frigid geyser]

what, it's on delivery

[final loom]

@sudden merlin

[final loom]

.

[humble venture]

true

[safe radish]

@sudden merlin Has your question been resolved?

[sudden merlin]

lol

well thx for the help

ig its fruitless to keep it open any longer

please let me know if you figure out an easy way of doing it

.close

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[low nest]

how do i show that $v$ in some complex inner product space $V$ and ${\varphi_1,\dots,\varphi_n}$ being an orthonormal basis in $V$ that
$$v=\sum_{j=1}^n \langle v,\varphi_j\rangle\varphi_j$$

[flat frigate]

Beous

[low nest]

i tried to show that $\left\langle\sum_{j=1}^n\langle v,\varphi_j\rangle\varphi_j,v\right\rangle=\Vert v\Vert^2$ but was not able to

[flat frigate]

Beous

[plucky elk]

these are two different things to show. which one do you want

[low nest]

this one

<@&286206848099549185>

nvm found a solution

.close

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[crisp linden]

What are you stuck on?

[junior wagon]

i tried just doing the expected value

but that wasnt the right answer

like 15000(1/6000) + 2500(3/6000) + ...

[crisp linden]

Well

If I paid $5 for a ticket, and got a car worth 15000

What is my actual profit

[junior wagon]

oh wait

Id have to subtract 5

is that correct?

[crisp linden]

Well depends on how you calculate the expected value

for the (6000-15)/6000 chance, do you say -5?

[junior wagon]

this should give expected value

then - 5

[crisp linden]

So you calculated E = 15000(1/6000) + 2500(3/6000) + 700(1/6000) + 250(10/6000)

[junior wagon]

yeo

yep*

[crisp linden]

mhm, and what's the answer on the answer sheet?

[junior wagon]

-0.7167

Which is the result we get

[crisp linden]

Great! There are two ways to calculate it

Either you calculate the expected value of the profit, by doing E = 14995(1/6000) etc.

Or you calculate the expected value of the win

and subtract the price

Since the price is paid regardless of whether you win or not

[junior wagon]

make sense

thanks

.close

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[wraith adder]

I need help with some trig homework so far I have

[wraith adder]

and I’m unsure of what to do next

[safe radish]

@wraith adder Has your question been resolved?

[wraith adder]

<@&286206848099549185>

[wraith adder]

oh im stupid

.close

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[safe radish]

Channel closed

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[fickle charm]

.close

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[marble forum]

Can someone explain perfect square rule for eg, (a-b) power to 2 please

[marble forum]

i dont understand this

[plucky elk]

that's unreadable in dark mode

[marble forum]

sorry imma get a better one

[ancient escarp]

just a ""shortcut"" for expanding?

[marble forum]

ig

im just trying to understand the perfect square rule

[ancient escarp]

$x^2=x\cdot x$

[flat frigate]

a disappointing son

[marble forum]

?

[ancient escarp]

$(a+b)^2=$...

[flat frigate]

a disappointing son

[marble forum]

yes that

[ancient escarp]

what don't you get

[marble forum]

the whole thing

but i want a example for (a-b)powered by 2

[ancient escarp]

do you understand how they go from $(a+b)^2$ to $(a+b)(a+b)$?

[flat frigate]

a disappointing son

[marble forum]

not really

[ancient escarp]

does this make any sense?

[marble forum]

yes

[ancient escarp]

but this doesn't?

it's the same concept

squaring something is multiplying that thing by itself

[marble forum]

that makes sense but is it diferent with minus?

[ancient escarp]

it's the same concept, different result

[marble forum]

oh

[ancient escarp]

only slightly different though

[marble forum]

bc ive done everything from the internet and the awnser doesnt match up with my book anwsers

(d-1)^2

this one im struggling

[ancient escarp]

do you have work you can show?

[marble forum]

like the working out i have rn or the aqual question

[ancient escarp]

the working out you have

[marble forum]

Is it right?

[ancient escarp]

yeah

[marble forum]

then my book is a scam lmao

[ancient escarp]

wait what is that thing after the 2d

[marble forum]

it says that its

oh

its

d^2 - d1 - d1 -1^2

[ancient escarp]

i mean the last line

[marble forum]

oh

[ancient escarp]

something between the 2d and the +

[marble forum]

d^2 - 2di +1^2

[ancient escarp]

what's the i?

[marble forum]

its the 1

my bad lmao

[ancient escarp]

i think you misunderstood the notation

[marble forum]

oh

dammit

[ancient escarp]

1d = d with a coefficient of 1

[marble forum]

yeah

\

[ancient escarp]

which is commonly just written as the variable, d

since multiplying by 1 doesn't change anything

[marble forum]

oh

so its just 2d

[ancient escarp]

so "d1" should be written "1d" or just "d"

yes, just 2d

[marble forum]

oh ok

but why does my book say

1^2 = 1??

[ancient escarp]

because that's true

[marble forum]

really?

[ancient escarp]

1 to any power is 1

remember what raising 1 to that power means

you're just multiplying that number by itself

1 times 1?

[marble forum]

but when i put it on the calculator it said 2

oh wait

nvm

i did 1 * 2

my bad

thanks for helping and pointing out the obvious lmao

.close

[safe radish]

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[marble forum]

.repoen

.reopen

[safe radish]

✅

[marble forum]

wait aminute

can you explain why theres a plus at the end part

d^2 - 2d + 1

i dont understand the plus part

ok nvm

.close

[safe radish]

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[plucky elk]

@valid grove did you figure this out?
#help-10 message
https://cdn.discordapp.com/attachments/903430332324405288/947359899346673694/IMG_10C2F8567CF9-1.jpg

[valid grove]

Yup

Arctan 1

[plucky elk]

Nice. Induction?

[valid grove]

Is it correct?

[plucky elk]

Yea. Did you need to factor polynomials?

[final loom]

[valid grove]

What does that mean?

[plucky elk]

Mine looked like uhhh
$\frac{2n^3 + 4n^2 + 3n + 1}{2n^3+6n^2+5n+2}$

[flat frigate]

riemann

[plucky elk]

did some cheating before i realized that was equal to $(n+1)/(n+2)$

[flat frigate]

riemann

[final loom]

Ask your math question in a clear, concise manner

The link is randomly teleporting me lmao

[valid grove]

Yeah

[crisp linden]

That's the purpose of it

[plucky elk]

https://cdn.discordapp.com/attachments/903430332324405288/947359899346673694/IMG_10C2F8567CF9-1.jpg

[crisp linden]

teleport you to the scene of the crime

[final loom]

ohh

[plucky elk]

i shoulda done that in the first place

[final loom]

well mwahahaha

[crisp linden]

I just got the worst error you can get in LaTeX

Missing } inserted

😭

[plucky elk]

happens to me hourly

[plucky elk]

Ansh solve this and tell me my solution is inefficient

[final loom]

isn't it just

(2i +1 ) - (2i - 1)

/

1 + (2i - 1)(2i + 1)

and telescope?

[plucky elk]

fuck

[final loom]

👀

most of those arctan summation question are surely just telescopes of this form

UwU

[plucky elk]

my induction was inefficient then

just solving for partial sums
$\arctan(r_n) = \sum_{i=1}^n \arctan\left(\frac{1}{2 , i^2}\right)$
I let $a_i = \frac{1}{2 , i^2}$, so I had to prove $r_n = \frac{n}{n+1}$, for $n\ge 2$.
E.g. using that sum of arctan identity $r_2 = (a_1 + a_2)/(1+a_1 a_2) = 2/3$ and $r_3 = (r_2 + a_3)/(1+ r_2 a_3)$.

[flat frigate]

riemann

[final loom]

to be able to come up with such a strong induction hypothesis

omg

[plucky elk]

The inductive hypothesis step $(n+1)$ was to show

[flat frigate]

riemann

[plucky elk]

$$r_{n+1} = \frac{\tfrac{n}{n+1} + \tfrac{1}{2(n+1)^2}}{1 - \tfrac{n}{n+1}\tfrac{1}{2(n+1)^2}} = \frac{n+1}{n+2}$$

[flat frigate]

riemann

[plucky elk]

That was pain. highly don't recommend

[crisp linden]

STEM is pain

But we're all here aren't we

[final loom]

Anyways, here's the alternative

$$\tan^{-1} \frac{1}{2i^2} = \tan^{-1} \frac{(2i + 1)-(2i -1)}{1+(2i+1)(2i-1)} = \tan^{-1} (2i + 1) - \tan^{-1} (2i - 1)$$

[final loom]

And telescope UwU

[crisp linden]

hm, I changed nothing in my LaTeX except write some text

and now magically the error went away

[plucky elk]

https://giphy.com/gifs/clapping-clap-pingu-QBC5foQmcOkdq

[plucky elk]

this gives me only $-\arctan(1)$ ?

[flat frigate]

riemann

[plucky elk]

$$\sum_{i=1}^\infty [\tan^{-1} (2i + 1) - \tan^{-1} (2i - 1)] $$

[flat frigate]

riemann

[plucky elk]

$$= \tan^{-1} (3) - \tan^{-1} (1) + \tan^{-1} (5) - \tan^{-1} (3) + \tan^{-1} (7) - \tan^{-1} (5) + ...$$

[flat frigate]

riemann

[plucky elk]

Oh forgot the last term

$\lim_{n\ra\infty}\arctan(n) - \frac{\pi}{4} = \frac{\pi}{4}$

[flat frigate]

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

[plucky elk]

.close

[safe radish]

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[final loom]

.mwahahaha

[ruby cloak]

Is this correct I solved with radicals

[safe radish]

Please don't occupy multiple help channels.

[ruby cloak]

@thin bridge

<@&286206848099549185>

<@&286206848099549185>

Please unmute me

I am struggling a lot

[safe radish]

@ruby cloak Has your question been resolved?

[wild cape]

@ruby cloak pls dont use multiple channels

.close

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[ruby cloak]

Please help me

[safe radish]

Please don't occupy multiple help channels.

[zinc crown]

did u not see what rokabe (a moderator of the server) and the bot just said?

use .close to close this channel and wait for someone to help you in your other one

#❓how-to-get-help

[safe radish]

@ruby cloak Has your question been resolved?

[ember bough]

you can literally use a calculator to check

[wild cape]

@ruby cloak again, pls dont use multiple channels

.close

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[snow robin]

if i have the radial vector $\vec{r}$ which when normalized is $\hat{r}$ and i have some vector $\vec{r_2}$ what can be said about $\hat{r} \cdot (\vec{r}-\vec{r_2})$

[flat frigate]

Katharine

[safe radish]

@snow robin Has your question been resolved?

[safe radish]

@snow robin Has your question been resolved?

[safe radish]

@snow robin Has your question been resolved?

[safe radish]

@snow robin Has your question been resolved?

[lean otter]

Radial vector meaning it's a vector in a spherical basis? Also do you know dot products distribute over addition?

[snow robin]

what would i get if r2 is on the z axis

[lean otter]

So you're saying we're in three dimensions yeah? Give it a try yourself and see what you get. Do you know what I was suggesting to you by saying that the dot product distributes over addition?

[snow robin]

yes

i have tried but i can't figure it out

[lean otter]

What do you have so far?

[snow robin]

well i thought if r2 is on the z axis

then r dot z

would be cos theta

if both r and z are unit vectors

but that doesn't give me the right thing

let me say it differently

that gives me an integral that not even mathematica can solve

[lean otter]

rhat dot r2 would be |r2| cos(theta) I would think right?

if r2 is on the z axis

oh they're unit vectors you said

but is that true of r2? you said it wasn't giving you the right thing maybe it's just off by a factor?

[snow robin]

well i have

$\int_{V} \frac{\hat{r} \cdot (\vec{r}-\vec{r_2})}{r^2 |\vec{r}-\vec{r_2}|^3} dV$

[flat frigate]

Katharine

[snow robin]

and previously i got $\hat{r} \cdot (\vec{r}-\vec{r_2}) = |\vec{r}|-|\vec{r_2}|\cos{\theta}$

[flat frigate]

Katharine

[snow robin]

but that gives me a very unsolvable integral

[lean otter]

wait shouldn't the left term be $1/|\vec{r}|$? since $\hat{r}\cdot\vec{r} = |\vec{r}|\hat{r}\cdot\hat{r}$

probably doesnt help lol sorry

wait oop

no Im buggin

[snow robin]

no because $\vec{r} = \hat{r} |\vec{r}|$

[flat frigate]

Katharine

[lean otter]

yeah fixed

[flat frigate]

zd

[lean otter]

i had it flipped lol

[snow robin]

nvm it's solvable

[lean otter]

nice!!!

[snow robin]

i just saw the way to solve it

[lean otter]

pls tell 👀

[snow robin]

so in spherical coordinates

the r^2 on bottom disappears against the r^2 from the jacobian

and we are left with $(r^2+{r_2}^2-2 r r_2 \cos{\theta})^\frac{-3}{2}$

[flat frigate]

Katharine

[snow robin]

and if you differentiate what's inside the perenthesis

you get what's on top

i think

it's 2 times what's on top

thank you for the help!

[lean otter]

;-; not much lol but yw anyway

[snow robin]

helping me think about the problem differently and with the dot product

thats quite a bit

[snow robin]

.close

[safe radish]

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[cinder breach]

I have a exam tmr I don't understand how to solve this

[cinder breach]

Nvm ik that one lemmi give another pic

<@&286206848099549185>

Almost 20mins

Also if y'all respond can y'all do it step by step and even write if u devide or multiply etc like multiply by 3 etv

[tame charm]

@cinder breach well we want to help you understand it, not do it for you

So do you know what “directly/indirectly proportional” mean

[cinder breach]

Not really

Kinda have finals tmr and gotta understand it lol

[tame charm]

Ok

[flat frigate]

abs_0

[cinder breach]

What about substitution and elimination method in linear equations

[tame charm]

Ah we can work on those after this

Those are different from proportions

[flat frigate]

abs_0

[tame charm]

Are you understanding this so far?

@cinder breach

[cinder breach]

Kk

Sorry late

[tame charm]

So you got it?

[cinder breach]

Yup

[tame charm]

Ok so here’s an example problem

A ten foot pole casts an eight foot shadow. How long is a pole which casts a twelve foot shadow?

[flat frigate]

abs_0

[tame charm]

Did that make sense

[cinder breach]

Can u explain how it's 15?

[tame charm]

Well we have [\frac{\text{pole}}{12}=\frac{10}{8},] so multiplying both sides by 12 yields $$\text{pole} = \frac{10}8 \cdot 12 = \frac{120}8 = \textbf{15}.$$

[flat frigate]

abs_0

[cinder breach]

Multiply 12 by 10 and then devide the answer by 8?

[tame charm]

Yeah

Well like

$$\frac{10}8 \cdot 12 = \frac{10}8 \cdot \frac{12}1 = \frac{10\cdot12}8$$

[flat frigate]

abs_0

[tame charm]

It’s all the same

[cinder breach]

Sorry my phone's glitching and taking while to see ur msgs and is hard to type with shitty internet

[tame charm]

Oh np

Just lmk when we can move on, or if u don’t get something

[cinder breach]

Ok just rebooted it might be fine rn so we can move on

Hey?

@tame charm

[tame charm]

Oh sry

[cinder breach]

Np

It's rn 11 30pm need to sleep in 30mins

[tame charm]

Ah

Wait ur test is tomorrow?

[cinder breach]

Math about proportion and substitution and elimination method

And inversely proportion

[tame charm]

Ok I’ll just send some stuff like before

Read all of it

[cinder breach]

Alright

[tame charm]

The ratio of boys to girls in a class is 3:2. If there are 35 students in the class, how many girls are there?

[flat frigate]

abs_0

[cinder breach]

3 boys :2 girls?

[tame charm]

Yeah

[cinder breach]

That shouldn't be too hard

[tame charm]

Ok so ur good with this

Here is an exercise for you

[cinder breach]

Can't I multiply 3 by 7 and 2 by 7 for answers?

[tame charm]

Why 7

[cinder breach]

Because 3 and 2 total is 5 and 7 5s is 35

[tame charm]

Yeah that works too

Nice

[cinder breach]

Kk

[tame charm]

Ok so here’s your exercise

In 4 games, Michael Jordan scores 124 points. If he continues scoring at this rate, how many points will he score in the next 6 games?

[cinder breach]

Hmm

Will usually need a paper for this brb

186

[tame charm]

Yep, nice

Ok one more, then we’ll talk about inverse proportions

[cinder breach]

Kk

[tame charm]

The ratio of wins to losses of the Yankees is 15:16. If the Yankees lost 64 games, how many games did they play?

(Slightly harder, be careful)

[cinder breach]

Wdym 15 16?

15 is wins and 16 loses?

[tame charm]

Yeah 15 wins : 16 losses

[cinder breach]

I don't understand?

[tame charm]

Wdym

[cinder breach]

4 games?

Like confusing me

[tame charm]

Actually when I did this, I got it wrong

So here’s the explanation

[cinder breach]

Alright

[tame charm]

The Yankees lose 16 out of every $16+15=31$ games. Thus losses$/$games $=16/31$. Since they have lost 64 out of $x$ games, $16/31=64/x$, so $x=124$. Thus, they have played \textbf{124} games.

[flat frigate]

abs_0

[cinder breach]

What

[tame charm]

What

[cinder breach]

Ah out of 31

[tame charm]

Ye

[cinder breach]

I was gonna do out of 31 but. I thought na that counts wins also lol

[tame charm]

Oh yea

[cinder breach]

And it only wants loses

[tame charm]

I think you could also do $$\frac{wins}{losses} = \frac{15}{16} = \frac x{64},$$ so they won $x = 60$ games and thus played $60 + 64 = 124$ games

[flat frigate]

abs_0

[tame charm]

^ that makes more sense for me personally

[cinder breach]

Alr ty

Inversely I think ik about

[tame charm]

Alright

[cinder breach]

It's if one decreases other increase or no?

[tame charm]

yeah

[cinder breach]

We use multiplication and devition?

[flat frigate]

abs_0

[cinder breach]

How do we know when to devide and when to multiply

[tame charm]

Wdym

Oh

[flat frigate]

abs_0

[cinder breach]

Kk

[cinder breach]

Just hope I get like atleast 40/50 tmr test

[tame charm]

Ye

Wait so

Can you post some problem from your book again

Cause these problems I’m sending might be kinda different from what ur class is doing

[cinder breach]

Alright

Inversely proportional?

[tame charm]

Yeah

Direct or inverse

[tame charm]

Ok so it looks like they’ve got you doing some sort of table

Do you think you can do those now

They seem to have a slightly different flavor than the problems I’ve been sending you

[cinder breach]

Alright

It got me confused ngl lol

2 and 9 that's not possible in multiplication table I think

[tame charm]

Nvm nvm

If x = 2 and y = 9, and they’re inversely proportional

Then xy = 2*9 = 18

So 18 is the constant of proportionality

[cinder breach]

U multiply them?

[tame charm]

We’re talking about #5 right?

[cinder breach]

Yeh

[tame charm]

Yeah they say x and y are related by the equation xy = constant

They tell you when x = 2, y = 9

So the equation is xy = 18

[cinder breach]

Kk

[tame charm]

So now can you figure out the other blanks in that table

[cinder breach]

Kinda confused

How can I use 18 for others?

[tame charm]

Oh

So for the second column it says x = 3 and y is unknown

We know that xy = 18

[cinder breach]

Ye

[tame charm]

Substituting 3 for x gives us 3y = 18, so y = 6

[cinder breach]

Ahhh

[tame charm]

Ye ye

[cinder breach]

Multiply by 2

[tame charm]

What? Lol

[cinder breach]

Kuz 9 u multiply by 2 and this also u multiply by 2

Wait

Stoopid

[tame charm]

Do u get this explanation tho

[cinder breach]

Not really srry

How do u substitute 3 for x when it's 3x?

[tame charm]

x equals 3

xy = 18

[cinder breach]

Yeh

Ok

[tame charm]

so write “3” in place of “x” in the equation “xy = 18”

[cinder breach]

You wrote 6 because 6x3 equals 18?

[tame charm]

Yeah

3 • y = 18

So that means y = 6

[cinder breach]

Ye

[tame charm]

Ok cool

So for the last column we have xy = 18, so x(5) = 18, which means x = 18/5

[cinder breach]

I think it's somewhere between 3-4 since those are closest to 18

[tame charm]

Yup 18/5 = 3.6

[cinder breach]

Oh

Each .2 is 1?

And .1 is 0.5

?

[tame charm]

Yeah

Cuz .2 * 5 = 1

5 * 3 = 15

We need to append something to 3 so we hit 18

[cinder breach]

What's * for?

[tame charm]

Multiply

[cinder breach]

Ah

[tame charm]

Anyway

[cinder breach]

Alr

[tame charm]

We don’t need to worry about decimals

18/5 is a completely fine answer

[cinder breach]

Alright

[tame charm]

Man these table problems are kinda dull

So do you think You could do the other ones

[cinder breach]

Lemmi try

First box needed be multiplied by 9 and second by 4 how?

[tame charm]

Well what’s the equation

[cinder breach]

6

[tame charm]

What?? Lol

[cinder breach]

?

[tame charm]

I mean like the equation xy = something

[cinder breach]

It's inversely proportional

Just did 7 wrong or.

[tame charm]

Yeah that’s not it

For 7 the equation is PQ = 0.5

Cuz in the second and third columns, when u multiply them it gives you 0.5

For the first column it says P = 0.25

[cinder breach]

Oh

[tame charm]

So 0.25Q = 0.5, which means Q = 2

[cinder breach]

How can 0.25 turn into 2?

[tame charm]

It isn’t

0.25 • Q = 0.5

1/4 times Q equals 1/2

So Q must be 2

[cinder breach]

Oh

[tame charm]

Yeah you’re just using what you know to solve for Q

For the last column, we have P • 0.4 = 0.5

So P = 0.5/0.4 = 1.25

[cinder breach]

Can we move on to substitution and elimination

[tame charm]

Yeah

Lol

[cinder breach]

It's already 12 30 would need to wake up pretty early

[cinder breach]

Kk

[cinder breach]

I don't think those examples are same as in the book

They look kinda different

[tame charm]

Certainly not the same example

But same idea

[cinder breach]

Alr

[tame charm]

Yeah

[cinder breach]

That's confusing af

[tame charm]

It should be

You’re probably wondering how they decided to multiply by -3 and 2

And substitution is just an ugly mess

[cinder breach]

I only know that you need same number and symble to eliminate and u may multiply other place with the number you have if it works

Like 3 x 2 is 6 and 2 x 3 I'd 6 so it's same so u can eliminate I think

So then those are gone and all left is the 9y 8y and 3 and 14 or?

[tame charm]

Exactly

Yeah you figure out what to multiply by so you have the coefficients lined up

[cinder breach]

I have no idea what to do next after that tho lol

[tame charm]

Add the equations

[cinder breach]

18 y and 17?

[tame charm]

(Do you know what adding equations is)

-18y = 17

That’s what u get

[cinder breach]

Oh right I always forget signs which can make whole answer wrong

[tame charm]

Ye

So you have the equation -18y = 17

Then you can solve for y

[cinder breach]

How?

[tame charm]

Divide both sides by -18

y = 17/-18

[cinder breach]

I hate devition ugh lol

[tame charm]

Lol

Yeah not fun

Oh wait wait

Nvm nvm

[cinder breach]

My brain doesn't work at divition

[tame charm]

When you add the equations it should be -17y = 17

So y = -1

[cinder breach]

How?

Oh wait

[tame charm]

Yeah lol

[cinder breach]

17 devided by 17 is 1 lol

[tame charm]

Yep

[cinder breach]

Can't wait next year we can use calculators

[tame charm]

Haha yeah

I don’t like arithmetic

[cinder breach]

Then my brain won't need to die in few questions

[tame charm]

Yeah silly mistakes cost a lot

[cinder breach]

I fucking hate we have 3 terms instead of the normal 2

[tame charm]

Terms?

[cinder breach]

Sections of the school year

[tame charm]

Oh

[cinder breach]

Example like 3 months in first term then a exam

[tame charm]

“Trimesters” or some nonsense

[cinder breach]

Yeh

Anyways back on topic

[tame charm]

Okay so

[cinder breach]

Y is -1 right?

[tame charm]

Yep

[cinder breach]

So what about x

[tame charm]

So now you can sub that into one of the equations to find x

2x + 3y = -1
2x + 3(-1) = -1 — sub in y = -1
2x - 3 = -1 — simplify
2x = 2 — add 3
x = 1 — divide by 2

[cinder breach]

Can u explain

How

[tame charm]

Yeah which step

[cinder breach]

How to find x

[tame charm]

Yeah that’s just algebra

[tame charm]

These are the algebraic steps

[cinder breach]

Like can u explain what u did at each step

[tame charm]

Jus edited it

[cinder breach]

Ok I'm now fucked on substitution that is still confusing

[tame charm]

Yeah that example is super confusing for substitution

They took the first equation and solve for x in terms of y

2x + 3y = -1
2x = -1 -3y
x = (-1 - 3y)/2
x = -1/2 -3/2 y

So now it’s in the form
x = something not involving x

[cinder breach]

It's now more confusing

My brain is exploding

Too much info in like 2 hrs lol anyways continue can u do another example?

[tame charm]

Yeah you’re doing well for how much info I’m throwing at you lol

This is a special case

Does that make sense

[cinder breach]

Gimme sec

-4x =22

[tame charm]

How’d you get that

[cinder breach]

Oh waait

Elimination or sub?

[tame charm]

Do it by elim

[cinder breach]

Gotta hit the engine a few times to get it working yk like my brain

[tame charm]

Lol

[cinder breach]

Nope I have no idea how :/

[tame charm]

Ok multiply the second equation by -2

Then add the equations together

[cinder breach]

Kk

X * 2 is x? And 2y * 2 is 4y and 2 * 2 is 4 so x - 4y = 4

[tame charm]

Should look like

2x - 4y = 7
-2x + 4y = -4

[cinder breach]

And to add em x and x x 2? 4y and 4y we elim?

[cinder breach]

O

[tame charm]

When you add these, you get 0 - 0 = 3

[cinder breach]

Oh

Um

[tame charm]

Ye

[cinder breach]

So no x and y

[tame charm]

Yeah they cancel out

Both of them

So you get this nonsensical equation 0 = 3

[cinder breach]

Damn 1am lol

[tame charm]

Hahaha

1am we’re looking at 0 = 3

[cinder breach]

Guess who's gonna get 4hrs of sleep lol

[tame charm]

😵‍💫

Lol

Ok I gtg

[cinder breach]

Anyways continue if any more questions left?

Oh alr

[tame charm]

I’ll send this last thing

And you should have all the info

Go to sleep, get some rest,

[cinder breach]

Alr will do before I go to school for the exam I'll try to revise

[tame charm]

And in the morning ( if you want) you can reread some of this stuff lol

[cinder breach]

Sso I don't forget lol

[tame charm]

Ok lol

This is the other special case

[cinder breach]

My memory is like a goldfish

Shitty

[tame charm]

Lmao

Ok get some sleep

I’m gonna head out

[cinder breach]

The answer is 0

Cya

[tame charm]

See y’a dude

And good luck on ur test

[cinder breach]

Yeh ty

Lemmi download all these tstuff before I go lol

[safe radish]

@cinder breach Has your question been resolved?

[safe radish]

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[proper pier]

it says my vectors are linearly dependent

[proper pier]

what does that mean

and why isn't this right

[broken yew]

🤔

show your working for attempting to solve the system

the row elimination

[crisp linden]

Judging by the s, there will be one free variable that you have to parametrize.

[proper pier]

I mean yeah but

does -1 0 0 not literally solve the system

let me run through the elimination again

[broken yew]

It might do

But when solving systems, you are looking for all solutions

not a solution

[proper pier]

excellent point

Well that looks kinda right

oh but it is a garbage picture

[crisp linden]

Looks about right

[proper pier]

yeah so what do I do with that 0 row

[crisp linden]

Ignore it

You now have two equations

$x-8z=-1\$
$y-12z=0$

[flat frigate]

Remavas

[proper pier]

right

there's something here I'm not getting 🤔

[crisp linden]

Well, now we need a parameter s right

Let $s = z$.

[flat frigate]

Remavas

[crisp linden]

Can you find y(s) and x(s)?

[proper pier]

hmm

[crisp linden]

(Also, I prefer x,y,z to x_1,x_2.x_3, so forgive me)

[proper pier]

no worries

yeah so x(s) = -1+8s

and y(s) = 12s

oh

well thank you

[crisp linden]

yw

[proper pier]

.close

[safe radish]

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[steel elbow]

I recognize that this is easily shown to be true by Divergence theorem, however I was told that the point of this problem was to show the theorem is correct by computing LHS and RHS individually to show that the two expressions are in fact equal. I worked out that the LHS is equal to 4pi(r^3) (not even sure if that is correct), and am currently struggling to figure out how to evaluate the RHS expression

[safe radish]

@steel elbow Has your question been resolved?

[steel elbow]

<@&286206848099549185>

[safe radish]

@steel elbow Has your question been resolved?

[safe radish]

@steel elbow Has your question been resolved?

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[spark lagoon]

hi i dont know how to solve this problem. this is what i have so far

[lean otter]

try to write the sqrt(x)/x as fractional power and recall the power rule

[stoic dune]

,w integral of x^n

[flat frigate]

Results provided by WolframAlpha

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Requested by Kaynex#7520

[stoic dune]

Should know that rule as the integral power rule

[spark lagoon]

Then i would have this

How do i get it out of the fraction

[lean otter]

1/x = x^(-1)

[spark lagoon]

And i have 3 x’s how do i make it 1?

[stoic dune]

Remember exponent properties

[safe radish]

@spark lagoon Has your question been resolved?

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[inland latch]

Hello. If I have 2 Discrete Random Variables X1 and X2 and I define X=dX1 + bX2, then how does the gen function of X look like? I know that if X was X1+X2 then G(X) would be G(X1)*G(X2). But how do these constants alter it?

[safe radish]

@inland latch Has your question been resolved?

[inland latch]

I guess its okay to ping <@&286206848099549185> now

[safe radish]

@inland latch Has your question been resolved?

[woven kiln]

where d and b are constants, well.
if g = dX1
h = bX2
X = g + h
i think you see where i’m going with this

@inland latch

[inland latch]

So gen function of X just equals Gx = d*Gx1 * b * Gx2?

[woven kiln]

i’ve not done gen functions but i’d presume so considering they’re discrete variables.

G(dX1) * G(bX2) = G(X)

[safe radish]

@inland latch Has your question been resolved?

[woven kiln]

this suggests it’s linear and not linear at the same time so i’d go with my arrangement

because they’re still new discrete variables

idk enough to give a definite answer tho, but it seems intuitively right

[inland latch]

Okay well I'll go with that right now

.close

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[jolly minnow]

do i just express the dy/du in u and v

if i do it get i +/-

<@&286206848099549185>

[safe radish]

@jolly minnow Has your question been resolved?

[safe radish]

@jolly minnow Has your question been resolved?

[safe radish]

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[civic dove]

why are they independent if the probabilities arent the same

[civic dove]

nvm

.close

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[fresh hill]

Where did I go wrong ?

Am I not allowed to square root it cause + or -

[lean otter]

Didn't knew that $e^x=tan^2(x)$

[flat frigate]

CatHashira

[fresh hill]

I let that = that

Like trig sub

[lean otter]

I see

[fresh hill]

Idk if that’s Lloawe

Allowed

But tried it 🤣

[lean otter]

On that triangle yea tan^2 is indeed e^x

[upbeat swan]

just wondering what grade is this?

[fresh hill]

Calculus 2

Uni yesr 1

For most

Majors

[lean otter]

Weird u got $\frac {1}{sin \theta}$ but isn't that what written there just $csc$

Oh wait not cot

[fresh hill]

Csc

[flat frigate]

CatHashira

[lean otter]

Yup

I mean the integral is just that of hypotenuse divided adjacent which is csc why even make it so complicated

Oh wait

The hypotenuse is $\sqrt {e^x+1}$

[flat frigate]

CatHashira

[lean otter]

And adjacent is $1$

[flat frigate]

CatHashira

[lean otter]

That's just adjacent over hypotenuse or cos theta?

[fresh hill]

Wdym

[lean otter]

What's the hypotenuse sqrt(e^x+1)?

As per the diagram u showed

[fresh hill]

Yes

Cause of a^2 + b^2 = c^2

[lean otter]

Ya

And the adjacent or base is 1

[fresh hill]

Ye

[lean otter]

So 1/sqrt(e^x+1) = adjacent / hypotenuse = cos

[fresh hill]

Yes

[lean otter]

And the integral of cos is sin?

[fresh hill]

Yes

[lean otter]

And what's the value of sin

Opposite / hypotenuse

What's the value of opposite?

[fresh hill]

Sqrte^x

[lean otter]

So ur answer is sqrt(e^x) / sqrt(e^x+1) ig?

[fresh hill]

Wait figured what I did wrong lmao

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[keen kettle]

Hi

[keen kettle]

Let X be a random variable taking the values {1, 3}. Define the event
A = {X = 1}. What is true
(a) E (X) = Pr(A) + 3(1 − Pr(A));
(b) E (X) = 3;
(c) E (X) = 4 Pr(A);
(d) E (X) = 4.

Is the answer C

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[keen kettle]

<@&286206848099549185>

[slender coral]

Do you know what the formula for expected value is?

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@keen kettle Has your question been resolved?

[keen kettle]

yes

@slender coral The answer is A

I figured it out

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[gleaming juniper]

just how to start it please!

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[queen parcel]

(some term in the sequence) / (the term before in the sequence) = common ratio

[gleaming juniper]

so like this?

[queen parcel]

Yup!

Exactly

[gleaming juniper]

thank you!

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[vast temple]

Whats Latex coding amd how do I do it?

[pulsar condor]

Google, or #latex-help

[vast temple]

Thanks!

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