help-42

[minor sleet]

and for r i just did the difference like

[ancient helm]

ok do you know how to do like polynomial long division?

[minor sleet]

yeah

[ancient helm]

if i asked you to find say (3x^2 - 7x +4) / (4x + 3)

ok duplicate the entire procedure

but with gaussian integers now

[minor sleet]

okie

so

wait but this is odd no

we want q to be an integer

[ancient helm]

no

we want q to be a Gaussian integer

[minor sleet]

yeah

but that has to be an integer

[ancient helm]

no

[minor sleet]

i mean

a and b

in a+bi

[ancient helm]

is 1 + 2i an integer?

[minor sleet]

have to be integers

[ancient helm]

oh yes

[minor sleet]

like if i divide the long division way

i'll get 1/34(191+293i)

wait

i mean

1/25(184+187i) or smthn like that 💀

[ancient helm]

uhh

how are you getting that

[minor sleet]

wait no i was right before

[minor sleet]

its this

[minor sleet]

doing long division

[ancient helm]

please show work

[minor sleet]

okie

[ancient helm]

nvm i see what you're doing

ok so

[ancient helm]

if you evaluate this, you get like 5.6 + 7.03i

you can round these and use that as your value of q

so you can do q = 5 + 7i, or q = 6 + 7i as the two most natural choices

and see what you get for r in both cases

[minor sleet]

Damn that really sucks lol

So if u wanted to find the Gcd of two Gaussian integers

That would rlly take a while

To do by hand

Like for part b) in the original question

You’d start with q=(5+7i) and then r = 2-3i

And then u do (3-5i) = (2-3i)(1) + (1-2i) or smthn
(2-3i)=(1-2i)(1) + (1-i)
(1-2i) = (1-i)(1) - i
(1-i) = i(-1-i) + 0

Feel like I’m most definitely doing this wrong

[ancient helm]

yeah it's slightly more tedious for sure

[ancient helm]

and uh im too lazy to check your work but if those are valid divisions then just substitute everything back and it will be your bezout identity

[calm coral]

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[mossy wigeon]

why is x^x lim to 0+

[mossy wigeon]

an indeterminate form for LH

[ancient helm]

x^x = e ^(x lnx)

x lnx is a case of 0 * infinity

[calm coral]

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[crystal skiff]

The problem is to find the moment of inertia of a square plate along the diagonal from one corner to the opposite. I split the square into four right angle quarters and then found the moment of inertia of that. The answer should be $\frac{1}{12}Ma^2$ but I got $\frac{1}{4}Ma^2$

[potent lotus]

𝒥𝒶𝓈𝓂𝒾𝓃𝑒

[mighty bronze]

im not sure but im just wondering if you really need a double integral for this

[blazing coyote]

Do you have to use integrals?

won't the parallel axis theorm be enough here?

[crystal skiff]

it's what I'm trying to learn to do

[blazing coyote]

ah

[mighty bronze]

i think there should be integration and parallel axis

[blazing coyote]

ok

[mighty bronze]

i mean personally i'd try to not split it in half and just find the moment of intertia from one corner to another

that should work unless im forgetting smth

dm/da=m/(2a)^2

then bounds would be 0 to a√2

not sure if that's right but logically i think i should be good

[crystal skiff]

i thought $dm = \rho dA = \frac{M}{a^2} dx dy$

[potent lotus]

𝒥𝒶𝓈𝓂𝒾𝓃𝑒

[crystal skiff]

we didn't cover parallel axis theorem so idk what to do there

[mighty bronze]

oh i see

ngl i dont know the way to do this without parallel axis other than 1 integral

sorry

[crystal skiff]

np

[calm coral]

@crystal skiff Has your question been resolved?

[crystal skiff]

<@&286206848099549185>

[calm coral]

@crystal skiff Has your question been resolved?

[calm coral]

@crystal skiff Has your question been resolved?

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[remote mural]

any ideas ?
ive found the radius of small circle but i dont know what to do next

[timid oracle]

if you have the small radius then use the area difference to get the big radius

[remote mural]

how to use that

[timid oracle]

pi (big radius)^2 - pi (small radius)^2 = 75pi I think

[remote mural]

worked

thank u

.close

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[near finch]

hey, I have trouble solving a differential equation of the form

y' + cos(x)y = 1/2sin(2x)

I tried finding a integrating factor which resulted in e^sin(x) which gave me a pretty annoying integral?

I tried doing variation of parameters but got a very similar integral, any ideas if I'm on the right track?

I got Ae^-sin(x) for my homogenous solution, which after plugging in and making A a function of x yields:

A'(x)e^-sin(x) = 1/2sin(2x) (A(x) is the previous constant that is now a function that I need to find to get the particular solution)

[near finch]

the integral is smg along the lines of cos(x)*sin(x)*e^sin(x)

It seems like I have more trouble with the integral then with the differential lol...

[untold locust]

try using substitution

for the integrating factor

I have a nice result

But maybe I made some mistakes

[near finch]

substitution inside the integral or differential?

[untold locust]

The final result is y = sin x - 1. Maybe too nice

Substitution is for the integral

On the right side

After integrating both sides

I can write it down nicely if you want

But it will take a few minutes

[near finch]

I tried it, for cos(x)sin(x)e^sin(x), but that leaves me with the integral of sin(x)*t dt which cannot be solved

[untold locust]

Substitute sin(x) = t

and I have an integral of t*e^t

[near finch]

omg indeed

hold on I'll try to solve it

ok

I got e^t*(sinx -1)

[untold locust]

?

[near finch]

sorry

e^sin(x) ( sin(x) -1)

[untold locust]

You do not want both x and t

[near finch]

forgot te replug the sin(x) lol

[untold locust]

No, wait. What about the left side?

There is an e^sin as well

simplify

[near finch]

oh right the A(x)e^-sin(x), I forgout about it completly

so like you said, sin(x) -1

wait no

[untold locust]

Check the answer with wolfram

[untold locust]

What's wrong?

[near finch]

so the left side is A(x)e^-sin(x) after integrating the derivative

and the right side e^sin(x)*(sin(x) -1)

[untold locust]

why to the power of negative?

[near finch]

I solved for the homogenous first,

so y' = -cos(x)y

woops

which gave me y = Ae^-sin(x)

[untold locust]

All correct?

[near finch]

and then I made A a function x and replugged it in the original equation

so now I ended up with Ae^-sin(x) = e^sin(x)*(sin(x) -1)

[untold locust]

Still, why ^-

the integrating factor is e^int(cos x) dx, right?

[near finch]

oh yeah, but I'm doing variation of parameters

[untold locust]

I have no idea what that is

[near finch]

oh rly? it's pretty usefull when integrating factors gives annoying integrals

basically when you have a diff eq of the form a(x)y' + b(x)y = c(x)

replace c(x) with 0 and solve for the homogenous solution

[untold locust]

Seems complex, but this question is more highschool stuff. So I tried it

[near finch]

then you take the constant inside your homogenous solution and make it a function of x

and plug it in your original equation

y' + cos(x)y = 1/2sin(2x)

so in this case you get

d(A(x)e^-sin(x))/dx + cos(x)(A(x)e^-sin(x) = 1/2sin(2x)

[untold locust]

Well, gl determining whether I did a mistake or if you did something wrong. I have no idea 🙃

[near finch]

which gives A'(x)e^-sin(x) - cos(x)A(x)e^-sin(x) + cos(x)(A(x)e^-sin(x) = 1/2sin(2x) and the terms on the left cancel

so now I was trying to solve for A(x) which is the integral of e^sin(x)cos(x)sin(x)

which is the same integral as if you would integrate with the integrating factor?

[near finch]

I think you were right

[untold locust]

No

$y(x) = c_1 e^{(-sin(x))} + sin(x) - 1$

[near finch]

hm

[potent lotus]

mlysikowski

[untold locust]

like this

[near finch]

oh wait I got smg else

A(x) = e^2(sin(x))*(sin(x)-1) and replugging it inside of the homogenous solution gives e^(sin(x))*(sin(x)-1)+C

[untold locust]

Ok

I got it

A small mistake

I will rewrite it

[near finch]

alright

oh wait nvm I made mistake it's $y(x) = e^{(sin(x))}* (sin(x) - 1)+c_1/e^{(sin(x))}$

[potent lotus]

Mephisto

[untold locust]

no

[near finch]

oh wait

[near finch]

I think you have the same result as me

just written differently

[untold locust]

No

[near finch]

I wrote my result wrong with the LaTeX thing

[untold locust]

You have an extra mutliple of e^sinx

[near finch]

the division sign had to be for the whole equation

[untold locust]

Ah, it works then'

[near finch]

{e^{(sin(x))}* (sin(x) - 1)+c_1}/e^{(sin(x))}

bruh

[untold locust]

$\frac{e^{(sin(x))}* (sin(x) - 1)+c_1}{e^{(sin(x))}}$

[potent lotus]

mlysikowski

[near finch]

huh, it seems like it doesn't work that way

yeah basically this

[untold locust]

Ok, I am going back to other work

gl

[near finch]

thanks mate !

[untold locust]

$\frac{e^{(sin(x))} \cdot (sin(x) - 1)+c_1}{e^{(sin(x))}}$

[potent lotus]

mlysikowski

[calm coral]

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[lone ember]

Please help me with 3rd question

[lone ember]

I have done so far

[timid oracle]

the hint should help you

try dividing the top and bottom of the original fraction by 2

[surreal patio]

the hint would make the question simpler then if u were to differentiate normally

[lone ember]

i am still confused

how do i simplify

[timid oracle]

recall that sin^2(x)=(1-cos(2x))/2 while cos^2(x)=(1+cos(2x))/2

[lone ember]

should i use 2sin^2x

[timid oracle]

yes

[lone ember]

i did

[surreal patio]

u could other wise use
$\newline sin^2 \theta + cos^2 \theta = 1$
$\newline cos2\theta = cos^2 \theta - sin^2 \theta$

[potent lotus]

JustToPro

[timid oracle]

so the inner fraction is tan^2 theta

then sqrt of that is just tan theta right

[lone ember]

yes

but then the answer is only sec^2x

[timid oracle]

what’s the derivative of tan theta

[surreal patio]

apply chain rule correctly

[timid oracle]

I should really be saying x

not theta

what’s the derivative of tan x

[lone ember]

sec^2x

[timid oracle]

yeah

[dusky flax]

Big hint:\\
$=\sqrt{\frac{1-(1-2sin^2(x))}{1+(2cos^2(x)-1)}}$

[timid oracle]

that’s it

[dusky flax]

$=\sqrt{\frac{2sin^2(x)}{2cos^2(x)}}$

[dusky flax]

now you can see that this simplifies very easily

can you find what this simplifies to?

[lone ember]

what do i do about 1/2

[dusky flax]

what 1/2

[lone ember]

i don't get it

[dusky flax]

look

[surreal patio]

@lone ember

u didnt take derivative of tan^2 x correctly

u should have

[dusky flax]

$cos(2x) = 1 - 2sin^2(x) = 2cos^2(x) - 1$

do you know this formula?

[surreal patio]

$$\frac{1}{2tanx} \times 2tanxsec^2x$$

[potent lotus]

JustToPro

[timid oracle]

guys I’m pretty sure they already got it

[dusky flax]

im just telling a very fast method

[timid oracle]

also sqrt(tan^2(x)) is literally just tan(x)

[dusky flax]

yes

thats what im saying

the expression can be simplified to tan x

[timid oracle]

then why on earth is the chain rule needed

[dusky flax]

and then the answer is very very straightforwars

[surreal patio]

they took the derivative of it , so im just telling that both ways lead to same conclusion

[dusky flax]

it isnt

thats what IM SAYING

[lone ember]

??

[timid oracle]

Uh yeah I guess that works

long winded but it works

[lone ember]

okay thanks

.close

[calm coral]

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[proud osprey]

can anyone give me a hint? im really stuck

[proud osprey]

by counting the number of ways to choose a team with a designated captain from
among n players

[timid oracle]

I would use induction

although there are probably other ways

also the summation index is probably supposed to be k

[proud osprey]

I know how to prove this with induction, i dont know how to prove it by counting in 2 ways ]

[ivory lion]

think about what the left side is counting

well you have a summation, and sums hint that you are counting different cases

so ignore the summation now and just look at the term inside it

because they are multiplied, this means you need to make multiple choices

n choose k is easy, imagine n people and you choose a team of k

now we ALSO have to make a choice from a k amount

any ideas how?

@proud osprey

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[limpid flame]

my channel got closed so im opening a new one: here are the rules so far for superpositional algebra, can anyone spot any holes in my logic

[limpid flame]

if not im going to try and evaluate some limits for an arithmetic infinity

to see if it holds there

i guess its flawless

mods can i get a channel for my new field

superpositional algebra

S is really what makes the infinities work so its a good name

i invented it bungo so i want a place to discuss it

i had many students a moment ago but i lost them all when the channel closed

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[olive blaze]

help please

I have like a few questions to do

<@&286206848099549185>

[pallid moth]

I'm not a helper but

Do you know how to setup the two matrices?

[olive blaze]

uh yeah

matrix right?

[pallid moth]

Yeah, there's two of them

A is the number of items

Z is the price of the items at that store

[olive blaze]

tryna figure out B

[pallid moth]

Oh alright sorry

[olive blaze]

nah ur good

i just don't know how to get the matrix for it T-T

[pallid moth]

Sorry if I'm misunderstanding

What's 'it'?

Matrix A, matrix Z, or matrix AZ

[olive blaze]

for part a

oh im slow

[pallid moth]

Oh okay I see

[olive blaze]

thats a Z not a B

my bad

im looking for the matrix Z so i can do part A and B

hold up lemme pull it bacck up because the screenshot is blury

[pallid moth]

No worries

[olive blaze]

is this better

[pallid moth]

Is this a question done by hand or with calculator?

Yeah

[olive blaze]

well i typically do it by calculator

but its possible to be done by hand

[pallid moth]

So it tells you that each stand at all 3 locations sells the stuff at the same price

So all numbers in matrix Z column 1 is 0.70

Column two is 0.60 in all 3

[olive blaze]

ohhh

so fill in those same numbers?

in the matrix

3 collums and 2 rows?

[pallid moth]

3 columns 3 rows

1.20 in the 3rd one, as it tells us hotdogs (column 3) is that price

[olive blaze]

and whats baout the other collums

[pallid moth]

Wdym?

Look at your data

You already have matrix A

Matrix C has 3 Numbers that repeat for 3 rows each

[olive blaze]

ohhh

alright

<@&286206848099549185>

[calm coral]

@olive blaze Has your question been resolved?

[olive blaze]

T-T help

[calm coral]

@olive blaze Has your question been resolved?

[calm coral]

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[sly bone]

need help for c and f (ik its wrong)

simplify algebraic expressions

[calm coral]

@sly bone Has your question been resolved?

[calm coral]

@sly bone Has your question been resolved?

[sly bone]

<@&286206848099549185>

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[sly bone]

.close

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[lusty turret]

Hey any hints on how to approach this problem? I used MATLAB and the sum of the sum of 1/k^2 is similar to just the sum of 1/n, but don't know how to approach the problem at all.

[untold compass]

it diverges right? I assume it wants you to get sum 1/n out of it somehow

[lusty turret]

Ye it diverges

pretty sure it diverges

[untold compass]

the hint is saying to look at k=n to k=2n, and do a comparison test

[lusty turret]

that would only work if it converges, since 1/2k and 1/k would be larger than 1/k^2

[untold compass]

I mean like the whole sum in between those

you have like k+1 terms of at least 1/2k

[lusty turret]

Do u mean the sequence a_n or the total sum of a_n

[untold compass]

this bit

oops these should all be squared

[lusty turret]

ye

I was gonna say

[untold compass]

so like k+1 terms that are at least 1/(2k)^2

[lusty turret]

ok ye

[untold compass]

will act sort of like 1/4k

[lusty turret]

ok

why is that true

[untold compass]

I also used k instead of n

[lusty turret]

ah its fine, notation smotation

[untold compass]

there's something similar in the usual 1/n proof, where you compare everything to 1/2^n

here it's (1/n^2+...1/4n^2) >= (n+1)*(1/4n^2) >= n/4n^2

[lusty turret]

wow that's so smart

omg that's so smart

holy shiet

thank you so much I get that connection between (1/n^2+...1/4n^2) >= (n+1)*(1/4n^2) is super hard to c

[untold compass]

yea you sort of just have to think in O(n) notation of having n*1/n^2 and hoping it works

[lusty turret]

wow, I've never thought of applying that here

[calm coral]

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[ruby goblet]

where do i even start

i know pi LOL

[thick sinew]

calculate the sun area

then calculate the small semicircle that left of the dotted line

SunArea - 2 * SmallSemicircleArea = VisibleSunArea

VisibleSunArea / SunArea = result

[calm coral]

@ruby goblet Has your question been resolved?

[thick sinew]

@ruby goblet are thee there?

.-.

[ruby goblet]

mb

i was away

[ruby goblet]

pi x r^2?

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[calm coral]

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[weak rapids]

We can find minor and co factor of only a square matrix right??

[swift laurel]

yes

[weak rapids]

Ok pnop

Thanks

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[unkempt drift]

Sorry, that's correct

You just remove the column and row passing through that specific entry

This can be done with non-square matrices

But you can't calculate the determinant of a non-square matrix so it's not useful

[warm warren]

I think you just get zero

[weak rapids]

.reopen

[calm coral]

✅

[unkempt drift]

Wait a minor isn't talking about the matrix; it's the determinant

Woops

[unkempt drift]

Yes then that's why

[weak rapids]

what's going on lmao

im confused

[unkempt drift]

They were correct, sorry for the confusion

[weak rapids]

but this is a square matrix

[unkempt drift]

Technically you could do the same process with non-square matrices

But then you can't find the determinant

[weak rapids]

so what you're saying is we can calculate the minor a of non square matrix but since the minors gonna be a non square matrix we cant find its determinant

so its useless

[weak rapids]

yea

[unkempt drift]

Yeah

Also the minor refers to the determinant of the matrix and not the matrix itself

[weak rapids]

ah ok thanks for the correction

also is this definition correct? @unkempt drift

[unkempt drift]

,rotate

[unkempt drift]

Yes, this is correct

[weak rapids]

ok can you please make me a one for co factor using similar wordings?

i tried googling but there are just some definitions which r not making sense to me

ok wait I'm making it by myself u just tell me if its making sense and if its appropriate

,rotate

[weak rapids]

@unkempt drift

[unkempt drift]

Yes that's correct

[weak rapids]

yay ok

thanks

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[spark stratus]

n is an odd natural
u = e^(1/(2n) * 2πi)
can someone verify this identity real quick (i.e. both are equal for all complex x)

work shown in https://www.desmos.com/calculator/omvx6dqycn, but refrain from revealing spoilers until I verify that this "approach" is valid

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@spark stratus Has your question been resolved?

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[vocal sorrel]

Can someone further explain this. Im not rlly understanding the question

[calm coral]

@vocal sorrel Has your question been resolved?

[vocal sorrel]

<@&286206848099549185>

[remote mural]

It is saying basically what type of another set u need to union with this one to make it transitive relation

And the smallest set

[vocal sorrel]

So would it be d,d?

[untold drum]

do you know what a transitive relation is?

[vocal sorrel]

like 0,1 -> 1,2 -> 2,3

like they are connected in a sense

[untold drum]

yes. so take your example. what could 0,1,2 be?

[vocal sorrel]

like an example?

[untold drum]

(d,e) -> (e,d) -> (d,d)

[vocal sorrel]

uhhh

[untold drum]

so you would need to add (d,d) to be transitive.

[vocal sorrel]

ok so whats the difference between adding d.d and e,e

[untold drum]

why should there be a difference? we started with (d,e) and we see we need to add (d,d). now start with (e,d).

[vocal sorrel]

so we need both?

[untold drum]

to be transitive means if a is in relation with b and b is in realation with c then a has to be in relation with c. (a.b) -> (b,c) -> (a,c). If you are asked to make a relation transitive you have to check this condition for all passibiliities you can start with (a,b).

in your example it means to add both, (d,d) and (e,e).

[vocal sorrel]

ohhh ok

so transitive you should be able start anywhere?

[untold drum]

yes

[remote mural]

Yes 💯

[vocal sorrel]

ty

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[vale lintel]

is this right

[calm coral]

@vale lintel Has your question been resolved?

[untold locust]

Seems ok

[vale lintel]

becausze its diameter

so u need to do radius/2

[untold locust]

Yes, I see it now

[vale lintel]

and when you square 1/2 its 1/4 times 1/2 = 1/8

[untold locust]

I have the exact same way of thinking

I forgot that its a semi-circle

[untold locust]

Seems ok

I checked it step by step with desmos

[vale lintel]

alr thank you

[untold locust]

And I have an identical answer'

I am no calc expert tho

[vale lintel]

thanks for the help tho

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[viral night]

Stuck in Statistics

[viral night]

trying to find the lower and upper limits but I'm stuck on how to find them given the data sets

<@&286206848099549185>

[charred citrus]

Yea

[viral night]

please help me 😭😭

[charred citrus]

What is it 💀

[viral night]

.

[charred citrus]

What the question

[viral night]

how do I find the lower and upper class limits

[charred citrus]

The upper class is the amount of class before the main class

Wait no

The upper class is the most

And the lowest is the least, I think…

[viral night]

yeah but how do I find it in the sense of the equation

[charred citrus]

Wait

[charred citrus]

I’m not sure right now but the upper limit is the highest in range and the lower is minimum

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[stuck cipher]

can you please show the steps of working this out by hand?

[blazing coyote]

are you sure that has a closed form?

oh, nvm

you'd want to start with a substitution

[stuck cipher]

do u mind showing each line of working pleasee

[plain harness]

have you tried following along the steps of https://www.integral-calculator.com/

[north grail]

real

[plain harness]

that looks painful to do

[north grail]

its a simple substitution but the compution is a pain in the ass

[plain harness]

it may be to blurry to read

[stuck cipher]

omg thank you i didnt know that site existed

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[vernal basin]

Can I have some help making sense of this picture for the question?

[vernal basin]

I believe the general solution it asks for would be

200 = x1 + x2
40 = x1 - x3 -x4
60 = x4 + x5
100 = x2 +x3 - x5

is that all I have to do for i)

and then put this into a matrix

to know its shape, I could rref it, right?

[calm coral]

@vernal basin Has your question been resolved?

[vernal basin]

<@&286206848099549185>, is this on the right track?

[vernal basin]

.close

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[mossy cliff]

<@&286206848099549185>

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[remote kettle]

Do you know the answer? Also, what have you tried?

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[dense rapids]

a(r) = a + (r-1)d

*general term of an AP

where d is common difference

and a is the first term

[abstract cipher]

ok

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[visual tinsel]

If I launch a bowling ball at 400 feet per second from the top of a building that is 100 feet high, would the equation that models this siutation be:
-16t^2 + 400t + 100

[gloomy spindle]

Launch as in downwards?

Wait nvm

What angle

[visual tinsel]

I have no clue but its supposed to use the -16t^2 +v0t + h0 kinematic equation

oh wait I got it now nevermind

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[lyric bay]

why is there suddenly an 8

i mean i get that its still equal 1

but howd it come to that lol

[dull wagon]

preknowledge and/or partial fraction decomposition identity

you don't need to follow exactly what they did

you could approach the decomposition as you would normally with
$$\frac{1}{(x-4)(x+4)}$$

[potent lotus]

ℝαμΩℕωⅤ

[lyric bay]

ohh i see

oh wait question then

so it would be like

A/ [(x-4)(x+4)] + B/ [(x-4)(x+4)]^2 + C/ [(x-4)(x+4)]^3

right

ahh nvm

there are diff types lol

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[idle fractal]

F(x)=x/x

[idle fractal]

G(x)=1

Both are not same

Because of 0

Am i right?

[alpine current]

yea

because f(0) = 0/0

which is undefined

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[surreal patio]

So, we can find tangent of a conic by differentiating and then plugging it into the formula to find the equation but when do we use these
y = mx + a/m
y = mx ± √(a2m2 + b2).
y = mx ± √(a2m2 - b2).

[surreal patio]

nvm idk what im trying to ask

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[violet solar]

How do I find the domain of:

xy-e^y = 0 algebraically

Or the same thing, find the range of

y = (e^x) / x

[violet solar]

I'm thinking diff but not sure

Or is there another nicer way to find it?

Without graphing

Is it even possible?

[hasty fiber]

I don't see why not

Seeing as there's a /x, you should probably investigate what happens around 0

Checking the behaviour as x grows to +/- infinity also seems like a good starting point

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If you are done with this channel, please mark your problem as solved by typing .close

[violet solar]

You get the domain by that... but not the range

[hasty fiber]

I'm talking about the behaviour of y

The domain is hopefully obvious

[violet solar]

yea... y goes to infinity when approaching from pos, and -infinity when approaching from neg

I don't see how it would help finding the range

<@&286206848099549185>

[hasty fiber]

Ok that's around 0. What about as x goes to the infinities?

[violet solar]

y also goes to infinity

[hasty fiber]

In both directions?

[violet solar]

oh yea, as x goes to -infinity, y goes to 0

@hasty fiber

[hasty fiber]

Alright, so for negative x, it's hopefully clear that the range is (-inf, 0)

For positive x, you get infinity twice, so more investigating needs to be done

I have to go now, good luck

[violet solar]

Alr

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[cloud flume]

Hi, does this seem okay?

[remote mural]

Yeah

thats correct

[cloud flume]

Great then thank you for checking

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[plucky elbow]

hi

[plucky elbow]

i need help with this

question b

i did 15/50 = 50n/50

but the answer said its

750 dollars

[north grail]

which part?

[plucky elbow]

question b

[north grail]

okay part a, what did u get?

[plucky elbow]

i got l = 50n

[north grail]

correct!

[plucky elbow]

:3

[north grail]

now lets see part b

so l=50n
now in 15 months, what is the value of n?

[plucky elbow]

idk i did it as this, but the answers said its $750

[north grail]

what does l represent in your equation>?

from the question

[plucky elbow]

interest

earned on in an investment

[north grail]

and n?

[plucky elbow]

number of months

[north grail]

so if the number of months is 15, then n is 15 too right?

[plucky elbow]

OH

omg

so like times by 15?

[north grail]

yupss!

[plucky elbow]

im still confused...

[north grail]

15*50

you are essentially substituting the values in

you already know l=50n
n=15, so l=50*15

[plucky elbow]

😊

[north grail]

yess

[plucky elbow]

omg okay

tysmmm

bc this specific question is calculated differently

than how i was taught

so im confused

[north grail]

hahah its okay

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[dark zenith]

no 23 pls

[minor knot]

what have you tried

[dark zenith]

not anything really, just really confused and donbt know what to do with this one

[dark zenith]

ping

[minor knot]

alright

always start with defining x

[remote mural]

Say you deposited x php into an account

[minor knot]

x represents ammount deposited in 4% interest
(1800000 - x) represents the remainder deposited at 6%

[dark zenith]

with that info,

? maybe?

[remote mural]

Yes

[dark zenith]

nice

then find x subtract then find that 4%?

[minor knot]

correct

[remote mural]

What

[dark zenith]

aight wait imma do it before i close this

[remote mural]

Just solve for x bro

[minor knot]

once you find x tho, that will be the ammount deposited at 6%, not at 4%

since its 0.06x

[dark zenith]

ahh ok

x=-35510000

what do i do now

[dark zenith]

done

[minor knot]

leme see what you did

[dark zenith]

multiplied all to 100

[minor knot]

i mean your solution

[dark zenith]

wrong message sry

[minor knot]

you multiplied 100 twice

on the left side

[dark zenith]

ah thats why i messed up

wait

[minor knot]

ye

$4(1800000-x) + 6x = 9800000$

[potent lotus]

Renz

[minor knot]

should be that

since the 100 went to the 0.04 and 0.06

[dark zenith]

OHH

i thought it goes to the brackets too

[minor knot]

nah

if you put it in the brackets too, you will be multiplying by 100*100

[dark zenith]

ah aight ty my guy imma try again

[minor knot]

alright

[dark zenith]

soo the answer is 500%?

wait i mean 500k

the answer is right

TYY

imma close now

.close

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[shy iron]

hello quick question

[shy iron]

when t=0 shouldnt dx/dt = 0

then when t=1 dx/dt is 6

since it says it goes up by 6 per day?

[unkempt drift]

No

It's saying when t = 0, x = 3, and dx/dt = 6

They chose a time when the reservoir is partially full and the water level is rising to start, as t = 0

[shy iron]

oh ok

also for the c part

why when t = 7 e^-t is 0

when i thought it was only when t = infinity

also i dont get where 3-(-3) comes from

[unkempt drift]

e^(-7) = 1/e^7 and e^7 is fairly large

The graph of e^(-x) is the same as e^x but you flip it across the line x = 0

So there is an asymptote y = 0 as x tends to +infinity for e^(-x)

[unkempt drift]

So basically you can substitute e^(-t) = 0, so only the 3 cos t term survives from the expression in part b

The max value of 3 cos t is 3

The min value of 3 cos t is -3

[shy iron]

oh

ok that makes sense

thank you for the help

[unkempt drift]

No worries

[shy iron]

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[marsh girder]

I need help with part b lol

[marsh girder]

In the answer it just sort of assumes u know to work in mod3, but idk why u would do that

[timid oracle]

basically you just guess a factor and check if it works

i dont see a very slick way to do it otherwise

unless some other helpers good at number theory can figure that out

[marsh girder]

lol ill post the mark scheme to the answer one sec cos i dont get that either

Im quite confused on why 34^34 turns to (1)^34 unless its some number theory rule or smthn lol

[timid oracle]

34 is congruent to 1 mod 3

[marsh girder]

o yeah

[timid oracle]

if thats not convincing, then you can consider 34^34 as (3*11+1)^34 and use binomial theorem

[marsh girder]

nah i get why that is

but then on the next part what are we actually trying to do?

[timid oracle]

the next part tries to manipulate the expression to factorize it

[marsh girder]

so basically im trying to factor as many high power numbers to simplify it enough i can work out a factor easily?

[timid oracle]

kinda yeah

[marsh girder]

right okay thank you

[timid oracle]

its hard to see these things in advance but it gets easier with practice

[marsh girder]

yeah im quite new to number theory and the resources for my school are very small

[timid oracle]

im also relatively new lol

but the number theory questions that i have to face are much easier so i cant be bothered to do harder ones

so im bad at number theory

[marsh girder]

fair enough the rest of my topics arent too bad this is probably the hardest part of the spec

and we dont learn a lot of the useful techniques for mod arithmetic either

[calm coral]

@marsh girder Has your question been resolved?

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[gilded drum]

how would i start with the integration of artanhx

[marsh girder]

if u let y = artanh(x)

wait hold on

[glass heart]

write it as 1*arctanh and then ibp

[timid oracle]

^ and this is also how you integrate majority of inverse trig functions

[gilded drum]

ok thank you let me try this

so would it just be xartanhx-artanhx + c?

[glass heart]

no

[gilded drum]

wait

i realised

what i did

1 sec

ok is it xartanhx + (1/2)ln(1-x^2)

[modern peak]

you mean

??

[gilded drum]

ya

wait no

it's hyperbolic

so artanhx

[modern peak]

so

[gilded drum]

instead of arctan

[modern peak]

$x \arctanh x + 0.5\ln(1-x^2)$

huh

$x \artanh x + 0.5\ln(1-x^2)$

no?

tf

[gilded drum]

😭

[gilded drum]

but yh this is what i mean

.close

[calm coral]

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[calm coral]

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[mental scarab]

i got the curl F and the limit bounds but i dont understand how to get N

[untold compass]

the surface is some cut out of the plane z=x, and it's good to know that the normal vector to the plane equation ax+by+cz=d is (a,b,c)

[calm coral]

@mental scarab Has your question been resolved?

[mental scarab]

so could i just say z=x=0 and y =1?

[untold compass]

you'd move around the equation z=x to -1*x+0*y+1*z=0

[mental scarab]

im sorry but im blanking really hard, so does that mean by normal vector is then <-1,0,1>

[untold compass]

yea

[mental scarab]

so then i multiply <1,1,y> (what i got as curl F) and <-1,0,1> to get -1 +y? then use that for my integral?

[untold compass]

that's right

[mental scarab]

okay thank you, my math burnout is killing me right now so i cant even think

.close

[calm coral]

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[mental scarab]

.reopen

[calm coral]

✅

[remote mural]

you didn't read anything i said lol

[mental scarab]

also just a quick question, to make sure my limits are right. z=x and z=x^2+y^2
x=x^2+y^2
rcosu=r^2cosu^2+r^2sinu^2
cosu=rcosu^2+rsinu^2
cosu=r(cosu^2+sinu^2)
cosu=r
thus 0 <=R<=1?

[untold compass]

if this is for the surface you have z <= x^2+y^2

[mental scarab]

it would still get me the same but cosu<=r instead right? my bound for r is still 0<=R<=1

[untold compass]

yea that's good

[mental scarab]

ok 👍 and since its a circle, the other bound is just 0<=theta<= 2pi

[untold compass]

it goes around the circle twice with that (r negative half the time) so you can do 0 to pi

[mental scarab]

okay thank you

.close

[calm coral]

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[calm coral]

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[analog heart]

Can someone help?

[calm coral]

@analog heart Has your question been resolved?

[quaint sphinx]

the question asks for which values of b where f and g intersect... but also do so on line y = 0

normally a question like this does not have a nice solution so you'd expect the process to be unnaturally clean here

with that in mind, it may help to perform f(x) - g(x) to first find the x-values where the functions intersect (not necessarily on y = 0)

[calm coral]

@analog heart Has your question been resolved?

[calm coral]

@analog heart Has your question been resolved?

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[fallow spruce]

Yoo, does anyone have a good latex template for research proposals?

[calm coral]

@fallow spruce Has your question been resolved?

[calm coral]

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[graceful dust]

@upper sparrow

[upper sparrow]

Hiii

[glad sinew]

hiiii!!😄

[graceful dust]

this is gonna be a good channel

[glad sinew]

staying here for hmmm

8 hours more

okie

[upper sparrow]

A whole work day

[north grail]

what are we proving

[graceful dust]

sqrt is continuous

[north grail]

like rigorously?

epsilon delta is out of my paygrade

enjoy

[serene harbor]

isnt sqrtx already continous tho
we can say by graph 🤓 🤡

[upper sparrow]

Anyways, as for here

[glad sinew]

my question is, I see |x-c|/|sqrt(x)+sqrt(c)| is already smaller than delta..

[olive umbra]

hi

can someone help me find the lenght of l6

[brisk pelican]

Just take delta = epsilon * |sqrtx+sqrtc|

[brisk pelican]

No

[olive umbra]

i pay

[glad sinew]

but like Im not really understanding this part

[graceful dust]

note that for any x,c, we have sqrt(x) + sqrt(c) > sqrt(c)

[brisk pelican]

oh

silly me

[upper sparrow]

"Smaller denominator, bigger fraction" and all

[glad sinew]

well like if |x-c| < delta, then a denominator of |sqrt(x)+sqrt(c)| should work too?

[graceful dust]

what does “work” mean?

[serene harbor]

why dont they teach epsilon delta nd sequential limits in school

[serene harbor]

integral F (dot) ds

i think yall should help me out in my help channel

im dying

[graceful dust]

this channel is so fire

how could we leave

[serene harbor]

so i just put my ques here

🗣️ 🔥

number theory approach pls

i'd pay you all

[glad sinew]

since |x-c| < delta, it follows |x-c|/|sqrt(x)-sqrt(c)| < delta/|sqrt(x)-sqrt(c)| = epsilon
so set delta equal to epsilon (|sqrt(x)-sqrt(c)| )

[graceful dust]

delta depending on x is no good

[glad sinew]

oh ok

[graceful dust]

you need to find one, constant delta that “works”

[upper sparrow]

Remember that x comes after you've chosen your delta, after all

[glad sinew]

ok

theres actually smth wrong with me

I did learn epsilon delta proof but for limits, and thats basically same idea

and I just dont know still

[upper sparrow]

It can take a while

[glad sinew]

once again, i’m stuck

how to get rid of x in the 4x

Ill try and find some more examples..

@upper sparrow

oh

hm wait

actually no

nonono

pleaseee if anyone can help

Im stuck

@graceful dust

wait

alright, so few things, I noticed I subbed in values wrong

I end up getting |(x+1/x-1 )-2 | instead

now I have |x-3/x-1| < delta/|x-1|

and I need to get rid of x somehow

and so since delta/|x-1| is a fraction, and x is positive

then I can say delta/|x-1| < 1

so |x-3| <1

right? because if I have x=2, then delta < 1

and all other values of x works, where x != 1

[upper sparrow]

Sorry was out for a while

[glad sinew]

nw

delta = min(eps, 1)

wait but for eps

Im literally setting delta to eps

wait what

[upper sparrow]

that's not always bad tbf-

[glad sinew]

ok

yea ok I got it I think

this was ehh

oof an hour to solve

now Im gonna do b

[upper sparrow]

That one should be a bit easier to work with hopefully

[calm coral]

@glad sinew Has your question been resolved?

[glad sinew]

yea lol

hopefully

i chose eps/3 to be delta

Idk ill see if its right

oh

its |x-2||x-2| |x+2|

and x+2 even if you factor a negative its -(-x-2) not -(x-2) so its not another delta

[upper sparrow]

is it?

[glad sinew]

yea?..

i need to get rid of x

[upper sparrow]

Noooo

There are ways to deal with that, up to here is fine-

if you instead had x^2 - 4 then your factorisation would've been fine, but cause it's not

[glad sinew]

|x| = |x-2+2| <= |x-2|+|2|

< delta + |2|

[upper sparrow]

That's the idea

[glad sinew]

but then if plug into |x-2||x-2| | |x-2|+|2| +|2| |

like my original factorisation but with x substituted..

i get 2delta (delta + 4)

[upper sparrow]

Hmmm, that's kinda sus what you have there but anyways...

[upper sparrow]

similar as to here, if you set delta such that it's less than 1, then you get that you can write that as being less than 10delta