help-4

[glass kelp]

,rccw

[glass kelp]

how did we get here

[maiden panther]

Tanx/2 substitution and partial fraction decomposition

I probably did it wrong tho

Here

[heady pawn]

For the love of god please just use integral-calculator.com

[heady pawn]

it gives all the steps and substitutions

[astral spoke]

your handwriting is so nice

[timber swan]

Uhh

[maiden panther]

First time getting that🙏💔

[glass kelp]

id do u = pi/2 - x first

[astral spoke]

no way bro.... anyways good luck with the maths, I'm not good enough to help but you got this

[glass kelp]

then t = 1 - sinu

[maiden panther]

Appreciate it

[maiden panther]

What the hell that mean

[dire cloud]

Can't you substitute 1-cosx=t and get everything in terms of t

[timber swan]

Or just kings 🗣️

[heady pawn]

expand

use u = cos(x) - 1

[timber swan]

Theres a really trivial way of doing this

[maiden panther]

Let me try that

[heady pawn]

then use u = tan(x/2)

GL

if you think u solved it you can look up if ur solution is correct on integral-calculator.com

[maiden panther]

Aight

[heady pawn]

you can see the steps

[dire cloud]

I can figure out substitutions but idk integration

[glass kelp]

Using ${u = \frac{\pi}{2} - x}$, ${\dd u = -\dd x}$. So,
[ -\int_{0}^{\pi/2} \frac{1-\sin(\pi/2 - x)}{1 - \cos(\pi/2-x) \dd u} = -\int_{0}^{\pi/2} \frac{1-\cos(u)}{1-\sin(u)}\dd u ]

[rocky lotus]

k

[glass kelp]

Then, let $t = 1-\sin(u)$

[rocky lotus]

k

[glass kelp]

should get a simple 1/t

[maiden panther]

Hmm let me do that

[timber swan]

Hmm

[dire cloud]

Did my method work or no?

[maiden panther]

Im restarted

I didn't get it

[maiden panther]

But I think this is yours too

Let me try again

[glass kelp]

sub u first, then sub t

[timber swan]

Idk why all this is needed

[glass kelp]

u used a jee trick?

[heady pawn]

kings rule?

[timber swan]

Nahh

U could just break it into two very easy integrals

😩

[maiden panther]

Oh hell nah how do I do this

[dire cloud]

Yeah that's what I'm trying

[timber swan]

1/1-cos

[glass kelp]

oh i see

[timber swan]

Sin/1-cos

[timber swan]

This is trivial

[heady pawn]

yeah That's what hthey're doing right

[dire cloud]

Wait I forgot to crop

[timber swan]

As u can just convert to cosec

[dire cloud]

Ignore the physics on top

[heady pawn]

1 is easy and the other one is annoying

[glass kelp]

i think sub is faster tho

[timber swan]

Which

Both are easy

[heady pawn]

Keep smellin ur own farts bro

anyways

[dire cloud]

Does this work yall?

[timber swan]

Splitting is wayy faster

[maiden panther]

Reasonable response

[heady pawn]

how about you give a detailed answer to this question and label it with the spoiler

[timber swan]

Uhh

[heady pawn]

🥀

[glass kelp]

anyhow

back to the op

[maiden panther]

Yessir

[glass kelp]

progress?

[maiden panther]

Im not getting shi

[dire cloud]

Anyone?

[maiden panther]

Im so bad at integration

[glass kelp]

sub u = pi/2 - x first

Using ${u = \frac{\pi}{2} - x}$, ${\dd u = -\dd x}$. So,
[ -\int_{0}^{-\pi/2} \frac{1-\sin(\pi/2 - x)}{1 - \cos(\pi/2-x)} \dd u = \int_{-pi/2}^{0} \frac{1-\cos(u)}{1-\sin(u)}\dd u ]

[timber swan]

||1/1-cos can just be converted to cosec^2(x/2)||

[glass kelp]

are u following this

[maiden panther]

How did

Limits change

[rocky lotus]

k

[timber swan]

And ||sin/1-cos is just sub cos||

[glass kelp]

when u do a substitution the bound change

[dire cloud]

ok I'm just gonna go then 🥀

[glass kelp]

ts aint trivial imo 💔

[timber swan]

[glass kelp]

but works

[maiden panther]

Why zero tho

[timber swan]

💔

[timber swan]

Seeing 1-cos makes sin^2x instantly pop up

[glass kelp]

pi/2 - pi/2 = 0 and pi - pi/2 = pi/2

[glass kelp]

fair

[timber swan]

So yes

Its trivial

🗣️🗣️

[heady pawn]

lol

[maiden panther]

Now but why did you substract pi/2

Normally when making substitution do we change limits too?

[timber swan]

Yes

[maiden panther]

Damn bruh nvm yeah

[maiden panther]

Ok after this what

When I use another substitution will the limit change again?

[timber swan]

😔

[glass kelp]

Let ${t = 1-\sin(u)}$. So ${\dd t = -\cos(u) \dd u }$

[glass kelp]

ye kinda

thats the way my brain works

i have bias against trig

like childhood trauma

[maiden panther]

So what will it change to?

[glass kelp]

u should also consider this if u wanna practice integration

[maiden panther]

Nah i feel you fr

[glass kelp]

[ \int_{0}^{1} \frac{1}{u}\dd u ]

[rocky lotus]

k

[maiden panther]

I see

[glass kelp]

which is poggers

[timber swan]

🗣️🗣️

[maiden panther]

Indeed

Alright I'll try it

Thanks for the help

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[glass kelp]

ive realized i make a mistake

[glass kelp]

Using ${u = \frac{\pi}{2} - x}$, ${\dd u = -\dd x}$. So,
[ -\int_{0}^{-\pi/2} \frac{1-\sin(\pi/2 - x)}{1 - \cos(\pi/2-x)} \dd u = \int_{-\pi/2}^{0} \frac{1-\cos(u)}{1-\sin(u)}\dd u ]

[rocky lotus]

k

k

[glass kelp]

@maiden panther Use this method

i made a mistake

[maiden panther]

Grahh

.open

[glass kelp]

its open

thats why one shouldnt integration in their head

[maiden panther]

1-sinu differentiation

[glass kelp]

anyhow

[maiden panther]

Won't give

Numerator

[glass kelp]

yes

[maiden panther]

1 is gone

[glass kelp]

that was the mistake

[maiden panther]

Ok so now what

[glass kelp]

[ \int \frac{1}{1-\cos(x)} \dd x - \int \frac{\sin x}{1-\cos(x)}\dd x]

[rocky lotus]

k

[maiden panther]

Got it

[glass kelp]

right one is trivial u = 1-cos(x)

[maiden panther]

Don't look that trivial to my dum ahh😭🙏

[glass kelp]

left one use this

[maiden panther]

Oh yes

I get the right one

Bet let me try that

By the way it's sinin denominator

[glass kelp]

?

[maiden panther]

Sin In denominator

[glass kelp]

just ignore my method

ignore the former subs

i thought pi/2 -x would work out nicely

my apologies

[maiden panther]

Oh forget the previous ones?

[glass kelp]

ye

[maiden panther]

Alright got it

[glass kelp]

$ \cos^2x + \sin^2 x = 1$ and $\cos^2 x - \sin^2 x = \cos(2x)$. Adding both sides yields $2\cos^2 x = + \cos(2x)$. So, ${2\cos^2 x - 1 = \cos(2x)}$. Thus, ${1 - 2\cos^2(x) = -\cos(2x) = 2 - 2\cos^2(x) = 1-\cos(2x)}$. Therefore, ${2\sin^2\left( \frac{x}{2} \right) = 1 - \cos(x)}$

[maiden panther]

We get cosec square x/2 divided by 2

How does that help

[glass kelp]

LHS becomes
[ \frac{1}{2}\int \sec^2\left(\frac{x}{2}\right) \dd x]

[rocky lotus]

k

[glass kelp]

anyhow this

[maiden panther]

No it's cosec look

[glass kelp]

i derived it

[maiden panther]

So

Cosx = 1-2sin square x/2

[glass kelp]

OH SHIT

the negative sign

[maiden panther]

So 1-cosx becomes 2sin square x by 2

Oh yeah integration cosec square is just cot

Minus cot

Thanks a lot

Got it

.close

[rocky lotus]

k

[glass kelp]

.close

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[glass kelp]

@maiden panther Once again srry

🙏

ill be off discord now

[maiden panther]

Oh hell nah why sorry

You was the most helpful

.close

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[terse chasm]

what happens between the last and the penultimate step?

[viscid spade]

Factor out the fraction (y2-y1)/(x2-x1)

[terse chasm]

could you help me through that? factorization is unintuitive for me with fractions

[viscid spade]

If you let Z= fraction does it seems more natural?

[stark wedge]

^

with the abbreviation yaku suggested, the step becomes:

$zx + (y_2 - zx_2) = z(x-x_2) + y_2$

[rocky lotus]

Ann

[terse chasm]

oh i can see

[viscid spade]

Also you can write without the parentheses for the first one if it helps

[terse chasm]

yes that is very useful

thanks

[viscid spade]

Do it each time then ! One day you won't need to anymore

[terse chasm]

i think i will take it from here

thanks for the advice

!close

[viscid spade]

.

[terse chasm]

whats the command again?ç

[viscid spade]

!done

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

[terse chasm]

!done

[vale dock]

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[terse chasm]

.close

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[hearty fern]

1. Determine and in such that and

\text{lcm}(a, b) - \gcd(a, b) = 7

2. Determine and in (the set of positive natural numbers) such that

2 \cdot \text{lcm}(a, b) - 7 \cdot \gcd(a, b) = 11

3. Determine and in such that and

\text{lcm}(a, b) - 3 \cdot \gcd(a, b) = 4

[hearty fern]

Need help plz

[glass kelp]

holy

$\text{lcm}(a, b) - \gcd(a, b) = 7$

[rocky lotus]

k

[dusk stirrup]

i think you missed a few characters

[hearty fern]

Here's the original it's in french so I tried to translate it

[glass kelp]

what is that notation

anyhow

i would consider prime factorisation of a and b

(not sure if its the right approach tho)

[hearty fern]

Already tried that but didn't work

[glass kelp]

8 and 1 are prolly trivial soln

hmm i dont know much no. theory

i will double it and give it to the next helper

[hearty fern]

<@&286206848099549185>

[stark wedge]

gcd and lcm are both divisible by the gcd, and therefore so is their difference

so gcd(a,b) divides 7

[hearty fern]

Thank you ☺️

[glass kelp]

$2 \cdot \text{lcm}(a, b) - 7 \cdot \gcd(a, b) = 11$

[rocky lotus]

k

[hearty fern]

But then we get GCD=1 or GCD=7. What's the next step to find a and b ?

[stark wedge]

two cases then

if gcd=1 then lcm has to be 8

therefore the product of your two numbers is 8

[hearty fern]

Ah ok I got it from here

Thanks

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[worldly prawn]

Use the functions provided to solve each of the following. Let the region R be the area bounded above by f(x) and bounded below by g(x) in quadrant I. Write, but do not solve, an integral expression that can be used to find the volume of the solid generated when region R is rotated about the y-axis.

f(x) = 4-x^2
g(x) = x+2

I keep getting conflicting results whenever I do it, and some help would be appreciated

[stark wedge]

show your latest attempt

also show your graph if any

if you don't have a graph then you should make one

[worldly prawn]

alr ill send that

basically waht i did is

f(x) = 4-x^2
g(x) = x+2

so

sqrt{4-y} = x
y-2 = x

Then, washer method, but i dont know what to use as my bounds....

[red tulip]

i think its easier to use cylinders

instead of putting it in terms of y

[stark wedge]

shells are easier yeah

[half wind]

Cannot use shell method

[stark wedge]

do you have a graph

[half wind]

Hes required to use disk and washer method apparently

[stark wedge]

you're not op??

[worldly prawn]

nah shes one of my friends

its for AB

so we cant use shell

[stark wedge]

can't = forbidden?

[half wind]

Yes

[stark wedge]

bruh

[worldly prawn]

yup

[stark wedge]

ok well do yall have a graph though

[half wind]

Send graph?

[worldly prawn]

alr

[red tulip]

but then you are getting a function that is symmetrical across the y axis so it would mess with the result

[worldly prawn]

the problem said quad 1 only i tihnk

so we just use that section

heres a graph i spun up

i have a paper one but unfortunately no way to photograph it rn

[red tulip]

yes but its difficult to get a function with just the quadrant you need

wait actually

in this case its a sqrt function

so you would only get the top part

nvm

[worldly prawn]

\int_{2}^{4}\pi\left(\left(y-2\right)^{2}-\left(\sqrt{4-y}\right)^{2}\right)dy

.latex

sry is there a way to render latex?

this is the solution i got so far

i just need help w finding the bounds im pretty sure its 2->4 and then when i rewrote the functions i jst chose the one on top to be the first function and then the one on the bottom to be the subtractant but im not 100% sure abt that as well

[red tulip]

if you translate everything into being in terms of revolving around the x axis it might be easier

[worldly prawn]

true...

so take the inverse of each function?

[stark wedge]

if shell is banned you need two integrals

i think

[worldly prawn]

wait whys that?

[red tulip]

maybe unless you can subtract

[stark wedge]

let me illustrate

[red tulip]

no wait what if you just take the integral from 2 to 3 for the x+2 function and from 3 to 4 for the quadratic function

and add them together

[stark wedge]

you have to split the solid into the revolution of the green stuff and that of the purple stuff

[red tulip]

yeah take the green and purple parts individually

[worldly prawn]

ohh

alr i see: since what curve is on top changes

,help

[rocky lotus]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[worldly prawn]

so if i do that then i should get

\int{3}^{4}\pi\left(\left(y-2\right)^{2}-\left(\sqrt{4-y}\right)^{2}\right)dy+\int{2}^{3}\pi\left(\left(\sqrt{4-y}\right)^{2}+\left(y-2\right)^{2}\right)dy

,tex

this essentially

[red tulip]

you dont need to combine the functions

just take the individual integrals of each function and add them

[worldly prawn]

ohh so i can integrate over 2->4 for both of them and then sum those?

wait that doesnt quite make sense i think i misunderstood u

[red tulip]

no only 3 to 4 for the quadratic and 2 to 3 for the linear

[stark wedge]

no

just pi(y-2)^2 and pi(sqrt(4-y))^2

[worldly prawn]

okay i got what you mean

so then i get 23/6 * pi

like this

[red tulip]

other way around

[worldly prawn]

A;R

alr

wait but dont we want 2->3 for the linear (4-y) and 3->4 for the quadratic?

[red tulip]

yes you got them the wrong way around

[worldly prawn]

alr then we get 5pi/6 which is far more reasonable

thank you so much!

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[scarlet sandal]

I don't understand how the author could combine the limits (subtly)

[scarlet sandal]

The two subtracting log terms put in one expression implies that the limits were combined, but I thought you can only combine the sum of two limits if both of them exist, which in this case is false

but the author's goal is to justify that $\int_0^{\infty} \frac{1}{x^3-1} dx$ is convergent

[rocky lotus]

sed

[scarlet sandal]

here's the content immediately before and after the pinned screenshot

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[scarlet sandal]

<@&286206848099549185>

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@scarlet sandal Has your question been resolved?

[scarlet sandal]

(i'm headed to sleep now please don't close the channel automatically ;-;)

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[last shuttle]

pa solve guys

[haughty jackal]

Did I form the equation correctly? If so, how do I compute this? What does it even mean for the power of a matrix to approach infinity?

[last shuttle]

help to solve this problem 😢

[junior jay]

!occupied

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[dapper dagger]

lim_{n to inf} A^n converges if each of the 4 sequences, corresponding to the 4 entries of A^n, converge

however, the entries of A^n are not just the n'th powers of entries of A, so that's a bit of a drag

[haughty jackal]

Um, so is there a way to compute this

[dapper dagger]

well, if it was diagonalizable, that'd be cool

[haughty jackal]

What does it mean to be diagonalizable

[dapper dagger]

firstly, have you encountered diagonal matrices?

they're just (square) matrices in which every entry that is not on the main diagonal is 0

for a square matrix A to be diagonalizable, there must be an invertible matrix P and some diagonal matrix D so that A = P^(-1)DP

it's a really nice trick if you know it; power series become much nicer. for instance,

A^2 = AA = (P^-1DP)(P^-1DP) = P^-1DDP = P^-1D^2P

[haughty jackal]

I have

[dapper dagger]

consequently, A^n = P^(-1)D^nP (why?)

[haughty jackal]

Then $A^n=P^{-1}D^nP$?

[rocky lotus]

Will

[dapper dagger]

that's what I wrote, yeah

[haughty jackal]

Yes

[dapper dagger]

it makes limits involving A^n's nicer

so that's the appeal

[haughty jackal]

Time to learn diagonalization

Thanks

[dapper dagger]

yee

[haughty jackal]

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[old finch]

Can someone tutor in me in pre algebra

[old finch]

I know nothing

And I would like if we are vc

[tidal terrace]

just go for khan academy

[old finch]

So no one will actually tutor

Me

[junior jay]

Well it's just not really the point of this server, at least not the help channels. You might be able to find someone still, but with the ressources available on the internet you can go a long way

[old finch]

Ok recommend any

[tidal terrace]

.

[old finch]

The thing is I just get so unfocused when doing stuff alone

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[sharp harness]

Hi guys i need help with solving these questions

[sharp harness]

I solved a and b

But I don’t understand c and d

[mystic harness]

How many solutions if:
x1 = 0?
x1 = 1?
x1 = 2?
...

[sharp harness]

Hm?

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[copper willow]

<@&286206848099549185>

[tidal terrace]

!status

[vale dock]

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

[copper willow]

OMG SORRY I JUST FIGURED IT OUT

i need some sleep

[tidal terrace]

lol 1+6(n-1) or 1+6n ?

yeah ?

[copper willow]

im at a place where i keep getting stuck everywhere

the first one is just 6n cuz i thought it was like all cells in all circles

omg

[tidal terrace]

you still need to account for the initial cell

[copper willow]

in the second one tho only right

i mean part b

[tidal terrace]

even the first one because it needs to show the sequence

[copper willow]

is there a formula needed for part b tho

[tidal terrace]

it's the formula you found in part a

[wraith heart]

!15m

[vale dock]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[copper willow]

isnt it just asking how many in this ring

and question b is asking for total honeycombs

[slate basin]

Is it 1 + 6 ( summation 1 to n)

[tidal terrace]

a says find a rule in other words a formula that describes the number of honeycombs to n rings then you use that to find the number of cells in b

[copper willow]

but i tested and 6n works for both rings

and the next one

oh also for part b will i need to do 6n for every ring then add the initial or is there another formula

if anyone knows if i need to do 6n for every number up to 9 and add them up let me know

thanks

im in grade ten taking grade 11 functions right now in the YRDSB school board btw if any of yall had somethin different

[stark wedge]

yes, the total is 6*1 + 6*2 + ... + 6*9 + 1

there are several ways to calculate this

do you know how to do any of the following?

• find the sum of the first n natural numbers, i.e. 1+2+...+n
• find the sum of an AP given its first term, term count, and common difference

[vale dock]

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[vale dock]

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[drowsy atlas]

what did I do wrong bruh

[drowsy atlas]

Is the answer not C and E

hi @glass kelp lol

you’re always here

[glass kelp]

v'(t) is the acceleration

not velocity

[drowsy atlas]

yea ik

does any of the answers I chose contradict that

[glass kelp]

the graph of v'(t) isnt D

[drowsy atlas]

wait

Oh

I’m so blind

[drowsy atlas]

wait so why’d you send this

Did I do smth wrong there too

[glass kelp]

i cant read it

its small

firstly

lets decide what is the correct of v'(t)

[drowsy atlas]

Oh

B?

[glass kelp]

ye

[drowsy atlas]

nice

[odd jackal]

I thought cat was @primal crest

[drowsy atlas]

who is that lol

[drowsy atlas]

yea

is D vt graph

Cuz I use D to find B

[glass kelp]

how abt F

[drowsy atlas]

wdym F

F as VT?

[odd jackal]

Is this a test?

[drowsy atlas]

no

[glass kelp]

.

[glass kelp]

here

[drowsy atlas]

imagine having online tests 🔥🔥🔥

[glass kelp]

why is it wrong

[drowsy atlas]

oh

[glass kelp]

infinite grade glitch

[drowsy atlas]

cuz

it’s only slowing down if

VT and V’T are opposite signs

if they’re the same sign

It speeds up

Or else im teipoinf

teipoinf

trooping

[glass kelp]

is there any point where VT is negative

[drowsy atlas]

real

[drowsy atlas]

Based on the graph given? No

but

it was talking abt the graph given?

[glass kelp]

so is there any instance that when v'(t) is negative it causes v(t) to increase in magnitude

[drowsy atlas]

yea

only when v(t) is also negative

[glass kelp]

.

[green urchin]

hi

[glass kelp]

hello

[drowsy atlas]

im so confused

[green urchin]

where are you stuck at

[drowsy atlas]

based on the statement F I thought it’s saying

[green urchin]

drop it

[glass kelp]

fidning out whether F is corrrect or not

[drowsy atlas]

Car is always slowing down if v’(t) is negative

[green urchin]

v' is acceleration, so when it's negative, it indicates retardation

So its slowing down

that is, F is korecht

[drowsy atlas]

korecht

ok

[green urchin]

real

[drowsy atlas]

L wording

[glass kelp]

i mean

[green urchin]

its always korecht or inkorecht

[glass kelp]

this is true since there is no instance where v(t) is negative

is the point im trying to convey

[drowsy atlas]

ok ty guys

Where’s your pfp from @green urchin

[green urchin]

Its from tbate

[drowsy atlas]

sick

[green urchin]

ty twin

you can read tbate aswell its good

[vale dock]

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[floral lily]

An acquaintance of mine needs help with this, and since i’m an english major I have no clue how to solve it 😭

[marsh forge]

similar triangles

[floral lily]

Could you explain?

DE = BC or is it like 1/3 of BC? I’m trying to recall some rules but idk..

[marsh forge]

DE=BC is congruence (meaning the two triangles are exactly equal size)

[floral lily]

[nocturne flower]

hey! i can help you with this

[marsh forge]

similarity between two triangles just mean two triangles have the same shape, not necessarily the same size

[floral lily]

Would be good to receive your input as well

[marsh forge]

and also the ratio between two corresponding sides are equal

[floral lily]

The person who needs help said this

Taking AE = x
EC = 4.8 - x

In ABC =>
DE || BC

Therefore, using BPT (Thales Theorem)
AD:DB = AE:EC

Putting in values
3x:5x = x:4.8-x

Therefore, x = 9/5 = 1.8

Something something

[slate coyote]

1.8?

[floral lily]

So the answer is 1.8?

[floral lily]

He did those steps to get 1.8 I believe

[marsh forge]

thales theorem💔💔🥀

[floral lily]

Yeah whatever that thing is …

[marsh forge]

its just similarity

[nocturne flower]

AD//DB=AE//EC=DE//BC

[marsh forge]

i dont think this is right

let AD = 3x, DB = 5x and by similarity we should have AD/AB = AD/(AD + AB) = 3/8

then the "scale factor" of ADE wrt to ABC is 3/8

[nocturne flower]

use the basic proportionality theorem

that is AD= 3X DB=5X

[marsh forge]

wait is DE even findable

[manic hatch]

You just find it via similarity?

[marsh forge]

they gave AC

[nocturne flower]

then AB= AD +DB = 3X+5X=8X

[marsh forge]

;-;

[manic hatch]

Ah, I messed the drawing up

[marsh forge]

so cant you just vary DE

sighh

[manic hatch]

But still, similarity?

[marsh forge]

by similarity you can only deduce AE,EC

maybe i suck

[nocturne flower]

use the ratio now to find DE

[vale dock]

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[fluid pecan]

Oh shit

[safe fulcrum]

why did you close?

[stark wedge]

yeah just go take a new channel

[timber swan]

Do division

[fluid pecan]

The remix was messed up srry

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[rare ridge]

in a mapping diagram, does a many-to-many relationship exist? 😭

[quick rose]

well sure whats the problem

why wouldnt it exist

[rare ridge]

can i get a visual to what it looks like

just wanna check i didnt get it mixed up

[lyric sundial]

!original

[vale dock]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[deep cedar]

hello

[stark wedge]

hello, welcome to the server. this channel is currently occupied by someone else (ie has someone else's name on it)

if you want to ask your own question, open your own channel -- see #❓how-to-get-help for instructions
if you just want to chat, go to #discussion or #chill

[deep cedar]

oh mb didnt notice

[rare ridge]

i think i got it now, thank youu

.close

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[north scarab]

how do i solve this first quesiton

[north scarab]

dont tell me the answer

i know that this is repciprcal function of cos(theta)

it gave me an undefined answer

defintely messed up this question

[sick sparrow]

you forgot to divide by 1.8939 to isolate cosβ

[north scarab]

which line

[sick sparrow]

the last one

for some reason you included it inside the cosine operator

[north scarab]

u mean the second last line

[sick sparrow]

second last, sure

the inside of the inverse cos operator should be the reciprocal of 1.8939

$ \frac{1}{\cos\beta} = 1.8939 \implies 1 = 1.8939\cos\beta$

good bot

[vale dock]

@north scarab Has your question been resolved?

[north scarab]

@sick sparrow

We need to make

Beta the subject right?

[sick sparrow]

as in, solve for it? Yes

That's what I meant by isolating it

[north scarab]

so

1/cosB = 1.8939

B=arcos(1.8939)

[sick sparrow]

no

[sick sparrow]

what you wrote here is the same as secβ

so you would need to use the inverse operator of secant to solve for beta, arcsec

[north scarab]

wait this is correct right?

[sick sparrow]

yes, but this isn't

[north scarab]

yep let me fix that

so i got

1=1.8939(CosBeta)

[lyric sundial]

No need of parentheses

But yes

[north scarab]

1=1.8939cosBeta
this better?

[lyric sundial]

cos not Cos

[lyric sundial]

Which is exactly what's written here

[north scarab]

"the inside of the inverse cos operator" what does that mean again

[lyric sundial]

What is the inverse cos?

It's secant, right?

[north scarab]

inverse cos is

cos^-1?

right

[lyric sundial]

So the inverse of that is sec^(-1)

[north scarab]

u mean reciprocal

[lyric sundial]

No no

Reciprocal ≠ inverse

[sick sparrow]

cos^-1 is an operator that is applied to some quantity or variable, eg. cos^-1 (xyz) means that the inverse cosine operator is applied to xyz

dw too much about that just focus on the procedure

[north scarab]

because ik that inverse is used to

get the value of the angle

but i never heard the phrase how u said it

oh so u mean

xyz

this is the answer but i have written 58 degrees, 8 minutes

should i have written 08?

or 8 is fine

[stark wedge]

this is quite the hairsplit

i don't think it's criminal to write 58°8' but it is very slightly atypical.

is this a current homework or did you get docked points for it

[north scarab]

just a bit behind im catching up

idk why and how i got a bit behind

do u have any advice for staying ahead of the content

including doing asigned homework ahead of time

[sick sparrow]

it really comes down to discipline imo

I took trig online through covid and I made it my routine to never miss a deadline and always study at minimum 1 or 2 nights before exams

[north scarab]

always study at minimum 1 or 2 nights before exams?

[sick sparrow]

also known as cramming

[north scarab]

isnt that bad/

?

[sick sparrow]

but if you're in a position where you can spread out your study time then that's better

[north scarab]

yea but ngl im jsut a bit slow in completing my maths work

discipline works even if ur slow right?

[sick sparrow]

then make sure not to put yourself at a disadvantage by starting assignments early and studying in a timely manner

[north scarab]

like the questions are faily easy but it just takes my some time to complete them

[stark wedge]

how much work do you have left to do atm

[north scarab]

I dont have a schedule so its hard and even if i did, i wont stick to it

[sick sparrow]

that's the point of discipline

[stark wedge]

like how many assignments

[sick sparrow]

is that if you do it enough times it becomes your routine

[north scarab]

i dont have assignments i just got homework

and then a test next week

[stark wedge]

well how much homework

i wanna know the like

volume of work that you've got left to get through

[north scarab]

well I have to complete the practise questions first (fairly easy), then attempt the tutorail questions which picks up the difficulty but can still be easy

[stark wedge]

so one set of practice questions and one set of tutorial questions?

about how many in each?

[north scarab]

yea for each topic

[stark wedge]

...how many topics

maybe you could screenshot your online homework system or whatever if you got one?

anyway ok let me just go straight to my suggestion

since you say you have no schedule i will assume you have no time-dependent classes or other shit atm

[north scarab]

yea im a bit flexible atmp

atm

[stark wedge]

so what i suggest you do is this:

on your phone's calendar app, make an event titled "Math Homework Grind" or anything else you wanna call it
for one hour, each day Mon-Fri

whichever time you think is best

maybe 13:00 if you can't decide

and then, when the time comes, your phone will flash the reminder

and this is your signal to sit your ass down for an hour and do math stuff

if you think you can't manage an hour, start with a shorter time, like 30 minutes

[north scarab]

the thing is i allready do study every day especially for maths

[stark wedge]

you study math but you can't fit in time to dedicate to homework backlogs?

[north scarab]

for hours

[stark wedge]

or like idk, you could have your first such scheduled session be a test-run to see how much you can handle
point is if you dont have a schedule but have shit that needs doing, you make yourself a schedule

[north scarab]

yea apparently

[stark wedge]

and during math homework grind you put away all your other study materials, and focus solely on the backlog of work you're going through.

if you think you can't handle an hour you can start with 30 minutes. idk if i can recommend going any further down in duration given i don't know how big/hard/laborious the questions are.

but my point remains the same:

dedicate some time to the shit that you want done.

and do it externally such as through your calendar app.

[north scarab]

like rn it took me like 50 mins to solve all this

[stark wedge]

ok, so one page takes you 50 minutes.

so dedicate 1h each day to do one page of questions.

do i need to remind you of the heart of my suggestion? or did you hear and understand it

[north scarab]

but note that this is prac questions, the tutorail questions takes me like a whole day

to complete them

[stark wedge]

one tutorial question takes you an entire day?

[north scarab]

no the whole tutorial questions for a topic

[stark wedge]

so you're saying the entire tutorial takes you a full day to complete.

[north scarab]

but it depends ofcourse, some tutorials are easier than others and range between 2-3 sometimes 4 pages

[north scarab]

yeah because there will be some tricky questions ive spend time on

[stark wedge]

ok

in that case:

dont set any goals for volume of completion

just sit your ass down for an hour and do as much as you have time for.

then, once the hour is up, bookmark the question you finished at.

and then don't touch it until the next session.

where you will do the exact same thing again.

[north scarab]

ive been doing that, i use the pomodoro sessions (30/5 or 50/10)

well not entire day exactly maybe

7-12h

[stark wedge]

keep doing it then ig

interleave with your other studies

though i do recommend you take multi-hour breaks so as not to burn yourself out

and reserve at least one day a week for no studying at all

unless you have abusive/toxic parents who will abuse you for doing that ig

[north scarab]

[vale dock]

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[sage kestrel]

limit x tends to 0, find (cosx-1)/x

[timber swan]

Hmm

Just apply L hospital

[proven wasp]

Yes.

[young lion]

🏥

[sage kestrel]

if i first put value of x in cosx, then we'll get (1-1)/x

[timber swan]

0/0

[proven wasp]

💊🩺

[timber swan]

Which is an indeterminant form

Which u can solve by lopital

[sage kestrel]

in theory, we should put the limit in both the numerator and denominator, right?

[young lion]

yes

[timber swan]

Yea?

[proven wasp]

In theory?

Wdym

[sage kestrel]

then it becomes (1-1)/lim x

[timber swan]

🥀

[timber swan]

What

🙏

[sage kestrel]

i know l' hospital rule

but i am trying to do it manually

[timber swan]

U know l hospital but are struggling with putting in the limit

[glass kelp]

isnt (1-cosx)/x a known limit

[young lion]

squeeze theorem?

[glass kelp]

its derived from sinx/x

[proven wasp]

💀 🙏

[sage kestrel]

the eqn becomes 0/(lim x tends to zero, x)

[timber swan]

x^2 in the deno

[glass kelp]

for op or for the known lim

[timber swan]

bri

Bro

Use the taylor series

🗣️🗣️

[timber swan]

Known

[sage kestrel]

wait

[glass kelp]

nah its (1-cosx)/x

idk

prolly urs is (1-cosx)/x^2

[timber swan]

What

The known limit is

1-cosx/x^2

which equals to 1/2 when x tends to zero

[glass kelp]

||0. its the less known sibling of sinx/x||

[timber swan]

How

Im intrigued

[timber swan]

[sage kestrel]

yes, you're right

[timber swan]

Taylor expansion

[glass kelp]

conjugate with 1+cosx

[timber swan]

😔

[glass kelp]

sinx/x and its lost sibling

[timber swan]

What will u get

[glass kelp]

im too lazy to latex it

[sage kestrel]

thanks

.clsoe

.close

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[tardy sinew]

I dont really understand what ln is and how 1 ln 1 is equal to ln1^1

[turbid summit]

it says it there... ln is the natural logarithm

and $a\ln{b}=\ln{b^a}$

[tardy sinew]

yeah im not sure what a natural logarithm is tbh

[rocky lotus]

parabolicinsanity

[tardy sinew]

is there a simple way to explain it?

[turbid summit]

$\ln{x}=\log_{e}{x}$

[rocky lotus]

parabolicinsanity

[turbid summit]

logarithm base e

and logarithm is the inverse of exponentiation

so eh, do you get it now?

[tardy sinew]

so what do i do if theres a number before it/

[turbid summit]

this

[tardy sinew]

oh

mb

im blind

lol

thanks

.close

[vale dock]

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[turbid summit]

happens more than you think

good luck 🫡

[tardy sinew]

tanks

do i have to close this again?

[hot tide]

no

[vale dock]

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[lone reef]

actually genuine question for me who doesnt even like touching complex world

• is complex world natrually smaller than Real numbers?

[heady pawn]

define smaller

[lone reef]

Just wanna know that what does complex world actually do like in real life applications tho?

Complex plane isnt it all imaginary just for calculation?

[pure mirage]

The complex numbers include the real numbers

[heady pawn]

honestly too many to count

[lone reef]

examples?

like the vague mostly used for?

[pure mirage]

They're used all the time in physics

[heady pawn]

water flow, time, schrodinger

[lone reef]

i see.. math genuine getting harder for science stream students. Im just a business and commerce just plainly for calculation...

[lone reef]

how does time messure in i tho?

[heady pawn]

imaginary numbers are also handy in business

[lone reef]

something plus i?

[lone reef]

hmm? i didnt know that..

[heady pawn]

probability, differential eqations, browning motion in finance

[gleaming kindle]

complex can be used to solve special PDEs

[pure mirage]

They make your life easier a lot of the times. For example, when you want to solve the differential equation for electromagnetic oscillations, the basic approach is x(t) = e^(lambda * t)

[lost marlin]

I don't know of any applications of complex analysis

[heady pawn]

or to solve some matrices in complex systems

[lost marlin]

But people still do it

[pure mirage]

You will end up with complex numbers but it's many times easier than doing it without them

[lost marlin]

Because they find it intriguing

[heady pawn]

the most important part is pretty much the fact that e^ix = cos(x) + isin(x)

[lone reef]

oh ye thats the fact i know

[lost marlin]

Almost all of quantum mechanics

[lone reef]

like imaginary also explains why sin cos tan cosec sec and cot existed

in Complex worl

[lost marlin]

Stems from imaginary numbers

[pure mirage]

Here a slide from a presentation of mine on this

[lost marlin]

Also I haven't done complex analysis so you should ask someone who has maybe it does have applications

[pure mirage]

After this, you can transfer e^(lambda * t) into an expression with sin and cos using Euler's formula/polar coordinates

[lone reef]

Pretty good.

Thanks yall

.close

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[summer parrot]

Hey sorry but i have a maths test tomorrow morning and I cant get understand this at all. AI says the the north line is closer to point A when its not? why cant i just write 055 degrees as my answer? any help appreciated

[summer parrot]

true bearings are only answered from North clockwise right? point A is 55 degrees from North

[heady pawn]

reflect the point a over the line N in ur head

and then measure the distance from A (reflected) to W

which is 035

oh wait

I don't get it

[summer parrot]

i didnt understand that

[heady pawn]

I i get it

ur trying to measure the angle beteen a and the origin?

from A

[summer parrot]

A from O yeah i think

From O pretty sure

Bro this makes no sense at all

[heady pawn]

!nogpt

[vale dock]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[heady pawn]

the ai is confused about OA

please don't use it for mathematics

[summer parrot]

What do you suggest

[heady pawn]

look it up on google urself

[summer parrot]

Can you help me with this question

[vale dock]

@summer parrot Has your question been resolved?

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[latent mortar]

Find side AB

Know the diagonal AC = 2*sqrt13

So I used Pythagoras to find CD

Then now I’m trying to find BC so I can subtract it from AD to find the triangle

Hypotenuse

Well not the hypotenuse

Yet

But to find the bottom part of the triangle

[slate folio]

you don't need to

[latent mortar]

How

[slate folio]

oh nice idea

[latent mortar]

I know this so far

Idk how to find AB now though

[slate folio]

have you learnt trigonometry yet?

[latent mortar]

Yes

I tried doing Tg

[slate folio]

ok

[latent mortar]

But I need 2 sides at least right

Oh I have all the angles

[slate folio]

yea

[latent mortar]

Idk what to do with that though

[slate folio]

∆ABE clearly has a right angle at E

[latent mortar]

Mhm

[slate folio]

you know the length of the side opposite of the 60° angle

you want to find the hypotenuse

[latent mortar]

Yes

[slate folio]

what trig function should you use?

[latent mortar]

Don’t I need 2 sides though

[slate folio]

no?

[latent mortar]

Like u need to relate 2 sides

Tg is opposite/adjacent

[slate folio]

what about other functions

[latent mortar]

I don’t have adjacent

[slate folio]

what about sine?

[latent mortar]

Sin is opposite/hypotenuse

But I don’t have hypotenuse

[slate folio]

the hypotenuse is what we're trying to find here

[latent mortar]

I can u just label it x?

[slate folio]

and we know that ratio has an exact value which is sin(60°)

[latent mortar]

I don’t remember what the value is 1 sec

Sqrt3/2

[slate folio]

(√3)/2

alr what is the length of the opposite side?

[latent mortar]

4

[slate folio]

actually you might have an idea of how to get the length of the hypotenuse now

[latent mortar]

I think I know now

Lemme try

I got it

(8*sqrt3)/3

[slate folio]

nice job

[latent mortar]

Thanks man

[slate folio]

!done

[vale dock]

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

[latent mortar]

I have another question

Should I close and open another if I need later?

[wraith heart]

ideally just close and open new one

[latent mortar]

Ok

.close

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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fervent perch]

we need to find all discontinues points

[frozen ledge]

ok and what do you think they are

have you tried picturing this

like a graph

[fervent perch]

no

[frozen ledge]

do you have any idea where this might be continuous?

[fervent perch]

maybe 3

[frozen ledge]

mhm

anywhere else?

[earnest ingot]

Why?

[fervent perch]

dont think so but dont know how to prove it

[frozen ledge]

what condition must be satisfied for f to be continuous at say x = a?

[fervent perch]

limf(x) = f(x0) as x approches x0

[frozen ledge]

mhm so we assume a ≠ 3 then show that lim x -> a f(x) ≠ f(a)

if a ≠ 3 you could simply show the limit doesn’t exist so i’m assuming you’re familiar with the epsilon delta definition of a limit yes?

[fervent perch]

yes

.close

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• Show your work, and if possible, explain where you are stuck.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[haughty jackal]

My logic:
$$n(R)=8^4$$
We'll calculate the number of cases when $ad-bc$ equals zero (i.e., the number of non-invertible matrices in $R$) and then subtract it off from $n(R)$ to get the number of invertible matrices in $R$.
\newline\newline
Case 1: $ad=bc=0$
$$\text{Number of matrices }=(2!\cross 8)^2$$
\newline
Case 2: $ad=bc\neq 0$
$$\text{Number of matrices }=2(7^2)$$
Therefore,
$$\text{Number of invertible matrices }=n(R)-[(2!\cross 8)^2+2(7^2)]=3742$$
But the answer is apparently $3780$. Where am I going wrong?

[rocky lotus]

Will

[storm perch]

Oh I've done this before

This was a recent JEE question right?

[haughty jackal]

2023, yep

[storm perch]

Lemme get my document

Ok so I'm not even CLOSE to that shit

[haughty jackal]

Ouch

I'm probably overcounting the number of non-invertible matrices

[storm perch]

Wait why ad-bc

Wouldn't it just be when the determinant is equal to 0

[haughty jackal]

$M$ is invertible $\implies\det M\neq 0$

[rocky lotus]

Will

[haughty jackal]

Yes

[storm perch]

Lemme recalculate rq

Ok yeah you're right

[graceful zinc]

yes

[haughty jackal]

Where

[graceful zinc]

in both cases

[haughty jackal]

Can you elaborate

[storm perch]

Could you explain how you got each number of matricies?

[graceful zinc]

ok tell me how did you come up with (2! X 8)^2 numbers

[haughty jackal]

Okay sure

Case 1:

Let's fix a = 0, then d can be chosen in 8 ways. Now fix d = 0, then a can be chosen is 8 ways. Therefore, ad = 0 can be formed in 2 * 8 ways.

Similarly, bc = 0 can be formed in 2 * 8 ways.

Therefore, to form ad = 0 and bc = 0 simultaneously, it can be done in (2 * 8)² ways.

[graceful zinc]

"Let's fix a = 0, then d can be chosen in 8 ways. "
one of those 8 ways also include the case where d=0
it gets re counted when you say "Therefore, ad = 0 can be formed in 2 * 8"

[haughty jackal]

OKAY