help-4

[astral depot]

this indicates we cannot improve the flow anymore and reached the maximum

[prisma flint]

for iterations 2 first line why is it 10 and not 0?

[astral depot]

bcause you only filled the 1-3-4-6 path with flow of 10

[prisma flint]

it gets improved by 0?

[astral depot]

in which case?

[prisma flint]

in iteration 2 6 gets improved by 4 and 1 by 0?

[astral depot]

wdym 1 by 0?

1 is the source right?

[prisma flint]

why is the knoten 1 still 10 and ht eknoten 6 is 4? in iteration 2?

[astral depot]

knoten 1 is infinity right?

[prisma flint]

right i meant the 2

[astral depot]

knoten 6 gets improved by 4 along the 1-2-4-6 path, you are right.

[astral depot]

yeah that is the maximum it can improve (theoretically)

compared to previous iteration

[prisma flint]

but how can 2 improve by 10?

[astral depot]

why not

after iteration 1, the flow through 1-2 is 0

because we only filled the 1-3-4-6 path with flow of 10

[prisma flint]

i get it, but how can in iteration 3 it still improve by 6?

2 amd 3 are lareaady fully used?

[astral depot]

because after iteration 2, we fill 1-2 -4 -6 path with flow of 4(which is the maximum among all other flows in iteration 2)

so removing 4 flow from max flow of 10, we get 6(for 2)

[prisma flint]

and in iteration 4? i dont get it at all

[astral depot]

after interation 3, we fill the 1-2-5-6 path with flow of 4 (maximum among athers in 3)

then we find that nodes 5 and 6 have maxiumum possible flow(we cannot improve further)

i hope i'm not overwhelming you😅

[prisma flint]

thanks for trying but i dont understand it at all
the iterations 2 3 and 4

[astral depot]

oh i'm sorry i couldn't help

do you have any other examples of this?

[prisma flint]

no, sadly not

[astral depot]

oh okay

gimme 2 mins ill check for mine

[prisma flint]

thank you 🙏

[astral depot]

found it

okay, lets start fresh

what is the first thing you will do here?

[prisma flint]

go up top with 8,9 and 2?

[astral depot]

uhh but the S-D-C-T will have max flow of 3 right (which is better than S-A-B-T)

[prisma flint]

ah got it

s-d-c-t, right?got it

[astral depot]

yeah

[prisma flint]

because 3 is more than 2

[astral depot]

try analyzing this as people on islands and the maximum flow to be similar to maximum people who can cross the bridge between islands

[prisma flint]

so in iteration 2 i would do what then?

[astral depot]

first you got that s-d-c-t is better right

[prisma flint]

yes

[astral depot]

then fill s-d-c-t with flow of 3 (which is max permittable flow)

[prisma flint]

yep

[astral depot]

now, do the same thing again

[prisma flint]

s-a-b-t with 2?

[astral depot]

bingo

now fill s-a-b-t with 2 and repeat it again

[prisma flint]

finished, right?

s to a is complety used and b to t is fcompletly used

[astral depot]

well the d-b arrow is a bit misleading, flow can be in any direction

try again but take the d-b arrow to be on the other side this time

[prisma flint]

what?

a got it

so we can take 1 more through the new line

but only 1

[astral depot]

just because the d-b arrow is pointed towards b, flow need not be from d to b it can be otherside too

yeah you got it

[prisma flint]

thank you very much!

[astral depot]

now fill the path s -a -b -c- t with 1 flow and do it again

[prisma flint]

you are my hero!

[prisma flint]

but now everything is filled, right?

[astral depot]

yes

you got it

everything is filled

[prisma flint]

you are my hero, thank you very much!

[astral depot]

so max flow is 6

[astral depot]

well i hope the table makes more sense now

[prisma flint]

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[final lance]

<@&286206848099549185> Any kind hearted soul pls help me solve this

[static birch]

ok, well.

[odd jackal]

add and subtract 1 to the numerator

[static birch]

the epic method, is epicly algebric.

[final lance]

so what i did is divide num and denom by x^2

[static birch]

you can change x^2 to x^2+1-1

np, have a good day.

[final lance]

to get in the form of tan-(x) derivative

[final lance]

so what i did was wrong?

[static birch]

no no relax yourself.

first, do the plus one minus one

which yields

int(1+1/(x^2+1))dx

[final lance]

ohhh

ur right

[static birch]

yes

why do you torture yourself over it

[dry trail]

-1

[static birch]

yes sorry

[final lance]

ok tysm guys

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[lilac lintel]

Guys do u know anything Abt IGCSE?

[plush crater]

yes

[lilac lintel]

Can u explain to me how it works

[plush crater]

well this isnt really a math question

but sure

u ideally spend 2 years preparing for a syllabus given by a system called the IGCSE

u can take core programs (basic level, not really worth it) or extended programs (more standardized levels)

if u r asking how useful it is to uni applications, its pretty useful so i do recommend it. the rest u should ask ur school about

[stone aspen]

(some details differ from zone to zone)

[lilac lintel]

Oh sorry

[lilac lintel]

I'm an anxious kid 💔

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[upper yew]

help

[vale dock]

@upper yew Has your question been resolved?

[upper yew]

NOOO

[errant fulcrum]

Take all the equation given equal to a constant

[errant fulcrum]

And then try to solve it

[upper yew]

okok trying

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[civic goblet]

For a question like this

Prove that 97 is a prime number.

I've got to do it by proof of exhaustion. I attached the solution on screen, I am just confused why, because of the fact that there are no factors less than root 97, that just means all possiblities have been exhausted? I understand why in the demo, we skip 4, 8, 9 and 10 (because we have the multiples of them), i just don't get why he ends it there.. what about the rest of the prime numbers like 11, 13, 15, 17?

[ebon glade]

factors come in pairs

one will be smaller than sqrt97, the other bigger than sqrt97

(or both equal to sqrt97 but 97 is not a square)

[civic goblet]

ohhh i see

how about if its prove that 201 is a prime number (as an example)
do i do factors until sqrt 201? because then if there are no factors before that, that must mean there aren't any factors after that?

[ebon glade]

yes

[civic goblet]

okay, thank you 🙂

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[sly torrent]

Help, I am not familiar with err function concepts so try to avoid them, I am fresher so I have knowledge about just basics of fourier transform

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[sly torrent]

<@&286206848099549185>

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[cerulean mesa]

Can you find the Fourier cosine transform of e^-(t^2/2)?

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[thorny sandal]

is this right? i dont understand how x=11 cuz if u substitute the value of x in the table i dont think it matches up..

[drowsy kraken]

In present , x is Danilo and y is cora
Hence ,( x-4) = 2(y-4)

[thorny sandal]

so is the solution wrong.. i got this from a video

[drowsy kraken]

No sorry I got the q wrong

[drowsy kraken]

This Is ur first eq

Now three years from now, Danilo is x + 3 and cora is y +3

The sum is 47
X+3+y+3 = 47

This is eq 2

Hence eq 1 : x-2y = -4

And eq 2 : x+y = 41

[thorny sandal]

41 or 47?

[drowsy kraken]

41

Cause x+3 + y+3 is 47

=> x + y+6 = 47

X+y => 41

[thorny sandal]

but then how do i get coras age 8 years from now using that equation if both x and y is missing 😓

[drowsy kraken]

Umm how are they missing??

Do u know elimination method ??

Or substituon method

[thorny sandal]

i know a bit but im not quite sure how to use it there sorry

[drowsy kraken]

What method do u know

[thorny sandal]

elimination

[drowsy kraken]

Alright

[drowsy kraken]

Now we can right this as -x-y = -41

Right?

[thorny sandal]

yes

[drowsy kraken]

This will help you

Now u get cora is 15

So what will her age be 8 years from now

[drowsy kraken]

Please tell if you don't understand a step

umm

did you get the answer??

[thorny sandal]

hi sorry i get it much better now but can i ask why do we need to multiply this by 2

[drowsy kraken]

so when we do elimination

we can eliminate the y

and since in eq 1

we have x-2y then

having 2y in eq 2 can help us cutting the 2y when we are adding the two eqs

[thorny sandal]

ah okay gets, thank you very muchh

[drowsy kraken]

npp

gtg now bai

[eternal osprey]

@thorny sandal if the question has been answered satisfactorily, please type .close

[thorny sandal]

.close

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[sharp yacht]

im stuck on this question

[sharp yacht]

i believe my bounds are incorrect

i am trying to do the integral from 0-3pi/4, and 5-7

of sqrt(5)r^2/4 dr dtheta

[dawn fox]

that sounds right to me

[sharp yacht]

hold on ill try to redo this question then

109sqrt(5)/6 for the first integral

so int 0-3pi/4 of 109sqrt(5)/6 dtheta

109pisqrt(5)/8?

its showing that its wrong, im not sure why

i double checked that answer with WA as well

[dawn fox]

oh that's not correct, wait a sec

you can't assume the angle is 3pi/4

make a triangle and try to find it out by hand

[sharp yacht]

but how do i find the x-value of that point at c

[dawn fox]

well you know the distance from the origin to the corner y = 4 intersects is what

[sharp yacht]

5

[dawn fox]

yep

that should help you find the distance you mentioned

[sharp yacht]

and then i could do 16+y^2=25

so y=3

[dawn fox]

what is y representing here

[sharp yacht]

distance from the y-axis to the corner?

aka the length of that dashed line

[dawn fox]

okay sure yeah

just making sure

[sharp yacht]

or actually i dont even need that right

[dawn fox]

not particularly lol

but what is the angle then

(it's not a pretty value)

[sharp yacht]

arccos(4/5)?

[dawn fox]

yep that works

[sharp yacht]

so it would be int from 0-arccos(4/5), 5-7 of that

that makes sense

unfortunately i ran out of attempts for this question but at least i understand it now 😭

[dawn fox]

wiat not 4/5

arccos(3/5)

or arcsin(4/5)

either way, sorry you ran out of attempts

[sharp yacht]

why arccos(3/5)?

arent we using the angle formed by the y-axis and that diagonal to the corner?

[dawn fox]

oh i was measuring the angle inside the triangle 🤦

[sharp yacht]

oh wait also

it would be pi/2 plus that right?

[dawn fox]

not exactly

it would be pi - arccos(3/5)

OH you made a different triangle

yes your way is fine

[sharp yacht]

yeah now i see what you were trying to say lmao

[dawn fox]

you're completely correct, i made the other triangle lol

[sharp yacht]

i made the triangle formed by the diagonal from the origin to the corner, the y-axis, and the dashed line

im assuming you did the one with the diagonal, x-axis, and the vertical up to the corner?

[dawn fox]

yes lol

so we were both correct

[sharp yacht]

okay okay that clears it up haha

anyways appreciate the help!

[dawn fox]

yep np!

[sharp yacht]

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[ornate dragon]

Hey folks. It's a physics problem but maybe someone's able to give me a hand

"A cylinder of mass M and radius R is on an incline, with an angle theta respective to the horizontal plane. This incline has static and kinetic friction, mu e and mu c respectively. Determine the longitude the center of mass travels when the cylinder has performed its first full rotation. Consider first whether or not the cylinder slides."

So I'm a bit confused on the forces at play

Do I take friction into account for the center of mass?

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@ornate dragon Has your question been resolved?

[steady charm]

well if you assume that it is rolling without slipping, then there would need to be sufficient static friction for the required angular acceleration

if that exceeds the static friction coefficient then it would slip

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[quick agate]

how do i do either of these

[tepid wind]

C) This means that, to get to Q from P, you must go along the vector [5, -1].

[tepid wind]

D) This means that, to travel along [5, -1] and end at M, you must start at L.

[quick agate]

i still dont get it

how do u go along vector v

what does that even mean

[restive aurora]

Assuming I'm understanding "equivalent vector" right, if v is a line from (0,0) to (5,-1), what would Q need to be for the line from P to Q to look like v?

[quick agate]

3 and -8 i think

[restive aurora]

Yeah!

[quick agate]

so u just take x and y of PQ and subtract by x+y of vector V

[restive aurora]

Uh, I'm not quite sure what you mean ^^;

[quick agate]

like this

and that would be coordinate of q

[restive aurora]

Yeah, though I think it should just be P in that equation instead of PQ (again, assuming I'm following this terminology right)

[quick agate]

oh true

[restive aurora]

And maybe it should be + instead of -

[quick agate]

yea

i was gonna correct myself my bad

[restive aurora]

All goodo

d) is similar but you've gotta do it backwards

[quick agate]

wdym

do i subtract instead of add or smth

[restive aurora]

You're getting the final point and need to find the start point

Well if you had P and PQ then wanted Q you had to add to P to get there

Now you have M and LM and want L

So you've gotta find the point L that will get you to M

I'm not gonna tell you the exact operation you need to do, sorry ^^;

[quick agate]

is it just the same as before but instead of adding its just subtracting

cuz ur going backwards from the point or smth

[restive aurora]

Ye

[quick agate]

fire

so L would be (0,-9)

if i did the math right

[restive aurora]

Mmhmm! Though I think you added that - sign there by accident

[quick agate]

yea that was my bad idk why i did that

ty for help

[restive aurora]

No problem!

[quick agate]

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[tough maple]

Hello 👋, Could someone explain why the answer is ♾️ ? What did I do wrong in my answer cause I js know we get the highest powers with each other and stuff that’s how we get the answer

[tough maple]

@ me when someone answers thanks !

[fringe lava]

1 > 1/2

It is not x/x

Just try l'hopitals

[fringe lava]

That will result in the numerator growing much faster than the denominator

[tough maple]

The question is in the first pic if that’s what u mean

[steady charm]

i was circling the = sign because i don't see any justification for that equality

[tough maple]

Oh yeah

[tough maple]

Wait I just realized I mixed up two concepts

Where it said radical(x^2) can be x or -x depending on the infinity sign

[tough maple]

Can I divide both the numerator and denominator with x?

I’ve been kinda struggling with limits and all their ways

So can I just isolate x/radical(x) and it’ll equal radical(x) then I substitute the x value so that’ll just be infinity?

[fringe lava]

I'll make it clearer

Just divide all sides of the fraction by sqrtx

Then certain terms go to 0

[vale dock]

@tough maple Has your question been resolved?

[tough maple]

Oki so we divide each part by whatever power is the largest in the denominator always?

Also I have another question if that’s okay, over here in this question it tells us to find the increasing and decreasing intervals in the first quadrant

f(x)=x+cos2x ; 0<x<pi/2

Whys 5pi/12 included? Is it because the x is 75 ?

But it’ll be 150

Idk I’m confused

@ me when someone answers, thank you

[vale dock]

@tough maple Has your question been resolved?

[oak crane]

yes because $\frac{5\pi}{12} \text{ radians} = \frac{5 \times 180^\circ}{12} = 75^\circ$

[rocky lotus]

Aκιρɑ

[tough maple]

But how do I make sure I’ve gotten all the angles/ numbers possible ?

I just thought it didn’t go with it because 75x2 is 150° and 150° isn’t quadrant I.

[oak crane]

,, \dfrac{5\pi}{12} \text{ is within the interval } 0 < x < \dfrac{\pi}{2}

[rocky lotus]

Aκιρɑ

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@tough maple Has your question been resolved?

[weary swan]

how do you write these out

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[autumn silo]

Can someone check if I did this correctly pls? Thanks

[autumn silo]

I need to find the area of the region

[stark wedge]

looks correct but you could simplify those logs a bit

at the very end

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[pseudo python]

does anyone know how to do 36? i know i can get the ans by trialling with each option and i know its b but i dont get the theory behind how to get that answer

[pseudo python]

i tried this with the general sum formula for ap but i dont get the ans

[fickle rose]

is that 5 supposed to be squared, or to the power of 5?

[pseudo python]

i think its supposed to be squared just a typing error in the question

[stiff fossil]

Try rewriting the sum as a series with only whole numbers

So evaluate all squares of the series

[pseudo python]

but then the difference wont be constant ?

[noble anvil]

1,3,5 are in AP yes, but if they're all squared you can't say they're in ap (since 1,9,25 are not in AP)

[stiff fossil]

The series has AP and GP at the same time

Like from 1 to 9 thats + 8, 9 to 25 thats +16, and in between that there's a common ratio of 2

[pseudo python]

so how am i supposed to find the sum ??

[stiff fossil]

So finding the sum using AP sum quite frankly won''t work.

[stiff fossil]

Do you know the summation system?

[noble anvil]

Knowing the sum of squares of n natural numbers seems a good way to start

[pseudo python]

i only learned n/2 (2a + (n-1) d) formula

[stiff fossil]

Yeah

[stiff fossil]

This progression doesn't seem to be using AP by itself

Ever heard of $\sum_{k=i}^{n} k = i + (i+1) + (i+2) + ... + (n-1) + n$?

[rocky lotus]

pix

[pseudo python]

is this that k = n(n+1)/2 and k^2 = n(n+1)(n+2)/6 thing

[noble anvil]

Yes

Sigma k^2, not just k^2 btw

[stiff fossil]

Ik

[noble anvil]

No im telling him

[stiff fossil]

Oh

Alr

[pseudo python]

i had it in the book but i didnt know hwo to use it

[stiff fossil]

Notice how odd numbers starting from k = 1 is always 2k-1

[noble anvil]

You did get the formula wrong tho

Its 2n+1 not n+2

[stiff fossil]

Can you use that as a function for the summation system?

[pseudo python]

sorry i didnt memorise it 😅 im learning these new subtopics for my entry test so im just cramming it i didnt get time to learn it properly

[noble anvil]

This is a different thing

The series in the qn is not of the k^2+k form

[pseudo python]

i mean the formula in the box on the side

is it that

[noble anvil]

Ah ok

Yes

[pseudo python]

so how am i supposed to use it for this question

?

[pseudo python]

can u explain how to use the formula here bc we don't know how many terms there r

[stiff fossil]

@pseudo python

By this, we can determine that the amount of terms in the series 1^2 + 3^2 + 5^2 + ... + n^2 should be 2n-1

And since the series is T1^2 + T2^2 + T3^2 + ... Tn^2, then the sum would have a function of (2n-1)^2

Thus the sum becomes $\sum_{k=1}^{n} (2k-1)^2$

[rocky lotus]

pix

[stiff fossil]

Can you figure it out from here?

[pseudo python]

like this then use summation thingy?

[noble anvil]

Just remember that the (2n-1) here is "n" in the question

[stiff fossil]

Not quite

[noble anvil]

No not ap again

[stiff fossil]

Are you able to determine the common difference from here? No, right?

I'll give you a hint: If you have a sum that includes a polynomial function, seperate each term to become a summation of its own.

[pseudo python]

i didn't get how u found number of terms 😅if we cant use the Tn firmula

[stiff fossil]

All even nuumbers are multiples of 2, which means an even number is 2k

And since odd numbers are always 1 less than an even number, then an odd number should always be in the form of 2k-1

[noble anvil]

Maybe you should write 2r-1 so the variables aren't mixed up

[pseudo python]

ohh ok i get it now

[stiff fossil]

Yeah we could do it that way too

[pseudo python]

so now the number of terms are (2n - 1 ) ^2 ?

[stiff fossil]

No that's how you get each term

[noble anvil]

No, no square

[stiff fossil]

The number of terms that exist in the series is just 2n-1

[pseudo python]

ok then to find sum we put number of terms as (2n -1) in the summation formula thing ?

[noble anvil]

Have you been taught how to work with $\sum$ yet?

[rocky lotus]

Executor

[noble anvil]

Or to be clearer put the last term as 2r-1 where 2r-1 = n

[stiff fossil]

Yes

[pseudo python]

means this ?

[stiff fossil]

And the summation function should be (2r-1)^2

[noble anvil]

Basically when we have a polynomial within the $\sum$ sign, we need to split the sigma between the terms of the polynomial and solve them individually

[rocky lotus]

Executor

[stiff fossil]

Yes

[noble anvil]

Ye

However

You cannot use that formula

Since that would include the sum of all the even squares as well

[stiff fossil]

Now try to determine the sum of k, and sum of 1

[noble anvil]

You need to do it by this

[noble anvil]

.

[pseudo python]

im getting abit confused between whats n and whats k

n is number of terms or k

[stiff fossil]

n is the last term

[pseudo python]

k is number of terms ?

[stiff fossil]

k is the nth term

[noble anvil]

Ok let me rephrase it
$\sum {k=1}^{k=r} (2k-1)^2$ where 2r-1 = n

[rocky lotus]

Executor

[noble anvil]

Uh

Im bad at latex

[stiff fossil]

Wait

[noble anvil]

Pix can you do it if you understand what i mean

[stiff fossil]

$\sum_{k=1}^{r} (2k-1)^2$ such that $2r - 1 = n$

[fickle rose]

put a _ after \sum and before {k=1}

[noble anvil]

Perfect

[rocky lotus]

pix

[stiff fossil]

From here, k is the nth term starting from 1, and r is the last number plugged on a term

[pseudo python]

so solve liek this then what do we do ?

[noble anvil]

Perfect

[stiff fossil]

Notice that you can place the "4" from the two functions in front of the summation

[noble anvil]

Now just use the individual sigma formulae

[pseudo python]

for k ^2 and k ?

[noble anvil]

Also u put n instead of r in the upper bound

[noble anvil]

Yes, and also for $\sum 1$

[rocky lotus]

Executor

[pseudo python]

what willw e put for the one summation

[noble anvil]

A sigma 1 just means the upper bound "r"

Since its just 1+1+1....r times

So its 1*r

Which is r

[noble anvil]

You should fix that or you'll get confused

[pseudo python]

@noble anvil @stiff fossil okay thank you i got it now.. sorry it took me so long to get it : -) ❤️

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[noble anvil]

Pog

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[uneven token]

hi again

[uneven token]

its asking about asa saa stuff like that

but if im beinghonest i dont understand at ALL

i know its showing 2 sides and 1 angle

[stark wedge]

what are the answer options here

[uneven token]

but idk where the s or a goes

[uneven token]

im thinking its neither?

but

i need to make sure i understand lol

[stark wedge]

ASA means a side between two angles

SAA means a side not between two angles

however, notably, both the second and third answer options call for one side (S)

while you have two sides and only one angle

[uneven token]

yeah but wdym between? how do i know where to start?

[stark wedge]

so you are correct

[uneven token]

okay

how do i know which angle i start the postulate at?

[stark wedge]

yeah but wdym between?
like if you imagined walking along the triangle like an ant...

it doesnt really matter where you start

[uneven token]

where would i start out as the ant

but wouldnt that change the answer?

[stark wedge]

idrk how else to explain it cause it's rather visual

like you should be thinking about the "shape" that the known sides and angles form inside each triangle if that makes sense

[uneven token]

okay so like

if i start at the circle and go theway the arrows are going

wouldntthat be ssa?

[drifting hornet]

Yes, it is SSA

[uneven token]

ph my god thank you for that chart sm

[drifting hornet]

oh it doesnt have SSA apparently

[stark wedge]

yeah that would be SSA or maybe ASS if you wanna be funny

those are not congruence criteria anyway

[uneven token]

i was taught that ASS was bad for that reason LOL

oh my god i think i get it now tysm

ok im cosing it thank yall

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[stark wedge]

ASS is ass

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[modern wren]

can someone help me

[modern wren]

on why is it D and C

[lost marlin]

It is an isosceles trapezium

So

[modern wren]

yes im aware

[lost marlin]

Adjacent Angles are equal

[modern wren]

for no. 4 its 135 because its equal to

the other side

?

[lost marlin]

Yes

[modern wren]

ohh

[modern wren]

wait so is the bottom left also 45

or 135

[lost marlin]

That is 45

[modern wren]

I thought that the cross is equal

[lost marlin]

I would recommend trying to prove why they are equal

[modern wren]

for example bottom left is 135 and top right is 135

vice versa

[lost marlin]

That is for parallelogram

[modern wren]

OHHHH

but for trap its same line

[stark sable]

(bad drawing but hopefully it makes sense)

[modern wren]

kind of

[lost marlin]

Only for isosceles trapezium

[modern wren]

oh alright

is it alr if I show another example

[lost marlin]

Yes

[modern wren]

okie thank you

for MEN I answered 50

turns out it was 125

why tho

[lost marlin]

The 2 lines are parallel

What does that tell us ??

[modern wren]

that we can subtract 20 and 35 from 180?

[lost marlin]

Yes

[stark sable]

(same concept )

[modern wren]

OHHH right I get it now so if theres a line

and its asking for the angle congruent to it

I just add the top angles to get the bottom angle

like in the pic

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[modern wren]

my bad I kind of need help again

how do you get g

i dont get how its 70

[mortal token]

.reopen

[modern wren]

.reopen

[vale dock]

✅

[mortal token]

are you given that the horizontal lines are parallel?

[modern wren]

naw

well its kinda obvious tho

[modern wren]

only 5 was given

[mortal token]

i see

well considering that the lines are parallel, g and 70 are corresponding angles

[modern wren]

oh

but how was 70 achieved

50+20?

[mortal token]

sum interior angles

if you look at the biggest triangle, so basically the entire triangle, it has 3 angles right? the 50, the 60, and that bottom right corner

we know that the sum of a triangle's angles must equal 180

[modern wren]

yes

[mortal token]

so 180-50-60=70

[modern wren]

OHHH

[mortal token]

then g and that 70 are corresponding angles

[modern wren]

i get it now thank you\

[mortal token]

cool, np

[modern wren]

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[rocky lotus]

brandon

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@tawdry lintel Has your question been resolved?

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[shell raptor]

hello its only a small one but im stuck on this

[viscid spade]

what is the formula for rectangle's area

[shell raptor]

just time eachother

the side

s

i did root80 over 2 + root5

but now im confused

[ivory valley]

you may rationalize

[viscid spade]

you can multiply by 2-sqrt(5) top and bot

[shell raptor]

why do i change the symbol to -

[viscid spade]

conjugate quantity

to make the root disapear

[shell raptor]

uhm wait

big words

wait so like

[viscid spade]

you want the denominator as (a-b)(a+b)

so that you do a² - b²

[shell raptor]

OH

i see now thank you

[viscid spade]

will 95% be this thing when you have to rationnalize

and there is sqrt

[shell raptor]

okay thank you

im going to close now

i think i can do the rest

maybe

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[tender orchid]

I need help, i don't understand how how it concluded to true?

[fringe lava]

proof by induction.

[noble anvil]

It says "induction"

[tender orchid]

what is induction

i

what does it mean

[fringe lava]

probably something you would learn about watching a video online

more efficient than having one of us explain

[noble anvil]

Induction is when you set up a base case, then prove another subsequent case, and then show that the statement is valid for all numbers

Thats just the gist

[noble anvil]

So yeah this.

[tender orchid]

i didn'

t find this type of question on youtube

that's why i'm asking here

[tender orchid]

in this case the base case is k? and the subsequent is k+1 right?

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[noble anvil]

Yes

[tender orchid]

we suppose that 2^k+1 >= k+2, and then we want to prove 2^k+2 >= k+3, i don

[noble anvil]

Yes

[tender orchid]

I don't see what we can substitue to get the answer

it's a bigger/smaller than question which i don't completely understand

[noble anvil]

Just rewrite 2^k+2 as 2^k+1 * 2

[tender orchid]

after multiplying with two we got 2k+4, which is greater than k+3, that's why we got true am i correct?

[noble anvil]

Yes

[tender orchid]

Thanks

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[tough maple]

Could someone explain this statement?

[tough maple]

@ me when someone answers thanks !

[ivory valley]

where x is in some interval I containing a

cos(t) is continuous and sin(x) is differentiable

basically you are differentiating the antiderivative

[tough maple]

That’s it? That’s what it says 🧍‍♀️

[ivory valley]

Yes, it's FTC 1

[tough maple]

Wait what abt this question why isn’t this f’(x) ( I think it’s related to this statement)

Q3

Doesn’t that equal to the definition of differentiation

[ivory valley]

Say F(x) was the antiderivative of f(x) then by FTC2 you get F(x+h)-F(x)

and the rest is the limit definition of derivative for F(x)

[tough maple]

But how do I know it’s F(x) it shows f(t)

[ivory valley]

FTC2

[tough maple]

Is it in this

[tough maple]

What’s FTC2

[noble anvil]

Fundamental theorem of calculus 2

[tough maple]

Oh

[ivory valley]

Well it's a bit weird notation

but you kinda say okay if f(t) is my integrand then let's denote an antiderivative of f(t) as F(t)

[tough maple]

All the time?

[ivory valley]

well most authors do so

[tough maple]

pls I’m gonna have this test that’s gonna decide my whole future so 😭

[ivory valley]

some may call the antiderivative diffrently

like g(t) or whatever but F(t) is the most common

[tough maple]

why do they denote it f(t) but not F(x)

is it because integration gives the notation F(X) mostly

[ivory valley]

You integrate with t

[tough maple]

I thought it would look like dis that’s why it looked like f’(x) to me

[ivory valley]

Your mindset is like, okay f(t) is an arbitrary function. Now to continue with this limit, we would use FTC, so say okay, let an antiderivative of f(t) be denoted as F(t).

[tough maple]

oki then when would the answer be f’(x) ?

[ivory valley]

Well the integral of a function f is not always itself

Integral of x isnt x

[tough maple]

Oooo yeahhh

Omg wait I get it now

[ivory valley]

It's in full most cases a very different function and that we would call for example big capital F

[tough maple]

What if the question was something like this

Everything’s the same except f(t) is f’(t)

[ivory valley]

An antiderivative of f'(t) would be f(t), yes.

And that's a more intuitive notation.

Q3 is basically stating that this limit returns the integrand function back with respect to x.

[tough maple]

Is it written as f(t) because it represents the change in time or like rate of change and it corresponds to time or something like that?

[ivory valley]

It could represent anything, if there is no context.

[tough maple]

oh oki

[ivory valley]

But you only need to know that f(t) is just some function

[tough maple]

f(t) is a derivative of some function or just a function

[ivory valley]

What f(t) is supposed to do or be is not of your concern

[ivory valley]

Well now it depends for which you ask

[ivory valley]

For this f(t) is an antiderivative.

[ivory valley]

In the original as it was written, it was the derivative of the antiderivative F(t).

[toxic ridge]

Antiderivative is a fancy word for integral

[tough maple]

Oki I get it now

Thank you 🙏

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[tall rivet]

Im having troubel with finance

[tall rivet]

Math

[solid vale]

accounting is math right?

[tall rivet]

I think so

Im having trouble with deductions

[solid vale]

okayy help with accounting my final is today i just have one tiny issue

idk how to solve a fifo

[tall rivet]

What is fifo

[solid vale]

first in, first out

i just know how it works but i have no clue how to solve it

[tepid wind]

!occupied

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[tall rivet]

Im really confused with step 6

[tall rivet]

Can you help?

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[shrewd chasm]

can anyone explain and help me understand this

[stark wedge]

that notation for ratios looks odd to me tbh

maybe they meant $-y_1 : y_2$?

[rocky lotus]

Ann

[stark wedge]

the x-axis only ever crosses the segment PQ if y_1 and y_2 (the coordinates of P and Q) are of opposite signs.

@shrewd chasm are you still here?

[shrewd chasm]

yes, they are ratios..
and they expressed - as externally cut...
i don't understand the later

[stark wedge]

ok then let's maybe try to understand wtf they meant together.

do you see the diagram i drew here?

[shrewd chasm]

yes

[stark wedge]

ok, do you agree that:
\begin{enumerate}
\item triangles $PAR$ and $QBR$ are \textbf{similar}
\item $PA = |y_1|$ and $QB = |y_2|$
\end{enumerate}

[shrewd chasm]

do you want to see a question on this

[rocky lotus]

Ann

[stark wedge]

i don't think we'll need one, but if you want to share one after i'm done with my explanation then we can take a look

[shrewd chasm]

ok

now

[stark wedge]

since $PA : QB = |y_1| : |y_2|$ and triangles $PAR$ and $QBR$ are similar, we get that $PR : QR = |y_1| : |y_2|$ also.

[rocky lotus]

Ann

[stark wedge]

agree or disagree?

[shrewd chasm]

sorry, i don't understand the point no. 2

what is y1 and y2

[shrewd chasm]

.

[stark wedge]

omg why did you say nothing earlier

[shrewd chasm]

sorry

[stark wedge]

y1 and y2 are the y coordinates of P and Q respectively

you have to read and re-read and re-read everything i say at least 5 times.

[shrewd chasm]

I am completely new to geometry...but let's conitnue..

so

PA : QB = |y_1| : | y_2|
and PAR and QBR are similar..
therefor the proportions are equal

am i right?

[stark wedge]

PAR and QBR are similar

but yes

[shrewd chasm]

ok

now..next step

[stark wedge]

there is no next step, we're done

i will say again the book's notation $-(y_1, y_2)$ looks plain wrong to me

[rocky lotus]

Ann

[stark wedge]

maybe you could show a question with worked solution from the book now

[shrewd chasm]

questions are not very calculative..they are MCQs..

[stark wedge]

so there are no worked solutions in the book at all?

in this chapter or this topic specifically

[shrewd chasm]

this topic

[stark wedge]

so it's all just MCQs and answers without explanation.

[shrewd chasm]

no..

i do have a video solution..
but didn't understand fro. there as well

[stark wedge]

unfortunate

[shrewd chasm]

i can watch the video 3 times only

btw what is the point of externally?

[stark wedge]

i have no idea what you mean by "the point of externally".

[shrewd chasm]

my internet slower than a turtle..
wait..the picture is being sent

[noble anvil]

External ratio?

[shrewd chasm]

yes

[stark wedge]

external division means that the "dividing" point lies outside of the segment it's dividing

(it is kind of strange to speak of it that way, but we can still speak of the ratio of its distances to A and B)

[shrewd chasm]

the point can be anywhere on x or y axis?

[noble anvil]

Anywhere along the line joining the two points it's externally dividing.

[shrewd chasm]

but it also cuts internally

[noble anvil]

How does an external point cut internally?

[noble anvil]

this is in no way related to what we are talking about. Does your external point lie on the line you say it's dividing? No. Also, how are you comparing an external point to an internal one and saying that it divides both externally and internally? They're different points.

I suggest you read midpoint and section formula once so your concepts are clearer.

What was your original question?

@shrewd chasm

[shrewd chasm]

.

[shrewd chasm]

.

[noble anvil]

Ok yeah this is just wrong lol.

The minus cannot be outside of the bracket.

[shrewd chasm]

ann taught me very well..and i understood what she saying..
but now confused between externall and internal

[noble anvil]

Just read your notes, it'll be clearer.

Me explaining the concept from ground up wouldn't be as helpful as reading from where you learnt it since I believe it always helps more if the first explanation comes from a familiar source so you don't get confused. After reading that if you still have doubts, you can come and ask them here.

Also, i suggest asking whoever gave you the original question what it means, since their (-) might imply something different than what we're used to (though i believe it doesn't, but anyway).

[shrewd chasm]

it meant external

[noble anvil]

That doesn't really make sense since for that to be true, one of the ratio's components y1 or y2 would need to have the minus sign attached.

[shrewd chasm]

let me ask my mates

[noble anvil]

My suggestion would still be this despite the question being wrong btw.

[shrewd chasm]

.close

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[smoky osprey]

i’m sorry where do i begin for this question

[smoky osprey]

is y intercept y=mx+c

[stark wedge]

y = mx + c is gradient-intercept form

the y-int is the c in y = mx+c.

and the gradient is the m.

so actually all you have to do is read them off.

you do not need that formula you wrote out. (and you miswrote it slightly anyway)

[smoky osprey]

oh.

so do i just state what c and m is

OH WAIT I SEE

[stark wedge]

yes

[smoky osprey]

so m = 1/2

and y = -3

[stark wedge]

c = -3

[smoky osprey]

is that it

oh

yeah c

y=mx+c -> 6=1x3+c and then would it be 6/3 or 6-3

to find c

[stark wedge]

$6 = 3 + c$

[rocky lotus]

Ann

[stark wedge]

what do you think you need to do to get c?

[smoky osprey]

6-3?

[stark wedge]

yes, you need to subtract 3 from both sides to get c = 6 - 3.

[smoky osprey]

from both sides?

like 6-3 and 3-3

[stark wedge]

im talking about the exact same procedure you would use to solve an equation

c + 3 = 6
c + 3 - 3 = 6 - 3
c = 6 - 3

(c = 3)

[smoky osprey]

ohh thank you i get it now

.close

[vale dock]

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[open niche]

Hello i cant find the proof of continuous functions based on riemann integral being integrate able on the part where we need to show that the limit Sum is independent to how ξ1, ξ2, ...., ξn are picked when n goes to infinity

[vale dock]

@open niche Has your question been resolved?

[open niche]

Anyone

[wet night]

tag helpers

I would but i havent done that yet

[frail rock]

I responded in #calculus

[vale dock]

@open niche Has your question been resolved?

[vale dock]

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[zealous plover]

can someone double check my work real quick?

[vivid fiber]

It's probably correct

[zealous plover]

made me doubt myself more with that probably

[vivid fiber]

Nah I'm pretty sure

[zealous plover]

so i aint cooked?

[vivid fiber]

Like how important is that qn?

[zealous plover]

all or nothing

[vivid fiber]

Oof

Meh u ain't cooked bro

It's safe

[zealous plover]

aight bet

LESSSS GOOOO

thanks for the help

.close

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[woven ivy]

Flux

[woven ivy]

I have a question

Basically

the exercise gives me S which is a surface in R3 that's has a domain that surrounds it called E that's which is inside Z=0 and Z=4-x^2-y^2

you have F(x, y, z) =(xz, yz, z^2)

And it asks: to Calculate the flux of F that's coming out the closed surface S And it also asks to calculate the flux of F coming out the superior part of S surface

[vale dock]

@woven ivy Has your question been resolved?

[woven ivy]

<@&286206848099549185>

[woven ivy]

Welp

[pliant halo]

okay what comes to mind when you've been specified a volume, flux and vector field?

(they want you to apply a theorem)

[woven ivy]

Yea

[woven ivy]

So basically

For the first one they want me to apply

The divergence theorem

[pliant halo]

yep

[woven ivy]

For the second I'm not sure

Because it's open

So i can't use that again

But idk the formula

💀

Like I'm not sure

[pliant halo]

For the second you can think of what the flux through the closed surface actually is

it's the flux through top surface+flux through bottom surface

so if you can find flux through bottom surface (relatively easy)

[woven ivy]

Oh ok

So i have the total flux

Now

[pliant halo]

yeah

[woven ivy]

So i just gotta take away the bottom one?

[pliant halo]

yep

[woven ivy]

The bottom one has z=0

But it's still open

Or nah

Cuz I'm in R3

Or do i look at it like 2d or something

[pliant halo]

it's open but you can find the flux easily no?

[woven ivy]

I haven't done many flux stuff yet

I started 2 days ago

[pliant halo]

yeah, flux through a 2d surface

[woven ivy]

Ok

So double integral

[pliant halo]

I mean, even for the "3d flux" you're still taking it over the surface

[woven ivy]

0 2 pi

And the r

[pliant halo]

it's just that the divergence theorem relates it to a volume integral

[woven ivy]

Ohhh

Wait can i apply divergence on 2d

Or nah?

[pliant halo]

No

divergence is for a closed surface

[woven ivy]

Gotcha

[woven ivy]

Makes sense

[pliant halo]

like a sphere or smth

[woven ivy]

Yea

So wait

[pliant halo]

for open you just gotta rawdog it

[woven ivy]

Crap

Well

I didn't do opens yet

[pliant halo]

but it'll be easy in this case since z=0

[woven ivy]

I kept finding closed ones lol

[pliant halo]

like you pointed out

[woven ivy]

Yea?

I mean it's a double one

But i don't put divergence in

Wait

Lemme write

[pliant halo]

plug in z=0 to your vector field 🙏

[woven ivy]

Wait it's 0 isn't it

[pliant halo]

yes

[woven ivy]

Or something

Damn

Welp

If i had to do the upper one

Would it be complicated?

[zinc spear]

more complicated than the lower disk

but still doable

[woven ivy]

True 💀

[pliant halo]

I will point you to some resources on how to take flux

[zinc spear]

however you can shortcut with the divergence theorem

[woven ivy]

Thanks I'm still figuring them

[woven ivy]

Gotchu thx

Also for range

[zinc spear]

total flux = flux trough bottom + flux through top = 0 + flux through top

[woven ivy]

In the integral

Radius*

Mb

I noticed he does z=0

And then radius

[woven ivy]

Gotchu

So to find radius he unites the 2 conditions

Z=0

And the function

Or am i wrong

Cuz in another one he had a cone and sphere

And just used sphere radius

Without caring about cone

He was asking the total flux cuz he did divergence

[zinc spear]

yeah this way, you will find the curve at the intersection of the plane z = 0 and the surface z = 4-x^2-y^2

[woven ivy]

Wait lemme send it

[zinc spear]

you can then deduce the radius from the curve you find

[pliant halo]

https://www.youtube.com/watch?v=phk05iSMezA&list=PL4C4C8A7D06566F38&index=27
https://www.youtube.com/watch?v=WfEQabCGAqI&list=PL4C4C8A7D06566F38&index=28
https://www.youtube.com/watch?v=wu8kXZSAp20&list=PL4C4C8A7D06566F38&index=29

[woven ivy]

Wait it's messed up

Ima send original too

[woven ivy]

Thanks!

[woven ivy]

Ok thx

one last thing

For the third integral

He used spherical coordinates

And shoved them into cone

Then found the intersection

But in the one i showed at the begin

He does that to the original function because

You got only that basically

Right?

You don't count the sheet on z=0 at all

[zinc spear]

i'm not sure what you mean but I can say that if you want to compute the integral of the divergence inside the volume in the og problem to use the divergence theorem, then using cylindrical coordinates would be more appropriate than spherical imo

[woven ivy]

Makes sense

[zinc spear]

that's where the radius comes from when i tried it

[woven ivy]

I meant for this

He uses spherical coordinates for the cone

But for the other

He uses cylinder coordinates