help-4

[storm coyote]

Other one might cause some issue tho.... still a lengthy one. I will try it nonetheless

[stark wedge]

uhhhhhh god. ish? i think?

defintely would not write it in this way

[storm coyote]

Idk combinatorial proof means?

[storm coyote]

Ohh ok.

[keen tundra]

you find a combinatorial problem that this expression is the answer to

and then you solve that combinatorial problem in a different way

[pearl nacelle]

I did it out

It’s not so bad

[storm coyote]

I am bad at combi problems 😭

But sure

Lets try

[storm coyote]

Ok. I will try after my current integration method....

[pearl nacelle]

I don’t think a combinatorial proof will be easy because the answer isn’t very nice

[storm coyote]

True

I am sending the answer I got btw

From the long method

|| $\frac{2^{n-1}(n^2-n+2)-1}{(n+1)}$ ||

[rocky lotus]

Double_mytrouble

[storm coyote]

I guess spoiler doesnt work for texit....

[storm coyote]

oh nice. It actually isnt that bad.

Thanks.

[pearl nacelle]

You can reduce it further too with (n-k+2-2)

[woeful trench]

,texsp $\frac{2^{n-1}(n^2-n+2)-1}{(n+1)}$ will hide what you write, at least

[rocky lotus]

@woeful trench

[storm coyote]

I think I got it. These are much easier.

I am just waiting for the combinatorial argument

@keen tundra Brother, what question does it become in Combinatorial proof?

[keen tundra]

didn't think much about it

[storm coyote]

oh

np

[keen tundra]

but this is a possible solution to problems like this

[storm coyote]

Our next question is to prove Hockey Stick Identity using PnC

Lets see if I can use generating functions there

and then backtrack

Because I solved the no of integrals solution in a range using hockey sticks previously lol

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[pearl nacelle]

Nice

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[hexed tangle]

If you have

VBB >= IB(RB + (1+ Beta)) + 0.7
and
VCC > (Beta)*IB(RC + RE/alpha) + 0.2

Then you can not simply solve for IB in the top expression and plug it into the bottom one with VCC right? Since one is >= and the other is >

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[hexed tangle]

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[crystal flax]

No context, what could be the purpose for this?

[crystal flax]

I don'þ get it either ;v;

[mild kernel]

why no context

[stark wedge]

no context no explanation tbh

[mild kernel]

don't you think that'd be helpful

[crystal flax]

p is always 0 or 1

[stark wedge]

this is a jumble of symbols

[crystal flax]

I was given no context either bro 😭

[mild kernel]

by who

[crystal flax]

a frend ;v;

[mild kernel]

sure

[crystal flax]

gurl 😭

idk dude 😭 I can't figure out what i() or I() do so ;v;

[mild kernel]

you dont figure it out, you get told

[crystal flax]

it's basically useless w/o it

[crystal flax]

huh

[mild kernel]

if you read whatever paper you pulled this from on google image search you'd probably be able to figure it out

[stark wedge]

yeah just leave this crap alone tbh

[crystal flax]

I didn't thoughh 😭😭😭

[crystal flax]

fs it's sucky 💀

how do you end this ticket

[tropic reef]

dot close

[crystal flax]

tyyy

.close

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[dull cipher]

what do i do here

for the infinite

[stark wedge]

the graph keeps going on forever to the right

[dull cipher]

yeah

[dull cipher]

in a straight line?

[stark wedge]

yes

[dull cipher]

ah

thanks

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[dull cipher]

.reopen

[vale dock]

✅ Original question: #help-4 message

[dull cipher]

what are the zeros

[stark wedge]

where does your function intersect the x axis if at all

[dull cipher]

it doesnt

that means no zeros?

[stark wedge]

are you 100% sure

you didn't draw its graph completely yet

which x, if any, makes -3x+36=0?

[dull cipher]

-3(8) + 36

y = 12

[dull cipher]

which is here

and then the inifnite

it goes in straight line

[stark wedge]

... which you didn't draw.

and also you did not answer my question

[frozen ledge]

croissant

[dull cipher]

i did

.close

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[eternal hawk]

Why is this question this way.

The product of 3 and a number increased by 4 is 26

[eternal hawk]

I checked this with quite a few of my friends who are far farther into math than I and they all agreed that this question is written really shitty and ambigiously

I feel like if it was written "The product of 3, and a number increased by 4 is 26"

[frozen ledge]

yea

the comma is important

[eternal hawk]

I need to like avoid missing that on the Test due monday

So I guess I'll just make a note page specifically catered to this problem

[frozen ledge]

the product of 3 and a number, increased by 4 is what you did

[eternal hawk]

since it'll be written the same

but with different numbers

Pearson so fun..

Not everyone fell for that question though that I asked.

My friend and their SO each had different answers.

My friend who's been helping me study had the same answer as me while their SO came up with the correct answer when asked so like

there must be SOME way to tell

Also this problem since the denominators are already identical all we're doing is solving the fraction?

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[silent dune]

got a question about an alternative proof of the MCT, and wondering if im not missing anything in my reasoning

[silent dune]

before stating it, i make use of a key idea that's a sort of MCT but for sets that i proved; which i use as a lemma

And here's my attempt for a proof of the MCT using this^

[rocky lotus]

Aslan

[silent dune]

thx

[tepid swan]

i see no problem here

after deep read

it's precesly the monotone convergence theoreme

[silent dune]

yeah thats what MCT stands for

[tepid swan]

ye

you're skipping anything

i guess

[silent dune]

hm?

[tepid swan]

hm ?

[silent dune]

not sure what u mean by this

[tepid swan]

i mean

that i think you are skipping anything

like

its perfectly wrotes

[silent dune]

oh i see, youre saying im writing it clearly and in a way that doesnt make any big leaps in reasoning?

[tepid swan]

yes

your 5 steps are clear

and nothing can be discussed

[silent dune]

damn okay

well i guess i felt the proof "must be wrong" in the sense that other proofs make use of measure theory and mine just uses simple properties of sups and limits

[tepid swan]

checked it

step by step

[tepid swan]

its make the proof way more elegant

[silent dune]

well thanks

[tepid swan]

np mate

[silent dune]

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[dull cipher]

guys how can i find the a of this square root function?

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[restive void]

Question:
I am hearing different people saying different things on whether a midpoint rule Riemann sum overestimates or underestimates and integration of a function from a to b . . .
Some said that you can determine that by knowing the function's monotonicity over an interval along with the interval's respective concavity . . . others said that you can't easily determine whether it'd be an under/over-estimate based on the graph's features . . . so what'd be correct?

[violet needle]

well maybe you should think about what you're even asking

in the limit the riemann sum will approach the value of the integral

but for finite sums you could see all kinds of things depending on which function you have and how you've chopped up the x-axis into parts

[restive void]

Yes, I mean a finite sum

I will show you a generalization I found from one of the resources on YT

[violet needle]

if you do equl sized intervals then the function f(x) = -x+1 from 0 to 1 will not overestimate nor underestimate

but you can come up with other scenarios that behave differently

actually no

I mean one way to think about this for evenly spaced intervals is whether the average of your function values is the true average of the function

[violet needle]

and actually yeah I think this will work as I said for that reason

but why are you trying to figure this out

maybe just do some numerical exploration

and interpret the results

that's the best way to learn

[restive void]

I see, so does this conclusion stand for finite sums of rectangles of equal widths:

[violet needle]

no not in general

[restive void]

That's right, but I was caught off guard when I saw people saying conflicting things on it

[violet needle]

people like to open their mouths without thinking

myself included

[restive void]

and that if I applied it to some examples maybe they wouldn't apply as a generalization

[restive void]

It's for the calc course I am taking

They ask for justifications on whether a finite sum of an over/underestimate

[violet needle]

You could have a small number of intervals and see what happens

[restive void]

assuming the function is continuous - like they could give you a table of values to work with

[restive void]

Right and left sums are simple in that regard

[violet needle]

surely if you only have one interval then it undershoots for x^2

[restive void]

so are trapezoidal sums

[violet needle]

you have access to a computer and limitless time to work on your assignment

I'm just a guy on my phone on a metro whose going to get sushi

you're asking a very open ended question

[restive void]

Haha, by the way the help is much appreciated . . . I'll do some experimenting with some graphs and see what I come up with, perhaps a generlization

[violet needle]

Given a partition you can always design a function that disrespects your partition in whichever form you like

but if you increase the amount of points you approach the true value

[restive void]

Yeah like $||P|| \to 0$

[rocky lotus]

SeaSamak

[violet needle]

Monotonicity is a weak constraint because you can still disrespect a partition by forming steep slopes at just the right points

[restive void]

Right, thanks! I do some more searching on this and'll do some testing

Have a nice day! bye

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[violet needle]

you too!

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[glad minnow]

c.

[glad minnow]

is this the right way to do this?

[honest stone]

Is that GPT?

[glad minnow]

yes

[honest stone]

!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).

[glad minnow]

how do i do this?

<@&286206848099549185> im gonna be away but if u can give a starting point b4 the channel closes, thanks

[pure pulsar]

Alright

[pure pulsar]

First and second derivative of the proposed solution

And then substitute it into the differential equation and solve

To check lhs and rhs

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[burnt mulch]

dimension of a cubiod is a,b,c
then prove 1/V = 2/S(1/a + 1/b + 1/c)

[burnt mulch]

S = surface area

total surface area*

S = 2( ab + bc + ca )

[storm coyote]

S=2(ab+bc+ca)

[burnt mulch]

V = a x b x c

yes

[storm coyote]

You could just open it....

Bad idea...

[burnt mulch]

hm?

i would recipropal it ig?

[storm coyote]

wait it cancels out

see 2/S =1/(ab+bc+ca)

[burnt mulch]

how

[storm coyote]

(1/a +1/b +1/c) =(ab+bc+ca)/abc

cancel out (ab+bc+ca)

you are left with 1/V

[burnt mulch]

is it not supposed to be S/2

[storm coyote]

no.....

[burnt mulch]

V = abc

[storm coyote]

its in denominator

[burnt mulch]

wdym

its S = 2(ab+bc+ca)

[storm coyote]

wait I am sending in writing...

Sorry for bad handwriting

[burnt mulch]

you recipropal both side?

[storm coyote]

No

I just solved

Do not reciprocal

[burnt mulch]

Doing S/2 to 2/S is recipropal

wait lemme explain

[storm coyote]

Send the question original

I think you might have wrote it wrong...

[burnt mulch]

S = 2(ab + bc + ca)
(divide both side 2)
S/2 = ab + bc + bc
recipropal both side
2/S = 1/ab + bc + ca

[storm coyote]

Well yeah

[burnt mulch]

thats what i am saying

[storm coyote]

I mean I just put S

[burnt mulch]

recipropal.

[storm coyote]

In 2/S

[storm coyote]

Misunderstanding sorry

[burnt mulch]

idk what you did after that

[storm coyote]

Btw which grade are you in if you dont mind me asking...

[burnt mulch]

9th

[storm coyote]

Ok

See 1/a +1/b +1/c =(ab+bc+ca)/abc

You understand this ?

[burnt mulch]

but it was 1/ab + bc + ca

[storm coyote]

....

[storm coyote]

That was 2/S

I took the other part

[burnt mulch]

yeah

[storm coyote]

That was not 2/S

Get it?

[burnt mulch]

no

[stiff vale]

hello every one

[burnt mulch]

hey

[burnt mulch]

i am trying to understand what you did after this

[storm coyote]

There are two product of fraction. One 2/S and other (1/a +1/b +1/c)

First you found out 2/S

Which was 1/(ab+bc+ca)

Then you take the other one

[burnt mulch]

its 1/ab + bc + ca bro

[storm coyote]

Bro thats 2/S

[burnt mulch]

yeah

[storm coyote]

I am talking about the other think you wrote

[stiff vale]

can talk me what the question

[burnt mulch]

V = abc
1/V = 1/abc

[stiff vale]

yes is true

[burnt mulch]

welp

[storm coyote]

,rccw

[rocky lotus]

Couldn't find an attached image in the last 10 messages.

Oct 19, 2025, 3:06 PM

[stiff vale]

but the group that abc includ

[burnt mulch]

how do i prove 1/V = 2/S(1/a + 1/b + 1/c)

[burnt mulch]

from where the 1/a + 1/b + 1/c cane from

[storm coyote]

See the photo. What I wrote is your question. Then I am saying that part

[storm coyote]

From your question....

[burnt mulch]

wait

i get it now

[storm coyote]

Lock in bro

[burnt mulch]

You put the 2/S value we found earlier

maybe

[storm coyote]

Yes

[storm coyote]

I put that in place of 2/S

In other words I substituted it

[burnt mulch]

(1/ab + bc + ca)(1/a + 1/b + 1/c)

[storm coyote]

Yes

[burnt mulch]

what to do after this?

[storm coyote]

You understand fraction addition of variables? (Geneuinly asking)

[burnt mulch]

yes

finding LCM to add

[storm coyote]

Yes

[burnt mulch]

of variables

[stiff vale]

ok

[storm coyote]

You know LCM of a,b,c is abc

[burnt mulch]

how can i find variables LCM

[storm coyote]

?

[burnt mulch]

oh right

[storm coyote]

You multiply them

[stiff vale]

yes

is trues

[burnt mulch]

alr so

[storm coyote]

|| (Good lord, I wont be a good teacher) ||

[burnt mulch]

@storm coyote I GOT IT

[storm coyote]

Ohh ok

[burnt mulch]

i find lcm then its cancel out

[stiff vale]

and aftre that you find the

[storm coyote]

Yes

[storm coyote]

Then you are left with the required value

1/V

[burnt mulch]

(1/ab + bc + ca)(3/abc)?

i forgot how to do addition on variables

with lcm

well

[storm coyote]

No need to add

wait how did you get 3?

[storm coyote]

Where did this come from ? 💀

[burnt mulch]

1/abc + 1/abc + 1/abc

[storm coyote]

Nope

Nope

1/a

[burnt mulch]

you gotta tell me how to do addition man

[storm coyote]

Then multiply bc with both numerator and denominator

[burnt mulch]

i kinda forgot

[storm coyote]

Watch a youtube video man

[burnt mulch]

why bc

why not abc

[storm coyote]

9th grade, you shouldnt forget these. Geneuine advice

[storm coyote]

Because a is already there

1/a

A is there

So multiply bc

[burnt mulch]

i never put much attention to math

since it was easy

it kinda become lil hard in 9th, i just need to lock in a bit

[storm coyote]

Lol, you will get it eventually. My brother is like this (although he is in 7th standard)

[burnt mulch]

oh right

how i can forgor ts

i get it now

[storm coyote]

1/a =bc/abc

Similarly....

1/b =ac/abc

1/c=?

You write

[burnt mulch]

1/c = 1 x ba/c x ba

ba/abc

[storm coyote]

Yes

Now add them

(They have common denominator)

[burnt mulch]

ab + bc + ca/abc

cancel out

[storm coyote]

Put a bracket it looks neat (suggestion)

[burnt mulch]

i know i just trying to msg fast

[storm coyote]

Only ab+bc+ca cancels out

[burnt mulch]

yes

it became 1/abc

[storm coyote]

You have 1/abc left

[burnt mulch]

and 1/V = 1/abc

hence proved

[storm coyote]

Yes

Exactly

Good j*b

[burnt mulch]

@storm coyote you should gimme a tip to increase math analysis man

[storm coyote]

Some people get triggered by || job||

[burnt mulch]

i got 98 in 8th but in 9th half yearly i got 79

[burnt mulch]

hah brainrot ig

[storm coyote]

You live in which country?

[burnt mulch]

india

[storm coyote]

For the curriculum.

[storm coyote]

CBSE i am guessing

[burnt mulch]

yes

[storm coyote]

You will take science in 11th?

[burnt mulch]

yes

since i dont like lang 2

i mean i atleast got 96 in science

[storm coyote]

I dont too but math after 11 is different.

[storm coyote]

Take PCB

In 11th

[burnt mulch]

and 79 in math because i got stuck on a very simple question and lost marks because of time

[storm coyote]

Understandable

[burnt mulch]

is there even pcb?

[storm coyote]

Yes

[burnt mulch]

i thought it was pcm and pcmb

[storm coyote]

There is just a simple math paper (not on advanced level)

[storm coyote]

Math is there but not on Joint level

[burnt mulch]

but i really wanna take math instead of biology

[storm coyote]

Well if you are in 9th, you giving NSEJS?

[burnt mulch]

i suck at drawing diagrams and memorizing

[storm coyote]

Or SEHSS?

[burnt mulch]

no idea what they are

[storm coyote]

Diagrams disapper after 11th

Only memorisation

[storm coyote]

These are olympiads

[burnt mulch]

i believe i will get good in math, i never study math even a bit after all

[storm coyote]

Bro, beleive me it doesnt work like that

I also was like that

Never studied math till 10th

In 10th also I beleived math will be easy

But then in pre board I got 92 ( tripped up a Geometry question of 5 mark and rest silly)

I am weak in geometry still because I didnt study

[burnt mulch]

i see

it always silly mistakes in math

[burnt mulch]

atleast for me

[storm coyote]

Btw if you dont mind we can talk in pc, dont occupy this help channel....

[burnt mulch]

sure

how do i close again?

[storm coyote]

Write .close

[burnt mulch]

.close

[vale dock]

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[exotic jay]

I need help @everyone

[pallid oasis]

what are you having issues with?

[exotic jay]

Leave it

It’s done

[pallid oasis]

OK

[tawny moon]

so you no longer need help?

[tawny moon]

!done if so

[vale dock]

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

[pallid oasis]

yup

[exotic jay]

.close

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[glad minnow]

ty i overcomplicated it haha

[glad minnow]

.close

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[digital vine]

Anyone know how to make the answers appear normally?

[cinder creek]

,rotate

[jolly ledge]

I think it might help if you gave the context of this calculation. is it a polynomial, or a matrix calculation, or something else?

[digital vine]

Polynomial simultaneous equation

The answers correct but i need to subtract and add the 2 numbers they give for each x

Might be cause its solved in a matrix but wonder if theres a setting that solves it automatically

[past forge]

what do you mean?

[spark hamlet]

In the answers it gives it shows like

A decimal minus a decimal or a decimal plus a decimal

I think ranes asking how do you make it automatically do that

[digital vine]

Yeahh

[past forge]

ohhh

[spark hamlet]

I can’t find anything

[past forge]

i have not worked with that calculator so i am clueless

[digital vine]

I hate it

[past forge]

what model is it?

[digital vine]

[spark hamlet]

I have that calculator I’ll look at it really quickly

[digital vine]

Ti84 gdc

[digital vine]

Appreciated 🙏

Dont wanna get stuck on that during a test

[spark hamlet]

Yeah

Ok so I tested it on my calc

And it auto added everything.

How did you make the x y and z into x1 x2 and x3?

Or is that a different mode

I’m on the systems of equations solver

Are you sure you’re in normal mode and not scientific or engineering?

That’s all I can find

[digital vine]

I have 4 unknowns it automatically made me do a matrix

Maybe thats why

[spark hamlet]

Yeah

[vale dock]

@digital vine Has your question been resolved?

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[restive void]

Hi, I asked this question previously here and just wanted more clarity as my research still has led to conflicting results . . .

[restive void]

I was hearing different people saying different things on whether a midpoint rule Riemann sum overestimates or underestimates and integration of a function from a to b . . .
Some said that you can determine that by knowing the function's monotonicity over an interval along with the interval's respective concavity . . . others said that you can't easily determine whether it'd be an under/over-estimate based on the graph's features . . . so what'd be correct?

[restive void]

Based on my initial knowledge and my textbook, answering that question wouldn't be very concrete. Based on other resources, I found the following conflict:

I will send the other source shortly

[keen tundra]

Makes sense

[restive void]

Yes this is the other source

[restive void]

It seems that the sources actually don't conflict

[keen tundra]

Yeag

[restive void]

but then why is it that when we have a concave up function, the midpoints sum is an underestimate?

and overestimate for concave down?

Are those sound generalizations?

[vale dock]

@restive void Has your question been resolved?

[restive void]

But we can't we suppose the area outside is equal to the gap area inside

[keen tundra]

I was trying to find a way to explain this

formally

[restive void]

Like these

I recall in some case there was a finite middle sum that equaled the exact area under a curve

[keen tundra]

the point is that the second derivative determines if its an under estimate or an over estimate

[restive void]

It was like a parabola

[keen tundra]

but that is not monotonic

[restive void]

oh, right

[restive void]

I think I am getting it now

[keen tundra]

actually it might not matter

because the second derivative of a parabola is constant

[restive void]

So the area outside the concave down rectangular estimate is >= the gap under the curve for the rectangle?

So the blue area is greater than the red one due to down concavity?

[keen tundra]

yeah

you can prove it formally

I am a bit busy right now for that

sorry

[restive void]

I wouldn't know how to prove it, that'd be interesting . . . but as for the intuitive part I might've got it

The weird thing is that there isn't much of a mention of this online . . . like the proof

But only some mention the rule without state of reason

or proof

I really thank you for the help perhaps we can potentially have the proof from you if ever possible and feasible another time

but as for now the question is cleared!

Thanks!

.close

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[stark wedge]

!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

[signal chasm]

god knows

what i have to do

1

and sorta 3

idk what method the ms has

its really weird

[stark wedge]

do you want to look over the mark scheme or would you like to hear me yap about it independently

[signal chasm]

u yap abt it independently

[stark wedge]

ok

[signal chasm]

ms makes no sense

[stark wedge]

1-interval trapezoid rule is exact when the function you're integrating is linear

thus when you subdivide, trapezoid rule is exact if the function is linear individually on each piece

that make sense so far?

i got more yaps to come but i want to make sure you follow @signal chasm

[signal chasm]

yh

makes sense

[stark wedge]

yeah ok

so when you divide [0,1] into n equal intervals

the partition nodes are 0, 1/n, 2/n, ..., (n-1)/n, 1

ie these are the endpoints of the subintervals

now the key thing is that your function has "breaking points" at 1/3, 1/2 and 3/4

[signal chasm]

yes

[stark wedge]

what you want is that these breaking points should each be a partition node

as i will illustrate shortly

[signal chasm]

ik the trapezium rule formula

if ur deriving it

[stark wedge]

no im going visual

[signal chasm]

yh

[stark wedge]

this is what happens when 1/3 isn't a partition node

you get this bullshit with trapezoid rule overestimating the area

we DONT want that shit happening

same story at the other marked x points

[signal chasm]

wait what

😭😭

but i thought

it was inevitabke

[stark wedge]

no here's the thing

[forest topaz]

Is it darboux sum?

Sorry for interpreting

[stark wedge]

if 1/3 is a partition node, everything's A-OK

[signal chasm]

but

what is the partition node

like

thats just the height right

[stark wedge]

the partition nodes are 0, 1/n, 2/n, ..., (n-1)/n, 1
ie these are the endpoints of the subintervals

[signal chasm]

like yk the h/2

[signal chasm]

ok

so a constant gap

in the x axis

right

[stark wedge]

when i say "partition nodes" i mean specifically the x-values at which you sample your function

[signal chasm]

do they need to be equally distanced

[stark wedge]

the question says yes

[signal chasm]

ok yh

i understand

[stark wedge]

but that wasnt even my point really

my point was for you to see what happens when 1/3 is not a partition node (red) vs when it is (green)

capisce?

[signal chasm]

i really dont get

why theres an overestimate

when its not at 1/3

is it jst

a memorisation thing?

[stark wedge]

no it isn't a memorization thing

i want you to LOOK AT the sketches i have drawn for you

do you see this little sliver of extra area

[signal chasm]

yes

but why does that occour

like

well

acc

i can kinda see why

[stark wedge]

you should see the little trapezoid right there that i went and drew for you 😭

[signal chasm]

no i knew that 😭

i meant like

why it goes

from red to red

and then i realised

its making trapeziums

[signal chasm]

and like with the other its perfectly

alligned

[stark wedge]

yes exactly

the whole point of all this was that in order to be exact we need 1/3 to be one of our partition nodes

and the story is the exact same for 1/2 and 3/4

do you understand now

[signal chasm]

ok

yed

yes i understand

[vale dock]

@signal chasm Has your question been resolved?

[vale dock]

@signal chasm Has your question been resolved?

[barren sierra]

Help I

Hello

I need help

[vale dock]

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[dull cipher]

A artist invested 2000$ to purchase necessary equipment for the fabrication of wicker basket. The produciton cost of each wicker basket costs him 5$

[dull cipher]

What is the rule of this rational function that helps him calculate the average of production of a wicker basket according to the amount of wicker baskets produced

f(x) = 2000/x + 5 ?

.close

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[zealous gust]

i need help understanding the last exercise i need to do like dilatations reflection and shifts but i don't understand under what circumstances, how and when to do it

[zealous gust]

my test is in Tuesday i spent one week trying to understand this but as of now i only understood the exercise 2 (all of it) and the 1a 1b nd 1e
i didn't get exercise 3 and 1c and 1d

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[ornate dragon]

Need to draw the set V and write in three different "orders" the integrals

so I know 1 will be a tetrahedron so we need to figure out the vertices

we'll need four points

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@ornate dragon Has your question been resolved?

[wintry raptor]

Why is it in another language???

[ornate dragon]

...because i speak spanish?

[wintry raptor]

Could you translate to English

[ornate dragon]

.

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@ornate dragon Has your question been resolved?

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@ornate dragon Has your question been resolved?

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[radiant raft]

Is finding the limit of a function just finding at which x y points its at? Ex. lim on pics + graph. The x is nearing 5 at y 1/2

[cinder creek]

In a way, yes, but the limit is the value when you approach a point, not the exact value at the point.

[steel edge]

it means to find an approximate value of the function approaching a certain value in the domain

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[vernal jewel]

Hi, is there any other way to solve this integral besides using this kind of substitution because it would never come to my mind to substitute it like this and I need to solve other integrals of similar type so I need another way? Thanks in advance.

[glass kelp]

uhh

why are there two questions here

[meager verge]

.close

[glass kelp]

hm

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[umbral falcon]

Two things:

1. Is everything I’ve done up to this point valid?
2. Where do I go from here??

[modest hemlock]

Yo

Bruh why u writin csc😂

so ur trying to find

integral of arccosec

,w integrate arccosec(x)

[rocky lotus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ripico

[umbral falcon]

Bro

Anyone can plug anything in a calculator

I’m here trying to do the derivation

<@&286206848099549185>

Anyone???

<@&286206848099549185>

[modest hemlock]

yo

[modest hemlock]

i know lmao

[modest hemlock]

which part exactly

gng yk u can js ask gpt right

[umbral falcon]

It’s the last row

[modest hemlock]

u think ur wrong or ur asking to double check?

[umbral falcon]

I don’t trust that digital duffus

[umbral falcon]

For double checking

[modest hemlock]

ok wait

uh yeah

its correct

[umbral falcon]

So the next part of my question was how do I continue from my last line of math?

[modest hemlock]

Alr

idrk tbh

[wild linden]

Note sgn(cosh(θ)) is a constant.

So essentially this reduces to an integral of the sign function

Which is not correct

[modest hemlock]

oh fr

[wild linden]

I assume the second substitution is the problem. cosh is not a surjective function.

@umbral falcon

In other words there are values of x for which no real value of θ satisfies the equation x = cosh θ

[umbral falcon]

I see, so what would the fix be? Should I do a trig sub instead of hyperbolic sub?

[wild linden]

If you can manage with a tan sub, sure, but a sin or cos sub will have the same problem

[umbral falcon]

I’ll see if I can manage a tan sub

Wait

How would that even work in this scenario?

I have x^2 - 1

So x=sec(t) would be my only option

But sec(t) isn’t surjective

I might have to scrap trig subs entirely at that point

So there must be another viable method

[wild linden]

@umbral falcon looking at the form of the WA answer, after integration by parts try taking the 1/|x| and changing it to x/sqrt(x^2). Now perform a u-sub u = x^2

That might just get you somewhere

[umbral falcon]

I’ll try that out

Wait

Do you mean x/|x| becomes
x/sqrt(x^2) or 1/|x| becomes
x/sqrt(x^2)?

[wild linden]

Sorry 1/|x| becomes 1/√(x^2)

[umbral falcon]

Ahh ok let me give it a go

[wild linden]

||You should wind up with du/√u√(u-1), after the u-sub is done, it looks like you'll need to do the substitution v = u^2 + 1. Then you'll get to dv/√(v^2+1), which is ln|√(v^2 + 1)+v|+C||

After that it's just a matter of rewinding the definitions.

Need to go afk

Marked my work as a spoiler

[vale dock]

@umbral falcon Has your question been resolved?

[umbral falcon]

The u-sub is helpful, but then that v-sub, where are you getting u^2+1 from, if sqrt(u) • sqrt(u-1) is equal to sqrt(u^2-u)?

[umbral falcon]

Wait

I have an idea, but correct me if my thinking is wrong

We still do the whole 1/|x| being equal to 1/sqrt(x^2)

But then we multiply sqrt(x^2) by sqrt(x^2-1)

We get sqrt(x^4 - x^2)

Then we can do a trig sub

Where a=x^2 so a^2 is x^4

Thoughts?

[pearl nacelle]

Id try an integration by parts

The derivative of csc^-1 has a nice closed form so

Oh you tried it, but your derivative is wrong

[umbral falcon]

How?

Oh shoot, forgot the negative sign

But that shouldn’t be a big deal

I’ll just write a plus sign for the integral instead of minus

Ok just fixed it

But other than that, it’s still a problematic integral

That pesky x/|x|

[pearl nacelle]

No I don’t think there should be an absolute value at all

[umbral falcon]

Actually, it’s very necessary to ensure the derivative is always negative, because f’(x)<0 for all values of x

Meant negative

[pearl nacelle]

csc(x)=1/sin(x), this means arccsc(x)=arcsin(1/x)

Then it’s just chain rule

Youll always have a negative f’

Itll be what

-1/(x^2(sqrt(1-(1/x)))

Something like that

[umbral falcon]

But if you take the derivative of both sides, it should be possible to reduce everything to the same appearance, hence there isn’t any benefit in finding the integral of arcsin(1/x) and then applying int by parts.

[pearl nacelle]

I think your derivative of arccsc is just wrong lol

Arcsin(1/x) and arccsc(x) are the exact same function

[umbral falcon]

And so too are their derivatives

[umbral falcon]

We can either go through a lengthy process and prove that the derivative is correct, or you can do Google or even YouTube searches

The absolute value is included

[pearl nacelle]

But the derivative you have is NOT the same as what I got which I just looked up

[umbral falcon]

Show me please

Send an image

[pearl nacelle]

You have an x term in the numerator

There shouldn’t be

[umbral falcon]

Ohhhh

That’s correct

Because it’s int by parts

[pearl nacelle]

You can factor out x^2 in the root and youll have |x|xsqrt(1-1/x^2)

Which then you can integrate more easilt

[umbral falcon]

.close

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[plush stirrup]

hii! so i have a test tomorrow and theres some things on the practice sheet that i dont quite understand. i would appreciate if somebody could help me understand them a bit. its mainly just inequalities that i dont understand

[plush stirrup]

do i have to send a picture when i ask btw?

[plush sentinel]

Helloo, it would be best if you sent the practice sheet itself ^^

[plush stirrup]

tyty

dont mind the doodles when i send. its hard for me to focus a lot

[stark wedge]

it's not obligatory but encouraged

[plush stirrup]

its numbers 2-6 that i dont get

once again dont mind the doodles 😭

[plush sentinel]

For number 3 i'm like 51% sure you combine like terms

[plush stirrup]

ty. ill try it and see what i come up w

the answers are on the back but i wanna see if i can get the same answer

[plush sentinel]

That's great!

[timber venture]

number 1 is factorising

i think

[plush stirrup]

ty. i think i did smth wrong 😭

[plush stirrup]

oh ty! i think i kinda got the hang of that. i just gotta memorize which one is which

[plush sentinel]

How so?

[plush stirrup]

i suck at math and did this

the answer is 10

[sage gust]

Which question are you on?

[plush sentinel]

I think you should write it on a separate piece of paper yes

So like

[sage gust]

Oh..

\begin{align*}
\frac{2}{3}x - \frac{1}{2} = \frac{7}{6} + \frac{1}{2}x
\end{align*}

You can isolate x on one side and a constant on the other..

[rocky lotus]

Maddie

[plush sentinel]

2/3x - 1/2x = 7/6 + 1/2 and start from here

[plush stirrup]

im stupid 😭 thats what i did wrong

tyall

[sage gust]

all good

Before you go, let us know your answer

[plush stirrup]

okay

i got 1/6x=5/3. i subtracted 2/3x - 1/2x and added the other side. did i do smth wrong or do i just have to simplify or smth?

[sage gust]

I think there may be something wrong.

You got x/6 correct

but 5/3?

Oh you simplified

Okay, ure good

[plush stirrup]

ohh okay

[sage gust]

so to find "x"
multiply 6 on both sides

because x/6 = 5/3, you can remove the fraction from the x by multiplying by 6.

And normally, its well, what we do if we want to find x.

[plush stirrup]

OH okay i just put it in on my calculator

and got 10 :D

[sage gust]

Thats right :3

[plush stirrup]

sorry but how would i know which number to multiply by? is it the greatest common multiple?

[plush sentinel]

lcd

lowest

[plush stirrup]

mb i get it mixed up

[sage gust]

Consider 1/2.

To make up 1, you need 2 halves right?

So simply multiply by the denominator on both sides.

This is also quite common if you're solving for x where x exists on a denominator on one side.

For example

\begin{align*} \frac{x^2}{2} = \frac{5}{3x}\end{align*}

ok im very rusty with latex

[rocky lotus]

Maddie

[sage gust]

ill leave it at this.

[plush stirrup]

XD its okay. tyall so much!

[sage gust]

Have you learnt uh polynomials, if you haven't its okay.

[plush stirrup]

i have at some point but forgot 😭

[sage gust]

To isolate x on one side, simply multiply 3x on both sides. Then you'd have 3x^3/2 = 5

Then just multiply 2 on both sides to remove the half fraction.

then you'd get 3x^3 = 10

divide by 3 on both sides x^3 = 10/3

You'll get the hang of it.

[plush stirrup]

tyty!

does this channel delete the messages? i might have to look over it again tmrw

[timber venture]

This one's factorising (for when you do it)

[plush stirrup]

tysm!

[sage gust]

No. Just do .close

If you need any help you can still DM me if needed

[timber venture]

no, they stay here, just remember the number (you may have to scroll a bit)

[plush stirrup]

okay :D thats good to know

so would number 4 be kinda similar to this too?

[timber venture]

I think it's like an equation

except if you multiply of divide by a negative number, you flip the symbol

[plush stirrup]

ohh okay i see

i just learned this not too long ago. idk how i forgot

[sage gust]

Inequalities simply have you find where on the numberline can x exist.

if x > 5, it means that x can exist only AFTER 5 (it excludes 5).

of course if you have a greater or equal sign, ≥, then x is inclusive of 5.

[plush stirrup]

oh okay that makes sense

im looking at my notes for a problem thats kinda like that one. would i multiply by the 8?

[sage gust]

In an interval,

a < x < b

x exists within a and b, but exclusive of a and b itself. The image is a graphical representation (Note that the boundary at x = a and x = b are actually dotted lines, meaning that they are not included.)

[sage gust]

Im not following.

[plush stirrup]

oh im sorry

ill take a picture of it

[sage gust]

all good

[plush stirrup]

the 1st picture is the problem im working on. the 2nd one is from notes i took in class

[sage gust]

-2 <= 3-5x/8 <= 1

Multiply everywhere by 8 to remove the denominator.

so -16 <= 3-5x <= 8

Now, lets actually split this up.

case 1:
-16 <= 3-5x

case 2:
3-5x <= 8

So we must satisfy both uh.... cases.

[plush stirrup]

ohh okay ty. i didnt know i had to split them

[sage gust]

You dont,

It's just to help you visualize better.

You could just mentally do it, but at this starting stage, you could split it up (IF it helps you).

[plush stirrup]

okay ty ^-^ this is what i got from reading the notes. looks like i kinda got it right

[sage gust]

Fun, youre given the function forms.

[plush stirrup]

XD ye

[sage gust]

If you wanted,

you could flip the signs to remove the negative.

[plush stirrup]

wait wrong pic

it would be this instead?

[sage gust]

No,

-19/-5 >= -5x/-5 >= 5/-5

[plush stirrup]

oh mb forgot the line under

no i still did it wrong

[sage gust]

Initially, you have -19 <= -5x <= 5

If you divided all by -5
then
-19/-5 >= x >= -1

[plush stirrup]

tyty!

on the back it says the answer would be [-1, 19/5] so would the -19/-5 change to positive then?

[sage gust]

yes

Dividing 2 negatives = +ve
Multiplying 2 negatives = +ve

-6/-2 = 3
-6 * -2 = 12

[plush stirrup]

gotcha. tysmm!

[sage gust]

meow :3

[plush stirrup]

XD

the rest i think i should have or at least i can find smth in my notes

tyall again 😭

.close

[vale dock]

Channel closed

Closed by @plush stirrup

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[vale dock]

Available help channel!

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[modest hemlock]

what u want

the derivative?

like

deriving that

i mean

if u factor it out

u shld js

factor out

a+b+c

[vale dock]

@stark inlet Has your question been resolved?

[timber venture]

You find the GCF to put outside the brackets

[safe fulcrum]

the hint is that one bracket is (a + b + c)

which terms must you have in the other bracket to make a^3, b^3, c^3?

[vale dock]

Channel closed

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[safe fulcrum]

.reopen

[vale dock]

✅ Original question: #help-4 message

@stark inlet Has your question been resolved?

[vale dock]

Channel closed

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[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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[fair wing]

Need help with q4

[fair wing]

Q3*

I don't know where to begin?

[modest kettle]

you can use conservation of energy

after it launches from the incline

and equations of motion until it reaches the tip of the incline

[vocal whale]

find the velocity at h = 10 and then use xf - x0 = v0t + (1/2)at^2 to find the maximum

[vocal whale]

.

[fair wing]

Yes

Idk why I am getting 10m

[vocal whale]

its 12.5 ig

u messed up some signs

[fair wing]

Can you notice it

[vocal whale]

I

and it cant be 10 because u just calculated that the velocity at h = 10 that is not 0

wait lemme write

[fair wing]

The. How to determine that acceleration

when block is on incline

[vocal whale]

its - the acceleration

[fair wing]

How to determine the acceleration when block is on incline 🙏

[vocal whale]

wait i went to grab some food

is it confusing?

[fair wing]

I'll ask exactly if I don't understand your solution

[vocal whale]

u get the y component of acc on incline which is contrary to the motion of the object

[fair wing]

Minus in -gsin30 shows that acceleration axis is taken negative?

[vocal whale]

yes

the - on acceleration means that the force that is “creating” that acceleration is pointing to the negative way of the reference frame

but it doesnt mean that its automatically deaccelerating, if v and a have the same sign (++ or - -) the object is accelerating

[fair wing]

Explain this step

[vocal whale]

its the y component of the velocity

like

if you throw an object at an angle, it will have velocity that will make him go upwards or downwards (y component) and sideways (x component)

[fair wing]

Explain the last second step

👋

[vale dock]

@fair wing Has your question been resolved?

[modest kettle]

you took acceleration as +10 m/s as it is travelling up the incline

you could reason it for a bit and realize that it cant just accelerate as it goes up the incline

both the magnitude and the direction are wrong

[vocal whale]

hm you just use the movement equation without t because theres no time

u used it to calculate the distance of the inital movement in the ramp

[vale dock]

Channel closed

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[vocal whale]

theres only 1 equation of movement, the other variants are just deduced from manipulating v = v0 + at

[vale dock]

Available help channel!

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[open thistle]

what is the best way to cram for PSAT math?

[open thistle]

I'm taking it tmr🥲

[jovial edge]

Grinding problems

That's it, if you already know the materials