help-38

[empty tulip]

my answer is 4.6units^2

[trim lichen]

show work

[empty tulip]

Thanks for
checking

[trim lichen]

your sketch looks incorrect to me, for one... should be a parabola opening upwards

this arithmetic looks very sus too

4/3 - 3...

[twin storm]

this yellow area u have to find

verify if u have made correct sketch

[trim jolt]

@empty tulip Has your question been resolved?

[empty tulip]

So my sketch is wrong

[twin storm]

yes

[empty tulip]

Ah okay

[empty tulip]

Apart from the sketch which I have fixed is the working correct?

[trim lichen]

.

didn't check the others

[empty tulip]

I'll re add

[trim lichen]

but would suspect arithmetic to happen to you there as well

[empty tulip]

Ah okay

I'm still getting thr same

[trim jolt]

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[covert laurel]

Hey

[covert laurel]

Can anyone help me at this:

Find the quantity of positive integers that do not belong to C = {ax + by | x, y ∈ N} for a, b ∈ N

I proved before that any number greater than ab - a - b belongs to it and that ab - a - b does not belong to it

But i really just kept staring this for a long time and i have no idea about what to try

[left oriole]

are you given something about a and b?

[covert laurel]

Oh yeah sorry

They are positive integers

[sharp heart]

Well, C is basically all ways to take a multiple of a plus a multiple of b.

[covert laurel]

Yeah indeed, its a linear combination between a and b

But non-negative variables only

[sharp heart]

Do you know about Bezout coefficients?

[left oriole]

do you know the theorem that relates gcd(a,b) to linear combinations of that form?

[covert laurel]

I know about Bezout theorem idk if thats the same

[covert laurel]

Yeah

[sharp heart]

Well, the Bezout coefficients are where you have ax + by and that equals the GCD of a and b.

[covert laurel]

But that does not guarantee that the variables are non-negative does it?

[sharp heart]

x and y are the coefficients.

No, but it's a start.

[covert laurel]

Oh okay

I really thought that we only called coefficients the constant numbers

[left oriole]

x and y are constrained to be positive integers, though?

[covert laurel]

Non-negative

Anyways, how would it help?

[sharp heart]

OK, so you can factor the GCD out of a and out of b, right?

[covert laurel]

Because that would only be for 1 value of ax + by

[covert laurel]

Correct

Yeah the gcd must divide the other side

[sharp heart]

So, the answers will be multiples of the GCD.

[covert laurel]

Yeah

[sharp heart]

So, any nonmultiples of the GCDs aren't included in the set, obviously.

[covert laurel]

Indeed

So euler function of their gcd?

Plus other numbers of course

But for this part

[sharp heart]

OK, let's talk about the LCM.

Is the LCM of a and b in the set?

[covert laurel]

Hmmmm

Yes

[sharp heart]

Why?

[covert laurel]

Since its either a times something plus 0 or 0 plus b times something

Thats a way to get it right

[sharp heart]

Well, lcm(a, b) = ab/gcd(a, b).

[covert laurel]

Oh yeah

But?

[sharp heart]

Let's say we have 21 and 35 for a and b.

The GCD will be 7. The LCM will be 15.

If we factor out the GCD, we have 3 and 5.

2 * 3 + -1 * 5 = 1

Those are the Bezout coefficients there.

Now we can add 3 to the -1.

2 * 3 + 2 * 5 = 16.

This is one over the LCM.

Let's say we want to get 17.

4 * 3 + -2 * 5 = 2

We can add 3 to -2.

4 * 3 + 1 * 5 = 17.

Let's say we want to get 18.

6 * 3 + -3 * 5 = 3.

We can add 3 to -3.

6 * 3 + 0 * 5 = 18.

Let's say we want to get 19.

We can take 16 and add 1 to the first coefficient.
3 * 3 + 2 * 5 = 19.

And we can keep going like that.

So, anything above the LCM is in the set.

The question is what elements at or below the LCM are in the set?

[covert laurel]

Hmmm

[covert laurel]

How will the lcm be 15?

15.7 no?

[sharp heart]

No, that's their product.

[covert laurel]

Their product is 15.7.7

[sharp heart]

After you take the GCD out of a and b's factors, the product of the remaining factors is the LCM.

[covert laurel]

How is a multiple of 21 and 35 going to be 15

[sharp heart]

Hmm, one second.

Oh, right.

Yes, you're right.

You can do the same sort of thing with the LCM.

[covert laurel]

Alright, but how do u ensure you can get to 16, 17 and so on forever?

Can u still?

Instead of lcm we have the product of their coprime factors

[sharp heart]

2 * 21 + -1 * 35 = 7.

[covert laurel]

So for every number greater than or equal lcm/gcd we guarantee its in the set?

[sharp heart]

0 * 21 + 3 * 35 = 105

2 * 21 + 2 * 35 = 105 + 7

4 * 21 + 1 * 35 = 105 + 2*7

1 * 21 + 3 * 35 = 105 + 3*7

3 * 21 + 2 * 35 = 105 + 4*7

And so forth.

So, every multiple of the GCD that's the LCM or over is possible.

The question is what about the multiples of the GCD that are less than the LCM.

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[void grove]

Need help on part B of each of these (my answers were 7.090 and 0.474, respectively)

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[void grove]

No dawg

[limpid dawn]

this?

[void grove]

Yes

I got 0.474 for part b

[limpid dawn]

you need to show your workings

[limpid dawn]

seems about right

[void grove]

In my problem?

Or is ur question unrelated

[covert laurel]

Sorry i had no Internet for a while guys

It just got sent from before

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[wraith hinge]

i have done the part A im looking forward to the part b and c so for part B i did like
sin(1/x) - cos(1/x)/x x > 0 and 0 , x < 0

is that looking fine

[bright quarry]

sure

[wraith hinge]

and for C its not diff on x = 0 because those two are not same for

x approaching 0

[bright quarry]

maybe say for x > 0 and for x < 0 for clarity btw

[wraith hinge]

true

[ionic pendant]

that shows that the derivative is not continuous at x = 0

to show it's not differentiable you have to calculate the derivative from definition

[wraith hinge]

sorry i meant function as x approaching 0 from both sides

for left and right diff

[wraith hinge]

right

so like derivative for x < 0 and x > 0 are not same

that implies at x = 0 its not differentiable

for this
f' = 2xcos(2/x) + 2sin(2/x) for x != 0

is the f' = 0 for x = 0?

how do i do the part D

[ionic pendant]

for part D, you just show that the derivative is not equal to the limit of the derivative (or the limit DNE), as you would for any other function

[wraith hinge]

ahhhh

i see

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[round minnow]

can someone explain this part please

how does it go from A to B

[cyan zinc]

let's use a simple example.

sqrt(20) = sqrt(4 x 5) = sqrt(4) x sqrt(5) = 2sqrt(5)

[round minnow]

okay, that's clear

[cyan zinc]

so then:
sqrt(x^2(3+2/x^2)) = sqrt(x^2) * sqrt(3+2/x^2)

and what is sqrt(x^2)?

[round minnow]

x?

[cyan zinc]

wrong

try again

[round minnow]

uh

+-x?

|x|

[cyan zinc]

there you go

it's the absolute value of x

[round minnow]

okay, noted

[cyan zinc]

but why does the absolute value sign not matter here?

[round minnow]

x^2 always positive?

[cyan zinc]

not quite

we're taking the limit as x goes to +infinity

[round minnow]

oh

so it's pos regardless

bc + infinity

[cyan zinc]

exactly

[round minnow]

okay, A makes sense

wait

like this?

[cyan zinc]

yes

[round minnow]

ohh

so then the x at the top cancels out

and leaves top with 1

bruh

[cyan zinc]

yes. precisely!

[round minnow]

nice! would you be up to check out this problem with me?

similar concept I think

what's confusing me is the problem we just did together didn't use the Dominant Term Rule (I think that's what it's called)

but I think my prof used Dominant Term Rule for this one

idk what to do

[cyan zinc]

for sure

let's rewrite

$$\frac{-x+2}{\sqrt{4x^2+5}}$$

[solid kiln]

Mr. Gamer 🇵🇸

[round minnow]

okay

[cyan zinc]

then can you do the thing with the square roots we did earlier?

[round minnow]

alright

[cyan zinc]

almost

you should still have a square root in the last line around the 4 + 5/x^2

[round minnow]

oh yeah, right

now what?

the negative infinity and the -x and the |x| kinda scaring me

[cyan zinc]

but the bigger issue is this: you can pull out more than just the x^2

2-x / sqrt(5 + 4x^2)
-x + 2 / sqrt(4x^2 + 5)
-x + 2 / sqrt(4x^2 (1 + 5/(4x^2))

[round minnow]

so would it just end up being |4x| outside?

[cyan zinc]

almost\

what's the sqrt of 4

[round minnow]

2

so

is sqrt(4)*sqrt(x^2) the same as sqrt(4x^2)?

[cyan zinc]

yes.

[round minnow]

okay so |2x|

wait no

2|x|?

[cyan zinc]

yep, looks good

[round minnow]

which one, the 2|x|?

[cyan zinc]

those are the same

|2x| = 2|x|

[round minnow]

oh yeah right

okay so we end up getting

[cyan zinc]

now this is important:

we are taking the limit as x goes to?

[round minnow]

-infinity

[cyan zinc]

so then |x| becomes -x

so we have -2x in the denominator

[round minnow]

ohhh

so we get

$$\frac{-x+2}{-2x\sqrt{1 + 5/(4x^2)}}$$

[solid kiln]

Paul04

[cyan zinc]

yep

[round minnow]

-infinite + 2 / +infinite * sqrt(1+ 0) ?

[cyan zinc]

haha

almost

you're right that the term inside the square root becomes 1

[round minnow]

lol

okay

and I'm right about the numerator 2 surely

so what happens with the -x and -2x?

[cyan zinc]

(-x+2)/(-2x)

just use the dominant term rule now

[round minnow]

do I divide by -x or x?

also is the 2 included or nah?

I don't think the 2 is included

[cyan zinc]

just take the ratio of leading coefficients

so you get:
(-1) / (-2)

[round minnow]

I don't follow, sorry

ur saying I divide by -1/-2?

to evertything?

[cyan zinc]

let's break this down into steps

do you agree that we have:
(-x+2)/(-2x)

[round minnow]

yeah

ik where u got -1 and -2

I just dk what u do with it

[cyan zinc]

just divide

-1 divided by -2

[round minnow]

so just 1/2?

[cyan zinc]

yep

[round minnow]

okay

now what?

[cyan zinc]

we're done

let's check our work

,w lim_(x to -infinity) (2-x)/sqrt(5+4x^2)

[solid kiln]

Results provided by WolframAlpha

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

Requested by the.mr.gamer

[round minnow]

my brain is loading gimmie a sec

$$\frac{-x+2}{-2x\sqrt{1 + 5/(4x^2)}}$$

[solid kiln]

Paul04

[round minnow]

so the sqrt ends up being nothing

we have -x+2 / -2x

[vagrant kraken]

another finding asymptote question

alright here is a simplier way of looking at this question

first of all, tell me is the answer positive or negative?

[cyan zinc]

if it helps, just rewrite this as:
-x/(-2x) + 2/(-2x)

[vagrant kraken]

I guess you can also do it like that

[round minnow]

okay

[round minnow]

mr gamer found it to be positive

[round minnow]

wdym?

idk why my brain is lagging so hard

[vagrant kraken]

alright

so this is my method:

first of all, when x approaches -infinity

the top is positive, so is the bottom

therefore you will get a positive answer

then you expand the sqrt upwards

you get sqrt((x^2-4x+4)/(4x^2+5))

[round minnow]

I understand up until this part

[vagrant kraken]

now if you know how to find the asymptote of a function like that

the asymptote of the inside is 1/4

after sqrt

it is 1/2

and the answer is positive

therefore the answer is 1/2

[round minnow]

ig that sorta makes sense

I think I'm gonna practice a few more problems and I might open new help channels

I think I understand this one but we'll find out once I do more ig

[vagrant kraken]

alright, good luck

[round minnow]

thanks

.close

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[round minnow]

,w lim_(x to infinity) (3x^3 -8)/(x^2 +4))

[solid kiln]

Results provided by WolframAlpha

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

Requested by paulonsteam

[round minnow]

you get 3x/1 = 3x when solving this

but like why is 3x slant asymptote?

[empty orchid]

Because it is a specific case, when the degree of the numerator is one more than that of the denominator

[round minnow]

is there like a list that explains all this or smth you can refer me to

so everyh time numerator is one more than denominator, whatever the output, it's the asymtpote?

[empty orchid]

Yes

By polynomial division

[empty orchid]

There is a limit list, and polynomial asymptote list

[round minnow]

okay, I'll look into it. thanks

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[real herald]

Can this work in the other direction? Like if there exists a index that meets those requirments, then the orbit is that size?

[real herald]

is it an iff theorem?

[amber python]

what does that mean?

what do you expect the other direction to even be

[real herald]

Lets say im trying to prove that G_x is = G, can i show that using this theorem in the backwords direction showing that an orbit exists with order 1?

[amber python]

that doesn't make sense

what's the backwards direction?

once you have an x, G_x and O_x are fixed

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[pine iron]

The sum of two strictly positive integers is equal to 11,552.
The sum of the reciprocals of these two numbers is equal to 1 / 2006.
What are these two numbers?

[pine iron]

This question is supposed to be solved without an calculator

[undone forge]

What did u try

[pine iron]

i was trying something but then realzied that i had to find the square root of a number in the 20 millions

so whateve i tried is completely pointless

though there i one thing i think i know

the sum of the reciprocals is like 1/x + 1/y = 1/2006, suppose x is 5 and y 6, 1/5 + 1/6 = 2/11, therfore 1/2006 was originally 2/4012

but still two numbers adding up to 4012 as the denominator, can't give 11552

[undone forge]

1/5 + 1/6 = 2/11 are u sure about this?

[pine iron]

wait mb

omg

mb

well that just makes it even more complicated

[undone forge]

Start by transforming the question into equations first

[pine iron]

right x+y = 11552

1/x + 1/y = 1/2006

[undone forge]

Do u know substitution

[pine iron]

yeah

[undone forge]

Two equations and two unknowns, its likely they can be solved through substitution

[pine iron]

xy/2006 = 11552

[undone forge]

Right

[pine iron]

xy = 23217312

but remember i cant use a calc

[undone forge]

Yeaa

Dont multiply them just write it as product

U can do the computation later

[pine iron]

so just keep this for now

.close

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[still widget]

Calc 2, Surface Areas of rotated curves

Not sure what I did wrong, probably a small mistake somewhere. Attached picures are of the problem and my work

[dense jasper]

you forgot to apply chain rule while differentiating

@still widget

[still widget]

Thanks

That was it,
152pi/3 is the correct answer

thanks!

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[crimson halo]

I need help on 9-19 how do you get the equation?

[orchid wagon]

there are a couple of things you can deduce from this table, can you find them?

in particular, i want you to look at y when x = -2 and x = 5

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[crimson halo]

i got it sorry my phone died tysm

[chilly bobcat]

close the channel using .close then

[crimson halo]

.close

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[ebon flare]

find range of $f(x) = \frac{x^2 - 5x + 6}{4x^2 - 5x - 6}$

[solid kiln]

Eren

[ebon flare]

I have done this

[trim lichen]

,rccw

[ebon flare]

I'm getting ans Real - {-1/11}

[ebon flare]

But the answer is real - {1/4,-1/11}

[trim jolt]

@ebon flare Has your question been resolved?

[ebon flare]

<@&286206848099549185>

[tawny carbon]

w h a t

[ebon flare]

^

[tawny carbon]

classic

fuckin hate range and domain

let me turn on my lamp

ok

im mathing

dnd

ok epic

you range

want range

so paper is useless 🙏

demos 🔥🔥🔥

stroke

range

is

-infinity to infinity

@ebon flare

@ebon flare

😡

[ebon flare]

Wrong ans dude

[tawny carbon]

how

there is no other answer

wdym

[ebon flare]

Correct ans is - real - {1/4,-1/11}

I have already gotten real - {-1/11}

If you don't know how to solve just don't help

[tawny carbon]

bro wdym 🙏

[ebon flare]

You don't know anything about it I got it

[tawny carbon]

it goes on forever 🙏

how tf is the range right below 0

[ebon flare]

Idk what you're talking about

[tawny carbon]

it goes on forever in all directions 🙏

[trim jolt]

@ebon flare Has your question been resolved?

[buoyant marsh]

y=1/4 is the horizontal asymptote as x approaches infinity on both sides

[buoyant marsh]

wait

[buoyant marsh]

your approach is very interesting

you have solved for when the discriminant (b^2-4ac) is nonnegative (square root of a negative number is undefined)

however you missed the key part where 2a =/= 0

because that will also lead to an undefined result (division by zero)

simply you need to solve 2(4y-1) =/= 0

and you will find that y =/= 1/4 should complete your answer

@ebon flare

heading to bed soon

[ebon flare]

Wait dude

[buoyant marsh]

whats up

[ebon flare]

Yeah I wrote x in terms of y

And founded the domain of y

The domain of y will be the range of x

[buoyant marsh]

yes

but you used the quadratic formula to solve

[ebon flare]

I don't understand from here

[ebon flare]

I used the condition D>=0 because in this condition I'll get real x

[buoyant marsh]

hmm

so i believe D refers to the "Discriminant"

[ebon flare]

Yes

[buoyant marsh]

the formula is undefined if D < 0 (negative)

so it must be D >= 0

but also

the formula is undefined when 2a = 0

so it must be 2a =/= 0

[ebon flare]

I haven't used quadratic formula

you know like for a quadratic equation
D=0 then roots are equal
D>0 roots are real
D<0 roots are imaginary

I just used the condition D>=0 to get real x (to find domain of y) and the domain of y will be range of x (of question)

[buoyant marsh]

hmm

[buoyant marsh]

you can't always find the range of the function just by solving the domain (if i understand you correctly)

but following your approach, there is one definite mistake

[modest scarab]

hey guys great job

[buoyant marsh]

if you multiply out the denominator, you must include the condition, or else your implication (->) is false

[modest scarab]

continue the work

and never give up

❤️

[buoyant marsh]

alright

i think the problem is simply the definition of discriminant

https://proofwiki.org/wiki/Discriminant_of_Quadratic_Equation

wait

[buoyant marsh]

you need to find where you got this from

my bet is that the caveat is "if a=/= 0 ..."

So your answer is still incomplete

I'll stick to my original explanation

We started with y=f(x) (which allows us to easily see which values of x are undefined) but now we have x = g(y) (which allows us to easily see which values of y are undefined)

you simply need to find where it is undefined (there are two places to look)

[buoyant marsh]

there is certainly a deeper meaning to this

however, since we are dealing with a strictly real numbers (your original question), it is not necessary to dive into the complex roots.

D < 0 simply indicates that it does not exist

in our case (real numbers only)

Otherwise, you will need to perform complex analysis

Which unfortunately I am not equipped to do

I'll leave you off with the somewhat unsatisfying answer that "seemingly by coincidence, 2a =/= 0 and a^2 =/= 0 both give you the missing piece that y =/= 1/4". Perhaps someone can provide a cleaner answer if you throw in which textbook definitions/theorems you are using.

@ebon flare aight im actually heading to bed now, hopefully i didnt open a whole new can of worms lol

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[median iron]

Is this question badly worded or am I crazy?

[undone forge]

(1-28%) x = 62.5

[zenith nexus]

yes it is

[median iron]

couldn't it be argued that 28%=0.28 so x-0.28=62.5?

[median iron]

this is how you get the correct answer

but is that other way arguable?

[undone forge]

It's common to associate percentages as of something than just them only.

[median iron]

ahh okay, i was going to do that

but then i was like

maybe its a trick question

well thank you for double checking

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[halcyon jolt]

I don’t know if I’m meant to ask this here but does anyone have any good study methods be it YouTubers websites etc for learning and understanding functions (composite compound linear quadratic cubic)

[empty orchid]

Yes

For one, Khan Academy

For two, OCT (organic chemistry tutor)

These are general foundational youtubers

[hidden linden]

I was actually recommended Khan Academy the first time I asked it openly. It's pretty good in my experience.

[slender shard]

For any level of math, it helps to write out your thought process while solving a math problem, preferably in a different color ink. When it comes time to study for a test, all you really need to do is look over the notes you wrote. If your notes don't make any sense to you after a day or two, rewrite them until they do make sense.

[hidden linden]

Also it's generally good to write down anything as you read. It is a reinforcing strategy.

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[fleet niche]

how do i find the vertices and foci?

[dusty sleet]

By completing the square, for instance

[fleet niche]

so in ellipse form this becomes (x-5)^2 / b^2 and (y-5)^2 / b^2 = 4

[fleet niche]

uh huuuuuhhh

[broken pilot]

what is b?

[dapper swift]

search up the actual equation of an ellipse

[empty orchid]

That is not actually ellipse form? The right hand side has to be 1.

[fleet niche]

to find the foci, the formula is a^2 - b^2 = c^2

[fleet niche]

yeah yeah so divide both sides by 4 but then what

[empty orchid]

Could you write the whole equation out with numbers?

[fleet niche]

um ignore the 2. exercise

[dapper swift]

yeah then (5, 5) is just the centre

[fleet niche]

1. reorganize the equation
2. complete the squares
3. add the same amounts to the other side
4. end up with the correct x and y coordinates
   but now what

so here's what symbolab says

it found a^2 and b^2 but it locks the explanation behind a paywall

[dapper swift]

does it really ask you for the vertices and foci later on?

[fleet niche]

oh i didn't read the question, i thought it was going to ask for everything

[dapper swift]

oh yeah you have a slight mistake, should be (x - 5)^2 / 1^2 there

[fleet niche]

i can do math just fine but i don't pay attention to the questions

ugh

[dapper swift]

the values of a, b don't affect the centre

I see

[fleet niche]

allllright

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[fleet niche]

.reopen

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✅

[fleet niche]

ok this is freaking me out

divide both sides by 784 to get = 1???

[empty orchid]

Yes

You know how to find major axis right

[fleet niche]

yea yea

i just thought 49 and 16 were multiples in 784 but not the other way around

[empty orchid]

Lol

[fleet niche]

so i can cancel out the nominator coefficients

49x^2 / 784 becomes x^2 / 16

[empty orchid]

Yep

[fleet niche]

alright i got it

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[empty tulip]

the asnwer is 4units^2

however i got 4.6

can someone check my workings please

[exotic tundra]

,calc 1/3 - 4*1/2 + 3

[solid kiln]

Result:

1.3333333333333

[exotic tundra]

that area 1 is wrong

[empty tulip]

okay which part

[exotic tundra]

the calculation after you plugged it in

[empty tulip]

so

the actual answer is wrong?

not the working out

[cyan snow]

There seems to be a bit of calculation error

In your part

In area 1

And probably area 2 as well

[empty tulip]

im not sure which part

[exotic tundra]

this part on number 5

[empty tulip]

should it be 4/3 - 0

[exotic tundra]

yeah

[empty tulip]

i did that initially

ill write that again and see

[cyan snow]

If you put 1 instead of x it won't be 5/3 but 4/3

(It's Area 2 btw)

[empty tulip]

so

final area

is

4/3 +4/3

+4/3

[cyan snow]

Absolutely

[exotic tundra]

and that's also not 0

[modest scarab]

hey guys great work

also NEVER GIVE UP

❤️

[empty tulip]

yeah okay i see

[empty tulip]

is it not?

[cyan snow]

It is zero tho

[empty tulip]

lol

area has to be positive

[exotic tundra]

oh it is

I read that as 3^2/3

mb

[cyan snow]

Oh understandable

[cyan snow]

Yup

[empty tulip]

that was my mistake before

[cyan snow]

It's 4 unit²

The total area

[empty tulip]

i did -4/3 + 4/3 +4/3

[cyan snow]

Yeah exactly

[empty tulip]

yeah

ive got it now

thanks

❤️

[cyan snow]

no problem

Have any more of it?

[empty tulip]

yeah

[cyan snow]

If you need assistance do share it then

[potent frost]

Yo yo

Is this integration

[cyan snow]

Yup

Rather application!

[potent frost]

Did u get ur answer

Or u still don’t know it

[cyan snow]

Fortunately yes

[potent frost]

Cool

[empty tulip]

i dont understand how theyve got the bit in red]

[potent frost]

Integration by parts

[empty tulip]

YES

[potent frost]

Show the question

Il try it

[potent frost]

Gimme 2 mins

Working it

[limpid dawn]

they did ibp twice

[potent frost]

Ok this is confusing

I’m at sinxex - integral of cosxex dx

[limpid dawn]

they always chose e^x as v' btw

[empty tulip]

which part have they intergrated twice

[potent frost]

@empty tulip

What country r u in

[limpid dawn]

second line the integral of sin(x) * e^x

[potent frost]

What level maths is this

[limpid dawn]

that's also done by ibp

[potent frost]

Bc I do the same thing

[limpid dawn]

If you let u = sin(x) and v' = e^x

[empty tulip]

oh ok

[empty tulip]

uk

[potent frost]

A level or advanced maths

[empty tulip]

both

[potent frost]

What is it called for u

[empty tulip]

foundation maths

[cyan snow]

I can give you a small trick, it might not work every time but I do remember this phrase called ILATE - Inverse>Log>algebraic>trigonometric>exponential

This can help you identify what may be the first function and what can be the second.

Like for this question we made the trigonometric function the first function and exponential function the second. Following the order of ILATE

If you're completely clueless it might help!

[empty tulip]

yh ive heared of that

[potent frost]

Can somebody tell me why it went to 5 and 15

On the left side

[limpid dawn]

tbag this channel is occupied

[potent frost]

How to get help

[limpid dawn]

!Help

[trim jolt]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[potent frost]

!help

[trim jolt]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[limpid dawn]

there are 3 channels available rn

[empty tulip]

.close

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[empty tulip]

.reopen

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✅

[empty tulip]

here they have substituted

but

[limpid dawn]

yes that's the second time applying ibp

[empty tulip]

and im confused fater that

[limpid dawn]

The called the intial integral I to make it better readable

[empty tulip]

ohhh

i get it now

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[strong delta]

how do i even start with this...

[nocturne sky]

i will give u a hint

[nocturne sky]

a positive int t can be written as a^2 + b^2 as if only every prime p ≡ 3 (mod 4) on its prime factorization with an even opponent

[strong delta]

even opponent???

[vagrant prism]

exponent

lol

[strong delta]

oh

so if the factor is p^2 i can write it as a^2+b^2

[vagrant prism]

no

if all primes that are p = 3 (mod 4) have even power then you can do it

[strong delta]

ohhk

got it

thank u

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[visual tree]

Can the three dimensional heat equation be used to calculate the heat loss for the earth ( without atmosphere perhaps)

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[frozen nacelle]

would it be because $h''\left(\frac{4}{2}\right) = h''(2) > 0$ by the graph, but $f''(4) < 0$?

[solid kiln]

Hailinqs‽

[frozen pumice]

Yeah

[frozen nacelle]

ok thanks

[frozen pumice]

2nd derivitive is negative

[frozen nacelle]

.close

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[frozen pumice]

Supposed to be positive because its increasing

The rate is

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[vale pelican]

What is the apostrophe on a doing there?

[nova spire]

they interpreted a as a function

[empty orchid]

It represents the derivative of a with respect to x.

[nova spire]

so its derivative comes into play

[vale pelican]

But I meant for the a to be a constant

[nova spire]

then replace the a'(...) by a

[vale pelican]

Ohhhh

[empty orchid]

Alternatively, you can just take the a out of the WA calculation and multiply at the end.

[vale pelican]

The AI assume that a is a function cause I put it that way

Thanks, everyone

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[empty orchid]

WA isn’t exactly AI. It is a computational machine.

[vale pelican]

Ouh

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[storm trellis]

This question is saying "The figure shows four linear lines which intersect with eachother. Find the sum of p + q + r + s."

Any tips? No answer, just what i should look out for

[storm trellis]

Ive tried a very weird and clutterly approach, like trying to find the individual angles, which lead to something weird

for example the 180* on the s angle, where the other angle would be 180 - s etc.

But that feels like the wrong approach

since i really cant do much more after that

[tired marsh]

Just use the formula in which the sum of internal angles of a triangle is the opposite external angle

With 180-r, 180-s and the opposite external angle

[storm trellis]

I dont quite follow here

[tired marsh]

Then take the small triangle, and apply the angle sum property(sum=180)

Ok, let me explain it in a more understandable way

So, you see the point at which angle r is, right? Let's call it A

[storm trellis]

The enitre angle that r resides on?

[tired marsh]

The point is A

[storm trellis]

Yes i see

[tired marsh]

At which the two lines meet, and angle r is subtended

[tired marsh]

Yess, now, the same for angle s is B, let's say

Tip, write it down on a piece of paper, because it's pretty difficult to visualise

[storm trellis]

Like this?

[tired marsh]

Yes

The point at which q is subtended is D, and for p, it's E

Now, the angles of ∆ABC are 180-p, 180-s and some angle x, let's say

[storm trellis]

Yes

[tired marsh]

For ∆ABC

[storm trellis]

Ah i wrote a little wrong in my sketch

I thought x was the missing angle in the triangle

Otherwise yeah that would be true

[tired marsh]

Uh, it is

[tired marsh]

X is angle(ACB

[storm trellis]

Did you mean it like this?

[tired marsh]

Yeahh, ok

Yes, your diagram is right

[storm trellis]

sorry if its a little clunky

[tired marsh]

So, we take ∆CDE

[tired marsh]

Nah , chill... I'm way too messy, I'm used to it

[tired marsh]

And apply angle sum property of the triangle with 180-q , 180-p and 180-x

Substitute 180-x as this in the equation

Ahh, shit

[storm trellis]

X = 180 - s + 180 - r

[tired marsh]

Mb, i confused the variables

[storm trellis]

right?

[tired marsh]

Yes, mb

[storm trellis]

i was a little confused over that but nice to know im thinking correctly :x

[tired marsh]

So, you do a bit of algebra, and you get 180-s + 180-r +180-p + 180-q =180

[storm trellis]

yess i was about to type that

[tired marsh]

So, p+q+r+s = 540

Was it correct?

[storm trellis]

i dont have a answer shee 😭

its from an entrance exam

[tired marsh]

Oh, ok

Which grade are u in?

[storm trellis]

so they dont post the answer for "intuitve" thinking

[storm trellis]

8

or 9 depending on where you live

however geometry isnt my strong suit

i mostly do calculus

[tired marsh]

Hehe, I'm 9th, but 10th, technically, since classes started

[storm trellis]

Good to know i can learn from you

[tired marsh]

Hehe, same... I'm in Sri Chaitanya , from india

U?

[storm trellis]

Im from Sweden, and this is for a high school

[tired marsh]

Damn, nice

[storm trellis]

its mathematics specialised

exams tomorrow 😭 idk if im getting in

[tired marsh]

Oh wow, my school is like for preparing for IIT-JEE entrance exam

[storm trellis]

Really? already??

[tired marsh]

Do u know what it is?

[storm trellis]

Yeah

very tough exam that can determine ur future

[tired marsh]

Yeah, started in 8th

[storm trellis]

its like gaokao

[tired marsh]

For sure, ig is the 2nd toughest

[storm trellis]

Might be tougher depending on how you perceive it

What are you studying now?

[tired marsh]

Oh, just 12th syllabus

[storm trellis]

Any special areas?

[tired marsh]

I'm almost done with the syllabus 😬

[storm trellis]

Woahhh

[tired marsh]

In maths, I'm currently learning continuity and differentiability, though I know some basic calc

[storm trellis]

you will do great on exams

then

[tired marsh]

Lol, idk abt that

[storm trellis]

yee u will

if ur almost done with 12th syllabus

[tired marsh]

I always screw up in the exam

[storm trellis]

me too but we hope for the best

[tired marsh]

What are you learning?

In maths,i mean

[storm trellis]

Im currently learning calc 2, but i need to retrace to learn some ground things

like formulas or ideas

[tired marsh]

It's 12 am rn 😬

[tired marsh]

Damn

[storm trellis]

Talk about time zones heh

[tired marsh]

My parents will beat me up if they find out

I'm chatting rn

Lol

[tired marsh]

Ikr

[storm trellis]

what is bro doing awake 😭

buddy sleep before you sleep forever

[tired marsh]

I was sick, so i slept mid day

[tired marsh]

Lol

K, gtg, peace man

[storm trellis]

Me too, exams tomorrow

wish me luck 🙏

[tired marsh]

Good luck, hope u top

K bye

[storm trellis]

Cyaa

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[midnight vessel]

What is "Algebraic Expressions and Equations" again? I have a test on this tomorrow and I feel so sleepy I can't understand nothin
I don't really have work to show, but I can try if you ask me for some specific work or I just provide the textbook thing the teacher put

[midnight vessel]

I found some questions the textbook already has maybe u guys can show me how they are solved

[kind elm]

Which ones

[midnight vessel]

1, 2, 8, 11

[kind elm]

Alr

1a

The highest degree is 2

So that would be a quadratic

Linear= No exponent aka 1

Quadratic = 2 as exponent

Cubic = 3

Quartic = 4

Quintic = 5

To count the number of terms

You can look at it as being seperated by operations

So why dont you try number 2

[midnight vessel]

will try

like 2 or 1b

[kind elm]

1b

mb

[midnight vessel]

1b has 2 terms so like binomial and it has only this W so Linear

Binomial, 1 degree, Linear

[kind elm]

Nice

Keep going until there's a problem

[midnight vessel]

1c, has 4 terms so its this polynomial I think

[kind elm]

Yeah

Anything past Quintic is just a polynomial

[midnight vessel]

degree is 16?

[kind elm]

Degree would just be 6

Cause thats your highest exponent value

[midnight vessel]

ohhhh

So it's polynomial, degree 6, polynomial again or smth like that

[kind elm]

Yeah

[midnight vessel]

1d, monomial, degree 3, linear

[kind elm]

It wouldnt be linear though

[midnight vessel]

What would it be?

[kind elm]

^

[midnight vessel]

Wait how?

[kind elm]

What

Look at the terms i said

If the exponent is 3 what would it be?

[midnight vessel]

cubic

[kind elm]

Nice

[midnight vessel]

So now 2

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[kind sandal]

I need help understanding the central limit theorem

If I had some distribution heavily skewed to the left such that the most extreme 10 or 20 values on the right hand side have the highest possibilities. How is it that the distribution of the sum is normal?

If I took many many sums, sometimes il get that extreme value which is small causing the sum to be smaller but their histogram bars will be much shorter simply because the values needed to make those small sums require you getting some of those unprobabl smaller values.

On the flip side the most largest values are the most probable and so therefore wouldnt the largest sums also be the most probably sums as well meaning that the majority of the mass of our sum distribution will still be to the right and be left skewed?

[kind sandal]

this is the first question I have il say the 2nd after I crack this

[trim jolt]

@kind sandal Has your question been resolved?

[digital bolt]

it's not the distribution of the sum, it's the distribution of the mean

i think this is important

[kind sandal]

yes this would make sense but they were also saying that the distribution of the sum also follows a normal distribution and when I looked online they confirmed this as well

[digital bolt]

anyway to answer your question, basically if you take the mean of the sum of variables with the same distribution

then the distribution of that mean is more symmetrical

it just is

[kind sandal]

like in the Khan academy video he said the same thing and that the greater the n was (ie the number terms in your sum) the more the distribution of the sum will approximate

[digital bolt]

and if you take lots and lots and lots then it turns into the normal distribution which is completely symmetric

like, you're right that it will still be skewed

but it will be less skewed

[kind sandal]

no my reference to the skewness was just when we were making a distribution of the sums

not of the averages, if we kept taking the averages

it will bea normal

[digital bolt]

?

i mean the sum is the same

the sum will still be skewed, but it will be less skewed

[kind sandal]

but the point I was making is that if we repeated it ininfinity times the distribution of the average would be normal but the distribution of the sums would be still left skewed

[digital bolt]

no it wouldn't

the more you repeat it

the less skewed it gets

like, it's just a limit

1 is positive

1/2 is positive

1/3 is positive

1/n is always positive for any n

but the limit as n -> infinity of 1/n is 0

which is not positive

[kind sandal]

yes that for the average though, when we take the sum there is no 1/n

[digital bolt]

no

i'm just using a limit example

let's go like this

1/2 is less than 1

3/4 is less than 1

7/8 is less than 1

1 - 2^-n is always less than 1 for any n

but the limit of 1 - 2^-n as n goes to infinity is just 1

which is not less than 1

similarly, for your question, any finite sum of the variables will still be skewed, but the limit as you go to infinity is not skewed

[kind sandal]

where are you getting this from

1-2^-n

[digital bolt]

1/2 = 1 - 2^-1

3/4 = 1 - 2^-2

7/8 = 1 - 2^-3

like

it's just the formula

do you know what limits are

btw

[kind sandal]

yeah

[digital bolt]

ok

[kind sandal]

but I dont what formula you are referring to

when you said 1/n I assumed you meant the fraction outside the summation

for the average equation

[digital bolt]

i am taking a completely separate example

as a metaphor

i'm not getting it from anywhere

i'm making it up

to show the point

the point is that, even if a property holds for every term in a sequence, it doesn't mean the property will hold for the limit of that sequence

[kind sandal]

why does this relate to your context

[digital bolt]

if you take the sum or the mean of 1, 2, 3, however many terms of your skewed variable

the result will still be skewed

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[kind sandal]

the less skewed it gets , but I dont see how itl becoem normal though

[digital bolt]

but the limit distribution is not skewed

ok can you accept that the limit distribution may not be skewed

does that make sense to you

.ropen

.reopen

[trim jolt]

✅

[kind sandal]

you want me to accept that it can be normal ?

[digital bolt]

no

i want you to accept that it may not be skewed

it might be symmetric

[kind sandal]

but thats just normal

[digital bolt]

no

there are symmetric distributions that are not normal

[kind sandal]

but we are approximating here

even if it is discrete values

we are using the properties of the normal

[digital bolt]

look

for finite values, it's only approximately true