help-23

[cunning zephyr]

🤦🏾‍♂️

I'm stupid

x(5x) ?

[charred saffron]

youve js somehow

worsened it lmao

[charred saffron]

x(5x) = 5x²

not even close

[cunning zephyr]

oh my god

[charred saffron]

look

ill try to go about factorising x³ - x² step by step

ur okay with that?

[cunning zephyr]

sure

[charred saffron]

alr so

x³ is same thing as x^(2 + 1) right?

[cunning zephyr]

yea i think so

[charred saffron]

3 is 2+1

so the exponent 3 i write it as 2+1

[cunning zephyr]

okay

[charred saffron]

using this formula

i turn x^(1+2) = x¹ × x²

any issue with it?

[cunning zephyr]

yea

[charred saffron]

u didnt understand?

[cunning zephyr]

no issue

[charred saffron]

oh

[cunning zephyr]

i said the wrong thing

mb

[charred saffron]

so

x³ = x¹ × x²

meanwhile x¹ is the same thing as x

[cunning zephyr]

yea

[charred saffron]

anything to the power of 1 is the same as it is

so

x³ = x × x²

and thus

x³ - x² = x × x² - x²

yes?

[cunning zephyr]

yeaa

[charred saffron]

cant u factorise the expression x³ - x²?

isnt it obvious now?

[cunning zephyr]

x (x * x ² - x) ?

[charred saffron]

no

not quite

[cunning zephyr]

damn

[charred saffron]

the way to factorise is

ab + ac = a(b+c)

[cunning zephyr]

yes

[charred saffron]

find smt commonly multiplied between those two terms

and get it out

in x×x² - x²

whats the commonly multiplied term?

[cunning zephyr]

x²

[charred saffron]

yes

so factor it out

[cunning zephyr]

wdym

[charred saffron]

x²(x - 1)

[cunning zephyr]

where did you get the 1 from

[charred saffron]

no?

because x² is x² × 1

[cunning zephyr]

oh

[charred saffron]

x² × x - x² × 1

say x² = a, x = b, 1 = c

it becomes ab - ac

ab - ac = a(b - c)

if u put back in the values u substituted

u get x²(x - 1)

[cunning zephyr]

okay

[charred saffron]

feels like u didnt understand

[cunning zephyr]

took me a while but i think i did

[charred saffron]

ok so

going back to ur old expression

x³ - x² + x - 1

that would be equal to x²(x - 1) + x - 1 yes?

weve js discussed it

[cunning zephyr]

I'm sorry bro

[charred saffron]

the factorisation of x³ - x²

[charred saffron]

aw

[cunning zephyr]

yes

[charred saffron]

if x³ - x² = x²(x - 1)

then substitute that in x³ - x² + x - 1

becomes x²(x - 1) + x - 1

no?

[cunning zephyr]

yes

what about the other part

[charred saffron]

im starting to doubt your yes's

[cunning zephyr]

lol I'm not surprised

I'm a mess

[charred saffron]

x - 1 = (x - 1) × 1 aint it?

anything multiplied by 1 is itself

[cunning zephyr]

yes

[charred saffron]

so

x²(x - 1) + 1(x - 1)

can u

like

try to impress me

[cunning zephyr]

no?

[charred saffron]

and factorise it?

[cunning zephyr]

I'm incapable

[charred saffron]

.

[cunning zephyr]

actually

[charred saffron]

shit

[cunning zephyr]

wait lemme try

[vernal brook]

bros literally me

[cunning zephyr]

(x-1) * (x²-1) ?

[charred saffron]

ALMOST

[cunning zephyr]

is it the x²

[charred saffron]

its not -1 in the second bracket

[charred saffron]

nuh uh

[cunning zephyr]

huh

why not

[charred saffron]

cause ab + ac = a(b + c)

ab - ac = a(b - c)

signs are a bit of a mess when it comes to factorising

[cunning zephyr]

no I'm the mess here 😭

[charred saffron]

dw

ur a different type of mess

thats okay

[cunning zephyr]

thanks? 😭

[charred saffron]

it says +1(x - 1)

not -1(x - 1)

see?

[cunning zephyr]

OOHOHOHOHHO

YES

[charred saffron]

that has to be the most

sarcastic

yes

ive ever heard from a person

[cunning zephyr]

I'm not being sarcastic

[charred saffron]

OKAY

WELL THEN THATS IT

X³ - X² + X - 1 = (X - 1)(X² + 1)

[cunning zephyr]

sorry for everything

but thanks

[charred saffron]

its okay

im js joking around with ya

[cunning zephyr]

I'm gonna ask my teacher tomorrow too

so help me God

[vernal brook]

goodluck

[cunning zephyr]

i need it

[vernal brook]

dw uve got it

[cunning zephyr]

😭

[safe radish]

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[mortal aspen]

honestly need help perfecting exponential functions

[fathom adder]

Ask your problem

[mortal aspen]

well theres multiple

[plucky elk]

do one problem at a time

[mortal aspen]

alright this one is where i dindt fully understand it from my teacher

one sec il send it

ok so it was

y = log3^x

If i recall correctly we leave the log behind and turn into exponenteal form?

[fathom adder]

So $y = \log(3^x)$

[flat frigate]

YakuBros

[fathom adder]

Right ?

[mortal aspen]

Yes

we use the function table

[fathom adder]

I don't see what you mean by fonction table

Like you can get the power down

$\log(a^b) = b\cdot \log(a)$

[flat frigate]

YakuBros

[mortal aspen]

x -2 -1 0 1 2
y

y = 1/9 | 1/3 | 1 | 3 | 9

[fathom adder]

What are you trying to do ?

[mortal aspen]

wdym?

[fathom adder]

With the function table ?

Whats the purpose

[tidal pasture]

Good question

[mortal aspen]

these exponential functions we

show graphically

[tidal pasture]

And?

[mortal aspen]

and this table Helps us connect the numbers in the graph to how should i say it

hold on im gonna send u a ful method

[tidal pasture]

What is the problem again?

Like the question?

[mortal aspen]

.

[fathom adder]

$\log_3 ^x$

[flat frigate]

YakuBros

[tidal pasture]

Thats an expression

Not a question

[mortal aspen]

hmm?

[tidal pasture]

What are you suppsoed to do with that?

Like what a problem asks you to do

[fathom adder]

Indeed you have that $\log_3(x)$ is the inverse of $3^x$

[flat frigate]

YakuBros

[mortal aspen]

graph the function

[tidal pasture]

Oh

Well

Thats not a problem

Just graph like 3 points

And connect them

[fathom adder]

Do a symetry wrt the y=x line

Once you have one of them

[mortal aspen]

yeah i know thay

Do other people not use the function table?

[tidal pasture]

No

People use desmos

Or some braincells and basic analysis

[mortal aspen]

that supposed to offend me 😭

our teacher requires us to do ot

it

[safe radish]

@mortal aspen Has your question been resolved?

[fathom adder]

Well, there is no real pedagogy point to make students graph function imo, it will always be wierd, its just unsatisfying job and they learn nothing more except at least that inverses are symetric etc... imo ofc

[tidal pasture]

I agree

Much easier to just think about the function like this

Based on a shape of simple functions

[mortal aspen]

yeah

never liked this professor holy she has made me fall behind in math

[safe radish]

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[mortal aspen]

ok

[mortal aspen]

did i show the graph and connect the dots correctly

[safe radish]

@mortal aspen Has your question been resolved?

[lean otter]

I can't see well

[safe radish]

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[green ginkgo]

How do i start solving b? If I need to find the angle of one of these how do I find it

[obtuse kettle]

find this angle first and then do cosine rule

[green ginkgo]

How do I find that angle?

[obtuse kettle]

you know that angle and 130 are part of a co-interior angle so it adds up to 180 degrees

[green ginkgo]

Ohh I get it thank you

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[candid ocean]

I came back

[plucky elk]

mila's comeback

[candid ocean]

Max = 56
Min = 0 (right??)
Amplitude = 28

I just wanna start off by making the equation for this question

I got:
y = -28cos(0.8pix) +28

[junior smelt]

How did you get this?

[candid ocean]

isnt it kinda like this

[candid ocean]

ok 28 is amplitude and its negative bc its flipped

p = 2pi/k

k = 2pi/2.5

[junior smelt]

(In particular, when time is 0, do you get the height as 0.5m above ground? )

[swift loom]

ur c value is not 28

[candid ocean]

yeah im confused then whats my max and minimum

is minimum 0.5

[swift loom]

min is 0.5, max is 56, ur midline would be 0.5+56/2

[candid ocean]

but isnt the diameter 56 itself

[junior smelt]

You could get on and go underneath the height that you get on

[candid ocean]

so max is 56.5

[swift loom]

nvm misread

[candid ocean]

oh ok so minimum is still 0

wait whats wrong in my equation

[junior smelt]

(assumedly so, I will say, it isn't entirely clear imo )

[candid ocean]

so reall

[junior smelt]

@candid ocean

[candid ocean]

[junior smelt]

Regardless, you want when time is zero, you get a height of 0.5m

[candid ocean]

yeah

sooo

[junior smelt]

If you're using x for time, then you get a height of 0 at time 0

[candid ocean]

hm

[candid ocean]

wait so thats not right?

or wait

👩‍🦯

[junior smelt]

It isn't

[candid ocean]

BRUHHH

[swift loom]

56 + 0.5/2

[candid ocean]

why did the answer key show this then

[fathom jewel]

https://tenor.com/view/kitty-eat-lava-minecraft-kitty-kitten-lava-gif-12699880907047808909

[junior smelt]

They really should at least say e.g. the minimum height you're at

[candid ocean]

bruh the text book sucks

[swift loom]

is this not the canadian

textbook

[candid ocean]

it is.....

😑

i cant do this

[fathom jewel]

🥁

[candid ocean]

ugh

[swift loom]

u forgot to subtract

by fifty

since ur solving for the intervals

[fathom jewel]

https://tenor.com/view/50cent-smile-driving-oh-well-im-out-gif-5135974

[swift loom]

this matches the textbook

[candid ocean]

why - 50

[swift loom]

ur base function is -28\cos\left(\frac{2\pi}{2.5}x\right)+28.5-50

but its asking when its above 50

[candid ocean]

ohh

yeaheyah

[fathom jewel]

$code$

[swift loom]

didnt know this server had latex

thats nice

[fathom jewel]

i thought you were trying to use it

[swift loom]

how does the command work

[candid ocean]

ziming is it u no way

[fathom jewel]

put $ before and after

[candid ocean]

from data

??

[swift loom]

yes

[candid ocean]

BRUH

LMFAOO

[swift loom]

no way man

[candid ocean]

LKMFAOO

ignore me pls i ask the stupidest questions here

[fathom jewel]

hahaha

[candid ocean]

thats crazy

anyways

ok ty

😝

.close

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[fathom jewel]

https://tenor.com/view/varya-gif-23367099

[swift loom]

💀

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[lament steeple]

Well I guess the question explains itself. I tried using shell method but I can't figure out x, no calculator allowed

[lament steeple]

this is homework so i graphed it anyways and now i'm not even sure you can use shell method

[fathom jewel]

So you are given x = e^ycos(y)?

[lament steeple]

yeah

i transformed the problem to a y based function

[fathom jewel]

I see

[lament steeple]

and its prolly correct i graphed it seems to be right

[fathom jewel]

What you got

[lament steeple]

i cant get it

to use shell method i need to determine upper bound of x

i can't

because that requires solving e^xcosx=pi

[fathom jewel]

Let me write down what we have I am kinda confused

between $y = e^x \cos x$ and $x = 0$ around the $y$-axis from $y = 0$ to $y = \pi$

[flat frigate]

𝔸dωn𝓲²s

[lament steeple]

umm...lets look at the y graph i made which is essentially the same thing

[fathom jewel]

yea send it

[fathom jewel]

there are infinite many intersection with y = pi

something isn't making sense

[lament steeple]

just the green part

[fathom jewel]

how come

https://www.desmos.com/calculator/f0t1o733jp

[lament steeple]

lets just kinda say it is otherwise the is completely not solvable

[fathom jewel]

ok

[lament steeple]

or just infinity making this question too easy

[fathom jewel]

also are you supposed to get the upper bound algebraically

[lament steeple]

this volume is given by Integral (pi-y)2pixdx between x=0 and the upper bound

2 problems

one i cant get the upper bound

second if you take in y you'll find yourself intergrating xe^xcosx

which is prolly solvable but hard

[fathom jewel]

washer method also exists

you can turn cos²(x) into (1+cos(2x))/2

and then you have e^(2x)

[lament steeple]

huh i dont see cos^2

[fathom jewel]

R²

[lament steeple]

then you gotta find y=e^xcosx what is x in terms of y

and also still you need to find the intersect which is the upper bound

the first part is hard because e^x cosx is not one to one

[fathom jewel]

,w N[e^xcos(x)=pi]

[lament steeple]

lmao can't even find first section how am i supposed to even consider more

[fathom jewel]

I just find that problem pretty weird

[fathom jewel]

well it seems the solutions are just pi-periodic

,, xe^x\cos(x) = xe^x \cdot \frac{e^{ix}+e^{-ix}}{2} = \frac{x}{2} \cdot \left (e^{x(1+i)}+e^{x(1-i)} \right )

[lament steeple]

yes this also crossed my mind

[flat frigate]

𝔸dωn𝓲²s

[lament steeple]

lets just find the smallest positive x

[fathom jewel]

well you found the numerical value

[lament steeple]

"no calc allowed"

[fathom jewel]

lol

[lament steeple]

lol

this better not be on my test

[fathom jewel]

you must have a special teacher

,w Integrate[2pi * x * (pi - e^x * cos(x)), {x, 0, 4.739}]

[flat frigate]

Results provided by WolframAlpha

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Requested by adoniszgks

[fathom jewel]

I've got no clue

,w x/2 * (e^(x(1+i))+e^(x(1-i))) = pi

[flat frigate]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by adoniszgks

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[flat blade]

I'm really confused how the first expression simplifies to the second one, i always get a (x-1)^3/2 when i try to simplify

[quiet plume]

$(x-1)^2 = |x-1| \cdot |x-1|$

If you bring one inside the sqrt, you get $$|x-1| \sqrt{(x-1)^2 \frac{x+1}{x-1}} = |x-1| \sqrt{(x-1)(x+1)} = |x-1| \sqrt{x^2 - 1}$$

[flat frigate]

Azyrashacorki

[flat blade]

Ah i see

What about when i do it as $$(x-1)^2 = \sqrt{(x-1)^4}\sqrt{x+1} / \sqrt{x-1}$$ and simplify to get sqrt(x+1)(x-1) ^ 3/2

[flat frigate]

Tobbib
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[flat blade]

oh formatting

*to get this

but it only gives me half of the actual true result

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@flat blade Has your question been resolved?

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[lean otter]

can someone walk me through this I'm horrible with graphs

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

[humble fossil]

a linear function and a piecewise function??

[lean otter]

yes

just the piecewise function

is

what i need help with

[humble fossil]

okay so like this is how my teacher taught it

she usually calls it a breaking point

she would set whatever is inside of the absolute value symbol equal to 0

and she does a number chart

like this @lean otter right

[lean otter]

yes sorry

I'm here

To find the vertex correct? @humble fossil

[humble fossil]

yeah vertex

after finding the vertex u just apply the signs from the sign chart to whatever is inside the absolute value symbols

so like

-(x-2) and (x+2)

[lean otter]

wait

but

why

like

I don't understand, where do the signs come from and why are they important to the function

[humble fossil]

when u plug in any number left of 2 into x

u get negative

right of 2

u get positive

[lean otter]

Oh yes

[humble fossil]

they’re considered important to the function bc ur just turning the absolute value function into a piece wise function

[lean otter]

Oh okay I see

[humble fossil]

so yeah when u see an absolute value function just think of this

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[stark bear]

how can i find how much solutions this has

[safe radish]

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[magic vale]

Hey. Just wanted to check that the correct working out in this situation is to find the minimum value of faces not visible (15) and subtract this from the total possible value if all faces were visible(3*21 = 63) to get an answer of 48. Thanks in advance.

[tidal pasture]

Yes

Correct

[magic vale]

Thanks

Apreci8 it

.close

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[soft folio]

Where am i going wrong, the black is my first attempt pencil is second im supposed to get 2a - b = 4

[safe radish]

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[vagrant ice]

,rotate

[vagrant ice]

what's the original question?

[vagrant ice]

ah you're just solving $k + \lambda k = 2k - \mu k$

[flat frigate]

south, just south

[soft folio]

yeah i tried by getting mew and lambda in terms of a and b then subtituting it into that

[vagrant ice]

great and then the other equations are:
$3 - \lambda = 4 + (1 - \lambda) a$

$-2 + 2 \lambda = 4 + (1 - \lambda) b$

I see now

[flat frigate]

south, just south

[vagrant ice]

ah so $a = \frac{-1 - \lambda}{1 - \lambda}, b = \frac{-6 + 2 \lambda}{1 - \lambda}$

[flat frigate]

south, just south

[vagrant ice]

now just compute 2a - b from this

[soft folio]

im not following, do i multiply a in terms of lambda by 2 and b in terms of lambda by -1?

[vagrant ice]

yes

[soft folio]

if im subsituting something in for a and b how is my final answer gonna haev a and b

[vagrant ice]

it won't

you just want to show 2a - b equals 4

[proud dust]

buongiorno

[vagrant ice]

hi?

[vagrant ice]

okay so you want to replace a and b with lambda
not the other way around

[soft folio]

oh then if i add 2 of a in lambda terms and -b in lambda terms ill get 4?

[vagrant ice]

yes, you have to show your work for that step though

but yes the numerator and denominator cancel

[soft folio]

oh okay, 1 thing tho why wouldnt it work if i were to replace lambda with a and b?

[vagrant ice]

it just doesn't work; you're going in the wrong direction with those steps you wrote

[vagrant ice]

.

[soft folio]

hmmm okay

thansk for the help

[vagrant ice]

no worries

don't feel bad that it didn't work out btw!

at least you came here to ask for help when you couldn't figure it out

[soft folio]

yeah i kinda got a new way of thinking about questions like that, usually i just mindlessly do stuff and get the same equation it was weird that it didnt wrok out this time but ur method made more sense

[vagrant ice]

yeah worked solutions also help a lot at times

[vagrant ice]

alright type .close if you're done

[soft folio]

alr!

.close

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[regal violet]

My thought process is to use induction followed by Vieta's formula, nevertheless I am still confused

[eternal carbon]

descartes' rule of signs

[regal violet]

That only identifies the number of roots though?

The number of positive / negative roots

There are 3 sign changes so there are 3 / 1 +ve roots?

[eternal carbon]

oh wait

factor f_n(x)

[regal violet]

Factorize?

[strong oar]

couldn't you just find explicit expressions? $C_n = (-1)^{n-1}5^{2^n}$ for example

[flat frigate]

aeiou

[strong oar]

sorry

[eternal carbon]

C is always negative

[regal violet]

Yeah C is always negative

[eternal carbon]

yes

[regal violet]

And C100 is just -5^2^99 but I don't think that helps

[eternal carbon]

to get an idea, consider f_1(x) and maybe f_2(x)

this question is solved by factoring

[regal violet]

So I factorize it with An, Bn and Cn in it?

[eternal carbon]

knowing that your goal is to factor should be enough of hint

[regal violet]

ALr I'll try that rn

Gimme a min

[regal violet]

Do I use induction to prove that all of fn(x) can be factorized?

[eternal carbon]

to prove a given factorization is valid yes induction is fine

though i would recommend actually finding the factorization before trying to prove it...

probably you know this. just saying though

[regal violet]

Like finding the numerical values and stuff?

[eternal carbon]

finding the factorization

[regal violet]

Oh the general expression for the factorization

?

f1(x) = (x-1)(x^2 -5x +5)

[eternal carbon]

yes ofc what else lol

[regal violet]

Mb

[eternal carbon]

but to do that it would be very wise to look at f_1 hence why this msg

looks like you already looked at f_1

should give you an idea on how to proceed

[regal violet]

WAIT I just realized

All of them have 1 as root

[eternal carbon]

indeed

[regal violet]

For some reason

[regal violet]

How did you find out so quickly

Is there some sort of trick

Or technique

I'm missing out on

[eternal carbon]

no, i just have calculator open when helping people. so when looking for these patterns it does not take too long

[regal violet]

Oh ok

[eternal carbon]

but for this (type of) problem, intuition is very important

[regal violet]

So now I know fn(x) = (x-1)(x^2 ......)

[eternal carbon]

playing around with problem before attempting it

[regal violet]

Got it

[regal violet]

Wait but

Don't I have to show mathematical proof that 1 is always a root

I can't just trial and error and then make this claim right

[eternal carbon]

no

claim: f_n(x) = blah

proof of claim: etc

you don't need to motivate your claim, so long as your proof is correct

the idea is you find a general form for f_n(x) and prove that form is correct using induction

[regal violet]

So do I substitute x=1 into the general fn(x)

And prove that it is 0

[regal violet]

Ah

[eternal carbon]

no

[regal violet]

OH

I get what u mean now

So I claim fn(x) = (x-1)(x^2.....)

Is true

And then inductive step

?

When n = k+1

[eternal carbon]

no

because you have not found the full factorization

[regal violet]

Oh yeah.

[eternal carbon]

your goal is to prove that f_n(x) has 3 real distinct positive roots

& just proving f_n(x) = (x - 1)(x^2 ...) does not help you do this

[regal violet]

Can't I just show the discriminant of the quadratic >0

Oh wait

It might also be x=1 as a root tho

Hence repeated

[eternal carbon]

find the full factorization

[regal violet]

Alr let me have a shot

fn(x) = (x-1)(x^2 -?x +5^2^(n-1))

Tryna figure out what ? is

-5, -15, -175 for the first 3 of f(x)

[regal violet]

I think I need a small hint

[eternal carbon]

note you might not be able to find a closed form expression for ?

everything given in the problem is defined recursively, so ? is also probably defined recursively

[regal violet]

Ohhh

Kinda difficult to determine the recursive expression

[regal violet]

I don't think I can find the recursive expression...

Any hints

[eternal carbon]

given the quadratic is in form (x^2 - d_n x - c_n), the d_n can be expressed in terms of d_{n - 1} and c_{n - 1}

[regal violet]

Isn't c_n just 5^(2^(n-1))

I verified for the first 3

Of f(x)

Nvm they are the same thing

Question already gave recursive for c_n

[regal violet]

I don't think I can find a d_n recursive with the numbers -5, -15 and -175 tho

I don't see a pattern in any way

[eternal carbon]

also consider the values of c, which you have -5 and -25

[regal violet]

Still don't see a relationship...

Do I need to use vieta's formula?

[eternal carbon]

d_n = d_{n - 1}^2 + 2c_{n - 1}

[regal violet]

Hm

[eternal carbon]

for my equation

might not be the same for yours

could be plus or minus flipped somewhere

wtv, but the idea is there

[regal violet]

Oh yeah you are right

Dude how are you so good bruh

Ok now that we have the general expression, I have to prove that quadratic has 2 distinct positie real roots that are not 1?

[regal violet]

Wait

I can just use vieta's formula now

For the quadrati

Cuz their sum and product must both >0

Their product is -Cn, which is positive

@eternal carbon How do you prove d_n is always positive?

Help please

@eternal carbon Hello?

[safe radish]

@regal violet Has your question been resolved?

[regal violet]

Anybody help pls

[safe radish]

@regal violet Has your question been resolved?

[safe radish]

@regal violet Has your question been resolved?

[safe radish]

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[jaunty root]

How do I solve part C?

[plucky elk]

Did you do a and b?

where even is the definition of S and how it relates to Q?

[safe radish]

@jaunty root Has your question been resolved?

[jaunty root]

yes i have done a and b

[jaunty root]

i dont know? this is all the information i was given

[plucky elk]

what does the footnote number 1 say

[jaunty root]

"The function S is called the entropy. That δQ/T is an exact differential (and not just for ideal gases) was
discovered by Clausius in 1867."

[plucky elk]

alright i see now

You could use integration by parts on $\int \frac{\delta Q}{T}$

[flat frigate]

riemann

[plucky elk]

after dividing this by T and integrating, you should be able to write it as the integral of a single function instead of a sum of two functions

[jaunty root]

If I divide by T, I should have 1/Vdv+k/TdT?

[plucky elk]

yea integrate those separately then combine

[jaunty root]

ln(V)+kln(T)+C?

is there some way i could solve for C?

[plucky elk]

derivative of a constant = 0

[jaunty root]

mm

true

so yeah we cant really get C then

so that would just be the answer to the question then?

[plucky elk]

The answer is the function S which you haven't shown yet

[safe radish]

@jaunty root Has your question been resolved?

[jaunty root]

is this not S?

[plucky elk]

no

[jaunty root]

hm

ok then

[jaunty root]

would it be the integral of this?

[plucky elk]

no

you should be writing down all your steps next to each other so you can see

[jaunty root]

yes i did this. so if i integrate SQ/T I should get S right?

[plucky elk]

try it

[fathom jewel]

It's essentially like an exact differential equation

[plucky elk]

write S here and show delta Q / T = dS

[fathom jewel]

Johnn #❓how-to-get-help

[jaunty root]

im not sure i know what that is

[fathom jewel]

dont mind

[fathom jewel]

Why is this not S?

[plucky elk]

pain

[plucky elk]

i was wrong, i thought you had to integrate one more time

[fathom jewel]

I am sorry

[plucky elk]

$\int \frac{\delta Q}{T} = \int dS = S$

[flat frigate]

riemann

[plucky elk]

Thanks @fathom jewel

[jaunty root]

ok, so this is my answer then?

[fathom jewel]

yes

[jaunty root]

cool. Thank you

.close

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[somber fern]

can i get sum help

[safe radish]

@somber fern Has your question been resolved?

[acoustic mulch]

lines are usually in the form of y=mx+b, in which m is the slope of the line and b is the y-intercept (where the line crosses the y-axis)

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[rain pine]

anyone know where ive went wrong here

[plucky elk]

2x + 8x = 10x

[keen plank]

just about to say that

[rain pine]

ohhh i see

rookie mistake

thank you

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[round dove]

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

smh i got it to 496 instead of 16

[round dove]

when f'(2)

[narrow ridge]

😭

[round dove]

f'(2)=42^3-2*4(2)^1

f'(2)=8^3-16

where did i do wrong?

[narrow ridge]

first of all

differentiate x^4

[flat frigate]

w

[narrow ridge]

4x^3

then differentiate 4x^2

you get...

[narrow ridge]

wrong process

[round dove]

yeah the thing did not work when onsertinf the *

its hidden discord thing

wait ima type what i did again looks messed up

[narrow ridge]

power rule

[round dove]

f'(2)=4 * (2)^3 - 2*4(2)^1 dont mind the spaces

[long copper]

that gets you 16 doesnt it

[round dove]

emm 4*(2)^3 is 16? but the - makes it 0? please write it out i cant make it smh

-2 * 4(2)^1

-2*8
-16
??????

[long copper]

4*(2)^3 is 32

2^3 is 8

8*4 = 32

[round dove]

damn bro im really slow

lol

[long copper]

happens to the best of us

[round dove]

thanks Tom and Capybara

.close

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[cedar lance]

I think I kinda get it, but I'm not sure. Why is 2.1% turned into 1.021?

[lean otter]

2.1% is 0.021 in a decimal

[cedar lance]

Yeah, but the answer given to us was D. I don't get why. 🥲

[lean otter]

since we are increasing, it is 100% + 2.1% = 102.1% = 1.021

think about the parent function of exponential growth

[cedar lance]

OHH OKAY. THAT MAKES SENSE! thank youuuu

[lean otter]

oki

[cedar lance]

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[lean otter]

yw

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[proper nexus]

the question basically asks which one of the options is true

I think either c or d

i'm pretty sure d) is true but i don't know for which values c) would be false

[safe radish]

@proper nexus Has your question been resolved?

[blazing sigil]

b and c i think

[safe radish]

@proper nexus Has your question been resolved?

[proper nexus]

can't be b) because x^2 + x can't = something negative in R

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[sacred ferry]

would it be fair to say in cases of common notation "n!!" is to be interpreted as "(n!)!" as opposed to a double factorial?

[tawdry hamlet]

depensd on ur level of math lol

but typically, i dont think so

n!! is the double factorial, if a paper said (n!)! specifically, then read it as that

and vice versa

[toxic stratus]

nobody interprets n!! as (n!)!

[sacred ferry]

yeah, my friend was auguring with me about stupid notation things

[lime dust]

Sometimes notation has interpretations, is not the case

[solid shell]

Point of inquiry from someone who is ignorant to this: what's the difference?

[toxic stratus]

well it is believable that one might interpret n!! as (n!)! if one were not aware of double factorials

but generally people in the know do not do this

[sacred ferry]

i think they weren't

[lime dust]

I meant correct interpretation

[toxic stratus]

whether it's correct is another matter

one writes notation to their audience

if the audience is familiar with one thing, then that shall be the notation

[toxic stratus]

6!! = 6*4*2 for example

one skips every second number

[solid shell]

So it's factorial but only of same parity

[toxic stratus]

yes

[mortal sandal]

230 - 220 / 2

[solid shell]

Interesting

Thanks

[mortal sandal]

you may not believe it but this is equal to 5!

[lime dust]

That is not notation

Is format

[toxic stratus]

i believe it

[lime dust]

[mortal sandal]

Snow knew what was coming lol

[toxic stratus]

[lime dust]

But enthusiastic 5 or factorial 5

Up to the audience

[mortal sandal]

20 - 10 / 2

you may not believe it but this is equal to 5!!

[toxic stratus]

[safe radish]

@sacred ferry Has your question been resolved?

[safe radish]

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[jovial sentinel]

tried doing implicit differentiation, what did i do wrong?

[jovial sentinel]

unless i should instead be doing product rule

[noble kiln]

you dont even need to differentiate implicit

just $y = \frac{4}{x}$ should be ok

[flat frigate]

Death.cv

[noble kiln]

then quotient rule @jovial sentinel

[jovial sentinel]

the problem asks me to solve using implicit

[noble kiln]

oh

we dont use dx/dx i think

[jovial sentinel]

i did 1 * y + x * 1 dy/dx

[jovial sentinel]

yea probably not

its just equal to 1

[noble kiln]

wait i see a bit

its multiply

cant d/dx one side and d/dx the other two

[solid shell]

You're on the right track

[noble kiln]

so d/dx 4 = d/dx xy i think

[jovial sentinel]

do i move the y to the other side

so its x * 1 dy/dx = -y

[noble kiln]

product rule i think

I might be wrong

[solid shell]

You've differentiated the left side, now just need to differentiate right wrt x and then solve for y'

This channel is taken

Read #❓how-to-get-help

[noble kiln]

!occupied

[safe radish]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[solid shell]

Then open your own help channel and ask it

[noble kiln]

dude just go to #help-24

or smth

[jovial sentinel]

have i not already differentiated the left and the right?

[solid shell]

Ah I forgot to look at your picture

So you have $y+x(\frac{dy}{dx}) = 0$

[flat frigate]

m. frost

[jovial sentinel]

ok yes

[solid shell]

Now just solve for dy/dx

[noble kiln]

oh

im so stupid lol

[jovial sentinel]

is the 1 connected to dy/dx or is it like on the x

poorly phrased question

[solid shell]

$\frac{d}{dx}y = 1\frac{dy}{dx}$

[flat frigate]

m. frost

[solid shell]

I just shorthanded it

[solid shell]

That's what I got

[jovial sentinel]

what if instead of 1 it was 2?

[solid shell]

like x(2y)=4

$2y+x(2\frac{dy}{dx})=0$

[jovial sentinel]

like that

?

[solid shell]

yeah

[jovial sentinel]

okay i think i got it

thank you

[solid shell]

[jovial sentinel]

👍

.close

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[jovial sentinel]

.reopen

[safe radish]

✅

[jovial sentinel]

another question, i need to find the 2nd derivative d^2y/dx^2
of x^2+y^2=36

i got dy/dx = -x/y

so how would i get the 2nd derivative from here

[lapis shadow]

Differentiate it again

[jovial sentinel]

differentiate -x/y?

[lapis shadow]

Yes

[jovial sentinel]

(-y-x dy/dx) / y^2

👍 or 👎

[crude condor]

polynomials explained

[lapis shadow]

Yes but the signs are not correct

[jovial sentinel]

where?

it seems right to me

[lapis shadow]

It's not two minuses

[jovial sentinel]

is it not the derivative of -x

and then subtracted by -x* derivative of y

[lapis shadow]

The quotient rule says (f/g)' = (f'g - g'f)/g²

Yes

Subtracted

[jovial sentinel]

-(-x)?

[lapis shadow]

Yes

[jovial sentinel]

okay

so -y+x dy/dx / y^2

[lapis shadow]

Yes

U can replace dy/dx by what u know it is equal to

[jovial sentinel]

so -y+x(-x/y) / y^2

[lapis shadow]

That's a + not *

Yes

[jovial sentinel]

oops ye

a

-x^2 - y^2 / y^3

[lapis shadow]

Yea

🥳

[jovial sentinel]

okay, great!

thank you

.close

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[viscid fulcrum]

I am trying to compute the total number of ways that teams can choose heroes in a Deadlock game. There are a total of 23 heroes. Each game consists of 12 heroes (where each team has 6 heroes). There can be no duplicate heroes in a game. What are all the possible ways you can pick heroes for a Deadlock game?

[viscid fulcrum]

Combinatorics break my head

Apparently the answer is 23C12 * 12C6 / 2

I get that we have to divide by 2 to avoid double counting team symmetry

(Swapping teams represents the same game)

[safe radish]

@viscid fulcrum Has your question been resolved?

[safe radish]

@viscid fulcrum Has your question been resolved?

[swift loom]

couldnt you just do 23C12, I dont think you are overcounting. You are picking a game of 12 heroes from a pool of 12.

theres many ways to do this

you can also just do 23C6 * 17C6

but to adjust for overcounting it devides by 2

since you consider swapping around team a and b as 2 different configurations

[viscid fulcrum]

How did you get to this result?

[swift loom]

first choose the first team

you choose 6 out of a pool of 23

then you choose the second team

[viscid fulcrum]

Ah I see

[swift loom]

then have 23-6 choices

[viscid fulcrum]

And this doesn't accidentally pick a duplicate member?

[swift loom]

it depends if you consider swapped teams

duplicates

if team a has (1, 2, 3) and team b has (4, 5 6), do you consider the same game composition but swapped teams as 2 games

[viscid fulcrum]

Yeah that would be considered the same game

Which is why we divide by 2

[swift loom]

then u would have to devide by 2

[viscid fulcrum]

I meant that

Okay, we first choose 6 from 23. Then you have 17 left over to choose from, but how does this ensure that you don't include the same member in that 17?

Or is that already implied?

I guess it kind of makes sense... Somewhat tricky to think about

[swift loom]

because you are not counting them

we shrinked the sample

[viscid fulcrum]

You exclude them?

I see

Thank you :) @swift loom

[swift loom]

all good

[viscid fulcrum]

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[lapis shadow]

For sqrt[(z - a)(z - b)], why is the branch cut from a to b?

[lapis shadow]

a, b are real numbers, and b > a

With z - a and z - b as above, if I write theta1 + 2npi and theta2 + 2mpi

Instead of just theta1 and theta2

The function value is $\sqrt{r_1r_2}\cdot e^\frac{i(\theta_1+\theta_2)}{2}\cdot e^{i(m+n)\pi}$

[flat frigate]

SilverSoldier

[lapis shadow]

If I'm at a point to the right of b, I can still find m and n so that it is multivalued

Why don't I branch there?

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@lapis shadow Has your question been resolved?

[lapis shadow]

@<@&286206848099549185>

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@lapis shadow Has your question been resolved?

[azure delta]

Every point potentially multivalued. The issue is whether you can choose one of the values and preserve continuity.

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@lapis shadow Has your question been resolved?

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[clever forge]

can someone help me with this limit

where should i start

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[blissful flicker]

How is this possible?

[blissful flicker]

This is how I have done this, I don't get what's wrong in my method

[hollow kiln]

should be -1.3333…

missed a negative

[blissful flicker]

but (-2)/(-1.5) = 1.333

[hollow kiln]

row 3 should be 0 0 1 0

[hollow kiln]

R3 —> R3 + R2

will return 0 0 1 0

[vagrant ice]

no the issue is that -1.5 + -1.5 * -1 = 0 right
but -2 + -1.5 * 0 = -2

[blissful flicker]

oh shoot, i did the wrong question

anyways thank you guys 🙏

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[lean otter]

Making the rectangles smaller and smaller would eventually give us points

And then we choose only those rectangles(points) which lie inside and on the boundary

[desert pasture]

which book is this

[lean otter]

Yo

Thomas calc

[desert pasture]

ah

cool

hi

[lean otter]

I get this

But

[lean otter]

So matter how small we make the polar rectangles (which eventually give us arcs) we cant get the exact area

https://tenor.com/view/head-scratch-confused-gif-14061427

That's the problem

[limber jewel]

@lean otter what do you mean by you cant get the exact area?

[lean otter]

Prof Leonard didn't go in depth explaining it, in the video

Like

[lean otter]

😅I'm not able to explain it properly

[limber jewel]

@lean otter so u mean to tell that the polar rectangle's area becomes a curve and does not have an area?

[lean otter]

Yes

But that's not the problem I'm facing

[limber jewel]

@lean otter then what is your issue here?

[lean otter]

Here for the first curve which touches g1(x(

We should be considering it right

[limber jewel]

@lean otter what do you mean?

[lean otter]

Lol

We are making the polar rectangles really small

Which leaves us with arcs

[limber jewel]

Yes

[lean otter]

So we sum up these arcs to get the area

But the sum of arcs always gives us a region like this

Assume the top is an arc of a circle

[lean otter]

But this is nothing like the region we have

I'm here please dont ping 😭

[limber jewel]

think of it this way, we are making the polar rectangles smaller and smaller but they never collapse to form an arc , they remain small polar rectangles, if they are small enough u can approximate these polar rectangles with small trapezoids , when we mean adding areas we mean that we add the areas of these small trapezoids