help-41

[verbal lark]

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[quick ridge]

😟

that’s not the first derivative test

[verbal lark]

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[quick ridge]

fine

if f has a relative maximum at x = c if for x < c f’ > 0 and for x > c f’ < 0 and at c f’ = 0

basically just saying increases to max stops then decreases thereafter

and for min it’s the opposite

decreases to a min, stops, then increases thereafter

just look at what happens just before and after

the critical point

for min you’ll have f’ < 0 for x < c and f’ > 0 for x > c

[verbal lark]

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[quick ridge]

yea it’s like interval testing

sign chart

like you did back in precalc

[verbal lark]

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[quick ridge]

🤔

bro thinks ik what unit 5

is

as if that’s some universal shit

[verbal lark]

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[quick ridge]

fun

[verbal lark]

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[verbal lark]

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[quick ridge]

what’s up

do what

same thing?

EVT

is that the question

[verbal lark]

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[quick ridge]

what’s confusing

[verbal lark]

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[quick ridge]

whet

that’s y

not y’

[verbal lark]

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[quick ridge]

y’ is never zero brother

[verbal lark]

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[quick ridge]

ya cant

[verbal lark]

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[quick ridge]

😹😹😹

[verbal lark]

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[quick ridge]

you can’t sir

[verbal lark]

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[quick ridge]

hint : ||read the damn question especially the part in italics||

[verbal lark]

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[quick ridge]

💀

[verbal lark]

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[quick ridge]

brother

stationary point is where f’ = 0

but not all critical points are stationary points

🙈🙈

[verbal lark]

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[quick ridge]

but but what

[verbal lark]

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[quick ridge]

a critical point is a value of x such that f’ = 0 or f’ doesn’t exist

[verbal lark]

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[quick ridge]

wont even

[verbal lark]

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[quick ridge]

😭😭🙏🏻

where is this function not differentiable

meaning f’ doesn’t exist

[verbal lark]

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[verbal lark]

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[quick ridge]

yea

because -2≠1

[verbal lark]

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[quick ridge]

let’s consider an easier piece wise ok

$f(x) = |x|$

[grizzled pagoda]

knief

[verbal lark]

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[quick ridge]

do you know the piecewise definition of this

[verbal lark]

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[quick ridge]

how would we write |x| as a piecewise function

[verbal lark]

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[quick ridge]

close

but not quite

[verbal lark]

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[quick ridge]

yea

so what’s the derivative for x < 0

[verbal lark]

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[quick ridge]

^

[verbal lark]

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[quick ridge]

no

[verbal lark]

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[quick ridge]

for x < 0 we have -x

not x

[verbal lark]

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[amber wasp]

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[amber wasp]

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[river spade]

can somebody explain why its inreasing on (-2,2) even though f'(x) is less than zero on that interval

[ivory river]

It's plain wrong

[river spade]

is it switched??

like the answer for the first part goes into the second cause thats what im thinkjing

[ivory river]

Do you see anything wrong in the right box?

[river spade]

no

just the first one

[amber wasp]

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[split sail]

I can't evaluate the integral to get 8/3 unit^3 for the volume of two cylinders x^2 + y^2 = 1 and y^2 + z^2 = 1 and z = 0

[patent raptor]

I think I have dealt with something like that before

and it didn't get me anywhere lol

#calculus message

[split sail]

i guess im not the only one, ty

[patent raptor]

But let's see

maybe this time

we can cook something up 🔥

,, \int_0^{2\pi} \int_0^1 r\sqrt{\cos^2\theta+\sin^2\theta(1+r^2)} : \dd r \dd \theta

[grizzled pagoda]

𝔸dωn𝓲²s

[patent raptor]

Maybe u = 1+r^2 works

,, \int_0^{2\pi} \frac{1}{2} \int_1^2 \sqrt{\cos^2\theta+\sin^2\theta \cdot u} : \dd u \dd \theta

[grizzled pagoda]

𝔸dωn𝓲²s

[patent raptor]

nah you would divide again by sine

[split sail]

what tool you using to make equations appear like that

[patent raptor]

,w 0.5 * Integrate[-2((cos(theta)+sin(theta))^(3/2)-(cos(theta)+2sin(theta))^(3/2))/ (3sin(theta)), {theta,0,2pi}]

TeXit

,w simplify -2((cos(theta)+sin(theta))^(3/2)-(cos(theta)+2sin(theta))^(3/2)) / (3sin(theta))

[split sail]

Integrate[1, {x, -1, 1}, {y, -Sqrt[1 - x^2], Sqrt[1 - x^2]}, {z, 0, Sqrt[1 - y^2]}]

[patent raptor]

,w Integrate[1, {x, -1, 1}, {y, -Sqrt[1 - x^2], Sqrt[1 - x^2]}, {z, 0, Sqrt[1 - y^2]}]

[grizzled pagoda]

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Requested by adoniszgks

[patent raptor]

I am also surprised with this integral, that polar coords dont make it work

because of the division with sine

0 and 2pi become critcal

[split sail]

I have the same struggle, so im skipping it for now and visit it later

[patent raptor]

,, \frac{1}{2} \int_0^{2\pi} \frac{1}{\sin^2\theta}\int_{\cos^2\theta}^1 \sqrt{u} : \dd u \dd \theta

[grizzled pagoda]

𝔸dωn𝓲²s

[patent raptor]

,, \frac{1}{3} \int_0^{2\pi} \frac{1-\abs{\cos^3\theta}}{\sin^2\theta} : \dd \theta

[split sail]

is it possible to use trig substitution instead

[patent raptor]

actually

[grizzled pagoda]

𝔸dωn𝓲²s

[patent raptor]

let's split it

0 to pi/2 +
pi/2 to pi -

,, \frac{2}{3} \int_0^{\frac{\pi}{2}} \frac{1-\cos^3\theta}{\sin^2\theta} : \dd \theta + \frac{2}{3} \int_{\frac{\pi}{2}}^{\pi} \frac{1+\cos^3\theta}{\sin^2\theta} : \dd \theta

[grizzled pagoda]

𝔸dωn𝓲²s

[patent raptor]

indefinite integral

,, \frac{2}{3} \lim_{a \to 0} \int_a^{\frac{\pi}{2}} \frac{1-\cos^3\theta}{\sin^2\theta} : \dd \theta + \frac{2}{3} \lim_{b \to \pi} \int_{\frac{\pi}{2}}^{b} \frac{1+\cos^3\theta}{\sin^2\theta} : \dd \theta

[grizzled pagoda]

𝔸dωn𝓲²s

[patent raptor]

,w Integrate[(1+cos(x)+cos(x)^2),{x,0,pi/2}]

[grizzled pagoda]

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Requested by adoniszgks

[patent raptor]

,, \frac{2}{3} \left ( -2 - \lim_{a \to 0} (\cos a - 2) \cot ( \frac{a}{2} ) + \lim_{b \to \pi} (\cos b - 2) \cot ( \frac{b}{2} ) + 2 \right )

[grizzled pagoda]

𝔸dωn𝓲²s

[patent raptor]

ok the a -> 0 is the problem

,w 2/3 * Limit[Integrate[(1-cos^3x)/sin^2x),{x,pi/2,a},a->0]

maybe one can take advantage of periodicity

,w Limit[(cos(a)-2)cot(a/2),a->0^-]

,w plot (1-cos^3x)/sin^2x between-pi+0.1 to pi-0.1

it seems to exist actually hmm the area unter the curve

[grizzled pagoda]

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Requested by adoniszgks

[patent raptor]

it's crazy

,, \frac{1-\cos^3\theta}{\sin^2\theta} = \frac{(1-\cos\theta)(1+\cos\theta+\cos^2\theta)}{1-\cos^2\theta} = \frac{1+\cos\theta+\cos^2\theta}{1+\cos\theta} = 1 + \frac{\cos^2\theta}{1+\cos\theta}

this should work

theta = 0

and theta = pi/2

so crazy

,, \frac{2}{3} \int_0^{\frac{\pi}{2}} 1 + \frac{\cos^2\theta}{1+\cos\theta} : \dd \theta + \frac{2}{3} \lim_{b \to \pi} \int_{\frac{\pi}{2}}^{b} \frac{1+\cos^3\theta}{\sin^2\theta} : \dd \theta \

[grizzled pagoda]

𝔸dωn𝓲²s

𝔸dωn𝓲²s

[amber wasp]

@split sail Has your question been resolved?

[patent raptor]

,w Integrate[cos^2(x)/(1+cos(x)),x]

[grizzled pagoda]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by adoniszgks

[patent raptor]

damn pretty ugly

but I think it should work out lol

yup
https://www.desmos.com/3d/5gt1rv6u8c

crazy

[ \frac{2}{3} \left ( \frac{\pi}{2} + \frac{1}{2} \left ( -2\frac{\pi}{2}+3\tan ( \frac{\pi}{4} ) + \sin( \frac{3\pi}{4} )\sec(\frac{\pi}{4} ) \right ) + \lim_{b \to \pi} (\cos b - 2) \cot ( \frac{b}{2} ) + 2 \right ) ]
[ = \frac{2}{3} \left ( \frac{\pi}{2} + \frac{1}{2} \left ( 4-\pi \right ) +2 \right ) ]
[ = \frac{2}{3} (2+2) ]
[ = \frac{8}{3} ]

[grizzled pagoda]

𝔸dωn𝓲²s

[patent raptor]

@split sail

[patent raptor]

I basically used here a³-b³ = (a-b)(a²+ab+b²)

but this is a crazy integral, i dont know if there is a simpler way

[split sail]

omg, ty

but imma go ask my teacher bc im crying

[amber wasp]

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[fervent axle]

Hello, was helping someone out but I also couldn't figure it out, the correct answer is 30°. How should I approach this?​

[fervent axle]

It's 2x btw, sorry if it's blury

[nova cloud]

uh waht is it asking

cuz thats not english

[fervent axle]

There's an English translation down

Below the other language

[lusty pine]

any inscribed angle where they are inscribed from the diameter is 90

[fervent axle]

Mhm, that I know. The answer however seems to have something to do with the 2x. I couldn't figure it out since I obviously can't divide 90° by 2 since it equals to 46°

45° woah

[amber wasp]

@fervent axle Has your question been resolved?

[fervent axle]

<@&286206848099549185>

[mint nacelle]

angle in a semicircle is?

yeah

so x + 2x = 90

[fervent axle]

so basically x + x = 45

Sorry for being dumb

Guy..s

[mint nacelle]

no.......

[fervent axle]

😭 😭 😭 😭

HELP

I can barely understand math in English mb

Having a hard time translating it with Google translate

https://cdn.discordapp.com/emojis/1267562446998999061.png?quality=lossless&name=sigh&size=48

[mint nacelle]

what happened to your algebra

[fervent axle]

Oh yea I eman

[fervent axle]

Thsi is unfinised

I forgot what x+x equals to

😭

What does it..

[mint nacelle]

https://www.youtube.com/watch?v=Z-ZkmpQBIFo

[fervent axle]

HELP😭 😭 😭 😭 😭

Hey...

I'm just really tired, it's supposed to be my English and malay exam tomorrow but someone suddenly asked me for help

What's x+x 😭

I need to let this go..

YEAH LMAO 😭

[mint nacelle]

2x

[fervent axle]

Rahh....

[mint nacelle]

3 + 3 = 6
4 + 4 = 8
5 + 5 = 10
6 + 6 = 12

get the pattern?

[fervent axle]

IM RIGHT

[mint nacelle]

but that's not at all related to x + 2x = 90

what's x + 2x?

[fervent axle]

Oh wait

OH...

3x...

https://cdn.discordapp.com/emojis/1267562446998999061.png?quality=lossless&name=sigh&size=48

[mint nacelle]

yes

[fervent axle]

I'm dumb

[mint nacelle]

3x = 90

[fervent axle]

Sorry y'all...

Thanks yall

Also I recognise your main pfp @split sail

Shhh

Gonna close the channel now, thanks 😭 😭 😭 😭 😭 😭

Especially because of how dumb I was

.close

[amber wasp]

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Use .reopen if this was a mistake.

[fervent axle]

https://cdn.discordapp.com/emojis/1267563775318560940.png?quality=lossless&name=alfmabrook&size=48

Yes, the person straight up asked me on my birthday too 😭

Thnak you!!!

Doing exactly that 😭 😭 😭 😭

I'M NOT EVEN SMART

[cobalt sandal]

Y'all who do I ask for help in integration?

Ok so imma send you my previous test paper cause I legit got a zero

[molten shale]

which q

[cobalt sandal]

All bro 😭

I'm sorry I really got no clue

[molten shale]

its to small for me to read

[cobalt sandal]

I'm cooked in my next test

At this point I might as well take all the help I can get

[molten shale]

point m is the turning point

do u know how to find a turning point

question 1a

ok

[cobalt sandal]

Preferably on paper

[cobalt sandal]

I don't think we have to find the turning point right?

[molten shale]

well

[cobalt sandal]

Cause I'm pretty sure everything on this paper is strictly integration

[molten shale]

point m is the local max

u can find point m by differentiating this function and let it =0

[cobalt sandal]

So Max point is differentiation of y?

[molten shale]

yeah

when it = 0

[cobalt sandal]

Np bro

[cobalt sandal]

@molten shale

So like this?

I used the product rule

[molten shale]

yes

and let it =0

and solve for x

[amber wasp]

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[split sail]

no

[split sail]

I need urgent help

[ripe vale]

With?

[split sail]

I have no idea, and I don't know how to speak English.

[ripe vale]

Sine of the angle is just AC /BC

Opposite side by hypotenuse

Can you do that?

[split sail]

I don't understand seriously

:c

[ripe vale]

8/17 is sinß

[split sail]

and I only have 10 minutes to solve it

[ripe vale]

Is that a test?

The question is in a language i dont understand:( I'm afraid i cant help you

[split sail]

geometry homework which I couldn't do for medical reasons

[ripe vale]

I'm sorry :(

[ripe vale]

Well see , in the 7 th question they have given tanß as 1/3

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@split sail Has your question been resolved?

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[split sail]

The region in the first quadrant bounded by
x = 0, y = sin(x ^ 2) and y = cos(x ^ 2) is rotated around the y-axis. The resulting rotational volume is?

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@split sail Has your question been resolved?

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@split sail Has your question been resolved?

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[thick canyon]

<@&286206848099549185>

Ik im not supposed to ping helpers until 15 mins but im kinda in a hurry

[gilded nebula]

FD x DC = BD X DE

go from there

and the question is not clear

[thick canyon]

find the area

it cant get clearer than this

absolutely nothing else is given

triangle ain't isoceles or anything

just a regular triangles

the lines from the angles aren't perpendicular or bisectors

they are random

just random

[ivory river]

draw AD, denote parts by x and y and make ratios from the available areas.

[thick canyon]

ignore the left part

the area is?

ik im not supposed to ask u for the answer but

i just dont understand at damn thing

[ivory river]

its up to you to find

[thick canyon]

would u mind solving it for me, I've been at it for 3 days and I still have no solution and the other helpers refuse to help me out

Im trying to use online but no matter what I type I cant get this type of question

I cant find out online how to solve it and I dont understand what the other helpers are saying

do you understand my frustration?

[ivory river]

I do mind ) I've given almost entire solution. You just need to look at it. and solve the system

[thick canyon]

ok

just tell me this

[thick canyon]

why does x/9 = (y+6)/11

just tell me that

is it similarity in triangles?

crogruety?

like what property are u using

[ivory river]

thats the propertie of the areas and fractions

[thick canyon]

never heard of that

what does it state

[ivory river]

Areas of the triangles with the same height

[thick canyon]

you are aware that F and E are not parallel to the base

[ivory river]

absolutely

[thick canyon]

my god I still dont see it

[ivory river]

do you agree with that?

[thick canyon]

ok so all I need to keep in mind is x/9 = y+6 / 11 and I can solve the rest

[thick canyon]

nah I need to check myself back into 8th grade I can see why I'm not following...

wait

yes I agree

I agree

[ivory river]

then just expand that fraction on the right

[thick canyon]

ys = tx

mhm

ys + yx = tx + yx, so ys = tx

[ivory river]

so, s/t=x/y

[thick canyon]

mhm

but theres no perpendicular here

[ivory river]

by the way, the height does not matter

the same is true for this

[thick canyon]

ohh right

I see

[ivory river]

that's what i applied in your problem

[thick canyon]

OHH

I JUST NEEDED TO FLIP THE TRIANGLE

TO SEE IT BETTER

[ivory river]

yep

[thick canyon]

ok so

what is this called

I have a maths comp coming up and I guess i gotta revise

so uh what is this called

[ivory river]

I don't know. It is just a property of areas of triangles with common height

their ratio of areas equals the ratio of their bases (bc they have a common height)

[thick canyon]

hmm

I'll research into this

[ivory river]

bc S1=ah/2, S2=bh/2, so S1/S2=a/b.

[thick canyon]

thx

@ivory river

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@thick canyon Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[mint siren]

how do i do this?

[mint siren]

assume region D is

i kinda just need checking

is it

int(0)(pi)(sinx)^2?

thats it?

[deep ice]

im not a helper but im very sure that region is an integral

[mint siren]

yea

[deep ice]

yes

it should be sin(x) and not sin(x)^2 though

[mint siren]

why?

[deep ice]

the function you're given there is sin(x), not sin(x)^2

[mint siren]

yes

i know

but

were finding volume

not

area

unless

[deep ice]

you can't find volumes with integrals

[mint siren]

roated to the x axis doesnt mean find volume

dude what

now youre jsut wrong

you draw like

a circle

a loop around the integral

like this

[deep ice]

oof, my dumbass🤣

[mint siren]

ok wait like

i actually do have a

area question

ims tuck one

If 𝐷 is the region in quadrant I bounded by the parabola 𝑦^2 = 2𝑥 and line 𝑥 − 𝑦 = 4, then area 𝐷 =

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[devout ledge]

im sorry im really struggling here 😅

[devout ledge]

u = x^-11
du = -11 x^-12 dx

- 1/11 int cos (pi u) du

[full sand]

you could u sub again if you only know how to solve cos(x)dx

or v sub

[devout ledge]

ooh would there be chain rule or smth since it's pi u

I only know

[full sand]

yes its chain the chain rule

[devout ledge]

Yeah I don't think I know chain rule for anti -derivative

[full sand]

you know that cos(something) = sin(something)

[full sand]

you can just do the substitution that you let v = pi*u if you want

[devout ledge]

I got it

I had to divide by pi

[full sand]

yeah

[devout ledge]

since
sin(pi x) - derivative -> pi cos(pi x)
cos (pi x) - int -> sin(pi x) / pi

makes sense

thank you

.close

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[exotic olive]

am i right with D here?

thank you guys

.close

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[exotic olive]

am i right with A here?

<@&286206848099549185>

[naive pivot]

no

[opaque maple]

A is correct unless i'm missing something

[naive pivot]

f^-1(x)

[opaque maple]

oh wait yea

[exotic olive]

oh

[little widget]

The range of the inverse function is the domain of the function

[opaque maple]

i guess keyword is the graph, not the function

[exotic olive]

oh wait

let me send graph

[opaque maple]

so the domain and range are linked as kaue said above

only note i'd make is the question is careful to say graph and not function

otherwise you'd need restrictions

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[weak zenith]

how did they find the 18th derivative of f(x)???

[pallid canopy]

they read the power series

[formal yacht]

by the uniqueness the coefficients are determined by f with a handy formula, try and recall what it is

[weak zenith]

wait sorry, i rlly dont get. they were able to find the derivative cuz its a power series?

also I thought it was a geometric series...

[tranquil shale]

its not geometric bc u got that n^2 in there

and they didn't find 18th derivative $f^{(18)}(x)$, but just $f^{(18)}(2)$, far easier

[grizzled pagoda]

rain

[tranquil shale]

because if you take $x=2$ everything disappears but the constant

[grizzled pagoda]

rain

[formal yacht]

So the power series is written in a special type of form, now since f(x) is equal to this power series within some radius of convergence then the coeffcienets are uniquely determined by a formula depending on f

[formal yacht]

hm?

oh nvm

[weak zenith]

wait I thought power series is like 1/n where n is raised to some power

but in my series, n is in the numerator

[formal yacht]

a power series is a series with terms of the form c_k (x-a)^k

where c_k is any sequence not depending on x

so in ur case c_k = k^2/3^k

now since f(x) is equal to this series and converges within some radius > 0, then c_k = f^(k)(a)/k!

this is the formula i wanted u to recall

[weak zenith]

ohhhhh ok yeah I think i got it

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[hasty junco]

can anyone tell me what i did wrong? i was looking for the turning points

[mint patio]

,rotate

[hasty junco]

WP just means turning point

wait.

nvm

<@&286206848099549185>

[mint patio]

,w 6+8

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[midnight pond]

Can some one explain the basics in quantum physics

[hasty junco]

??

[mint patio]

Sorry that was a bit rude in retrospect

By W do you mean the point where concavity flips

[hasty junco]

yes

[hasty junco]

not at all

[mint patio]

if you dont consider it rude it is because you did not get it

but i really dont know how to do this equation

[hasty junco]

.close

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[mint siren]

i need help udnerstading what my teacher did (thi s is a pre recorded thing)

[mint siren]

specifcally this part

ok see

how did

rt(px)

intergrate into

(2/3)*(Px)^(3/2)/p

how did the p get down there?

at the denomenator

was this a mistake?

thing is this was a multiple choice and he got an answer

[naive pivot]

Ig cuz they simplified weirdly
Sqrt(px)=sqrt(p)*sqrt(x)
sqrt(p) is constant
So it’ll be 2/3sqrt(px^3)
and their answer also simplifies to 2/3sqrt(px^3)

[mint siren]

the question is this

[fast badger]

Because of chain rule but reversed. If you were to differentiate (px)^(1/2) you would multiply by p, so to reverse that you divide by p

[ \frac d{dx} \left( \frac 23 \cdot (px)^{3/2} \right) = \frac 23 \cdot \frac 32 \cdot (px)^{1/2} \cdot p ]

[grizzled pagoda]

shsgd

[fast badger]

You can see there is a p left over, so you need to divide by p to cancel it

@mint siren

[mint siren]

(yea im here im writing all this down to try to understand sorry for not replying)

ok wait uhh something doesnt add up

[mint siren]

i wrote this guys equation down it began as

rt(p)*rt(x)

intergrate it

oh

wait

[fast badger]

Im guessing you didnt simplify

[mint siren]

p is a constant

yeah

[fast badger]

[ \int (px)^{1/2} , \dd x = \sqrt p \int x^{1/2} , \dd x = \frac{2\sqrt p}{3} \cdot \sqrt{x^3} + C ]

[grizzled pagoda]

shsgd

[mint siren]

ok yeah

so is the p being in the denomenator like just a weird way of writing it?

[fast badger]

Not necessarily weird way of writing it. Just not yet simplified

[mint siren]

i understand this like yeah

this is wat i did

ohhh

[fast badger]

Since p^(3/2) / p = p^(3/2 - 1) = p^(1/2)

[mint siren]

so if i plug it in it should end up as the same

[fast badger]

Yes it’s the same outcome

[mint siren]

alright ty

[mint siren]

oh yeah yeah taht makes sense

ty very much

/close

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[split sail]

i need help with this

[amber wasp]

@split sail Has your question been resolved?

[vestal hatch]

what's the slope of AB?

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@split sail Has your question been resolved?

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[split sail]

.reopen

[amber wasp]

✅

[fluid prism]

Where have you gotten so far?

@split sail

[split sail]

I got someone else helping me

[fluid prism]

👍

[split sail]

I get it now

But ty for asking

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[frigid python]

I just need help with this problem

[signal furnace]

Hint: Rewrite $2^x=\frac{1}{2} \cdot 2^x+\frac{1}{2} \cdot 2^x$

[grizzled pagoda]

Civil Service Pigeon

[signal furnace]

Then, it's a one liner by ||am gm||

[frigid python]

How do I use the am-gm with 4^(17-x)?

[signal furnace]

.

[frigid python]

Oh, do I rewrite 4^(17-x) into 2^(34-2x)?

[signal furnace]

yeah that would make it more obvious

[frigid python]

So the answer is $3\cdot2^{\left(\frac{32}{3}\right)}$

[grizzled pagoda]

DT

[signal furnace]

,w minimize 2^x + 4^{17-x}

[grizzled pagoda]

Results provided by WolframAlpha

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[signal furnace]

,w 30722^{2/3}=32^{32/3}

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[frigid python]

so it is correct? Thanks!

/close

!clos

!close

[signal furnace]

!done

[amber wasp]

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

[frigid python]

.close

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[verbal lark]

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[ebon glacier]

just chatgpt it

[split sail]

!nogpt

[amber wasp]

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).

[verbal lark]

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[split sail]

!noping

[amber wasp]

Please do not ping individual helpers unprompted.

[quick ridge]

💀

[verbal lark]

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[quick ridge]

great idea bro

[quick ridge]

really

[verbal lark]

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[quick ridge]

who

anjali?

[verbal lark]

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[quick ridge]

they don’t know calculus

they’re in algebra

[verbal lark]

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[quick ridge]

to tell that guy off

[verbal lark]

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[quick ridge]

for suggesting chatgpt

[verbal lark]

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[quick ridge]

what

no

lol what

she’s not a mod

she just wanted to tell that guy to fuck off for saying that

[verbal lark]

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[quick ridge]

using the factoids is fun

[verbal lark]

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[quick ridge]

bingo but also this is still EVT yea?

[verbal lark]

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[quick ridge]

so what do you have to check

🙈

[verbal lark]

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[ebon glacier]

use am

[verbal lark]

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[ebon glacier]

use am gm

[quick ridge]

how do you figure

[verbal lark]

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[quick ridge]

i was being sarcastic

[verbal lark]

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[quick ridge]

good idea

they don’t know that though

they’re an ap calc casual

[verbal lark]

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[ebon glacier]

that aint calc

[quick ridge]

i’m aware

[ebon glacier]

it says its the sum of sides of a rectangle

use that

[quick ridge]

i’m saying he only knows how to use calculus methods

he doesn’t know any inequalities

[ebon glacier]

u said this tho

[verbal lark]

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[quick ridge]

i was referring to @split sail

because he asked why she wouldn’t help him

[split sail]

lol

[verbal lark]

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[ebon glacier]

oh idk calc, good luck thats too complex

[quick ridge]

negative perimeter

🔥🔥

[verbal lark]

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[quick ridge]

i like his solution of AM-GM tbh because it shows that this is the minimum perimeter but you can use calculus and do some derivative test shit if you want

[verbal lark]

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[quick ridge]

but this isn’t a derivative test

use the second derivative test

[verbal lark]

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[quick ridge]

$P’ = 2 - \frac{200}{x^2} \implies P’’ = \frac{400}{x^3}$

[grizzled pagoda]

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

[quick ridge]

plug the 10 in

you’ll see that P’’ > 0

hence minimum

by second derivative test

[verbal lark]

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[quick ridge]

minimum maximum etc

explain in words

[verbal lark]

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[quick ridge]

you didn’t do part a entirely

[verbal lark]

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[quick ridge]

think about it a bit more

[verbal lark]

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[quick ridge]

look at the endpoints of the interval

🙈🙈

[verbal lark]

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[quick ridge]

what

[verbal lark]

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[quick ridge]

you’re not doing what the problem asked

[verbal lark]

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[quick ridge]

yea that’s the thing though

[verbal lark]

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[quick ridge]

you cant plug it in

[verbal lark]

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[quick ridge]

explain to me why you can’t

[verbal lark]

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[quick ridge]

and what else

[verbal lark]

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[quick ridge]

yep

[verbal lark]

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[quick ridge]

mhm

limits

[verbal lark]

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[quick ridge]

mhm

so is there a maximum

[verbal lark]

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[quick ridge]

yep

[verbal lark]

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[quick ridge]

can i show what he was talking about btw

[verbal lark]

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[quick ridge]

AM-GM

$\frac{a+b}{2} \geq \sqrt{ab}$

[grizzled pagoda]

knief

[quick ridge]

the left hand side is called the arithmetic mean

[verbal lark]

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[quick ridge]

the right is called the geometric mean

we can prove this easily

just square both sides

[verbal lark]

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[quick ridge]

you’ll end up with (a-b)^2 >= 0 which is true for all a and b

[quick ridge]

bruh

$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$

[grizzled pagoda]

knief

[quick ridge]

but if you want you could multiply the 2 over first

[verbal lark]

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[quick ridge]

as a quick exercise can you derive this

like show me how i got that

[verbal lark]

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[quick ridge]

\

\

[grizzled pagoda]

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

[verbal lark]

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[quick ridge]

and what’s the right hand side

[verbal lark]

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[quick ridge]

$\frac{(a+b)^2}{4} \geq ab$

[grizzled pagoda]

knief

[quick ridge]

multiply by 4

expand the (a+b)^2

then the 2ab - 4ab will become -2ab

[quick ridge]

and you’re left with

[verbal lark]

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[quick ridge]

a^2+2ab+b^2 >= 4ab

subtract 4ab

a^2-2ab+b^2 >= 0

[verbal lark]

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[quick ridge]

(a-b)^2 >= 0

which is true of course

any real number squared is nonnegative

so let’s apply that to this problem

a = 2x and b = 200/x

[verbal lark]

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[quick ridge]

$\frac{2x + \frac{200}{x}}{2} \geq \sqrt{2x \cdot \frac{200}{x}}$

[grizzled pagoda]

knief

[verbal lark]

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[quick ridge]

$x + \frac{100}{x} \geq \sqrt{400}$

[grizzled pagoda]

knief

[quick ridge]

$x^2+100 \geq 20x$

[grizzled pagoda]

knief

[quick ridge]

(x-10)^2 >= 0

x = 10 for equality

[verbal lark]

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[quick ridge]

but just from this step we can see this hold on

$\frac{2x + \frac{200}{x}}{2} \geq \sqrt{2x \cdot \frac{200}{x}} \iff 2x + \frac{200}{x} \geq 40$

[grizzled pagoda]

knief

[quick ridge]

because the sqrt simplifies to 20

sqrt(400)

and i multiplied by 2

but notice 2x + 200/x is just our perimeter

[verbal lark]

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[quick ridge]

so our perimeter is greater than or equal to 40

which answers the question

[verbal lark]

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[quick ridge]

yea you should

[verbal lark]

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[quick ridge]

it means when x = 10 you have an absolute minimum

40 is the minimum perimeter

and you can say that as x->0 and as x->infinity the perimeter goes to infinity because in each of those cases one of the pair of sides grows without bound

[verbal lark]

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[quick ridge]

yea they just wanted you to explain it

[verbal lark]

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[quick ridge]

never

[verbal lark]

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[amber wasp]

Channel closed

Closed by @verbal lark

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
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[fierce edge]

hi, if the standard deviation of 140-a, 140, 160, 160+a is 50 then what is a given a > 0

[coral wigeon]

[fierce edge]

i can not do it the mainstream way bc of using the formula bc that takes too long

[fierce edge]

heyyy nice pfp

[coral wigeon]

loll thx

[fierce edge]

depressed kuromi 😔

i'm depressed too

[coral wigeon]

😭

[fierce edge]

all kuromis are depressed

[shrewd elm]

do it the non-mainstream way

[fierce edge]

which is?

[shrewd elm]

no idea

[fierce edge]

Lol

[coral wigeon]

all girls really

[shrewd elm]

hi slayla ma'am

[coral wigeon]

hi bungo sir

[fierce edge]

all girls till bf 😔

[sullen knoll]

Well standard deviation is just the square root of the variance which you can find

[fierce edge]

i'm aware

but that takes a long time to solve

long time = 3 minutes

[sullen knoll]

Variance is E(X^2)- EX^2

Thats the only way to solve ot

It

Sadly

[fierce edge]

U mean E(X^2) - (E(X))^2

[sullen knoll]

You will have to do it that way

[sullen knoll]

Yeah

Mb

[fierce edge]

not really

idk the other way

[sullen knoll]

?

[fierce edge]

but i'm sure there's another way

my test tmr has 100 questions with like 100+ minutes

if i do this the formulaic way lol i'll lose like 3 mins+

[sullen knoll]

lmao

[fierce edge]

which is 3-4 questions

[sullen knoll]

It doesn’t take me 3 minutes

Maybe 1-2 minutes max

The numbers are quite nice aswell

Not decimal or anything

[fierce edge]

first calculate the mean which is 10-20 seconds

u get mean = 15

then you write out the whole variance thing bit

[shrewd elm]

you can see in like 2 seconds that the mean here is 150

[sullen knoll]

Yes

[shrewd elm]

and the four given numbers are symmetric around 150

[fierce edge]

that's already 40 seconds

[shrewd elm]

so that might speed things up

[sullen knoll]

It wont take

That lomg

😭

[fierce edge]

the quadratic isn't that nice either though?

[sullen knoll]

Well I guess you could try something else

You can keep this channel open until you get something then

[fierce edge]

someone said something yesterday about the dispersion being equal to the left and right

[sullen knoll]

No it should not take long to expand

Do you get to use calculators

[fierce edge]

no

[sullen knoll]

But 14 squared is just 196, and add the zeros

It shouldn’t take that long

Alright

[fierce edge]

u don't need 14 squared though

i think

okay then i guess this is the only way

[sullen knoll]

I’m not sure you can try to google the other wau

[shrewd elm]

it's not hard at all, notice that 140 and 160 are both distance 10 from the mean

[fierce edge]

I guess

okay thanks

[coral wigeon]

btw i used your trick again a few days ago

i love it sm

it's so helpful

[fierce edge]

omggg 😭

😔 i don't deserve ownership of that trick

but yasss i keep trying to do that with all combinatorics questions i see of that form

[coral wigeon]

too bad, it's the mmmm7 trick forever to me

[fierce edge]

yours too

[coral wigeon]

[fierce edge]

there was another thing which i found interesting

maybe u know it already but like

if you're drawing only 2 things

u can just assume that you're drawing how many ever things

but only look at the first 2 things