help-27

[fast river]

yup

[urban spindle]

I am sorry, I cannot help with differential equations; I didn't understand them myself :D. I'll let someone with more experience help.

[fast river]

lmaooo

thank you

no its all good i know the answer

[devout snow]

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[small urchin]

but if I have to integrate a function on a square of side $l$ , what changes between writing $\int_0^l f(x) l dx $ and $\int_0^l \int_0^l f(x) dx dy?$

[woven radish]

Bob Pancakebutter

[small urchin]

because in theory one integral represents an area and another the volume, but it gives the same result

But in theory I could immediately make the second one the same as the first

But I don't know

[low holly]

“integrating over a square” makes sense if your function depends on two variables

[small urchin]

Thanks very much

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[abstract ginkgo]

how did we pass from that line to that line

i understand how we get everything before teh first -

are we factoring it as a trinome?

what am i missing

[tall knoll]

they added and subtracted 2sin^2(theta)cos^2(theta)

hence this new bit

[abstract ginkgo]

I DONT GET IT

srry caps

doesnt that ruin the equality

[tall knoll]

you can add and subtract the same thing as much as you want

[abstract ginkgo]

but you just added 2 sin^2cos^2

out of nowhere

[tall knoll]

and subtracted

[abstract ginkgo]

how do we go from -sin^2cos^2 to
-2sin^2cos^2 - sin^2cos^2

or does the term from before factor into this

[tall knoll]

$\sin^4(\theta) + \cos^4(\theta) \color{red}{+ 2\sin^2(\theta)\cos^2(\theta) - 2\sin^2(\theta)\cos^2(\theta)} \color{black}{- \sin^2(\theta)\cos^2(\theta)}$

oh boy

[abstract ginkgo]

its all g take ur time

[tall knoll]

i'm not sure why the spacing is so cringe

[abstract ginkgo]

so the +2 and -2 cancel out eachother'

i see that

but

[abstract ginkgo]

where did the +2 go here

i only see the -2

[woven radish]

Steakanator

[tall knoll]

good enough

now what is $\sin^4(\theta) + \cos^4(\theta) + 2\sin^2(\theta)\cos^2(\theta)$?

[woven radish]

Steakanator

[abstract ginkgo]

hmm

we cant add them since they have different exposants

so idek

can i break it down int]\

[tall knoll]

let a = sin^2 and b = cos^2

[abstract ginkgo]

yes

[tall knoll]

a^2 + b^2 + 2ab = ?

[abstract ginkgo]

oh its a perfect squaer

trinominal

or wtb u call it

= (a+b)^2

[tall knoll]

quite so

[abstract ginkgo]

i see

[tall knoll]

so you get (sin^2 + cos^2)^2, and have an extra -2sincos left over

[abstract ginkgo]

we just created and added our perfect sqaure trinomial (or wtvc the temr is)

yes

i see

[tall knoll]

(omitting writing theta because i'm lazy)

[abstract ginkgo]

thanbk you

[tall knoll]

[abstract ginkgo]

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[tender crag]

I need help with 5-3

[rotund umbra]

Suppose I have the line y = mx + c. How would you make ANY perpendicular line to this?

[tender crag]

Make the slope -1/2

[rotund umbra]

if you mean, make the slope -1/m, then yes
https://www.desmos.com/calculator
https://www.desmos.com/calculator/wmj0iw7xlz (I posted the wrong link, sorry)

So. step 1; can you rewrite the second equation (i.e. x = ay + c) in slope-intercept form (i.e. y = Mx + C)

hmmm, actually having looked at this, there isn't a real number a that would make the two graphs perpendicular.

1. what's the slope of the first equation
2. what's the slope of the second equation
3. why can't there exist an a such that would make the slopes perpendicular?

[devout snow]

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[wicked hawk]

There is

-1 for a

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[pale sapphire]

can someone write out the equation when you plug in the bounds? im trying to put it in the calculator and its not giving me the write answer

[devout snow]

Please don't occupy multiple help channels.

[pale sapphire]

<@&286206848099549185>

[spare dust]

What answer are you getting

[pale sapphire]

-1. something

[spare dust]

Your getting a negative number?

[pale sapphire]

yeah

thats why i need help writing out the function to know how to properly put it in a calculaor

[spare dust]

You didn't multiple the 2000 by both sides

[pale sapphire]

i didnt need to do that for the last integral problem i did

wait im still getting a different number

wait no im not

[spare dust]

That's the thing the constant is multiplied by the whole integral

[pale sapphire]

how would you know whether to multiply the constant by each bound or by the entire integration

how would that look like to know the difference

[spare dust]

Well it's given in the original integral the constant is multiplied by the integral you can take it out

[pale sapphire]

if you look at this you only multiply 1/5 by the entire thing

[spare dust]

That's the same thing

[pale sapphire]

no cuz you multiply 4/5 (F(5)-F(0))

here you do 2000(F(5)- 2000f(0)

[spare dust]

Also in original integral they forgot the t in the power

[pale sapphire]

i know

i realized that

does that make sense?

[spare dust]

Yes

That's what I meant

[pale sapphire]

but how do you know when to do either scenarios

like whats the difference in the equation

[spare dust]

It's the same it's a property of integrals

[pale sapphire]

omg

i didnt add another set of parenthesis

[spare dust]

That's what I said

[pale sapphire]

however you spell it lol

where lol

[spare dust]

Did you get the right answer?

[pale sapphire]

yes

[spare dust]

!done

[devout snow]

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[wispy geyser]

am i right with D here?

[wispy geyser]

<@&286206848099549185> please help

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[distant kite]

is there a way to do this without it being by infinite descent?

[distant kite]

it's supposed to be an easy question and infinite descent... is easy but not so easy

[wise storm]

idk I guess reduce mod 5 and notice 2 is not a quadratic residue, but maybe that leads to the infinite descent arg

[distant kite]

yea, mod 5 leads to the infinite descent

it is what i did

[wise storm]

well let's try poking at reducing mod 2 or powers of 2

hmm, maybe trying mod 3 or mod 101 based on the exponents can get us somewhere too

[distant kite]

it's okay, maybe there isn't an easier method

thank you btw

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[autumn helm]

cannot solve for x 😦

[delicate river]

what have you tried so far?

[scarlet shoal]

34-9.52= pi x + 2x

4.76117

[scarlet shoal]

you can find area of the half circle using 2pi r
just do pi r instead cause its half. and you know r is x
thus 2y+ 2x (bottom side) + xpi (top curve) =34

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[prime whale]

my linear algebra sucks ass and dick

can someone tell me about hint II in this screenshot?

[nova glen]

It has something to do with the determinant

[acoustic leaf]

can you post the original question?

[prime whale]

i mean sure i can but im not sure how helpful its gonna be lol

This is the preceding context

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[royal radish]

what was plugged in for t for the magnitude to get 3?

[acoustic leaf]

they used a trig identity

[royal radish]

r u sure? the dude didnt show any work\

since its a circle its just a radius anyway

but i dont wanna not know how it works

[acoustic leaf]

it's known that sin^2 t + cos^2 t = 1

[royal radish]

oh so thats reallly all they did

alr

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[royal radish]

.reopen

[devout snow]

✅

[royal radish]

for the x= and y =

why is it +3t

and +4t

i get +3 and +4

but why is t also there cause u plug t=1 in so its just not there anymore right?

is that just how it works

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[mortal plinth]

Looking for some help on this! if anyone has a moment to take a look at this that would be great

[mortal plinth]

<@&286206848099549185>

[lusty sapphire]

!15m

[devout snow]

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[mortal plinth]

my appoligies

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[rotund sail]

need help understanding what to do here

[supple knot]

Are you supposed to integrate to get g(x)

[rotund sail]

i think so

[rotund sail]

i tried finding the original function of the derivative but am stupid and i couldn't find anything to go after that

[misty crest]

integrate g’

[alpine python]

it's not easy

[misty crest]

$\int_2^{10} g’(x) dx = g(10) - g(2)$

[woven radish]

knief

[misty crest]

it is

[rotund sail]

this is part of the fundamental theorem of calculus right?

[misty crest]

yes

[alpine python]

i can't integrate it

[rotund sail]

but still how do i find g(2)

[misty crest]

🤔

[misty crest]

you’re solving for it

$g(2) = g(10) - \int_2^{10} g’(x) dx$

[woven radish]

knief

[misty crest]

pending postgrad

[rotund sail]

with the equation you just shown?

ok lemme try

[misty crest]

is this a calculator question or no calculator

[rotund sail]

calcualtor

[misty crest]

that makes it easier for sure

[rotund sail]

i've barely done integrals on calculators tho

[misty crest]

but you can integrate it

[misty crest]

ti-84?

[rotund sail]

yea

[misty crest]

math 9

[rotund sail]

ah

ok

so 4.107?

wait no

1.33?

if i use the formula you've shown me i get 1.32

[rotund sail]

but where did you get this from

[misty crest]

literally just rearranged

[rotund sail]

oh im stupid

ok thanks

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[alpine python]

i just took that to get access to the group theory channel

[misty crest]

you can just get the advanced role

[alpine python]

how can i get that?

nvm i asked in a more appropriate channel

[misty crest]

https://discord.com/channels/268882317391429632/790023107708583946

[misty crest]

you only need undergrad

[devout snow]

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[finite echo]

Can this phenomenon be explained? When graphing y = log_n(x) and y = nx, making a flipped L between the points of n on them, and making a diagonal line between those points, you get a triangle. https://www.desmos.com/calculator/00wduaaqwj

[brisk raptor]

p sure this just shows that n^x and log_n(x) are inverses

a point (a, b) one curve means there is a point (b, a) on the other

that's where the straight line comes from, and a triangle can be formed from any diagonal with just its x and y components

[finite echo]

Thanks! Fascinating.

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[steel escarp]

can someone help me understand what to do to solve this? I have no clue

[wicked hawk]

start with two points in F(Ω) and call them y1 y2 for example

[devout snow]

@steel escarp Has your question been resolved?

[steel escarp]

Okay, and F(Ω) is what? is F(Ω)=(1-t)A+tB? and instead of A and B, we use y1,y2?

[celest valve]

No, F(Ω) is defined in the picture you sent here:

[wicked hawk]

Yep it's defined

If you really can't solve it ping me

[devout snow]

@steel escarp Has your question been resolved?

[steel escarp]

so then F(Ω)=(1-t)y1+ty2, what's next? I'm sorry, im just not seeing it

[wicked hawk]

No

Let m think of how I can easily explain

Now idk but you can very easily prove it if you understand it

F(x) = Ax+b is an affine function Ω is a convex set

We know that affine functions preserve convexity right?

That's fundamental

You understand that so far?

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[fringe smelt]

I want to do Pythagoras is the right side 4 m?

[fallen harness]

are u talking abt the unknown side?

[fringe smelt]

Nvm

It’s the same

[fallen harness]

yea

but it isn't 4

[fringe smelt]

I thought it was different

[solar goblet]

it will he helpful to label the vertices for ease of reference

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[frail osprey]

how do i show this is odd

[solar goblet]

tell me definition of an odd function

[frail osprey]

f(-x) = -f(x)

[solar goblet]

cool, now show that

[frail osprey]

i tried to, im pretty sure my algebras wrong

im not able to show that theyre equal

[solar goblet]

show your work?

[frail osprey]

Oh wait, nvm i got it 🤦‍♂️

thanks regardless

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[solar goblet]

[royal radish]

what if x isnt a whole number and is a decimal couldnt it be even than

[frail osprey]

huh, how

[solar goblet]

?

[royal radish]

idk

[frail osprey]

it would still fit the definition right

[royal radish]

its just a thouight

prob wrong

[frail osprey]

decimals can be negative

[solar goblet]

they meant even as in even function, not an even number

[royal radish]

o

[frail osprey]

yes, im aware

[solar goblet]

odd*

[frail osprey]

even if it was a negative number it would fit the definition right

[royal radish]

im p sure he was talking to me if not im completely lost

[solar goblet]

yeah i was talking to them, youre good

[solar goblet]

yeah

[frail osprey]

Ahhhh alright haha

[frail osprey]

hopefully you got why that isnt the case man, cheers, have a good one

[royal radish]

you too

[devout snow]

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[undone tide]

yi

[undone tide]

how does my teacher get the Y and X values in the blue chart?

[urban spindle]

It looks like your teacher gave values to x: 0, 1, 2, 3...

then plugged it in the formula for y

[undone tide]

noo

[urban spindle]

to get the y for each x

[undone tide]

i mean this

how does she get these numbers

[devout snow]

@undone tide Has your question been resolved?

[undone tide]

<@&286206848099549185>

[oblique carbon]

So I think if the equation of the x-4 is for example 0, the y is 1 then

so x is the product of x-4

[undone tide]

can you explain it in an example please

im slow sorry

[oblique carbon]

so for example if you need to find whats number makes the x-4 equal to 0 you first need to find the number

that is what the teacher did in pink

so if the x is then 0

then the y needs to be 1 because y^0 is always 1

[undone tide]

im talking about the blue tho

[oblique carbon]

yea the x is the product that is made by x-4

and if you get the x you can find out y by y^x

and in the equation y is 3 in the first one

So x=-(x-4) any y=y^x in the first one

[undone tide]

so would it be 3

[oblique carbon]

if x is 1 then yes

but everything by the power of 0 is always 1

[undone tide]

can we go through one of the questions? pleaseee

[oblique carbon]

so the other one?

[undone tide]

this one

but instead of 4 we do 3

[oblique carbon]

so instead of -4 you want -3?

[undone tide]

yes pls

[oblique carbon]

so x is always the number that makes the product x

so if you put for example 3 in there you'll get x=1

so and then to find y you just put 3^1 and thats 3

so then its x=1 and y=3

so if its a different equation when x is asked, you'll need to put another number to get the same x

understood?

[undone tide]

Yes

so will it always go from 0 to 3 in this?

[oblique carbon]

yes it can

but remember it could get further

the 0-3 are just a few examples

So any questions?

[devout snow]

@undone tide Has your question been resolved?

[devout snow]

@undone tide Has your question been resolved?

[undone tide]

Thanks!

/close

!close

[supple knot]

.close

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[lean void]

How did this simplify into this?

[sinful rover]

You can cross out the n in the numerator of the first and and then distribution- I think they also multiplied by the lcm

[alpine python]

1/n^4=(1/n^2)^2 and you can now bring it into the square brackets in the first image

[devout snow]

@lean void Has your question been resolved?

[lean void]

.close

[devout snow]

Channel closed

Closed by @lean void

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Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[deft steeple]

Can someone help me find the domain of this?
$f(x)= \sqrt[3]{cosx}*log(\sqrt{3}+cot(2x)-\sqrt{log_\frac{1}{2}\frac{3x}{x+2}}

[livid scarab]

could you resend that?

[deft steeple]

f(x)= \sqrt[3]{cosx}*log(\sqrt{3}+cot(2x)-\sqrt{log_\frac{1}{2}\frac{3x}{x+2}}

[sinful rover]

Need the $ on the ends and front

[livid scarab]

i'm sorry, but i think you'll have to use $ sign before and after for it to work

[deft steeple]

$f(x)= \sqrt[3]{cosx}*log(\sqrt{3}+cot(2x))-\sqrt{log_\frac{1}{2}\frac{3x}{x+2}}$

Oh ye

Thx

Waittt

[woven radish]

Astreaf ♡

[deft steeple]

Now correct

[sinful rover]

Ok so what would make this irrational? Or undefined?

[livid scarab]

I suggest that you look at the functions individually and think about how and where they are defined

[deft steeple]

Cosx is R
First log has to be >0 and 2nd one aswell

[sinful rover]

Yes

Good job

[deft steeple]

Yyyy

[sinful rover]

What about the log tho

[deft steeple]

Does wolfram has the thingy for domain calculator?

[deft steeple]

Wdym

[livid scarab]

i think it does

[sinful rover]

And sqrt is greater than or equal to 0 not just greater than

[livid scarab]

i just looked it up

[sinful rover]

The log in the second sqrt

[deft steeple]

But on this server

[livid scarab]

I don't know about that

[deft steeple]

It has to be >0

,help

[woven radish]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[deft steeple]

.help

[devout snow]

Commands:

• clopen: .close, .reopen
• consensus: .poll
• factoids: .tag
• help: .help
• version: .version

Type .help <command name> for more info on a command.

[sinful rover]

Yes

Good job

[deft steeple]

,help cmd

[woven radish]

You really shouldn't take it literally :upside_down:. Please type ,help ping, for example!
The full command list may be found using ,list.

[livid scarab]

think about what the periodic function, cot, means when it is used inside a log

[sinful rover]

Ok so now you can write the domain

Cot is equal to cos x/sin x

[deft steeple]

So sin has to be ≠0?

[livid scarab]

yes, otherwise it would be undefined.

[sinful rover]

Yes and it also cot has to be greater than 0

Because it’s in the log

[livid scarab]

-√3*

[sinful rover]

?

[livid scarab]

remember the whole thing is greater than zero, not just cot

[sinful rover]

Yes

What

[livid scarab]

i took this as if you meant that cot(2x) > 0, but correct me if im wrong

[deft steeple]

Cot2x >-sqrt(3)

[livid scarab]

that's correct, but don't forget that some values that still satisfy that, might still not be part of the domain, as you've got more to deal with

e.g. the other log term

[deft steeple]

Yyy

[deft steeple]

What

[sand quarry]

,w domain of cbrt(cos(x)) * log(sqrt(3) + cot(2x)) - sqrt(-log_2((3x)/(x+2))

[woven radish]

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[deft steeple]

Ooga booga

Thanks

[sand quarry]

its a smort bot

[deft steeple]

Its not log_2

But 1/2

,w domain of cbrt(cos(x)) * log(sqrt(3) + cot(2x)) - sqrt(-log_1/2((3x)/(x+2))

[woven radish]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by astreaf

[deft steeple]

,w domain of cbrt(cos(x)) * log(sqrt(3) + cot(2x)) - sqrt(-log_(1/2)((3x)/(x+2))

[woven radish]

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Requested by astreaf

[deft steeple]

Ughhhhhhh

,w domain of cbrt(cos(x)) * log(sqrt(3) + cot(2x)) - sqrt(-log_0.5((3x)/(x+2))

[woven radish]

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Requested by astreaf

[deft steeple]

Nooooo

[swift yoke]

Astreaf relaks. This is ... you gonna get better.

[deft steeple]

,w domain of cbrt(cos(x)) * log(sqrt(3) + cot(2x)) - sqrt(log_0.5((3x)/(x+2))

[deft steeple]

...

[woven radish]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
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Requested by astreaf

[deft steeple]

Ky....

[deft steeple]

Ye sure buddy

As we can see

:33

[swift yoke]

I hate cos and V

[deft steeple]

Wdym

[swift yoke]

Never paid attention to understand them.

[deft steeple]

What V

[swift yoke]

like V___

[deft steeple]

Sqrt?

[swift yoke]

yes

[deft steeple]

Bruh

Its more like v--

But sure

[sand quarry]

log_1/2 = -log_2

[deft steeple]

Ohhhhhhh

Peppo smart

Thats why u r the smart one

Oh no

My nitro edned

Cant use emotes

Gg

(Dont buy me)

[sand quarry]

lmao insane timing

[swift yoke]

pf... way to spend money... face dissapointment...

[deft steeple]

Nah

Discord gave it to me for free

So i took it

And there is no vencord on mobile 😿

So no free nitro plugin

Cot2x is sin2x/cos2x?

@sand quarry

[sand quarry]

cos(2x)/sin(2x)

[deft steeple]

Yeye

So how does it affect my domain?

@sand quarry

[runic grove]

cbrt(cos(x)) => x belongs to R

neglect that

log

log(x) is defined when x > 0
So -√3 > cot(2x) ------------ 1

and underroot so value has to be >= 0

log_0.5 (whatever) >= 0
whatever =< 1
3x / x+2 =< 1 ----------- 2

The inequality flipped cause base of the log was < 1

Take x1 intersection x2

[devout snow]

@deft steeple Has your question been resolved?

[deft steeple]

What does the --------1 mean?

[runic grove]

equation 1

lol

[deft steeple]

What??

Ahhh

Okay

$f(x)= \sqrt[3]{cosx}*log(\sqrt{3}+cot(2x))-\sqrt{log_\frac{1}{2}\frac{3x}{x+2}}$

[woven radish]

Astreaf ♡

[deft steeple]

So it's closer

[deft steeple]

How to calculate this?

Like for x(?)

[runic grove]

try making a graph

notice a pattern

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[deft steeple]

.reopen

[devout snow]

✅

[deft steeple]

Give me a moment

[deft steeple]

Are sure, not cot2x > -sqrt(3)??

[runic grove]

ur right

sorry

[deft steeple]

,w domain of sqrt(3) + cot(2x)

[woven radish]

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Requested by astreaf

[deft steeple]

,w domain of sqrt(3) + cot(2x) > 0

[woven radish]

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Requested by astreaf

[deft steeple]

Is that right? @runic grove ?

[runic grove]

yea

no

no

,w solve √3 + cot(2x) > 0

[woven radish]

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Requested by _nofluxgiven

[deft steeple]

Ohh okok

[deft steeple]

Can it show all steps?

[runic grove]

if you pay

hold on

plot something like this @deft steeple

[deft steeple]

Wdym plot?

[runic grove]

make the graph

[deft steeple]

But how

Yyyyy

[deft steeple]

Just draw it?

[runic grove]

yea

[deft steeple]

But my teacher gonna have problem thatbi didnt provide calculations

[runic grove]

what else

[runic grove]

u calculate to find the Points of intersection

you use the graph to help you generalize the solution

[deft steeple]

Ye but she gonna have problem where did i got this graph

[runic grove]

didnt you do graphs of functions?

like sinx

cosx

[deft steeple]

But it's cot2x

[runic grove]

okay so what

[deft steeple]

So it's stretched

Or smthing

[runic grove]

it's just squeezed

[deft steeple]

Yeyeye

[runic grove]

yeah

simple

[deft steeple]

But how does the sqrt3

Work

[runic grove]

just plot y = √3

a vertical line at y = 1.73205

now

our inequation was √3 + cot(2x) > 0

so √3 > -cot(2x)

take the graph of cot(2x)

flip it to get -cot(2x) btw

[deft steeple]

Its just tan then?

[runic grove]

no

flip the graph

not the function

[deft steeple]

Yee

[runic grove]

so whatever portion of the graph is below √3 will be our answer!

since √3 should be greater

[deft steeple]

I mean its gonna be like tan

The graph

[runic grove]

now just find the points of intersection

and ur done

[runic grove]

not really, tan goes asymptomatic on (2n+1)pi/2

cot does that on n pi

[deft steeple]

Ye but the shape

Nvm

[runic grove]

anyways im going to bed

seeya

[deft steeple]

Bay bay

Wai

Yyyy

One mor

[runic grove]

if you need help ping helpers

[deft steeple]

Thingi

[runic grove]

ok

[deft steeple]

Gays

[runic grove]

ask

"Gays"

[deft steeple]

How?

[runic grove]

😭

[deft steeple]

Idk how it happened

[runic grove]

for what v cot(v) gives you -√3

hint: it's v = 150 degrees

or 5pi/3

so we have -√3 = cot(2v)

cot(5pi/3) = cot(2v)

or 5pi/3 = 2v (IM NOT BEING GENERAL HERE)

and solve for v

[deft steeple]

Where did it came from?

[runic grove]

just know ur trig values

[deft steeple]

V=2x?

[runic grove]

v is just a random variable i chose

to illustrate

,w cot(150 deg)

[woven radish]

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Requested by _nofluxgiven

[runic grove]

yippee

[deft steeple]

Ahhhh

The table

I look for value closest to sqrt3

Or

Ugh

[deft steeple]

How

Yyyy

[deft steeple]

So it's gonna be 5pi/6???

[runic grove]

yea

[deft steeple]

Sooooo

Yyy

My domain

Out of it?

[deft steeple]

Dont qiqi me here

[deft steeple]

Hutao bette

Whats my domaijnnnnnnnn

[runic grove]

I'm sorry I'm just very tired right now

[deft steeple]

Me to buddy

Dw

[runic grove]

I wanna help but I physically and mentally and emotionally can't

[deft steeple]

Okay so

[runic grove]

Going to bed, take care </3

[deft steeple]

<@&286206848099549185>

[deft steeple]

Aha

</3????

Okay

Not nice

[runic grove]

That was a prediction

for this exact scenario

</3

[deft steeple]

Goo

Too

Bed

And stop

Pls

Bay bay

[runic grove]

ok

[runic grove]

bye bye

[deft steeple]

<@&286206848099549185>

[deft steeple]

How do I get a domain out of it

.close

[devout snow]

Channel closed

Closed by @deft steeple

Use .reopen if this was a mistake.

[deft steeple]

.reopen

[devout snow]

✅

[deft steeple]

Help

,w calculate xrt((arcsin(1/5^x))/x)

[woven radish]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by astreaf

[deft steeple]

No i chuj

[deft steeple]

<@&286206848099549185>

Pls hekp

Help

.close

[devout snow]

Channel closed

Closed by @deft steeple

Use .reopen if this was a mistake.

[deft steeple]

.reopen

[devout snow]

✅

[lusty sapphire]

What is your question?

[deft steeple]

.

[deft steeple]

.

[supple knot]

what is "it"

[deft steeple]

Look at what i anwsered with the .

[deft steeple]

I did the close thing

So this channel gonna be higher on the list of opened channels

My question is still the sams

[deft steeple]

How do I get the domain out of this

[supple knot]

domain of what function?

[deft steeple]

You gonna have to read a little of the chat history to understand how we got ther

[deft steeple]

Sqrt(3)+cot(2x)>0

[supple knot]

,w domain sqrt(3) + cot(2x)

[woven radish]

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Requested by rie.mann

[deft steeple]

I can do the same

[supple knot]

that's an inequality, not a function.

[deft steeple]

Vut i have to calculate it

[deft steeple]

.

.

Here

I got most of it

[supple knot]

,w domain log(sqrt(3) + cot(2x))

[woven radish]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by rie.mann

[supple knot]

find the domain of each summand individually then domain for f(x) will be the intersection of all the domains of the summands

[deft steeple]

Ufufhdhdhshaosjjsjdjskd

Ik

I k this

But

I need to calculate the domain for cot2x

[supple knot]

it's here

,w domain cot(2x)

[woven radish]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by rie.mann

[supple knot]

adding sqrt(3) does nothing to the domain

[deft steeple]

No

Cuz it's fucking

Under log

So it has to be grater that 0

Sqrt3 + cot2x

[devout snow]

@deft steeple Has your question been resolved?

[deft steeple]

No it has not

[devout snow]

@deft steeple Has your question been resolved?

Channel closed

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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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[naive mortar]

1/(x*y) = (1/x) * (1/y) right?

[wicked turtle]

yep

[naive mortar]

ty

lov u

apr u

kiss u

hug u

ty bb

.close

[devout snow]

Channel closed

Closed by @naive mortar

Use .reopen if this was a mistake.

[naive mortar]

Can I simplify this:
x/(y/z)

[spare zealot]

You can

[naive mortar]

(x*z)/y?

[spare zealot]

Correct

[naive mortar]

ty

lov u2

apr u2

kiss u2

hug u2

[spare zealot]

$\frac{x}{\frac{y}{z}} = \frac{x}{1}\times\frac{z}{y} = \frac{xz}{y}$

[woven radish]

al-jebruh

[spare zealot]

Utilizing the fact that division of a number is multiplication of its reciprocal, so dividing by y/z is the same as multiplying by z/y

[naive mortar]

so if 1/(y/z) its just z/y

[spare zealot]

Yup

[naive mortar]

I see now.

ty 🙂

[spare zealot]

[naive mortar]

Another question. Can I also simplify this:

wait how do I express that I want a sum of something?

The big E symbol.

[devout snow]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
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[tight ibex]

would indefinite intervals be the same as antiderivative?

[versed juniper]

pretty much but dont forget + C

[tight ibex]

Thank you

.close

[devout snow]

Channel closed

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[devout snow]

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• Show your work, and if possible, explain where you are stuck.
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[rich helm]

why is (a) here not a relative maximum?

[tight ibex]

We don't know if it's going up further as x decreases.

[acoustic leaf]

,rccw

[tight ibex]

End points can never be a relative extrema.

[rich helm]

but then why is (s) a relative extreme

[zealous oyster]

its not

[rich helm]

in the book it says its an absolute max

[acoustic leaf]

there does appear to be a local maximum at a

[acoustic leaf]

endpoints are always relative extrema if the function is continuous

[rich helm]

yeah but the answer sheet says that (a) is nothing, (b) is a loca min, (c) is a local max, (d) is neither (r) is absolute min, (s) is absolute max

[rich helm]

interesting

so in this case is it just like a go figure

[acoustic leaf]

but really it depends on how your textbook defines local minima/maxima. some textbooks specifically exclude endpoints from meeting the definition and some don't

[rich helm]

they go off the extreme value theorem

[a,b]

and it has brackets so it's including it

so it should be including endpoints

[acoustic leaf]

before they do that, they have to actually define what it means to be a local min/max. the answer depends on that

[rich helm]

oh shoot okay

so they state that it's not a local maximum because it occurs at an endpoint

buuuuut does that mean absolute maximum's can be endpoints?

🤔

[acoustic leaf]

that depends on how they define absolute maxima

but they presumably define it in such a way that allows for endpoints, unlike local maxima

[devout snow]

@rich helm Has your question been resolved?

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[burnt geyser]

guys i need

[burnt geyser]

is there a formula about geometric and arithmetic sequence

im struggling a lot

like

a) 5, 11, 17, 23,

and b is so brain dead

b) 100, 92, 84, 76,

the teacher js told us which pages will help a lot to the upcoming exam

i have js been soloing this sh

[dense jay]

arithmetic follow the form
nth term: a+(n-1)d where a is the first term and d is the common difference

geometric follow nth term: ar^{n-1}, a is the first term, r is the common ratio

both starting with n=1 in this case

[burnt geyser]

aha

okay thanks a lot

[dense jay]

looks like you have some quadratic sequences there though, one of them anyway

[burnt geyser]

i dont even know what i have

please tell me about quadratic sequences

[trail ridge]

hey i dont mean to pry but i havent gotten help for like over 30 mins, would u mind checking my question out after u finish up here? https://discord.com/channels/268882317391429632/359052604149465088

[dense jay]

quadratic sequences have a constant second difference

oh nvm i was hallucinating

i just cant count

[burnt geyser]

thank god you were

i was gonna have to study more

thank you for your time

[dense jay]

no worries

[dense jay]

im sorry i swore off statistic outside of physics, poor soul

[trail ridge]

all good, could u get someone else in to help tho i really need it

[devout snow]

@burnt geyser Has your question been resolved?

[devout snow]

Channel closed

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[maiden turret]

no idea how to do this

[lusty sapphire]

This is a related rates problem.

What do you know about related rates?

[maiden turret]

i just know i have to take the derivatie

hello

anyone there

<@&286206848099549185>

um

hello

<@&286206848099549185>

[devout snow]

@maiden turret Has your question been resolved?

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[steep jungle]

help

please

i dont understand

<@&286206848099549185>

[devout snow]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[steep jungle]

yes sir

can someone help

[thin fern]

It's not a function if one x-value maps to two or more different y-values

[steep jungle]

wait i just realised

wrong screenshot

mb

gimme a sec

[solid osprey]

💀

[primal jolt]

distribute

move the sum term from the LHS to the RHS

wait no nevermind you’re not given any critical points

[solid osprey]

just plug x and y into eaxh equation and see which one works

[primal jolt]

yeah pretty much

[steep jungle]

ok

thanks

[devout snow]

@steep jungle Has your question been resolved?

[devout snow]

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[lament schooner]

i guess theres a 0 space

[lament schooner]

but how can i determine

if its closed under addition

[wicked turtle]

wdym there's a 0 space?

[lament schooner]

sorry i meant 0 vector

[acoustic leaf]

can you simplify the given condition?

[lament schooner]

uh yeah thing is i have 0 idea where to start

for clusure under addition at least

usually i need 2 vector to figure closure under addition

[wicked turtle]

xy = 0 implies what about x and y?

[lament schooner]

ah wait

are x ywait no

is W a vector

or a set of vectors

[acoustic leaf]

W is a set of vectors

specifically the set of vectors in R^3 which satisfy the given condition

[lament schooner]

ok so

that means xyz are all vectors

or rather

ah no nvm

i can just do

u=(x1 y1 z1) v = (x2 y2 z2)

but then addition test doesnt work cuz

x1y1 = 0 and y1z1= 0

this must satisfy

x2y2 = 0 and y2z2= 0

this must satisfy

so addition

x1+x2 , y1+y2 must be 0

and this doesnt make sense so W is not subspace of R^3?

if im not on drugs rn

[acoustic leaf]

if it is false then you should be able to come up with a counterexample

[lament schooner]

x=1 y=0 z=2

thing is

can i put

1

for x1 and x2?

like

can i have them as parallel/colinear vectors

would it still work or not

[acoustic leaf]

a counterexample would be to come up with two vectors v1 = (x1, y1, z1) and v2 = (x2, y2, z2) where both v1 and v2 are in W but v1 + v2 is not in W

[lament schooner]

just to be sure

v1 = (1,0,2) v2= (3.0,4) is right right?

[acoustic leaf]

is it?

1. are v1 and v2 in W?
2. is v1 + v2 in W?

[lament schooner]

v1 + v2 would be (4, 0, 6)

uh

it still is

[acoustic leaf]

so then this is not a counterexample

[lament schooner]

v1 = (1,0,2) v2= (0.1,0)

ok

took me embarassingly too long to find it LOL

but yeah

[acoustic leaf]

it takes practice

[lament schooner]

this confirms its not clsoed

ty

.close

[devout snow]

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[devout snow]

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[hexed berry]

How do I integrate x/(x-6)?

[dense fulcrum]

-6 + 6 in the numerator then split the fraction as two fractions

[hexed berry]

Why -6 +6?

[delicate river]

try u=x-6

[dense fulcrum]

you want the numerator to remain the same, while also making the expression easier

(x - 6 + 6)/(x-6) = (x-6)/(x-6) + 6/(x-6)

you can cancel out the x-6 and it becomes a much simpler expression to integrate

[dense fulcrum]

this should also work

[hexed berry]

Thank you. It worked.

[dense fulcrum]

type .close @hexed berry

[devout snow]

@hexed berry Has your question been resolved?

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[queen bluff]

Hi everyone, can someone explain me definite integrals?

[prisma harness]

Yes - definite integrals represent signed areas (so if an area is under the x-axis, that is, is negative, area can be negative) under a function from a to b

[queen bluff]

So how do we solve?

Like indefinite ?

[prisma harness]

do you know the fundamental theorem of calculus?

[queen bluff]

No, first time hearing about that

[prisma harness]

there's some good videos on Khan Academy on it, would suggest watching.

basically, 2 parts:
FTC 1 relates derivatives and integrals. It says that the integral of f' is f (original function), +C. C represents any constant, and its because when you differentiate constants are lost and you dont know what they are.

FTC 2 is what you need right now, it says the integral from a to b of f'(x) is f(b) - f(a) (that is, the definite integral is equal to the original function evaluated at b and a)

[queen bluff]

I see, thanks for suggestion and explanation

[devout snow]

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[devout snow]

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[woven locust]

Could I also say that its going from -inf to 2

[woven locust]

or how would I do question 18 all tg

[rare merlin]

can u show a full picture of f(x)

[woven locust]

you arent given f(x)

you have to go off of f'(x)

[glad seal]

I believe the solution in the image is correct

[woven locust]

ok but why cant I include -inf to 2

its also inc

[glad seal]

f(x) is not increasing from -inf to 2, since the gradient of f(x), or f'(x), is negative from -inf to 0

When the gradient is negative, it means the function is decreasing at that point

[woven locust]

oh so if its below the x axis its negative and above is positve

[glad seal]

The graph given is labelled poorly (it doesn't even label the x-axis), but presumably, yes

[woven locust]

wait so is the question asking which is postive and inc or just where its inc

thats why i got confused from -inf to 2

[glad seal]

You cannot say where f(x) is positive or negative, you don't have enough information

From the information given, you can only deduce where f(x) is increasing, decreasing, or neither

[woven locust]

so its just where its above/below x axis

[glad seal]

For this question, yes

Below x axis => negative gradient => decreasing f(x)

Above x axis => positive gradient => f(x) increasing

[woven locust]

ok thanks, would you know how to do this

its the same graph

instead of negative it would be positive and the why is the f"(x) wrong

[glad seal]

That answer should be correct, if I remember all the definitions correctly

f''(x) at x=1 is negative, so concave down

[woven locust]

so the only thing wrong in my answer was the first part and looking from the graph since its concanve down that means f"(x) is negative meaning concave down

[glad seal]

I can't say for certain

For that, I'd say you should ask your teacher or whoever marked it

[woven locust]

oh alr

and for point of inflection i can only use 2nd dera

its asking if x=2 is a poi

[glad seal]

Yes, points of inflection is when the concavity changes

[woven locust]

so my reasoning would be that it changes from concave down to concave up? like f"(x)<0 to f"(x)>0?

[glad seal]

Yes, that should be right

And of course, concave up to concave down also counts as a poi

[woven locust]

ok yea

ok how would i do a b c d

[glad seal]

From the comments, it looks like you should be using a specific method?

[woven locust]

idk any method is there one you know?

i under stand b since thats where its both cc up and down

[eager agate]

the method to find these questions on the true or false is to understand what would be true of the original function if either the first or second derivative is greater than, equal to, or less than 0.

So, if f'(x) is less than 0, what does that mean for the original function f(x)?

[woven locust]

its decreasing

and b is the only one decreasing

ohhh

[eager agate]

correct.

i'm going to go ahead and say that i believe that all your answers are correct. but do you know why they're correct?

[woven locust]

ik theyre correct but i dont really understand them

i get a nd b now