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[tribal matrix]

Is n even?

[stiff nexus]

yes, it's even

[tribal matrix]

thx

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[winged light]

hey guys i need help with an optimazation problem

[winged light]

im stuck and im not too sure what to do

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@winged light Has your question been resolved?

[stiff summit]

check the third line of the left column

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@winged light Has your question been resolved?

[winged light]

what do you mean

[stiff summit]

in your work, there are 2 columns of equations

I think you might have made a mistake in the third line of the left column

[winged light]

but i fixed it the next line down

no?

[stiff summit]

no, you left it as 10 instead of 20

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[twin wolf]

Not sure how to start

[buoyant pewter]

use the formula for the distance of two points

then you get the function of the x variable and you easily find its minimum

you will be able, even , to use trinomial square theory

[twin wolf]

Trinomial square theory?

[hot herald]

are you on part a or b

[twin wolf]

I haven’t started on either

[hot herald]

the distance would simply be:
(the y-coord of the top curve) - (y-coord of bottom curve)
(at the same value of x)

[twin wolf]

Oh so I don’t use that square root x2-x1 stuff

[hot herald]

no need, overcompliating

you "could" if you want

but x-coords of the points are the same

so that component will be 0

[twin wolf]

I see

[twin wolf]

Wait if this is the case then how come there are still x terms in the answer for the distance? Wouldn’t it just be a number then

[hot herald]

the distance will be in terms of x

for any location, you subtract the y coords to get the distance

[twin wolf]

Hm, do I first differentiate both the top and bottom curve to find this?

[hot herald]

overthinking

simply subtract the y-coords,
top - bottom

[twin wolf]

Well the top y coord is 5 and the bottom is 0

5 -0?

[hot herald]

that's when x=0

and when x=0 that will be the distance

[twin wolf]

So it is 5?

[hot herald]

when x=0

that'll be the distance when x=0,
not what the question asks for

[twin wolf]

I’m confused then

Not sure what to do if not that

[hot herald]

literally subtract what you're given directly

[twin wolf]

x^2 - 2x + 5 - 4x - x^2?

oh

lol

[hot herald]

missing ()

[twin wolf]

i see

well that was easy

And for part b) all I have to do is take the distance and do simple derivative stuff right

[hot herald]

or consider vertex of a paroabla / vertex form

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[digital barn]

Hello I need help with this please

[digital barn]

Hoe can I find the f(x) and g(x)?

[torn jolt]

by logic

[digital barn]

can you further explain for me

I dont really get it

[torn jolt]

f(x) is sqrt(x) + 10

and g(x) is 5x³

[digital barn]

how did you get it?

I dont understand well

<@&286206848099549185>

I have another problem so i wanna know how to do it 2nd problem) h(x)=3^x^2 but i also want to the first one aswell

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[digital barn]

.reopen

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✅

[digital barn]

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[odd fox]

Let's z be a complex number satisfying the equation z²-(3+i)z+∆+2i=0

[odd fox]

∆ belongs to R

i=√(-1)

Suppose the equation has a real root, find the imaginary root

But don't imaginary roots come in pairs?

[odd fox]

So shouldn't both be imaginary roots?

[spice orchid]

That's only true if your equation has only real coefficients

[elfin stream]

i think thats not necessary

[odd fox]

Ok thanks

Makes sense

Does the quadratic formula still apply if a or b or c are imaginary?

Nvm

💀

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[lament garnet]

Let $\phi: \bR^{3} \to \bR$ be a continous single valued function of coordinates $x_i$ , it is said that under a coordinate transformation that sends $x_i$ to some $(x_i)'$ (the transformation acting on scalars is said here to not change the scalars i.e $\phi(x_1,x_2,x_3) = \phi '(x_1',x_2',x_3')$ it is then stated that: $\pdv{\phi '}{x_i'} = \pdv{\phi}{x_j} \pdv{x_j}{x_i'}$ (summing over j). My question is regarding where does the above equality come from ? The chain rule doesn't tell me much and leaves me rather confused, if i assume that the entire expression works because all the sum terms are 0, except $\pdv{\phi}{x_i'} \pdv{x_i'}{x_i'} = \pdv{\phi}{x_i'}$ , when $j = i$ , otherwise 0 it starts to make slightly more sense, but i'm not assured whenever that is true, also that doesn't fix my main source of confusion which is that coordinates out of the blue are now uhhmmmm transforming like this: $x_j = \lambda_{kj} x_k'$ which becomes ohio when a conclusion is reached that $\pdv{x_j}{x_i'} = \lambda_{ij}$ im sorry but my brain is not braining to understand where each step comes from nahhhhhhh.. ohio

[glossy valve]

Bishop

[lament garnet]

????????

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@lament garnet Has your question been resolved?

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@lament garnet Has your question been resolved?

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@lament garnet Has your question been resolved?

[limber sandal]

Hi you know what is differential of function

@lament garnet ?

[lament garnet]

Uh

I know what a derivative is

[limber sandal]

Its like function that linearly approximate it

[lament garnet]

..??

[limber sandal]

for our funcion it like

[lament garnet]

A derivative of a single variable function defined point wise is a limit value as x tends to c of (f(x) - f(c))/(x - c)

[limber sandal]

Yeah and its diferential is f'(x0)dx at point x0

And you can say

[lament garnet]

what is a differential

[limber sandal]

Its the best linear approximation

Like for one variable

[lament garnet]

What is a linear approximation

[limber sandal]

f(x)-f(x0) = df(x0)+o(x)
Where df=f'(x0)dx

[lament garnet]

what's o(x)

[limber sandal]

Okay you dont study this staff?

[lament garnet]

I don't know what df nor o(x) is

is that even in multivariable calculus class?

My multivariable calculus book doesn't have that

[limber sandal]

Its in analysis

You dont study it in calculus or analysis?

[lament garnet]

no

i'm very confident that df nor o(x) is not in my calculus or analysis book

[limber sandal]

Okay

[lament garnet]

So that thing is relevant for understanding each step ?

the differential thingy

[limber sandal]

I think no but its more rigorous

Okay for one variable you can say f(x)-f(x0)=f'(x0)Δx + error term

And error term is very small when (Δx = x-x0) Δx is small

????

@lament garnet

[lament garnet]

so

basically

[limber sandal]

And this type approach you can do with multivariable

[lament garnet]

what if i say lim x --> x0

i.e

$\lim_{x \to x_o} f'(x0) \Delta x := f(x) - f( x_0)$

[limber sandal]

Its 0

[glossy valve]

Bishop

[lament garnet]

oh

nvm im dumb

wel

it doesnt matter ik

[limber sandal]

But if you both part divides by Δx its more usefull

[lament garnet]

how so

won't it still go to 0

isn't $\Delta x$ just a real number $\in \bR$ ?

[glossy valve]

Bishop

[limber sandal]

Yeah very small

[lament garnet]

(well if we divide then non zero)

[lament garnet]

ok i get it

so what is that usefull for here

[limber sandal]

But its necesarry rigorously prove how error term is smaller then f'(x)dx

And for this question there is small o(Δx)

Buts okay

If you have multivariable the same thing you can do

[lament garnet]

i got to go

sry\

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[limber sandal]

Bye

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[glossy valve]

Retribution | ᵇᵃˢᵉᵈ

[torn jolt]

Someone please help

[gritty rose]

Do you have limits on the integral

[torn jolt]

integral from -4 to 4

[glossy valve]

Retribution | ᵇᵃˢᵉᵈ

[torn jolt]

wait i know why but idk why it didnt load in the _

i literally copy and pasted

<@&286206848099549185>

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[glad prairie]

where does this formula come from?

[fast peak]

what kind of answer do you want

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[limpid moat]

the blank answer

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[wicked frost]

Can someone help spot my mistake, I used Vb in my solution instead of Va but the equations are still correct but I ended up with the wrong answer

(The left is the solution key and the right is my work)

(we're using j as the imaginary number i)

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[wicked frost]

<@&286206848099549185>

[soft axle]

look at the difference

and provide the circuit schematic

and node analysiis diagram on the schematiic

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[wicked frost]

the equations I got which the solution labels as 1 and 2 are the same tho so it shouldn't matter, its just a matter of solving those 2 equations simultaneously

[soft axle]

show the schematic and analysis

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[eager crag]

g(x) isnt differentiable at x=a
because of an absolute value
lets say (f'(g(x))g'(x))' is defined in all x
that equals to f''(g(x))*(g'(x))² + f'(g(x))g''(x)
is f''(g(a)) = 0 because of the chain rule or is it irrelevant because of g'(x)²

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[eager crag]

<@&286206848099549185>

[torn jolt]

yea

wut is it?

[eager crag]

what

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@eager crag Has your question been resolved?

[tired mirage]

Can't you use KCL here

It's p much just a big system of equations

All the currents leaving each node

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[unborn moth]

Hi

[limpid moat]

🫡

[unborn moth]

How does one

Prove

That

[limpid moat]

what

[unborn moth]

The

[onyx glen]

send

[unborn moth]

That thing

[onyx glen]

more

than

one

[limpid moat]

word

[onyx glen]

word

per

[unborn moth]

Sorry

[onyx glen]

message

[unborn moth]

I’m

[onyx glen]

please

[unborn moth]

Sorry

I am

So sorry

For my

[limpid moat]

i will

close

[unborn moth]

Very inconvinience

[limpid moat]

this

[unborn moth]

Don’t do

it please

[onyx glen]

you can stop typing like this at any time btw

[unborn moth]

Yes that is correct

[onyx glen]

and you really should, bc it's a nightmare to read

[open igloo]

What is goin on

[onyx glen]

take a deep breath, collect your thoughts, and formulate your question.

[unborn moth]

How do you do the integral of the equation of the circle

I tried trig sub

U sub doesn’t work btw

[open igloo]

It doesn't?

[limpid moat]

can you wrote the integral under discussion

[open igloo]

What's the equation you have

[unborn moth]

Integral of sqrt(1 - x^2)

1 for the radius

I mean

r^2 = 1

[open igloo]

I get it

[unborn moth]

And from -r to r

[limpid moat]

cuz i mean an equation is not a real function

[unborn moth]

Oh

You get it bruhh

Expression

[open igloo]

$\int {\sqrt {1-x^2}}$

[glossy valve]

Lorentz

[covert siren]

u cant integrate it as a single function

it isnt a single function

[unborn moth]

Wdym

[covert siren]

like the eqn of a circle is really two functions

[unborn moth]

Oh yeah I know that

[hidden harbor]

$\int {\sqrt {1-x^2} dx}$

[glossy valve]

LordFelix

[covert siren]

soo

[unborn moth]

So

[open igloo]

x=sinθ should work, no?

[unborn moth]

Yes

[hidden harbor]

it should

[unborn moth]

what

[open igloo]

Won't it be cosxdx

Instead of cos^2

[hidden harbor]

no

[limpid moat]

no because you have dx=cos t dt to add

[open igloo]

Ah makes sense

I need sleep

[hidden harbor]

step by step:
x=sint
dx=costdt
sqrt(1-x^2) = sqrt(1-sin^2(t)) = sqrt(cos^2(t))=cost

[limpid moat]

however bradencher next time try to google your integral ... you will often get an instant answer

[unborn moth]

hmm

Ok

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@torn jolt Has your question been resolved?

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[halcyon sleet]

can I go any further than this?

or am I just wrong

if so how do I correct/continue

<@&286206848099549185>

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[halcyon sleet]

<@&286206848099549185>

[halcyon sleet]

.close

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[tribal sluice]

Hi, I want to know how to set up an integral to integrate this area.

[gritty rose]

Do you know the equation of the ellipse

[tribal sluice]

Yes, I do, but my teacher wants me to integrate this 😦

[gritty rose]

Yea do that

[gritty rose]

Correct

[tribal sluice]

?

[gritty rose]

What

[tribal sluice]

I tried to put this in the calculator but it doesn't work

[tribal sluice]

is this wrong?

[gritty rose]

It's the correct general formula

But to plug into a calculator you probably need values for a and b

[tribal sluice]

Okay let me try adding the values, are a and b in the right positions?

or should it be

smth else

[gritty rose]

I don't understand your question

You should be able to identify a and b from your graph

[tribal sluice]

its just 0

[gritty rose]

What are you claiming equals zero

[gritty rose]

b is most definitely not 0

[gritty rose]

.

[tribal sluice]

OH WAIT

OSRY

Ah I see, I got it, if I put it in a graphing calculator will I be able to see it?

[tribal sluice]

Om gee Riemann I solved it okay thank you 😄

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[delicate raven]

is it safe to assume that the bottom left triangle is a right triangle

[proper hawk]

you can check using Pythagoras theorem

[delicate raven]

alright

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[thorn fjord]

Long division, what am i doing wrong?

[thorn fjord]

I think m division itself is correct, except for doing mod 3

Oh, -x mod 3 = 2x

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[frigid axle]

can someone help me? i need to find a,
m represents the slope
and x is where the slope occurs

[frigid axle]

x = pi/6

<@&286206848099549185>

[torn jolt]

Hey

[frigid axle]

hello

[cosmic olive]

Hey

[frigid axle]

can you guys help?

[torn jolt]

Sure

Is it cos of (x+a)

?

In the numerator

[frigid axle]

int the question it showed just cosx + a

[torn jolt]

You want the slope of the function to be 4 at x=pie/6 right?

[frigid axle]

yes

[torn jolt]

Cosx + a or cos(x+a)

[frigid axle]

cosx + a

[torn jolt]

Okay

5 minutes

[frigid axle]

alright thanks

[torn jolt]

Is this what you wanted?

[frigid axle]

yes

thank you

[torn jolt]

I get a=6

[frigid axle]

that's correct

thank you very much

[torn jolt]

Do you understand the solution

[frigid axle]

just a sec

[torn jolt]

Sure

[frigid axle]

oh i haven't learnt about sec yet

[torn jolt]

Derivative of secx?

[frigid axle]

yes

[torn jolt]

Or secx as in trigonometry?

[frigid axle]

can you do it without it

i didnt learn it at all

[torn jolt]

Do you know u/v rule

?

For derivativw

[frigid axle]

i don't

can you explain it?

[torn jolt]

Uhm

What do you know about derivative?

[frigid axle]

i know how to do derivatives

i didnt learn how to do it with secx

i havent learnt about it yet

[torn jolt]

Oh so you are not familiar with u/v rule right?

[frigid axle]

maybe i am but i am not sure

[torn jolt]

Can you do derivative with limits?

[frigid axle]

i live in a different county

country

[torn jolt]

No worries

[frigid axle]

so i dont know the name

what do you mean by derivative with limits?

[torn jolt]

Like basic definition of derivatives

Lim h tends to zero

[frigid axle]

oh yes

i do

[torn jolt]

Yes so you can do this by that

Or I would suggest learn u/v rule

And then apply it for 1/cosc

1/cosx

As 1/cosx is secx

[frigid axle]

my class did this question differently

with secx

[torn jolt]

How so

[frigid axle]

and i am not sure if they would allow me to do sec in the test

[torn jolt]

Okay what was that method?

[frigid axle]

just doing the derivatives i think but i am not sure

[torn jolt]

No problem

Download my solution

[frigid axle]

yeah i did

thank you anyway

[torn jolt]

Ask your friends if they did by same method

[frigid axle]

ok

thank you very much

[torn jolt]

Otherwise download their solution

And send it here

[frigid axle]

ok

[torn jolt]

Ill try explaining

Welcome

.close now?

[frigid axle]

yes

.close

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[somber remnant]

find the total surface area of the combined solid

[somber remnant]

.close

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[knotty flower]

Could someone help with these? I need to compute the integrals using integration by parts but I keep getting stuck
B: every combination of f'(x) and g(x) seems to give a more difficult integral
C and D: I try to get rid of the x^3 by taking the derivative of it (multiple times) but as a result of the integral of (9+x^2)^-0.5 in C and e^x^2 in D I just get a new 1/x in the integral (see picture below)

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@knotty flower Has your question been resolved?

[knotty flower]

<@&286206848099549185>

[elfin stream]

do u know DI method for IBP? @knotty flower

[elfin stream]

I think u might be able to do some differentiation shenanigans in b

[knotty flower]

I don't

[elfin stream]

lemme send a vid

https://www.youtube.com/watch?v=2I-_SV8cwsw

[knotty flower]

yeah that makes the process a bit quicker if you need to do the integration by parts many times in a single question but it doesn't solve my problem

but thanks anyway

[elfin stream]

if we had - integral -(x+1)e^-(x+1) * 1/(1+x)^2 then it would have been simply e^(-x-1) evaluated from 0 to 2, i think

[buoyant pewter]

read this solution, I made a similar thing:

$\int_{0}^{1}x^{5}e^{x^{3}}dx=\int_{0}^{1}x^{3}e^{x^{3}}x^{2}dx=\frac{1}{3}\int_{0}^{1}te^{t}dt=\\=\frac{1}{3}\left( \left[ te^{t} \right]{0}^{1}-\int{0}^{1}e^{t}dt \right)=\frac{1}{3}\left( e-\left( e-1 \right) \right)=\frac{1}{3}$

[glossy valve]

Joanna Angel

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[odd fox]

My answer is pi/2

Wanted to confirm

[devout valley]

,w int 2 tan(x) sec^2(x) /(tan^4(x) + 1), x from 0 to pi/2

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[odd fox]

Goat

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[peak stone]

!help

[full forum]

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

[peak stone]

!status

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[peak stone]

1

[onyx glen]

x is the pollutant level in ppm

you are given dx/dt as 1.4 ppm/yr

find dF/dx and then from it dF/dt

[peak stone]

alright wait lemme process

so first I differentiate F = 32000/3+rootx right?

-(16000/rootx (rootx + 3)^2

is this correct

[onyx glen]

missing parentheses

[onyx glen]

write it on paper

[peak stone]

okay wait

this

@onyx glen

[onyx glen]

ok yeah that looks correct

[peak stone]

alright then after finding it

what do I do

how do I find dF/dt

@onyx glen

[onyx glen]

chain rule lol

dF/dt = dF/dx * dx/dt

unless this is your first ever related rates problem this should be at least a little bit familiar

[peak stone]

I know chain rule

but I don't understand why would I use it to find dF/dt

[onyx glen]

F is a function of x, and x is itself a function of t

[peak stone]

I am lost

[peak stone]

alright lets just say I use this

It should look like this

right?

[onyx glen]

yes

oh, you also need the value of x that results in F = 4000.

also you need to write your t's in a way that makes them distinct from plus signs.

[peak stone]

oh okay

this

then I find the value of x

[onyx glen]

no

F = 32000/(sqrt(x) + 3)

[peak stone]

what

[peak stone]

oh yeah

thats what I meant

like this

[onyx glen]

the second = sign is inappropriate

but yes x = 25

now plug that in

[peak stone]

alright

[onyx glen]

the rest is arithmetic

[peak stone]

so like this?

@onyx glen

[onyx glen]

not ppm/year

also you should write the middle stroke in your sevens

[peak stone]

oh its not ppm/year?

what should it be

[peak stone]

alright I'll keep that in mind

[onyx glen]

fish/year, what else

F is the number of fish

[peak stone]

ohh alright

[peak stone]

can you explain this a bit more into detail

I can't get it through my head

[onyx glen]

i don't know how to explain it in any other way, sorry.

F is a function of x and x is a function of t

dF/dt is naturally found via the chain rule

[peak stone]

alright

thanks for the help

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[cedar hawk]

Im stuck on c. refreshing my completing the square skills was fun because its been a long time but how the hell do you deal with that xy term. I'd assume it wants you to complete the square again and use that but doing that with the xy is beyond me. I could do this with derivatives but thats obviously not the point here

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[cedar hawk]

<@&286206848099549185>

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@cedar hawk Has your question been resolved?

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@cedar hawk Has your question been resolved?

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[full forum]

Please don't occupy multiple help channels.

[rotund pulsar]

how do i do z_bar ?

,w 1/30 * int_0^{x+y+5} int_0^1 int_0^4 z dxdydz

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[rotund pulsar]

what am i supposed to put for x and y in this expression?

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[obtuse orbit]

why is x=-1 an asymptote? or is the workbook wrong

[hollow zenith]

its not

[obtuse orbit]

okay

thanks

[full forum]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[obtuse orbit]

oh

[minor crater]

maybe they are indicating something else?

[obtuse orbit]

hmmm? do u think its just an error?

[minor crater]

most likely

[obtuse orbit]

just to clarify, you can find asymptotes by finding the intercepts right

[obtuse orbit]

?

[minor crater]

the x intercept of f(x) is the asymptote of 1/f(x)

if that's what you mean

[minor crater]

the same vice versa too

[obtuse orbit]

ohh right

okay thanks!

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[torn jolt]

what am i doing wrong

[torn jolt]

because i don't think i can equate those

so i must have screwedup somewhere

<@&286206848099549185>

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@torn jolt Has your question been resolved?

[torn jolt]

<@&286206848099549185>

[torn jolt]

<@&286206848099549185>

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[nova ice]

can someone walk me through this?

[torn jolt]

Sure @nova ice

[nova ice]

where do i start?

an example of a linear inequality in two variable

s

no clue what that means

[slim creek]

x + 2 > y?

Linear means the power of x is one

Inequality means it’s not equal so it’s either greater or lesser than

two variables means the presence of only x and y

As unknowns

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@nova ice Has your question been resolved?

[nova ice]

ok i still dont get it

can someone explain this like im a baby

i started a new adhd med and i am wacked out of my shit so please be patient

<@&286206848099549185>

[sleek salmon]

ok so lets imagine that you have two friends, let's call them "x" and "y." now we have some rules about how they can be friends and then you can express those rules using numbers and symbols.

ok so for example, you might say that 2 times the number of "x" minus 3 times the number of "y" is less than or equal to 6 the "less than or equal to" part means that as long as you follow this rule, everything is okay.

rule could be written like this:
2x−3y≤6.

now you can think of this rule as a magic line on a piece of paper. If you pick any point on or below this line, it means you are following the rule. but if you go above the line, you are breaking the rule.

idk if that helps but i tried

[nova ice]

ok ok

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[spring pagoda]

Please help I just cant understand it conceptually

[spring pagoda]

I knnow how to find a and b and ab but im not sure how to incorporate that into finding the other roots

all i need is an explanation of just generally how to solve one example and then i can do the rest 😄

[thorn lichen]

for example, having roots of $2\alpha$ and $2\beta$ means that the polynomial is $\gamma(x - 2\alpha)(x - 2\beta)$

[glossy valve]

marieme

[thorn lichen]

$\gamma$ can be any number such that the resulting polynomial has integer coefficients

[glossy valve]

marieme

[spring pagoda]

yes

[spring pagoda]

how would i do it without solving for the roots though?

@thorn lichen

[thorn lichen]

you kinda have to solve for the roots of the first equation

[spring pagoda]

oh

[surreal hull]

you can use vieta's formulas

if you don't want to solve

[spring pagoda]

💀 i thinnk i gotta use what i learn in class

[surreal hull]

do you know "product-sum"?

[spring pagoda]

like sin(a+b)?

[surreal hull]

for a quadratic

[spring pagoda]

oh thats sums to products

mb

[surreal hull]

there's a relation between the product and sum of roots and the coefficients of the polynomial

have you learned about that?

[spring pagoda]

uhh no lmao \

[honest night]

yk how to form an equation with product and sum of roots ?

[honest night]

yes

[spring pagoda]

i may have seen it just didnt know the name

is it about a+b and ab

[honest night]

1 min

[honest night]

ok

ya back

[spring pagoda]

like a+b = -b/a?

[honest night]

yes

[spring pagoda]

oh yes

[honest night]

should i send the sol ?

[spring pagoda]

i dont rlly knnow how to incorporate that

[spring pagoda]

yes please thank you

[honest night]

.

ok

give me 5 mins

[spring pagoda]

thank you

[honest night]

@spring pagoda

waitttt

i wrote formula wrong 😭

[spring pagoda]

oh ok thank you very much. for c i would just use a simultaenous equation right?

[honest night]

for everything u have to use the same method i did

and also rember this formula
a^2 + b^2 = (a+b)^2 -2ab

[spring pagoda]

oh yeah thanks 😄

[honest night]

[spring pagoda]

can i just leave this channel open to check my answers once im done

if thats alright with you

[honest night]

np for me

[spring pagoda]

for c i got 9x^2 + 18x + 29

[honest night]

should i try

[spring pagoda]

i think it should be good but yes please

and for d i got 9x^2 + 38x + 49

[honest night]

9x^2+18x+25

is what i got

[spring pagoda]

u got to the step where its 105/9-76/9 right?

[honest night]

9x^2 - 38x +49

[honest night]

no?

[spring pagoda]

oh i think its positive because to get the root -38/9 you have to have negative in the brackets and then in the square

[honest night]

its like a^2+b^2 right

[honest night]

and we know this

if u apply that

once check

[spring pagoda]

like its x^2 - (sum) + product

and sum is -38/9

[honest night]

send pic once

[spring pagoda]

oh sorry i cant use phonne im in school

[honest night]

oh

fuck

my bad my bad

i wrote - as = so i got confused 😭 💀

[spring pagoda]

all g tho im nnot rlly sure abt this one

(d)

bro i cant send images

OH DMAN it loaded

[honest night]

delete

spammed msges

[spring pagoda]

its all g i got it

thanks

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[honest night]

bruh 😭

[vagrant shell]

Blud really waited that good 10 mins just to promote a server

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[proven temple]

an app where you plug in points and it automatically graphs them?

[honest night]

demos

https://www.desmos.com/calculator

[proven temple]

hmm ok

ill learn how to use it

thank you 🙏

[honest night]

np

[proven temple]

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[torn jolt]

With questions that ask you to solve for x in 0=x=2pi, and have a root somewhere, would you solve for all quadrants?

[rocky vale]

Do you mean $0 \leq x \leq 2\pi$? or maybe $0 \leq x < 2\pi$?

[glossy valve]

tatpoj

[rocky vale]

or something like that, right? not with equal signs

in which case, yes, they're asking for every solution between 0 and 2pi, which is all four quadrants

[torn jolt]

because roots are both positive and negative, correct?

my emphasis was on the root in a question

[rocky vale]

I'm guessing from context that "x" here refers to the angle, and since you're letting the angle range from 0 to 2pi, that covers a full circle

[torn jolt]

so say you had your final answer be pi/4, but there was a root somewhere in the question

you'd have to solve for all 4 quadrants?

like add pi subtract from pi

2pi.. etcc

[rocky vale]

you mean like if you had a +/- sign from taking a square root?

and ended up with pi/4 and -pi/4?

[torn jolt]

let me give you an actual question instead

so I worked it out

up until here

[rocky vale]

ok

[torn jolt]

sqrt2 sinx = 1

divide both sides by sqrt 2

sinx=1/sqrt2

which rationalizes

to root2/2

or in other words, if you took the inverse sine

pi/4

you would include all 4 quadrants because there was a sqrt

and following cast rule

you'd have to solve for each quadrant since it is both when sin is positive and negative

[rocky vale]

no, there are only solutions to that equation in the first two quadrants

did you start from sqrt2 sinx = 1?

or did you introduce that sqrt yourself

[torn jolt]

I introduced it myself

[rocky vale]

what was the full original problem?

[torn jolt]

the start was $cotxcsc^2x=2cotx$

[glossy valve]

mj

[rocky vale]

okay, I see

when you take the square root of both sides, you must introduce the +/- sign, yes

so you have $\sin x = \pm \frac{\sqrt{2}}{2}$

[glossy valve]

tatpoj

[rocky vale]

sorry I'm bad at latex lol

at some point, you took a square root, and when you do that you must consider the positive and negative square root

[rocky vale]

also don't forget the solutions that come from cotx = 0

but none of this has to do with the restriction of 0 < x < 2pi

[torn jolt]

yeah, I'm unfamiliar with a positive/minus concept of square roots in this context

so I just wanted to verify

because sin is both positive and negative

and the RA is pi/4

it would be pi/4 RA in all quadrants

[rocky vale]

RA?

oh reference angle

[torn jolt]

reference angle

[rocky vale]

yes

yeah it would be all four

[torn jolt]

but this is exclusively because we ended up with an answer of sinx= some square root value

[rocky vale]

it has nothing to do with having a square root visible in the answer

it's because you took a square root yourself

just like

x^2 = 4
x = +/- 2

[torn jolt]

only other instance I've done +/- sqrt was in a quadratic formula

[torn jolt]

what would it differ otherwise

[rocky vale]

All you need to know is that when you take the sqrt of both sides of an equation, insert a +/- sign in front of it

like if you have $\sin^2 x = 1$

[glossy valve]

tatpoj

[rocky vale]

you would take the square root of both sides and get $\sin x = \pm \sqrt{1}$

[glossy valve]

tatpoj

[rocky vale]

but that's just $\sin x = \pm 1$

[glossy valve]

tatpoj

[rocky vale]

there is no square root sign visible in your answer, but you still added the +- sign when you take the square root

[torn jolt]

may I ask why is it a square root of a number can be both positive and negative?

[rocky vale]

yeah good question tbh, a lot of people misunderstand the sqrt sign

If I ask you, what is the square root of 4, what would you say?

[torn jolt]

I would say 2

[rocky vale]

and why is it 2?

[torn jolt]

because 2^2 is 4

[rocky vale]

but (-2)^2 is also 4

so 2 and -2 are both square roots of 4

[torn jolt]

you have a point

so cube root would not hold a plus or minus infront of it

for that very reason

[rocky vale]

correct, only even degree roots

[torn jolt]

fair enough

thanks for clarifying

[rocky vale]

the sqrt sign specifically means the positive square root, but both roots solve the equation, so we add the plus/minus sign to include both

[torn jolt]

I just assumed it was positive and did not understand the reason for having +/- infront of the root in a quadratic formula

[rocky vale]

ahh, yeah, formulas with sqrts in them often have a +- in front for that reason

we usually want to include both

[torn jolt]

welp, good to know

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[steel mason]

Can you multiply x and i

[left bone]

yes

but usually when you are working with complex variables you call them z not x

but you can multiple a (real) x times i anyways

[steel mason]

Ok my teacher wrote on the study guide that (x+5i)(x-5i)=(x^2- 25i^2) so is that wrong?

[left bone]

its right, but its not how one would usually write it

$i^2 = -1$, so $(x^2- 25i^2)=(x^2+25)$

[steel mason]

Oh never mind i just got it

Thanks sooo much for the help

[left bone]

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[light mantle]

Would that point be: (1/3, 7/4)?

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[vital wyvern]

Prove $U(p)={1,2,…,p-1}$ is cyclic.

[glossy valve]

splotchvan

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[fast marlin]

How do i write a formal proof for this:

[fast marlin]

hold on let me get the work i did

$x_1=2+0i=2\left(\cos(0)+i\sin(0)\right)=2\
x_2=-2+0i=2\left(\cos(\pi)+i\sin(\pi)\right)=-2\
x_3=0+2i=2\left(\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})\right)=2i\
x_4=0-2i=2\left(\cos(-\frac{\pi}{2})+i\sin(-\frac{\pi}{2})\right)=-2i$

[glossy valve]

Joshii

[fast marlin]

so clearly after the first equal sign we have standard, then after second we have polar

[gritty rose]

Looks right

[fast marlin]

right but is that all?

like its a proofs class so i assume they expect some sort of formality/explanations

[rare dock]

probably nobody can answer this better than you

[fast marlin]

wdym

bro?

my friend did this

hm?

[gritty rose]

Your answers were right. If you wanted your work checked too, you should show it

[fast marlin]

thats all i did

[gritty rose]

Uhhh

Up to your teacher then if that's all you need to show

I'm not your teacher

[fast marlin]

alright

i assume its fine then

ill figure something out if it comes up on paper

[rare dock]

i meant how much you need depends on your class standards which nobody will really know better than you

?

your pronoun bar in your bio is not appropriate for this server

[fast marlin]

i just dont see how else it would even be expanded

maybe through graphing?

and stating the definition of polar form

okay now that i'm thinking abt it more, i think i know what i would do on paper

thank you, i think i mostly just needed to talk to someone abt it

[rare dock]

idk, you could verify x^4 - 16 = 0 is true for each x value

[fast marlin]

yeah i thought about doing that too lol

alright i have a whole bunch of stuff if it comes up

praying i get a question that easy on the exam lol

thank you

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[pale peak]

struggling with d/dx x^(x-1) in the context of trying to use it for chain rule to solve the problem y' = (1-x)^(x-1)

[pale peak]

above is the problem from the practice test, below is the answer as found in the answer key

y=(1-x)^(x-1) C.R: f=x^(x-1), f'= ?, g=(1-x), g'=-1

[left bone]

what does the first part of your question have to do with the second part

why are you doing d/dx x^(x-1)

[pale peak]

f' = ?

need f' for f'(g(x)) * g'(x)

can't find f' because don't know how to get d/dx x^(x-1)

yeah essentially

would it not be ln(x^(x-1))?

[left bone]

$\ln(a^b) = b\ln(a)$

[pale peak]

okay

1/y = x*ln(x) + (x-1)*1/x

[glossy valve]

help a sad kid | desi

[pale peak]

i'm not quite sure what that means

dy/dx * 1/y

[glossy valve]

help a sad kid | desi

[pale peak]

not sure i understand sorry

(ln(y)' / dx) = (ln(y)' / y') * (dy/dx)

not sure what dy/dx is representing exactly

1 = (x*ln(x) + (x-1)*1/x)*y ?

`` 0 = (x*ln(x) + (x-1)*1/x)*y -1 ?`

so you're saying it's

1/y * (x*ln(x) + (x-1)/x)y = x*ln(x) + (x-1)*1/x

i got that but what is dy/dx and how do i get rid of it to get to the actual answer

all i want is the derivative of x^(x-1)

f'

1/y * y' = x * ln(x) + (x-1) * 1/x

is this the same thing?

so y' = (x * ln(x) + (x-1) * 1/x) * y

so y = x^(x-1)

so now i have (x * ln(x) + (x-1) * 1/x) * x^(x-1) as f'

d/dx f(g(x)) = f'(g(x)) = (x * ln((1-x)) + ((1-x)-1) * 1/(1-x)) * (1-x)^((1-x)-1) * g'(x) = -1

wait what

f = x^(x-1)
f' = (x * ln(x) + (x-1) * 1/x) * x^(x-1)
g = (1-x)
g' = -1

is using chain rule just a bad idea here or something?

do what the same way?

i don't know where you are getting (1-x)^(1-x) from sorry

my friend from class solved the problem and says he'll go over it with me

thanks for the help

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[atomic vortex]

How do u do this

[atomic vortex]

I tried getting the rref and got

1 0 0
0 1 0
0 0 1

idk if that is suppose to help though

[mellow urchin]

for square matrixes, the vectors form a basis iff det(A)=/=0

so use det(matrix)=/=0 and solve for a

[onyx glen]

put these vectors as rows into a matrix then row reduce it, taking note of any and all times when you divide by a variable

[atomic vortex]

oh so it cant be 0 for sure

[onyx glen]

that's one thing it can't be. do you have the rest of your row reduction written out?

[atomic vortex]

I solved the determinant and got a =0, 3

[mellow urchin]

ok

[atomic vortex]

oh so thats correct

[mellow urchin]

i guess idk

[atomic vortex]

what do u mean rest

like after i divided by a?

[onyx glen]

no i mean all of your work

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[stuck summit]

I've gotten the parameterized <t+2, t+1, 2t-2>, but now i'm stuck

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[thorn fjord]

"What polynomials are reducible in Z{2}[x] ?

[thorn fjord]

I dont really know how to start this problem

my first guess is that Z{2} has 2 zero-points; 0 and 1

[neat bay]

f(k) = 0 iff (x - k) is a factor of f

There’s two options for k: 0 and 1, as you said

f(0) = 0 is false for all of these so that’s just irrelevant

f(1) = 0 iff (x + 1) is a factor of f

If it’s degree 2 or 3, and it splits, then it must have a linear term (x + k) for k either 0 or 1. Thus for the degree 2 or 3 cases, it suffices to check f(0) = 0 or f(1) = 0

Actually this f(1) = 0 check works quite well, see (d)

[thorn fjord]

What about f). f(1) is not 0 here

But it is reducible

[neat bay]

Good question

For degree 4, it suffices to show either a linear term or quadratic term divide it

We’ve already checked the linear term for (f), and it doesn’t work

[thorn fjord]

Yeah my teacher divides it

[neat bay]

So we have to check if a quadratic term divides it; this quadratic must be irreducible

Yes the only degree 2 irreducible poly is x^2 + x + 1

So divide it by that

And it splits evenly

Damn this is kind of a dumb exercise ngl

Like what are you learning

Besides just spamming factor theorem and ad hoc methods

[thorn fjord]

Thats literally it

nothing more