help-27

[true coral]

Out of those formulas, you mean? Sure, but you need some context like "You know this and this, solve that"

[hasty scarab]

no clue lmao

i mean look

i told the teacher

she said

you need to listen carefully

i said

teacher you just typing shit not explaining

i told her to explain slower and dont just type shi and keep quiet

now i have no clue

[true coral]

I say this with the utmost respect: There seems to be a fundamental misunderstanding here

You can't find P and V with the information given

The formulas only tell you know to calculate those things from certain other things

You can't use them without the other things in question

^

[hasty scarab]

wdym

ah

nah

too much

[true coral]

If you want to "solve" stuff you gotta have stuff to solve for

Right now all you have is tools

Formulas are just tools

You give them some info and out pops other info

In order to solve for anything you gotta have a problem to begin with

[hasty scarab]

no clue bro

by problem you mean

start the equation or?

[true coral]

Wait flip I got tunnel vision

I didn't even notice the board on the rigjt of the image

[true coral]

But to my defense it's cut off so I can't make much of it

[hasty scarab]

ye i was about to point but

ye sadly i only sneak took a picture of the table

cuz the teacher was gone

the proffesor

[true coral]

Does that day a=40, H=80? Or is there some more stuff cut off on the right?

[hasty scarab]

as i remember i think

4 cm 2

and 4 cm 3

the small 2 and 3

yk what i mean

^

i think this is in typing terms

[true coral]

Sneak? You think this is Skyrim?
Does she not let you just ask?

[hasty scarab]

cm^2

[true coral]

Yeah yeah we call that superscript

I got you

[hasty scarab]

nah not risking

[true coral]

You mean 8 cm^3?

[hasty scarab]

idk bro its just as i remember

let me see my notebook if i even wrote smth

[true coral]

@hasty scarab Also what's written at the very top of the cardboard?

Seems relevant to the question but I can't read that language

[hasty scarab]

here?

[true coral]

Yeah

[hasty scarab]

surface area and volume of a prism

this is "prism with a base that is a regular triangle"

prism with a right triangle base

[true coral]

Aight, and do you see which shape that corresponds to?

[hasty scarab]

regular quadrangular prism

[hasty scarab]

no clue lol

rhombus-based prism

prism with isosceles trapezoid base

regular hexagonal prism

[true coral]

Hold on, there's actually 2 it could be

[hasty scarab]

тхатс алл

thats all

[true coral]

Well, many more, but you can reduce it down to those where P and V can be calculated just from the information given

And that yields 2 of them

Meaning we still don't have enough info

Yep, Imma need the rest of the board

sorry, can't be of much help otherwise

[hasty scarab]

this is soooooooooooooooohard bro

sec

[true coral]

And we didn't even get to the math yet

[hasty scarab]

ok laby

i know a goated guy

he will send me his notebook

[hasty scarab]

you understand those right?

[true coral]

The formulas? Yeah

[hasty scarab]

you can solve this fully?

wow bro you are the goat

[true coral]

Would've been a mistake taking on this channel if I didn't

[true coral]

If I have the right info

[hasty scarab]

ok

[true coral]

Of course I'm not just giving you the answer

[hasty scarab]

you giving me knowlege too

right

[true coral]

Whole point of the server is we help you solve your math problems but in exchange you have to learn something

heheheha we are so evil

[true coral]

Yeah otherwise I'd be a trash helper

[hasty scarab]

love you no homo

[hasty scarab]

well ngl i hate math my only goal is making my dad proud cuz he wants me to have math a good grade and i got from B to D cuz i didnt do shit in some periods also here is the table

[true coral]

Just wait until bro realises almost every helper does it like that

[hasty scarab]

no clue lol

[true coral]

Understandable, school turns the literal perfect subject of math into a whole Sisyphean task

[true coral]

do my eyes deceive me

[hasty scarab]

im just learning scripting

[true coral]

that's literally just the answers lol

[hasty scarab]

i told him to send me everything he wrote

[true coral]

Aight, well... usually I'd complain about the answers being handed out on a silver plateau but whatever

Though now there's so much stuff I don't even know where to start

Is there like a sheet of exercises you were supposed to do or something like that?

[hasty scarab]

no someone from the other major send me what posibly was on the exam

ok he didnt solve shit

[hasty scarab]

let me see my notebook rqrq

laby

[true coral]

Yeah?

[hasty scarab]

i have writen a similar to this

yes

i wrote it in the notebook

no way bro

[true coral]

Like I'd love to help but I legit don't even know what to help with

[hasty scarab]

no way i found it

it says

regular quadrangular pyramid

i think i typed it pyramid

yes

pyramid

let me show you

[true coral]

Dang so it ain't even a prism

[hasty scarab]

H is a height

S is side edge

a is base edge

h is apothem (side wall height)

and there the shit begins

[true coral]

Oh flip I got distracted

right so uhm

@hasty scarab Honestly just plug the values and compute

I don't have any deep insight here

[true coral]

@hasty scarab Soo...?

Not enough directions?

uh alright, ping me when you're back

[devout snow]

@hasty scarab Has your question been resolved?

[hasty scarab]

Mb I was eating

[hasty scarab]

Yo laby

Can we talk in dms cuz I don't get pinged

[true coral]

Eh not sure whether that's considered good server etiquette but sure

Just close this channel before

Type .close

[hasty scarab]

K

.close

[devout snow]

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[devout snow]

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[sacred sedge]

After solving the PDE through seperation of variables.

I get f_n = summation from n = 1 to inf of Csin(npi x/2)
and g(y) = -Bcosh(npi y /2) tanh(npiy/2) + Dsinh(npiy/2)

Can I apply superposition principle to non -periodic solutions like g(y)?

[devout snow]

@sacred sedge Has your question been resolved?

[sacred sedge]

anyways I think I got the solution if any one was looking at this

$$u(x,y) = \sum_{m=1}^{\infty} \frac{96}{(2m)^3\pi^3} \sin(m\pi x) \left[\cosh(m\pi y) - \sinh(m\pi y)\coth(m\pi)\right]$$

[woven radish]

InterGalactic

[sacred sedge]

.close

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

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[devout snow]

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[summer stratus]

For any natural number m, we have 2×m=m+m. - I am trying to prove this on lean, but I am stuck here. please help

[viscid root]

i would use succ_mul at the start (after applying whatever is necessary to get it to that form)

from there it's just two more lines

[devout snow]

@summer stratus Has your question been resolved?

[summer stratus]

I have used mul_succ like this

[devout snow]

@summer stratus Has your question been resolved?

[devout snow]

@summer stratus Has your question been resolved?

[devout snow]

@summer stratus Has your question been resolved?

[raw nebula]

Are you still stuck

[velvet geode]

@summer stratus u can ping the helpers yk tht right?

[raw nebula]

I dont think she knows that

[devout snow]

@summer stratus Has your question been resolved?

[devout snow]

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[devout snow]

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[keen sundial]

can someone explain me this slide ty

[acoustic leaf]

which part are you having trouble with?

[keen sundial]

the area of the base

doesnt the cross product give you a vector, not area

and this part too

[acoustic leaf]

the cross product gives you a vector whose magnitude is the area of a parallelogram (which in the diagram is the base)

[keen sundial]

how does magnitude look like?

I know that it doesnt contain directions

[acoustic leaf]

it's the length of the vector

[keen sundial]

but to find the lenth of the vector, it should be pointing a particular direction right

or is it assumed to be pointing 360 degrees?

like anywhere in the 3d space

[acoustic leaf]

i'm not sure what you mean

why would we need to know the direction of a vector to find its length?

[keen sundial]

thats true

so it gives the area of the parellogram of just the

floor base?

[acoustic leaf]

parallelograms are 2d shapes. all of the sides of this 3D parallelepiped are parallelograms. the cross product gives us the area of the base parallelogram

[keen sundial]

fair enough

whats going over here

the second line

with the green color

[acoustic leaf]

they replace red and green with the same thing but in math

the fact that the altitude is |a| |cos theta| is from the right triangle in green on the diagram

[keen sundial]

isnt it supposed to be sin theta for the y axis

[acoustic leaf]

don't just memorize cos = x and sin = y

use soh cah toa

[acoustic leaf]

because that's only true for angles with the horizontal axis

[keen sundial]

here we have 3d

ok i can use soh cah toa but where is the angle of interest?

[acoustic leaf]

the angle inside the trig function

i.e. theta, which is labeled on the diagram

[keen sundial]

like this?

[acoustic leaf]

yes

[keen sundial]

but how did they get this

[wintry socket]

cos(theta)=h/|a|

cah

[acoustic leaf]

it's the same length as the one you marked as A

[keen sundial]

so those sides are equal to O and A because they are parallel

right

[acoustic leaf]

they are congruent triangles

[keen sundial]

I see

but i dont get how they got len(a)*costheta

also why is there an abs value sign

[acoustic leaf]

cos theta could be negative

but we can't have negative lengths

[acoustic leaf]

this would make theta obtuse, which invalidates our right triangle analysis, but it turns out to give the correct magnitude just with a - sign in that case

[keen sundial]

ah that makes sense

also one more thing

I dont understand this

I think im confused with how x vec - a vec = X

[acoustic leaf]

they just shortened $ax_0 + by_0 + cz_0$ to $d$ as a shorthand

[woven radish]

clôud

[acoustic leaf]

if you line up the x and a vectors to have the same starting point (O in the diagram), x - a is the vector pointing from the tip of a (which is A) to the tip of x (which is X)

remember that X is a point not a vector

[devout snow]

@keen sundial Has your question been resolved?

[keen sundial]

When it says passing through that point

How can it pass through that point if it’s a normal

[acoustic leaf]

the plane passes through the point A, not the normal

[keen sundial]

The only thing that is possible is the normal vector is on the plane

[acoustic leaf]

vectors only have a magnitude and direction, their starting point is not mathematically part of them

[keen sundial]

wdym

[keen sundial]

True

[acoustic leaf]

d is a new variable that they defined to be equal to ax_0 + by_0 + cz_0 in order to not write a whole bunch of constants

[swift lantern]

u guys doing vectors?

[keen sundial]

Bunch of constants? How

[acoustic leaf]

for the purposes of the equation of a plane x,y,z are variables and all else are constants

[keen sundial]

how is the rhs a constant?

[unkempt lichen]

i think it's best to understand what exactly this is saying

it is saying that if you fix a point A(x_0, y_0, z_0), and a vector n, then there is a unique plane P passing through A which is normal to n

[woven radish]

barnav

[unkempt lichen]

constant variable i.e. constant

[raw cedar]

Follow that.

[keen sundial]

the only confusion is this

x-a

why cant we do $\vec{n} \cdot \vec{x}$?

[woven radish]

ø

[unkempt lichen]

it's an equation

so if you want to do n dot x, what would the right hand side be?

[keen sundial]

it woudnt be 0

because they arent perpendicular

[unkempt lichen]

no so im not asking you in that way

what i mean is

your point of confusion is on why we cant have n dot x instead of n dot (x-a) = 0

[keen sundial]

yea

[unkempt lichen]

but im saying that you cant replace an equation with an expression

so what equation do you want to have instead of n dot (x-a) = 0

once you say that i can tell you why it may or may not work

that's what i meant

[keen sundial]

wdym

[unkempt lichen]

you want to use n dot x
in place of
n dot (x-a) = 0

you are suggesting an alternative for an equation n dot (x-a) = 0

im saying n dot x cant be an alternative because it's not an equation

so im asking what equation you are referring to

[unkempt lichen]

^

[keen sundial]

thank you I should be ok now

.close

[devout snow]

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[inland carbon]

if $a \in \mathbb{N}$, and $X$ is a discrete r.v. following a Poisson distribution of parameter $a$, what is the expected value of X mod $p$ for a prime $p$?

[woven radish]

bloubbloub

[inland carbon]

I'm trying to calculate $\sum_{l=0}^{p-1} l \sum_{k=0}^{+\infty} e^{-a}\frac{a^{kp+l}}{(kp+l)!}$ but I don't see how to do it

[woven radish]

bloubbloub

[sand dove]

Reminds me of roots of unity problem

if you take 1,omega,...,omega^(p-1) the p-th roots of unity

Then $\exp(\omega^l x) = \sum_{m=0}^\infty \omega^{lm} \frac{x^m}{m!}$

[woven radish]

Rafibirthday2003

[sand dove]

So name $S_l = \sum_{k=0}^\infty \frac{a^{kp+l}}{(kp+l)!}$

[woven radish]

Rafibirthday2003

[sand dove]

First we have $\sum_{l=0}^{p-1} S_l = \exp(a)$

[woven radish]

Rafibirthday2003

[sand dove]

Then $\sum_{l=0}^{p-1}\omega^lS_l = \exp(\omega a)$

[woven radish]

Rafibirthday2003

[sand dove]

Etc...

System of equations

[inland carbon]

I see

[sand dove]

Linked with inverse of vandermonde matrix I believe

[devout snow]

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[faint gorge]

[split pivot]

bruh

[faint gorge]

You need to reopen

this will autoclose

[split pivot]

.reopen

[faint gorge]

a new channel sry

[split pivot]

oops

sorry

[devout snow]

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[drowsy parcel]

can people here help me understand statistics derivation?

[vital sedge]

Sure

[drowsy parcel]

wait i am telling you what the basics are

[vital sedge]

,rccw

[vital sedge]

,rccw

[vital sedge]

Alright, so where are you stuck

[drowsy parcel]

i don't get it, how they put the arbitrary values u and v...and how they wrote the power of it

its the second line

[vital sedge]

They just considered a function in that form to prove the result.

[drowsy parcel]

@vital sedge what is this

determinant?

[vital sedge]

Yes

determinant

[drowsy parcel]

what it represent? and why they used here? @vital sedge

ok so this is some property for writing coefficients of a quad eq in determinant form

[devout snow]

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[keen sundial]

I need help

How do we solve it

[feral agate]

What's the condition for a matrix to be symmetric?

[keen sundial]

When they march across the diagonal

[sage lion]

Do you know what happens if you multipy the matrix K with its transpose?

[keen sundial]

No, what happens

[sage lion]

If you mutipy a matrix with its transpose the matrix will always be symmertric

So the last term is always an symmetric matrix

You need to show now that the alpha and beta term are added together an symmetric matrix

[desert tree]

Theres another property that symmetric matrices have. What happens when you take the transpose of a symmetric matrix?

[sage lion]

Here you can see exacly what happens

same is for addition of a matrix with its transpose

This means if you last term is always an symmetric matrix you need to look just at the alpha and beta term

the sum of the two matricies are only symmetric if and only if alpha = beta

[feral agate]

or K is symmetric

[devout snow]

@keen sundial Has your question been resolved?

[devout snow]

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[late cave]

determine the norm and orientation of the vector AB

of the hyperbola

[devout snow]

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

[late cave]

1

[inland carbon]

I think you can calculate the coordinates of A using the properties of the focus points of a hyperbola

[late cave]

found a which is 6

c^2 = a^2 + b^2

c^2 = 6^2 + 8^2

c = 10

[late cave]

so (0,10)

.close

[devout snow]

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[viral kernel]

$e^{z}=e^{\ln(i)}$

[woven radish]

Nibaloth

[viral kernel]

$z=\ln(i)+i2k\pi$

[woven radish]

Nibaloth

[viral kernel]

$z=i(\frac{\pi}{2}+2n\pi)+i2k\pi$

[woven radish]

Nibaloth

[viral kernel]

Is this correct?

$z=i(\frac{\pi}{2}+2(k+n)\pi)$

[woven radish]

Nibaloth

[viral kernel]

.

[devout snow]

@viral kernel Has your question been resolved?

[supple knot]

Looks fine. Factor out a pi to make it even nicer I guess

[devout snow]

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[sweet steppe]

So I end up with something that gives me these values:
0, 2, -3

The answer sheet says:
0, -2, -3

I do not think I am making any mistakes. Could anyone please work it out to confirm? AI bots seem to suck at doing these for some reason so its not worth it to stay following with them

To be pragmatic too, for it to be 0, -2, -3 it'd need to be something like: (x as lambda)

x^3 - 5x^2 + 6x = 0 right?

I am getting: x^3 + x^2 -6x = 0

[tough prawn]

Sel isn't this a test question

(cheating isn't against the rules?

[sweet steppe]

its a past paper exam yes

no i have the answers for them i am just practicing

[tough prawn]

Oh

[sweet steppe]

why did u interject to say that 😢

[harsh blaze]

This group hates that so

🤷‍♂️

[sweet steppe]

but like

i just need help on that

and when u guys msg u ruin the thread

[viral ledge]

If it was a test he sure as hell won't have access to discord lol

[sweet steppe]

so now i needto close this and open it again

[viral ledge]

I think

[sweet steppe]

i literally said i have the answers

[harsh blaze]

Well You guys can just continue

[sweet steppe]

sure but i mean like

[bleak tide]

you don't

you can keep messaging in here

[sweet steppe]

ok good thank you

[tough prawn]

Yes

[sacred sedge]

is there a characterisic wolfram function?

[bleak tide]

it's better not to mess with help channels too much, we don't want to overwork the bot.

[bleak tide]

why don't you post your work and someone can read through it?

[sweet steppe]

sure i can send a pic

so iend up with

[unkempt lichen]

you're correct

[sweet steppe]

thank you

can i ask further if you got the eigenvectors?

[unkempt lichen]

yeah sure

[sweet steppe]

because im not sure then if the answer sheet might have the eigenvectors based off the wrong results or the right ones

thank you <3

[unkempt lichen]

yeah go on what did you get

[sweet steppe]

i got for
= 0 , [1, 1, 0]
= -3 , [4, 1, -9]
and the other i didnt work out

[unkempt lichen]

wait wait

yo

1 sec

sorry that's my bad
i wrote the matrix down wrong

im really sorry

[sweet steppe]

for what

[unkempt lichen]

gimme a minute let me look at your work

[sweet steppe]

no its fine i appreciate the help

yes its ok

maybe if u got the same results i wrote it down wrong too

[unkempt lichen]

yes you did, that's a -2 in the bottom right

not a 2

i made the exact same mistake lmao

[sacred sedge]

,w CharacteristicPolynomial[[(-1,1,1),(2,-2,1),(3,-3,-2)], x]

[sweet steppe]

no way!!

[unkempt lichen]

funny how that happens

[woven radish]

Results provided by WolframAlpha

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

Requested by interstellar3398

[sweet steppe]

what the hell?

[sweet steppe]

there we go

ohhhh

exactly then

[sacred sedge]

you made a mistake

[sweet steppe]

yes i did

i thought

you were a random person

in the wrong channel

i didnt know that was a characteristic polynomial

sorry for not responding

[sacred sedge]

that was what I asked for

dw

[sweet steppe]

yes sorry i thought u were unrelated

thank u for that that makes sense

[sacred sedge]

barnov spotted the mistake on time

[sweet steppe]

yes

thank you i appreciate the help everyone

how do i close this

.close

[devout snow]

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[devout snow]

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[cunning beacon]

when transforming functions such as y=sqrt(x) to x=y^2, doesn't it introduce extraneous solutions?

in a recent quiz i took about volumes of solids of revolution, the ff problem came up:

" the plane region bounded by y=sqrt(x) and 2=x+y is revolved about y=1. Determine the volume of the solid generated. Use the circular Shell Method."

i get what my professor is trying to make us do, which is to transform the curve they gave "y=sqrt(x)" to become "x=y^2," but doesnt that make the question "wrong" or incorrectly asked?

[sand quarry]

It is fine as long as there is a domain restriction carrying over

Like your problem would be specifying thar your physical region exists from y = 0 to y = 1 or something

[cunning beacon]

yeah but there isntt

[cunning beacon]

all the info given is as pasted here

the way the problem is currently given, it should be considered a bonus right hahahah

[devout snow]

@cunning beacon Has your question been resolved?

[devout snow]

@cunning beacon Has your question been resolved?

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[sturdy fiber]

<@&268886789983436800>

<@&268886789983436800>

[devout snow]

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as this action is irreversible, and this channel may abruptly lock.

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[plush patio]

i need help on this one

[plush patio]

i tried to write a pair of inequalities like a < x < b and f(x) < y < g(x) but it apparently isnt possible

[trail eagle]

As it is it's not possible to write the region they're talking about using one integral. You'd have to split it into multiple pieces that are bounded nicely between function.

If you've seen that you may want to use a change of variable.

[plush patio]

i didnt see yet

[plush patio]

i tried to write something like $\int_0^\infty\int_x^{4x}x^2y^2dydx-\int_0^\infty\int_{\frac1x}^{\frac1x}x^2y^2dydx$ but it didnt work

[woven radish]

~raul.c

[timber violet]

try using a change of variable

[trail eagle]

They haven't seen it.

[timber violet]

u know the jacobian determinant? @plush patio

[trail eagle]

Hum yeah no. If you want to avoid change of variable at all cost, you would split the region.

[plush patio]

i know but the book didnt show it yet

[trail eagle]

Probably into three parts from what I'm seeing

[timber violet]

why r u doin smth u havent covered yet?

[plush patio]

so we're supposed to solve without a change of variable

[timber violet]

u know green theorem?

[plush patio]

i still dont see it

[plush patio]

nope

[trail eagle]

Well have you sketched what the region looks like?

[plush patio]

yeah

[trail eagle]

Right, it looks something like this

[timber violet]

what do u know

tell me everything u know from the course u are taking

[trail eagle]

You can split it in three by dropping vertical lines where the green and black lines intersect and where the red and purple line intersect. All three of those will have "nice" bounds.

[plush patio]

i'm learning from apostol vol 2, i do only know double integrals by repeated integration

[plush patio]

hmmm

maybe integrating from the intersection of the red with the black one to the intersection of the red with the purple one

and then green-red and green-purple

[timber violet]

can you rewrite that region in term of min & max in an inequality?

[lusty sapphire]

Yuck i wonder if a polar integral would be better

[trail eagle]

It's a shame since it really looks like it's made for a change of variable 🙁

[plush patio]

1 <= xy <= 2 can be rewritten as 1/x <= y <= 2/x, and we also have x <= y <= 4x

so

[timber violet]

yes

so?

[plush patio]

y lies in [max(1/x, 2/x), min(x, 4x)]

[timber violet]

wrong

[plush patio]

not really, the author didnt present change of variables yet

[timber violet]

$\max(x, \frac{1}{x}) \leq y \leq \min(4x, \frac{2}{x})$

[woven radish]

Kira

[plush patio]

oh

i see

[timber violet]

now find the intersection points

[plush patio]

x > 1/x for x > 1 and 4x > 2/x for x > 1/sqrt2, so maybe here i need to split the problem

[timber violet]

solve x = 1/x & x = 2/x and 4x = 1/x and 4x = 2/x

with this u'll find the region of x

[plush patio]

x = 1, x = sqrt2, x = 1/2 and x = 1/sqrt2 since we are considering x >= 0

[timber violet]

ye

[plush patio]

hmmm

[timber violet]

okey continue from here

think about y-bound change

after that u'll just evaluate integrals

[plush patio]

we consider three intervals:

1. for 0 < x < 1/sqrt2: [1/x, 4x]
2. for 1/sqrt2 < x < 1: [1/x, 2/x]
3. for x > 1: [x, 2/x]

then i need to evaluate three integrals

[timber violet]

for 1) its 1/2 < x < 1/sqrt(2)

[plush patio]

oh

because thats the intersection of xy = 1 and y = 4x?

and for 3) it is 1 < x < sqrt2

yay i did it

thank you guys

.close

[devout snow]

Channel closed

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[devout snow]

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[gloomy drift]

the shaded part is 58 degrees, we have to prove they are equal without using the alternate segment theorem. how would i approach this?

[gloomy drift]

sorry for the bad drawing

[hasty cargo]

Whats thr alternate segment theorem again? I mightve known it by another name

Also you cant take an angle of a line and an arc section or am i tripping?

[gloomy drift]

tangent chord theorem?

[gloomy drift]

its not accurate

if thats what you mean

[winged tapir]

Do you know any theorem which relates to angle subtended by an arc centre and angle subtended by an arc at any point on circle

[gloomy drift]

nope..

[hasty cargo]

Nvm i found it out.

[gloomy drift]

i know the centre is double circumference, chord is perpendicular is always 90 and alternate segment

i didnt get taught them all unforrunately

[winged tapir]

Try proving this first

[gloomy drift]

oh my god

im dumb

58 as shaded so bottom is 32, isoceles so equal sided so 180-64=116 for 2x.. 58 for x

my goodness i didnt tyink that way

thank you, i tried smth using tangent and chord originslly

[winged tapir]

No problem

[devout snow]

@gloomy drift Has your question been resolved?

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[delicate harness]

How to sketch $Arg(z^{2}+1)=\frac{\pi}{3}$

[woven radish]

T. S. Idiot ඞ

[pearl relic]

why not start with substituting z=a+bi

[delicate harness]

algebra bash?

its a massive pain

is there no geometric intepretation

[warped fulcrum]

Unfortunately no

[delicate harness]

ik theres the apollonious circle argument thingy

but thats a quotient

this is a product of (z+i)(z-i)

[delicate harness]

hmm

[warped fulcrum]

Doesn't work here

The efficient way is to deduce the imaginary and real parts

Then construct an equation

[delicate harness]

hmm aight thx

.close

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[cerulean berry]

simplify 5√59

[main gull]

Simplify how?

Because you can't

[sick jasper]

am I crazy or is that simplified already? Yeah

59 is prime

[cerulean berry]

EXACTLY

My school work says simplify 5√59

[sick jasper]

It’s a “trick” question maybe?

[cerulean berry]

Its on a site called get more math, I doubt it

[main gull]

It's a trick question

[cerulean berry]

Then what do you think is the answer

[main gull]

You just leave it as is

$5\sqrt{59}$

[woven radish]

dldh06

[cerulean berry]

Nvm turns out the site was glitching, I had it as that but it didn't work

.close

[devout snow]

Channel closed

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[devout snow]

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[hasty parcel]

@icy otter @scenic quest

@surreal mason @spare socket @celest dock

[main gull]

Next time message modmail and don't ping the mods individually like this

[night gust]

doest his even count as MATH question?

[hasty parcel]

I don't know what modmail is, but this guy is spamming all of the channels and also dming people

[main gull]

It's the bot

[hasty parcel]

i think he has 3 open "questions" that are all the same thing

[night gust]

just dm

[hasty parcel]

ok, got it, will do that in the future, this guy is spamming people and the help channels

[night gust]

yo ucould do that for this one as well

[wicked rover]

.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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[frosty grotto]

Hey guys

I am in grade 11 pre calc and am having problems solving these types of questions

Could someone please lay out the steps for me?

[runic storm]

And what is the question?

[frosty grotto]

Simplify that?

Is what is says

The unit is exponents and radicals

[main gull]

,rotate

[frosty grotto]

<@&286206848099549185>

[tiny ore]

so, you can see that y^3 = y . y^2, z^4 = (z^2)^2

and 72 is 2^3 * 3^2

then you can cancel out some powers with the square root

[frosty grotto]

Uhm ok so how did you get 72 down to 2^3 * 3^2?

[tiny ore]

factoring out primes

[frosty grotto]

Ok

This is the answer

Is that the way your teaching?

[devout snow]

@frosty grotto Has your question been resolved?

[strange nimbus]

@frosty grotto Now you have:

y = 6 |x| |y| z² sqrt(2y)

[frosty grotto]

Yes that is the answer to the question I asked earlier. I just have 0 clue how to get the answer

I don’t know the steps

[strange nimbus]

What do you mean by the answer?

That equation I gave?

[frosty grotto]

This is the question I started with

[strange nimbus]

OK, so here's what you do.

[frosty grotto]

Ok

[strange nimbus]

Do you know the prime numbers?

[frosty grotto]

I can’t really remember them so no.

[strange nimbus]

OK, so an integer multiple of a number is the number times an integer.

[frosty grotto]

Ok

[strange nimbus]

So, if the number is 3, 18 is an integer multiple of 3, since you can multiply 3 times 6 to get 18.

[frosty grotto]

Ok

[strange nimbus]

So, 18 is an integer multiple of 3.

Now divisor is the reverse order.

3 is a divisor of 18.

[frosty grotto]

Ok

[strange nimbus]

So, for example 7 is a divisor of 21 because 21 is an integer multiple of 7.

Now the nice thing about divisors is that when you divide the number by them, you get an integer.

21/7 = 3.

3 is an integer.

[frosty grotto]

Ok

[strange nimbus]

So, let's look at 2.

What are its divisors that are 1 or higher?

1 is a divisor of 2.

[frosty grotto]

Ok

[strange nimbus]

2/1 is an integer.

2 is a divisor of 2.

2/2 is an integer.

Any number higher than 2 can't be a divisor of 2 because you'd get like 2/3 and the number on top is less than the number on bottom.

So, 2 has two positive divisors: 1 and 2.

Now we can count the divisors.

There are two positive divisors of 2.

Primes are numbers that have exactly two positive divisors.

So, 1 isn't prime.

1/1 is an integer, so 1 is a divisor of 1.

And that's the only divisor of 1.

1 has one positive divisor, not two, so it's not prime.

Let's do 4.

[frosty grotto]

Ok

[strange nimbus]

1 is a divisor of 4.

2 is a divisor of 4.

4 is a divisor of 4.

3 isn't.

So, 4 has three positive divisors, so it's not prime.

[frosty grotto]

So a divisor is something that you can divide the given number into and come out with a whole positive number?

[strange nimbus]

Right.

[frosty grotto]

15
Has 5

[strange nimbus]

It has 1, 3, 5, 15.

[frosty grotto]

Oh 1 counts ok

[strange nimbus]

Yep, 1 through the number can count.

[frosty grotto]

Ok great

[strange nimbus]

Now, 1 and the number are always divisors.

[frosty grotto]

Ok

[strange nimbus]

So if there are any divisors at all in between 1 and the number, it's not prime.

[frosty grotto]

Ok

[strange nimbus]

So, 2 is prime pretty easily because there are no numbers between 1 and the number.

And 3 is prime because 2 isn't a divisor of 3.

[frosty grotto]

Ok

[strange nimbus]

4 isn't because 2 is a divisor of 4.

[frosty grotto]

So 5 is prime?

[strange nimbus]

Right.

[frosty grotto]

Ok I’m understanding this so far

[strange nimbus]

OK, now there's a shortcut.

[frosty grotto]

Ok

[strange nimbus]

You can check only the primes less than the number.

So, with 5, you don't need to check 4 because it's not prime.

But you do need to check 2 and 3 because they are primes.

[frosty grotto]

Ok

[strange nimbus]

And there's one more shortcut.

You can take the square root of the number.

,calc sqrt(5)

[woven radish]

Result:

2.2360679774998

[strange nimbus]

And you check the primes up to the square root.

So, you only need to check 2.

[frosty grotto]

Ok

[strange nimbus]

So, let's say you're checking 25.

sqrt(25) = 5

[frosty grotto]

5?

Ya ok

[strange nimbus]

So, you check 2, 3, and 5.

If they all aren't divisors, 25 is prime.

[frosty grotto]

Ok

2 isn’t 3 isn’t and 5 is

[strange nimbus]

Right.

[frosty grotto]

Ok

[strange nimbus]

And the last trick is that if the square root is an integer, that square root is a divisor.

Why?

Because 5² = 25, for example.

So, 25 is a multiple of 5.

So, 5 is a divisor of 25.

[frosty grotto]

Ok

[strange nimbus]

So, that's an introduction to primes and how to get a list of them.

So, the first few are 2, 3, 5, 7, and 11.

Now we have 72 in the square root.

[frosty grotto]

Ya

[strange nimbus]

sqrt(72) is about 7 something.

[frosty grotto]

Ya

[strange nimbus]

Sorry, 8 something. Because 8² is 64 and 9² is 81.

[frosty grotto]

Ya

[strange nimbus]

So, we have 2, 3, 5, and 7 maybe.

So, let's do something called factoring into prime powers.

We start with 72.

We take the next prime on our prime list: 2.

Is 2 a divisor of 72?

Yes.

[frosty grotto]

Yes

[strange nimbus]

72/2 = 36.

So, we take out that 2 and we have 36.

Is 2 a divisor of 36?

[frosty grotto]

Ok

Yes

[strange nimbus]

Yes, 36 is even, so 2 is a divisor of 36.

Now we're doing the square root.

That's the second power root.

Or the second root.

We want to group the primes into groups of two here.

[frosty grotto]

Ok

[strange nimbus]

If you had the cube root, you'd do groups of three. If you had the fourth root, you'd do groups of four. If you had the hundredth root, you'd do groups of 100.

[frosty grotto]

Ok

[strange nimbus]

We're doing square root, so groups of two.

We just found 2 was in 72 twice.

[frosty grotto]

Yes

[strange nimbus]

So, every group of two of a prime, we get one of that prime coming out of the square root.

72/2 = 36
36/2 = 18

[frosty grotto]

Ok

[strange nimbus]

We had sqrt(72)

Now we have 2 sqrt(18)

[frosty grotto]

So sqrt 18 • 18?

[strange nimbus]

No, 2 · sqrt(18).

[frosty grotto]

Ok

[strange nimbus]

We had sqrt(2² · 18).

And since the 2 is squared, we can take it out.

[frosty grotto]

Ok

[strange nimbus]

That's why we do groups of 2.

Because if you have it in there twice, it's squared, so you can take it out.

That's also why if you're taking the hundredth root, you need 100 of them.

[frosty grotto]

Ok

[strange nimbus]

So, we have 2 sqrt(18).

We're on prime 2.

Is 2 a divisor of 18?

[frosty grotto]

Ya

[strange nimbus]

18/2 = 9

[frosty grotto]

9

Ya ok

[strange nimbus]

Is 2 a divisor of 9?

[frosty grotto]

No

[strange nimbus]

OK, so we couldn't get another group of two 2s.

So, we're done with the prime 2.

[frosty grotto]

Ok

[strange nimbus]

We have 2 sqrt(18).

We go to the next prime.

Is 3 a divisor of 18?

[frosty grotto]

Yes

6

[strange nimbus]

Is 3 a divisor of 6?

[frosty grotto]

Yes

[strange nimbus]

OK, and we get 2.

[frosty grotto]

Ya

[strange nimbus]

Two 3s.

So, we take that out.

2 · 3 sqrt(2)

And then 2 is less than the prime we're checking.

[frosty grotto]

Ok

[strange nimbus]

So we're done.

Also, sqrt(2) is 1 something, so we only need to check primes up to 1 something.

[frosty grotto]

Ok

[strange nimbus]

But we're already past 1 something.

So we're done for that reason too.

[frosty grotto]

Ok cool

[strange nimbus]

So, that's how they got 6 sqrt(2).

[frosty grotto]

Ok and as for the letters?

[strange nimbus]

OK, now we need groups of two.

[frosty grotto]

Ok

[strange nimbus]

x² has two xs multiplied together.

So, we have a group of 2.

Now x is a variable.

When you have a variable and take it out, you do absolute value.

sqrt(x²) = |x|

Here's why.

sqrt(3²) = sqrt(9) = 3

sqrt((-3)²) = sqrt(9) = 3

[frosty grotto]

Ok

[strange nimbus]

See how both 3 and -3 squared and then square rooted both give 3?

[frosty grotto]

Yes

[strange nimbus]

And |3| and |-3| are both 3.

[frosty grotto]

Oh?

Ok

[strange nimbus]

Yes, absolute value gives the nonnegative version of a number.

So, |3| = 3.

|-3| = 3

And that gives the right answer.

[frosty grotto]

Ok

[strange nimbus]

sqrt(3²) = 3

sqrt((-3)²) = 3

So, sqrt(whatever²) = |whatever|

Do you have any questions about that?

[frosty grotto]

Yes

The end number is always positive?

Regardless of negative previously

[strange nimbus]

Yes, squaring a real number always makes a nonnegative number.

[frosty grotto]

Ok

[strange nimbus]

So, you get a nonnegative number in the square root.

[frosty grotto]

Oh ya I understand now

[strange nimbus]

And then square root of a nonnegative number is also nonnegative.

OK.

So, sqrt(x²) = |x|.

[frosty grotto]

Yes

Ok

[strange nimbus]

Now sqrt(y³) has three ys.

We can get one group of two out of that.

|y| sqrt(y).

Does it make sense what I did there?

[frosty grotto]

Kind of

So you put 2 y together and then the other y?

[strange nimbus]

Yes, we group two of them up together.

We take those out.

[frosty grotto]

Ok

[strange nimbus]

And then the last one is stuck because we can't get another group of two together.

[frosty grotto]

So the last one stays under the root?

[strange nimbus]

Yes.

[frosty grotto]

Ok

[strange nimbus]

Much like our 2 stayed under the root because we couldn't get two 2s there.

So, sqrt(z⁴).

We have four zs.

That's two groups of two.

[frosty grotto]

Ok

[strange nimbus]

So, that becomes |z|²

|z| comes out twice.

[frosty grotto]

Ya

[strange nimbus]

Now |z|² = z² if z is a real number because ² already makes it nonnegative.

So, sqrt(z⁴) = z².

[frosty grotto]

Ok

[strange nimbus]

So, altogether, sqrt(72x²y³z⁴) = 6|x||y|z² sqrt(2y).

[frosty grotto]

Ok

What I did when I approached that was this.

Split 72 down to 8 and 9

Then 9 into 3 squares

8 down into 2• 2squared

Is there anything right I was doing

[strange nimbus]

Yes.

That's a faster method.

If you can see how to factor it like that, you can do that.

[frosty grotto]

Ok

[strange nimbus]

My method works when you can't see how to factor it so easily.

[frosty grotto]

If not then factor the primes

[strange nimbus]

Right.

[frosty grotto]

Ok great

So thank you for teaching me that method.

[strange nimbus]

No problem.

[frosty grotto]

Sorry to be bother but I have another question here

After I get 2 • 2 squared
And 3 squared

How do I know what to put in root and out of root?

[strange nimbus]

Well, the squared parts you take out.

You get one group of two 2s.

You get one group of two 3s.

So, you get 2 · 3 outside.

2¹ because 1 group of 2s.

3¹ because 1 group of 3s.

Then the 2 gets stuck inside because it's not part of a large enough group.

So, you get 2 · 3 sqrt(2).

Which is the 6 sqrt(2) we got earlier.

[frosty grotto]

Ok

Yes I think I understand it better now

Ok how do I close my question?

[strange nimbus]

Just type .close

[frosty grotto]

Ok great thank you for the help

[strange nimbus]

You're welcome.

[frosty grotto]

Your awesome and I appreciate it

.close

[devout snow]

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[devout snow]

Available help channel!

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

Any hints on how to do this, im totally stumped. It makes sense that this needs to be ture but im not sure how to show that

[winged hazel]

I started by stating the vector equation of a line and a plane

so i presume its just a case of proving that t is a real number that exists in P2= P1=tD

but idk how to do that

I tried subbing into the plane equation to get n.(tD)=0 but im not exactly sure how that would help because then i just get the abs n * abs t * abs D *cos theta

if the vectors line in the same line cos theta is 1 so then i had n* t * D=0

if n and d are non zero that doesnt make sense so what do i do

<@&286206848099549185>

thats my work so far but that doesnt make sense, so where have i gone wrong

tbf it could easily be all of it

[devout snow]

@winged hazel Has your question been resolved?

[winged hazel]

anyone able to help?

<@&286206848099549185>

I feel like im on the right lines but ive been looking at this for hours and i just cant see where it comes out

because if n and d are non zero t has to be zero but i dont think that can be the case either

anyone? I know its a bit of a long question but just a nudge would be very helpful im loosing my mind over this

[devout snow]

@winged hazel Has your question been resolved?

[devout snow]

Channel closed

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Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[keen storm]

I can’t do question 7

[keen storm]

<@&286206848099549185>

[deft hemlock]

Why not?

[keen storm]

Because I don’t know how

Can you teach me?

[deft hemlock]

Well did you find the critical numbers

[keen storm]

Yes

[deft hemlock]

Okay so now plug those x values back into the function

Then list then like this

(x1, y1), (x2, y2), etc

[keen storm]

Uh

[keen storm]

How would I list them like this?

[deft hemlock]

with parenthesis

The y values are the function evaluated at x

For example

Let’s say y = x + 2

And we know 3 is a critical number

[keen storm]

Uh huh

[deft hemlock]

Then y = 3 + 2 = 5

So the ordered pair would be (3, 5)

[keen storm]

Ohhhh

[deft hemlock]

Then you can just separate them by commas

[deft hemlock]

Like I did up here

[keen storm]

I see thank you

[deft hemlock]

Happy to help!

[keen storm]

.clow

.close

[devout snow]

Channel closed

Closed by @keen storm

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Remember:
• Ask your math question in a clear, concise manner.
• 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!

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

[keen storm]

I can’t do number 3 I need help

[keen storm]

<@&286206848099549185>

[drifting python]

@keen storm you find possible points of inflection by setting the second derivative equal to 0

[keen storm]

Wdym

[keen storm]

How?

[drifting python]

same way you found critical points

find second derivative of f, set it equal to 0, then solve for x

[keen storm]

Would the answer be none?

<@&286206848099549185>

I keep getting it wrong

[drifting python]

i believe the answer is 11 if it’s not DNE

[keen storm]

Yeah I got that

[drifting python]

any other questions?

.close

[devout snow]

Channel closed

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