help-27

[worthy heath]

$\frac {\dd{x}}{\dd{y}} = \frac{y^2-x^2}{2xy}$

[woven radish]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[worthy heath]

$\mu = e^{\int -\frac{1}{x} \dd{x}} = \frac{1}{x}$

[woven radish]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[worthy heath]

okay nvm im stupid asf

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[restive river]

dont worry im also stupid af

[devout snow]

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[golden flower]

ez af question but here:

[golden flower]

my friends put the x^-1+x^-2 down which is illegal lol

[pseudo basin]

this is not an equation. there is nothing to solve.

[golden flower]

it means simplify

or idk

[pseudo basin]

well then it should say simplify.

[golden flower]

what word is suitable

sorry 😢

[pseudo basin]

ok, so disregarding your friends' incorrect step

do you have any other progress so far

[golden flower]

yeah i do

[pseudo basin]

show

[true coral]

Are you saying your friends gave this to you?

[golden flower]

yeah we review for an upcoming test it's not part of anything though

[true coral]

Tell them next time they write down "solve" there better be an equal sign

[golden flower]

yes 🙏

[pseudo basin]

ok almost

up to and including $\frac{x}{x^3(1-x)}$ it looks correct

[woven radish]

Ann

[pseudo basin]

but then what was your final thing

or like

your final answer

cause the way you wrote the cancellation is a bit sus

or maybe the last line is cut off?

[golden flower]

1/x^2(1-x)

[pseudo basin]

ok yeah

good then

that's correct

[golden flower]

thank you

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[rare kernel]

how do i find number of solution of x^2 = 2^x

[rare kernel]

without graphical approach

[supple knot]

probably intermediate value theorem

[rare kernel]

what is that

[supple knot]

https://www.mathsisfun.com/algebra/intermediate-value-theorem.html

[rare kernel]

that is not taught to us

so i guess ill have to do it using graphs itself

or maybe just do substitution method

thx anyway

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[hexed scaffold]

need help solving this problem:
Battleship Puzzle

You have a 10 × 10 grid.
You can place mines on certain squares to prevent a ship from being placed.

Question 1:
What is the minimum number of mines needed to prevent any ship from being placed if ships have the following shape (a horizontal line of 5 squares)?

🟦🟦🟦🟦🟦

Question 2:
Same question for the following shapes (or other shapes of your choice):

🟦
🟦🟦🟦🟦

  🟦

🟦🟦🟦
🟦
(A "+" shape)

and
🟦🟦🟦
🟦
🟦

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[feral agate]

where did the first inequality come from?

[woven radish]

kheer257

[supple knot]

u sub and then bound by largest value of f

and then factored that out of the integral

[feral agate]

what exactly are we subbing?

ahh I see

[supple knot]

it's kinda dumb, but z = z(t)

[feral agate]

[ \begin{aligned}
\abs{\int_{\gamma}f(z) \dd{z}} &= \abs{\int_a^b f(z(t))z'(t)\dd{t}}\
&\le \int_a^b \abs{f(z(t))}\abs{z'(t)} \dd{t} \
&\le \sup_{t\in [a,b]} \abs{f(z(t))}\cdot\int_a^b \abs{z'(t)}\dd{t} \
&= \sup_{z\in\gamma} \abs{f(z)}\cdot\operatorname{length}(\gamma)
\end{aligned} ]

[woven radish]

kheer257

[feral agate]

I don't understand why they added an inequality sign at the end though

is $\sup_{t\in [a,b]}\abs{f(z(t))}$ not necessarily equal to $\sup_{z\in\gamma}\abs{f(z)}$?

[woven radish]

kheer257

[supple knot]

could be = instead of <= yea

should be the same yea

[feral agate]

now that I think about it, how do you even define a supremum on a curve?

[supple knot]

just call g = f(z) and now you have sup(g) over [a,b]

and the bottom sup is just equal to the top sup by definition

[feral agate]

hmm

but gamma is also just a subset of C

[lilac moat]

@feral agate this is sometimes called the ML-inequality, if you want to google it

riemann is correct, but im just giving you a name for it

[feral agate]

thank you very much! I wasn't sure what name to search this by

[supple knot]

right and z smoothly parameterizes gamma

i'm not sure what is confusing you

[feral agate]

nothing actually

thank you for your help

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[vagrant shard]

<@&286206848099549185>

I’ve never used one of these before 🙏

Need help tho

[lilac moat]

do you know the definition of x-intercept

[vagrant shard]

yes it’s a point that intersects the x axis

[lilac moat]

that's true, yeah. so if you have a function y=f(x), in general, not just for this problem, how do you find the x-intercepts of f?
also called the roots of f

[autumn girder]

!15min

[devout snow]

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[autumn girder]

Think about when will the graph intercept the x axis

What does it mean for a line to touch the x axis

[vagrant shard]

I have to show my work with out using or showing a graph.

so like through a method ig

Idk how to do that

Like if I got out graphing paper I could find it but without graphing it I’m stuck

[autumn girder]

Oh, then desmos.com should help. Also, it'll prolly help to watch youtube video on tracing graph of a quadratic function

[vagrant shard]

Hmmm ok I’ll try it.

Thank uu!

[devout snow]

@vagrant shard Has your question been resolved?

[vagrant shard]

not really 🙏🙏

It’s fine tho

Close it

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[vagrant shard]

Can someone check my work and lmk if I’m on the right track or not

I think I’m getting it somewhat

My work seems a bit messy and untrue

[devout snow]

@vagrant shard Has your question been resolved?

[vagrant shard]

<@&286206848099549185>

😔

[mellow talon]

Is there a chance that x=-41 isnt right?

The 4 is correct but c/a which is x1.x2 should equal to - 1

Which in your case it is - 164

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[vagrant shard]

Thanks for ur help!

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[lucid mason]

ok ik what a unit vector is and ik how to do it

but im confused about c)

because the magnitude that I get for c) is 15.62

but why is the textbook using the same magnitude for a) in the answerkey

isnt the unit vector different since now its u + v so magnitude is different

textbook answer

[acoustic leaf]

a unit vector is a vector of magnitude 1

[lucid mason]

yup

[acoustic leaf]

if you multiply a vector by a scalar, then its magnitude is multiplied that amount

so if you were to divide by its original magnitude, the new vector would have to have a magnitude of 1

[lucid mason]

ya ik

i know how to form it

but im confused why my answer isnt right

yuhh

so isnt the unit vector 1/sqrt244 ( u + v)

(sqrt244 = 15.62, the magnitude)

[upper schooner]

I think they probably messed up

[lucid mason]

omg ok

😞

because this is what the texbook wrote..

But 2sqrt21 is the magnitude for u-v...

oh well

Thanks chartbit

[upper schooner]

Yea and I agree with what you have

[lucid mason]

https://tenor.com/view/happy-catto-cats-gif-2693509101423175460

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[ember mantle]

i need help passing the math section of the ged and getting a better score on my asvab. how do you retain information well? whats the best way to study? please help, everything just goes in one ear and out the other nothing really sticks. am i cooked?

[midnight echo]

There's a crayon joke to be made here.

But honestly its just repetition. The asvab isn't particularly tricky or unique, its just a fairly significant amount of material being tested. Don't try to memorize any topic, go through it monday, then wednesday, then friday, and so on. You'll manage to internalize it after enough iterations.

[ember mantle]

is there any particular trick to studying like i heard chewing gum and scribbling on a paper while reading or some shit like that

idk twin im just tryna become an academic weapon

i just quit smoking weed 3 days ago maybe my brain fog is too bad rn i dont really know 😭

[midnight echo]

Nah. No need to overcomplicate it. You literally just need to bash your head at as many questions as possible over a long time period and you'll ace it.

Make sure to get step-by-step solutions to the ones you can't do/get wrong though.

But yeah no magic tricks unfort

[ember mantle]

ive been working with chat gpt as like a sort of math teacher and its been helping i guess your right about it just being repetition. i gotta write down a schedule and stick with it because i kinda js study whenever i gotta be more specific with it

[midnight echo]

Yeah. Be a bit careful with using chatgpt since llms can get wrong answers while sounding confident, but it should be mostly fine for the asvab. Better if you find practice sets/textbooks etc. I guarentee that if you study for 2-ish hours a day for at least a month or two you'll score high enough to be the nuke guy

which apparently requires a really high score (?)

[ember mantle]

damn i didn't know chatgpt was buggy like that

but yeah i was thinking of getting textbooks and stuff

ik you can probably find them for free online

aightjamin twin i dont got anymore stuff to ask

ill lock in on studying every other day for 2 hours though

thanks g

,close

fuck

how u close it 😭

.close

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[granite bough]

Hello!

[granite bough]

the integral of 1/x dx = lnx + c

wait nevermind it clicked

as i was writing the question

😄

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[forest orchid]

how do I solve this? I specifically need more help with part B. I haven’t learned derivatives yet and must utilize the IROC formula but i’m unsure how to start.

[supple knot]

Did you try completing the square

[forest orchid]

How would I do that?

[supple knot]

,tex .cts

[woven radish]

riemann

[forest orchid]

ty I think i got it from here

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[urban shore]

How do you approach a question when the bases aren’t the same?

[proud perch]

log_b(x) = log(x)/log(b)

using this knowledge, can you rewrite the equation?

[urban shore]

Like this?

[proud perch]

yeah!

[urban shore]

Now I apply the addition formula and switch to exponential form?

Is that the way to do it or am I missing smth?

[proud perch]

you're overcomplicating things

[urban shore]

Alright how should I solve this then?

[proud perch]

it's just basic algebra from here

clear the denominators

multiply the equation through by log(2)*log(3)

[urban shore]

Oh

You use a common denominator

Alright I get it now

Thanks

[proud perch]

then you should get

log(3) log(x) + log(2) log(x) = 0

[urban shore]

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[wise nacelle]

help

[wise nacelle]

have question about how to write answer

for (a) i got (-1,-7),(3,6)

don’t know if that’s how to write it

[pseudo basin]

they want the interval between the x values

when we talk about a function's behavior on or over an interval or other set, we talk about stuff on its domain

[wise nacelle]

would i only state the x values then?

[pseudo basin]

yes

(-1, 3) is what they want

[wise nacelle]

for decreasing would i write (-inf,-1)

[pseudo basin]

yeds

yes*

[wise nacelle]

and (3,inf) for constant

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[hybrid maple]

how do i show that a particular point is the only limit point of a set E

[hybrid maple]

this is my function

[polar chasm]

whats your defn of limit point?

Point s.t. its every neighborhood has nonempty intersection with E?

[hybrid maple]

yes

[polar chasm]

,rccw

[polar chasm]

I'd try taking any limit point of E and proving that it equals 1

[hybrid maple]

my professor hadn’t done an example for a particular number before i didn’t know how to do it

[polar chasm]

Let's start with a simpler case then. Can you prove that any x < 1 is not a limit point of E?

[hybrid maple]

tbh i don’t know how to these proofs work at all wouldn’t know how to prove that

[polar chasm]

Okay, what does it mean if something is not a limit point?

Refer to the definition

[hybrid maple]

it wouldn’t have a anything in the set of its delta neighbour hood excluding itself

[polar chasm]

Yeah, to be more exact, there would exist such a neighbourhood

because there is no the neighborhood

there are plenty possible neighborhoods

[polar chasm]

Now returning to this, can you find a neighborhood for e.g. x = 0 which doesnt have anything in common with E?

[hybrid maple]

for my example if i choose x =3/2 and a small delta it will only include itself

[polar chasm]

Yep, that's right and that's also the core of the argument

to make the argument exactly right, you'd only need to choose the right delta

I'd split this up to 2 cases, x < 1 and x > 1

the x < 1 case is simpler, so let's start with that

[polar chasm]

what delta can you pick for x = 0?

What delta can you pick for x = 1/2?

what delta can you pick for x = 0.99?

What delta can you pick for general x < 1?

(i just gtg quickly do sth, ill be back in few mins)

[hybrid maple]

for x= 0 i can pick a delta that is less than 1 because no value of 1 + 1/n is between 1 and 2

[polar chasm]

Exactly

[hybrid maple]

in general for 1 i just need to pick a delta such that x + delta is less than 1?

[polar chasm]

Yep

[hybrid maple]

okay

[polar chasm]

one way to do that is just pick delta = 1 - x for x < 1

[hybrid maple]

hmm okay makes sense

[polar chasm]

now for x > 1

this one is a bit more complicated

you have the right intuition that by choosing a small enough delta for e.g. x = 3/2, you can ensure that the only point of E inside the neighborhood is x itself

now the question is just how small does the delta have to be

[hybrid maple]

the delta we choose is because of left of the point if that makes sense

we are closer to the left points

[polar chasm]

That's not always true

if you had x = 1.499 you'd be closer to 3/2 than to 4/3

luckily for us, we can just determine the distance between both the point on left and the point on right and just pick the smaller one for our delta

We just need to find the "point on left" for our x

there is a theorem which says that every nonempty subset of naturals has a least element

if we consider the set of all ns, such that 1 + 1/n < x, then we should be able to pick the least n0, which will result in the point of E which is right to the left of x

the point on the left would then just be 1 + 1/n0

1 + 1/(n0 + 1) must then be >= x (since it cant be smaller)

so it's either x itself, or to the right of it

can you finish the argument by choosing the right delta now?

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[frigid torrent]

Guys how do I solve

[frigid torrent]

Ended up getting these 2 linear equations but didn’t work out

Can someone show me the right one

[devout snow]

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@frigid torrent Has your question been resolved?

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@frigid torrent Has your question been resolved?

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[graceful oyster]

I'm currently working on understanding modal analysis, my starting point is a sdof system to understand the eigenvalue problem. As far as I've seen so far the eigenvalue Lambda was defined as: Lambda = omega², with omega as the angular frequency and omega² = m/k, both of which can be derived from the statements in the screenshot.
What I don't get is the middle part, which states that solving mlambda²+k=0 for Lambda gives you Lambda1,2 = iomega, which in my understanding gives you Lambda1,2 = omega (not squared).

What am I missing here?

[sullen island]

What I don't get is the middle part, which states that solving mlambda²+k=0 for Lambda gives you Lambda1,2 = i*omega, which in my understanding gives you Lambda1,2 = omega (not squared).
what squared part are you talking about from the picture ? I don't see what you're talking about

but anyway, you're solving m L^2 + k = 0 for L, i.e. L^2 = -k/m
k and m are positive most certainly so -k/m is negative

[graceful oyster]

Sorry, not a native, I'll try to explain better.

[sullen island]

L^2 = something negative, the solution has to be imaginary/a complex number

[graceful oyster]

yeah that checks out, this is not the part I'm looking for.
I'm looking at the solution for L:
It's either L = omega² or L = omega, whether imaginary or not.

I have learned so far that L = omega squared, which makes sense, given that L² = -k/m and k/m = omega²

Why is the omega not squared in the middle part?

[sullen island]

k/m = omega²
what's sqrt(k/m) then ?

[graceful oyster]

omega (not squared)

[sullen island]

yeah

sqrt(k/m) is omega

[graceful oyster]

Ok, I understand where I took a wrong turn in this context.
Still, I have learned that L = omega² when doing modal analysis
Is that only true for dampened systems with one or more degrees of freedom?

[sullen island]

can you post their explanation around that formula ? (gonna be in german I guess)

[devout snow]

@graceful oyster Has your question been resolved?

[graceful oyster]

I'm trying to find something useful. I originally got it from an online lesson about modal analysis for 3 dof system (bridge). But there it was just described as a given. Within that example it makes sense and replicating that with my FEM software gave me the right values as well.

Thats a screenshot from the given video. There he only states that Lambda = omega ²

Also another book that I have describes the eigenvalue problem as (K-omega²M)[Eigen vector] = 0 but also for a 2dof system

Sorry that I don't have more to tell you. It's kind of my problem rn

I have found this in english, also speaking about mdof systems.

[sullen island]

yeah I'm out of my depth here, sounds like they're talking about a different type of eigenvalue

you could try asking in #modeling if no one else comes here

also ping helpers @graceful oyster

[graceful oyster]

Alright, thanks for your time and effort regardless

<@&286206848099549185>

To specify for people willing to look into this:
In my understanding the eigenvalue Lambda is equal to the square of the angular frequency omega: L = omega²
for the sdof system without dampening I found that Lambda seems to equal omega (not squared) It's also not specifically described an eigenvalue (problem) in the book I'm working with. I'm confused as to why there seem to be two seperate connections between Lambda and omega

[devout snow]

@graceful oyster Has your question been resolved?

[devout snow]

@graceful oyster Has your question been resolved?

[devout snow]

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[devout snow]

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[west nova]

Is this correct?

[west nova]

I think this is right I just would like help understanding if it's not

[brittle badge]

It's ||not|| incorrect, ||non-||sadly.

[west nova]

thank you bro

LMAO

u scared me

[drifting sinew]

-# Never let them know your next move.

[west nova]

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[lunar depot]

How to do this correctly ?

[lunar depot]

Answer is 0, but how to get it

[supple knot]

what tools are you supposed to use

can you just plot cos(x)

[lunar depot]

As simple as possible probably

[supple knot]

for x between 75 + 15 and 75 + 135

[lunar depot]

Is there a way by solving something?

[supple knot]

plotting is as simple as possible

[lunar depot]

What does that mean ?

[supple knot]

,w plot cos(x)

[woven radish]

Results provided by WolframAlpha

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

Requested by rie.mann

[lunar depot]

Once we have [90:210] what then ?

[supple knot]

look at the graph only in that range

90 degrees = pi/2 radians

[lunar depot]

Thanks

.close

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[dusty fox]

Is the distance formula for radius the same as it is for diameter?
And..is it derived from the "equation of a circle"?
What's that formula supposed to be used for anyway?

[glacial wyvern]

What is the "distance formula for radius"?

[sharp kayak]

which formula exactly are you referring to

[dusty fox]

Uhh I'm not quite sure but it's just the distance formula.
I just specified it because i was unsure

[autumn girder]

Can you provide any additional context?

[glacial wyvern]

Can you formulate the distance formula for us?

[autumn girder]

Maybe the original problem

[dusty fox]

It's not a question problem, I just want to clear my concepts.

[jolly dune]

A square bee square sea square?

[glacial wyvern]

This formula gives you the distance between two points (x1, y1) and (x2, y2)

[dusty fox]

Yes

[glacial wyvern]

So where do your radius and diameter come from

[dusty fox]

So the same i guess

[glacial wyvern]

radius and diameter relate to a circle

[glacial wyvern]

this doesn't relate to a circle

[autumn girder]

You can use it with a circle

but what it does is gives you distance of two points

[dusty fox]

I don't know i read somewhere that it's (x-h)² + (y-k)² instead for radius

[autumn girder]

That's for offsetting the origin

[dusty fox]

Dude

[dusty fox]

So it's essentially the same thing just using the actual variables for the points as they are?

That seems to be derived from the circle equation.
What's it supposed to entail?

[glacial wyvern]

If we call the center of a circle (x0, y0) and the circle has radius r, then for every point (x, y) on the circle it holds that the distance between (x, y) and (x0, y0) is r.

If you write this algebraically, according to the formula you described, then you get that

sqrt( (x - x0)^2 + (y - y0)^2 ) = r

which is more commonly stated as

(x - x0)^2 + (y - y0)^2 = r^2

[autumn girder]

https://www.desmos.com/calculator/jckfamgeem
Play around with this desmos graph (use sliders for a, b and r) to get some intuition on the topic

[dusty fox]

Thanks

[glacial wyvern]

The last equation is called the equation of a circle, because every point on the circle described by r and (x0, y0) satisfies the equation

and conversely every point satisfied by the equation lies on the circle

[dusty fox]

How'd the r get squared at the end

[autumn girder]

Because of the pythagoras' theorem

[dusty fox]

But r isn't a point on the circle just the distance

[autumn girder]

Yes

[dusty fox]

Oh yeah

My bad didn't see the sqrt

Yeah that is what i was exactly asking

[autumn girder]

Alr, do you understand it now?

[dusty fox]

No i still don't understand what the equation of a circle is supposed to be used for

[autumn girder]

I mean for a few(a lot) things but, primarily for drawing a circle?

Like if you ever have a graph which needs a circle you use this equation

[dusty fox]

Hmm

[autumn girder]

or if you ever get an equation of this form

you'll find a circle somewhere in it

[dusty fox]

How exactly

Like, coding or?

[autumn girder]

Programming could be one use for it, ye

[glacial wyvern]

If a geometric object is of the form some expression of x y z = 0, then we can find its normal vector at any point very easily using multivariable calculus. If we have a temperature distribution T(x, y, z) in space, then we can restrict that temperature distribution to a sphere x^2 + y^2 + z^2 = 1, and find the biggest temperature on the sphere by using multivariable calculus

[dusty fox]

I can't code but how could i use it except solving problems from a maths book

[autumn girder]

That depends really, which field do you wish to pursue IRL

[dusty fox]

Is that 0 or just a variable

[autumn girder]

0

[dusty fox]

Like anything that i could do as a layman i guess to get a better understanding of it

[glacial wyvern]

that is actually zero because mathematicians find it aesthetically pleasing (and some other reasons too.)

For instance the equation of a circle can be turned into that form if you write it as

(x - x0)^2 + (y - y0)^2 - r^2 = 0

[autumn girder]

If you just wanna get a better understanding of the equation, play around in desmos

That's prolly the easiest way to get used to it, as for how that equation comes to be

Do you know the definition of a circle in math?

[mellow talon]

How can i find the intersections of 2 circles?

[autumn girder]

!occupied

[devout snow]

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

[mellow talon]

Mb

[autumn girder]

np

@dusty fox

[dusty fox]

Hi

[dusty fox]

Yeah i get it, y-axis, x-axis & circle big/small

[autumn girder]

mhm

[autumn girder]

?

[dusty fox]

Are x y z in "x y z = 0" all multiplied or..?

[autumn girder]

No, he's just saying some expression involving the variables x, y and z

[dusty fox]

Uhh a 2D shape? With all points equidistant from each other from a fixed point.
Is 360° etc

[autumn girder]

Yes, Good job

well each other not so much

but from the center

Formally speaking

[dusty fox]

Oh yeah

Didn't mean to write that

[autumn girder]

A circle is the path traced by a point travelling in a 2 dimensional space, such that it remains at a fixed distance from some fixed point.

[dusty fox]

I don't understand the entire msg tbh, don't know most of the things being talked about or why there are specially 3 variables x,y, & z

[autumn girder]

Yea, that would make sense considering you're just learning about 2d equations still, don't worry, it'll make sense in future

[autumn girder]

Do you understand this?

[glacial wyvern]

I could've written F(x, y, z) = 0 for some function F but some people are scared of functions so I occasionally try to avoid them, perhaps causing more confusion in the process. Sorry about that

[dusty fox]

Yeah

Kind of like a planet but not quite

[autumn girder]

Ok, so now think about it in terms of this definition

[dusty fox]

You are so motivating hiten thanks

[glacial wyvern]

Our everyday geometry is always three dimensional, and even if it's two dimensional, it's occasionally a good idea to embed a two dimensional object into three dimensional space. Of course three variables is somewhat arbitrary in the sense that these ideas do generalize nicely to higher dimensions as well

[autumn girder]

If a point is in a two dimensional plane, then we measure two things about it, it's vertical distance from the origin, (y), and it's horizontal distance(x) from origin,

[dusty fox]

Yeah

[dusty fox]

What do you mean by
Embed a two dimensional object into three dimensional space ?

[autumn girder]

then, let's fix a point, and let's call it $P_C = (x_C, y_C)$ do you know what this notation means?

[woven radish]

@autumn girder

[glacial wyvern]

A two dimensional object can rest on the xy-plane, even if we have a z axis. This can occasionally be helpful if we have a special technique that only makes sense in three dimensions, but we still wanna try to apply it to a two dimensional object

[dusty fox]

No but I'm guessing P stands for point & C = centre?

[autumn girder]

Yes

$x_C$ is the horizontal distance of the center point from the origin, also called it's x co-ordinate and same applies for $y_C$

[woven radish]

@autumn girder

[dusty fox]

$P_C = point of centre$ ?

[woven radish]

👾

[autumn girder]

Yep, it's the central point

of the circle

Does this make sense so far?

[glacial wyvern]

Sounds like a regional convention, I've never seen that before

[dusty fox]

So basically h & k ?

[autumn girder]

We can get to conventions later

[autumn girder]

Legit just made it up lol

[dusty fox]

Where do you guys get the roles from

[glacial wyvern]

At the end of the day it doesn't matter how you write things down, the mathematics is still nonetheless the same. For me (x0, y0) makes more sense as the center than (h, k), but obviously you can call your center (å, ö) if you want to

[dusty fox]

I thought you were talking about a cult

[autumn girder]

that said, yes they're h & k in the formula

[dusty fox]

Why's it called h,k?

[glacial wyvern]

I don't know. I wouldn't use h and k

[autumn girder]

No reason except one guy woke up one day and decided to call it that

[dusty fox]

Oh

[autumn girder]

Anyways moving on

We now have our center

[dusty fox]

Continue 🙏

Yes

[autumn girder]

now we say that every point on the circle must be at some fixed distance (radius) from this center

i.e. if some point $P$ is on the circle then $distance(P, P_C) = r$

[woven radish]

@autumn girder

[autumn girder]

So far we have just mathematically written down the statement of definition

I hope there is no confusion thus far?

[dusty fox]

Yeah none

[autumn girder]

Oki cool

Now, how do we measure the distance of two points

?

[dusty fox]

Um the formula

[autumn girder]

Yep, the distance formula

do you understand how that works?

or should we go into that for a bit

and then come back here

[dusty fox]

Well

[autumn girder]

?

[dusty fox]

I wanted to know about the equation for a circle not the distance formula

[autumn girder]

I mean from here the next step is just the equation of circle

[dusty fox]

Yeah let's move on

[autumn girder]

$distance(P, P_C) = distance\left((x, y), (x_C, y_C)\right)$, because recall that every point is defined with two properties, it's x and y co-ordinates

[woven radish]

@autumn girder

[autumn girder]

$distance((x, y), (x_C, y_C)) = \sqrt{(x - x_C)^2 + (y - y_C)^2}$

[woven radish]

@autumn girder

[autumn girder]

This is just the distance formula by definition

and from here, we put it equal to r, as we derived from the definition of circle a few steps ago

[dusty fox]

Yeah

[autumn girder]

$r = \sqrt{(x - x_C)^2 + (y - y_C)^2}$

[woven radish]

@autumn girder

[autumn girder]

and there you go, there you have your equation

[dusty fox]

What's it supposed to be used for tho? also a last step left lol

[dusty fox]

Also could i see how to get the formula, step-by-step

[autumn girder]

Which one?

We already did the circle step by step

[dusty fox]

Circle

[autumn girder]

What more steps do you want, we went from definition to formula, which step did you feel was missing?

[dusty fox]

Yeah this but where does the - minus & square root get added from

[autumn girder]

That's the distance formula

Do you wanna see how that works?

[dusty fox]

No there's circle one at the end too

It simplifies to it i guess

I'm just curious why the - & square root get added

[autumn girder]

They get added when you use the distance formula from here

[dusty fox]

How tho

I don't get how it gets added

[autumn girder]

Ok, let's go into distance formula then

Would you agree that the distance between P1 and P2 here is gonna be equal to the length of the line joining P1 and P2?

[dusty fox]

Well...yes

[autumn girder]

Ok good

[dusty fox]

It's like discovering formulas from scratch

An apple just fell on my head

[autumn girder]

A lot was added here but does it make sense?

[dusty fox]

Thanks for doing this

[autumn girder]

NP

[dusty fox]

Yes

[autumn girder]

You should already have some intuition where we're going with this

[dusty fox]

y-axis (0,a)
x-axis(a,0)
Yeah

[autumn girder]

Here I color coded it a bit so that it's easier to see and added more stuff, does this makes sense too?

@dusty fox

[dusty fox]

Ywqh

Yeah

[autumn girder]

Ok cool

So now you see the right angle triangle formed by the purple, blue and orange lines?

[dusty fox]

Yes

[autumn girder]

What would the length of purple line be, in terms of the orange and blue line?

Hint: ||Recall, Pythagoras' theorem||

[dusty fox]

Oh Pythagorean,
H² = P² + B²

[autumn girder]

mhm

So, from that what would the lengh of purple line be?

[dusty fox]

Sum of P2P3 & P1P3 ?

Sorry for keeping you busy for so long, tbh I will revisit this sometime later, getting a bit late.
You could propose a time sometime later to continue from here, how's that sound?

@autumn girder

[autumn girder]

Would you agree?

[autumn girder]

That's fine by me but you already have the answer at this point

[dusty fox]

Yeah

Is that pink letter p?

[autumn girder]

Yes

[dusty fox]

Oh yeah nvm

[autumn girder]

p = the length of line between P1P3

[dusty fox]

What's that in teal green

G and something else

[autumn girder]

$90^{\circ}$

[woven radish]

@autumn girder

[dusty fox]

Do i..? 🥲

[dusty fox]

Oh

Yeah

[dusty fox]

So when ?

[autumn girder]

h is the distance of P1P2, right?

[dusty fox]

I gotta adios for now

[autumn girder]

Oki

!done

[devout snow]

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

[dusty fox]

Yeah

Thanks G

[autumn girder]

You're welcome

[dusty fox]

WAIT

When

[dusty fox]

.

[autumn girder]

?

... $d^2 = x \implies d = \pm\sqrt{x}$

[woven radish]

@autumn girder

[autumn girder]

since length can't be negative

$d = \sqrt{x}$

[woven radish]

@autumn girder

[autumn girder]

aka

$h = \sqrt{p^2 + b^2}$

[woven radish]

@autumn girder

[dusty fox]

Oh

Damn you just summarized it real quick rn

[autumn girder]

Yea, and now this is an exercise for you

Calculate p and b from the graph

You can do it on your own

Later if you don't wanna do it rn

[dusty fox]

H?

The distance

[autumn girder]

Hint: ||Look at the grid lines which have been drawn for your convinience||

[dusty fox]

Is it not H

Nvm

You asked the value of the points not the distance

[autumn girder]

The length of p and b

They won't be h

[dusty fox]

Yeah half of H

[autumn girder]

No

[dusty fox]

Huh

[autumn girder]

Did you look at the hint?

[dusty fox]

Shucks I'm bad

No

[autumn girder]

Well, it happens lol

[dusty fox]

Ok I'll do it later

[autumn girder]

Oki

try it

and if you can't solve it, open another thread

[dusty fox]

thanks again, but when will u be available

[autumn girder]

You can just post the question here

There are other helpers and they'll also be willing to help

so dw

[dusty fox]

Well yes, will do it later

Can't manage to keep this open

[autumn girder]

Alr, for now since your original question has been solved

please close this thread with .close

[dusty fox]

But..the homework

[autumn girder]

?

[dusty fox]

U just gave for me to solve

[autumn girder]

You can post it here for people to check or, ask if you have any doubts

don't worry, you'll be fine

If you want you can dm me too

[dusty fox]

Well

I'm asking cause i don't want to be waiting after sending it.
And if i know when you'll be available then getting it done quicker would be easy

[autumn girder]

Unfortunately, due to the current circumstances I am in, I can't give you a fixed time

[dusty fox]

Ah

Ok that's fine

[autumn girder]

So your best bet will be to use the help channels

when you need them

Under any circumstances, I'll close this channel for now.

.close

[devout snow]

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[fresh rock]

I need some help generating the transition matrix for this problem... I have an answer I'm just not sure if its right. I would love to compare work with someone. Thanks!

[glacial wyvern]

that takes a nontrivial amount of work to do

but maybe someone will be up to the task

You should probably try to network with the students who are actually taking that course with you

[fresh rock]

I generated a 13x13 matrix with this... Im "pretty" sure its right though

But Q, and I, in the F=(I-Q)^-1

What would those be if i only have 1 absorbing state in my 13x13

Because i need to make a fundamental transition matrix from it

[graceful cosmos]

You can cheat it a bit by not including the bottoms of ladders and the tops of chutes, since you can't land on those squares

So we're down to an 8x8

[fresh rock]

Thats what I was wondering about tbh. I included them in my matrix, but would that change the outcome?

Like in my matrix, since 3 goes to 7 with a ladder, rows 3 and 7 are the exact same in my matrix

same with chutes

The only absorbing state should be the end im pretty sure, so Q would be all others in a 12x12 and I would be a 12x12 identity matrix?

Heres my fundamental transition matrix, what exactly do these numbers mean?

[devout snow]

@fresh rock Has your question been resolved?

[devout snow]

@fresh rock Has your question been resolved?

[fresh rock]

.close

[devout snow]

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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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[bleak arrow]

if we have something like z = sqrt(4-x^2-y^2) and we were to graph the domain (x^2+y^2<=4) and range (0<=z<=4) together on one graph, what would this represent?

[drifting sinew]

It would look like a hemisphere.

$$x^2 + y^2 + z^2 = 2^2$$

[woven radish]

@drifting sinew

[drifting sinew]

When you take the square root of both sides, you find that z cannot be negative, so the sphere turns into a hemisphere

[bleak arrow]

if i graph the domain and range at the same time, what would I get?

i know S is a hemisphere

[drifting sinew]

Try to figure it out yourself; I don't want to do too much of the work for you.

I can try to give you a hint if you're not sure.

[bleak arrow]

would I get a cylinder?

[drifting sinew]

But we already established the graph is a hemisphere.

[bleak arrow]

that's what i got from desmos

[devout snow]

@bleak arrow Has your question been resolved?

[restive river]

That is indeed your domain

In that, the whole upper hemisphere, z = sqrt(2² - x² - y²), having radius 2, is enclosed in (it's a hollow hemisphere note that)

[bleak arrow]

the domain is x^2+y^2<=4, right? but if i were to graph the domain and the range together, what would you call this in math?

[restive river]

Nothing really like this is still math. You just restricted your domain R³ to the cylinder

[bleak arrow]

if u have a curve instead of a surface, u can represent the curve by domain + range, right? if u have something like:

intersection of z = sqrt(4-x^2-y^2) and z = 1, u get for the domain: x^2+y^2=3, and for the range: z = 1, u can write the equation of the curve as "x^2+y^2=3 at z = 1"

is there something similar to surfaces?

[restive river]

Err note that this is not really a domain technically because z is not defined for the circular strip 2<x²+y²<4. I don't know a better word for this, but if it were a disk of radius 2 centred at 0, that is your domain

Like for a semicircle with radius r, the domain is [-r,r] (which can be restricted), for a hemisphere it's x²+y²≤r² (which can be restricted)

For semi-circle, your "workspace" is R² but is not your domain

So even in hemisphere, you're "working in" R³ but that isn't your domain

[restive river]

The intersection is a level curve that you get -- a circle with radius sqrt(3)

[bleak arrow]

isn't the domain x^2+y^2=3 and the range z = 1?

is it just a coincidence that the equation of the curve is x^2+y^2=3 at z = 1

[restive river]

When you say z = sqrt(4-x²-y²), your
domain is the natural domain where the whole function z is defined in (without any restrictions) which is x²+y²≤4 and your range is the z values [0,2]

In other words, z is a mapping from D(0,2) -> [0,2] where D(0,2) denotes disk of radius 2 centred at 0

When you show both of them in R³, you get the surface -- hemisphere

The cylinder is not technically your domain, the disk is

[bleak arrow]

for the example of the curve, is it always like this? the equation of the curve is the domain + the range?

[restive river]

Yes when you graph the curve both input values (domain) and output values (range) are there, plotted as points on the curve you get

[bleak arrow]

show what?

[restive river]

So when you plot a point on the hemisphere (x,y,z), (x,y) is a part of your domain (the disk) and z is a part of your range

[restive river]

The domain you're working in and the range you get

[bleak arrow]

so the domain for the surface is the disk x^2+y^2<=4, and the range is the region 0<z<2

[restive river]

0≤z≤2 yeah

[bleak arrow]

if i graph them together I get the surface?

[restive river]

Graph them together as in you're plotting z = sqrt(4-x²-y²)

[bleak arrow]

how?

[restive river]

.

If you plot the disk separately that's different

[restive river]

If you graph domain separately, you get disk

Graphing the range separately gives you a line segment over [0,2] on z-axis

Plotting the curve z=sqrt(4-x²-y²) gives you the hemisphere

[bleak arrow]

how do i graph it on desmos

oh nvm i see what u got

[restive river]

I never used 3D desmos but just typing in z=sqrt(4-x²-y²) should work

[bleak arrow]

ok thanks

[restive river]

You're welcome. Have a great day

[bleak arrow]

.close

[devout snow]

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[lusty tapir]

I'm getting -10/3

[lusty tapir]

Got a=1/2 by solving the inequalities

So the parabola is y²=2x

Then slope is -beta/alpha no?

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[errant tapir]

Hello

[errant tapir]

Next Tuesday I have my Algebra 1 class final

I was wondering if anybody had anywhere I could find a solid review to go over this material

And also our final will go over a portion of midpoints and distance

[supple knot]

https://www.khanacademy.org/math/geometry/hs-geo-analytic-geometry/hs-geo-distance-and-midpoints/e/midpoint_formula

[errant tapir]

Oh thanks

But I also need like an algebra 1 full course review

[supple knot]

https://www.khanacademy.org/math/algebra

[errant tapir]

Thanks brother 🫡

[warped marsh]

I need hlp

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[shut drum]

I’ll send my work shortly

There are just some details I need help on

Thank you

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[shut drum]

Here’s my work

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[lost crag]

why sqrt(2n)*q^n -> 0
as n-> inf
where -1<q<1 a constant?

[vital sedge]

something ^ n grows faster than something * sqrt(2n)

so its gonna go to 0

[lost crag]

not an argument i can use

in calc

[vital sedge]

Why not?

[lost crag]

rigorous class

[vital sedge]

ah.

[versed juniper]

have we tried taking limits

[lost crag]

how exactly

[wind mason]

since -1<q<1 we can write it in this format

[woven radish]

;(

[wind mason]

We know 1/q will be some positive number that is not a fraction (since |q|<1 implies that it is such)

For q in the rationals

q in R will be a bit difficult but we can still get there

Just show that an exponential equation a^x with |a|>1 grows faster than sqrt(x) and you're all set

[woven radish]

;(

[wind mason]

Sorry if this is a bit too much info

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[dense rampart]

is this valid for all discrete distributions?

[dense rampart]

they say its for discrete distributions with non negative support, but they all kinda do. you cant put negative numbers into NB or Hyp. its just that some supports dont include 0 and now idk what to do with it

[pseudo basin]

it's valid for distributions of RVs that only take natural-number values.

including 0 in the support is fine too as it kinda inherently contributes nothing to the expectation

like

0 * p(0) is just 0

[dense rampart]

what if its for NB where support starts at r?

[pseudo basin]

that's also fine, the probabilities for points less than r will just be zero

[dense rampart]

do i have to pay attention to the lower limit of the summation?

[pseudo basin]

as in...?

[dense rampart]

do i ever have to change x=0 into sth else

[pseudo basin]

no, you don't have to

but it is sometimes a good idea to prune known-zero terms from your sum if you see any.

but don't overblow the importance of that.

[dense rampart]

ok thx

.close

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[rare kernel]

first tried x=19

then x-1 = 19

wasnt able to accomplish much

[lilac crescent]

Is it f(X) + f(X-1) ?

[rare kernel]

yeah

and f(19)=94

huh

[cedar berry]

oh hol up

wait a min

aight think in terms of difference of 2 consecutive squares

see a pattern

?

[brittle badge]

f(20) + f(19) = 20^2
=> f(20) = 20^2 - f(19)
f(21) + f(20) = 21^2
=> f(21) = 21^2 - (20^2 - f(19)) = 21^2 - 20^2 + f(19)
f(22) + f(21) = 22^2
f(22) = 22^2 - (21^2 - 20^2 + f(19)
f(23) = 23^2 - 22^2 + 21^2 - 20^2 + f(19)

You can see the pattern.

[rare kernel]

ohh yeah

[cedar berry]

bro we r suppose to make people suffer and realise the answer tho

[rare kernel]

but isnt that kind of an ugly pattern

[cedar berry]

nope

[rare kernel]

cause i need to find the remainder right

[brittle badge]

It's an alternating sum.

[cedar berry]

yeah

[brittle badge]

It's not hard to find f(94) itself I guess.

[cedar berry]

wait is 10^2-9^2

[rare kernel]

are u kidding

[cedar berry]

no bro

it is ez

think

[rare kernel]

there would be so many squares

[cedar berry]

nah

see

[brittle badge]

(n+1)^2 - n^2 = (2n+1)

[cedar berry]

think in terms of difference of 2 squares

[rare kernel]

yeah so ill need to do this for many terms right?

[cedar berry]

yeah but then u get 94+93+92+91+.......+21+20+f(19)

which ig u can do

if u know wat to do

[rare kernel]

huh yes i can do this sum but im a bit confused to how u got it

[lilac crescent]

It should be + and - tho

[cedar berry]

deez

[cedar berry]

nah

94^2-93^=(94+93)(94-93)

[rare kernel]

how did 94^2-93^2+92^2 so on become 94+93+92...?

[eternal aspen]

This mf trollin

[rare kernel]

me?

[cedar berry]

92^2-91^2

[brittle badge]

94^2 - 93^2 = (94 + 93)

[eternal aspen]

Not u

[brittle badge]

no way

[cedar berry]

moye moye

[rare kernel]

ohhh ok got it

oh ok so i can use this to find f(94) directly

and then divide by 1000

[cedar berry]

yessor

yessororror

[rare kernel]

thanks man

[cedar berry]

np

[rare kernel]

.close

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[cedar berry]

👍

[brittle badge]

wait

[rare kernel]

.reopen

[devout snow]

✅

[rare kernel]

yeah?

[brittle badge]

it would be -f(19) for x = even

at the last

[rare kernel]

ohh ok yeah

ty both

.close

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[brittle badge]

np

[devout snow]

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[buoyant dragon]

I’ve been trying to get the derivative of this and have done several retries, the strategy I’m employing is taking the natural log on both sides to bring down the exponent of e. Any guidance informing me where I went wrong or how to get this derivative would be greatly appreciated. Thank you

[autumn girder]

,rccw

[autumn girder]

Do you know implicit derivatives?

[buoyant dragon]

It’s been a few weeks so I’m rusty

Get the derivative and when you derive y dy/dx becomes a coefficient and you isolate that

Afaik

[autumn girder]

Exactly

Yea you have to use those here

[buoyant dragon]

Yeah I’ve been trying to but all the answers I’m getting are different than what google says the answer is

Can I send some other attempts, I feel like there’s something I’m missing

Or maybe does the ln to bring down the exponent strategy not apply here? That’s my best guess

[autumn girder]

It can work, but there is a much simpler way to do this

$\exp(a-b) = \frac{\exp(a)}{\exp(b)}$

[woven radish]

@autumn girder

[autumn girder]

Where $\exp(x) = e^x$

[woven radish]

@autumn girder

[buoyant dragon]

So the exponent since it has subtraction, can be put into a quotient form basically? Am I understanding that right?

[autumn girder]

Yes

[buoyant dragon]

So how would I rewrite that in this case? 5e^(2x/y)?

[autumn girder]

Isn't it 5e^(2x - y)?

[buoyant dragon]

Yes

[autumn girder]

Oh you're asking if you got that

well again, you can use ln

[buoyant dragon]

I think im misunderstanding the formula my bad

[autumn girder]

$5e^{2x - y} = \frac{5e^{2x}}{e^y}$

[woven radish]

@autumn girder

[buoyant dragon]

Got it thank you

And then just do quotient rule and chain rule

Going to try it out and send a pic one sec

[autumn girder]

Ok

[buoyant dragon]

This feels wrong still I think

[autumn girder]

Uh... you should've taken e^y to the LHS and brought x to the rhs

so that you'd have all the y stuff on one side and all the x stuff on one side

This should still work, don't get me wrong

but let me check

[buoyant dragon]

Oh I see, but like shouldn’t it work out either way, can’t you isolate dx/dy after getting the derivative? (ok cool yes)

I did notice I moved a term and forgot to make it positive cause it was negative

Otherwise though not sure

[autumn girder]

Where did 2 go from, $5 \cdot e^{2x} \cdot 2 \cdot e^y$

[woven radish]

@autumn girder

[buoyant dragon]

You’re asking like where I got the 2 from? In the quotient rule section? Could you circle it maybe?

[autumn girder]

No, I'm asking where did it go in the next step?

[buoyant dragon]

Oh I see yeah I forgot to make it 10e

Thank you, amending that, otherwise how does it look

[autumn girder]

I think it looks correct yea

[buoyant dragon]

Thank you I appreciate your help

[autumn girder]

Though if you can, try seperating variables and then use implicit differentiation

this will give you an answer faster

and less error prone

[buoyant dragon]

Yeah going to redo the problem and do it in that method

[autumn girder]

Alr, cool

[buoyant dragon]

Got a final at the end of this week so gonna do a lot of review, wish me luck and best of luck to you and all your stuff

[autumn girder]

Best of luck for your exam

You'll do good, I'll cheer for you

!done

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[round gate]

Can anyone help me with the scalar product of [-28,0,18]*[-1,0,3] I keep getting 82 when I try it but that's wrong

[potent tusk]

are you sure you are wrong

[round gate]

Yes

It's 78

I don't know how though

[untold crag]

I think whatever answer sheet you are using is actually wrong though, (-28)*(-1) + 18*3 is indeed 82

if by scalar product you mean the dot product

[near jolt]

is the question just computing the scalar product of those two vector or is it more than that? maybe you messed up a previous step

[round gate]

I need to find the Volum of point A(1,0,0) B(4,-3,4) C(4,3,4) and D(0,0,3)

The Crossprodukt of AB and AC is [-28,0,18] and AD is [-1,0,3] and Whenever I dot those I get 78

But when I try to do it without calculator I get 82

[near jolt]

you messed up the cross product

[round gate]

I did? I double checked in the CAS calculator and got the same as when I did it on paper.

[near jolt]

can you show your steps?

[near jolt]

you mean the volume of the parallelipiped formed by AB,AC,AD?

[near jolt]

yeah -24

you wrote -28

[round gate]

Oh man, thanks. I was stuck on this while

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[primal shadow]

hello again,,,, I've been staring at this for a few minutes and can't figure it out😔

[primal shadow]

Unless it's just trial and error again where i need to list out every locker

[pseudo basin]

given a locker number, how can you tell which students will change it?

[primal shadow]

If it's a multiple of that number?

[pseudo basin]

what number is a multiple of what

[primal shadow]

Which student it is

Like the 4th student every multiple of 4

?

[pseudo basin]

call the locker number L and the student number S. which one has to be a multiple of which?

[primal shadow]

L has to be a multiple of S

[pseudo basin]

right

so a locker L will be toggled by precisely those students whose number is a divisor of L, yes?

[primal shadow]

yes

i kind of get the basic concept but it gets confusing when the numbers get bigger

[pseudo basin]

if a locker is left open what can we say about the number of times it has been toggled

[primal shadow]

is that like after the first student opens them all

does that count as one toggle

or is closing the locker the first toggle

[pseudo basin]

both opens and closes are toggles

what i mean is after all the students have gone and done their thing

and we look at one particular locker that got left open

[primal shadow]

so then its been toggled an odd number of times?

[pseudo basin]

how many times must it have been toggled

[pseudo basin]

yes

so we are looking for numbers with an odd number of divisors

[primal shadow]

i seee

[pseudo basin]

think about what those look like

think about how the divisors of a number can pair up

[primal shadow]

?

sorry i don't understand 😔

if they have an odd number of divisors does that include 1?

[near jolt]

ofc, because, in the question, "student 1 toggles every locker that is a multiple of 1"

[near jolt]

for a positive integer n, there is a natural correspondence between divisors (if "a" is a divisor of n, then naturally, one other number is a divisor of n)

[primal shadow]

i see

so like all numbers divisible by 6 are divisible by 2 and 3

[near jolt]

that is one way to think about it, but the correspondence they had in mind is the following: a is a divisor of n if and only if n/a is a divisor of n.

[primal shadow]

i see

so like 6/2=3 and 3 is a divisor of 6?

[near jolt]

so this is a natural way to pair up the divisors, a with n/a

[primal shadow]

i seee

[near jolt]

think about how this pairing can be used to prove that the number of divisors of cartain numbers is even.