help-10

[red ice]

try using similar triangles

you can let AB = 1 for example

then EA/AB = PC/CB

7/1 = h/CB so CB = h/7

then AC = 1 - h/7

so you can use similarity on triangles APC and ADB again to come up with an equation in h

[stoic smelt]

Is the one on the bottom a formula? Or just an example

[red ice]

it's a formula actually

here a = 7 and b = 12

the harmonic mean is 2/(1/a + 1/b)

[stoic smelt]

I’m so lost

Cp= 1/2x84

Like this?

[prisma hatch]

yeah

[stoic smelt]

12-x/12=7-x/7

Like this(?)

[red ice]

no

[red ice]

seems like you didn't understand what I wrote here

[prisma hatch]

Do you know what the similarity in a triangle is?

[stoic smelt]

Yes

But I have no idea how to make an equation for this

[red ice]

look at triangles APC and ADB

[prisma hatch]

name every intersection so we can communicate

[red ice]

can you come up with an equality relating the sides

like how since AEB is similar to CPB, EA/AB = PC/CB

named.

[stoic smelt]

AP/AC=AD/AB

[red ice]

yes, that is indeed correct

but try not using sides AP and AD

[stoic smelt]

PB/PC=EB/AB

[red ice]

I mean don't use the diagonals

only use the sides AC, AB, PC, DB

[stoic smelt]

Ohh

[prisma hatch]

EB/EA

[stoic smelt]

I’m so confused sorry 😭

[red ice]

anyway since you understand the concept I'll just tell you

PC/AC = DB/AB

but if you've been following through my working, you'll know that AB = 1

AC = 1 - h/7

DB = 12

and we let PC = h

so we want to solve for h

[stoic smelt]

H/1-h/7=12

[red ice]

nice

h = 12(1 - h/7) now you should be able to do this

[stoic smelt]

4.42 ft?

[red ice]

,calc 84/19

[warm shale]

Result:

4.4210526315789

[red ice]

yeah

[red ice]

you can also rearrange this to get 7h = 84 - 12h or 7h + 12h = 84

so if you now work with any pole heights a and b, not just 7 and 12

you actually find you get (a + b)h = ab

hence h = ab/(a + b) = 1/(a/ab + b/ab) = 1/(1/b + 1/a)

[stoic smelt]

Thank u sm

🙏🙏

[red ice]

no worries!!

[stoic smelt]

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[split drum]

how do we solve this one? i used l'hôpital's rule and my answer is coming out to be infinity

[split drum]

ah wait

never mind

not infinity

[willow crescent]

.-.

[split drum]

but still, how do we go with it

[worthy birch]

take k = pi-2x

[split drum]

how does that help?

[worthy birch]

x --> pi/2 , k --> 0

then use 1-cos(2A) = 2sin^2(A)

A = k/2

[split drum]

shouldn't it be (pi-k)/2?

[worthy birch]

I skipped a step

1 min lemme get tex

[split drum]

ah it's in third quadrant

[worthy birch]

$\frac{1+cos(\pi-k)}{k^2}$

[warm shale]

Alaska

[worthy birch]

thats - cos

-cos(k)

[split drum]

yeah, got it

the answer should be 1/2, no?

[worthy birch]

ye

[split drum]

thank you!

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[silk trout]

Let $Y_1,Y_2,…$ be an i.i.d. sample of random variables, each with a Uniform(0,3) distribution. Define a new sequence of random variables $X_1,X_2,…$ by $$X_n = \frac1{n} \sum_{i=1}^{n} Y_i^2$$
Using the Law of Large Numbers, determine the value of a $\in \mathbb{R}$ for which $\mathbb{P}(\lim_{n \rightarrow \infty} X_n = a) = 1$

[warm shale]

Xotiic

[obtuse pebble]

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[median tinsel]

i need help with geogebra

[obtuse pebble]

@median tinsel Has your question been resolved?

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@median tinsel Has your question been resolved?

[median tinsel]

i just wanted to know how to make a fractal in new geogebra

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[brittle sand]

is there any mistake here?

[dark stirrup]

On the first sentence of your proof: Is there some guarantee that the $p_0$ will be in $\textbf{every}$ open ball? Or is it more that, for every $r>0$, there will be some $p_r\in\Delta$ in the open ball $B_r(z)$? How you wrote $p_0$ implies that it is a fixed value.

[warm shale]

SWR

[brittle sand]

oh yeah, the point is proving that for a contradiction

should`ve stated that

[dark stirrup]

I get that you are proving by contradiction

[brittle sand]

there must be a point in every ball

[dark stirrup]

Yes. a point. But your proof implies that the same point would be in every ball, which is not necessarily true.

[brittle sand]

p_0 is a fixed value

[brittle sand]

nono

there is some point, the first one is p_0. then i shrink the ball and there must be another point, say p_1. then i shrink it again and there must be p_2 and so forth

p_0 is not in the second ball, p_1 is not in the third, p_2 is not in the forth and so on

[brittle sand]

oh yeah

i should`ve said "for some r > 0"

my bad

[dark stirrup]

sort-of

sigmund, is english your primary language?

[brittle sand]

for some r > 0 (since it should be true for all r > 0)

[brittle sand]

no

[dark stirrup]

Allow me to make a minor correction then, to better express your proof in more standard mathematical wording

"for some r > 0" reads as "there exists some r > 0".
You do want "for any r > 0".
However, "let p_0 for any r>0" reads as "this p_0 is fixed", which you do not want.
And "let p_0 for some r>0" reads as if you are only looking at one specific r>0, not generalizing to any r>0.
The best wording (in my opinion) would be "For any r>0, assume there is some p_0"

[brittle sand]

yeah i know what it means

[brittle sand]

,

[dark stirrup]

optionally, you could say "assume for contradiction" to be more clear.

[dark stirrup]

Which I addressed here

However, "let p_0 for any r>0" reads as "this p_0 is fixed", which you do not want.

[brittle sand]

p_0 is in some ball since for any r > 0 some point is in the open r-ball

[dark stirrup]

Maybe your edited proof makes the point clearer. But when I use the context of the remainder of your proof, this line still sounds wrong to me.

[brittle sand]

lemme rewrite it then

[dark stirrup]

appreciate it

[dark stirrup]

Overall, I like the idea of your proof though. And use of the word "absurd" is amusing to me

[brittle sand]

@dark stirrup

[dark stirrup]

Okay that reads better

At a glance, your proof looks sufficient, but let me be certain for you

[brittle sand]

hmmm i think thats quite common among portuguese speaking math students

[dark stirrup]

I know there's another way to prove this, so I'd not want to send you off with an incorrect proof

[dark stirrup]

I have seen it a few other times around here, so that makes sense. In English mathematics, we would often say "which leads to a contradiction" or "which is contradictory" or something like that. But I like yours. It adds some joy (for me as an English speaker) to see something a little different, but not detracting from the quality of the proof.

[brittle sand]

virgin "it's a contradiction" x chad "it's farfetched"

thatd be wild

[dark stirrup]

In English "absurd" is a more accusatory/hostile way of saying "contradictory". "Contradiction" has very little emotional connections, but "absurd" can be seen as criticizing someone.

But take my words lightly, as I do not represent every English speaker, and I have zero problem with your use here.

[dark stirrup]

I'm not brave enough to use farfetched in my proofs

Anyway, back to reading your proof

[brittle sand]

yup same goes for pt
when i first learned maths and saw ppl writing "absurd" like that it often felt a bit too much to call things absurd
but it really comes from latin, "reductio ad absurdum", which roughly translates to "reduction to absurd"

maybe the lusophone tradition inherited this latin logical habit

[dark stirrup]

You conclude that $p_1\in\Delta$ is in $B(z, d(z, p_0))$ from your assumption that every open ball around $z$ has some point from $\Delta$. Am I understanding that correctly?

[warm shale]

SWR

[brittle sand]

it's virtually the same as the previous statement

since for all balls theres some point in delta, there must be this point p_1 in this ball

[dark stirrup]

Ah. very true. Also why we end some proofs with "Q.E.D."

[brittle sand]

yeah

(im not too fond of this guy tho)

[dark stirrup]

I figured. At first, I didn't understand how you came to that conclusion, I had to fill that line in myself. It may not heart to mention how you got p1, but feel free to omit it if you want to keep your proof short

You go on to prove something by induction. You don't explicitly mention the proof is inductive, but I understand your work. In my style, I mention it explicitly, but your approach is fine.

[brittle sand]

oh yeah i swiftly hid the induction there

[dark stirrup]

Your proof is sufficient

gold star

🌟

Have you studied continuity in metric spaces yet?

[brittle sand]

nope!

not yet

continuity is chapter 2 of this book

its a shame you english native speakers dont have this book, its so good

and its not common for a book written in pt to be better than the english ones

so i think it does it

dude u have no idea how much i tried NOT TO THINK of M as a field while doing this problem

bc i could force the cartesian to be some vector space and dang vectors spaces are just nice

its much easier to think of balls in vector spaces

but M could be literally anything

[dark stirrup]

I had the same problem when I first started studying metric and topological spaces.

[dark stirrup]

The reason I asked is because there is a nice, more generalized proof to this problem that uses properties of continuity.

[dark stirrup]

I believe you. But I was very satisfied with my topology book. A little too introductory, but it covered the material well and the exercises were still challenging (to me).

so let's all be happy we all get a nice book

[brittle sand]

yesss

[brittle sand]

what book was it btw

im thinking abt maybe after i clear this book i start doing munkres topology book problems like crazy

[dark stirrup]

Introduction to Metric and Topological Spaces, by Wilson A Sutherland.

it's very introductory, but it's a great start

[brittle sand]

i see

[dark stirrup]

But that book had a problem which is a more generalized version of your problem

[brittle sand]

i'm curious now

i'm not a complete illiterate when it comes to topology so i might understand it

send it pls

[dark stirrup]

There is also a third proof that doesn't require continuity. I think it may be a simplification of your proof

[dark stirrup]

Sure, but it's built around toopics that I do not think you have studied yet

[brittle sand]

such as?

[brittle sand]

a hausdorff space is a space with the hausdorff property right

[dark stirrup]

topological spaces, topological product spaces, continuity, hausdorff spaces

[dark stirrup]

ah you know that

yes

[brittle sand]

in which for any two points in the space theres a disjoint neighbourhood containing x and another containing y

(iirc)

[dark stirrup]

correct

and every metric space is Hausdorff

[brittle sand]

wow

thats nice

[dark stirrup]

Yes it is.

And its easily provable, since at any two distinct points, there is a non-zero distance, and at each point you can make an open ball of less than half that distance

[dark stirrup]

The converse of this statement also has neat implications. If a space is not Hausdorff, then it is not metrizable. So you can definitively state that there exists topological spaces that cannot be expressed as a metric space.

[brittle sand]

oh i think you meant the counterpositive

not the converse

[dark stirrup]

yes I did. Thank you

contrapositive

[brittle sand]

interesting

[obtuse pebble]

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[split dew]

what does the given by u x v = [...] mean?

[pulsar terrace]

wdym what does it mean

isnt that just the cross product formula for a 3d space

[split dew]

oh ok so thats just the formula sorry i assumed it meant something else

thank you

[pulsar terrace]

👍

[split dew]

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[severe vessel]

I have to find the derivative of this equation and this is practice for the AP Calculus exam? Also it's question 9

[twin patio]

are you asking for the second derivative of $\frac{1}{\sqrt{x}}$ or all the exs from A to E?

[warm shale]

mercidy

[severe vessel]

The second derivate of 1/square root of x

But idk if that derivate is the correct derivate of the equation

[twin patio]

okai you should approach this ex by considering the square root simply x^(1/2)

[severe vessel]

So it would be 1/x^(1/2) as the first derivate?

Then take the derivate of this equation?

[twin patio]

it would be 1/2 * 1/x^(1/2)

[severe vessel]

Can u explain why I would multiply 1/2 to 1/x^(1/2)?

[twin patio]

yes sure, how do you take the derivative of x^2?

[severe vessel]

It would be 2x

[twin patio]

you take the exponent (2) and multiply it by x^(2-1) right?

[severe vessel]

Yes

[twin patio]

in this case you do the exact same thing, you take the exponent (1/2) and multiply it by x^((1/2)-1), so 1/2*1/x^(1/2)

[severe vessel]

So it would look something like this?

[twin patio]

oh ops i was wrong i thought we were taking the derivative of sqrt(x) sorry

okai the logic is always the same

but in this case you have x^-(1/2)

[severe vessel]

Wait I'm confused??

[twin patio]

1/sqrt(x) is x^-(1/2) right?

[severe vessel]

I'm getting myself confused

[twin patio]

yeah my fault I was totally wrong sorry

[severe vessel]

Am I suppose to take the derivate of square root of x while taking the derivate if natural log??

Oh no ur good

[twin patio]

no no, you are not taking the derivative of 1/x, but 1/sqrt(x)

basically you are taking the derivative of x^(-(1/2))

so, like before, you take -(1/2) and multiply it by x^(-(1/2)-1)

-(1/2)-1 = -(3/2)

so you get $-\frac{1}{2}*\frac{1}{x^{\frac{3}{2}}}$

[warm shale]

mercidy

[twin patio]

right?

[severe vessel]

So it would look like this?

Bc there is a choice that matches up with this answer

[twin patio]

okai i'm really sorry i promise you that now i know what are we searching for, f(x) = ln(sqrt(x)), i thought it was f(x)=1/sqrt(x) because i only say your pencil on the right ahaha

okai i'm making this more confusing than how it should be

restart

[severe vessel]

Lol oke

[twin patio]

in general, when we are taking the derivative of a ln, we get a fraction, in which the numerator is the derivative of the element in the logaritm, and the denominator is that element

so the derivative of ln(sqrt(x)) is (derivative of sqrt(x))/sqrt(x)

right?

[twin patio]

mhm, what is the derivative of sqrt(x)?

[severe vessel]

x^-1/2

[twin patio]

consider it x^(1/2)

[twin patio]

okay yes, but remember the *(1/2)

so its (1/2)*(x^-(1/2))

imma use latex cause of the impossibility to reading xD

$\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$

[warm shale]

mercidy

[twin patio]

so you have that for your numerator, and at the denominator you have only sqrt(x)

[severe vessel]

Oke

I'm starting to understand

[twin patio]

yeah i'm really sorry i created a lot of confusion xD

[severe vessel]

No it's oke ur good

[twin patio]

you're doing great, the important thing here is to remember to use the power rule for any root

so we have $\frac{d}{dx}ln(\sqrt(x))=\frac{\frac{d}{dx}\sqrt{x}}{\sqrt{x}}=\frac{\frac{1}{2\sqrt{x}}}{\sqrt{x}}$

[warm shale]

mercidy

[severe vessel]

Ooh okay it's starting to click now

[twin patio]

okai so did we figure out the first derivative?

[severe vessel]

Currently working on it

[twin patio]

good

[severe vessel]

Is this the next step to take?

And would u get rid of the fraction or no?

Wait nvm I messed up lol

Would the next derivate be set up as the question below the first derivative equation?

[twin patio]

okai wait, first, what do you think about the first derivative? How could you rewrite it?

[severe vessel]

Would it be x^(-1/2) / x^(-1/2)?

[twin patio]

well, dividing is equal to multiplying with the inverse right?

so you could muliply 1/2sqrt(x) * 1/sqrt(x)

obtaining 1/2sqrt(x)*sqrt(x) right?

[severe vessel]

omg yes

My brain is literally running slow today😭

So it would look like this?

[twin patio]

yes right

and that is equal to 1/2x right?

[severe vessel]

For 1/sqrt(x)?

[twin patio]

sorry?

sqrt(x)*sqrt(x) is x right?

[severe vessel]

I'm sorry calculus is confusing to me sometimes😭

Ooooh

U mean multiplying both fractions right?

[twin patio]

yess right

[severe vessel]

Lol oke

[twin patio]

dw completely normal

[sour spear]

Hey

Oh

Sorry idk why I’m here

[severe vessel]

This is what I got so far

[twin patio]

ok perfect

now

what is the derivative of 1/x?

[severe vessel]

Is it -1/x^2?

[twin patio]

yes

so the final answer is 1/2*(-1)x^(-2)

or $-\frac{1}{2x^2}$

[warm shale]

mercidy

[severe vessel]

Lol oke bc I was like huh at first

That's the answer I got aswell

[twin patio]

perfect

[severe vessel]

Thank u for the help!

[twin patio]

nothing!

[severe vessel]

.close

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[fervent comet]

why is zero vector orthogonal to every vector when it kinda has no vector at all

[fervent comet]

or like no magnitude

[lost fjord]

Zero vector is a vector

[fervent comet]

yea ik that

[lost fjord]

it has magnitude 0

[fervent comet]

like visually

how do you define it

infinite perpendicular angles?

[tranquil rover]

A point

[fervent comet]

right

[lost fjord]

Assuming you mean euclidian iner product?

[pseudo swift]

parallel to any direction even when you think about it

[fervent comet]

yes

yea exactly

so like

[lost fjord]

It is orthogonal because the inner product with any other vector is 0.

[fervent comet]

but isnt the definition of orthongal also that it is perpendicular

[lost fjord]

That isn't necessarily the definition.

[tranquil rover]

Because dot product is always 0

Implies costehda is 0 so angle is 90 degree

[fervent comet]

but there is no angle

[fervent comet]

ok i can see this

[strong pike]

Zero is a bit of a weird case, its hard to interpret visually, but this might help:

1. Dot product of the zero vector with any other vector is zero. A zero dot product implies orthogonality.
2. If two vectors are parallel, one vector can be expressed as a multiple of another. You can't multiply the zero vector by any scalar to get a nonzero vector.

[fervent comet]

this helps a lot

[strong pike]

Orthogonality is most rigourosly defined through the inner (dot) product. The "two vectors make a 90 degree angle" definition doesn't work outside of R2. And since the zero vector isn't really in R2, that def collapses.

[fervent comet]

so since its not parallel then its perpendicular

[pseudo swift]

but you can always say 0 = 0 * v for any vector v though

[fervent comet]

i think ur using 0 as a scalar

[tranquil rover]

One is scaled another is vector

[tranquil rover]

You can

V1 = kv2 implies V1 and V2 are parallel

[fervent comet]

like i get the formula definition

i just wanted more intuition

[strong pike]

Yeah, I guess it would have to go both ways for two vectors to be able to be parallel. Either way, go with the inner product. Once you decouple the idea that "orthogonal means perpendicular", and go with "orthogonal means zero inner product", it'll be more useful.

[tranquil rover]

But when you take V2 as 0 vector it won't work , so zero vector is just a special case

[pseudo swift]

myeah

[fervent comet]

thanks guys

.close

[obtuse pebble]

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[tranquil rover]

Oh btw I forget to mention when doing dot product

V1.V2 = | v1| |V2| costehda

Here both side will yield 0 in case of 0 vector , so the angle can be anything

Which is true because 0 vector is just a point at orgin

And you can't define angle between a point and line

[fervent comet]

that was my original gripe

but yea I get it now thanks

[tranquil rover]

Yeah and its correct

[fervent comet]

.reopen

[obtuse pebble]

✅

[fervent comet]

can someone give me a quick summary of the goal of linear algebra

[lost fjord]

uhhh

depends

[fervent comet]

not abstract lin alg

or diff eq

[lost fjord]

are you math major

or to be math major

[fervent comet]

so I have the right mindset

yes

[lost fjord]

okay

current major?

what year?

[fervent comet]

econ

3rd

[lost fjord]

econ + math?

[fervent comet]

yup

[lost fjord]

well

from a pure math standpoint

its fundemental for a lot of algebra

and has a lot of connections with general functions/mappings

[fervent comet]

i'm trying to do a more applied route

[tranquil rover]

Linear algebra? Like x+y= 10 stuff? Or with lines?

[lost fjord]

A lot of the properties are especially important as you abstratc away the connetcion with a "plane"

[lost fjord]

i see

[fervent comet]

partial diff and optimization

[lost fjord]

good for ML, calculating stuff fast

[fervent comet]

probability

statistics

[lost fjord]

IMO its just a general basis for a lot of math

[fervent comet]

i c

bc i have a final coming up and I wanted to know how I should like view the questions

[lost fjord]

The goal of linear algebra imo, is to develop a level of intuition for relations, functions/mappings/transformations

[fervent comet]

i c

that makes sense

[lost fjord]

A goal for what exactly a space is, what is a thing, a bunch of things, and a sub-bunch of things in things.

I think were the most important topics that (i wish) i understood.

[fervent comet]

interesting

I will try to keep that in mind

thank you

.close

[obtuse pebble]

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[obtuse pebble]

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[ember bolt]

My math teach wanted me to analyze a piecewise function and I can't decide if I need to mention a hasy that doesn't effect the function 2^x bc its cut off at x=2

[obtuse pebble]

@ember bolt Has your question been resolved?

[obtuse pebble]

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[timid silo]

hello?

[timid silo]

i need help with the last part

why do we take N to be the max of those numbers specifically?

nevermind i just got it

.close

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[tardy epoch]

#❓how-to-get-help

<@&268886789983436800> troll

[obtuse pebble]

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[severe pagoda]

I am confused with the pi variable I do not understand how I am supposed to horizontally stretch it.

[high lily]

you're not supposed to stretch

also pi isn't a variable

[severe pagoda]

i know I am not supposed to stretch it

pi/6...

[austere abyss]

pi/6 shifts the graph to the right by pi/6

[high lily]

,tex .transformation rules

[warm shale]

ℝαμOmeganato5

[severe pagoda]

yea i know

but how far

[austere abyss]

pi/6

[high lily]

stretch is not the same as shift, these are very different

[severe pagoda]

i said stretch??

he is ragebaiting 100%

[austere abyss]

"horizontally stretch"

[severe pagoda]

i just want to know how far its stretched

[austere abyss]

it's not stretched

[severe pagoda]

whats the period

[austere abyss]

2pi

[severe pagoda]

2pi/6pi

[austere abyss]

lol

[severe pagoda]

.close

[obtuse pebble]

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[tall owl]

How would I start this question?

[wary badger]

increasing corresponds to y’ > 0 and at a decreasing rate means y’ is decreasing so y’’ < 0

[tall owl]

So do I take the derivative and then see which values are greater than and less than zero when plugged into y? When do I take the second derivative? Or am I comprehending this all wrong

[wary badger]

well first you should find when y’ > 0 since you’re already given y’

then you’d take the derivative yes but tbh you should have this memorized since it’s a logistic DE

[tall owl]

Oh wait ok

.close

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[sharp locust]

Can someone explain the jump from these lines?

[pulsar terrace]

is that a partial derivitive

[sharp locust]

Yes

[pulsar terrace]

which part

[sharp locust]

Sorry wrong one

[pulsar terrace]

oh

and its gone

[sharp locust]

Two sec

[pulsar terrace]

alr

[pulsar terrace]

what part are you confused about

[sharp locust]

This is my professors working, I’m just working through this example myself

[sharp locust]

So here I’m confused about the second line

How does it become that?

Here’s what I’ve done so far

[wary badger]

x = rcostheta and y = rsintheta

[pulsar terrace]

polar

[sharp locust]

Oh

I forgot about that💀

My bad

[wary badger]

all good

[sharp locust]

I’ll keep channel open cus I might need help in the next step

But let me try that rn

[obtuse pebble]

@sharp locust Has your question been resolved?

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[tawdry thicket]

Hello

[tawdry thicket]

how do I close this thing

[sage geode]

By typing .close

[tawdry thicket]

no

I messaged that earlier

let me send the screenshot

This is a theorem

[sage geode]

It looks like the ABC triangle is isosceles, use that and the facts that BAC + ACB + ABC = 180, ACP = BAC

[tawdry thicket]

named as Alternate segment theorem

I can solve that question

but

I have some problem understanding something

[sage geode]

What is it?

[tawdry thicket]

I am typing it

[obtuse pebble]

@tawdry thicket Has your question been resolved?

[tawdry thicket]

Question:
In the given diagram, PQ is a tangent to the circle at C, and AB is a chord of the circle. We know that ∠QCB (the angle between the tangent and the chord) is equal to ∠ABC (the angle subtended by the chord on the opposite side of the circle), as per the Alternate segment theorem in circles.
Now, my confusion arises and this is what we have concluded from the diagram that PQ and AB are not parallel.
So I remember that if we draw 2 parallel lines and draw a straight line intersecting them then alternate angles are equal
If we assume that PQ and AB are parallel then even if we neglect the circle then we would know that ∠ABC is equal to ∠QCB and we also know that in current situation, they are not parallel so if we were to make this thing in image from our assumption of the parallelism here (Remember that we are neglecting the circle here) we will rotate our PQ or AB, Now I say that we will be rotating the PQ, earlier when they were parallel ∠QCB = ∠ABC for obvious reasons of what we know about parallel lines with a line intersecting them. This is obvious that when we rotate that thing, ∠QCB's angle will change but the Alternate segment theorem says that they are still equal.
How in mathematics 2 different angles have one magnitude or be equal to one angle?
I am doomed, can someone end my confusion?

This is refined question by ChatGPT, IDK if it can help you understand whats the main thing but something is better than nothing

Question:

In the given diagram, PQ is a tangent to the circle at C, and AB is a chord of the circle. According to the Alternate Segment Theorem, ∠QCB (the angle between the tangent and the chord) is equal to ∠ABC (the angle subtended by the chord on the opposite side of the circle).

Now, here’s where my confusion lies:

We observe from the diagram that PQ and AB are not parallel. However, I recall that if we draw two parallel lines and intersect them with a transversal, the alternate angles formed are always equal. So, if we assume that PQ and AB were parallel and completely ignore the circle, we can conclude that ∠QCB = ∠ABC because of the alternate angles property.

But, in reality, PQ and AB are not parallel. If we were to create this situation by rotating PQ (while still ignoring the circle), the orientation of PQ changes, which should also change ∠QCB. Despite this, the Alternate Segment Theorem states that ∠QCB remains equal to ∠ABC.

My confusion is as follows:

1. If the orientation of PQ changes and ∠QCB adjusts accordingly, how can it still remain equal to ∠ABC, whose value seems fixed within the triangle?
2. Mathematically, how is it possible for two different angles (∠QCB and ∠ABC) to have the same magnitude, even when the conditions affecting one angle change?

Can someone explain how this works and resolve my confusion? I feel stuck!

Someone help

.close

[obtuse pebble]

Channel closed

Closed by @tawdry thicket

Use .reopen if this was a mistake.

[tawdry thicket]

.close

.reopen

[obtuse pebble]

✅

[high lily]

Now, my confusion arises and this is what we have concluded from the diagram that PQ and AB are not parallel.
how did you make that conclusion

[tawdry thicket]

Every single diagram on the internet shows its not parallel

[high lily]

diagrams aren't necessarily drawn to scale

[tawdry thicket]

hmm

[high lily]

based on the values given they are parallel
from altenate angles on parallel lines theorem (converse)

[tawdry thicket]

let

[high lily]

there are two different theorems involved here

[timid silo]

That happens if BC = AC

[tawdry thicket]

They must be parallel for what is happening here buttttttttttttt, internet is confusing

[high lily]

alternate segment theorem relates <BCQ and <BAC
alternate angles on parallel lines theorem relates <BCQ and <ABC

[tawdry thicket]

Oh

I understand it now

but it would have been better if they would have put 5 more minutes making those diagrams

[high lily]

diagrams are deliberately made like this

[tawdry thicket]

Why tho

[high lily]

to prevent people to determine answers just by "looking at it"
or using protractors/rulers

but rather use logical reasoning / established theorems/identities

[tawdry thicket]

I asked ChatGPT before this and it was breaking mathematics

Thank you for help

.help

[obtuse pebble]

Commands:

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

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

[tawdry thicket]

.close

[obtuse pebble]

Channel closed

Closed by @tawdry thicket

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[obtuse pebble]

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[rich patrol]

Where does the completeness of a Laplace transform come from?

[obtuse pebble]

@rich patrol Has your question been resolved?

[obtuse pebble]

@rich patrol Has your question been resolved?

[rich patrol]

Pls help?

[slim cove]

What do you mean by "completeness" here?

[rich patrol]

Meaning within the region of convergence. Given a signal x(t): X(s) is the laplace transform of x(t).

For a given s=sigma+i*omega within the ROC, if I set my sigma, the line along the imaginary axis paramterized by omega has an inverse and if I take the inverse laplace transform it will give me back my function

How do I know I can not lose information transforming from s domain to t domain

And further its like I can pick any sigma as long as it's within the ROC

And then go along a line on the imaginary component and that will totally describe my original function in t domain

It's like if the ROC exists and has a range (a,b) of sigma, s domain gives me infinitely many lines along the imaginary axis (omega) that can map back to my original function in t domain, so although my function has 1 representation in t domain, it can have infinitely many representations in s domain, pretty fascinating.

I think. Maybe I'm wrong on this

[rich patrol]

<@&286206848099549185> 👉👈🥺

[primal vine]

Let me se

[rich patrol]

Ty boss

[pearl grotto]

It = 62/^9

[tardy epoch]

what's ROC

[rich patrol]

Sorry, region of convergence

Meant to say "Within the region of convergence, given a signal x(t)..."

[obtuse pebble]

@rich patrol Has your question been resolved?

[obtuse pebble]

@rich patrol Has your question been resolved?

[obtuse pebble]

@rich patrol Has your question been resolved?

[obtuse pebble]

@rich patrol Has your question been resolved?

[slim cove]

This is by a theorem in complex analysis known as Cauchy's theorem. The integral of any complex differentiable function around a closed loop is equal to zero.

If you imagine a loop that starts at (0,-M), then goes to (sigma,-M), then goes to (sigma,M), then goes to (0,M), then finally back to (0,-M)

You can split it into four parts, two horizontal and two vertical

In many cases as you take the limit as M -> infinity, the horizontal contributions become negligible, and the vertical parts have to cancel out; therefore they have to be equal

So that's how you can justify that integrating along the imaginary axis will give you the same result as integrating along some other vertical line within the ROC

Hope this helps

[rich patrol]

Ohhhhhh thank you!!!!

I'm going to take a mathematical methods class in physics for complex analysis soon

So I should see this there too

[slim cove]

Awesome!

[rich patrol]

Thanks so much

[slim cove]

You're welcome

[vapid halo]

use .close to close the chat if your question has been answered goodnight!

[rich patrol]

Ahhh right thanks 🙏

.close

[obtuse pebble]

Channel closed

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[timid silo]

can anyone help me with calc 1

[lament raptor]

Sure

[lament raptor]

What's your question

[timid silo]

i need help with a study guide

could u join a call or osmething>

?

[lament raptor]

Tomorrow maybe

It's 1:30 am here

[obtuse pebble]

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[finite anchor]

Can anyone help with this

[pulsar terrace]

isnt BCD a equilateral triange

so all interior angles must be the same

[finite anchor]

Well do we know that from the question

It looks like it but idk exactly if it is or not

[pulsar terrace]

yea its not

you can use the law of cosines

to help you with the problem

[grand yoke]

Cos rule I think

[finite anchor]

lemme try

doesn't work it needs the angle that correlates to the side

When I do law of sines it just doesn't work out

[oak bolt]

you can get the height

[vapid halo]

Law of cosines?

Diamond I haven’t read the problem yet spare me

[oak bolt]

you can make a point E below C

with EC perpendicular to AB

[finite anchor]

like this or

[oak bolt]

yeah

now you can find CE and ED

[vapid halo]

E?

[finite anchor]

How could I

[oak bolt]

ACE is a 30-60-90 right triangle

[finite anchor]

how do we know

[oak bolt]

because you know it has 30 and 90

[finite anchor]

how do we know it has a 30

[oak bolt]

the problem says it is

at angle EAC

[finite anchor]

o wiat

you are doing the whole thing I was just using a section of the triangle

[oak bolt]

yeah

[finite anchor]

How does this help us get DCA

[oak bolt]

oh i thought we were working on part (a)

sorry T_T

[finite anchor]

oh well we are doing both

mb

[finite anchor]

With that how would that get us BCD

is it a bisector?

[oak bolt]

use the cosine of angle ECD

[finite anchor]

?

We found

angle ECA

right

[oak bolt]

yeah that one was 60

[oak bolt]

i mean you can use the inverse cosine to find ECA

ECD i mean

sorry

[finite anchor]

don't we just have this

for

ECD

[oak bolt]

CD is given in the problem

[finite anchor]

oh

yeah mb

[finite anchor]

I got this

Could I just law of cosines

sines

with the 90 degree anglke

[oak bolt]

i think so

[finite anchor]

why are we even doing this again

[oak bolt]

because ECD is half of BCD

[finite anchor]

oh ok

[finite anchor]

I think I got it, but I have to go to bed

gn

.close

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[fallow reef]

I've been working on this too long.. I'm making simple mistakes and misunderstanding simple things...

[fallow reef]

But I can't figure out where this 2 is coming from

And also, when integrating drd(Theta), isn't there supposed to be an additional "r" in the integrand?

[worn yoke]

could you show the entire question?

[worn yoke]

note that you are not actually integrating in polar coordinates, you are integrating over a parameterized surface where the parameters are named r and theta because it's based on polar coordinates. the factor that you have to multiply by is taken care of by the cross product.

[fallow reef]

Well, they've included it in other problems 🤷‍♂️

[worn yoke]

i can't really comment on that if you don't give an example

[fallow reef]

Oh well.

I'm busy with this particular problem at the moment, and I can't understand why it's being inconsistent with what I think I know

Namely, where'd the 2 come from

[worn yoke]

could you show the earlier part of the solution?

[worn yoke]

it explains the factor of 2 in the second paragraph

[fallow reef]

This is so fucking frustrating

And.. how does the cross multiplication eliminate the extra "r"?

As in.. when do I include rdrd(Theta) and when do I just include drd(Theta)?

[worn yoke]

you include the factor of r if you are integrating in polar coordinates. you instead include the cross product if you are integrating a parameterized surface. although r and theta appear to be polar coordinates, (because the parameterization is based on cylindrical coordinates), really they are parameters. you could call them u and v if that makes it clearer

[fallow reef]

It doesn't. x and y were swapped for rcos and rsin...

if these aren't polar coordinates (ok, cylindrical, but that's just polar extended along the z axis) then what are they?

Again, rdr was included in other problems

[worn yoke]

they are parameters used in the parameterization of a surface. similarly "t" is used to parameterize a curve

[fallow reef]

When do I include it? When do I leave it out? I don't understand the difference

[worn yoke]

in this case it is not a surface integral at all, but rather a triple integral (that is equal to a certain surface integral according to the divergence theorem)

[worn yoke]

maybe think of it as a difference between surface integrals and double/area and triple/volume integrals. another way to think about it is that the cross product does the same "job" that the r does elsewhere

[fallow reef]

I don't know

It seems too arbitrary.. just like the 2... I have no idea when to include it and when not to

My material doesn't explain where a lot of these things come from, or where they go when they're missing

.close

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[bright gazelle]

What’s this called in English?

[opaque dome]

Euclidian division ?

[bright gazelle]

When do you use it?

[opaque dome]

To simplify sqrt and also in some number theory

[bright gazelle]

Is it true you use to to find

least common divisor and greatest common multiple?

<@&286206848099549185>

[desert light]

what do you do in this?

[opaque dome]

Greatest common multiple ?

You meant least ?

And greatest common divisor

[nocturne bronze]

Euclidean algorithm is used to find HCF

[bright gazelle]

There are two @opaque dome

Least common multiple - you can use it to find what number can both numbers be multiplied to.
greatest common divisor - you have 2 numbers and you want to find the biggest number that can divide between those two numbers.

<@&286206848099549185>

[opaque dome]

So euclidian division is also useful for gcd

[bright gazelle]

@hexed rover This is important, learn that.

[obtuse pebble]

@bright gazelle Has your question been resolved?

[scenic oar]

https://study.com/skill/learn/how-to-simplify-a-square-root-explanation.html#:~:text=Simplifying a Square Root,paired inside the radical sign.

'simplifying a square root'

there's a step by step description of the process on the site.

some vocabulary, some examples. 🙂

"Simplifying a Square Root
Step 1: Find the prime factors of the number inside the radical sign.

Step 2: Group the factors into pairs.

Step 3: Pull out one integer outside the radical sign for each pair. Leave the other integers that could not be paired inside the radical sign.

Step 4: Multiply all the integers outside the radical sign and all the integers remaining inside the radical sign to get the final answer."

Quote from the site I just posted.

[bright gazelle]

I know how to do it

She tells you how to do it 🙂

You just need to find a number that it can be divided to

$sqrt{90}$

Is this acoustic

[scenic oar]

You started the channel asking for the name of the process. You later voted that your question wasn't answered. I provided the answer.

[bright gazelle]

$\sqrt{90}$

@scenic oar

Divide that with 10

$\sqrt{9}{10}$

$\sqrt{9}(10)$

[warm shale]

Shea

[scenic oar]

nope. that's wrong

[warm shale]

Wild123

[scenic oar]

I am not sure anymore what you're trying to find/do, or what help you need.

[bright gazelle]

Let’s say I have

[warm shale]

Shea

[bright gazelle]

How’d I simplify this?

[scenic oar]

take a pen and paper, and find the prime numbers that have the product 171.

[warm shale]

Wild123

[scenic oar]

$171=... \cdot ... \cdot ...$

[warm shale]

Wild123

[scenic oar]

found a prime divisor?

[bright gazelle]

So 13

$13\sqrt{3}$

[warm shale]

Shea

[scenic oar]

don't think so

[bright gazelle]

Okay

[warm shale]

Wild123

[bright gazelle]

Nah buddy

,w sqrt(171)

[warm shale]

Results provided by WolframAlpha

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

Requested by sheapey

[scenic oar]

😄

[bright gazelle]

3 then…

It’s hard tho

That’s why you use this method

[scenic oar]

I think you're trolling. I am slightly amused.

[bright gazelle]

how

lmfao

[scenic oar]

I asked you; take 171, divide it in primes.

You tell me 13, then solve it with wolfram alpha

xd like, what's going on? how do you plan to learn if you search for the answer

[bright gazelle]

I KNOW

What to do

I understand that

I’d have to guess all square roots table to guess that

Or use that method

123

[scenic oar]

yes, use the method in the video.

if you know the method...what's the question.

[bright gazelle]

123:3
41:41
1

3sqrt41

[scenic oar]

nop

we have a single 3.

so nothing goes out of the sqrt.

[bright gazelle]

Oh yeah

[warm shale]

Shea

[bright gazelle]

@scenic oar

[scenic oar]

yes...

[bright gazelle]

Mhm this is ez

[bright gazelle]

If there was +, could we add and do sqrt44?

[warm shale]

Wild123

[bright gazelle]

What if it was 41 and 41

[scenic oar]

what answer would you get?

[warm shale]

Wild123

[scenic oar]

x+x=?

[warm shale]

Shea

[scenic oar]

nope

how do you write x+x

shorter

x+x=?

[bright gazelle]

2x

[scenic oar]

yes

[bright gazelle]

Ooooh

[warm shale]

Wild123

Shea

[bright gazelle]

JINX!

Ok this is so easy

I completely forgot square roots

Never understood them

[scenic oar]

It's never too late to give square roots another chance

[bright gazelle]

They’re kinda useless ngl

[obtuse pebble]

@bright gazelle Has your question been resolved?

[obtuse pebble]

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[rocky glen]

i was thinking cosine rule
but it could and did get messy

ig for C = pi/2

it becomes

(0,0) (1,0) as the diameter

[obtuse pebble]

@rocky glen Has your question been resolved?

[rocky glen]

<@&286206848099549185>

[rocky glen]

am i fr never gonna get help wth

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@rocky glen Has your question been resolved?

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[oblique oasis]

hello

[oblique oasis]

it says since V = span(b1, ..., bm), we know that the system must be consistent

how do we know that its consistent?

i think i missed a couple theorems in my lectures, does anyone know how the system is always consistent?

[worn yoke]

the matrix equation is equivalent to the linear combination above

and since v is in V, and V = span(B), and by definition span is the set of all linear combinations of b1, ... bm, then we must have that v is in that set and therefore can be expressed as such a linear combination

[oblique oasis]

oh i think i get it

wait a second

so v is already a part of the basis set?

so thats why we can write v as a linear combination?

[worn yoke]

v is in the larger vector space V

[oblique oasis]

oh right

[worn yoke]

the span of B is the set of all vectors which can be written as linear combinations of the vectors in B

and we have that V = span(B)

[oblique oasis]

oh so since v is apart of V, and V=span(B), v is apart of the linear combinations of B?

[worn yoke]

yes

[oblique oasis]

got it got it!

im also a little confused about the last bit

it says

the system made from (b1 ...bm | v) has to be consistent

wait

does this mean b1 ... bm equals v?

also i was also wondering about the term consistent, i think it means that there is at least one solution. but if it can also have infinitely many solutions, how do we know that there is a pivot in every column

sorry i have a bunch of gaps in my knowledge

[worn yoke]

if you take the linear combination equation at the top and set each entry equal one at a time, then you will get a linear system of equations that can be written as B x = v, where B is the matrix formed by taking each b as a column, x consists of all the coefficients of the linear combination, and v is the same vector as before

then they are writing that system as an augmented matrix

in other words, finding a linear combination that equals v is the same as finding a solution to the linear system

[oblique oasis]

ohh i get it

wait how do we know theres one solution?

could there be infinite solutions?

[worn yoke]

if the columns of a matrix are linearly independent, there is a pivot in every column

[oblique oasis]

oh youre so right

i forgot about that

[worn yoke]

and they say that Theorem 2.37 says that means there is only one solution

[oblique oasis]

thank you, you answered all of my questions

i really appreciate it

🙂

.close

[obtuse pebble]

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[dire veldt]

precalc:

h(x) = 29 + 48.8 log (x+1)

therefore x=12 would create
h(12) = 29 + 48.8 log (12+1)
Can you convert log (12+1) into an exponent?

[drifting wraith]

no

you can't convert it

log12 + log 1 = log (12 × 1)

but you don;t have even that

[dire veldt]

Would it cause problems for me to add 12 and 1 inside the parenthesis?

[drifting wraith]

no

[dire veldt]

hmm
and log(13) should be convertable

[drifting wraith]

i don't see how

[dire veldt]

Ok then, what's my next step?

29 + 48.8 log(13)

[drifting wraith]

you can do
48.8 log (13) = log(13^48.8)

that's pointless

so no next step

[dire veldt]

hmm

i suppose the calculator can solve it (Ans = 83.360. . .)
Just wasnted to see how far i can simplify it

How about this other one?
16ln (10) + 31?

.close

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[kind valley]

can someone please help with 13?

[kind valley]

,rotate

[mossy fjord]

Which parts are you asking about?

[kind valley]

I guess I’m just confused on like b c e and f because they’re not on the graph meaning that they’re undefined I think?

[mossy fjord]

Do you know what f^-1 means?

[kind valley]

yes it’s the inverse

which I graphed

[mossy fjord]

Another way of thinking of inverses is that if you see f^-1(5), you can think "When the y value is 5, what is the x value?"

[kind valley]

oh

[mossy fjord]

So instead of finding 4 on the x axis and tracing up like part a, you can find 5 on the y-axis and trace sideways.

When you have inverse functions, it's just switching the x and y. Like the inverse of y=2x+3 is x=2y+3 and then you rearrange to isolate y again.

[kind valley]

but if you graph the f(-1)x the maximum value is 7 and the domain is [-3,4]

[mossy fjord]

The maximum value is 7 because that's where your table ends. In theory, the xy plane extends infinitely.

You just ran out of room because your teacher gave you that much.

[kind valley]

but on the graph theoretically of the f^-1x it wouldn’t fit into the domain or the max values

that’s why I’m confused

[mossy fjord]

Looking at this graph, what is the domain and range of the original function?

[kind valley]

-5 to -12 inclusive

[mossy fjord]

Right, and the range?

[kind valley]

-4 to 7 inclusive

oh bruh idk what I’m doing

okay

[mossy fjord]

Right, so if you swap those for the inverses, the new domain and range would also be swapped.

[kind valley]

okay

[mossy fjord]

So -4 to 7 domain, -5 to -12 range

It's possible that this is one your teacher wasn't aiming for you to draw.

[kind valley]

makes sense

wait

so the maximum value then wouldn’t be 7 though

of f^-1x

[mossy fjord]

Correct.

That's why I said your teacher didn't give you enough room to actually draw it.

[kind valley]

it would be -5

cause the domain and range are flipped

and maximum value is looking for the maximum y

[mossy fjord]

What are the domain and range of the inverse?

We talked about it briefly, just wanted to check that we're on the same page.

[kind valley]

this

inclusive

[mossy fjord]

Right.

Oh.

It's -5 to 12 range, mb.

[kind valley]

oh yea

ok that threw me off

so the max value would be -12

12

[mossy fjord]

My apologies.

Right. The maximum would be 12.

[kind valley]

The minimum would be -5

[mossy fjord]

RIght.

[kind valley]

wait

can you help me understand something

about inverses

[mossy fjord]

Sure.

[kind valley]

ok I’m proving that these are inverses but I got like something else and they’re supposed to be inverses

ok I’ll rework it

[mossy fjord]

To prove a function is the inverse of the other, just take the inverse of either function, and they should be equal. It looks like you made a compound function.

[kind valley]

yea and then u have to solve it because u do f(g(x)) and g(x)) and see if they = x

[mossy fjord]

f(g(x)) sounds like a lot of pain.

Since you have two x's.

[kind valley]

if one equals x you have to do the other anyway

[mossy fjord]

So $f(g(x)) = \frac{4g(x)-1}{2g(x)+3}$

Then you gotta do some substitutions

[kind valley]

uhhh

yeah

[warm shale]

Narutoes

[kind valley]

I’m working through it rn

[mossy fjord]

Gotcha

[kind valley]

cause at first I did a different method

of like making the same denominator but it’s easier to keep track and not make mistakes when u multiply by the denominator

to get rid of the complex fraction

I think

[mossy fjord]

From that, you went to $f(g(x)) = \frac{4(-\frac{3x+1}{2x-4})-1}{2(-\frac{3x+1}{2x-4})+3}$?

[warm shale]

Narutoes

[kind valley]

yep

[mossy fjord]

That is quite nasty indeed.

What did you do from there?

[kind valley]

multiply by the denominator of the complex fraction

then multiply by the 4 and 2 respectively and then simplify

ok I got x finally so I’ll do the other one now

[mossy fjord]

Nice

[kind valley]

alright wait could u check my work on this problem

it’s not making sense

,rotate

[mossy fjord]

Oh we got square roots now

So $f(g(x)) = \frac{\sqrt{g(x)^2+7}}{g(x)}$?

[warm shale]

Narutoes

[kind valley]

yes

[mossy fjord]

Then $f(g(x)) = \frac{\sqrt{\sqrt{\frac{7}{x^2-1}}^2+7}}{\sqrt{\frac{7}{x^2-1}}}$?

[warm shale]

Narutoes

[mossy fjord]

Gross.

[kind valley]

yes

well not exactly

the sqrt goes on the 7

[mossy fjord]

It goes on the whole numerator

[kind valley]

yes

[mossy fjord]

LaTeX just makes the line short for some reason.

[kind valley]

no I mean on the complex fraction

the sqrt is only on 7

[mossy fjord]

g(x) is all under a square root

Unless I missed something

$f(g(x)) = \frac{\sqrt{\frac{7}{x^2-1}+7}}{\sqrt{\frac{7}{x^2-1}}}$

[warm shale]

Narutoes

[mossy fjord]

$f(g(x)) = \frac{\sqrt{\frac{7}{x^2-1}+\frac{7(x^2-1)}{x^2-1}}}{\sqrt{\frac{7}{x^2-1}}}$

[warm shale]

Narutoes

[mossy fjord]

$f(g(x)) = \frac{\sqrt{\frac{7+7(x^2-1)}{x^2-1}}}{\sqrt{\frac{7}{x^2-1}}}$

[warm shale]

Narutoes

[mossy fjord]

Did you get something like this so far?

[obtuse pebble]

@kind valley Has your question been resolved?

[tight path]

You should cancel root of x^2 -1 and root7

[mossy fjord]

The next step would be $f(g(x)) = \frac{\sqrt{7+7x^2-7}}{\sqrt{7}}$

[warm shale]

Narutoes

[mossy fjord]

Which obviously gives you $f(g(x)) = \frac{\sqrt{7x^2}}{\sqrt{7}}$

[warm shale]

Narutoes