help-10

[misty terrace]

The \emph{general} construction of the center of the circle that is tangent to two given circles $\Gamma_1$ and $\Gamma_2$ (with centers $M_1$ and $M_2$ and radii $r_1$ and $r_2$ respectively) and a line $\ell$ where $\Gamma_1$ and $\Gamma_2$ are on the same side of $\ell$ that I found is pretty complicated, but bear with me:
\begin{enumerate}
\item Construct the line $k_1$ as the line on the other side of $\Gamma_1$ with $d(k_1, \ell) = r_1$ and $k_1 \parallel \ell$, and construct $k_2$ analogously.
\item Let $p_1$ be the parabola with focus $M_1$ and directrix $k_1$, and define $p_2$ analogously.
\item The intersection point of $p_1$ and $p_2$ is the desired center of the circle that is tangent to $\Gamma_1$, $\Gamma_2$ and $\ell$.
\end{enumerate}
This works since all points $P$ on $p_1$ suffice $d(P,\Gamma_1) = d(P,\ell)$ and similar. \\ \quad \\
In your case, however, $\Gamma_1$ and $\Gamma_2$ have the same radius $r$ and are of equal distance to $\ell$. In that case, there's nothing holding you back doing some Pythagoras and get the radius $R$ of the wanted circle, then getting the intersection of the circles with centers $M_1$ and $M_2$ with radius $r + R$. (Now that I think about it, the same might be possible for the general case as well... Let me think about that.)

https://cdn.discordapp.com/attachments/757583324632973403/937862098753568798/unknown.png

[warm shale]

Lance

[misty terrace]

the 'you' here is not you

wait maybe you can replicate the parabola construction

[small kestrel]

well I mean it's a task in past finals of my geometry class and we didn't go over parabolas and hyperbolas yet, so i believe there should be another solution, but thank you very much for your contribution ❤️

[misty terrace]

nah you just use the idea behind parabolas

lemme see if i can turn this into a construction

oh yeah you can

nice!

@small kestrel here is one of the circles!

either you can ask what's going on or why it works or figure it out yourself based on this one picture alone :p

the other is analogous

[small kestrel]

hmm, i'll try to understand, but anyways i found an interesting property as well - it turns out the centerpoint is a vertex of a regular pentagon with one of the sides being O_a O_b

[misty terrace]

by the power of a point (or just 'symmetry' lol) the midpoints of the circles that suffice must lie on the radical axis of the two circles, so by considering the two cases internally and externally tangent you see that there are only two options for the circles only

that's surprising, woah

[misty terrace]

it does not seem to be true in the general case though :(

[small kestrel]

shoot, so it was just a coincidence 😦

[misty terrace]

let me give the explicit steps of the construction btw

you can choose to prove the construction yourself if you wanna

[small kestrel]

I think I'm fine with you explaining this construction for now

shoot

didnt mean to include the pic

[misty terrace]

0. starting condition

[small kestrel]

yup, that's basically the pic we start with

[misty terrace]

1. perpendicular bisector of the two midpoints of the circles

[small kestrel]

that's where I ended

[misty terrace]

2. circle with same radius of the two circles at the intersection point of the perpbis and line

for externally tangent: draw a tangent line with the perpbis through the intersection point with the circle opposite from the two given circles

wait that's not even needed

nvm that

3. perpendicular bisector between intersection point and one of the given origins

4. result

for externally tangent: take the intersection point at the same side of the two given circles

thanks for the nice problem!

https://cdn.discordapp.com/emojis/622030615923130408.webp?size=48&quality=lossless

[small kestrel]

ooh! thank you so much! i love you!!

[misty terrace]

funny degenerate cases

interesting how this all checks out in general form

[small kestrel]

however, I just found out the teacher wants us to prove our constructions afterwards at our oral exam, would you mind explaining why this works?

[misty terrace]

oh yeah ofc! it's very much based on the construction of some points on a parabola

let me use some good letters for once

[small kestrel]

of course, feel free to!

[misty terrace]

basically, what you want to find for the first circle is the point E such that length EC is the same as the distances to w1 and w2

so you want to find E such that EC = EB - r

but that's the same as EC + r = EB

so using the perpendicular bisector of BD, we find E such that ED = EB, so EC + r = EB

which gives us exactly the point we need!

you can see it like this: you want to find the isosceles triangle ECF but don't have E or F, but you do have enough information for the extended isosceles triangle EDB

[small kestrel]

i seriously love

you

so

much

you just blew my mind

[misty terrace]

<3

[small kestrel]

it looks so simple and yet so challenging

[misty terrace]

uh huh

one of the reasons i like euclidean constructions

[small kestrel]

do you study maths or is that just a hobby of yours?

[misty terrace]

i'm only in high school :p

so a hobby ig

[small kestrel]

are you kidding me

im at uni and wasn't able to do it

well tbh, im so bad in geometry

[misty terrace]

well, i do get invested in math a lot

[small kestrel]

and i have to take three semesters of geometry, this is just the first one 🥲

[misty terrace]

oof

can't imagine the next ones

[small kestrel]

well the next ones are more about calculating

this one is about constructions

[misty terrace]

i have mixed feelings about that

[small kestrel]

geometry has always been my nemesis

[misty terrace]

yeah people are very divided about their opinions on geo

one of the trainers (at the dutch math training camp) said that there is some kind of 'click'

between not liking geo and loving it

https://cdn.discordapp.com/emojis/853555811842261063.webp?size=48&quality=lossless

guess i'm in my click phase

[small kestrel]

i guess my click phase has yet to come

i'm more into algebra, you know

or combinatorics

[misty terrace]

yeah i see

[small kestrel]

or number theory, but I don't have as much knowledge about it

[misty terrace]

i like combi but only when i see the proof

i've only done elementary nt/alg/combi/geo though

[small kestrel]

i'm taking the number theory course in like two years' time

nt?

[misty terrace]

number theory

[small kestrel]

oh

im dumb

oops

[misty terrace]

eh, you didn't know an abbreviation that isn't a reason to be dumb

https://cdn.discordapp.com/emojis/853555811842261063.webp?size=48&quality=lossless

[small kestrel]

i also tried digging into game theory, it's such an interesting topic but quite challenging as well

and one more question! what is the cool software you are using for constructions?

the one you were sending pictures from

[misty terrace]

ah that's actually a game about euclidean constructions

https://cdn.discordapp.com/emojis/715701345092370544.webp?size=48&quality=lossless

https://euclidea.xyz

[small kestrel]

EUCLIDEA

YES

I FOUND IT TODAY AFTER FORGETTING ABOUT IT

[misty terrace]

i used https://www.euclidea.xyz/en/game/packs/Eta/level/Heron for the black line

it's the easiest way to assure your euclidean construction is actually a construction

[small kestrel]

I found it for the first time about five years ago when i was in high school

and stumbled upon it again today

[misty terrace]

https://cdn.discordapp.com/emojis/853555811842261063.webp?size=48&quality=lossless

well, have fun replaying it!

by the way, i think the construction is generalizable? for any two circles

okay maybe not

but well, at least the two circles don't have to intersect

[small kestrel]

oh, that's interesting!

[misty terrace]

such a circle does seem to exist

oops

i think of my next message while i'm typing it and i write the wrong term

[misty terrace]

if you cheat a little and let two parabolas intersect in geogebra

uh

colors :D

ok well yeah, you can prove existence by two parabolas intersecting

but how to do it with straightedge and compass?

https://cdn.discordapp.com/emojis/944581311354511390.webp?size=48&quality=lossless

[small kestrel]

dude, i'm glad to know how to construct a circle using a compass, don't start on me with parabolas

[misty terrace]

oh yeah lemme clean up a bit

https://cdn.discordapp.com/emojis/853555811842261063.webp?size=48&quality=lossless

[small kestrel]

the last time i saw them was two years ago when i ended high school

[misty terrace]

parabolas are just a tool helping us get the actual euclidean construction here!

i wouldn't have come up with the solution to the original problem if it weren't for parabolas

at least this quickly

if we just knew how to construct the perpendicular line through J we'd be done by our previous construction

[small kestrel]

dude, i feel so dumb, the only thing i know about parabolas is that they are described by a relationship between two points (i hope it's not a hyperbola) such that x^2-y^2 is constant

[misty terrace]

uh, x^2 - y^2 = c is a hyperbola sadly :(

you can define a parabola by a focus point F and directrix d

then a parabola is the set of all points (locus) such that it has the same distance to F as d

(the directrix is a line and the focus is a point)

if only we knew how to intersect two parabolas with each other given focus and directrix...

[small kestrel]

oopsie

well as you can see, I know nothing about these things

[misty terrace]

i google how to construct the intersection of two parabolas

[small kestrel]

also - what do you call them in english? i mean circles, ellipses, parabolas and hyperbolas?

[misty terrace]

but i get equalling two polynomials of degree 2 with each other

[misty terrace]

exactly so!

[small kestrel]

in czech we call them "kuželosečky", basically "the things that you get by cutting a cone"

no, no, i mean the group of those four, do they have a name as a group?

[misty terrace]

ah

conics

[small kestrel]

great! that will be a simple term to remember, thanks!

[misty terrace]

https://cdn.discordapp.com/emojis/947850838313951293.webp?size=48&quality=lossless

google doesn't find anything about intersecting two parabolas given focus and directrix..?

[small kestrel]

oh, i have no idea, i'm trying to jot down the proof of the construction

[misty terrace]

ah, turns out this is one of the types of apollonius's problem

[small kestrel]

i am allergic to that name

[misty terrace]

https://cdn.discordapp.com/emojis/853555811842261063.webp?size=48&quality=lossless

damn

[small kestrel]

HOLY MOTHER OF JESUS

[misty terrace]

8 solutions in general

[small kestrel]

include orientated circles and lines and this whole masterpiece is screwed

[misty terrace]

orientated circles?

that sounds

awful?

[small kestrel]

it is

[misty terrace]

what is it though

[small kestrel]

well it means that you add orientations to your circles basically arrows pointing in a direction, the same goes with lines
the tangent objects have to agree with their orientations

look at this

this is a valid tangent line to the circle, right?

this is a valid tangent ray to a cycle as well

(a ray is an orientated line and a cycle is an orientated circle)

[misty terrace]

hmm

are there any nice properties about them?

[small kestrel]

but this

is not valid

they dont touch

[misty terrace]

that feels so unnatural

[small kestrel]

i mean there are, but i don't think i'm able to explain it in english tho, that is some advanced maths vocab

[misty terrace]

aw :(

[small kestrel]

i can try tho

oh, okay, i can't

i thought i found a dictionary that has those terms in it

but it doesn't

[misty terrace]

https://cdn.discordapp.com/emojis/778966798653587467.webp?size=48&quality=lossless

[small kestrel]

it doesn't matter tho, i'm not that skilled a geometer to be able to explain it even in czech, i think

[misty terrace]

ah ok that's fine :p

[small kestrel]

so i think i'll just close this thread as we don't talk about the topic anymore so that somebody else can use it

.close

[obtuse pebble]

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[quaint tangle]

i keep getting 3 or 10 and i was wondering if anyone could walk me through this problem

[timid silo]

can you just clarify what the o denotes

haven't dealt with that notation for a long time

[quaint tangle]

its basically g(f(x))

[small kestrel]

oh, sure, you just take the 2 as the argument of f(x)

[quaint tangle]

at least that is my understanding

[small kestrel]

and then you take this result as the argument of g(x)

so you get g(f(2))

[timid silo]

what a weird notation, why not just write g ( f ( 2 ) ), what's wrong with these conventions

sorry just ranting here

[small kestrel]

then, when you put the 2 into the f function, you get $g(\frac{2+3}{2})$

[warm shale]

Smethisko

[small kestrel]

then, that is just $g(\frac{5}{2})$

[warm shale]

Smethisko

[small kestrel]

and as g(x) = x^2-1

[royal shard]

it is actually a quite common notation

[small kestrel]

you will plug in the $\frac{5}{2}$ to get $(\frac{5}{2})^2-1=\frac{25}{4}-1=\frac{21}{4}$

[warm shale]

Smethisko

[timid silo]

no thanks I hate that notation

but yeah what the others said is right

you just do f

and then g

with what you get from f

[hexed agate]

Ever wondered why f and g were chosen as the letters?

[small kestrel]

i mean when you have 3 and more functions, the f(g(h(x))) just gets confusing

[timid silo]

as a hobby programer I disagree

[small kestrel]

f as in function, g because it's the next in the alphabet

[timid silo]

I like my brackets

because mathematicians are fagg*ts ?

can I say that here?

[small kestrel]

lol, calm down

[hexed agate]

Mathematicians love to fuck

[fringe minnow]

just to ask u to find gf(x)

[cedar lichen]

I wanna try! Okay, ahem... Mathematicians are n-

ice people :)

[devout solar]

No

[fringe minnow]

friendly reminder, this isnt #general

[hexed agate]

Says the one finding a gf

[quaint tangle]

thank you

.close

[obtuse pebble]

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[cedar lichen]

Sorry, just wanted to mock the usage of a slur

[obtuse pebble]

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[glass parrot]

isith thy correct

[cedar lichen]

Yes

[timid silo]

ye

[fringe minnow]

yes

[glass parrot]

yes

.close

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[little jacinth]

hiya, could someone help me workout this problem?

[timid silo]

how would you approach this?

[little jacinth]

one sec, sending my work

ok there it is

[timid silo]

ok so the first thing I'd do

is complain to your teacher

for using this notation

a^6 : a^4 -6a

is ambiguous

does the 6a belong in the numerator

or denominator?

[little jacinth]

lmao its a practice test for an exam i have coming up

[little jacinth]

wait i flipped it didnt i

[timid silo]

based on the given answers

I would say it's supposed to be

(a^6 / a^4) - 6a

but notation like that

[little jacinth]

my guess is the a^4-6a is denominator

[timid silo]

drives me up the wall

since there's 3 ways of interpreting that

[little jacinth]

yeaaah im not exactly a fan either trust me

[timid silo]

ok so in the first part

we just add up and subtract the exponents

to get -a

and that squared gives a^2

right?

[little jacinth]

right

[timid silo]

and for the second part

assuming my notation

it would be 5 * ( a^2 - 6a )

so thats 5 a^2 - 30 a

• that one a^2

gives 6 a^2 - 30 a

we factor out 6a

and get

6a * (a - 5)

which is the correct answer

so yeah, you missinterpreted the notation

because that notation should burn in hell

not your fault though, I don't blame you at all for making that mistake

[little jacinth]

lolololol

thank you so much

its miles more clear now

ngl the notation made me really unsure on how to interpret this

and it turned into a hot mess

[hexed agate]

To be devil's advocate, there is a clear difference between a^6:a^4 -6a and a^6:(a^4-6a)

[timid silo]

with brackets it's fine

clear

well defined

without brackets

it's an abomination

a terror among fiends

[solar trellis]

$\int_{\partial S} \bf{F}\cdot \dd{\bf{r}} = \iint_{S} \nabla\cross\bf{F} \cdot \dd{\bf{S}}$

[warm shale]

giannis_money

[timid silo]

don't get me started on nabla

[hexed agate]

What is nabla?

[solar trellis]

what's wrong with nabla

[devout solar]

tf

[timid silo]

(resisting urge to say nablabals)

[solar trellis]

sup?

[little jacinth]

.close

[obtuse pebble]

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[timid silo]

The table below gives the literacy rate (in percent) of 30 cities surveyed. What is the lower limit of the median class, the median of the data, and the mode of the data?

[timid silo]

This is what I've come so far in accordance of all the notes I have

Lower limit of the median class: 59.5
The median of the data: 62.8 (Rounded off the nearest tenths)
The mode of the data: 63.5

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[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

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<@&286206848099549185>

[timid silo]

<@&286206848099549185>

.close

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[vernal knoll]

I’ve eliminated the 4th option but I’m unsure still

[vernal knoll]

. Close

.close

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[dense rune]

Somebody please help me solve for R

What do i do wi th all those exponents

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@dense rune Has your question been resolved?

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<@&286206848099549185>

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[tacit mist]

I need help with the following problem for real analysis:

Give a proof of the composition of continuous functions using the sequential characterization
of continuity (from Abbot's real analysis).

My idea is to show that since f is continuous at c, and g of f is continuous at f of c, we have a sequence xn that shows continuity f(xn) converging to f(c) and likewise g(yn) converging to g(f(c)). I'm not sure where to go from there though.

[gilded needle]

i think you mean "and g is continuous at f of c"

suppose x_n is any sequence that converges to c, then by continuity of f, you know that f(x_n) converges to f(c)

now what can you say about g(f(x_n))?

[obtuse pebble]

@tacit mist Has your question been resolved?

[tacit mist]

since we know that g is continuous at f of c, we can see that g(f(x_n)) converges to g(f(c)), which would mean that we have shown that g of f is continuous at c?

[gilded needle]

yes

that's pretty much all there is to it

[tacit mist]

wait is that it?

wow, thank you!

[gilded needle]

yep

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[obtuse pebble]

@timid silo Has your question been resolved?

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@timid silo Has your question been resolved?

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[zinc escarp]

i just couldn't differentiate the given formula

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[weak torrent]

I am a bit confused with this question, Is it Row Echelon Form(REF) or Reduced Row Echelon Form(RREF)

[gilded needle]

well it's neither one as written

row 1's pivot is to the right of row 4's pivot, that's not allowed in row echelon form

[weak torrent]

thanks man

.close

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[hidden bronze]

a) i got the zscore of klare which is 0.83

b) and of simons which is 1

why is C answer as: klare

shouldnt it be simon

<@&286206848099549185>

can someone explain why

[novel knoll]

ambigious what they mean with better time in comparison

but yes Simon higher z-score

[hidden bronze]

so

is the textbook wrong

or is it just weird they asked for a comparrision

and if so, how would i calcualte which has better time in comparison

[novel knoll]

.

[obtuse pebble]

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[static sun]

8!-7!

[obtuse pebble]

Please don't occupy multiple help channels.

[static sun]

.close

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[cursive scroll]

we were given this problem for implicit differentiation and I'm having trouble in figuring out the equation I'm supposed to be using and in general just trying to illustrate the problem in my head

[cursive scroll]

the previous problems given to us already had an equation, but most of them were functions that can be graphed, this problem awfully seems like a one-dimensional equation though..? correct me if im wrong

I need some help in just illustrating the problem so I can understand it better

.close

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[main cargo]

idk how to work out P(AnB')

a and not B

on a venn diagram, how would that be shaded?

like that?

although

i dont think you can use a venn diagram for this

[slim leaf]

P(B') = 1 - 0.6 = 0.4

[main cargo]

how do i get anb' from there

i cant multiply by a

because its not indipendent

[slim leaf]

Yes

[main cargo]

i know this formula

i could use this

but then i need to know P(aUb')

i dont even know what aUb' means

a or not B?

[slim leaf]

Yeah

[main cargo]

how would that be on venn diagram?

at a guess?

or am i wrong

wait,

i think it might include the middle?

idk

[slim leaf]

Ah

I think you have to use P(A' | B) = 1 - P(A | B)

[obtuse pebble]

@main cargo Has your question been resolved?

[twilit ivy]

isnt that just P(A)-P(A∩B)

[main cargo]

yes?

[twilit ivy]

so thats just 0.5-0.2

[main cargo]

oh, is that indeed correct?

the diagram is P(AnB')

[twilit ivy]

thats correct

[main cargo]

thanks

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[sonic cape]

hi could anyone suggest a reason for why the solution applied e^x to determine whether the estimate is accurate or not

[sonic cape]

The solutions for part a and b are here if it helps

[dull rune]

its showing how useless the upperbound is

cause ln(e^4) = 4

[obtuse pebble]

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[ember vigil]

I’m confused with everything, can anyone help?

[midnight topaz]

use the Pythagoras theorem

[ember vigil]

Bruh im stupid as hell this is fine

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[ashen sierra]

How do I make dy/dx the subject?

[grizzled shore]

you would first need an equal sign

[tulip thicket]

a bit of algebra I thinj

[ashen sierra]

yh it's after the (1-dy/dx)

[grizzled shore]

that's an equal sign?

[ashen sierra]

i would assume so but i can't work it out myself

[ashen sierra]

lmao yeah

[brazen saddle]

,rotate

[brazen saddle]

Reckon this is the question

[ashen sierra]

yep same thing

[grizzled shore]

Does this look good

[ashen sierra]

yessir idk why i didnt think of that

thank you vm

[grizzled shore]

usually when things are complicated

substituting simple variables make it much easier on the eyes

[ashen sierra]

thank u, ill keep that in mind for future questions 🙂

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[cinder quiver]

i dont get it :(

[high lily]

which part don't you get

[cinder quiver]

all of it

[high lily]

this question is all about basic definitions/types of triangles

do you know what a scalene triangle is?

[cinder quiver]

no

[high lily]

do you know anything about types of triangles listed?

[cinder quiver]

its our first classwork about it i didnt seem to pay a lot of attention lol

[high lily]

do you have notes?

or just google each name in that list

[cinder quiver]

i have some but not about that topic i think ima check tho

[cinder quiver]

oh ok

yeah i still dont get it

is this part i did right?

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[hexed agate]

very goo question you got

[timid silo]

yo

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[tulip thicket]

(x+x^-1)^2

that would be x² + 2 + x-²

etc

factor the top side

yes

no

• not -

now factor out the x'es

divide etc

factor the numerator into parts that can be divided by the denom

[tulip thicket]

using this

numerator becomes (x+x^-1)² -3

,w factor x^2 + x^-2 -1

[warm shale]

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Requested by mixolydian mode#3126

[obtuse pebble]

@timid silo Has your question been resolved?

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@timid silo Has your question been resolved?

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[ionic kite]

So basically this was a meme but I felt like answering it anyway bc why not

This was my answer/response to the question

Is it correct?

[kind hawk]

well that depends on how you interpret the question. would it take more players less or more time to play the same thing (if we are already ignoring that it doesn't make sense)

[ionic kite]

It's a question from my exam board, the topic is direct and indirect proportion

But yh ur right overall we won't know

[kind hawk]

well it's a picture from a tweet. with the context that a teacher asked a bunch of these questions with the note that one of them makes no sense so their students wouldn't just blindly do some stuff without thinking

[ionic kite]

True Idk I just felt some urge to answer it 😭 😂

Anyway might as well close this

.close

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[brave bramble]

There is an answer, it's still 40 mins. It's not like more players make it faster

[kind hawk]

no the real answer is that there aren't enough people to play it 😛

[brave bramble]

For a question like this that actually makes sense:
"6 people can dig 4 holes in 2 hrs. How many holes can 10 people dig in 4 hours?"

[kind hawk]

(i don't know how many players that piece of music "needs")

[brave bramble]

Oh, yeah that makes sense too I accept it

[ionic kite]

😂 😂

Logically that makes sense

But maths wise no

U need to use a fOrMuLa to work it out

[obtuse pebble]

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[lyric iron]

Does anyone wanna try work through this with me? I'm lost.

[obtuse pebble]

@lyric iron Has your question been resolved?

[drowsy girder]

Pretty sure you just gotta find the limit

$$e^{\frac{-1}{x^2}}$$ as x approaches 0

[warm shale]

Pluton

[lyric iron]

but with the amount of difficulty everyone is experiencing, i think its deeper than that

IK the value of a is 0 since the function has to be continuous everywhere, and f(x) = a if x = 0

but i cant explain it using calculus

i am not smart enough

[versed turret]

For $f(x)$ to be continuous, $∀a ∈ D, f(a) = \lim_{x \to a} f(x)$. ($D$ is the domain.)

Continuity is a condition for differentiability, so we can see that $a = \lim_{x \to 0} f(x) = \lim_{x \to 0} e^{-1/x^2} = 0$

[warm shale]

castroploiin

castroploiin

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[normal wind]

can someone please help me understand this
Is the error same as standard error ?
if so, ho is error = |x bar - u | / sigma / sqrt(n)
isn't should be only sigma / sqrt(n)

[obtuse pebble]

@normal wind Has your question been resolved?

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[warm gust]

Does my approach look correct? Just having trouble with my limits of integration.

[obtuse pebble]

@warm gust Has your question been resolved?

[rigid pine]

r(r, theta) = <2r cos(theta), 4r sin(theta), 4 - 4r sin(theta) > (here I used z = 4 - y and used y(theta)).
Your position vector is a curve rather than a surface and it seems your doing a surface integral.

Then you'd have 0 <= r <= 1 and 0 <= theta <= 2pi.

[warm gust]

I am a bit lost as to what I am doing. I am trying to find the outward flux on the surface so I used the left side of the divergence theorem. I am just a little lost with how r is [0,1]

[obtuse pebble]

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[warm gust]

.reopen

[obtuse pebble]

✅

[rigid pine]

If you substitute x(r, theta) and y(r, theta) into the equation of the ellipse you find that 16r^2 = 16.

0 <= r <= 1. For, you cannot have a negative radius.

[warm gust]

Oh i see what I did wrong

[rigid pine]

Nice. 🙂

[warm gust]

Wait why is there an r in your parameterization for X and y.

[rigid pine]

You're parameterising a surface.

You need to get the points within the intersection curve.

[warm gust]

I thought ellipse was X=acost Y=bsint

[rigid pine]

Yes. You are within this ellipse. The surface is the part on the plane z = 4 - y within the elipse.

[warm gust]

Oh wait am I just calculating the intersection?

Oh nvm graphed it in Georgebra I think I get it now

Thank you

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[timid silo]

your math question in a clear, concise manner.* ?

[cursive bridge]

alright give me a second so i can type it out.

or should I just take a closer picture for each problem?

[timid silo]

i can help with the first one now

[cursive bridge]

alright

?

[hexed agate]

Do you know equation of a straight line?

Lucid?

[cursive bridge]

whats that again?

[hexed agate]

y=mx+b

[cursive bridge]

oh yeah i know that

[hexed agate]

So what is b?

So you know what it represents? And also what is m?

[cursive bridge]

to be honest with you I don't remember what they specifically mean

[hexed agate]

No worries

[cursive bridge]

cause we went to a new unit but im doing corrections right now so its been since the beginning of the year since ive done it

[hexed agate]

m is the slope of the graph. Outside USA we call it the gradient of the graph

[cursive bridge]

alright let me take notes

[hexed agate]

Do you know how to find it?

[cursive bridge]

could you refresh me?

on how to

[hexed agate]

(y2 - y1)/(x2-x1)

[cursive bridge]

isnt that the one of the versions of the Pythagorean theorem?

[hexed agate]

Not really

[cursive bridge]

whats this called again?

[hexed agate]

You are finding the slope which may also be referred as rise over run

[cursive bridge]

the first question is asking me to write an equation for the graph

[hexed agate]

Yes

Now let's do that

[cursive bridge]

alright

[hexed agate]

Which one is the x variable and which one is the variable y here?

Between c and t

[cursive bridge]

Umm how would I find that here

like whats in the middle?

[hexed agate]

Well, y is always the vertical line, the dependent variable

X is the horizontal line, the independent variable

[cursive bridge]

alright just marked those down as X and Y

[hexed agate]

We call them x and y axis

So c is the y axis and t is the x axis

Get it?

[cursive bridge]

yes

[hexed agate]

So your equation would look something like c=mt+b

But we are not done

We need to find m and b

[cursive bridge]

alright

[hexed agate]

One more thing do you know how to write a point on the graph?

[cursive bridge]

yes

write a dot where the specific coordinates are right?

[hexed agate]

Yes

[cursive bridge]

alright then i do

[hexed agate]

You right like (x, y)

Like (0, 0)

[cursive bridge]

yeah i know that

[hexed agate]

(500, 5000)

Good

Now to find the slope take any two points on the graph

Any two points you like

[cursive bridge]

500,1000

[hexed agate]

Write in brackets clearly

In the form of (x, y)

[cursive bridge]

ok

[hexed agate]

Like (0, 0) is a point

On your graph

[cursive bridge]

so dot (500,1000)?

[hexed agate]

Just to be clear the points where the line pass through

In (500,1000) the line doesn't pass through

(500, 5000) it does

Shall we pick that ?

[cursive bridge]

alright yeah

[hexed agate]

Ok so let (x1,y1) = (0,0) and (x2, y2) = (500,5000)

Now according to the slope formula I showed you prior

5000-0/500-0

Which is 10

So we got m = 10

Got it?

[cursive bridge]

im confused

[cursive bridge]

what do you do with this

[hexed agate]

I used the formula (y2-y1)/(x2-x1), plugged in the valued and did math?

[cursive bridge]

no no no no i mean like what do you do with that

like do i plug it in something

or is it just there for when we do something else

[hexed agate]

So you understand how I got m = 10?

[cursive bridge]

no

[hexed agate]

5000-0 is 5000

Divide it by 500-0

Which is like 5000/500

Which is 10

[cursive bridge]

ooooohh ok I get it now

[hexed agate]

So we found m

Now b is the y intercept

Y intercept means what is y when x = 0

In this example it is very easy because when x=0 y is also =0

So b = 0

Get it?

[cursive bridge]

yes

[hexed agate]

So as we know c=mt+b

We plug what we found

c=10t+0

c=10t

That is the equation of the graph

[cursive bridge]

alright thanks

makes way more sense then it did earlier lmao

[hexed agate]

Sorry if I was annoying. I was trying to explain super slow so that you get it

[cursive bridge]

it wasn't annoying at all

[timid silo]

hi idk if i already wrote here but

where can i find someone who will help me in stereometry

[cursive bridge]

i can close this channel for you so you can use it

[timid silo]

what does it mean

im new here so

[cursive bridge]

or go to the category that says Math help available

and type what you need help with in the two channels that are available

[timid silo]

thank u

[cursive bridge]

np

[obtuse pebble]

@cursive bridge Has your question been resolved?

[obtuse pebble]

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[lament frost]

Hi, I need help with the following question: The expression x^3 + 7x^2 – 3x + 12 has a remainder of –9 when divided by x + k. Find the
possible value(s) of k.

[grizzled shore]

Do you know polynomial long division

[fierce lagoon]

This is certainly a moment

Let's see

$$\frac{x^3+7x^2-3x+12}{x+k}$$
$$\frac{x^3+kx^2}{x+k}+\frac{(7-k)x^2-3x+12}{x+k}$$
$$x^2 + \frac{(7-k)x^2-3x+12}{x+k}$$
$$x^2 + \frac{(7-k)x^2+(7-k)kx}{x+k} + \frac{(k-4)x+12}{x+k}$$
$$x^2 + (7-k)x + \frac{(k-4)x+12}{x+k}$$
$$x^2 + (7-k)x + \frac{(k-4)x + (k-4)k}{x+k} + \frac{(16-k)}{x+k}$$
$$x^2 + (7-k)x + (k-4) + \frac{16-k}{x+k}$$

I'm gonna edit this as I go; feel free to look at my work as I edit this

[obtuse pebble]

@lament frost Has your question been resolved?

[warm shale]

Umbraleviathan

[fierce lagoon]

I think you know what to do from there

Ping me if you have questions

[obtuse pebble]

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[obtuse pebble]

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[red lark]

how did we arrive at the third last step?

[rancid dust]

dont helpers

its not been 15 minutes yet

[red lark]

sorry

[slate cosmos]

Cross multiplied?

[red lark]

but like

where did the tan come from?

[warm canopy]

1/cotx = tanx

If I'm correct about where you're confused

[red lark]

are we cross multiplying it?

ahhh ye

[warm canopy]

If its line 3 to line 4 confusing you then that's the reason

[red lark]

ye

i get it now

thanks @warm canopy

.close

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[tacit briar]

What's the expected value of the product of a six-sided die times an eight-sided die?

[tacit briar]

i basically find the EV of the six sided die, and the EV of the eight sided die

<@&286206848099549185>

[tardy epoch]

Calculate the probability mass function

[tacit briar]

i than multiplied both ev of six sided die and eight sided die together

would that be right?

EV of six sided die is 21/6

EV of eight sided die is 29/6

(EV of six sided die)(EV of eight sided die) = 203/12

@tardy epoch

[obtuse pebble]

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[odd frigate]

I’ve tried to solve this problem but I keep getting the incorrect answer what did I do wrong?

[obtuse pebble]

@odd frigate Has your question been resolved?

[odd frigate]

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[lime cypress]

I would have done u-substitution

Wait nvm

[obtuse pebble]

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@odd frigate Has your question been resolved?

[past nexus]

[finite vector]

What's the answer ur getting?

As per ur photo the answer's correct

But u haven't highlighted the same in the options

Probably missed a - sign

From ur last step

[brazen saddle]

Messy work , do it again properly

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[timid silo]

$2x + \frac{2}{x} = 3,$ find $x^3 + \frac{1}{x^3} + 3$

[warm shale]

kube

[timid silo]

so first I turned it into a quadratic

$2x^2 -3x + 2 = 0$

[warm shale]

kube

[timid silo]

but when I tried to solve it using the formula the determinant becomes negative

[devout solar]

so we are supposed to find what x³+1/x³+3 is?

[timid silo]

yes

I'm wondering if I should solve it the obvious way (find x then substitute it for x^3) or if there's another easier way to do it

does x have to be real?

oh

I'm dumb

there's an easier way

[devout solar]

I know

[timid silo]

substitute x^3 = t

wait no

nvm

ok

I thought about functions

the left expression can be written as x + 1/x = 3/2

[timid silo]

factor out 2

which is very similiar to the right one

[timid silo]

hmm yes

find the value of x+1/x

so I thought maybe some sneaky sub or something

then find of x^2+1/x^2

multply them

and form an equation

u will observe a pattern

[fierce lagoon]

Bruh this is easy

[compact shadow]

$x^{3}+\frac{1}{x^{3}}+3=(x+\frac{1}{x})^{3}-3(x+\frac{1}{x})+3$

[warm shale]

Cogwheels of the mind

[fierce lagoon]

Ihold on hold on

[compact shadow]

And you know that x+1/x=3/2

[timid silo]

where did you pull out this identity from wth?

$x^2 + \frac{1}{x^2} = \frac{1}{4}?$

[warm shale]

kube

[fierce lagoon]

Divide the left side by 2, you get

$$x+ \frac{1}{x} = 1.5$$

The right side is

$$x^2\left(x+ \frac{1}{x}\right)+ 3$$

$$1.5x^2 + 3$$

Wouldn't this be easier

[compact shadow]

So (3/2)^3-3(3/2)+3=15/8

[fierce lagoon]

Oh wait

[timid silo]

you can't factor out that x^2 like that tho

[fierce lagoon]

Im actually

Im atcuslly@mentally stupid

I did not just say that x^2(1/x) = (1/x^3)

[fierce lagoon]

You be quiet lmao

[timid silo]

ahh I got so close

[compact shadow]

I don’t think x^2+1/x^2 is needed, you can solve it surely

[timid silo]

$x^3 + \frac{1}{x^3} + 3 = \frac{15}{8}$

[warm shale]

kube

[timid silo]

the thing is

15/8 is not one of the options

but 15/4 is

it's probably a silly mistake I'll check

[compact shadow]

I got 15/8

And didn’t solve for x

[timid silo]

,w 1/4*3/2-3/2+3

[warm shale]

Results provided by WolframAlpha

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

Requested by pi=e#5213

[timid silo]

15/8 as well

[compact shadow]

I don’t see how this gives another answer other than 15/8

[timid silo]

i get 21/4 lol

[nocturne minnow]

I got pi/e

[timid silo]

actually no I'm dumb

nvm

ok now I got 35/8

I just cubed ( x + 1/x)

to get x^3 + 3 x + 3 1/x + 1/x^3

and then simplified that to

I did it like

$x^2+\frac{1}{x^2}=\frac{1}{4}$

$(x+\frac{1}{x})(x^2+\frac{1}{x^2})=\frac{3}{8}$

$\implies x^3+\frac{1}{x^3}+x+\frac{1}{x}=\frac{3}{8}$

$\implies x^3+\frac{1}{x^3}=\frac{3}{8}-\frac{3}{2}$

$\implies x^3+\frac{1}{x^3}+3=\frac{15}{8}$

(x^3 + 1/x^3) + 3 * (x + 1/x) = 27 / 8

[compact shadow]

Why no one wants to take a look…

[timid silo]

plugged in x + 1/x = 3/2

[timid silo]

Trust me i have

I have

[compact shadow]

Thanks

[timid silo]

and then from there got x^3 + 1/x^3

where did I go wrong to get this answer tho?

[warm shale]

a^3+b^3=c^3 has a whole no sol.

[timid silo]

I think the standard way to do this in school

is to cube it

like u did

[timid silo]

but i always prefer this

ok now I get 3 as an answer after finding a mistake

[timid silo]

ur mistakes are going like numbers in colartz conjecture

@compact shadow doesn't the formula u used also comes from adding and subtracting the thing left outside the cube and factoring one part right?

I really just did

8/2 = 3

I need profesional help

[timid silo]

take a break

take some rest

highly reccomended

[timid silo]

yeah this is exactly how I did it

so are we all coming to an agreement that the answer in my book is wrong lol

[timid silo]

yea

oh crap

literally all of us

I didn't notice

came with the same thing

there is an option for none of these

lmao

[timid silo]

[timid silo]

I think cubing it is kinda tedious I always avoid cubing

[timid silo]

yes

yeah but in this case

it sorts out nicely

I think Cogs method

is most suitable

if it works it works

still tedious because you have to go through the whole identity again

anyways

thank you all

.close

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[obtuse pebble]

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[timid silo]

ok i have question

[timid silo]

W = 2/s squared + m

i’m having trouble making m the subject

the squared s is putting me off

ignored 💀

dumbass💀

[warm canopy]

? Rude

[timid silo]

trolling trolling

[warm canopy]

?

[high lily]

$W= \frac{2}{s^2} + m$?

[nocturne minnow]

You can close the channel, OP got muted, more info in #help-9

[warm shale]

ℝamonov

[nocturne minnow]

Or I think ramonov can maybe close it too

[warm canopy]

Aight

.close

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[warm canopy]

Wow help 9 was a mad read

[nocturne minnow]

Yep that user was rude af

And I literally linked the help channel 3 times

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[paper coyote]

If we assume there is a pdf such that
f(x) = a.g(x) + b.h(x)

where g(x) and h(x) happen to be exponential distributions with rate L1 and L2, and a and b are constants set such that the integral from 0 to infinity of f(x) = 1.

I asked a question about the expectation a couple of days ago, and was told to consider the unconscious statisician and linearity of integration.

I did this, and I'm happy that by E[kx] = kE[x] and E[x] = 1/lambda for exponential distr, I can say that in this particular case:
E[x] = a.1/L1 + b.1/L2.

I also did this by integrating from 0 to inifinity of x.f(x) in Wolfram Alpha and I get the same number, so I'm pretty confident that's right.

Now for a bit of an inductive leap. Seeing as Var(x) is defined in terms of E[X] (or E[X^2]) I thought I could use the same principle for the variance. Knowing that Var(x) = 1/Lambda^2 for the exponential distribution, and that Var(kx) = k^2.Var(x), I tried
Var(x) = a^2 . 1/L1^2 + b^2 . 1/L2^2

This gets me 1/2 of what I calculate by just generating lots of samples and working it numerically. So, should there be a 2x somewhere, and if so, why? Should I consider somehow the Covariance even though there's only one 'x' ?

[paper coyote]

That's perhaps unclear, I'll try to reformat in latex

$Var(x) = a^{2} \cdot \frac{1}{\lambda_1^{2}} + b^{2} \cdot \frac{1}{\lambda_2^{2}} $

[obtuse pebble]

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[radiant juniper]

2 +2

[dawn meteor]

👀

[radiant juniper]

mhm

[dawn meteor]

what base

[radiant juniper]

wdym'

.close

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[hexed agate]

yes simpify

what have you tried?

[viral orchid]

im blank

[hexed agate]

you add two fractions

how do you add two fractions?

like 5/7 +9/8?

[viral orchid]

yes

[warm canopy]

how

[obtuse pebble]

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[languid valley]

Integral arctan (8x-4) dx = x * arctan (8x-4) - 1/2 (8x-4) * ln (x²+(8x-4)²) is this correct?

[kind hawk]

check with wolfram alpha

you forgot the constant of integration

[languid valley]

Oh yeah

but idk Wolfram isnt really helping me

[kind hawk]

,w integrate arctan(8x-4)

[warm shale]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Jun 2, 2022, 2:36 PM

[languid valley]

uh

[kind hawk]

difference of that and your answer is not a constant

[languid valley]

no I think there is a difference

I totally forgot we got send an answers paper anyway but I'm still confused

the answer is 1/8 (8x - 4)*arctan(8x-4)-1/16 ln((8x-4)²+1)+c

Im just confused where the 1/8 comes from

[kind hawk]

inner derivative probably

[languid valley]

wdym exactly

[kind hawk]

well somewhere in that whole mess when differentiating we will get the inner derivative of 8x-4

to compensate for that we need a factor of 1/8

[languid valley]

hm well the easiest way to solve it is with a book anyway

[kind hawk]

or from the other direction when you do the u-sub u=8x-4 you also get 1/8

[languid valley]

I'm not studying mathematics I'm an engineering student so we're more than fine to use books if we can find the integrations in there

but this should be it right?

[kind hawk]

if it says it somewhere in a book, probably

I'm notgonna check it

[languid valley]

but thats still missing the 1/8

or even how you get to the term ((8x-4)²+1)

[kind hawk]

you use a u-sub u=8x-4 to get integral of 1/8 arctan(u) du

then you apply that formula with x=u and a=1

and then you sub in u=8x-4 again

[languid valley]

oh ok finally

now it makes sense

I had an issue with another one however

Integral (tan^3 x)/(1-sin² x) dx

[obtuse pebble]

@languid valley Has your question been resolved?

[kind hawk]

dunno. write 1-sin^2=cos^2. then maybe you have some integral of the form sin^n/cos^m in your book?

or u=tan(x) ?

yeah probably u=tan(x)

[languid valley]

Yea well it goes correctly into 1/4tan^4x+c

I just

cos isnt a derivative of tan

he does say it has something to do with addition theorems

[kind hawk]

well what is the derivative of tan(x)

[languid valley]

wasnt it something weird