help-0

[buoyant saddle]

then drop out

wait but you’ve been in uni for 5?

actually enrolled

[iron dragon]

I just come and go

I've never done more than 2 classes at a time in a semester

[buoyant saddle]

ahh

hmm there should be a credit requirement

my school has a minimum credit hour

per semester

anyways

i don’t think you should drop so soon

just because you don’t understand it immediately

[iron dragon]

but it's only going to get harder

it doesn't get easier

it builds on the early stuff

[buoyant saddle]

that’s not a good mindset though

if you haven’t gone through with it how would you know

you should add calc 2 back

[iron dragon]

I know sometimes I think if I just tried not to drop I might have ended up passing

[buoyant saddle]

maybe take 1 math class at a time

[iron dragon]

and some people fail and the re take it and pass

the thing is, if I really find out I can't do the math I'm going to have to do something else

[buoyant saddle]

when did you take calc 1

[iron dragon]

9 years ago

[buoyant saddle]

oh boy

so you probably forgot most of it

[buoyant saddle]

i mean how good are you at cs?

[iron dragon]

yea I remember a little bit of limits and derivatives, just the simple derivatives like cos, sin, ln(x)

[buoyant saddle]

have you taken cs classes

[iron dragon]

but when it comes to trigonometric derivatives and trigonometric identities it seems so much to remember

should I re do Calculus 1?

[buoyant saddle]

i wouldn’t pay for it

[iron dragon]

yes I have taken as many CS classes as possible, they won't let me take anymore

[buoyant saddle]

you can learn it quite easily

[buoyant saddle]

and you’ve done well?

without the math classes

are the math classes all you need

[iron dragon]

A's in all computer science classes

well I need the math classes to get into the higher level computer science classes

[buoyant saddle]

ahh right prerequisites

[iron dragon]

first of all, they changed the rules

before you could take CS classes without being in the department

now there is a high demand for CS classes so you have to be accepted into the department / program

[buoyant saddle]

yea my uni has a "cs major" requirement for some classes

[iron dragon]

but on top of that, even if there was no department rule, the higher level CS classes need algorithms classes

and the algorithms classes want Discreet Math as a pre req

and some need Linear Algebra

and the Calculus is just there because

they want Calc 1, Calc 2 and Calc 3

[buoyant saddle]

have you used khan academy

for calc 1 and 2 specifically

[iron dragon]

yea and youtube

but the class moves so fast I don't have time to learn it

i guess I should re do Calc 1 derivatives?

[buoyant saddle]

yea start there

[iron dragon]

so then it makes u substitution for integrals easier?

[buoyant saddle]

you can take discrete math though

you don’t need the content from previous courses

[iron dragon]

I know, but I don't know what's going on

[buoyant saddle]

watch dr trefor

on youtube

he has a playlist

[iron dragon]

yea

[buoyant saddle]

he’s awesome

have you watched his playlist?

[iron dragon]

yea tried

[buoyant saddle]

have you read books too?

[iron dragon]

yea

they give us a pdf which is the whole course

[buoyant saddle]

hmm idk

i have to head off to bed though

you should stick it out

for at least one of the math classes

i didn’t see

yea

i know

ok

[iron dragon]

but knief

[buoyant saddle]

yea

[iron dragon]

I see you on here a lot

[buoyant saddle]

mhm

[iron dragon]

and you've helped me before

[buoyant saddle]

i believe so

[iron dragon]

do you tutor

[buoyant saddle]

not professionally no

i should

[iron dragon]

what degree are you trying to do?

[buoyant saddle]

math/applied math

double major

and i might do cs minor

[iron dragon]

I could help you with the CS if you need

[buoyant saddle]

if i need help i’ll ask

[iron dragon]

k can I ask one thing

[buoyant saddle]

yea

[iron dragon]

how did you get good at math, were you always good from highschool?

[buoyant saddle]

i mean this isn’t the answer you want to hear but i was always unique as a kid

[iron dragon]

haha it might be the answer I need to hear though

[buoyant saddle]

math was always my gift so to speak

[iron dragon]

I'm thinking I might not be the math type

[buoyant saddle]

and i just enjoyed it

[buoyant saddle]

don’t need to be

[iron dragon]

because when I was young I never did math homework

I dropped out of high school

and then upgraded as an adult

[buoyant saddle]

oh damn at least you’re back now

[iron dragon]

did math 11 and 12

and then got into university

[buoyant saddle]

that definitely contributes to your problems now i’m sure

[iron dragon]

I know algebra though

it's the trig identities, and trigonometric integrals / derivatives

there's so much to remember

[buoyant saddle]

id recommend a workbook for calc

you might have seen it

it’s like 10 bucks

i used it as an intro

before i bought textbooks

chris mcmullen i believe

calculus workbook

there isn’t much you can do about trig other than memorizing it

and the best way to do that is just doing problems with them

try not to have access to the derivatives on hand

only look when you forget and can’t remember for the life of you

[iron dragon]

yea I was trying and the class moves so fast, they were on normal integrals, then u subtitution, then trig integrals then integration by parts

and I'm just trying to still do u substitution

[buoyant saddle]

hmm and khan academy wasn’t helpful?

and neither was youtube

[iron dragon]

I never tried khan academy or youtube

I was just reading the textbook and looking at the slides

[buoyant saddle]

oh that could be your problem

[iron dragon]

but also just doing questions and then using the integral calculator to show me steps on the internet

[buoyant saddle]

binge youtube videos

it’s the best resource

trust me

[iron dragon]

see when I was in Discreet Math + Calculus 2 + 2 electives

there's not enough time

like I wake up, make some food

and then 2 hours for Calc, still don' get it

[buoyant saddle]

do you work too

[iron dragon]

move to Discreet, 2 hours don't get it

no

just school

[buoyant saddle]

id recommend blackpenredpen for sure

of course dr trefor

khan academy

organic chemistry tutor isn’t bad

i’m sure there’s more

[iron dragon]

do you think for the next 3 months

I should just do Pre Calculus, Calculus 1, Calculus 2, Discreet Math

[buoyant saddle]

professor leonard

[iron dragon]

at home without university

[buoyant saddle]

idk about this

i think you should pick on math class at least

and stick to it

dedicate all your time to it

[iron dragon]

but then when the semester is over

say I pick Discreet Math

when the semester is over, it's time for Calculus 2 and I won't be ready

[buoyant saddle]

if you don’t enroll in uni and wait you’ll never be ready

[iron dragon]

should I only take 1 math class per semester or should I take 2 at a time?

[buoyant saddle]

take 1

if it’s not your strong suit then don’t overwhelm yourself

[iron dragon]

hmm maybe I should try getting back into Discreet Math then

I havn't lost any marks yet

[buoyant saddle]

i agree

[iron dragon]

I did well on the first quiz and 2 mini assingments

[buoyant saddle]

there are plenty of resources you can use

[iron dragon]

but now there's a bigger assignment due tomorrow and I'm struggling

[buoyant saddle]

there’s always math discord

[iron dragon]

if my Calculus 1 is weak, because it's been 9 years

I couldn't just hop back into Calculus 2 could I?

[buoyant saddle]

ask the uni

or you mean

with the knowledge you have

[iron dragon]

the professor says if I am struggling in Calc 2 I should re do Calculus 1

[buoyant saddle]

i disagree

do it on your own time

[iron dragon]

should I pay for the course agains at university or nah

[buoyant saddle]

don’t pay again

you can learn at home

[iron dragon]

I was thinking the tests and assingments might have me workiugn harder

[buoyant saddle]

i mean that’s true

there is incentive

but calc 1 and 2 aren’t much different

calc 1 is contained in calc 2

they’re both single variable calculus

[iron dragon]

to get ready for Calc 2, would you re do Calc 1 start to finish

like limits and everything

or just do derivatives

[buoyant saddle]

derivatives

all you need

no need for limits and all that

from what i understand calc 2 is just integration techniques and infinite series

[iron dragon]

when I get to Calculus 3 will I need limits

[buoyant saddle]

yea there are limits in calc 3

different from the limits you’re doing

limits with multiple variables

calc 3 is quite different from calc 1 and 2

multiple variables

i’m gonna head off to bed

[iron dragon]

k one last thing

you think I should try to get into Discreet Math at University and then learn the Calc 1 / 2 on my own time at home

[buoyant saddle]

yes

don’t waste any more time

[iron dragon]

and if I can't get back into the university class just learn them both at home?

and then take both classes immediately next semester?

[buoyant saddle]

hope you can

but if you can’t then yea

[iron dragon]

just one last thing, I was thinking that if I'm in the discreet math at university it will take a lot of time and focus

and I won't learn the Calculus all that much

and the thing is, the Calculus is the one needed to get into the CS Department

so if I can't do the Calculus the Discreet doesn't matter

[buoyant saddle]

then do calculus

either or

pick one

and dedicate your time to it

[iron dragon]

ok I'll start doing some practice problems

and then if I'm doing well I'll email the professor to see if I can get back in the class

but I just remember trying to do a integral and someone said you need a double angle trig identity for that

and that's what's discouraging, not only do you need to know the derivatives / integrals... but you also need trig identities

and the trig identities there's so many so it's not always easy to see

[buoyant saddle]

you don’t need that many

double angle is important

the pythagorean identities

and that’s about it

[iron dragon]

I just looked up

integral sqroot(1 - cosx)

and the integral calculator says you need a hyperbolic trig identity

but yea I'll keep working hard and see what I can do

[buoyant saddle]

was this from your assignment?

[iron dragon]

I hope I don't need to switch degrees

that was just a question on the tutorial

before the assignment

[buoyant saddle]

ahh

best of luck though man

[iron dragon]

haha thanks

I'll be around lots

thanks for the help

[buoyant saddle]

you’re welcome

[iron dragon]

@stark crater can you show me what you were saying for part c?

[lone heart]

@iron dragon Has your question been resolved?

[small lance]

damn, my major is cs too

ig we can help each others on calculus, since I'm poor with Math 😭

[lone heart]

@iron dragon Has your question been resolved?

[iron dragon]

yes for sure! hahaha

[lone heart]

@iron dragon Has your question been resolved?

[lone heart]

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[quartz hedge]

hello yes i have a problem that i need solving. its not a textbook problem but i ran into it while trying to calculate somthing out, i cant seem to get the hang of the question.

to recap the thing that i need to do
i have the radius of r4 (most outside circle)
and i need the radius of r1 (inside) r2 r3
all the circles are toughing and for reference its in a nonagon

also here is picture

[rustic coral]

Do trig on this triangle

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@quartz hedge Has your question been resolved?

[lone heart]

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[twilit frigate]

wht is this

[mild fractal]

this situation is identical to 'n' die being thrown together such that only 3 particular numbers show up, each showing up atleast once

[mild fractal]

uhh a probability question?

[twilit frigate]

no i mean wht kind of stupid question is this

nvm im going insane

[mild fractal]

why is it stupid?

[uneven isle]

well lets first break it down

how many ways can u choose 3 numbers out of 6

[mild fractal]

gimme a second am typing my approach

6c3

[uneven isle]

ok

[mild fractal]

so total combinations is 6^n

number of ways to choose a specific set of 3 numbers is 6C3

now let those numbers be x,y,z
we now have 'n' spaces such that we have to fill each exactly once, and x,y,z must have appeared atleast once;
so we take out 3 spaces, and fill them with x,y,z.
now we have n-3 spaces, each of which can hold either x,y,z, total combinations: 3^n-3

so answer should be (6C3 . 3^n-3)/6^n

but well this is wrong apparently

[deep plaza]

check dms

[uneven isle]

u double counted i think

[mild fractal]

where tho-

[uneven isle]

ok so if 3 numbers appear right, we know each one at least appears once

[mild fractal]

yup

so shd i jus go w 3^n?

[uneven isle]

im just going ot list the conditions we have to satisfy

[mild fractal]

yesyes

[uneven isle]

so that we can get our favourable outcomes we want

actually, does principle of inclusion and exclusion mea nanything to u

[mild fractal]

yeahhh it does

the solution used that only

[uneven isle]

we can use it to find the overcounted cases

[mild fractal]

it's just i dont like applying it as a formula, i wanna know alternate ways of doing this problem without it

[uneven isle]

hmm

ok

so if we only care about how many ways there are such that either 1, 2 or 3 of the numbers appear it is just 3^n so far

[mild fractal]

indeed

[uneven isle]

we have to subtract the follow overcounted cases to satisfy the question

1. only the 1st number appear (x)
2. only 2nd number appear (y)
3. only 3rd number appear (z)

• only 1st and 2nd, or only 1st and 3rd or only 2nd or 3rd

equivelantly, it is the union of (1) (2) and (3) that are overcounted

like

[mild fractal]

indeed indeed

yes you're basically going over why we need the inclusion exclusion principle

[uneven isle]

yeah

[mild fractal]

but like the entire point of seperating 3 spaces

was to avoid the overcounting of cases

if i took 3 spaces aside and filled one w x, one w y and one w z

then it doesnt matter if the rest of em have only x,only y only z or only x,y , only y,z and so on

why am i still gettin the wrong answer-

[fickle heath]

It does

[uneven isle]

oh actually

i think i realised what u are doing

[fickle heath]

Say you put 2,1,3 in the first three spaces

In the last three spaces, you count every possible combinations with {1,2,3}

But if you have a 2 in one of the last three spaces, it means you didn't need it in the first three

So you could have put 1,1,3 in the first three in the case you have a 2 in the last three

[uneven isle]

yeah u cant fix them at the start i htink

they occur anywhere

[mild fractal]

ohhh im kind of getting what ur tryna say

[fickle heath]

Are you familiar with stars and bars?

[mild fractal]

yeah

[fickle heath]

Do that

You have some number of 1s (at least 1), some number of 2s, etc.

6 total

So three bins with at least one element each, 6 elements total

[mild fractal]

well alright

thanks, you both!

.close

[lone heart]

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[fervent heron]

5. Write the left hand side of the differential equation
   (2x + y)+(x + 2y) dy
   dx
   = 0,
   in the form
   d
   dx (F (x, y)) = 0,
   where F (x, y) is a polynomial in x and y. Hence find the general solution of the equation.
   Use this method to find the general solution of
   (y cos x + 2xey) + ¡
   sin x + x2ey − 1
   ¢ dy
   dx
   = 0

[tight pier]

Solve for dy/dx and then multiply the fraction by (1/x)/(1/x) and substitute u = y/x

[fervent heron]

oh yeah nice

is there any reason why you try that sub?

[alpine sable]

bacc

[tight pier]

looks like it will lead to something

[fervent heron]

ok fair

[uneven isle]

it is just a known technique

(i think)

at least ive never been taught any intuitive reason why they derived this form to use

it just works

[tight pier]

No

(2x+y) and (x+2y) do have a pattern

the disturbance is that I cannot really separate it

So one idea is to see what happens if you divide by a variable, you seem to get rid of one

then you managed to bring y and x into one which can be substituted into one variable

[uneven isle]

sure it makes it algebraically nice but i dont really understand the validity of it

[tight pier]

wdym validity

i jsut tried it on my paper randomly

and it seems to work out

cause of 2x+y and 2y+x having this pattern

it's as factoring out one variable

[uneven isle]

like if u sub y = vx
is it saying there exists v for every x such that y = vx

ik it works algebraically but idrk whats actually happening in the sub

at least i havent been taught as a higherschool yet

[tight pier]

you are just defining the product/quotient as something new

say y/x gives any number and we call that now u

if we wanna work now with u, we have to change the differentials and stuff

[vale crag]

$(2x+y) + (x+2y)\dv{y}{x} = 2x + (y + x\dv{y}{x}) + 2y\dv{y}{x}$

[tight pier]

we are making a transformation of variables

[ocean seal]

aPlatypus

[vale crag]

if you look at the LHS this way, it's not hard to integrate really

prolly useless for the 2nd one tho

[uneven isle]

im probably misunderstanding u but again, it is still just algebraically valid no?

i never understood why we can virtually sub anything

[tight pier]

let me ask you this
why does y = f(x) make sense to you?

[uneven isle]

y is just defined as funciton of x

what else is there to say

[tight pier]

same thing you can do here

u is a function of x

you are having essentially u = f(x,y(x))

you can define it as that

and from there transform things

[uneven isle]

u is function of y/x no?

oh

nvm

[tight pier]

y is a function of 1/x?

or a function of e^x

u(x) = y/x where y could be any function like ln(x) or e^x or sin(x)

whatever

[uneven isle]

i kinda get what u trying to point to but i still don't rly understand subtitution in calc in general, it probably is just i haven't learnt enough to understand transformation of variables

like there are certain questions where the teacher has explicitly stated some substitutions are invalid

for whatever reason

[uneven isle]

well more specifically this was in context of integration

[lone heart]

@fervent heron Has your question been resolved?

[lone heart]

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[elder hamlet]

Here’s what I’ve done

[elder hamlet]

Doing the last questions of the chapter and idk how to approach them

Since it’s not asking for. I’m assuming n is irrational or something like that so ig the approach is to go straight to n cubed

[lone heart]

@elder hamlet Has your question been resolved?

[lone heart]

@elder hamlet Has your question been resolved?

[tacit arch]

the question itself is rotated. upload it separately and organize your work better

[elder hamlet]

,rotat

,rotate

[elder hamlet]

Question 11

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@elder hamlet Has your question been resolved?

[lone heart]

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[frail grove]

$card\left({\emptyset}\right)=0$ or $1$

[ocean seal]

Adam Ch.

[cinder tundra]

1

The empty set has one element

The empty set

[tiny sky]

you mean the set containing the empty set?

[frail grove]

the empty set has one elemnt ?

that's why i asked that quesiton actually

[cinder tundra]

Yes

[frail grove]

why is it called empty set if its cardinality is equal to 1

[tiny sky]

the empty set has cardinality 0

but what u've written isn't the empty set

u've written the set containing the empty set

[frail grove]

my prof and samuel said that it's also called empty set

[cinder tundra]

Oh right

My bad

Yes

I got confused because of that

[tiny sky]

the set containing the empty set is not the empty set

either you meant $\emptyset$ or ${}$

[ocean seal]

LY

[frail grove]

i think my prof also got confused then

so P(\emptyset) is not called empty set

$P(\emptyset)={\emptyset}=\emptyset$ ?

[ocean seal]

Adam Ch.

[frail grove]

P is paritition power set and not probability

[pseudo ice]

This here is not true (think of the right as an box with nothing in it, and the left as a box which has another box inside it, with the said "inner box" being empty)

[frail grove]

can we call the power set of the empty set an empty set ?

and $card (P(\emptyset))=1$ ?

[ocean seal]

Adam Ch.

[hushed locust]

the power set of the empty set is the set containing the empty set

[frail grove]

ok i think i get it now

got it thnx guys

.close

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[formal hornet]

i got stuck, idk where to begin

[cinder tundra]

,rotate

[cinder tundra]

Me neither, idk where to begin to read

[manic wyvern]

same lol

[formal hornet]

number 8

idk how to do number 8

[pseudo ice]

Well, you've got the partial fractions form right, at least (assuming that's the aim!)

[formal hornet]

yeah but

its hard to solve for A, D and E

[pseudo ice]

So assuming you solved for B and C then?

[formal hornet]

yea

i worite it ther

oh its out of the frame

i just let x = 0 or x = a for them

[pseudo ice]

You could always e.g. take this and rearrange for one of the variables, say A or D, then plug into the other equations

[formal hornet]

is there no

faster way?

or a more organised way to do it?

it feels so slow and you have all these annoying fractions

[pseudo ice]

Erm, well, if you know linear algebra, there's putting those in matrix form, but that's only slightly less pain really

[formal hornet]

ph my god

why is life so shit

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@formal hornet Has your question been resolved?

[formal hornet]

.close

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[gray heart]

Did i do 3a and 3b correctly?

[lone heart]

@gray heart Has your question been resolved?

[gray heart]

Added the parametric equations

<@&286206848099549185>

[lone heart]

@gray heart Has your question been resolved?

[lone heart]

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[hollow cloud]

Why can you equate exponents of the same base?

[hollow cloud]

Like in this scenario:

I've had a lot of people tell me that 2^x = 2^4 but how can you prove that?

[high cosmos]

it's one of the rules. When you have exponents with the same base, the only way to them be the same is when the exponents are the same.

[hollow cloud]

sure but wouldn't that only apply if the case was independent from any other terms?

like if the question was 2^x = 2^4, then what is x, there it applies but how does it work when it's not independent like the equation shown above?

[tiny sky]

if x is an integer then yeah you can because of unique prime factorisation

if x is allowed to not be an integer then no you can't

[hollow cloud]

the question didn't say x needed to be an integer

it just said what is x and y

i know the answer already, but i don't like the method

original translated

[mossy jasper]

Could you give the answer out of curiosity

[hollow cloud]

x = 4 and y = 5

but other than just stating that x = 4 without proof doesn't satisfy me

[mossy jasper]

Just making 100% sure, both x and y must be integers, right?

[hollow cloud]

no, they can be whatever

like 3.231293818926798713570918349021

or somth

[high cosmos]

I would solve it like this: 3^(7+x)/3(2y+1) = 2^4/2^x, but i don't know if is that the objective

[mossy jasper]

That's literally exactly what I was just typing out but yeah I can't work out where to go from there either

You could solve the equation to get it in the form of just y but that would be absolutely horrible, but doable

[hollow cloud]

how are we struggeling with the first year of gymnasium math?

[high cosmos]

Not that bad i think

[hollow cloud]

I just learnt trigonmetry and I gotta deal with this shii

[mossy jasper]

I get, like, a million list of lns

[high cosmos]

Because then it would be: 3^(7+x-2y-1) = 2^(4-x)

[mossy jasper]

I'm so confused, is this the question or is the original one the question?

[hollow cloud]

yh

I don't have the book but yh

[mossy jasper]

Oh, that parts the easy part, but if you were to solve for y you'd need to get rid of the powers with ln, which is hell

Bc one of them has a power with both y and x terms, while the other is just x terms

[hollow cloud]

The original question in swedish is : Bestäm x och y om

[high cosmos]

This is just true when both exponents are 0, so you just = the exponents to zero

[mossy jasper]

Then where did you get this from?

[hollow cloud]

well it's my simplification of it

if I did it right or not, idk

[mossy jasper]

oml, please always send the original question first

[hollow cloud]

i did

[hollow cloud]

here

[mossy jasper]

FIRST, the first thing you sent was your simplification

I thought that was part of the question 😭

[hollow cloud]

oh alr sowwwwwyyyyyyyyyyy

🥺

[high cosmos]

Yeah, i was using it to

[tiny sky]

then the ppl saying that you can conclude x=4 are wrong

[tiny sky]

that's not true

3^2 = 2^(log_2(9))

[hollow cloud]

yh ik, that's what I though too, and they are above my pay grade, but the book says 4 so I will still use their method on tests, but I wanna know the correct method

[mossy jasper]

wait, so the book absolutely says that the answer is 4?

[tiny sky]

,w 3^7 * 6^x = 48 * 9^y

[hollow cloud]

yh 4 and 5, you can test it in the origional question

[ocean seal]

Results provided by WolframAlpha

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

Requested by lyheart

[hollow cloud]

what am i looking at😭

[tiny sky]

yeah ok they meant integer solutions

[mossy jasper]

This is what I said earlier when I said you get a bunch of logs if you actually solve it

[hollow cloud]

well they didn't mean shit tbh, they just said solve for

but yh, kinda cringe

[tiny sky]

that's really bad of them, they really should specify that they wanted integer solutions because then the problem is very easy

[tiny sky]

but yeah unless there was some very strange reason why, i was expecting infinitely many solutions

[hollow cloud]

yh

[mossy jasper]

Nah, I calculated it a while ago and got the same, but since they said the answers were 4 and 5 I assumed they wanted something other than to rearrange for y

[hollow cloud]

cuz i don't even know what half of those things are

or rather, C

what is c in that equation, the rest I know tbh

[mossy jasper]

Yeah, I'm confused what they want you to do in the original question then

[hollow cloud]

prbly just find the integer solution cause i aint doing all dat

💀

So on the test, if I get a similar question, I just trust in the exponents of the same basese are equal?

[tiny sky]

in a test, probs assume they mean integer solutions

if u haven't come across logs before then all you can do is unique prime factorisation

[hollow cloud]

yh I know basic log, like what they do but not how to use them.

not enough to use them like the equation above

ty for all the help from everybody btw, appreciate it

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[gusty pivot]

how many well formed strings of n opening and n closing parenthesis are there, such that the first two parenthesis are open?

[ocean seal]

sibber

[gusty pivot]

(C_n is the catalan sequence)

so my question is why is this wrong?

[ocean seal]

sibber

[gusty pivot]

so why is that?

[pallid scarab]

problem is... you're overcounting in this scenario

just take n = 3

I can for example have ((())

and then insert ) so I get (()())

but I could have reached the same configuration

by taking (()()

and then inserting ) at the end

also giving me (()())

[gusty pivot]

ahh

got it, thanks

.close

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[barren silo]

how does an inequality of x is less than or equal to 5 become a compound ineqaulity im just really confused

[barren silo]

for example

i have a question

[torpid dirge]

just take that inequality itself

or nest it with itself

[barren silo]

bro

wdym

how do i make it a compound inequality

[torpid dirge]

u can nest it with itself like

or use a useless other condition

[barren silo]

explain pls

[torpid dirge]

x<=5

is probably enough

but if u really need to use a second one u can just add a useless condition like

x<=5 and x<=5

[lone heart]

@barren silo Has your question been resolved?

[barren silo]

nope

[lone heart]

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[nimble fern]

i remember i was taught that to plug in and check

what are you finding btw

lemme check

looks like (10,0) satisfies

[charred pilot]

You can either arithmetically find x in terms of y and then plug that into other equations or graph them all and find where they all intersect

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[river cobalt]

Hello ! How can we solve a combinatorial problem without using Integer programming ?
eg : I try to solve a graph problem with IP , I specified my constraints and requirements, and IP solver can get a solution very quickly.
However, for more complicated problem, it seems that IP cannot handle it.
I am considering formulate my problem into an optimization problem , is it possible ? Thanks !

[lone heart]

@river cobalt Has your question been resolved?

[lone heart]

@river cobalt Has your question been resolved?

[river cobalt]

.close

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[blissful bluff]

help pls

[blissful bluff]

<@&286206848099549185> any help is appreciated, its a quick problem

[simple wigeon]

use powers

[blissful bluff]

wdym

square it?

[lunar pier]

I’m scared

[blissful bluff]

hmm?

[raw jetty]

maybe would help for you to visualize it is to do like
$$\frac{2(\sqrt{\frac{1}{2}}-2\sqrt{2}+3\sqrt{8})}{2}$$ and multiply inward

[ocean seal]

Skill_Issue

[blissful bluff]

why r u multiplying it all by 2

[raw jetty]

to get rid of fractions

[blissful bluff]

ok

[raw jetty]

you can skip it if you want, just change all the roots to fractions

[blissful bluff]

did u end up getting 4 root 2 + 1 @raw jetty

[raw jetty]

lemme see

nope incorrect

[blissful bluff]

Mb + 1/2

@raw jetty

right”

instead of +1 + 1/2

[raw jetty]

this?

[blissful bluff]

yes

[raw jetty]

no

[blissful bluff]

what did u get and how

WAIT NO

6 root 2

• 1/2

@raw jetty that”

[raw jetty]

no

[blissful bluff]

why u dekete it

[raw jetty]

actually i dont think i should tell you :3 cause its againts the rules i forgot

[raw jetty]

basically can you show your work?

ill check it and point out where you went wrong :>

[blissful bluff]

no it isnt

[raw jetty]

$2\sqrt{\frac{1}{2}}=\sqrt{4\cdot\frac{1}{2}}=\sqrt{2}$

[ocean seal]

Skill_Issue

[blissful bluff]

2x route 1/2 is route 1

oh

nvm

i got 9 root 2 over 2

[blissful bluff]

is that right

[raw jetty]

yeah

[blissful bluff]

Now it makes sense

ok next one

psrt b

[autumn goblet]

Hello, how do I get help with a task that I don't understand how to solve, I need help

[blissful bluff]

Wrong channel

[raw jetty]

im not sure what form your teacher wants it to be, but i would put it like that since its the simplest for me

[blissful bluff]

#❓how-to-get-help

[blissful bluff]

same

for b do we multiply it all by 12? @raw jetty

[simple wigeon]

oh hi

[blissful bluff]

or by 3

[raw jetty]

you can do that yes

[autumn goblet]

Sorry, I speak in Spanish, I'm translating everything, sorry for interrupting.

[blissful bluff]

lo siento

hablo espanol tambien

soy un translator

vas a #❓how-to-get-help para ayuda

[autumn goblet]

yap

[blissful bluff]

💀

[blissful bluff]

which one

[raw jetty]

the 12

[blissful bluff]

ok

[raw jetty]

actually you can do either

[blissful bluff]

is it 12 over anything!

like is there a division?

[raw jetty]

you want to multiply by 12, so divide by 12 also. so in the end you multiply by 12/12

[blissful bluff]

5/4

i got that

as answer

i think im riht

am i right @raw jetty

[raw jetty]

no

[blissful bluff]

wait no

i know my mistake

@raw jetty what about 1/3

@raw jetty

[raw jetty]

er

wait

[blissful bluff]

wait no

help pls

i gtg in 4 mins

help 😭 @raw jetty please im pleading for u

welp 3 mins now

1 minute.

@raw jetty i do have a request for u if u dont mind, pls delete this

thanks

[raw jetty]

er

its incorrect

[blissful bluff]

I got it now

[raw jetty]

sorry something urgent came up

ok nice

[blissful bluff]

5 root 3 over 2 right

@raw jetty

[raw jetty]

yes

[blissful bluff]

can u show me how u did it

my way was goof

@raw jetty

please

@raw jetty

[raw jetty]

er

[blissful bluff]

please

im begging brudda

mine was a guess and check

until it seemed good

can u show me your process @raw jetty

[blissful bluff]

ok i got the same thing but,

cant you simplify it immedaitly

[raw jetty]

wdym

[blissful bluff]

whats the point of the left imagine bottom side

or no

right image

top side

wait nvm

im trippin

i just did less work

[raw jetty]

uh i accidentally duplicated it

[blissful bluff]

i see

nice @raw jetty

alr ty

greatly appreciate it

@raw jetty tysm have a good night

.close

[lone heart]

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Closed by @blissful bluff

Use .reopen if this was a mistake.

[raw jetty]

bye0

[blissful bluff]

bye

[blissful bluff]

👋

[lone heart]

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[tight badge]

Let $a < b$ be two real numbers, and let $K: [a, b] \times [a, b] \to \mathbb{R}$ be a continuous function. For each function $x \in C[a, b]$ we define a function $F(x) \in C[a, b]$ by

$$(F(x))(t) = \int_a^b K(t, s) x(s) ds \forall t \in [a, b]$$

Show that the mapping $F: C[a, b] \to C[a, b]$ is continuous with respect to the metric $d_\infty$

[ocean seal]

void

[tacit arch]

!show

[lone heart]

Show your work, and if possible, explain where you are stuck.

[tight badge]

My thought process is to use the definition of continuity for the mapping F

[tacit arch]

Is that the sequence definition of continuous?

[tight badge]

So do we want to show that if $||x_n - x||\infty \to 0$ then $||f(x_n) - f(x)||\infty \to 0$?

[ocean seal]

void

[tight badge]

I’m on mobile so apologies for my formatting… im also quite tired 😔

[tacit arch]

Yea

[lone heart]

@tight badge Has your question been resolved?

[lone heart]

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[outer river]

simplification ((x + 4)(x - 1))/(x + 7)(x - 3)) * ((x + 2)(x + 7)/(x + 2)(x - 1))

[nimble fern]

what have you tried?

[outer river]

well i factored it to here and multiplied by the reciprocal

[nimble fern]

so, many terms can be cancelled out

[outer river]

its (x+7), (x + 2),(x - 1)

but what happens after cancellation

[nimble fern]

you can group the terms into a single fraction, and it will be done

[outer river]

are thes the right terms to cancel out

why does i turn into division there is a multiplication sign in between

[nimble fern]

lemme try to type it out

$\frac{(x+4)(x-1)}{(x+7)(x-3)}\cdot\frac{(x+2)(x+7)}{(x+2)(x-1)}$

[ocean seal]

Biscuity

[nimble fern]

You mean this, right?

[outer river]

yep

[nimble fern]

and the (x+7), (x+2) ,(x-1) cancels out?

and become this?
$\frac{x+4}{x-3}$

[ocean seal]

Biscuity

[outer river]

ohh

[nimble fern]

sorry, typo

[outer river]

i get it now thanks

[nimble fern]

cheers!

[outer river]

.close

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[outer river]

simpification

[nimble fern]

!show

[lone heart]

Show your work, and if possible, explain where you are stuck.

[fresh wedge]

Do you know butterfly method?

[fresh wedge]

You can't cancel terms like that

You need to simplify numerator and denominator

Separately

[outer river]

I have it's x^2+3x-10 for the denominator and 3x^2-12x+30 for the numerator

[fresh wedge]

Well that's your simplified form then

[outer river]

can i simplify it more????????

[fresh wedge]

I don't see anything else to do

[outer river]

ok thanks

.close

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[hoary wolf]

I don't understand how I'm meant to input my answer I've read what the teacher wrote as to how something should be written but it doesn't really make sense to me in this scenario. (I realize I made a small mistake on negative for 2y-1, it's fixed but not in sc)

[tight pier]

So your R(y) = (y-5)/(2y-1)?

[hoary wolf]

I guess but I can't input the answer like that- has to be as a domain but it's not recognizing my answers as domain n stuff

[tight pier]

You cannot put in fractions

[hoary wolf]

I'm aware- I'm saying I don't know how to write a domain onto internet text

for this to register my answer

[tight pier]

How did you get (2y-1)(y-5)

[hoary wolf]

all my calculations are there on the right screenshot

[tight pier]

,, 2y^2-9y-5 = 2y^2-10y+1y-5 = 2y(y-5)+(y-5) = (2y+1)(y-5)

[ocean seal]

bacc

[hoary wolf]

right and I said on my comment that I made a mistake and I fixed it- it's just not in the screenshot because I saw that later

[tight pier]

oh

mb

[hoary wolf]

nw, I just wanna know how I need to write my answer

into the machine

for it to see my answer and go "yep is good, good job"

[tight pier]

so you cannot use / for fractions

(y-5) : (2y+1)

[hoary wolf]

oh alright ill give that a go

what does that mean?

[tight pier]

idk

but you can also try this

(y-5) * (2y+1)^(-1)

[hoary wolf]

closer I suppose?

[tight pier]

yea

2y+1 must not become 0

so I guess you do a

y != -2^(-1)

weird ass programm

[rose sigil]

The Magical (Ass Pudding) or The (Magical Ass) Pudding?

[livid sage]

[hoary wolf]

Latter, as a kid it was meant to be more like The Magical Ass (shaped) Pudding

never changed it tho cus it's funny

[rose sigil]

great

[hoary wolf]

[tight pier]

What profession has your teacher?

wait

I see it

[hoary wolf]

Uhh I'm not entirely aware of what her background looks like- but she's my math teacher for my remedial math college class

[tight pier]

we forgot y != 5

all because layla keeps asking weird questions

[hoary wolf]

...err how do I put that in

sorry for the trouble- I never learned how to input domains like this

[wintry wadi]

from the question it seems like you need to do
<expression> for <restrictions>

[tight pier]

metoo

[hoary wolf]

so it'd be like..

(y-5)/(2y+1) for y!=5 and y!=-1/2 ?

[wintry wadi]

yea, try that

[hoary wolf]

oh sweet it works- only my domain is wrong gawd daem it

[tight pier]

bruh what

[tight pier]

denominator of R(y) was (2y+1)(y-5)

OH

you divided by y=0

as well

[hoary wolf]

should I not have done that?

[tight pier]

yes you can

because y!=0

[hoary wolf]

mm... sry that's not working in my brain, how does that make my domain wrong?

[tight pier]

You forgot to include that too

In the beginning when there was light

[ocean seal]

bacc

[hoary wolf]

oh! you're right

[tight pier]

thank god

[hoary wolf]

.. now the question is how do I input that tho cus

[tight pier]

y!=0

not =!

[hoary wolf]

wrong screenshot

[tight pier]

!=

[hoary wolf]

[tight pier]

maybe another and

y!=0 and y!=5 and y!=-1/2

also wtf

why does -1/2

work

-2^(-1)

and (2y+1)^(-1) btw

WTF

[hoary wolf]

yipee

[tight pier]

so / works

[hoary wolf]

[wintry wadi]

why wouldn't it work?

[tight pier]

because the programm is ass

\

[wintry wadi]

ah

[lone heart]

@hoary wolf Has your question been resolved?

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[hoary wolf]

Thanks guys, I appreciate it!

Would've taken ages to figure out how to input that myself x.x)

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[wintry wadi]

How do I get the closest point on an ellipse to another point?
do I do it the same as a circle?

[exotic canopy]

how would you do it on a circle?

[wintry wadi]

draw a line to the center thing

[livid sage]

that does not work for an ellipse

[wintry wadi]

yea, figured
how'd I do it then?

I'm gonna try something out and ask again

.close

[lone heart]

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[exotic canopy]

.reopen

[lone heart]

✅

[exotic canopy]

i was trying it out rn and i am not sure if it can be generalized but it looks like it works for ellipse x^2 + 2y^2 = whatever

fix point A, you want to find point E on the ellipse that is the shortest distance away from point A
draw a line through A and origin, (0,0). the intersection of this line through the ellipse will not be the closest point but it will help up find a better one. let this intersection point be C
draw a circle with center at A that passes through C, i.e. the radius is |CA|. since the distance between A and I is not minimal, the circle passes through two points: C and another one, let's call it C'.
now take the midpoint between I and I' and that midpoint is your E

actually nvm

the midpoint won't be on the ellipse

but i am pretty sure that if you pass another line through that midpoint from A the interesection this time will be the shortest path

at worst this is a good approximation

but the point is that i am trying to make this circle intersect at two points that are as close to each other as possible, approximating the circle touching the ellipse at exactly one point, which is the absolute shortest distance

hope this makes sense

[wintry wadi]

yea that is a really good approximation
how did you find C' though

[exotic canopy]

you mean algebraically?

[wintry wadi]

oh wait

just set the equation equal right?

[exotic canopy]

yep

but idk if it simplifies completely, unfortunately

[exotic canopy]

https://www.desmos.com/calculator/avfqie6ue0 this is the graph i was using btw; it's very messy but still nice to play around with

[wintry wadi]

ooh, I'll take a look

[severe portal]

I think you could try using the parametric form of an ellipse

$$ (a \cos t, b \sin t) $$

[ocean seal]

StrangeQuarkAL

[severe portal]

Let the other we know point be (x, y)

So the vector from the (x, y) to the some point on the ellipse is:

[wintry wadi]

currently trying that out and getting the normal
then trying to set it equal to the slope of the line to the closest point to the point

[severe portal]

$$ \begin{bmatrix} x - a \cos t \ y - b \sin t \end{bmatrix} $$

[ocean seal]

StrangeQuarkAL

[severe portal]

Yeah, this seems fine

$$ \begin{bmatrix} x - a \cos t \ y - b \sin t \end{bmatrix} = \begin{bmatrix} b \cos t \ a \sin t \end{bmatrix} $$

[ocean seal]

StrangeQuarkAL

[severe portal]

Wait

This won't work because the magnitudes are different

Both vectors will be some multiple of the other

Instead of doing this, just dot the left hand side with the vector of the tangent at t

Knowing that it should equal 0 to minimise the distance between the (x, y) and the ellipse

[wintry wadi]

so (p - e(t))·e'(t) = 0?

[wintry wadi]

e being that thing

[lone heart]

@wintry wadi Has your question been resolved?

[wintry wadi]

how do I solve it though
I got to $axsin(t) - bycos(t) = (a^2 - b^2)sin(t)cos(t)$

[ocean seal]

Sepdron

[wintry wadi]

is i could square it, then cos² to sin² thing

I'm gonna do this later on, I'm having a headache (not related to this)

.close

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[livid sage]

consider n^4 (mod 5)

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.close

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[rose vortex]

Can you help me with this one please, i must write it as an product: (x + 3)^4 + 2(x
2 + 6x + 9)^2 + (3x + 9)^3

[tiny minnow]

$(x + 3)^4 + 2(x^2 + 6x + 9)^2 + (3x + 9)^3$

[ocean seal]

Newt

[tiny minnow]

u meant like that?

just to make sure

[rose vortex]

yeah

[balmy pendant]

i would suggest factorizing the quadratic

[glad root]

Try to express everything in terms of x+3

You can take (x+3)^3 common then

[rose vortex]

okay thanks

[balmy pendant]

yeah

cuz two of them are alrdy written tht way

[tiny minnow]

write that y is x+3

so x^2+6x+9 = (x+3)^2

=> y^2

agree?

and what the next step u answer

[rose vortex]

hm but then i get 2(x+3)^2^2 right?

and thats (2x+6)^4

[tiny minnow]

$ab^2 \neq (ab)^2$

[ocean seal]

Newt

[tiny minnow]

we put 2 in outside

2(y)^2^2

2y^4

[rose vortex]

okay so the new term is 2(x+3)^4

(x+3)^3 ((x+3)(2(x+3) is the Answer i guess but the last one is missing, i dont know what to do with it (3x+9)^3

maybe only 3?

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@rose vortex Has your question been resolved?

[rose vortex]

no

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[charred cloud]

Three dice are rolled twice. What is the probability that they show the same
numbers
(a) if the dice are distinguishable.
(b) if they are not.
[Hint: divide into cases according to the pattern of the first throw: a pair, a
triple, or all different; then match the second throw accordingly.]

im getting different answers each time i try, sometimes i get the favourable outcomes for part a as 816 or 1/216 as my answer and my friends get different answers too

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@charred cloud Has your question been resolved?

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@charred cloud Has your question been resolved?

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@charred cloud Has your question been resolved?

[charred cloud]

No

[prime badge]

part a is just 1/216 there's nothing to do

[stray spruce]

Is this question is solved??

[tiny sky]

*6 for the 6 possible cases (1, 1, 1), (2,2,2) etc.

[prime badge]

no it's like, the second roll is identical to the first

[tiny sky]

oh wait they're thrown twice