help-23

[lavish skiff]

something to do with implicit differentiation ?

[edgy breach]

if instead of siny = x it was y = sinx what would dy/dx be

[lavish skiff]

but idk the specifics

[lavish skiff]

dy/dx = cos x

[edgy breach]

yes so in this case what is dx/dy

as in you swap y and x

[lavish skiff]

uhh

like find the inverse ?

[edgy breach]

no literally if x = siny what is dx/dy

[lavish skiff]

dx/dy = cosy

[edgy breach]

ok now how can you rearrange that equation to get dy/dx

[lavish skiff]

idk tbh

[timid ridge]

reciprocal

both sides

[lavish skiff]

reciprocal is just flip right

so you like flip dx/dy to be dy/dx

and then cosy/1 = 1/cosy

[timid ridge]

ok cool

[lavish skiff]

and then boom 1/cos x? am i missing anything?

[timid ridge]

why you turn y into x

[lavish skiff]

good question

uhhhhh

thinking

yeah not sure

[edgy breach]

its a little bit confusing but essentially dy and dx can be treated as individual variables in most cases

what dy/dx = 1/cosy means is that the rate that sin^-1 x is changing at any point is cos of the reciprocal of the current value of sin^-1 x

if that makes sense

[lavish skiff]

well i understand that the derivative is the instantaneous roc

[lavish skiff]

how does it allow y to turn into x tho

when u take the reciprocal

[edgy breach]

https://tenor.com/view/thats-the-neat-part-you-dont-invincible-gif-27194608

[lavish skiff]

whaaa

[timid ridge]

He started with an example with flipped y and x to demonstrate that there is nothing special about the variables

we are back to your original question

[lavish skiff]

what

😭

[edgy breach]

dy/dx doesnt have to be in terms of x

[lavish skiff]

yeah i get that

[edgy breach]

what are you asking

[lavish skiff]

oh

bro i was AI hallucinating

i see now

i thought the steps lead to 1/cos x

but no its 1/cos y

ok thanks guys

.close

[safe radish]

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[safe radish]

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[verbal pecan]

I dont get it

[glacial cairn]

Well you have a statement P -> Q

If the proof starts with "assume P" then it's direct

If it starts with "assume not Q" then it's a contrapositive (going for not Q -> not P)

[verbal pecan]

okay

[glacial cairn]

If it starts with "assume not (P -> Q)" then it's by contradiction

[verbal pecan]

oh i got it thanks

.close

[safe radish]

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[safe radish]

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[solar hazel]

i saw ts problem in another server a while ago and i was able to compute the answer but it was allegedly just from someone's homework and my method could not have been expected of the student. i am still curious if there is a nice way to write the answer

we have 70 red beads, 70 blue beads, 70 green beads, and 70 yellow beads, and we make a circle out of them randomly (choose one of the 280 uniformly, then another out of the remaining 279, and so on, and connect the first and last). what is the probability there are no two beads of the same color next to each other?

[solar hazel]

ts is so tuff

,calc 35160752767483967793856895649187208658376924104964949185034223823701112860537559776144881711744175020305185933564751125817871/1193901580824623492029582902841551340430563387679240327772058617014835289592291160051553067139282835765793966673575963676266928049345932726243971753166210600000

[flat frigate]

Result:

2.9450294171818e-35

[hardy verge]

uhm

that is hard

[solar hazel]

this is the answer i got. which i am pretty confident in. but there was a lot of computation involved and my method cannot really be "written as a math expression"

[safe radish]

@solar hazel Has your question been resolved?

[primal bone]

Not that I'm in a position to check this (it's 2am over here), but an approach for a question like this, that I'd describe as canonical, would be to scale down the question considerably (say, 2 beads of each colour), then seeing if you can scale up an answer as you increase the number of beads; then the core of the answer would be to prove this scaling is valid

Then if it is valid you can just answer the 70-beads-per-colour case as a corollary

[solar hazel]

this is not really saying anything

[hardy verge]

he tried

its 2am for him

[vague phoenix]

wait so if the numbers of beads are significant enough that when you place a bead, then a 2nd bead next to it, the probability of the 2nd bead that differs from the 1st is like approx 3/4

[solar hazel]

sure

[vague phoenix]

then i think since we have 280 beads (70*4)

then the probability of placing is (3/4)^280

[solar hazel]

no, that is not the answer

[vague phoenix]

what is the answer?

[solar hazel]

it is a good heuristic approximation but it's not that simple

[solar hazel]

this is the answer

,calc (3/4)^(280)

[flat frigate]

Result:

1.0402883853216e-35

[vague phoenix]

hmm this weird

[solar hazel]

if you can't see why it's different, try with 2 beads of each color

[vague phoenix]

the probability become significantly smaller as i use higher amount of beads each color

how did you come up to your answer

[mortal sandal]

did you DP?

[solar hazel]

i computed it with dynamic programming

yes lol

[mortal sandal]

ok good that's all i can think of atm

[solar hazel]

it was still quite complicated and slow

[mortal sandal]

must have been annoying to case in the last bead matching the first one lol

this reminds me of equitable graph coloring

[solar hazel]

chromatic symmetric function moment

[mortal sandal]

@toxic stratus categorical graph theory

well ig a fun fact is if you had as many beads of each color as you wanted there are 3^280 + 3 ways to do it lol

in this case the (3/4)^n heuristic holds up extremely well, so it's curious it doesn't appear to work for the balanced case

[solar hazel]

,calc (3/4)^(278) * ((3/4) * (1/4) + (2/4) * (3/4))

[flat frigate]

Result:

1.0402883853216e-35

[mortal sandal]

i think you can factor a 3/4 out of the right and you get back to (3/4)^280

[solar hazel]

yea

that version of the problem is boring though

[mortal sandal]

i could definitely see an error of order 1/280 arising

whatever color i use first, i have less of, so i'm marginally less likely to end my chain with it

[solar hazel]

here are some example numbers btw. the tuple is the amount of each color. the next two numbers are the number of good circles and the total number of circles (i am counting rotations of circles as distinct, unlike is usually done for necklaces, but since it's a probability problem we can treat it either way). and the last two are the probability of a good circle, in fraction and decimal form

[mortal sandal]

oh its an e thing isn't it

[solar hazel]

also my program accepts any length tuple if you would like to see any other color amounts

[mortal sandal]

each pair of beads is marginally less likely than 1/4 to be a pair

69/279 ig

,calc (210/279)^280

[flat frigate]

Result:

2.832863239807e-35

[solar hazel]

oh also here are the numbers for 3 colors

they have an oeis page https://oeis.org/A110707

but the 4 color version does not

[solar hazel]

or is it a pi thing

[mortal sandal]

wow

[solar hazel]

that's for 3 colors though

i am convinced that ts problem is very tuff and should not have been put on that kid's homework

even the 3 color version is like... bruh

and 4 colors is so much harder

[safe radish]

@solar hazel Has your question been resolved?

[solar hazel]

i'm not sure ms bot

kinda

i declare my question resolved because i have convinced myself the problem is tuff

[safe radish]

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[scenic phoenix]

I need to find the value of {0|1|2}

[scenic phoenix]

i have no clue what to do

[manic radish]

I'm not opening a pdf. can you write it here, or screenshot it?

[scenic phoenix]

i can, but its quite big

[gusty flicker]

was js about to say that

[scenic phoenix]

ill just screen shot what should be nesessary

[gusty flicker]

not trusting any download links on discord lol

[manic radish]

pdfs are horrendous for hidden stuff XD

[manic radish]

https://github.com/ading2210/doompdf

[scenic phoenix]

compleatly fair

[gusty flicker]

alr had a feeling this was too advanced for me by the active role lol

is this game theory or something

[scenic phoenix]

combinatorial game theory

[gusty flicker]

interesting

well gl ig

[scenic phoenix]

i can explain more if anyone is willing to help me

[shut hound]

I think it comes out to (1) + (~2) + (-3)

[manic radish]

I'm not sure I understand the notation

[manic radish]

that looks plausable

[shut hound]

It definitely took me a hot minute to get too

I was staring at it like stare_wtf

[scenic phoenix]

it doesnt, o({0|1|2})=(L,L,L,L,L,L) while o(1~2-3)=(R,R,R,R,R,R)

btw, n~n-n=0 always

so you can remvoe 1 from all three of those to get ~1-2

[shut hound]

{0||} = 1 from the definition of the unit
{|1|} = 1~1 from theorem 2.1
{||2} = 2-1 from theorem 2.1

[scenic phoenix]

not quite

{~1|1|}=1~1

therom 2.1 says for any subtle integer, 1 is not a subtle integer

same goes for the second

[shut hound]

~1~1 then?

[scenic phoenix]

wdym?

[shut hound]

~1 is the subtle int unit right?

[scenic phoenix]

yes

so {|~1|}=~2

[shut hound]

Yea that's what i said initially though

1 ~2 -3

[scenic phoenix]

you said {|1|} which is diferent

[scenic phoenix]

also $G\ne G^L+G^M+G^R$ so this doesnt track

[flat frigate]

pinkishnova cgt queen

[shut hound]

$G := {G^L|G^M|G^R}$

[flat frigate]

@shut hound

[scenic phoenix]

yes

where G^X is player X's options

[shut hound]

Ye, I got that

Side note, was this made to be confusing?

There are so many overloaded symbols

[scenic phoenix]

in what

also it sometimes just does this

for example: games with activness

[shut hound]

Ok can we at least agree that it's {0||} + {|1|} + {||2}?

[scenic phoenix]

but its not tho

[shut hound]

It's not?

[scenic phoenix]

no

[scenic phoenix]

this

[shut hound]

I thought you meant that like it's not 0 + 1 + 2

[scenic phoenix]

also because like, i can tell just at a glance that the outcome classes dont work out

[scenic phoenix]

i mean it isnt

but its also $G\ne {G^L||}+{|G^M|}+{||G^R}$

[flat frigate]

pinkishnova cgt queen

[scenic phoenix]

think about 1/2

[shut hound]

Hm... fair

Hold on, Imma try to unfold this on paper, I don't think anything I can do in the textbox is helping

[scenic phoenix]

i know its a bit weird

[scenic phoenix]

alright

good luck!

[shut hound]

It isn't so much so that it's weird

It's hyper-overloaded

[scenic phoenix]

yeah sorry about that

its just a very dence subject

[shut hound]

It's not your fault

It also makes it kinda a challenge so it's fun

[scenic phoenix]

alright

[safe radish]

@scenic phoenix Has your question been resolved?

[scenic phoenix]

here is a game diagram if this helps

I will not elaborate

[shut hound]

lol

How does $G + \neg G = 0$ work, how do you remove elements by performing disjoint addition?

[flat frigate]

@shut hound

[scenic phoenix]

You aren’t removing elements and what is disjoint addition?

Like it’s just notational shorthand for the additive inverse

[shut hound]

No I got that, but, $1 + \neg 1 = 0$, correct?

[flat frigate]

@shut hound

[scenic phoenix]

Yes

[shut hound]

$1 + \sim 1 + -1 = 0$

[scenic phoenix]

Yes

You have to use \sim

[flat frigate]

@shut hound

[scenic phoenix]

Yes that’s true

[shut hound]

Thanks I forgot that earlier

$0 = {||}$

[flat frigate]

@shut hound

[scenic phoenix]

Yep

[shut hound]

$1 = {0||}, \sim 1 = {|0|}, -1 = {||0}$

[scenic phoenix]

Yep

[flat frigate]

@shut hound

[shut hound]

$1 + \sim 1 -1 = { \sim 1 - 1 | 1 - 1 | 1 + \sim 1}$ though

[flat frigate]

@shut hound

[shut hound]

@scenic phoenix did I make a mistake?

[scenic phoenix]

Yes

[shut hound]

Where?

[scenic phoenix]

There are multiple ways to notate the same number

[scenic phoenix]

No I mean that’s all correct

[shut hound]

Oh ok

[scenic phoenix]

No you did not

[safe radish]

@scenic phoenix Has your question been resolved?

[shut hound]

I'm sorry, I don't think I'll be able to hand compute this

Do you have the permission to use scripts?

@scenic phoenix ?

[scenic phoenix]

Sure

[scenic phoenix]

Also sorry, I’m just a bit busy at the moment

[shut hound]

In that case I'd just write a BFS

[shut hound]

np

I actually also need to leave

which is why I sort of gave up on hand solving this

for the BFS you'll wanna try adding the units to the states and maintaining and unifying the states as you go I suppose

[scenic phoenix]

yeah

[safe radish]

@scenic phoenix Has your question been resolved?

[safe radish]

@scenic phoenix Has your question been resolved?

[safe radish]

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[sour wigeon]

how do I solve that integral?

[proud tree]

if i substitute a value for t, do i have anymore variables?

[sour wigeon]

I don't understand

[proud tree]

also remeber what log_e e^x means

[sour wigeon]

ohh

t=e^x

?

[proud tree]

i have log_e t, where t is a variable but if i sub a particular value for t
like 3 i have a constant now which

[slim lion]

Consider $f(e^\frac{\pi}{2}) = \int \left(\frac{1-\sin(\ln e^\frac{\pi}{2})}{1-\cos(\ln e^\frac{\pi}{2})}\right)$

[flat frigate]

Flatus

[proud tree]

no

[slim lion]

use this type of idea

[sour wigeon]

this is zero

ohh

[slim lion]

nope

[sour wigeon]

nahh i don't get it

[slim lion]

it's the integral of 0

[sour wigeon]

yea

[slim lion]

which is..?

[sour wigeon]

0

[proud tree]

u sure

[winged flare]

this feels wrong

[sour wigeon]

am I wrong?

[slim lion]

hm?

[slim lion]

yes

what's $\int x , dx$

[flat frigate]

Flatus

[sour wigeon]

x^2/2

+c

[slim lion]

no

yes

[slim lion]

+C

so integral of 0 is..?

[sour wigeon]

C

[slim lion]

right

so i can equate it

$C = -e^\frac{\pi}{2}$

[flat frigate]

Flatus

[sour wigeon]

yeaa

[winged flare]

that integral is not C

[slim lion]

now u can work with other given information

using this idea

[sour wigeon]

but it is not working

$f(e^{\frac{\pi}{4}}) = \int 1 dt$

[slim lion]

yeah nvm u gotta solve the integral lmao

[flat frigate]

Prathmesh

[sour wigeon]

bruh

<@&286206848099549185>

nahh okay I can solve it

.close

[safe radish]

Channel closed

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Use .reopen if this was a mistake.

[slim lion]

yeah soz i kinda trolled

[safe radish]

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[half comet]

can someone explain 10b please

[hard crest]

find real numbers (i think they'll be integers) a, b, and c such that that equation is an identity

such that $n^2 = a(n+1)(n+2) + b(n+1) + c$ for all $n$

[flat frigate]

schrödinger's kitten

[half comet]

i understand a

but can you explain b?

liek which value of b do u pick

[frozen marlin]

thats what you have to figure out

try writing the LHS as a general quadratic and expand the RHS

[half comet]

can you show me how to do it if u don't mind

[frozen marlin]

i mean

my role is to guide u thru the problem

[half comet]

wait

[frozen marlin]

i can help you think of what to do but i wont do it for you

[half comet]

one second

[frozen marlin]

yeah sure

[half comet]

i just realised one thing if i do that then im gonna get a=a b=b

it won't work

[frozen marlin]

!show

[safe radish]

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

[half comet]

wait let me draw out what ur trying to make do

one second

wait no nvm i think i just answered my own problem

so sorry but i think iget it

[frozen marlin]

alr

[half comet]

i just realised i was over complicating this so badly

[frozen marlin]

whats ur answer

@half comet if ur done do .close

[half comet]

1(n+1)(n+2 -3(n+1) +1

.close

[safe radish]

Channel closed

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[frozen marlin]

ye

[safe radish]

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[long yoke]

The answer here is incorrect, i mean the question is incorrect aswell in that case no? the final universal y should be existential y?

[long yoke]

My solution, confirm if im not going insane, cuz chatgpt gemini, claude all said the other solution is correct

[glacial cairn]

That De Morgan's application is incorrect

But so is yours

[long yoke]

in what sense

[glacial cairn]

Oh never mind

Just yours

Not sure what you did tbh

[long yoke]

how can it just be mine its the same thing

they both look the exact same

bottom right is where it confuses me

how did they grab the negation from R(x,y) and not from negation from Q

[glacial cairn]

Oh damn I got confused by your parentheses

[long yoke]

if they want it to be on the outside

[glacial cairn]

Disregard what I said

[long yoke]

if you really wanted an universal it woukd be something like negation of universal y (R(x,y) ^ Q(x)) no?

[winter whale]

Good handwriting

[glacial cairn]

The answer here is incorrect
Why do you think so?

[long yoke]

.

[glacial cairn]

Let me rephrase

Did anyone tell you it was wrong?

[long yoke]

and stuck it on the outisde

no

[glacial cairn]

Ok so you think it's incorrect

[long yoke]

yes

[glacial cairn]

Alright

The point is:

,, \neg \forall x P(x) \equiv \exists x \neg P(x)

[flat frigate]

Nel

[glacial cairn]

"not for all x is P(x) true" is equivalent to "there exists an x such that P(x) is false"

As in "not all x yield P(x)" and "there is a counterexample to P(x)"

In your case x is y and P(x) is R(x,y)

[long yoke]

ye but

ah shit i cant do this latex thing

[glacial cairn]

I think it's this part that confuses you

[long yoke]

∃y¬R(x,y)∨¬Q(x) is this or that
¬∀yR(x,y)∨¬Q(x) or ∃y¬(R(x,y)∧Q(x)) which is then ¬∀y(R(x,y)∧Q(x))

[long yoke]

ye i said, the negation can be taken out only from the first R(x,y) and not the second?

[glacial cairn]

No, the problem is that you moved the "exists y" outside the parentheses

,, \forall x (\textcolor{orange}{(\exists y \neg R(x,y))} \lor \textcolor{green}{(\neg Q(x))})

[flat frigate]

Nel

[glacial cairn]

The "exists y" does not apply to anything other than the "not R(x,y)"

[long yoke]

ah shit i got it

lowk brains fried

tysm

atleast i manipulated claude and gpt

[glacial cairn]

Love how they used the double negation law just to undo it 4 steps later

[glacial cairn]

Well yeah LLMs will tend to say whatever they think will please you

[long yoke]

btw you get taught this in undergrad math?

[slim lion]

yes

well it obviously depends on where you live

[long yoke]

oh interesting

[glacial cairn]

Idk, I think it may depend on the country

[slim lion]

we do this in Australia

[long yoke]

because i think it isnt taught here for undergrad math usually its other discrete chapters

im doing CS so we take different version of Discrete

[slim lion]

it's one of 3 subtopics of 5 major topics in the hardest math course in australia

it's more popular in uni tho

[glacial cairn]

Well even within a country I'm pretty sure you could have different kinds of undergraduate education

Some topics are more general but first-order logic tends to be specialized

[slim lion]

needs brackets, doesn't it?

[glacial cairn]

Why?

[slim lion]

isn't it implying (neg forall x) P(x)

but we want neg (forall x P(x))

[glacial cairn]

That's the same thing?

[slim lion]

wdym

negations are distributive

$(\neg \forall x) P(x) \equiv \exists x P(x)$, right?

[flat frigate]

Flatus

[glacial cairn]

No

"it's not true for all x, that P(x)"
"it's not true that, for all x P(x)"

That's the same sentence

[long yoke]

you dont need to take out the negations in the first place dough nah?

[slim lion]

wait so $\neg (\forall x P(x)) \equiv (\neg \forall x)P(x)$

[flat frigate]

Flatus

[slim lion]

that feels wrong to look at

[glacial cairn]

Well you don't usually write $(\neg \forall x)P(x)$

[flat frigate]

Nel

[glacial cairn]

It looks like a grammatical error

[slim lion]

ya

so it's kinda like having it like
-ab = (-a)b

[glacial cairn]

Like $(\neg \forall x)$ doesn't mean anything by itself

[flat frigate]

Nel

[slim lion]

i get it

[glacial cairn]

Don't quote me on that though, I'm not an expert logician

[slim lion]

algs algs

[glacial cairn]

(maybe ask in https://discord.com/channels/268882317391429632/496785905474994186 to be sure)

[safe radish]

@long yoke Has your question been resolved?

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[ocean rover]

.

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[potent nacelle]

.close

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[tiny geyser]

Any clue?

[tiny geyser]

Montrer que =prove that
On pose= we let
Determiner le rest de la division Euclidian de N par 27= find the rest of the division of N by 27

[open wedge]

,rccw

[winged flare]

do you find remainders of large numbers using binomial theorem in general?

[tiny geyser]

what...?

[winged flare]

ok uh how were you taught to calculate remainders in general

[tiny geyser]

congruence

[winged flare]

try using the result you got from question 1

$10^100 + 100^10 = 10^99 \cdot 10 + 100^9 \cdot 100$

[open wedge]

$10^{100} + 100^{10} = 10^{99} \cdot 10 + 100^9 \cdot 100$

[flat frigate]

1 divided by 0 equals Infinity

[tiny geyser]

i actually didnt even do one

[open wedge]

prove that $\forall n$ that $10^{3n} \equiv 1 \mod 27$

[flat frigate]

1 divided by 0 equals Infinity

[tiny geyser]

no shit

i can read

[open wedge]

that's the first question tho

[tiny geyser]

but 27 is so freakishly large

[open wedge]

you didn't even do this?

oof

[tiny geyser]

that i dont know how to compute it

[tiny geyser]

sorry

[open wedge]

what's 1000 mod 27 btw?

[weary whale]

1

[tiny geyser]

its 1 but how will you prove it?

[weary whale]

there's your answer

[weary whale]

you need to prove it form scratch or can you use properties

[open wedge]

if $a \equiv b \mod m$ then $a^n \equiv b^n \mod m$

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

boom

[weary whale]

maybe they need to prove this too lol

[tiny geyser]

WE START FROM 10 CONGRU -17[27]

we square them both

but 17² is where im stumped

[weary whale]

use a calculator 💀

or you can do 10^3n = 1000^n and 1000 = 1 mod 27

[tiny geyser]

not allowed twin

[weary whale]

is this a joke 🥀

[tiny geyser]

no im serious

[open wedge]

that's 289

[tiny geyser]

whats the rest by 277?

27*

[open wedge]

some olympiad questions does NOT allow you to use a calculator

i only remember 17^2

🥀

[onyx valley]

@tiny geyser27 = 3*9

[open wedge]

or you can prove this

if $a \equiv 1 \mod 3$ then $a^3 \equiv 1 \mod 27$

[onyx valley]

you should manage to get a 999 i believe

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

because i did notice that there's $10^3$ and $27$ is $3^3$

[flat frigate]

1 divided by 0 equals Infinity

[weary whale]

yeah this is way easier

[open wedge]

the way we can prove this is kinda simple

let $a = 3k + 1$ for some $k \in \mathbb Z$

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

and cube that thing

[open wedge]

so try to cube this

[tiny geyser]

thanks guys

think i got it

thanks @open wedge @weary whale @onyx valley @winged flare

.close

[safe radish]

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[open wedge]

np, i appreciate it

[safe radish]

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[brisk pine]

x^2+y^2+z^2&=-2026\ x^3+y^3+z^3&=123456789\ x^5+y^5+z^5&=987654321\ \sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}

[plucky elk]

Need $ around your tex

[mellow lodge]

$x^2+y^2+z^2&=-2026\ x^3+y^3+z^3&=123456789\ x^5+y^5+z^5&=987654321\ \sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}$

[flat frigate]

Mr. BananaHead
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[olive cloud]

No real solution for x,y,z atleast because in 1st eqn it is given sum of 3 >=0 numbers is negative

[tiny geyser]

complex?

whats even the question

[olive cloud]

ya man idk how to solve for complex solutions somebody else help

[olive cloud]

find the value of the expression given at bottom i guess

[tiny geyser]

@brisk pine come

[safe radish]

@brisk pine Has your question been resolved?

[brisk pine]

no

[summer coral]

if you dont want help then dont create a channel for it, this server isnt for giving answers.

[brisk pine]

i want direct answer

[summer coral]

thats not going to happen here good luck with that

[brisk pine]

ok so how to close it

.close

[safe radish]

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[fervent pulsar]

guys]]

[fervent pulsar]

part b

so

i know the formula

so

we can say

CD=5

i got this for line L

idk what to do after

[stoic torrent]

find 2 more sides in vector notation, AC and BC, and then just apply the formula for volume of tetrahedron?

[fervent pulsar]

yh

but u dont have C

its jst like

wtv

[fervent pulsar]

how wld u do it

wld u have variables in AC and BC

[safe radish]

@fervent pulsar Has your question been resolved?

[main zodiac]

?

Who pinging me

[safe radish]

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[coarse yarrow]

hey how could i figure out for which values of λ the lines are perpendicular?

[edgy breach]

what is the rule for perpendicular lines

[coarse yarrow]

m1 * m2 = -1?

maybe my math is just wrong im gettingλ=0 or λ=1 which is apparantly wrong

[plucky elk]

!show

[safe radish]

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

[edgy breach]

did you treat the slopes as lamda and 3lamda + 1?

because thats wrong

[coarse yarrow]

i got m1 = -λ/(λ-1) and m2 is -(3λ+1)/-2λ but ill snap a picture of my work real quick

[edgy breach]

that should be right

[coarse yarrow]

Apparently the solutions are λ=0 or λ=-3 I'm so confused

[edgy breach]

you accidentally forgot to put brackets around 3λ+1

you made it -3λ+1

[coarse yarrow]

damn knew it was gonna be some bs 🙏 thanks for the help

[plucky elk]

cancelling negative signs here in m2 will help with that

[edgy breach]

its annoying when you get a perfect whole number answer like 1 and then its wrong lol

[coarse yarrow]

fr alright thanks guys 🙏

.close

[safe radish]

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[safe radish]

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[soft lava]

Could somebody tell me if I got these true or false questions right

[plucky elk]

,rotate

[red raft]

All are correct acc to me

[safe radish]

@soft lava Has your question been resolved?

[safe radish]

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[worldly briar]

can anyone help me

not sure how to do part b

[stoic torrent]

okay so angle between two given vectors in space is

90 iff their dot product is 0

[worldly briar]

whats a dot product?

i think im used to a different term

[stoic torrent]

okay uh which grade?

[signal crane]

scalar product?

[worldly briar]

y12

[stoic torrent]

do u use scalar product maybe

[stoic torrent]

yeah i figured

[worldly briar]

no

1 sec ill try find it

[round egret]

is this gcse?

paper looks like gcse

[worldly briar]

no

yea i think it is scalar product

the teacher just sucks at teaching vectors he only taught us the i and j notation thing

[round egret]

you can do it without dot product I think?

[stoic torrent]

oh thats alrt so u gotta scalar product th two vectors PQ and QR

[frigid spruce]

yajant

[stoic torrent]

if it comes out to be 0 then you can say that Angle between 2 is 90

[stoic torrent]

say

[frigid spruce]

similar names

[stoic torrent]

oh not the first time

[worldly briar]

what does scalar product mean

like how do u do it

is it like factorising it or something?

this is what i did

[stoic torrent]

basically for any two vectors, scalar product/dot product is simply the multiplacation of respective i, j, k coefficents and then adding them all

[worldly briar]

adding all of them together? like the i values + j values

[stoic torrent]

hold on

[worldly briar]

ok

[round egret]

For vectors $\vec u= (u_1​,u_2​)$ and $\vec v = (v_1​,v_2​)$, the dot product is given by $u \cdot v= u_1 v_1 +u_ 2 ​v_2$ in 2D$

[flat frigate]

MxRgD
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[round egret]

that's the dot product

[worldly briar]

like that?

pq was -6,3

[stoic torrent]

yeah but how did u get -18

[worldly briar]

-6x3

[stoic torrent]

no its -6 x 9

[worldly briar]

ohhh

ok

-6x9 add 3x18

[stoic torrent]

corresponding coefficients of same axis

[stoic torrent]

yup

[worldly briar]

ah i see

[stoic torrent]

and then add the 2 which gives 0

[worldly briar]

ye got that

[stoic torrent]

need help with C too?

[worldly briar]

so is that all you have to show?

[stoic torrent]

yeah

[worldly briar]

im kinda confused, how does that show that pqr is 90 degrees

[stoic torrent]

any two vectors having scalar/dot product 0 have to be perpendicular

[worldly briar]

oh ok

another thing, how did u know which vectors were perpendicular

not ssure how to tell

[stoic torrent]

reason for this being that their is a term of cos($theta)

[stoic torrent]

that was asked in the question!

[worldly briar]

i meant which vectors

ohh

so the triangle would look something like that?

[stoic torrent]

see Q is the pivotal point here, and 2 vectors emerging from Q have angle 90 between them so i think its pretty obvious afterthis right?

[stoic torrent]

yeah just make the angle Q 90

[worldly briar]

yep

my class hasnt been taught vectors properly

[stoic torrent]

.

[worldly briar]

my teacher is so bad, he only taught us the i and j stuff and magnitude

[worldly briar]

sure

[stoic torrent]

oh thats alright happens

[worldly briar]

could u like walk me through the question

[stoic torrent]

first try on your own

[worldly briar]

ok

heres what i wrote so far

not sure how i could get rs

[stoic torrent]

this is more of a diagramatic question

approaching it using diagram would be easier as one of the angles (as prev proved) is 90

[worldly briar]

i drew the 90

where would s go

i dont know how to draw these lines ngl

[stoic torrent]

wait ill send a pic

[worldly briar]

ok

[stoic torrent]

wait i am actually not very fluent in using texit sorry

i could send a picture of my working if thats fine?

[worldly briar]

i did a bit of working out

[worldly briar]

yea sure

[worldly briar]

not sure if its right

[stoic torrent]

oh hey thats right good job

[worldly briar]

oh nice thanks

but idk where to put s

because i dont know how to draw vectors

do they work like coordinates or soemthing

[stoic torrent]

oh do you know how to use cartesian system?

[stoic torrent]

yes thats right,

just put i direction as x

[worldly briar]

one sec let me put it in some graphic calc

[stoic torrent]

and j direction as y

[worldly briar]

i got this

[stoic torrent]

thats correct

now just apply the area formulae and get the answer

[worldly briar]

is it possible to do this without a graphing calc

[stoic torrent]

yeah

diagram is only so that you understand the shape of the quardilateral better thats all

[worldly briar]

would u just draw it like plotting coordinates on a smaller cartesian plane?

[stoic torrent]

even a rough sketch is fine

[worldly briar]

ah ok

so it can be smaller

[stoic torrent]

not even to scale i would just draw rough figure just so i know what position the points are in

[worldly briar]

ah alr

do u need distance formula as well?

[stoic torrent]

yup

[worldly briar]

wow this is a TON for 4 marks

[stoic torrent]

yeah it sure is

but then again i am not really familiar with your solving methods and teacher style so i wont suggest my shortcuts/formulae but just know there are easier methods too

[worldly briar]

Oh alr

270

[stoic torrent]

seems right

.close

[safe radish]

@worldly briar Has your question been resolved?

[safe radish]

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[true venture]

"A 48-year-old person decides to improve their pension security. Starting on their 49th birthday, they intend to deposit €6,000 into an account that earns an annual return of 7.5%. They continue making these identical deposits every year until they turn 65. How much can the person withdraw from the account annually for ten years after retiring? Assume the account is completely emptied during this ten-year period."

I'm having trouble with the second part of this question specifically concerning how you get the exact amount you can take each year when after each withdrawal the remaining principal grows at the 7,5% rate. I know the way to calculate it is **beginning balance *** (i (1 + i) ^years) / ((i + 1)^years -1) but I don't understand how to get there from the given information or how each part relates to the question.

[kindred pier]

i only see one part of the question?

[covert yoke]

@true venture You'll have two parts to this problem, the first is getting the future value of the money given an annualized investment, this is generally called the A/F formula, which is what you gave. This is the easy part. Next you need to figure out the second part, which is another, slightly different application of A/F

The second part is tricky because you are using the formula in a bit of a non-standard way

Actually, I think that might be the A/P formula

[true venture]

What I mean by 2 parts is the first part you figure out how much you have after 17years of investing 6000 per year. the second part is how much can you take out each year for 10 years when the amount has to be the same each withdrawal.

[covert yoke]

For the second part, you have to solve for it using a negative deposit, (but the interest rate stays the same)

[true venture]

I guess what I'm having a hard time with is grasping how the equation takes into account the fact that you withdraw the same amount each time and the rest keeps gaining interest maybe I'll ask my prof next class or just hope I don't have to understand it fully just recognize the problem and remember the formula :D

[true venture]

I'll look into A/F and A/P maybe that'll give some insight

.close

[safe radish]

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[safe radish]

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[deft furnace]

Number 8, idk what to do with the ^3

[glacial cairn]

I don't see an 8

[hard crest]

its the first thing with the f'(5)

[hard crest]

idk what it's called in arabic but you would use the chain rule here

if $f(x) = (\psq)^3$ then $f'(x) = 3(\psq)^2 \cdot (\psq)'$

[flat frigate]

schrödinger's kitten

[deft furnace]

Oh thank you so muchhh

I’m stuck

[hard crest]

some corrections

[deft furnace]

Ya I’m half asleep ignnoreee

[hard crest]

now start evaluating, so log(5-1) becomes log(4)

[deft furnace]

(Log4)^2

?

[hard crest]

yea but work from the inside out

so next would be like
$3(\blue{\log_2(2^2)})^2 \cdot \f1{\ln2\cdot4}$

[flat frigate]

schrödinger's kitten

[hard crest]

focus on the blue thing, that can be simplified

[deft furnace]

4…?

[hard crest]

hmm yes but $\log_24$ can be simplified

[flat frigate]

schrödinger's kitten

[deft furnace]

So log2

?

[hard crest]

no, just 2

[deft furnace]

Whats

[hard crest]

thinking of rule 3 here

$\log_22^{\purple2} = \purple2\log_22$

[flat frigate]

schrödinger's kitten

[deft furnace]

let me stare at it gimme a moment

[hard crest]

okie

remember what log actually means as well

[deft furnace]

Oh then we removed the 2 and 2

[hard crest]

there are several 2's :)

[deft furnace]

The big ones

We remove them?

[hard crest]

ummm well the next thing we do is notice that $\log_22 = 1$

[flat frigate]

schrödinger's kitten

[hard crest]

so then we get $\purple2\cdot1 = \purple2$

[flat frigate]

schrödinger's kitten

[hard crest]

so we simplify $\log_22^2 \to 2$

[flat frigate]

schrödinger's kitten

[deft furnace]

I’m sorry gimme a moment my heads in a wall rn 😭

[glass egret]

rule 6

[hard crest]

oh yeah thats a more straightforward way lol

[mighty mango]

كانت بتكلمك علي log_2(2^2)

من خواص اللوغاريتمات (logs) انك ينفع تنزلي اي اس (exponent) جوا الlog ل بره الlog

يعني, log(a^b)=b*log(a)

ينفع تستخدمي نفس الخاصيه مع log_2(2^2) ب انك تنزلي ال exponent بره الlog
log_2(2^2) =2 log_2(2)

و بردو من خواص اللوغاريتمات ان log_a(a)=1

2 log_2(2) = 2 x 1 = 2

thats what hayla said.

[hard crest]

couldnt have said it better myself

[safe radish]

@deft furnace Has your question been resolved?

[glass egret]

😭

[safe radish]

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[thorn junco]

i am a bit confused on the way an exponent effects whether or not equations qualify as functions or not. i am not understanding why/whether an exponent on top of the x in y=mx+b means that the value for x has to be positive

[fleet willow]

what must a function satisfy?

[thorn junco]

a function has to have exactly one y value to each x value in the domain

[fleet willow]

wait just to be clear this is a function f : R -> R right

[thorn junco]

the relation has to satisfy that there is only one y value for each x value

oh my, i haven't learned what that means yet 😭 i'm on very basics of functions, im sorry

[fleet willow]

are x and y real numbers lol

[thorn junco]

yes, so far they've all been integers

[fleet willow]

ok, so we know there's one and only one y for each x

now consider x = -3

do we have such a y?

[thorn junco]

yes, that equals -243 if i multiply 3(-3^4)

[fleet willow]

i'm not saying y = -3

i'm saying x = -3

[thorn junco]

that would be -3/3 which is -1 and then moving the exponent i assume makes it -1^4.. im not familiar with moving exponents across the equal sign, does it change?

[fleet willow]

well, actually, you're partially correct

if x = -3 then y^4 = -1

but does such a y exist?

(sidenote: ||y = -1 doesn't work||)

[thorn junco]

on a graph, y can be -1. why is it that it doesn't work on paper?

[fleet willow]

if y = -1, then x = 3

lol

[thorn junco]

oh, so you're saying that if you plug it into the equation it comes out janky?

[fleet willow]

yeah

well uh

x = -3 has no solution

for any y you choose

(proof: ||trivial inequality on y^2||)

[swift fjord]

you’re correct

oh nvm

[thorn junco]

so it's not a function because it can't have negatives plugged in due to the exponent?

[fleet willow]

ye

also

x = 3 has multiple solutions—what are they?

[thorn junco]

1=y^4?

is that one of them? i still have no idea what to do with exponent but ill look up a youtube video on that HAHA

[fleet willow]

did you learn about difference of squares

[spiral birch]

for it to be a solution you would have to have y = something
but you are correct in that y is a solution if and only if 1 = y⁴

[thorn junco]

if you showed it to me i'd probably recognize it, but i don't recognize it by name

[fleet willow]

$a^2-b^2=(a-b)(a+b)$

[flat frigate]

nadat12

[thorn junco]

ah! i havent utilized that in my class yet, my algebra education was severely lacking in highschool. so no, i'll look that up as something to learn as well!

i think that i should close this since i got the basic answer i needed and will be looking into the concept more. thank yall!

.close

[safe radish]

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Closed by @thorn junco

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[safe radish]

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[trail heron]

Had an issue computing partial derivatives; didn't exactly know how to describe it, show I'll show you the images of my attempts alongside my worksheet- since my work felt so big, I have trouble finding errors. What am I missing?

(I noticed that the help forum is supposed to be precalc, I can close it if that's not what you want)

[warm hedge]

One good suggestion:

At $f_{xy}$, for your final step, don't factorise $z^9$ out.

[flat frigate]

Restarter

[warm hedge]

Leave it in the form: $e^{xyz^9}\left(xyz^{18}+z^9\right)$.

[flat frigate]

Restarter

[warm hedge]

Then you differentiate with respect to z.

This would be much easier.

[trail heron]

Alright, I'll try to make some changes, thanks!

.close

[safe radish]

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[safe radish]

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[safe radish]

Please don't occupy multiple help channels.

[empty gyro]

<@&268886789983436800> underage

[junior smelt]

Yea, sorry, we can't have you here ToS and all

.close

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[lone void]

i read that as ln(12)

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[eager sable]

Can somone help me make a gokart track in desmos

[eager sable]

it needs 8 or 9 functions

and atleast 5 equations

• Parabola hyperbola, cubic, linear, quadratic, circle.

thi sis teachers exampelar

my teacher asked us to start on paper

before implementing in desmos

[stoic torrent]

okay lets start

[eager sable]

ok

wait

i made a rough sketch on how i want it to look

[stoic torrent]

to make a gokart track, we need it to be continous at all points, aka a loop

[stoic torrent]

sure send it over

[eager sable]

Is this viable?

[dull sequoia]

Think about what functions and shape each part would be

[eager sable]

alr

Im not sure about the others

[stoic torrent]

okay have you divided it into parts?

[eager sable]

Here

[stoic torrent]

okay so the bottom most part left to the hyperbola is a parabolic path right?

[eager sable]

huh

[stoic torrent]

quick question are you allowed to use sinusoidal functions too?

[eager sable]

is that like the wave

sine wave

i dont think so

the orange one?

i dont think so

[round egret]

yeah the orange is a sin wave

[eager sable]

thank you

[stoic torrent]

hello there

[round egret]

hello yajant

[eager sable]

ok so

[eager sable]

does this look good?

should we start working them out

and putting it in desmos?

[stoic torrent]

yup

[eager sable]

should we start with a linear?

the on on the left?

[stoic torrent]

yes

[eager sable]

wait what are the blu eones

[stoic torrent]

prbola

[eager sable]

alrr

[stoic torrent]

ur maths teach seems pretty cool

good way to learn functions

[eager sable]

ye

he is

I'm stuck

ik the equation is y=mx+c

[stoic torrent]

okay so the first line looks like vertical right?

so for the first line, you need a line parallel to y

[eager sable]

uh

30,40

to

30,70?

do i calculate the gradient

with those

[stoic torrent]

yup

[eager sable]

then sub into y=mx+c

to find c?

[stoic torrent]

but if you calc gradient, its just infinite

[eager sable]

so we need domain?

[stoic torrent]

this is an excpetional case as such lines cant be put in y=mx+c format

[eager sable]

oh

could you show me how we would calculate this?

[stoic torrent]

do you know how they are represented?

whats the equation of y axis?

[eager sable]

uhh

idk

[stoic torrent]

its x = 0 right?

[eager sable]

oh

yeah

[stoic torrent]

as along the y axis everywhere x=0

[eager sable]

yes

[stoic torrent]

similarily if i make a line which passes thru these poijts,

what would the equation be?

[eager sable]

i dont get it

sorry

im running low on sleep

oh wait

i think i get what your saying

vertical lines cant be written as y=mx+c

cuz its just x

so cant we just do 30x

[stoic torrent]

yes!

[eager sable]

and have a domain

[stoic torrent]

yes that right

[eager sable]

ok ok

[stoic torrent]

so you could just write it as x= 30 right?

[eager sable]

yeah

[stoic torrent]

as x would be 30 everywhere

[eager sable]

yes

[stoic torrent]

oky good we have our first function

now lets do one more linear

[eager sable]

x=30 (40 ≤ y ≤ 70)

is this how i would put it in desmos?

[stoic torrent]

yup

[eager sable]

but with curly brackets

[eager sable]

shouldnt we move on to the hyperbole below or above the linear?

thats how my teacher did it

[stoic torrent]

okay then lets do that

[eager sable]

ok

y=a/x-b?

[stoic torrent]

okay sure

[eager sable]

y=a/(x+h)+k

[stoic torrent]

sure this is bteer

[eager sable]

ok

i have jno idea on how to do this one

[stoic torrent]

okay lets see the equation of hyperbola

y = a/(x+h) + k

[eager sable]

ok

[stoic torrent]

one sec

[eager sable]

no worries

[stoic torrent]

hey sorry abt that

[eager sable]

its all g

[stoic torrent]

so for hyperbolas equation, we need to know what a, k and h would shift the hyperbola

[eager sable]

ok

[stoic torrent]

https://www.desmos.com/calculator/2ge9jdzdue

See for changing a, k and h change the position and nature of hyperbola

[eager sable]

yeah i see

how would we work this out

on my book?

and sketch this

[stoic torrent]

on the book its a little tough

i would advise to use desmos only

so first of all we need to shift it up right?

[eager sable]

yeah

[stoic torrent]

see that increasing k shift the hyperbola upwards

so first increase the k to appropiate height

well?

[eager sable]

back

to 40 right?

wai tno

70

[stoic torrent]

put in desoms and see

[eager sable]

?

[stoic torrent]

yupp

now input the limits of y in paranthesis

[stoic torrent]

nice

put limits for x too

now start working on the linear n once ur done with that ping me

[safe radish]

@eager sable Has your question been resolved?

[warm hedge]

Interesting.

[warm hedge]

Oh must be hyperbola?

Because a circle would make that easier to be honest.

[warm hedge]

Why is there a broken segment over there?

[stoic torrent]

good now go for the linear

[stoic torrent]

his assignments says there must be some hyperbolas too

[eager sable]

i think we can make 2 of the turns circles

[warm hedge]

Oh okay.

[eager sable]

and hte bottom right a hyperbola

[stoic torrent]

okay go for it then

[eager sable]

ok