help-0

[alpine sable]

So I'm not sure if I should say it

[north hemlock]

I went from 3^103 to 3^503

which doesn't help

[alpine sable]

I have no idea how we could've figured it out from scratch

[north hemlock]

shit

[alpine sable]

Should I tell you?

[north hemlock]

sure

idm

[alpine sable]

3^100==1(mod 1000)

[north hemlock]

o

[alpine sable]

No freakin' idea how we could've figured that out

[north hemlock]

hmm

so really just 9 mod 1000 is the answer?

[alpine sable]

27 mod 1000

[north hemlock]

3^3 =9 ?

oH AIGN

RGN

AMEPINGR

shhh

LOL

yes

I def know exponents 🤡

[alpine sable]

📸

Now that I'm done getting free social points for your mistake, let me get back into the problem

Like we have the answer but didn't use Euler's totient theorem any bit

[vapid shuttle]

Euler has too many things anyways

we need to stand up every once in awhile and reject his theorems

use new methods

https://tenor.com/view/1984-gif-19260546

"use Euler's totient theorem"

Oh crap, sorry guys I thought this was the helpers-lounge channel

Sorry for interrupting your help channel so much

[north hemlock]

we do for the 3^100 part actually

I just realized

1000 = 125 * 8

so

$3^{100} \mod 1000 \implies 3^{\phi(1000)}\equiv 1 \mod 1000 \implies 3^{400} \equiv 1\mod 1000$

[ocean seal]

Juke | ping me if no response

[north hemlock]

since

400 = 4 * 100

$3^4 \equiv 1\mod 8 \text{ and } 3^{100} \equiv 1\mod 125$

[ocean seal]

Juke | ping me if no response

[north hemlock]

and multiplying those gives 1 mod 1000 ofc

as I wrote above

[alpine sable]

@north hemlock Is this meant to be proof that 3^100==1(mod 1000) or are you not finished?

[north hemlock]

yes

that's just that part

I think

[alpine sable]

I don't get it

Okay right so "multiplying those gives 1 mod 1000", sure

But...

Hmmm

Oh wait

So 3^4==1(mod 8)

Meaning (3^4)^25=3^100==1(mod 8)

Hence we have 3^100==1(mod 8) and 3^100==1(mod 125), meaning 3^100==1(mod 1000) since 1000 is the LCM of 8 and 125... Well I think you can justify it like that? That LCM part made intuitive sense in my head but I haven't seen proof of that

@north hemlock Does what I just said make sense?

[north hemlock]

yesh

to me at least kekw

[alpine sable]

Good

[north hemlock]

time to move onto even more cringe problems that'll suck

[alpine sable]

I guess the general strategy is to split the number into coprimes, apply Euler's totient function to both, then conclude 3^(LCM of powers)==1(mod LCM of coprimes, which is just the original number)

[alpine sable]

ono

[north hemlock]

yeha

[alpine sable]

@north hemlock Where are the even more cringe problems that'll suck?

[north hemlock]

I'm doing this one rn

[alpine sable]

ono

[north hemlock]

I think true, but I haven't finished

wait

this should be obvious bc gcd of any two primes is always 1?

[alpine sable]

This is the Mersenne number of a prime

Not necessarily a prime itself

[north hemlock]

oh shit

$2^{2^n-1}-1$

[ocean seal]

Juke | ping me if no response

[north hemlock]

this thing hmm

[alpine sable]

What the fudge

Why is it nested

[north hemlock]

mersenne primes are 2^n-1

[alpine sable]

The prime itself is not necessarily a Mersenne prime

[north hemlock]

so this is what we're working with

[north hemlock]

yeah that's fine

[alpine sable]

So it's not always gonna be of that form...?

Plus you're reusing n in a way that doesn't match with the original question

Now n means something else

I have a feeling this is gonna be a proof by contradiction

Say there exists m and n such that 1<=m<n and M_(p_m) shares a factor with M_(p_n)

Actually wait, we don't even have to set m<n like in the question

So we'd actually be assuming that there are 2 distinct primes who's Mersenne numbers share a factor then show how that leads to a contradiction

[north hemlock]

hm

[alpine sable]

Alright hmmm

This is a case where I feel like playing with the values might help

Let me see what the prime factorization of the Mersenne numbers up to 2^10-1 is...

[north hemlock]

2^10- 1 crazy

[alpine sable]

ikr

[north hemlock]

$p_n = 2^a- 1 \implies M{p_n} = 2^{2^a-1}-1$

um

[ocean seal]

Juke | ping me if no response

[north hemlock]

Idk why but I feel like this is

nah idk, this is weird

[alpine sable]

I on the contrary have a feeling this is irrelevant

[north hemlock]

yeah

but for some reaosn I keep gravitating to it

yeah I'm gonna do this later

I have no idea

.close

[lone heart]

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[alpine sable]

Deleted his epic ,close failed attempt

Like come on

[lone heart]

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[icy nymph]

how would ones self complete this question without trial and error:

[woven tiger]

I am not too sure but turn it into an equation so 2n plus 1 equals 3n -1

[ocean hawk]

that won't help here

[limpid turret]

I would just write out the values between 20 and 40 and match them

[icy nymph]

you wont have time for in an exam is there not easier way

[limpid turret]

Or you can solve 2n+1=3m-1 and find integer solutions, but that requires a lot of mental math

[ocean hawk]

you can just note which set of numbers can be represented by "2n + 1", and see which are also in 3n - 1 between 20 and 40

[icy nymph]

.close

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[dull nymph]

i felt like it was but for my p value when i input it in my calculator i got 0.0000

[lone heart]

@dull nymph Has your question been resolved?

[dull nymph]

<@&286206848099549185> sorry its been 15 min and im still stuck </3

[ocean hawk]

what do you mean by "1 propztest"?

[dull nymph]

the tests the calculator runs

like z test and t test and such

im in a college statistics class

[dull nymph]

Its under stats on calculator 84

[ocean hawk]

well you have 2 groups, 2 proportions

[dull nymph]

for that other one

[dull nymph]

bru

[ocean hawk]

for that other one, then yes it would be the 1 prop z test I think. with appropriate H1

[dull nymph]

Thank you I'm being killed by this assignment

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[humble heron]

@alpine sable can i dm?

[lone heart]

Please don't occupy multiple help channels.

[alpine sable]

!onechannel

[lone heart]

Please stick to your channel.

[alpine sable]

.close

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[solar maple]

fastest way to find sin(110)

[solar maple]

it can be in any form

I was thinking of splitting it but sin of 20 is stilll hard

[alpine sable]

maybe sin(3x) can help u

[solar maple]

wdym

sin(330?

then splitting it

then dividing by 3?

or something?

also i'm looking for the fassstest way

like zoom speed

[alpine sable]

using 110 = 90 + 20

[solar maple]

but what's sin of 20

[alpine sable]

u can have sin(90+20) and use the sum

[solar maple]

ik

I was thinking that irst

[alpine sable]

yes and now u use sin(60) = sin(3 * 20)

[solar maple]

but idk what sin of 20 is

Oh

interesting

maybe

[alpine sable]

u will have a cubic equation

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[final badge]

Whers did i go wrong on number 7?

[red hedge]

hi guys how do i solve this question?

the options are 2,3,4 and 8

[limpid turret]

number 7 what?

[limpid turret]

!help

[lone heart]

Please read #❓how-to-get-help

[worn fox]

<@&268886789983436800>(if that wasn't deleted by a mod then check deleted spam message of Kitty....)

[final badge]

Sorry, it didnt load.

load.inernet issues

Brub this it still not loading

[final badge]

FINALLY

IT LOADS NOW

[limpid turret]

Understandable. Sorry about your slow load times

[final badge]

Yeah thats life. Anyways, what did i do wrong with oh... its number 6 lol

What did i do wrong on number 6?

I have a feeling its its something with the division/fraction sign it came as

[limpid turret]

7 looks perfect

[final badge]

...

[limpid turret]

,w plot y=2(x+1)^2-3

[ocean seal]

Results provided by WolframAlpha

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

Requested by SWR#8588

[final badge]

Oh wait i meant 6

[limpid turret]

Are you solving $\frac{7^x}{7^2}=343$?

[ocean seal]

SWR

[final badge]

Yes

[limpid turret]

Mistake here

-2, not 2

[final badge]

Figures

So, whenever its 7^x/7^2

Its always gonna be a negative?

[limpid turret]

yup

$\frac{a^b}{a^c}=a^{b-c}$

[ocean seal]

SWR

[final badge]

Ah. Kay

Thank you!!

:)

.close

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@tiny mango Has your question been resolved?

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[dire gale]

It would be a bar chart because it's discrete?

.close

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[lone heart]

@ember mango Has your question been resolved?

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@ember mango Has your question been resolved?

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[low scaffold]

are these right?

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[drowsy fractal]

∫ (|x-2|+1)dx
b = 3
a = 0

[lone heart]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[drowsy fractal]

I'm not really sure how to integrate with the absolute value

[jagged cobalt]

i recommend splitting the integral into 2;
one where the bounds will have x-2 to be positive so its just (x-2)
and another where the bounds have x-2 to be negative so its -(x-2)

[drowsy fractal]

So along the the lines of this?

[bright prawn]

(x-2) is positive if x is between 2 and 3, and negative if x is between 0 and 2

So those should be the bounds of the two integrals respectively

[drowsy fractal]

Ohh, thank you that makes sense

.close

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[lone heart]

Please don't occupy multiple help channels.

[foggy knot]

can someone help me with this

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[foggy knot]

Yes

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[visual depot]

halo

[visual depot]

@wary stream

,rotate

[visual depot]

.close

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[rustic juniper]

Goodmorning, can anyone help me answer this problem

[scarlet oriole]

bro I'm lance

[rustic juniper]

My name is Lance too

[rustic juniper]

Can you help me with this problem?

[scarlet oriole]

what's variable x representing?

since it's not determined I'm going to say that x represents just the number of pink items out of 20 mothers

meaning probability that out of 20 mothers 11 get a female is 0.160

still something seems a bit off I don't get what the problem is asking

[rustic juniper]

P (4 pink nursery kits needed out of 20) = 20C4 * (0.60)^4 * (1-0.60)^(20-4) = 0.0003 is this right?

[scarlet oriole]

yes

binomial probability law

[rustic juniper]

So this will be the answer?

[scarlet oriole]

answer to what

[rustic juniper]

To this problem

[scarlet oriole]

problem had no question

like literally problem had a solved table of values

unless you filled those up

[rustic juniper]

The variable X should be until 20

[scarlet oriole]

here's a thing

if it asks ≥4 you need to add

each probability p(1)+p(2) and so on till 4

and when it comes to other cases X=x

you just use binomial formula you posted above

if it's asking of normal distribution then gl

[rustic juniper]

So if it's asking for ≤4 i will use Binomial Distribution formula?

[scarlet oriole]

you will but you use it 4 times

for each case of success 1/2/3/4

cause it's essentially asking what's the probability to get any of those specific scenarios

not just 1 scenario

for example lance what's the probability of me getting ≥20 kits?

it logically is 1

i expect on the trial that 1 of those scenarios will happen

i might get 1 kit only

or even 20

so chances are 100 that ≥20 will happen

makes sense?

[rustic juniper]

Yes thankyou for that explanation i got the point for this binomial distribution thing

[scarlet oriole]

no problem

[rustic juniper]

Thankyou very much Lance and sorry for wasting your time

[scarlet oriole]

no problem Lance 💀

[rustic juniper]

.close

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[random latch]

I'm stuck on this example work on stokes' theorm. My first issue is that I would have to take curl of F=<x,y,z> which is just the zero vector, and I tried parametzing the boundary of the graph which is a circle and doing a line integral from that but I also get zero because again F=<x,y,z> which makes F(r(t)) orthogonal with r(t) which gives me zero, maybe I have this all wrong, I can send an image of the problem to help. They state the answer is -pi which is my issue

[mossy burrow]

young modulus

here’s what i have so far

[mossy burrow]

sorry u posted same time i did

[random latch]

My bad g

[mossy burrow]

it’s ok

[random latch]

I can close if you need the help

[mossy burrow]

no i’ll move lollll thank u tho

sorry ahaha

[scarlet oriole]

24 post the question again

[random latch]

I'm stuck on this example work on stokes' theorm. My first issue is that I would have to take curl of F=<x,y,z> which is just the zero vector, and I tried parametzing the boundary of the graph which is a circle and doing a line integral from that but I also get zero because again F=<x,y,z> which makes F(r(t)) orthogonal with r(t) which gives me zero, maybe I have this all wrong, I can send an image of the problem to help. They state the answer is -pi which is my issue

I meant r'(t)

Bad with LaTEX

Here's an image for context

[lone heart]

@random latch Has your question been resolved?

[random latch]

No

<@&286206848099549185>

[random latch]

.close

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[hollow tartan]

I need help

Not sure what to do after this

Supposed to solve as simultaneous

[molten pivot]

What is log_3(1)

[hollow tartan]

Log_3(1) is 0

Wait shit

Yea it is

[lone heart]

@hollow tartan Has your question been resolved?

[chrome salmon]

who made jungkook for you

[hollow tartan]

??

Can help?

[chrome salmon]

your profile pic

[hollow tartan]

Can help first?

[chrome salmon]

yes

[hollow tartan]

Ya then?

[chrome salmon]

For $a,c >0$ when we have $\log_a b = \log_a c$ then $b=c$

[ocean seal]

fäf

[hollow tartan]

Wait

Then is it

a=b2

[chrome salmon]

yes

[hollow tartan]

I see

[chrome salmon]

say pfp

[hollow tartan]

Not yet

Then what do I do next ah

[chrome salmon]

nothing more

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@hollow tartan Has your question been resolved?

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[proper creek]

hello! i just wanted to know whether in this formula, if we put x=1, does 1/(n!) converge to e?

[vale wigeon]

the sum of 1/n! for n from 0 to infinity converges to e, yes.

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[proper creek]

thank you!

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[ornate condor]

can i think of a tensor as a n dimensional matrix

[heady pollen]

Not always

Numbers are 1d
Vectors are 2d
Matrices are 3d
Tensors can be nd

Notice that vectors can also be 1d
Matrices can be 2d and therefore also 1d
Tensors can be 3d therefore also 2d and also 1d

Every number is a vector, not every vector is a number

Every vector is a matrix, not every matrix is a vector

Every matrix is a tensor, not every tensor is a matrix

An example would be if you have a fluid and want to know the temperature everywhere. You would have a 3d tensor, a cube i guess

[ornate condor]

when u say is a, u mean there exists some isomorphism right, from x to (x), but not from say (x,y) to any number

like a tensor is a generalised matrix?

[heady pollen]

Ah yes, isomorphism is better said

[ornate condor]

also sry im new to tensors, what operations exist

like what are these

[heady pollen]

Mind you i never used tensors except for the once that are below 3d

[ornate condor]

oh these are 3d

oh wait below

um ok lol

https://www.nature.com/articles/s41586-022-05172-4

i realised i cant quite understand this as they perform some operations on tensors which i dont quite get?

what are these operations called and what do they do

nvm sry i found it earlier in the paper, outer tensor product

ill b back

thanks martin

💕

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[alpine sable]

I’m stuck

Nvm

I got it

.close

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@alpine sable Has your question been resolved?

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[sharp plaza]

I don't understand how this integration is done. I would like to know the steps.

[mortal trellis]

try differentiating the resulting expression

see what you get from the chain rule

normally you'd do a u-sub to get there but they skipped that

maybe they thought it was obvious enough, not sure

[lone heart]

@sharp plaza Has your question been resolved?

[thick wyvern]

Please help me with this question:

Sarah and Eric are skating and holding a constant speed. Their position after t seconds is ls:{x=10+6t, y=8+3t} and {x=18+3s, y=25-5s}. Find the smallest distance between Sarah and Eric

[sharp plaza]

Why would the answer be important?

[thick wyvern]

It is my homework

[sharp plaza]

this is not how this problem solving works type !help to find it out

[thick wyvern]

!help

[lone heart]

Please read #❓how-to-get-help

[sharp plaza]

.close

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[spiral basalt]

if i use u substituon of u = sin^2x will my answer be wrong?

[echo socket]

No, it won't

Would even be better to pick the entire denominator as u

[vale wigeon]

you deleted your original message, now you must abandon this channel and claim a new one

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[silent roost]

Is there a way to solve this without using u-substitution? The tip is that the numerator is the derivative of the denominator

[silent roost]

by my understanding, i am supposed to use

along with

where G is the integral of the factor of y, above

I get

but I can't solve that integral, using just the tools above

[ocean hawk]

partial fractions maybe? substitution is much simpler

[pseudo ice]

Well if I gave you f’(x)/f(x), do you notice that as the derivative of something?

[silent roost]

hmmm, not really? tbh

It's been a few years since I did any calculus

and this specific problem seems unlike all the others before which i solved effortlessly

crap I gotta go apparently - if anyone has any ideas PLEASE dont hestitate to PM me for later, I WILL read and appreciate it! I have spent hours on this shit

.close

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[hollow root]

hey

[hollow root]

oh a second

but this is what I get for the first graphical question:

[tame tangle]

In first equation,

$x=1\implies y=-7$

[ocean seal]

Void Walker

[hollow root]

oh

alright is anything else right?

[tame tangle]

I'm checking

[hollow root]

okayy

[tame tangle]

Yes

And also, a tip is to set x=0 once and then y=0 next

[hollow root]

I did it in one equation and I remember getting fractional values :/

[hollow root]

okay let me redraw then

[tame tangle]

It doesn't matter, you can plot approximately

Graph paper makes it easier to approximate

[hollow root]

my teacher told me to not do that, but that is true, yes.

[tame tangle]

Then it's a different story

[hollow root]

btw you have time until I redraw, right?

[tame tangle]

Yes, absolutely

[hollow root]

tysm

[tame tangle]

Np fellow human

[ocean seal]

aadi_mone

[tame tangle]

Which equation?

[hollow root]

I mean its the point of intersection?

[hollow root]

the previous question, that we were talking about

[tame tangle]

Yes, looks to be it

[hollow root]

btw

[tame tangle]

Yes?

[hollow root]

I knew the answer before, but sometimes when I draw lines straight it doesn't meet to the real answer

the line moves a bit forward the solution changes

(-1, -2) but real answer is (-1,-1)

any tips to avoid this?

[tame tangle]

How off the mark is it usually?

[hollow root]

the question is based on 5 marks

[tame tangle]

Because there can be a number of reasons

[tame tangle]

I mean like how far away is your answer to the "real" answer

[hollow root]

oh like this

or sometimes

if its 3

I get like 3.5

[tame tangle]

That is most probably just some sketching error, when you do things by hand it usually occurs

[hollow root]

yeah :/

so I can't really do anything about that?

anyway lets move on to the second question

[tame tangle]

I did use to get that but can be solved by just being a bit more careful

[tame tangle]

Let's go!

[tame tangle]

First step: plot them

[hollow root]

yeah I am finding x and y

do you think my speed is okay?

[tame tangle]

Yes it is

[hollow root]

is this correct?

[tame tangle]

Looks to be

[hollow root]

okay let me plot

I plotted

both equations have a same point

[tame tangle]

Show me please

[hollow root]

it looks a bit ugly, but I hope you can bare with me

[tame tangle]

I shall

Looks good tbh

[hollow root]

is it correct?

please check

[tame tangle]

Your line for the first equation is slightly off at (1,4), but otherwise all good

[hollow root]

okay

now we should find the area of the triangles

and form the ratio

what is the height of the triangle in the y axis?

[tame tangle]

Did you identify the triangles first, and if so, which are they (shade them lightly and send me a pic)

[hollow root]

there are 2 triangles btw

[tame tangle]

Correct

Now here, for the y-axis triangle, you can take the length on the y-axis as the base and the perpendicular to (1,4) from the y-axis as height

(just rotate the triangle 90⁰ anticlockwise and you should see the reason)

The height is thus 1 unit

[hollow root]

oh

and base is 5 units?

[tame tangle]

No, the base is 5-2=3 units

[hollow root]

5-2?

[tame tangle]

No, sorry, 6-2

=4

[hollow root]

uhh

[tame tangle]

Just calculating distance from (0,2) to (0,6)

That's since the triangle has vertexes (0,2), (0,6) and (1,4)

And we're taking the side from (0,2) to (0,6) as our base (the y-axis)

Was that clear enough?😅

[tame tangle]

Correct

For a reminder, you can write base = DH

[hollow root]

what is the base of the second triangl?

the line is supposed to hit on 3 so

[tame tangle]

CB

Bcoz the 2nd $\Delta$ is ABC

[ocean seal]

Void Walker

[hollow root]

what is CB?

[tame tangle]

You marked in your graph

[tame tangle]

Here

[hollow root]

oh that

yes what is the value of CB?

is there no numerical value for that?

[tame tangle]

There is

It is the distance between (3,0) and (-1,0)

Which is 3-(-1) =4

[hollow root]

so all the time bigger value - smaller value?

[tame tangle]

IF AND ONLY IF you're measuring on an axis (or a line parallel to the axis)

Otherwise there's a formula

[hollow root]

I don't know the formula, but mind sharing it?

[tame tangle]

One min

[tame tangle]

Distance between $(x_1,y_1)$ and $(x_2,y_2)$ is
$\sqrt{ (x_1 - x_2)^2 + (y_1 - y_2 )^2}$

[hollow root]

oh crap

I guess I'll study it later

but thanks

[ocean seal]

Void Walker

[tame tangle]

Np

[hollow root]

thank you my questions are solved

[tame tangle]

Actually it's just the Pythagorean theorem in disguise

[tame tangle]

Yaay! 😄

[hollow root]

yeah I know it I guess, but I never used it

thank you so mucuh

[tame tangle]

Happy to help

[hollow root]

I hope everything gets sorted for you 🙂

[lone heart]

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[rigid mesa]

could someone help with this geometry problem?

[quasi vector]

what is p?

[mellow grail]

find p..?

[rigid mesa]

p is perimiter

[mellow grail]

i see

[rigid mesa]

like ive literally tried everything

the furthest i get is

[mellow grail]

I amma see

[rigid mesa]

finding OA and that's it

and the way i do it is very very tedious

[mellow grail]

well how are you calculating cosine of alpha?

what triangle are you considering

also, let AN = x, AM = y.
y = x-5

[rigid mesa]

im not

its given by the problem statement

[mellow grail]

also whats beside 2 and 3? or is that 12/13 (i think its an arrow)

[mellow grail]

oh

[rigid mesa]

the drawing is a summarizatio nof the problem statement

[mellow grail]

got it 👍

[rigid mesa]

btw its parallelogram

[mellow grail]

yes, i have spotted that

[mellow grail]

(in cosine expression)

[rigid mesa]

wait wdym

its just cos(alpha) = 12/13

[mellow grail]

ah yes, i was confused

[lone heart]

@rigid mesa Has your question been resolved?

[mellow grail]

well i havent been able to do this

i figured out alpha

[lone heart]

@rigid mesa Has your question been resolved?

[lone heart]

@rigid mesa Has your question been resolved?

[rigid mesa]

<@&286206848099549185>

[pearl void]

hi

[alpine sable]

Ok

I got it

@rigid mesa

[rigid mesa]

how

[pearl void]

a very special number, 4533333345, and no, it's not sheesh

[rigid mesa]

tell me 😭

[alpine sable]

Is the answer 52?

[rigid mesa]

yeah

[alpine sable]

Ok

Just a min I'll send a pic to explain this

[rigid mesa]

ok sure

[alpine sable]

Do the same thing for the other two parallel sides

[rigid mesa]

holy shit

its right there

omg

im so dumb

[alpine sable]

[rigid mesa]

its literally right there

[alpine sable]

It took me long as well lol

[rigid mesa]

how much time

did u think on this

[alpine sable]

Idk like 10min

[rigid mesa]

yeah thats not long

it took me 30 mins before i asked the question😂

ah anyways

thank u

[alpine sable]

Np np 👍

[rigid mesa]

im so dumb... smh

.close

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[brittle sun]

Is the following function:

f(x) = x for any x < -1
      -x for any x > 1
      undefined for any x in between -1 and 1

bound?

[brittle sun]

Essentially, is a function that is undefined in some area still considered bound

Like is f(x) bound in [-2,2] because -2 <= f(x) <= 2?

[marsh rapids]

a function can only be bounded on some part of its domain

that's the only kind of set for which the question can be asked

[plain flame]

in ordered to be bounded it needs to be defined

undefined is neither less nor greater than any number

its bounded on [-2, -1) union (1, 2]

[brittle sun]

I see

I couldn't quite recall the definition

Thanks fellas

.close

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[vagrant dagger]

I´m trying to calculate the intersect of two lines, both lines are made up of measured values.

Basically trying to figure out how to do interpolation with 2 missing values.

[lone heart]

@vagrant dagger Has your question been resolved?

[vagrant dagger]

.close

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[brave stratus]

hey, can someone help with finding inflation points for y=^3√(x+1)-^3√(x-1), second derivate is ( 2/9)*(1/(x - 1)^(5/3) - 1/(x + 1)^(5/3))

[alpine sable]

inflection points?

equate the second derivative to 0

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@brave stratus Has your question been resolved?

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@brave stratus Has your question been resolved?

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[ionic sedge]

if f' has a maximum at x=c then f switches from concave up to concave down at x=c

[lone heart]

Please don't occupy multiple help channels.

[ionic sedge]

i believe it is true because if f'(x) has a local maximum at x=c, then the second derivative test tells us that f''(c) is negative, meaning f is concave down at x=c, which implies that f switches from being concave up to concave down at x=c.

can anyone verify if im correct

[vapid shuttle]

!15mins

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[ionic sedge]

ok

<@&286206848099549185>

[late sage]

sup

[ionic sedge]

hey can you verify if its true or not

if f' has a maximum at x=c then f switches from concave up to concave down at x=c

[shell widget]

@ionic sedge If c is an endpoint then it is possible that the statement is false, otherwise it's true I believe. For c in say (a,b), f'(c) is a local max, so f''(c) > 0 for x < c and f''(c) < 0 for x > c so there is a change in concavity at x = c.

Better to show first derivative test which shows the change in concavity.

[ionic sedge]

oh

so is it still true?

@shell widget

[shell widget]

Yes

[ionic sedge]

thanks!

the statement is: **
if f''(c)=0 then f has an inflection point at x=c**

i believe it is true

<@&286206848099549185>

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@ionic sedge Has your question been resolved?

[ancient saddle]

Not true for all functions

try a constant function

[ionic sedge]

ah, so it must be false

[ancient saddle]

tes

[lone heart]

@ionic sedge Has your question been resolved?

[pallid scarab]

constant functions are both concave up AND concave down, so I'm not sure whether this can be considered as false

[ionic sedge]

you say its true?

[lone heart]

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[frosty prawn]

i've tried to put this in the coordinate system

there are just so many cases for this problem

and i have to figure out the probability of each one

is there a simpler way of doing this?

[fallen verge]

Consider the bottom face with relation to the height

[lone heart]

@frosty prawn Has your question been resolved?

[frosty prawn]

that gives me the volume, yeah. is that it?

there are just so many different cases depending on the position of I

actually, maybe there are only two?

either I is inside of WXYZ or I is outside of WXYZ

okay

let W=(1,a), X=(b,0), Y=(0,c), Z=(d,1)

let the area of WXYZ=m and let the area of the right triangle with hypotenuse WZ=n

the probability that I is inside WXYZ=m/(m+n)

<@&286206848099549185>

[alpine sable]

Hey there.

How can I help?

[frosty prawn]

can i please have some guidance as to how the find the area of an irregular pentagon given 5 side lengths?

[frosty prawn]

i know the areas of the triangles that surround the sides of the base pentagon (which i can then subtract from the area of the unit square), but i don't know how to calculate the area of the quadrilateral with points WIZ and the corner of the cube

if i knew the area, i would be able to find the area of the base pentagon

[waxen flame]

Do you know how to calculate the area of a triangle and the area of a four-sided pyramid? That shape is just three irregular pyramids.

[frosty prawn]

yes

[waxen flame]

Read or watch a video about the Shoelace formula.

[lone heart]

@frosty prawn Has your question been resolved?

[wild skiff]

what is happening to the server

[waxen flame]

We're not sure but we have the best scientists in the world working on the problem as we speak.

[lone heart]

@frosty prawn Has your question been resolved?

[wild skiff]

ye

[prime crown]

hawo

[lone heart]

@frosty prawn Has your question been resolved?

[dense jacinth]

if f''(c) is 0 and concavity changes

as in f''(c) goes from negative to positive or vice versa

[lone heart]

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[quartz granite]

https://cdn.discordapp.com/attachments/268882317391429632/1091580625414279168/image.png

[quartz granite]

could anyone categorize these 10 questions? like for example, 1 combinatorics, 2 probability, 3 calculus, etc etc?

[jagged jewel]

wheres it from

?

[lone heart]

@quartz granite Has your question been resolved?

[quartz granite]

HMMT

[lone heart]

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[wheat isle]

What are the reciprocals of trig functions used for?

[vale wigeon]

are you asking why they were historically important enough to warrant their own names, or specifically for places where they appear

[wheat isle]

im not asking why they have their own name. im asking what is its uses

[last ether]

Finding the angle between two vectors

Bearings

Mapping

Momentum

Anything with physics

[wheat isle]

like if regular trig ratios are used to find the angles and sides of an unknown triangle

what would the reciprocals of that do

[vale wigeon]

same thing but with slightly different input data perhaps.

[last ether]

Well if you know the side lengths of a right triangle, you can use those rations to find the two non-right angles

[vale wigeon]

but then again you could rewrite sec and csc in terms of cos and sin

[last ether]

Oh wait reciprocals, not inverses?

Kek I'm blind

[vale wigeon]

there are no use cases for csc and sec and are not served also by cos and sin.

[gray isle]

reciprocals are a type of inverse, specifically the multiplicative inverse

[last ether]

I mean volume of a cone using spherical coords uses sec. To me, it's just nicer to write

Also helps cheese trig ratios like instead of writing 2/x = cos(23), I can say x/2 = sec(23) and I know immediately x = 2sec(23)

[vale wigeon]

x/2 = 1/cos(23) => x = 2/cos(23)

same difference

[wheat isle]

so in what i guess examples would you use reciprocal trig functions instead of the normal one

[serene junco]

sec is convenient sometimes for how it's related to tan through derivitive identities and pythag identity. Makes trig subs nicer

But that's very specific

[vale wigeon]

none

[serene junco]

The real answer is basically none

[wheat isle]

okay

i was also wondering

where we get those values

the 2 and root3 or is that just a property of a triangle

[last ether]

Well

Really it should be x, x sqrt(3), 2x

[wheat isle]

oh

[gray isle]

doesn't really matter

anyway those values can be obtained from splitting an equilateral triangle (with side length 2) in half

[wheat isle]

and also, is there a reason for the specific 30-60-90 and 45-45-90 triangles

[gray isle]

the relations between the sides of those triangles can be determined with ease

[vale wigeon]

do you mean "a reason why they exist" or "a reason why we care about them"

[wheat isle]

kinda both

[vale wigeon]

you're not gonna like the answer

the reason these triangles exist is that for any triple of positive numbers adding up to 180, a triangle with angles measuring that many degrees exists.

the reason why we care about those in particular is that their side lengths can be easily calculated by elementary geometric considerations, which lets us calculate trig functions of a few special angles

[wheat isle]

oh okay

so for example if we try to find the co secant of 45 degrees we would use the 45-45-90 triangle?

[gray isle]

yes

[wheat isle]

so then its xsqrt(2)/x ?

and you cancel to just get root 2 im assuming

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[grim swift]

Does -x^4 have any coefficients?

[gusty gorge]

coefficient of -1 yeah?

[grim swift]

Yeah cause -x can substitute for -1 right

Just making sure

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[tender latch]

Given a collection of points (2-dimensional), how would I get a circle that is as small as possible but contains all points within it? So basically, trying to find the "center" of all points and then the radius would just be the distance from that center to the furthest point in the collection. Perhaps a drawing might help if I'm not clear:

[tender latch]

I guess my example drawing isn't even the correct solution to what I'm proposing as my problem I'm trying to solve, but I hope it paints an idea.

[vapid shuttle]

hmmm

just throwing some ideas at this

would maybe finding the two furthest points

and then drawing a line in between them

choosing the center of that line as the center of your circle

would that work?

[gleaming mantle]

Average of x and average of y would be the “centre of mass” of your points

[tender latch]

I guess but to what point should I sort on distance though

[tender latch]

to this center of mass ig?

that could probably work

[gleaming mantle]

Are you given coordinates of the points?

[tender latch]

Yes

[gleaming mantle]

Okay yep

[tender latch]

Alright thanks guys.

How do I close this?

/close

woops

[gleaming mantle]

I think .

[tender latch]

.close

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[oak walrus]

Hello, can I get advanced python math help in any of these channels? :3

[vale wigeon]

you can certainly try

[gusty gorge]

what in the world is advanced python math LOL

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[oak walrus]

I dont know, im gonna find out

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[alpine sable]

Hi all, i have a doubt about this

[alpine sable]

$a\sqrt{ab}$

the book says that the result is $sqrt{a^3b}$

[pallid scarab]

The LaTeX bot may be down

[alpine sable]

but why?

[vale wigeon]

a = sqrt(a^2)

[crimson carbon]

development

[alpine sable]

that's unfortuned

[vale wigeon]

[since a is under the square root to begin with, it must be that a ≥ 0, and so it is legal to say sqrt(a^2) = a]

[alpine sable]

the first a it's outside the square root

[pallid scarab]

Your result is correct, the textbook just put everything under the square root

[alpine sable]

it's a*sqrt(a^3b)

[vale wigeon]

now you are confusing yourself

your expression was a * sqrt(ab)

[pallid scarab]

Nonono, this is correct

[vale wigeon]

not a * sqrt(a^3 b)

and what i am saying is that, BECAUSE a (times something else) appears under the square root, it must be that its value is positive.

[alpine sable]

yes sorry, i copied the wrong line from the book. the expression is a * sqrt(ab)

[alpine sable]

but why a^3 and not a^2?

[modern sedge]

||what if B was negative as well as A?||

[vale wigeon]

ok you got me there

[vale wigeon]

a * sqrt(ab) = sqrt(a^2) * sqrt(ab) = sqrt(a^2 * a * b)

[alpine sable]

oh now i understand, i just saw another square root property

[alpine sable]

wait, why a * sqrt(ab) = sqrt(a^2)? because it multiply a by the square index?

[vale wigeon]

i did not say a * sqrt(ab) = sqrt(a^2).

i said a * sqrt(ab) = sqrt(a^2) * sqrt(ab).

[alpine sable]

i know, i'm just focusing on the sqrt(a^2) part

[modern sedge]

this is true for a>0

[vale wigeon]

then don't cut off part of what i say

if you wish to ask me why a = sqrt(a^2) then ask why a = sqrt(a^2)

[alpine sable]

ok, this is what i wanted to understand, thank you. i just saw this property on another page on my book, it's strange that they didn't put it before other properties, but anyway

[modern sedge]

note that it's true only for a>0. So this might also work only for a>0

[alpine sable]

i've asked twice, but you got mad right away without reason. i don't cut off anything 🫤

[pallid scarab]

Don't worry, nobody's mad. You can ask your questions safely

We don’t judge your questions nor your answers

[vale wigeon]

you did.

[alpine sable]

thank you. i know that this things are really elementary, but i have to study all the highschool maths after more than 15 years, i'm rusted

thank you again for the help

[vale wigeon]

i give off the appearance of someone who is mad, therefore i am clearly a madwoman and i am angry and mean and an asshole to everybody on planet earth regardless -- or even DESPITE -- any statement of intent i may make

[pallid scarab]

Yeah that resonates with me on an entire level

[vale wigeon]

as in you consider me a madwoman?

or you share the sentiment i expressed

[pallid scarab]

The latter one of course

You can ask in another channel

[vale wigeon]

@hazy quest open your own channel

[alpine sable]

i clear the channel for other people. thank you again for the help and patience, have a nice day

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[alpine harness]

I’m having trouble understanding question 3 c and confused about how to figure it out

Also an explanation on terminology of what a hat and unit vectors mean

[vale wigeon]

a unit vector is simply a vector whose length is 1.

[alpine harness]

And hat is what represents it right?

[vale wigeon]

not quite.

putting a hat on top of a vector represents normalizing it, i.e.:

[alpine harness]

What does that mean? 😮

[vale wigeon]

$\hat{a}$ is the vector that points in the same direction as $a$, but has length 1

[ocean seal]

Ann

[alpine harness]

Ahhhh that makes sense

[vale wigeon]

its actually very easy to do c

you just find $\frac{1}{2} (\hat{a} + \hat{b})$ and normalize it

[ocean seal]

Ann

[alpine harness]

Ooh ok ok

How would I normalize it?

[vale wigeon]

you have already done so on two occasions

normalizing a vector is simply dividing it by its own norm

[alpine harness]

Oh I got confused cause of wording 😭

Ok ok

Thank you

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[marsh sedge]

Hello I have a question about part b of this question. I solved part a and got it right but in part b the mark scheme used the same quadratic as we worked out in part a to find the invariant lines. How come they can use this same equation when we don’t know that all the invariant lines pass through the origin in part b?

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[marsh sedge]

No

<@&286206848099549185>

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[marsh sedge]

.reopen

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✅

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[marsh sedge]

.reopen

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✅

[marsh sedge]

<@&286206848099549185>

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[errant dagger]

ω=1+2k (1)
3/2+ω/2=2+k (2)
-1/2-ω/2=-1-k (3)

[errant dagger]

if I have these three simultaenous equations

hmm

actually nvm

if i want to show that two parametric versions of a line are the same line

how would I do that?

[sour verge]

I guess there are several ways you can do that.
One of them :
Show that the two lines are parallel (direction vectors are proportional to eachother) and then show that they share a point (try whichever one is the easiest). It follows that both lines are the same.

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[oblique knot]

I’m a bit confused as to how to find the answer to a) when I don’t know TS and the radius of the smaller semi

I tried doing like (6+r)^2=6^2+(TS+r)^2 but got stuck

[upbeat gorge]

Find another way to express TS in terms of r and constants

[oblique knot]

Im not sure what to express it as besides just a variable

[upbeat gorge]

Note that TS and two radii of the smaller semicircle form the radii of the quartercircle

Based on the diagram

[oblique knot]

Ohh right but how do I find that out?

[upbeat gorge]

wdym

[oblique knot]

Because there isn't the line for TS that SR has

that states they're equal

[upbeat gorge]

Yeah they may not be equal

[oblique knot]

ohhh 12-2r

[upbeat gorge]

yes

your pytha strategy should work now

[oblique knot]

Tysm

[upbeat gorge]

np!

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[upbeat gorge]

Follow the hint

[alpine sable]

it means write 4 in terms of a power of 2

yeah

[upbeat gorge]

Laws of exponents

[paper maple]

try to answer these questions
what is

1. (2^2)^5
2. (2^2)^1/2
3. (2^2)^5/2

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[paper maple]

now you should be able to solve your initial question....

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[wicked spear]

pls help

[upbeat gorge]

Look for similar triangles

Then use proportions

[earnest panther]

(Going to label the point where the weather balloon is as D.)
Just to be sure, did you notice that the angle at point D is a right angle? (the right angle symbol is very small and potentially hidden by the perpendicular line from station B)

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@wicked spear Has your question been resolved?

[wicked spear]

no

<@&286206848099549185>

[wicked spear]

ya

i noticed

but i quite dont understand how to solve it