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[frozen willow]

Thanks very much

[twilit loom]

np ^^

[frozen willow]

.close

[obtuse pebble]

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[west warren]

Hi, what do we call this in english ? ($f(x)=x^2
f'(x)=2x$)

[warm shale]

ramey

[west warren]

i study math in arabic & french lmaoo

[quaint glen]

f'(x) is called the derivative of f(x)

is that what you wanted?

[west warren]

tysm

.close();

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[timid silo]

tried it twice, couldn’t get the answer

solution is 9/2

[timid silo]

q just asked for the gradient

to the tangent

[short spruce]

derivative is incorrect

[timid silo]

nvm, got it

[short spruce]

alright

[timid silo]

wait what

[short spruce]

lol

you got it with the wrong derivative?

wait i can’t read nevermind

[upbeat island]

m = 5 - 1/2sqrt(1) is correct

you subtracted incorrectly

[timid silo]

i just realised that 5-(1/2sqrt1) gives 9/2

[timid silo]

yep

just looked over it and noticed

[short spruce]

yeah don’t have my glasses on i’m smart 👍

[timid silo]

hahaha it’s no problem cheers for the help 👍

attempted this a few times and got the same answer

the answer is 5/64🤷🏻‍♀️

[upbeat island]

first term of derivative should be -3/2 x^{-5/2}

exponent -3/2 - 1 = -5/2

[timid silo]

which line?

[upbeat island]

when you compute the derivative
very bottom left

[timid silo]

oh yeah that’s right.

also this too.

[upbeat island]

you plugged in positive two

[timid silo]

i did?

oh.

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[uneven scaffold]

how to solve this question

and this one

[upbeat island]

in the first question, the triangle is drawn a little misleadingly, i would start with annotating everything they give you and see what it forces as an outcome (don't trust how it's drawn)

for the second i would start with cubing both sides of the given equation

[obtuse pebble]

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[idle condor]

how would I go about solving this question?

[soft sail]

Are you familiar with the fact that the collection of (x,y) pairs in the plane satisfying each equation defines a line?

[idle condor]

I'm not sure what that means

[soft sail]

Have you worked with equations of the form y=ax+b?

[idle condor]

yes of course

[soft sail]

The define a line in the plane yes?

[idle condor]

yeah makes sense now

[soft sail]

Would you agree that each equation in the problem you gave defines a line then? Since you can rearrange the terms to make them look like "y=ax+b"?

[idle condor]

yes

[soft sail]

So, solving the systems of equations stands for finding the points at which the 2 lines ( 2 equations ) intersect, agree?

Since each equations defines a line so if I have an intersection that point "satisfies" both

[idle condor]

yeah

I can't solve for m though

[soft sail]

With all that in mind, what do you think is going on geometrically when the system has no solutions?

[soft sail]

We'll get there

[idle condor]

the lines are parallel when there are no solutions right

[soft sail]

Bingo-ish

2 lines being parallel stands for the a in y=ax+b being equal right?

[idle condor]

yes

[soft sail]

If we have 2 lines with equal a but different b then they never intersect

If the B's are equal they intersect everywhere since the equation is the same

That's the infinitely many solutions part of it

So if you reduce both equations to y=ax+b with equal a and different B's it solves the no solution case, if the B's are qual it solves the infinitely many case

[idle condor]

yeah

[soft sail]

Can you make each equation look like y=ax+b?

The first one is $y=\frac{8}{2}-\frac{m}{2}x$

[warm shale]

Brun。

[soft sail]

Can you do the second?

[idle condor]

yeah I got that one

y = (-2m -4x)/(2-m)

[soft sail]

You got the signs wrong

On the 4x

[idle condor]

meant to type y = -(2m -4x)/(2-m)

anyways the gradient is 4/(2-m) right

[soft sail]

Yup, and you want that to be equal to -m/2

For them to be parallel

[idle condor]

m is -4 or 2

[soft sail]

There's a problem at 2 tho

[idle condor]

oh yeah

[soft sail]

We've got ourselves a nasty division by 0 since we rearranged

But looking back at the original equation you see that m=2 just makes the y term vanish

That means we have a vertical line at x=1

[idle condor]

so it has to be -4 for there to be no solutions

and -2m/(2-m) needs to equal 4

[soft sail]

Well, let's see what happens to the constant term

[soft sail]

Yup

[idle condor]

which m = -4 satisfies

[soft sail]

It doesn't

8/6

Different constants => no solutions

[idle condor]

oh so it is correct

[soft sail]

And this one has one solution

And that's it

[idle condor]

so for infinitely many solutions the two lines need to be the same right?

[soft sail]

Yeah, same a and same b

[idle condor]

is that even possible?

[soft sail]

Not in this one

No value of m makes that happen here

[idle condor]

doesn't 4?

[soft sail]

One will be 4 and the other will be -2

The slopes

[idle condor]

you sure?

[soft sail]

You can verify it on desmos, it's a great way to catch errors lol, I apologize if I've made any mistakes but I think I am right on that

Write the 2 equations there and let m=4 to see if they don't intersect

[idle condor]

4 does work

[soft sail]

Wdym by work tho? Let's clear that up first

[idle condor]

the two lines are the same when m = 4

[soft sail]

Huh, that's a thinker

I'd go back and redo the quadratic from before

Maybe there's something there?

[soft sail]

This

[idle condor]

I think I was wrong before it should be m = 4 or -2

[soft sail]

Oh nice

And -2 doesn't make the B's equal, so that's the no solution one

[idle condor]

it makes sense then

so what does the last questions even mean to solve the equation in terms of m for suitable values of m?

[soft sail]

m that doesn't satisfy the quadratic

you want the lines to not be parallel

To have a meaningful unique solution

If it lets you chose an m to work with I'd go for a 2 to get rid of the y in the second equation

[idle condor]

the solution to part b is

I don't really understand how this works

[soft sail]

You will have to sub-in the y given by the first equation into the other

That forces it to satisfy both "rules" at once

Then you just isolate for x and you should get the expression in the answer there

[idle condor]

did I do something wrong I'm not sure?

[obtuse pebble]

@idle condor Has your question been resolved?

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[hybrid moat]

can someone explain me what is linear map ?

[hybrid moat]

in simple worlds

[sage geode]

Linear map is a mapping satisfying the following two properties:
$f(u + v) = f(u) + f(v)$ and $f(c\cdot{v}) = c\cdot{f(v)}$, also $f(0) = 0$

[warm shale]

Touch Our Beans

[sage geode]

Shortly, a mapping which preserves addition and multiplication

[hybrid moat]

mapping is like a function?

[sage geode]

Yeah

[hybrid moat]

so it is a function

so what the diffrence betweein functions and linear map ?

[sage geode]

Function represents mapping between two sets (domain and codomain)

[hybrid moat]

and linear mapping?

[sage geode]

Again, for a mapping to be linear it needs to satisfy these properties:

[hybrid moat]

and transformation matrix ?

[sage geode]

Linear mappings are represented with corresponding matrices

[hybrid moat]

and are there other ways to represnt linear mappings?

[sage geode]

For example in real functions we have an identity function f(x) = x (it leaves the input unchanged), and in linear algebra there's an identity matrix (which similarly leaves the input unchanged)

For example this is the identity matrix in 2 dimensions

[hybrid moat]

what is identidy matrix

why is it so confusing

[sage geode]

A matrix which leaves the vector unchanged

[junior inlet]

*when multiplied to a vector, that is

[sage geode]

So $\mathbb{I}\cdot{v} = v$

[sage geode]

Yup

[warm shale]

Touch Our Beans

[sage geode]

Where I is the identity matrix and v is an arbitrary vector

[junior inlet]

same with v * I

it's also v

[junior inlet]

a linear map is a map that satisfies:\
$f(a+b) = f(a)+f(b)$\
$f(na) = nf(a)$

[warm shale]

just bob

[sage geode]

Yes I already told him that

[junior inlet]

oh ok

[hybrid moat]

but how can i represent it in

[junior inlet]

didn't notice that

[hybrid moat]

matrics

[hybrid moat]

thats a function

i want to do it with matrics

[sage geode]

Then view that as $f(v) = A\cdot{v}$ for some matric A

[warm shale]

Touch Our Beans

[junior inlet]

matrices transform vectors

if you want a function with 1 variable, don't use matrices

[junior inlet]

also, it's matrix btw

(redundant also)

but whatever

[hybrid moat]

i will upload something

what is the meaning of what is written

[sage geode]

There's a lot of different things going on there, which part interests you?

[junior inlet]

it's talking about some matrix

which isn't shown in the picture

so i have no clue what this means

[sage geode]

For example in first lines there's a matrix $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ and the writer is trying to find the unknown vector given in the equation, I suppose

[warm shale]

Touch Our Beans

[sage geode]

This is something about using different basis in the vector space, can't tell in details

[hybrid moat]

thats true

thats what he told

something about changing basis

[junior inlet]

maybe it's

1 -1 0
1 1 0
0 1 1

that's the matrix, right?

[hybrid moat]

but what does it even mean

[sage geode]

Every nxn (n by n dimensions) matrix has n^2 entries (numbers) in it

[hybrid moat]

and something about ker

[junior inlet]

ker is the kernel (the set of vectors that gives the 0 vector when multiplied with the matrix)

[sage geode]

Ker stands for kernel of a matrix

[hybrid moat]

kernel of linear map

[sage geode]

In terms of linear transformations, kernel is always just a set with the zero vector

As far as I'm concerned

[hybrid moat]

wait so it is the same

about linear transformation and matrix ?

the kernel

[junior inlet]

the defn is the same, yes

[sage geode]

Linear transformations are matrices, in some sense

[hybrid moat]

what does it mean L (v)

[sage geode]

L is a linear transformation applied to some vector v

[hybrid moat]

and v belong to v

[sage geode]

Capital V is some vector space

(set of all vectors)

[hybrid moat]

but this definition is multiply with some vector in order to get 0 vector

[hybrid moat]

why in the formual there is no multiplation/?

[sage geode]

This is the same definition

[hybrid moat]

L(v) it means to multiply with some vector in order to get 0 vector

[sage geode]

Again, L(v) means that you're multiplying v by some matrix

[hybrid moat]

matrix which is L

right ? ?

[sage geode]

Yes

[hybrid moat]

is there a diffrence between nullspace and ker ?

i think i started got it

a little

[sage geode]

"Kernel is also known as the nullspace"

[hybrid moat]

so it is the same thing

[sage geode]

Yes

[hybrid moat]

linear operator

is te same ?

[sage geode]

Linear operator is the same as what?

[junior inlet]

it seems like you have no knowledge of linear algebra whatsoever

[hybrid moat]

why is my teacher told me that ker=nullspace only in this case

[hybrid moat]

and he told me that L(v) is not multiplication it is like a function it can be multiplaytion but also addition

[junior inlet]

it's matrix multiplication, which can result in various different things

[hybrid moat]

how much time will it take me to learn it

is it possible to master it in 2-3 days from the begining and know how to solve problems ?

[junior inlet]

personally it took me 1 month at 10 hours of math / day

[hybrid moat]

wow, why so much time

thats pretty a lot of time

is there a lot which I still dont know ? ? ?

[junior inlet]

yes

[hybrid moat]

Inner product space
is the last subject of mine

eigenvalue

Eigenvalues and eigenvectors

[junior inlet]

wow this is pretty complicated (not that complicated, it's a vector field with the inner product on it) (but still kinda complicated)

[junior inlet]

these are properties of matrices

[hybrid moat]

are you kidding ?

[sage geode]

For every matrix there are some vector(-s) such that $A\cdot{v} = \lambda\cdot{v}$

where lambda is some scalar

[warm shale]

Touch Our Beans

[sage geode]

The solutions to that equation are eigenvectors

And values of lambda are corresponding eigenvalues

[hybrid moat]

wat is eigenvalues?

[junior inlet]

basically

v is the eigenvector
A is your matrix
lambda is the eigenvalue

[sage geode]

Imagine solving equation f(x) = kx, where k is some constant, this is similar thing

[junior inlet]

lambda that satisfies beans's thing

[hybrid moat]

beans's thing?

[junior inlet]

this

[hybrid moat]

what about dimension of matrix

[sage geode]

For nxn matrix there are n eigenvectors/eigenvalues

(sometimes eigenvalues are repeated)

[hybrid moat]

(1 1)

dimension 2?

thats not about 3D ?

LIKE MOVIES 3D 2D GAMES ?

THERE IS NOT RELATION BETWEEN THE TWO ?

??

[sage geode]

What do you mean

[hybrid moat]

how do i know the dimension

if i have matrix 2x3

is it just doing multiplaction for knowing the dimension?

2*3=6

thats the dimension??

[sage geode]

You just write the dimension as 2x3 (2 by 3)

I think

[hybrid moat]

i mean is there a diffrence between dimension of matrix and dimension of vector spaace?

[junior inlet]

dimension of vector space = number of vectors in its basis

[hybrid moat]

and dimenson of matrix

[hybrid moat]

is like that

[obtuse pebble]

@hybrid moat Has your question been resolved?

[obtuse pebble]

@hybrid moat Has your question been resolved?

[obtuse pebble]

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[steady bay]

Hey guys,
I have this function, a solution to a differential equation
I1 = ]-inf; 0[
I2 = ]0; +inf[

[steady bay]

mb not finished

We want to even find the solution when t=0

So we find the limit when t --> 0

The problem is I don't understand how they got to the limit exists if C1=C2=-1

I used l'hopitals rule and I got the limit equals 0 no matter what constant we use

So thats wrong

Can someone help me with determining C1 and C2?

[obtuse pebble]

@steady bay Has your question been resolved?

[steady bay]

oh sorry

<@&286206848099549185>

[obtuse pebble]

@steady bay Has your question been resolved?

[obtuse pebble]

@steady bay Has your question been resolved?

[steady bay]

No

[slim leaf]

@steady bay For the limit to exist, we need the right hand limit to equal the left hand limit. So that's what they did.

The right hand limit is lim t-->0^+ [(c2(t^2 +1) + 1)/t], left hand limit is lim t-->0^- [(c1(t^2 +1) + 1)/t]

since t-->0^- in the left hand limit, we have lim t--> 0^- [-(c1(t^2 +1) + 1) / |t|]

[steady bay]

Yes i understand that part

[slim leaf]

since t-->0^+ in the right hand limit, we have lim t-->0^+ [(c2(t^2 +1) + 1) / |t|]

equating these, terms cancel out

you get c1+c2 = -2

if u dont understand the absolute value stuff, then there is another way. lim t-->0^- means t approaches 0 from the left so it's gonna be a smaller and smaller negative number

you take the negative sign out, send it to the numerator

[steady bay]

This one i understand

[slim leaf]

and what u have left is a positive number

[steady bay]

This is to find the value of c1 which we replaced by c2

[slim leaf]

thats why i wrote |t|

[steady bay]

But how do we know that the limit exist to begin with

What I dont understand is that if I try to get the limit with whatever C i get 0

[slim leaf]

the limit is 0 yes

[steady bay]

Did you get it using lhopitals rule?

Because i get it like that and on top if t goes to 0 it doesnt matter the coefficient so even if c1 is different than C2 we get the same limit.

Thats why i dont know what i did wrong

[slim leaf]

lets take c1 = 5

so you have lim t-->0^- [(-5(t^2 + 1) +1)/t]

and this limit doesn't exist

,w limit of [(-5(t^2 + 1) + 1)/t] as t goes to 0 from left

[warm shale]

Results provided by WolframAlpha

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Requested by Sup?#7960

[slim leaf]

same for lim t-->0^+

so if we want the limit to exist

we want it to have the form 0/0 or inf/inf to apply LHR

what u are doing is just applying LHR(not thinking whether it even is of the form 0/0 or not)

so u get the first equation c1 + c2 = -2 by equating left and right hand limits

For second equation, we see that the limit is of the form (constant/0) which means the limit doesn't exist, so for it to exist, we must have the form 0/0 so that we may apply LHR

so applying lim t--> 0 to the numerator of either limit, you get c1 + 1 or c2 + 1 which must equal 0 so we have c1 or c2 = -1

and then from the first equation, we find the value of the other constant

[steady bay]

Okay I see that makes sense

Thank you very much

[slim leaf]

np

Close the channel if u are done, or ask any more questions that u may have.

🙂

[obtuse pebble]

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[obtuse pebble]

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[storm wagon]

sinx+1/√3cosx=1
What is the smallest positive value of x?

[storm wagon]

I find sin(30+x)=2/√3 and this looks so strange, and I don't know what to do next.

[sage geode]

$\sin(30 + x) = \frac{2}{\sqrt{3}} \iff 30 + x = (-1)^k\arcsin(\frac{2}{\sqrt3}) + \pi{k}$ for some whole number $k$

[warm shale]

Touch Our Beans

[storm wagon]

I understand this but, I still can't find the smallest positive value of x :/

[sage geode]

Oh

Then you need to find some value of k

[sage geode]

Btw it should technically be 180degrees instead of pi there since we're working in degrees

Here I'd consider cases k being odd/even and looking for lowest positive values of x in both of the cases

And then compare the results

[storm wagon]

I agree, thank you for your help Bean. I made it :3

.close

[obtuse pebble]

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[candid orchid]

Can someone help with these here?

[candid orchid]

B especially

is there an identity I'm missing?

[formal atlas]

Take tangent of both sides, use the fact that $\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\cdot\tan(b)}$

[warm shale]

Dmytr

[dapper geode]

try using tan(a+b)

oh

do arctan(p) = a and arctan(q) = b

[obtuse pebble]

@candid orchid Has your question been resolved?

[candid orchid]

this is exactly what I needed

[obtuse pebble]

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[candid orchid]

thank you both!!

[obtuse pebble]

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[neon grove]

What is the probability of a 2% event occurring 4 times out of 10?

[next violet]

any arrangement of this occuring will have the same probability

so the first step is to find how many arrangements are possible

[spiral maple]

is each time independent?

[neon grove]

question just came out of the blue, i just high-rolled on a game

don't know the difference :^)

[spiral maple]

So.... I asked a yes or no

[neon grove]

.

[spiral maple]

If you had the event happen, will that affect the next time it happens?

[neon grove]

no

[spiral maple]

So yes, they're independent

Anyway you can model it w/ a Binomial distribution

[neon grove]

https://tenor.com/view/roxy-migurdia-gif-20808668

Don't really know how to do that :^)

[spiral maple]

$X\sim Bin(n,p)\implies P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$

[warm shale]

Mosh

[neon grove]

wat

[spiral maple]

That's binomial distribution..

[neon grove]

don't know how to use that :^)

[spiral maple]

n=10, p=0.02...

x=4..

[neon grove]

what does the squiggly mean

[spiral maple]

Follows.

[neon grove]

and that means?

[spiral maple]

and tilde

X the random variable follows a binomial distribution w/ parameters n and p.

[dull forge]

Here you go! This is an excellent explanation of the basics. With visual aid and everything 🙂
https://www.youtube.com/watch?v=8idr1WZ1A7Q

At least this will give you a really fun starter and if you have any specific questions there, it might be more helpfull to come here. Mosh probably tries his best explaining this topic to you, but since it could take a while i thought this might help.
Also some games are programmed so that the results of each attack (or whatever) aren't independent. But that's probably not helping you with your question.

[neon grove]

I do know that some games are programmed to not be independent, but would like to know the answer if it were independent.

[dull forge]

Then the buzzword Mosh gave you with this video should help you a lot! I think its great that you are trying to get into math with real world problems you encounter 🙂

[neon grove]

I have no clue what the big x and the small x in that equation means

[dull forge]

Did you try wathcing that video? It might be hard to explain this whole topic of binomial distributions in an easy to understand and short way here 🙂

[neon grove]

don't really understand what this means / how i'm supposed to use it

[dull forge]

do you know what the ! stands for?

[neon grove]

factorial idk

[dull forge]

yep!

Do you know what that is?

[neon grove]

n x n-1 x n-2...

[dull forge]

until n-whatever reaches 1. Correct 🙂

Oh you were asking what the n and x stand for right?

[neon grove]

yeah, it's not something I've ever seen

don't know how to use it

[dull forge]

Well its called "n over k" (at least i think it is. english isnt my first language)
To calculate it, you use the formula above. But what do you think the n and k could mean?

[neon grove]

the guy in the video says choose

n over k is like n/k

I don't really know how it can be used or what it would mean

[dull forge]

https://youtu.be/8idr1WZ1A7Q?t=418
Here he explains what n over k is.
So out of 50 tries, 48 "went right"
what does it mean in your case?

btw its the same video just with a timestamp

Oh and just as a small motivation:
At the end of this you will be able to solve lots of problems you will encounter in games! You could basically calculate if some builds or bossraids are worth your time 😉

[neon grove]

yeah, that'd be cool

out of 10 tries, I got 4 high-rarity drops

[dull forge]

perfect thats n over k!

[neon grove]

i don't really know where to go from here even after watching the videfo

can i get a really rudimentary explanation as to what I would do with knowing n over k?

[dull forge]

what did you get out of the video is this

$$p^x(1-p)^{n-x}$$

[warm shale]

Arne

[dull forge]

i mean you got the n over k part down, now you are just missing this thing from the equation right? A little hint: It's in about the same timestamp i send you earlier.

[neon grove]

no no

i don't know how to apply

n over k

I know how to plug everything in

but I don't know how to use n over k

[dull forge]

Does this help?

[neon grove]

oh i kinda like

didnt look at that very closely lOL

(210) x (0.02)^4 x (1-0.02)^6

[dull forge]

looks about right

[neon grove]

so approximately a 0.003% chance?

[dull forge]

i got 0.0003%
But i (probably) used the same numbers.

[neon grove]

oh am i not supposed to multiply by 100 to turn it into a percent?

[dull forge]

oh and i guess its worth mentioning that many calculators use 10 nCr 4 as 10 over 4

[neon grove]

oh really?

that's helpful

[dull forge]

I had to look it up, since it was a while ago i used it. So sorry for only telling you now

[neon grove]

thanks for the help

[dull forge]

I hope that helped you!

[neon grove]

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[minor warren]

hey so according to the right hand rule should the magnetic force be pointing upwards or downwards in this problem?

I've gotten the magnetic force to be pointing ||upwards|| but it is ||downwards|| in the book

[minor warren]

<@&286206848099549185>

[obtuse pebble]

@minor warren Has your question been resolved?

[minor warren]

okay got a solution for my answer
it is pointing downwards cuz the charge is negative and the right hand rule is supposed to be used for positive charges

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[latent canyon]

well i thought it was the 5000 1000p vector + t(2.5cos70 2.5cos20) but ms says this

[quaint glen]

your answer is manjit's position vector

you want raj's position vector

[latent canyon]

Ohhh

How do we do that then?

I'm guessing you gind the velocity from the 2m/s speed?

[quaint glen]

yes. His speed and the bearing he's travelling at.

[latent canyon]

So the speed is the hypotenuse in this case?

[quaint glen]

yuh

[latent canyon]

This right.

?

Angle 40 not 20

Oh yes I see now

Tysm

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[novel inlet]

Ha car accelerates from 0 km/h to 100km/hr in 2.5 seconds. if the acceleration is uniform how much does it accelerate in 1 second

[sage geode]

a = (v1 - v0)/deltaT

[novel inlet]

Can yiu tell answer i can't understand what yiu gave

[sage geode]

And convert the km/hr to m/sec

[sage geode]

First do this tho

[novel inlet]

Did

0 m - 10,000 m

[high lily]

wheres 10000m coming from

[cedar flicker]

change km/h to m/s not just km to m

[novel inlet]

Ohh

[royal basin]

do we need to do any unit conversions here though

[novel inlet]

27.7778/s

[royal basin]

ew decimals

[novel inlet]

Lol

[royal basin]

(also missing units? idk)

we aren't asked to find the acceleration, we're asked to find by what amount we accelerate in one second

and answering "the car speeds up by ___ km/h" is perfectly fine

[novel inlet]

My brain is dead

[royal basin]

okay then maybe you could revive it and come back here once it's back up and running?

[novel inlet]

No no its running

[cedar flicker]

(final velocity-initial velocity)/time is the formula

for acceleration

plug in the numbers

[sage geode]

Well the question asks for acceleration per second but the velocities are given in km/hr

[novel inlet]

My brain not undertaking formula

[cedar flicker]

what is final velocity of the car

[royal basin]

it asks "how much does it accelerate in 1 second"

as-stated, giving an answer in (km/h)/s is perfectly fine

[sage geode]

You weren't given the formula before this problem?

[royal basin]

UNLESS we are also required to convert it into m/s^2, of which OP has made no indication

[novel inlet]

Isnt thia formula

[cedar flicker]

yes

can you tell me the final velocity of the car

[novel inlet]

I just need answer idk whrre the mam gave this question from

Tbis is the question the q:2

[cedar flicker]

well you did the q1 wrong

but nevermind that can you just tell me the last speed of the car

[novel inlet]

100km/h

[cedar flicker]

YES

what was its first speed

[novel inlet]

Qhat the q1 is wrong

0

[cedar flicker]

yes

so what is final speed-initial speed

[novel inlet]

I used my brain in q:1

What final initial speed idl you can see the question

[cedar flicker]

final speed is last speed

initial is first

you just said 100 and 0

so what is the difference

[novel inlet]

100 is difference

[cedar flicker]

yes

it goes from 0 to 100 in 2.5 seconds

[novel inlet]

Yes

[cedar flicker]

so how much will it go in 1 second

[novel inlet]

Idk thats shy i am here

Wait i will divide by 2.5 i think

[cedar flicker]

yea good idea

[novel inlet]

Time to find my calculator

[cedar flicker]

you can just

[novel inlet]

40

[cedar flicker]

okay

[novel inlet]

100÷2.5=40

So is 40 answer

And also what i did wrong in q:1

[cedar flicker]

you figure that out

I believe in you

[novel inlet]

I can't my brain dead

I gave my brain power to you remember our contract

[cedar flicker]

if there is no force acting on it

it will fly

[novel inlet]

An object with no net forces acting on it which is initially at rest will remain at rest. If it is moving, it will continue to move in a straight line with constant velocity. Forces are "pushes" or "pulls" on the object, and forces, like velocity and acceleration are vector quantities.

Got from google

[cedar flicker]

yea if it is initially at rest

[novel inlet]

Ohhhhh

[cedar flicker]

otherwise it will fly like a bird you should write that down

[novel inlet]

Big brain

Lol

[cedar flicker]

im serious

[novel inlet]

Ok

[cedar flicker]

like it will keep going up and down

it makes sense right?

[novel inlet]

Ohk

Yea

.close

[obtuse pebble]

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[cedar flicker]

and if you dont touch it

it will turn into a black hole

[novel inlet]

Ir

It will rain money $$$$

[cedar flicker]

so when you see an object going up and down you should touch it

[novel inlet]

$_$

[cedar flicker]

in real life too

[warm shale]

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

[cedar flicker]

its important

[obtuse pebble]

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[obtuse pebble]

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[brisk arrow]

how i write the symbol theta

[twilit loom]

$\theta$?

[warm shale]

dino dik

[royal basin]

do you mean write on a computer or on paper?

[brisk arrow]

paper

does it look like theta

[royal basin]

yes

[brisk arrow]

which is better

[royal basin]

the one on the right looks better

[brisk arrow]

becuz the line on the middle of the circle? it should be horizontal right

[high lily]

slightly curved is fine

[ruby ravine]

yeah, simply make a zero, and put a line across it

[high lily]

I start from the mid left
draw the horizontal line
and continue with the ellipse, anticlockwise

(without the writing implement leaving the page)

[ruby ravine]

but a lot of people wont mind how you write the symbol, just make sure it resembles theta

[brisk arrow]

got it

thank you guys

[royal basin]

that looks like it's easy to mistake for e if you're not careful

[high lily]

I never had that issue

[obtuse pebble]

@brisk arrow Has your question been resolved?

[obtuse pebble]

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[trail vector]

Hello! I’m having a lot of trouble proceeding with this exercise, I don’t know how to get off from here. Can someone give me a hint ? Thanks !

[obtuse pebble]

@trail vector Has your question been resolved?

[obtuse pebble]

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[leaden vortex]

Can anyone see how i can go from the bottom line in my equation to the below equation

[leaden vortex]

to this

[obtuse pebble]

@leaden vortex Has your question been resolved?

[obtuse pebble]

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[brazen glen]

hello i have a question please on this drawing i need to find the possibilities/numbers of rhombus to create a star and the angles need to be between 10° and 160°. To find the possibilities do i need to take the numbers divisible by 360 and that are between 10 and 160 like 20 ? thx

[junior inlet]

yeah you should do that

[brazen glen]

❤️

thank you very much

https://tenor.com/view/the-office-bow-michael-scott-steve-carell-office-gif-12985913

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[torpid lintel]

Prove that for any real numbers a, b, c, so that a + b + c = 3 the inequality a^2 + b^2 + c^2 >= 3 is true

[torpid lintel]

the only thing that comes to mind is
((a+b)+c)^2 = a^2 + 2ab + b^2 + 2ac + 2bc + c^2 = 9

so

[royal basin]

and you have shown in a previous exercise that a^2 + b^2 + c^2 ≥ ab+bc+ca, so...

[torpid lintel]

seriously?

i don't see it

nvm

thanks

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[obtuse pebble]

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[torpid lintel]

.reopen

[obtuse pebble]

✅

[torpid lintel]

actually give me a sec

[stable breach]

have you heard of rearrangement inequality?

[torpid lintel]

no, but i've solved this q

[torpid lintel]

what is rearrangement inequality?

[stable breach]

nvm

your issue is that you dont know a^2 + b^2 + c^2 ≥ ab+bc+ca right?

[torpid lintel]

nah, i've proven it b4

just didn't realize it

[stable breach]

then whats ur issue

[torpid lintel]

nothing

i've solved this q

.close

[obtuse pebble]

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[stable breach]

oh ok

[torpid lintel]

.reopen

[obtuse pebble]

✅

[torpid lintel]

@stable breach would you help me with another question?

[stable breach]

ok

[torpid lintel]

i have to show that for nonnegative reals such that x>= y the inequality x^4 + y^4 >= 2xy^3 is rtue

here's what i did

(x^2+y^2)^2 >= 2xy^3 + 2x^2 + y^2

(x^2+y^2)^2 >= 2xy^2 (x+y)

and don't know what to do now

maybe write it as

((x+y)^2 - 2xy)^2 >= 2xy^2 (x+y)

and rty to divide by x+y

what do you think

[stable breach]

i know at least 2 ways to do it but they both involve inequalities that might not be suitable for your course

1 is by rearrangement which you do not know

the other is by muirhead

have you heard of muirhead?

[torpid lintel]

no, i'm in highschool, and an esl, so i might know rearrangement

so let's try it

[stable breach]

all right

wait is this school work

or olympiad maths

[torpid lintel]

no

it's just a highschool graduation exam preperation book

[stable breach]

oh then ig you dont need to muirhead

right

so let $\sigma_1, \sigma_2, \sigma_3, \cdots, \sigma_n$ be a random permutation of the numbers $1, 2, 3, 4, 5, \cdots, n$

[warm shale]

A Fellow Human

[stable breach]

so e.g. for $n=3$ you could have $\sigma_1=3, \sigma_2 = 1, \sigma_3=2$

[torpid lintel]

permutation was the one where order matters, right?

[warm shale]

A Fellow Human

[stable breach]

yeah order matters here

does this make sense?

[torpid lintel]

yeah sure

[stable breach]

so you could visualise this as getting a list of the first $n$ integers, then randomising the order, then $\sigma_i$ is the ith element of the list

[warm shale]

A Fellow Human

[stable breach]

if that makes sense we'll go ahead

so let $x_1\geq x_2 \geq x_3 \geq \cdots x_n$ and $y_1\geq y_2 \geq y_3 \geq \cdots y_n$

[warm shale]

A Fellow Human

[stable breach]

so rearrangment says $x_1y_1+x_2y_2+\cdots+x_ny_n\geq x_{\sigma_1}y_{\sigma_1}+x_{\sigma_2}y_{\sigma_2}+\cdots+x_{\sigma_n}y_{\sigma_n}\geq x_1y_n+x_2y_{n-1}+\cdots+x_ny_1$

[warm shale]

A Fellow Human

[stable breach]

take some time to take that in

for the case $n=2$ that more or less says $x_1y_1+x_2y_2\geq x_2y_1+x_1y_2$

[warm shale]

A Fellow Human

[torpid lintel]

take some time to take this in?

this just says that if you have some order

then terms which are products of numbers in this order are greater than terms which are products of numbers in this order but randomized

[stable breach]

yeah

[torpid lintel]

so it seems like some obvious fact, i'm not sure how would you go about using it

[stable breach]

right

so if $x\geq y$ then $x^3 \geq y^3$

[warm shale]

A Fellow Human

[stable breach]

so by rearrangement $x\cdot x^3+y\cdot y^3\geq x\cdot y^3+y\cdot x^3$

[warm shale]

A Fellow Human

[stable breach]

and $\frac{xy^3}{yx^3}=\frac{y^2}{x^2}\leq 1$ so $yx^3\geq xy^3$

[warm shale]

A Fellow Human

[stable breach]

so that gives us $x^4+y^4\geq xy^3+yx^3\geq 2xy^3$

qed

[warm shale]

A Fellow Human

[torpid lintel]

let me think about it

thanks anyways

i might dm you in a bit

.close

[obtuse pebble]

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[stable breach]

sure feel free

(this should also be provable with am-gm tho i will need to think about it)

[obtuse pebble]

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[dense dirge]

find the number of pairs (a,b) of natural numbers such that b is a three - digit number, a + 1 divides b - 1 and b divides a^2 + a + 2

[dense dirge]

I would appreciate a step by step explanation coz its not that easy to wrap my head around...

[obtuse pebble]

@dense dirge Has your question been resolved?

[dense dirge]

<@&286206848099549185> ?

[obtuse pebble]

@dense dirge Has your question been resolved?

[dense dirge]

sad

just sad

[clear girder]

you have to solve it with no computer help right?

[dense dirge]

yep

[crisp depot]

Sorry, I am not entirely sure how to solve this

Or what field of math this would go in...

I know though we can limit 'b'

And try to see if we can narrow it down

[final tartan]

just 900 numbers to try

[clear girder]

okay it is pretty ez, just choose for b 1 and for a 1.

$2|0, 1|4$

[warm shale]

Armin

[clear girder]

or not XD?

[final tartan]

they are not 3-digit

[clear girder]

trueeeee XD

[crisp depot]

I am curious on this solution, can you explain the what appears to be absolute value signs?

[void pelican]

2 divides 0

1 divides 4

I believe

[crisp depot]

2 divides by zero?

[void pelican]

no

[crisp depot]

That doesn't seem right

[void pelican]

it basically means 0/2 is an integer

and 4/1 is an integer

it's kinda a backwards division

[crisp depot]

Oh

[void pelican]

at least that's how I look at it

[crisp depot]

How would that lead us to the answer?

[void pelican]

idk

[timid silo]

trial and error

[crisp depot]

I would approach it by making $c = a + 1$ and $d = b - 1$

[warm shale]

Bman

[crisp depot]

Which would eliminate numbers just above or below prime numbers

And all other numbers who are both not made of the same prime factors

Which narrows it down quite a bit

Then eliminate based on the last rule

[final tartan]

a and b are even or a and b are odd based on the first rule

i honstly don't think it has a sol

[final tartan]

I added the equations on the right but ended up with a=<0
so either my method is wrong or it really has no sol

[obtuse pebble]

@dense dirge Has your question been resolved?

[next violet]

@dense dirge you asked for a step by step solution?

[dense dirge]

yep

[next violet]

here?

ok here let me try

let there be some integer n such that

b-1 = n(a+1)

thus b=na+n+1

it is known that b|a^2+a+2

or na+n+1 |a^2+a+2

then from further simplification na+n+1 is divisible by -a+2n

now na + n + 1 must be divisible by 0, and then you can just plug in 2n for a and plug that into b and find all numbers of that form

thats a bit tricky

is this a competition problem?

[dense dirge]

yep

[next violet]

ok

that makes sense

[dense dirge]

the given answer was way too hard

[twilit knot]

But then how do you know there's an easier solution? Can you perhaps provide the source and the given one? Maybe someone can explain it if you're stuck

[next violet]

was my solution above similar to the given one

[obtuse pebble]

@dense dirge Has your question been resolved?

[obtuse pebble]

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[timid silo]

$\int_{\gamma} f(z)\abs{dz}=\int_{-\gamma} f(z)\abs{dz}$

[warm shale]

Cüneyt

[timid silo]

is this correct?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[royal basin]

i remember you asking this very same question multiple times and the conclusion being that nobody here has any fucking idea what this notation is supposed to mean

[timid silo]

Lol

But I need to find it out

It may be in my exam

[royal basin]

then why ask here and not your professor?

[timid silo]

Because he told us if it is correct and I didn't listen to him at that moment

I don't wanna ask

[royal basin]

well you will have to

[timid silo]

Okay one more question and Ill close the channel

Is this correct

[royal basin]

????

[timid silo]

Check real-Complex anaylsis

.close

[obtuse pebble]

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[honest jewel]

how would i do b>?

[junior inlet]

rewrite as y=mx+c and use formula for perpendicular lines

[honest jewel]

what if formula for perpendicular lines

[junior inlet]

slope of the line perpendicular to y=mx+c is -1/m

[honest jewel]

so y=mx + b form would be

x = 1/2y - 2

[junior inlet]

no

well

actually, you can continue with that, but it will be a bit different

[honest jewel]

oka

[obtuse pebble]

@honest jewel Has your question been resolved?

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[timid silo]

i am working with natural logarithms and would like to know how in this workings out how the -t becomes positive?

[hushed moat]

the -ln(1/2) becomes ln(2)

then t=10ln(2)

[timid silo]

so from -t/10 i understand that we take -10 from both sides

hence the -10 ln1/2

[hushed moat]

yes

[timid silo]

but that would leave -t

-t = -10ln1/2?

how does the -t become positive?

[hushed moat]

nono, the minus sign cancells out

$-10 \cdot -\frac{t}{10} = t$

[warm shale]

Frosty

[timid silo]

wait

so

[hushed moat]

when you multiply by -10

[timid silo]

i am not multiplying i am subtracting

[hushed moat]

the "take" here should mean multiplying

otherwise you wont get t on it's own

[timid silo]

i just thought of something

can i divide both sides by -1

[hushed moat]

yea

[timid silo]

that will give +10.....

[hushed moat]

but that's not the right equation

$-\frac{t}{10} = \ln(\frac{1}{2}) \ -t = 10\ln(\frac{1}{2})$

[warm shale]

Frosty

[hushed moat]

that is just multiplying by 10

now you can divide by -1 or whatever

[timid silo]

right

i think i got

multiplying both sides by -10 worked

thanks man i knew i was missing something simple

.close

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[timid silo]

hello

[timid silo]

quick question

if a triangle has three equal angles does that mean all sides are equal

?

[little basin]

yes

[timid silo]

okay thanks

.close

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[timid silo]

Suppose I have:
sin(2x) = sin(x)

[timid silo]

So I say:
2 sin(x)cos(x) = sin(x)

can someone explain why dividing both sides by sin(x) will hide one of the values of x?

[high lily]

you lose the solutions from sin(x)=0

will hide one of the values of x
in fact you're lose infinitely many solutions

[timid silo]

You're right. I was thinking about a certain domain of values of x but yes in a general sense that's right.

[timid silo]

Why is this?

Because I'm dividing by zero or because I no longer look for sin(x) = 0?

[high lily]

consider approaching the question with factorisation

you're dividing by something that could potentially be 0

[timid silo]

Okay that makes sense

[high lily]

like lets take a cubic equation in factored form
x(x-2)(x-7) = 0
you wouldn't blindly divide both sides by x, (or (x-2), or(x-7)) would you?

[timid silo]

That's what my teacher said, referring to the principle of not dividing by sin(x) so I came and asked for why that's a principle. This could also apply to cos(x), right?

[high lily]

generally applies

[timid silo]

That's right

This brought the idea closer to me so thanks

[high lily]

don't divide

by something that could potentially be 0
unless you really know what you're doing

[timid silo]

Alright 👍

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[obtuse pebble]

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[static patio]

for all n >= 0

is it a subspace ?

i dont know how to check subspace properties when its a sequence

[runic trout]

For closure, pick (a_n) and (b_n) from the set. They satisfy the recurrence relation. Consider (a_n + b_n) and (k a_n). Check whether they satisfy the recurrence relation.

[static patio]

hmm

i dont get it tbh

can you show me little example ?

[runic trout]

Prove that V = {(a_n) in R^N : a_i = a_{i+1} } is closed under scalar-multiplication. Pick (a_n) in V and k in R. Consider (k a_n) = (k a_0, k a_1, k a_2, ...). For any subscript i, since a_i = a_{i+1}, we have k a_i = k a_{i+1}. This shows that (k a_n) is in V.

For closure under addition, pick another (b_n) in V. Consider (a_n + b_n) = (a_0 + b_0, a_1 + b_1, ...). For any subscript i, we check that a_i + b_i = a_{i+1} + b_{i+1}.

We also need to check that the zero sequence is in the subspace.

[static patio]

thanks a lot

i got it

.cloose

.close

[obtuse pebble]

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[dapper geode]

can someone explain the steps here

[dapper geode]

when I_1 =

and I_2 =

[twilit loom]

The first term $\frac{ae^{-il_2a}}{-il_2}$ of $I_1$ and the first term $\frac{(1-a)e^{-il_2a}}{il_2}$ of $I_2$ cancel each other out when summed

[warm shale]

dino dik

[twilit loom]

Then the factor of ${1\over l_2^2}$ is factored out, ending up with the top expression

[warm shale]

dino dik

[obtuse pebble]

@dapper geode Has your question been resolved?

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[dapper geode]

first but x-1 in g(x)

[dark steeple]

ok

[dapper geode]

then put that in f(x)

ah wait

is that

g*f

or g(f(x))

[dark steeple]

gf(x)

is

g(f(x))

yeah

but

[dapper geode]

ah ok

[dark steeple]

how do u do it with x + 4

i dunno know that

[dapper geode]

put f(x) in the x of g(x)

[dark steeple]

wait lemme do it

how do u do the first part

so

x^2-25 * x+ 4 *X+1

[scarlet gale]

@dark steeple What's fg(x + 1) simplify to?

[dark steeple]

(x+1)

[scarlet gale]

No.

[dark steeple]

OH

[scarlet gale]

fg(x + 1)
f(g(x + 1))

[dark steeple]

so

[scarlet gale]

Now, to do g(x + 1), you do this:
g(x) = x + 4
g(x + 1) = (x + 1) + 4

See how you replace x with x + 1 on the left?

[dark steeple]

so is it x+4 * x+1

g = x+4 * x+1

[scarlet gale]

Do you see how I replaced x with x + 1 on the left?

[dark steeple]

yeha

yeah*

[scarlet gale]

So because I replaced x with x + 1 on the left, I also did it on the right.

And I put it in parentheses to avoid mistakes.

g(x) = x + 4
g(x + 1) = (x + 1) + 4

Does it make sense what I did?

[dark steeple]

yess

so it will become x^2

• 1x

[scarlet gale]

Stop working ahead please.

I'm trying to show you a method to solve these.

[dark steeple]

oh

sory

my bad

[scarlet gale]

OK, so

f(g(x + 1))
f((x + 1) + 4)

Now we do the same thing with f.

f(x) = x² - 25
f((x + 1) + 4) = **((x + 1) + 4)**² - 25

See how we had x on the left and replaced it with (x + 1) + 4?

[dark steeple]

oh

ok

[scarlet gale]

Then you do the same on the right, but with parentheses around it.

[dark steeple]

kk

[scarlet gale]

So, you do the same thing to both sides.

[dark steeple]

right

[scarlet gale]

You replaced x with something on the left, so you replace x with it in parentheses on the right side.