help-13

[astral shard]

for the rsa

or is it just convience

[trail jackal]

if you look into the algorithm, nothing changes if you change which one is p and which one is q, so it doesn't matter at all

[astral shard]

thank you

.close

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[bold compass]

Hello, I've been stuck on this question from a real analysis book for a bit. It's question b from the following picture, I've already taken a bit off and have now come back to it but I still don't see it. Next to my question for a hint about how to continue the question, do you have any advice of how to continue these type of questions when stuck? I'll also show 2 things I've tried, I tried more then what I showed here but didn't see anything that could come of those attempts.

[bold compass]

hmm not sure it's readable

in the first attempt I tried to substitute but got stuck when I have

||a|-|b| | <= |a| + |b|

the 2nd part fell flat quite fast, as I thought I could maybe prove it for different relationships between a and b seperately

but with a,b both being smaller then 0 I saw that didn't work either

[cedar kiln]

@bold compass Has your question been resolved?

[cedar kiln]

@bold compass Has your question been resolved?

[elfin hemlock]

For a) write |a-b| as |a+(-b)| and use triangle inequality

[bold compass]

yeah I'm purely stuck at b

I got a

[elfin hemlock]

For b) square both sides and work on that or split up in cases;
|a|-|b|<=|a-b| and |b|-|a|<=|a-b|

[bold compass]

oooh alright

I'll try that

[bold compass]

the squaring one makes sense

how did you get the split up cases?

ooooh nvm

absolute value

ofcourse

completely forgot to use that :I

[bold compass]

as for any general tips, I don't know if you have anything, but I feel like I'm getting stuck a lot , is there anything you do when stuck on a proof to maybe think in another direction? Or is it purely just something that comes with time and practice after seeing enough proofs being done similar to the one you are trying to proof? As I feel like I don't know how to come up with the direction of where to go when trying to solve something like this sometimes.

[elfin hemlock]

Mostly just practice, but a lot of proofs is just unpacking definitions. Like for b) I either got rid of the annoying abs value by squaring or by using definition

[bold compass]

hmm, alright

that helps

thanks 🙂

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[sick sparrow]

how would i solve this?

i know that the median must be 90

but i have no idea about the mean

and answer said 97 but idk how tto get to it

[modern compass]

@sick sparrow You know how to find the mean, yes?

in general, not necessarily for this problem

[cedar kiln]

@sick sparrow Has your question been resolved?

[sick sparrow]

yes

add all the data then divide by the number of data points

[modern compass]

ok, so what happens if you add the data but also add 200 to the last point?

so instead of your data being $x_1,x_2,x_3,...,x_{100}$ it's $x_1,x_2,x_3,....,(x_{100}+200)$

[wraith dagger]

Zybikron

[sick sparrow]

but whats the sum all all the data points?

[cedar kiln]

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[sick sparrow]

.reopen

[cedar kiln]

✅

[sick sparrow]

9,000+200 / 100?

9500+200/100*

off i see now

9700/100

97

thank you so much

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[hidden magnet]

so this math is rlly easy for yall prob but im kinda stupid

[hidden magnet]

it says "simplify the following"

3x-(5x-6)

[twin dune]

$3c-5x+6$

[wraith dagger]

macharicky

[hidden magnet]

it has the parentheses in it though

its like some thing u gotta do

its sum algebra thing

[twin dune]

nah

it's just opening the parentheses

[cedar kiln]

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[hidden magnet]

..

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[crimson sedge]

Hi I should show if A is strict subset of B then f(A) is strict subset of f(B) for arbitrary f

[crimson sedge]

But I don't think this is true since f is not injective

We can have x in A \ B such that f(x) = f(y) for some y in the intersection of A and B

[mighty drift]

this property is obviously wrong as any constant function is a good counter example

then if A is not empty, f(A) = f(B) is not a strict subset

[crimson sedge]

Yep

I guess they're abusing the notation

ig

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[tender moss]

I done it trial and error

[tender moss]

How would i prove this algrebraicly?

[rigid ridge]

what was your trial and error method?

[trim sentinel]

Calculate how much time it takes car

Everything else will be clear then

[tender moss]

I just picked values

But the hits in black

At the bottom

It just happened to be it was my first pick

[rigid ridge]

lucky

[rigid ridge]

anyway, you should follow pluton's advice

[tender moss]

We cant thi

We just know

Car goes X mph

on a 240 mile road

How should we know how fast it went

It couldve went in 20 hours or 5

[trim sentinel]

Oh i see

Just calculate how much time it takes lorry

[rigid ridge]

you have a nice relationship between the time of the lorry and the car

if you can express these times in terms of x you are done

[trim sentinel]

You dont even have to. Just calculate time it takes lorry , then time it takes car , then cars speed

[tender moss]

It says use an algebraic method tho

[rigid ridge]

in terms of x right?

[trim sentinel]

$$t = \frac{s}{v}$$

[wraith dagger]

Pluton

[rigid ridge]

but v is in terms of x

[trim sentinel]

$$v = \frac{s}{t}$$

[wraith dagger]

Pluton

[trim sentinel]

Ooop i seee

I thought lorry went 12 mph

[tender moss]

Its x : x+12

[rigid ridge]

well, would've been way easier

[trim sentinel]

Yes i was incorrect you write it in terms of x

[tender moss]

The regular x goes one hour slower

[rigid ridge]

https://cdn.discordapp.com/emojis/776170462732353556.webp?size=48&quality=lossless

anyway, can you solve it now @tender moss?

[tender moss]

Not particularly

My imagination is kind of unrich as you can see

[trim sentinel]

I hate solving physics with math. (Aka using variables x , y , z)

If you are doing physics use v

[tender moss]

Can you show or?

[rigid ridge]

have you considered this yet?

[tender moss]

Oh

240/12 is 20

4 x 5 is 20?

[rigid ridge]

wdym

[trim sentinel]

12 is not speed

Unfortunatly

[rigid ridge]

yeah, these speed problems are getting a bit boring now

[tender moss]

Oh

Its the different my bad

Difference

Umm

[rigid ridge]

fun how it's "oh wow when trying to work with rates for time you use harmonic mean" (or smth like that) the first time, then you get bored of it

[tender moss]

Idk blad

[trim sentinel]

I just feel this question has too many ways to solve

I cant think of the best/easiest one

[rigid ridge]

hmm?

[tender moss]

Are you saying its so easy

Its obvious?

[rigid ridge]

i only see one

it becomes the more obvious when you work more with algebra :)

[trim sentinel]

I mean maybe its just my mind doing same steps just in different order

But i found 1 solution that has a 3 , 3 variable equations

[rigid ridge]

hmmmm

sounds like redundant variable definitions then

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

[trim sentinel]

When you do this type of questions having a pen and paper is much easier

[rigid ridge]

yup

[tender moss]

My guys

[rigid ridge]

icekxrmaa do not open this until you have solved
this is the one variable way kekw

you could open it but that would ruin the fun

[trim sentinel]

Comment : ||i mean ye thats probably one of the better solutions just define t in terms of x||

[tender moss]

Just show me the obvious way

I havent got time for all these new methods

I have an exam tommorow

[trim sentinel]

Physics or math?

[rigid ridge]

yeah, but i feel like defining new variables when only using them once is a bit redundant

[tender moss]

Its maths

Paper 2

[rigid ridge]

ok so

you know the distance (240 miles), a relation between the two speeds (one of 12 mph faster than the other) and a relation between the times (one finished 1 hour earlier than the other)

[tender moss]

Yes

[rigid ridge]

can you turn these relations into equations with x?

[trim sentinel]

I mean other way would probably be ||define speed in terms of time||

[tender moss]

X:x+12 = x: x-1

[trim sentinel]

Everything else is just complicating this 2 ways

[rigid ridge]

what is this notation?

oh

: means dividing by

but no brackets

[tender moss]

Isnt it a ratio?

[rigid ridge]

what have you done here?

[tender moss]

Im saying the car is 12 mph higher

[rigid ridge]

and please fix your brackets

[tender moss]

And the time is 1 hour less

[trim sentinel]

solution dont open
||
t_1
t_2 = t_1 + 1
v_1 = 240/t_1
v_2 = 240/(t_1 + 1)
240/t_1 = 240/(t_1 + 1) + 12
||

[rigid ridge]

I am reading this as \[ \frac xx+12 = \frac xx - 1 \]

[trim sentinel]

Should be a valid solution too

[wraith dagger]

Lance

[tender moss]

No loo

Lol

[rigid ridge]

yup that works too

[tender moss]

Whatever

Im just gonna look

[rigid ridge]

all the same though :P

[trim sentinel]

Defining time in term of speed is probably better

[rigid ridge]

hmmm

the morally correct choice is to get to an eq with just x yeah

[tender moss]

This doesnt even explain anything

How does this show what x is?

[rigid ridge]

of course it doesn't to you

[trim sentinel]

Why is there a physics question in math in the first place

[rigid ridge]

that too

[tender moss]

Blad i cba

If i fail tommorow idc

[rigid ridge]

bruh

[tender moss]

Im a bit

[rigid ridge]

it is an equation in terms of x from the givens

[tender moss]

Im getting cancer from tryna get all these questions down

[rigid ridge]

i just didn't explain how i got here

[trim sentinel]

Im so confident when i solve math questions but when i solve physics questions my confidence falls so much

[tender moss]

Its gonna be morning until this happens

I have many other questions and my exam is tommorow cant waste time like this

[rigid ridge]

what i did here is take the time relation and turn the two given times in terms of x

[trim sentinel]

Well giving you solution is same as not doing anything

[rigid ridge]

yeah ;-;

[trim sentinel]

If looking at solutions actually help i would just look at all IMO solutions and see ya there in 2022

[rigid ridge]

LOL

i should start grinding imosl soon™️

[trim sentinel]

Im just doing competitive math lately. And making a mistake in basic math just kills me

[rigid ridge]

i know the feeling

messing up multiple choice q's in the early rounds of MOs sucks

or short answer

[trim sentinel]

Wait some MO have multiple choice q's

[rigid ridge]

yeah

the netherlands has these

also amc (usa)

the netherlands just for the first round tho

and it's mixed in with short answer

i guess multiple choice is easy to check :)

[trim sentinel]

Probably

[rigid ridge]

oh well the second round is this friday

https://cdn.discordapp.com/emojis/893806491029618689.webp?size=48&quality=lossless

[trim sentinel]

Well gl

[rigid ridge]

thx

https://cdn.discordapp.com/emojis/842306526776524800.webp?size=48&quality=lossless

[trim sentinel]

I litteraly failed on addition on my last test

[rigid ridge]

rip

[trim sentinel]

Costed me 3 points

[rigid ridge]

out of?

[trim sentinel]

6 that you can get from that question

[rigid ridge]

bruh

was it shortanswer/multiple choice then

[trim sentinel]

Nah it was pretty simple

Sum of 456 - 555. And i failed on addition

[rigid ridge]

huh

[trim sentinel]

That question litterally screamed free points

[rigid ridge]

just 456-555 on a math test??

[trim sentinel]

Divided by some other thing

[rigid ridge]

ah

[trim sentinel]

Its only first round and first question

[rigid ridge]

oh an MO

what country?

[trim sentinel]

Croatia you probably never heard of it

@tender moss you still here

[tender moss]

Nah g i cut

I put my books back i cant be asked

[trim sentinel]

Well dont give up just like that

[trim sentinel]

Sum from 456 to 555 not subtraction

[rigid ridge]

ah

hmm croatia mo

[trim sentinel]

I mean only thing that is actually supposed to be hard is that they dont teach the formula

[rigid ridge]

lmaooo

you're expected to become gauss

[trim sentinel]

$$\frac{(n_e - n_s + 1)(n_e + n_s)}{2}$$

[wraith dagger]

Pluton

[trim sentinel]

Ik the formula and still failed

[rigid ridge]

:(

was it like last 3 digits or smth

calculating the entire product sounds hard

(for a first q)

[trim sentinel]

It just becomes
25 * 1011

I think

[rigid ridge]

uwuw sum from 456 to 555

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Lance#8018

[trim sentinel]

,w calc 25*1011

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Pluton#6301

[rigid ridge]

alright that's not too much

[trim sentinel]

,calc 25*1101

[wraith dagger]

Result:

27525

[trim sentinel]

Oh ye its 50

,calc 50*1011

[wraith dagger]

Result:

50550

[rigid ridge]

nice

[trim sentinel]

But i failed in addition of 555 and 456

[rigid ridge]

https://natjecanja.math.hr/wp-content/uploads/2022/02/skolsko_zad_2022_Bvar.pdf
i suppose?

[trim sentinel]

Ye

I mean first one was so easy

Second one will be easy too but you actually need to be better than other people xD

[rigid ridge]

Dvije cijevi napune bazen za 4 sata. Samo jedna od tih cijevi napuni bazen za 5 sati i
15 minuta. Koliko vremena treba da se bazen napuni samo drugom cijevi?

well i can't read croatian

i'll use google translate

[trim sentinel]

Ye that one was so easy

[rigid ridge]

Two pipes fill the pool in 4 hours. Only one of these pipes fills the pool in 5 hours and
15 minutes. How long does it take to fill the pool with just another pipe?

LOL

IT'S ANOTHER RATE QUESTION

[trim sentinel]

Those hour questions are really just everywhere

[rigid ridge]

ikr

[trim sentinel]

And this one was easy as every other before

[rigid ridge]

uwuw 1/(1/4-1/5.5)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Lance#8018

[rigid ridge]

right

[trim sentinel]

I gurantee that wasnt the answer

[rigid ridge]

oh

uh

two pipes A and B
A and B -> 4 hours
A -> 5.5 hours
B -> ? hours
that's the question right

[trim sentinel]

Ye

[rigid ridge]

OH NO

it's 15 minutes

i goofed up

1/(1/4-1/(5+1/4))

uwuw 1/(1/4-1/(5+1/4))

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Lance#8018

[rigid ridge]

this?

[trim sentinel]

Nope

[rigid ridge]

[trim sentinel]

$$4(a + b) = 1$$
$$a = \frac{1}{5.25}$$

[wraith dagger]

Pluton

[trim sentinel]

Then just solve for b

[rigid ridge]

and then take the reciprocal right (since you're working with rates now)

[trim sentinel]

Ye

[rigid ridge]

uwuw {1/b, 4(a+b)=1, a=1/5.25}

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Lance#8018

[rigid ridge]

huh

[trim sentinel]

Huh

[rigid ridge]

16.8h is what rolls out

(= 16h48m ofc)

[trim sentinel]

It should be 6

[trim sentinel]

Oh i guess it is 16

[rigid ridge]

:/

[trim sentinel]

Well whatever i got it right xD

[rigid ridge]

hopefully you didn't get it wrong on the actual MO

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

[trim sentinel]

Nah i didnt

[rigid ridge]

ah nice

this one seems like a nice number theory

[trim sentinel]

I just thought it was 6 probably from some other question that has litteraly same things

[rigid ridge]

ah

Determine all natural numbers 𝑛 for which the fraction 35 / (2n-5) is an integer.

[trim sentinel]

That one was also easy just put 2n - 5 = 1 , 35 , 7 , 5

[rigid ridge]

oh yeah

the translator converted it into a dull divisibility problem

there really isn't much to say except +/-(1, 5, 7, 35)

then check if you get a valid n

[trim sentinel]

Yes but some of them arent natural

[rigid ridge]

yeah

is 0 natural btw?

in croatia

[trim sentinel]

Nope

[rigid ridge]

https://cdn.discordapp.com/emojis/835743928572968973.webp?size=48&quality=lossless

[trim sentinel]

I mean where is it?

[rigid ridge]

idk

[trim sentinel]

You can really just ask that and they will tell you

[rigid ridge]

yeah

anyway

i kinda feel guilty hijacking this channel tho

[trim sentinel]

Lol

[rigid ridge]

maybe move to #competition-math? seems empty

[cedar kiln]

@tender moss Has your question been resolved?

[cedar kiln]

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[patent flame]

I'm drawing a graph of resistance against distance from light source for a Light Dependent Resistor. My teacher has asked us to try and draw a linear graph, but I don't know what would go on the $x$ axis and the $y$ axis for the graph since Resistance and distance from light source do not have a linear relationship. The only equation I know that could be applied is $$I_l = \frac{P}{A}$$ where $I_l$ is light intensity, $P$ is power and $A$ is cross sectional area, and i broke that equation down into $$I_l = \frac{I^2 \cdot R}{4\pi \cdot r^2}$$ where (as far as i know) $I$ is current, $R$ is resistance and $r$ is distance from the light source. I know this is basically physics but I'm completely stuck as to which 2 variables would have a linear relationship (bearing in mind my dependent variable is Resistance, $R$ and independent variable is distance from light source, $r$).

[wraith dagger]

I can't believe you've done this

[patent flame]

the problem that I have is that changing $r$ would change both $I_l$ and $R$ which means that the graph would not be a straight line since the gradient would not be constant. The only other useful information which i think i can get from that equation is $$I_l \alpha \frac{1}{r^2}$$ (sorry about the spacing). Any help would be very much appreciated.

[wraith dagger]

I can't believe you've done this

[cedar kiln]

@patent flame Has your question been resolved?

[cedar kiln]

@patent flame Has your question been resolved?

[fast crater]

if R is dependent on r then it would be good to have a formula showing how R depends on r (same for I_i)

also what is A the cross sectional area of?

if it's of the resistor then r cannot be the distance to the light source

and if it's the percentage of light rays that hit the resistor, it makes no sense for that to increase as r increases

[cedar kiln]

@patent flame Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Dont know if anyone can help for this one since its about physics but.

[crimson sedge]

Why do Hollow objects take a longer time to fall rather then dense objects?

Lemme find a gif real quick

https://m.imgur.com/chk4EeL

This ones about roations but mines about falling

Im guessing its similar

From what i know its because its hollow and has a bigger surface

[upper abyss]

They're not falling, they're rolling

[crimson sedge]

I knkw

But thats not the experiment

I csnt find a gif for it

This one gives an ides though

[upper abyss]

And the "resistance to rolling" is called the second moment of inertia

[crimson sedge]

I just know it takes a greater force for it to be able to get dragged down

[upper abyss]

The wider hollow sphere is harder to rotate than the little red one.

Note the GIF is kinda hiding that fact, that the radius does matter

[crimson sedge]

Because the mass is only concentrated on the outside axis?

[upper abyss]

Yeah, since there's more further away, it gains "rolling resistance"

[crimson sedge]

Verus dense object where its the same everywhere

[upper abyss]

But you were saying that the other experiment had no rolling in it?

[crimson sedge]

No

No rolling

Just dropping

It still has to do with the center of gravity right?

[upper abyss]

So all objects fall at the same rate until you bring air resistance into the problem

[crimson sedge]

All the objects fell at the same time exept the hollow ball

[upper abyss]

Really light, buoyant objects like paper will catch the air and "hover"

[crimson sedge]

I just need to find the correct why

[upper abyss]

Perhaps the ball was that light?

[crimson sedge]

Yeah thats the surface

[crimson sedge]

I mean probably

But that shouldnt have much impact

It is a pretty big ball for it to be that light though

Does that make a difference?

It has at least double of the surface of the other balls

But it weights half of them

Say palm size

[upper abyss]

A crazy video I remember learning a lot from was from an Apollo mission, where they dropped a hammer and feather while on the moon. They both drop at the exact same speed and hit the ground together

For srs, mass really doesn't matter

[crimson sedge]

Yeah i know that part

Difference is the energy

Is it energy?

Eh

How strong the impact at the end will be

[upper abyss]

Yeah that's a good thing to notice, even on the moon, the hammer will put a bigger dent

[crimson sedge]

So hollow ball..

[upper abyss]

Without seeing the video, the only thing I can think of is air resistance, haha. Sorry if I seem off the mark

[crimson sedge]

Probably is air resistence

Welp google it is

.close

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[torn hare]

I really feel like I am missing something. I am also not sure how to expand (lnx)^2.

[jaunty mural]

I see a quadratic inequality.

[torn hare]

yeah, that is one way to solve

[jaunty mural]

one?

I don't see any other myself 🤔

[jaunty mural]

They factored the quadratic, yes.

[torn hare]

oh my goodness, i see it now

Theoretically, is there a specific way to figure out/ expand (Ln x)^2?

[dusty marsh]

i don't think so

[torn hare]

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[void kiln]

Hey uhh

[void kiln]

I’m confused

on my math problem

It asks what’s 100 to the 29th

but I don’t know what it means by that

it’s not very specific

[vagrant breach]

as in 100 to the power of 29?

[void kiln]

I’m not even sure

It just asks 100 to the 29th

[vagrant breach]

ohhh

[void kiln]

but that’s it

[vagrant breach]

well I can't really help

the only thing that comes to mind is powers

[void kiln]

Oh bruh

thanks for the help anyway

[vagrant breach]

sorry

[cedar kiln]

@void kiln Has your question been resolved?

[cedar kiln]

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[polar pasture]

hey

[polar pasture]

Express vector x=(-11,7) as a linear combination of vector y=( -3, 1 ) and vector z =(-1, 2 ).

can anyone help me on this

i have no idea where to start

[vagrant breach]

you know what they mean by linear combination right?

[bold vine]

#❓how-to-get-help

[vagrant breach]

so make a guess for a and b such that
x = ay + bz

[cedar kiln]

@polar pasture Has your question been resolved?

[polar pasture]

yeah ik that's how that works

i just dk how tf do to it lmfao

[vagrant breach]

ohh right

it's just a bit of guesswork

[polar pasture]

oh fr?

[vagrant breach]

yep

so try a=1,2,3 and b=1,2,3

[cedar kiln]

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[polar pasture]

.reopen

[cedar kiln]

✅

[cedar kiln]

@polar pasture Has your question been resolved?

[vagrant breach]

what's wrong?

[polar pasture]

wait

[vagrant breach]

you know what I was being stupid

there is a way to solve it

[polar pasture]

yea i just got the answer ty

[vagrant breach]

simultaneous equations

oh ok

[polar pasture]

mhm

yea ty

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[vagrant breach]

sorry

[cedar kiln]

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[winter ingot]

Is anybody here that could help answer some questions?

[chrome lantern]

#❓how-to-get-help

[cedar kiln]

@winter ingot Has your question been resolved?

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[timid ice]

what is this

[timid ice]

why does this function look like that

[crimson sedge]

it's because the equation is too complex for desmos to plot in a reasonable time

so it spits out the equation plotted at a lower resolution

try tan(x*y!)=13, and scroll out a bit, every few moments the graph shifts, it's just something desmos and other graphing softwares can't handle i guess

[timid ice]

yeah its also making chrome lag

anyways thanks, this seemed interesting

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[crimson sedge]

Hello. How would you go about turning:
a( sin⁴(t)cos²(t) + cos⁴(t)sin²(t) )
Into:
asin²(t)cos²(t)

[bold vine]

Factoring the inside

[gentle lintel]

sin²t + cos²t = 1

[bold hinge]

let's ignore the a for right now

what is the highest term that can be factored from the sum?

[crimson sedge]

sin²(t)cos²(t)?

[bold hinge]

yep

let's try that

[crimson sedge]

oh

that was easy

[bold hinge]

yep

nice job

[crimson sedge]

sin²(t)cos²(t) . [sin²t + cos²t]

Alright thanks people

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[deep flicker]

Hi I just wanted to make sure, is my logic correct?

[cedar kiln]

@deep flicker Has your question been resolved?

[cedar kiln]

@deep flicker Has your question been resolved?

[cedar kiln]

@deep flicker Has your question been resolved?

[deep flicker]

<@&286206848099549185>

[sweet vapor]

Yes

That makes sense @deep flicker

[deep flicker]

Oh okay thanks!!

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[empty smelt]

Solve Simultaneously
a=b+c=0, a-b+c=-4, 3a-3b-c=-1

[empty smelt]

With substitiution method btw

[upper garnet]

since a=b+c

sub a into second equation

find c

then sub a and c into third equation

find b

then u can find

a

$\qed$

[wraith dagger]

The Fractalogist

[cedar kiln]

@empty smelt Has your question been resolved?

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[crimson sedge]

what is b here?

[crimson sedge]

Im learning this and for things like f(x) = x^2 i like inputting x to get the y values

but because of b

i dont know how to try this one out

[true totem]

b is the basis to your function so for example b could be 5 then f would f(x) = 5^x

[crimson sedge]

what do you mean basis of my function?

if b is 5

how do i know what x is?

[true totem]

then you see that b > 1 (first graph) and take a look at the first pic

x is the input you enter and you get a result

[crimson sedge]

so am i supposed to do anything with b?

b doesn't change right?

if b is 5, i can try x = 0, 1, 2, 3, -1, -2, -3 etc?

[true totem]

if your function is f(x) = 5^x then f(2) would be 5^2 so 25

[crimson sedge]

right so once i designate b, it doesn't change right?

[true totem]

yes

[crimson sedge]

i try different x values for that b?

okay thanks but why is it > 1?

why does it have to b > 1?

[true totem]

it doesnt, the first and second graph show how you how it would look when b is > 1 and < 1

[crimson sedge]

ahh got it,

thank you

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[honest bobcat]

manee idk for a

[crimson sedge]

what is C / {0}

[honest bobcat]

the set of non-zero compelx numbers?

[crimson sedge]

oh

[upper garnet]

complex number without zeros

[crimson sedge]

hmm

[tropic oxide]

without zero*

[upper garnet]

:(

[tropic oxide]

anyway

show that if z ≠ 0 then the equation w + 1/w = z has a solution for w

[honest bobcat]

yep

[tropic oxide]

tbh tho. even 0 can be expressed like that

as i + 1/i

but ok

[honest bobcat]

do i gotta quadraticize this

[fierce ermine]

Hi

[honest bobcat]

@tropic oxide

i uh got solutions for w

w = (z +- sqrt(z^2 - 4)) / 2

i dunno what significance that is supposed to have though

[tropic oxide]

technically you did more than what was asked for... but yes

that's fine

as long as you know that any complex number can be square-rooted

[crimson sedge]

im pretty sure you can

[honest bobcat]

wait what

[crimson sedge]

i remeber working out the square root of a+bi a few years ago for fun

[honest bobcat]

but how do i know that that cant be 0

cus z is not zero

I didnt understand the solution for that

[tropic oxide]

??

[honest bobcat]

How do i prove that those solutions dont make z = 0

[crimson sedge]

just exclude zero from your solution?

if you have a soluton that works for all number including zero it aslo works for all number excluding zero

[honest bobcat]

This is what im confused about

[crimson sedge]

you cant devide by zero, we found what w equals and z = w + 1/w so w cant be 0

[honest bobcat]

I dont reallyt understand

what are we dividing

[crimson sedge]

z = w + 1/w

if w = 0

then z is not defined

[honest bobcat]

were saying w cant be 0?

huh 😦

[crimson sedge]

yes w cant be zero

[honest bobcat]

why not?

doesnt the equation show that they literally can be 0?

the part that i highlighted

[crimson sedge]

so z can be rewritten to the form w + 1/w as as long as
w^2 -zw + 1 = 0
but if we take w = 0
then
0^2 -z*0 + 1 = 0
0 - 0 + 1 = 0
1 = 0
if w = 0 then out rule is violated.
leading to the conclusion that
z cannot be re written as 0+1/0 which makes sense

though thew question seems to be oddly written

cause while w cant equal 0, z can

unless im missing something

[cedar kiln]

@honest bobcat Has your question been resolved?

[cedar kiln]

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[amber cipher]

Compute the equation of the tangent to y = x^2 in the points (1/2, 1/4) and (0,0)

[amber cipher]

Given graph

[crimson sedge]

You take the derivative

[amber cipher]

i did that

the derivative of

y = x^2

y= 2x

[crimson sedge]

Yup

[amber cipher]

i got dy/dx thats 0.5

and got stuck there

[crimson sedge]

You got the function of the derivative

Now you plug in x

[amber cipher]

2 * 0.5 = 1

[amber cipher]

like this?

[fierce ermine]

Is this all DLP

[amber cipher]

dlp?

?

[amber cipher]

I plugged it in and after that?

[crimson sedge]

The output is the y

[amber cipher]

yeah 1

[crimson sedge]

The input is the x

[amber cipher]

0.5 , 1

x , y

[crimson sedge]

So know y=f'(a)(x-a) +f(a)

[amber cipher]

I dont understand this

[crimson sedge]

Ok

Now the tangent is the slope right

Of a line

The line passes through 0,0

[amber cipher]

no?

wait yeah

[crimson sedge]

It might not pass through the point you want

[amber cipher]

no wait it doesnt

[amber cipher]

it doesnt pass at 0,0

[crimson sedge]

I'm talking about tangent

[amber cipher]

it passes through x = 0 but not y - 0

[amber cipher]

yeah the line

[crimson sedge]

Like the linear function f'(a) *x

Where f'(a) is the tangent

Now we want this function to pass through the desired point

Are you understanding

[amber cipher]

I follow what you mean

But

You mean the red line?

i dont see it come near 0 , 0

[crimson sedge]

That's the function we want to reach

[amber cipher]

yeah

[crimson sedge]

I want you to understand how we got it

[amber cipher]

but youre saying it goes through 0 ,0 i dont quite get that

how is it going through 0, 0 if it is not near it

[crimson sedge]

I meant the function f'(a) *x

Where f'(a) is the derivative evaluated at a

[amber cipher]

This one?

[crimson sedge]

No

[amber cipher]

so which one there are only 2

[amber cipher]

the derivate of this fucntion?

y = x ^2?

[crimson sedge]

This is another function I'm using so you can understand

It isn't on the graph

[amber cipher]

Okay like that okay

[crimson sedge]

So the function is y=f'(a) *x right

[amber cipher]

yeah right

[crimson sedge]

We want it to pass through the point you want

[amber cipher]

yeah

[crimson sedge]

The line will either be below the point or above it or goes through it

If it is above it or below it we fix that by adding the distance or removing the distance so that it passes through it

[amber cipher]

okay..

[crimson sedge]

We do that by adding the f(a)

Then removing f'(a)

[amber cipher]

okay

[crimson sedge]

That's like saying if the function is above it then lower it, if it is below it make it go up

Are you following

[amber cipher]

bit hard but i am following it

[crimson sedge]

That's it

We have that distance that we want to fix the function with

We then simply add it to the original function

[amber cipher]

thats

y = x ^2

[crimson sedge]

That is y=f'(a) *x

[amber cipher]

okay

[crimson sedge]

So it becomes y= f'(a) *x + f(a) - f'(a)

And that's it

You now have the function of the tangent that passes through that point

[amber cipher]

okay yea

so you will get two answers?

[crimson sedge]

It will be one answer

[amber cipher]

really because the answer sheet shows 2 answers

[crimson sedge]

There can be other lines that pass through the point but their slope won't be the derivative at that point

[amber cipher]

weird

wait

is it y = 0?

[crimson sedge]

No

[amber cipher]

y = x - 1/4

[crimson sedge]

No

[amber cipher]

i am lost than

[crimson sedge]

y=4*x+4-8

y=4*x-4

[amber cipher]

the answer sheet has

y = x - 1/4 and y = 0

[crimson sedge]

Wait I got confused, sorry I thought the question was the equation of the line that passes through 2,4

[amber cipher]

no

the equation is already given

it was: Compute the equation of the tangent to y = x^2 in the points (1/2, 1/4) and (0,0)

[crimson sedge]

Oh

[crimson sedge]

.

Use that

[amber cipher]

what should i use as a inpout

the derivative of

y = x^2

[crimson sedge]

Replace a with the input

[amber cipher]

okay one second

with the coordinaties

so

f'(1/4) * 1/2 - f'(0)

[crimson sedge]

No you don't change the x and you only use one input at once

[amber cipher]

so how do you choose what to input

[crimson sedge]

You are given the x and the y

The x is the first one and the y is the second one

[amber cipher]

owh the coordinates okay

yeah i understand

so its f'(1/2) * 1/2 - f'(1/2)

-1/4

[crimson sedge]

The second 1/2 should be x

Because the tangent is in function of x

[amber cipher]

so thats 0

1/2 * x - 1/2

nevemmind so

1/2x - 1/2 is the answer

[crimson sedge]

1/4x +1/4-1/2

[amber cipher]

you just said

to keep the x

and use one input at once

[crimson sedge]

I did

The y is f(a)

[amber cipher]

= f'(a) *x + f(a) - f'(a)

[crimson sedge]

Yup

[amber cipher]

so the f'(a) are x and the f(a) are y

[crimson sedge]

f'(a) is the derivative evaluated at a

[amber cipher]

i meant the coordinates

that you choose for it

[crimson sedge]

No

[amber cipher]

...

you said this

y = f(a)

[crimson sedge]

f evaluated at x is y

[amber cipher]

could you explain it a bit simpler

[crimson sedge]

A function is a set of instructions to output a number

f applied at x is equal to y

[amber cipher]

yeah makes sense

I understand that

[crimson sedge]

So now you input the x

[amber cipher]

yeah thats 1/2 and not 1/4

[crimson sedge]

Yup

The function y=x^2

Input 1/2

[amber cipher]

you used 1/4

[crimson sedge]

And you get 1/4

[amber cipher]

wait

you first need to put in the y = x^2?

[crimson sedge]

Yeah

That's what f is

y=f(x)

[amber cipher]

okay

[amber cipher]

so why the 1/2

[crimson sedge]

The 1/4 is f(1/2)

[amber cipher]

okay yeah i understand

so thats

1/4x-1/4

[crimson sedge]

Yes

[amber cipher]

so how does that beome y = x- 1/4

[crimson sedge]

f'(1/2)=1

1*x-1/4

y=x-1/4

[amber cipher]

wait one second

you used

y = x^2?

f'(x) = 2x

[crimson sedge]

The second one

[amber cipher]

f'(1/2)= 2*x

y = 4x-4?

[crimson sedge]

No that's wrong

[amber cipher]

wait

[crimson sedge]

f'(1/2)=1

[amber cipher]

first you have

f(x) = x^2 > dy/dx = 1/2 > f'(x)= 2x > f'(1/2) = 2 * 1/2 = 1

are these steps correct

no dy/dx = 1/2

1/4-0 / 1/2- 0 = 1/2

[crimson sedge]

dy/dx=2x

[amber cipher]

i dont quite get it how you got that

[crimson sedge]

dy/dx=f'(x)

[amber cipher]

okay yeah

but wouldnt the answer still be 1/2?

[crimson sedge]

The derivative will be 2x

[amber cipher]

2x or 2?

[crimson sedge]

2x

[amber cipher]

where does the x come from

[crimson sedge]

Y=x^2

The derivative of that is 2x

[amber cipher]

yeah okay understood

but how does y = x -4 and y = 0 come from

[crimson sedge]

.

[amber cipher]

so this is the first one

and second you just use 0 as a input and that gives you y = 0

is that correct?

[crimson sedge]

Yes

[amber cipher]

I understand thanks!

appreciate the help

.close

[cedar kiln]

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[buoyant veldt]

Hello! Can someone teach me how to solve this?

[buoyant veldt]

<@&286206848099549185>

[zealous crystal]

Draw a diagram

[steady trench]

FE=2FA, perimeter is FA+AC+FE+CE, FA=AC so perimeter is just 6FA

yea draw a diagram and then its straightforward

[zealous crystal]

Just dont give away answers

[buoyant veldt]

Oh ok

My answer is
FE= 10 inches

Is that correct?

[steady trench]

yes

[buoyant veldt]

Thanks for the help!

.close

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[crimson sedge]

when trying to inverse this matrix

i am getting two solutions

one is right

which is this one

but here if i instead do C2=C2-3[C1]

i am getting 10 below and 1 above as answer

lol

is that operation illegal ?

[marsh pecan]

i dont see the issue this is correct

[crimson sedge]

yes

that is correct

[marsh pecan]

wdym by "here" can you send me how you did it and got a wrong answer

[crimson sedge]

so like in this image i reached till first thing

then instead of what he did i did this

[crimson sedge]

THIS

the answer that comes

10 is below

instead the computers answer..

where 10 is above

[marsh pecan]

thats because you are using both row and column operations

[crimson sedge]

HUH

[marsh pecan]

im pretty sure you have to stick to 1 while getting the inverse

[crimson sedge]

but it works in determinant calculation

[marsh pecan]

thats a different situation

and i can dive into the theory of it but that might be a bit too confusing rn

[crimson sedge]

you mean graphical representation

so the other things like "multiply row 2 by K both sides" also have to be done in row or column only ?? whichever we choose to do

like if we do C2=C2+2C3

we cant do R3=K (R3) next

and we can do C3=K (C3)

[marsh pecan]

no you missunderstood me

[marsh pecan]

im saying while using this method to find the inverse of a matrix

you cant interchange using column and row operations

you have to stick to 1

[crimson sedge]

but i can multiply any row or coulm with K

[marsh pecan]

as long as k is not 0

[crimson sedge]

ok

theories of such things arent taught in school

[marsh pecan]

you'll learn them eventually

[crimson sedge]

but wont i learn higher things like use of matrix later

rather than deeper things like proofs and intuitions

[marsh pecan]

learning the intuition and proofs is part of learning how to use them

[crimson sedge]

DAMM

ok

close

this

.close

[cedar kiln]

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[cobalt ether]

Hey I'm struggling a bit with my first steps into the world of equation system solving and I'm unsure what it is I'm being actually asked by exercises in my textbook

[cobalt ether]

the way the exercise is laid out is

x1 + 2x2 = 5
2x1 + kx2 = t

[dire geode]

show the original question

[dire geode]

this format is vague

[cobalt ether]

uh alright two seconds

it's in Norwegian though so uh

[dire geode]

equations are enough. you should translate

[cobalt ether]

ok this is the section I have issues with

so I already starting putting my Ax = b form down and attempted to use Gaussian elimination on it

https://media.discordapp.net/attachments/846128265932701737/950869516605345853/20220308_223539.jpg?width=526&height=701

but I'm unsure if

a) I'm even using Gaussian elimination right here

b) what the exercise is asking me concretely when asking for which values of k and t I have

infinite solutions, a fixed solution or no solution

wait

my Gaussian thing is flawed apologies

ok I think I have something better

which is to to the same thing I did

eliminate the first collumn 2nd row 2

but have the 2nd column 2nd row be k - 4

which means my output system should be

This right?

but still unsure how from there I'm supposed to find the number of solutions?

[cedar kiln]

@cobalt ether Has your question been resolved?

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@cobalt ether Has your question been resolved?

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[lethal knoll]

trying to prove this statement, not really sure where to go with it. tried a few approaches.

given that: a, b >= 2, gcd(a,b) = 1, and ab = (x+y)^n, prove that x+y is not prime
(where a,b are natural numbers, x,y are any integer)

[rigid ridge]

what does gcd(a,b) = 1 say about the prime factorization of a and b?

[lethal knoll]

they share no common prime factors

[rigid ridge]

so why can x+y not be prime?

[lethal knoll]

hmm, still not sure

[rigid ridge]

i have no idea why it's x+y so let's call x+y = z for simplicity

if z were prime, then the prime factorization is ab = z^n for a prime z

do you see why this gives problems with gcd(a,b) = 1?

[lethal knoll]

not quite tbh

yeah, played around with it a bit but not sure @rigid ridge

[buoyant perch]

if z is prime then its a contradiction because either a = 1 or b = 1 or a divides b or b divides a non of which are true

[rigid ridge]

yeah that

[buoyant perch]

say ab = p^n then a = p^r and b = p^s where r + s = n but then the gcd(a,b) = p^min(s,r) or one of a,b = 1 which is a contradiction since the gcd(a,b) = 1 and a,b > 1

[rigid ridge]

if ab = z^n with z prime then the prime factorizations of a and b can only have z (and so have at least one by a,b >= 1), but they also cannot share any prime factors

[lethal knoll]

ah i see. i think i just got the same with FTA (?); a and b having disjoint prime decompositions means that p would have a factor other than itself and 1

though idk, mine seems poorly worded even if it is true

[buoyant perch]

if you ever get a question about proofs thats prove something isnt this then you can probably just do it easily with contradiction by saying it is and showing why thats wrong

so just prove it by contradiction thats probably the easiest way

[cedar kiln]

@lethal knoll Has your question been resolved?

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[remote tusk]

Can somebody help give me a clearer explanation of SOH CAH TOA? Im not confused on the hypotenuse part but telling where the adjacent and opposite parts are isn’t making sense

[tight herald]

Opposite is the side opposite of the angle, so the side that the angle is not touching

Ignore my bad drawing

[remote tusk]

This is how it is on the review

[tight herald]

Do you understand adjacent?

[remote tusk]

Not at all

[flint plinth]

adjacent is the one that is not opposite or hypotenuse 😆

[remote tusk]

Hypotenuse is across from the 90 degree angle but I’m still not getting how that opposite works

[flint plinth]

hypotenuse is easy, it's the longest side and the one that does not touch the right angle

[remote tusk]

Would it change where your adjacent and opposite are placed depending if it’s SOH, CAH, or TOA?

[flint plinth]

no, the placement of opposite and adjacent only depends on which angle you are considering

[remote tusk]

On here for Cos S it’s 16/34 and wouldn’t 16 be opposite? and tan S is 30/16 but wouldn’t that mean the 30 is adjacent?

[livid hound]

no

Cos: Adjacent /Hyp:

[flint plinth]

this is why "soh cah toa" exists, to help you remember

[livid hound]

for cos(S),
r=16 is the side adjacent to S

and t=34 is the hypotenuse

resulting in cos(S) = adj/hyp = r/t = 16/34