help-0

[dry zephyr]

i see

[hushed scroll]

now what do you think is the last thing you need?

[dry zephyr]

normal

a point

mb

[hushed scroll]

you know if a vector in parametric form
(a+b+c)+k(a'+b'+c')
then (a',b',c') is the direction vector of that vector?

[dry zephyr]

yes i do know

[hushed scroll]

you know Direction vector of both vectors that lie on the plane, so their cross product will give bormal vector

[dry zephyr]

yes

[hushed scroll]

so you have everything

[dry zephyr]

im not quite getting it for some reason

[hushed scroll]

try and tell me if you get the answer

[dry zephyr]

so lets say im taking t=0 right?

[hushed scroll]

yes that will give coordinates (x0, y0, z0) of a point on the plane

[dry zephyr]

i am then plugging it into the point form equation

?

[hushed scroll]

yes

[dry zephyr]

i got -x+5y+22z-62=0

[dry zephyr]

thats exactly what i got doing my method also lmao

[hushed scroll]

so like just tell me what were the two direction vectors you cross multiplied?

or tell me what (a,b,c) you got

its probably calculation mistake

[dry zephyr]

oh wait actually it might be a calculation error

hold on

[hushed scroll]

so you had already got the answer but these little calculation mistakes

[dry zephyr]

ok i should get (9.75,-4.75,-5.5)

as my cross product right>

[hushed scroll]

no

[dry zephyr]

damn

[hushed scroll]

what are two vectors you are multiplying

[dry zephyr]

ok so i am cross the two direction vectors of the two vectors we generated right?

[dry zephyr]

(5/2,1/2,4) x (3/2,5/2,1/2)

[hushed scroll]

5/2,1/2,0 it should be

[dry zephyr]

your right

damn it

[hushed scroll]

[dry zephyr]

so

[hushed scroll]

i too took 4 first lol but i saw the mistake

[dry zephyr]

(-0.25,1.25,-5.5)?

[hushed scroll]

yess

you got it

[dry zephyr]

omg

yea i finally got it

yea that was pretty intuitive

i think its just the 3am is getting to me

[hushed scroll]

lol

[dry zephyr]

i also think i found a hybrid between standard and your way also

okay thank you

[hushed scroll]

probably because i used fixed points and standard way they had used some variables p,q,r then simplified it lol

ok bye

[dry zephyr]

.close

[lone heart]

Channel closed

Closed by @dry zephyr

Use .reopen if this was a mistake.

[lone heart]

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[raw jetty]

let $a,b,c\in\mathbb{R}$, if $a^4+b^4+c^4=1$ find the minimum of $\frac{a^3}{1-a^8}+\frac{b^3}{1-b^8}+\frac{c^3}{1-c^8}$

[ocean seal]

skissue.in.a.teacup

[raw jetty]

status 1

[proud obsidian]

at first glance maybe harmonic mean?

eh nvm

[rancid marten]

what did you try

[raw jetty]

!status 1

[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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[rancid marten]

do you know Lagrange multipliers

[lone heart]

@raw jetty Has your question been resolved?

[raw jetty]

no

[median oar]

[raw jetty]

you hated geometry and number theory time to add algebra to the list :p

[fallen verge]

algebra always been my least favorite

NGCA, any other order is wrong

cant see how this helps but you can do
3-a^4+...=2
3+a^4...=4

then add then and use am gm on each part to get 3>=sqrt(1-a^8)+sqrt(1-b^8)+sqrt(1-c^8)

nvm thats obvious

actually i can do it better

[empty mason]

First make a substitution and work with one individual part. Like work with only a^4 for example.

[fallen verge]

sum>=3cbrt((a^3b^3c^3)/((1-a)^8...))
using amgm we get 1/3>=(abc)^4/3
(1/3)^(3/4)>=abc
and using what i did earlier
1>=((1-a^8)...)^1/6

so we multiply that together while adjusting powers and you get a number

(1/3)^(3/4) i think

its an upper bound but i dont think ive proven its a minimum

[raw jetty]

uhh how did you get the 1>=((1-a^8)...)^1/6 part

[fallen verge]

3-(a^4...)=2
3+(a^4...)=4

add and average

then amgm

actually the signs switch because we need the 1-a^8 stuff in the denom and 3/4 is a power less than 1

and then multiplying together

doesnt change signs

still think thats just a lower bound tho

because lowest i can get is (9/8)(1/3)^(3/4) by setting a=b=c

[empty mason]

Like say a^4 = x, b^4 = y and c^4 = z. Then x+y+z=1. Then only look at the a^3/(1-a^8) = x^(3/4) /(1-x^2) part and prove that it’s always going to be greater than or equal to something in terms of x. Then you can apply the fact that x+y+z = 1 to obtain a minimal value

[raw jetty]

oh yea i missed this

[fallen verge]

im going to bed

[raw jetty]

what did you even do there cause wouldnt addinf them get 3+3=4+2 lol

[median oar]

🤮

[fallen verge]

just dont cancel

im tired

ill think about it again when i wake up

but ive always been bad at ineqs, so no practice is saving me rn

[raw jetty]

what can you even do to get a lower bound on that?

[empty mason]

Use AM-GM. Actually work with a instead of x so we want to get a bound on a^3/(1-a^8) in terms of a^4. Let this quantity be greater than or equal to c * a^4 for c is some constant. Then do some rearrangement, and notice the AM-GM you can apply

[raw jetty]

if you do am-gm wouldnt the denominator always have that 1- part or no

cause if it has that 1- i dont think it can be in the form c*a^4

[empty mason]

Yeah it would but upon rearrangement you have the (1-a^8) * a is less than or equal to 1/c. Also i should’ve mentioned before that c is just a random constant, not the one referring to c = z^4. Those two are different. Then from here you just raise it to the 8th power and try use AM-GM

[raw jetty]

what rearrangement what

oh

bur c wouldnt be a constant tho wouldnt it

[empty mason]

It would be a constant. You just have to try make it a constant by doing what i said before

[raw jetty]

sorry i still dont really get what yoh mean

[lunar saddle]

NO

[raw jetty]

NGAC icl

nt is goafed

[lunar saddle]

check my bio

[empty mason]

check my bio too

[raw jetty]

check my bio too

[lunar saddle]

||c u l t u r e d||

[raw jetty]

i cant believe you put combi at top

[empty mason]

Wait a min, one of my friends has the same as yours except G and A swapped around

[raw jetty]

A used to be my top

[lunar saddle]

my best friend is the same as you

[empty mason]

Yes, I love combi. Honestly how I did it was how much I liked the area. But still I do better in combi ngl

[raw jetty]

then functional equations hit.

[lunar saddle]

i do best in FE, todays training was on FE

fakesolved IMO 2023 A4 then i depressed

sry if yap too much off topic stuff

[raw jetty]

stfu carbonite i get it your fucking cracked dont rub it in me 💔

/j

[empty mason]

All good. Just we will assume for now that a^3/(1-a^8) ≥ c * a^4 ⇔ 1/(1-a^8) ≥ c * a ⇔ 1/c ≥ a * (1-a^8). Raise both sides to the eighth power and find maximum value of RHS by using AM-GM

[raw jetty]

how do youe ven get good in FE's

[lunar saddle]

t h o n k

sub shit and conjecture

[raw jetty]

why the eight power specifically?

[empty mason]

When raising to the eigth power, then we can use the fact that a^8 + (1-a^8) = 1 in our AM-GM.

[raw jetty]

1/c^8>=a^8*(1-a^8)^8
like this??

[empty mason]

Yeah

[raw jetty]

[lunar saddle]

smart

[raw jetty]

wouldnt a^8*(1-a^8) be the GM

[empty mason]

How??? I need to know the strat

[empty mason]

Yeah

[raw jetty]

so like 1/c^8>=a^8*(1-a^8)^8<=something or do you want to squeeze the am between 1/c^8 and a^8.... somehow?

[dense plover]

whats FE?

[lunar saddle]

functional equation

[dense plover]

ah

(like i get this stuff)

[empty mason]

Actually by AM-GM you determine exactly what the c is. No need to combine the two

[raw jetty]

ohhhhhhhhh

wouldnt there be aton of extra 1-a^8 terms tho after am-gm

[empty mason]

Also people who put A before C are doing a crime. Who even puts A before C? That’s despicable. C will always be the winner. At least we all show our appreciation for how great N and G is (and btw A sucks, I hate A a lot). A is seriously not nice at all, jk

[dense plover]

what is ngac 😭

[lunar saddle]

get out

the muirhead cult will find you

[empty mason]

Nothing, it’s not a thing

[dense plover]

okay

[lunar saddle]

we will send the biggest bulldozer of lineland to your house

it's 2015 imo c1 btw

[raw jetty]

i love how this convo is going

just a buncha oly nerds

[empty mason]

Yeah, so we want to change that so we multiply a^8 * (1-a^8)^8 by 8 before applying AM-GM

[empty mason]

Whoever joined this cult clearly has no head

[lunar saddle]

look at my status

"muir your head"

[empty mason]

No you shall not

[empty mason]

Interesting references

[empty mason]

I don’t get it, maybe I have no head

Also controversial opinion but I like 2024 IMO P5

[raw jetty]

with what black magic

[lunar saddle]

No turbo

I feel bad for the china team lmao

[empty mason]

Yes, black magic is used. But if you see that we want to use a^8 + (1-a^8) = 1, we need 8 of the a^8 since by AM-GM 8 (1-a^8) are already used

Pls don’t ask me too many questions, this is exactly why I put A at the bottom

[empty mason]

Yes turbo. That problem was needed. Turbo is needed

[empty mason]

Same

[empty mason]

Just edited this message. Should make sense now (the number of brain cells I have decreases during the day)

[raw jetty]

aaa i still dont understand how

like for AM-GM if one of the terms of GM is unbalanced, then you can duplicate some AM terms to balance it, and it will only mess with the coefficient

but if you try to duplicate GM it messes with the exponent but thats the part we dont want to change

[empty mason]

Uh, let’s try from the top, shall we? We want to find the maximum value of a(1-a^8).

This ^8 part is pretty annoying so we raise this to the power of 8 for the exponent of a to match it such that we find the maximum value of a^8 * (1-a^8)^8

Again this looks close to our GM, but we’re not done because this is not AM-GMable by how you said before (the extra (1-a^8) terms). We know that by AM-GM (regardless of how many terms we have), the AM will have 8 of the (1-a^8)’s. To counteract this, we multiply the a^8 by 8 such that the a^8’s cancel out in the sum of our AM. Hence why we multiply 8 to our original expression to get (8a^8)(1-a^8)^8.

Then we see that the number of terms used in our inequality is 1+8 = 9 so we now can use AM-GM with 9 terms to arrive at an answer

I’m so sorry if this isn’t making sense, I’m trying to also derive the motivation behind the solution

We just try to manipulate it into an AM-GM inequality if that makes sense

[raw jetty]

ohhhh

so c=sqrt4?

oh wait

c=3^(9/4)/8?

[empty mason]

Correct, you can simplify it more if you like

You’re nearly done now :)))

[empty mason]

Numerator can be written as 9 * 3^(1/4)

[raw jetty]

so original sum>=k*(a^4+b^4+c^4)=3^(9/4)/8?

and its equality is at a=b=c=1/sqrt4 which holds for the a^4+b^4+c^4=1?

[empty mason]

I think that’s incorrect. We only found that inequality for one term, you have to sum it with the other two terms

[raw jetty]

yeah for the term with a its c*a^4 right

[empty mason]

c(a^4 + b^4 + c^4) is not the same as c(a^4)

[empty mason]

Yeah

[raw jetty]

summing with the rest and factoring the constant out you get c(a^4+b^4+c^4)

[empty mason]

Correct, but you left the c as the same. It should be 3c

[raw jetty]

what

(lets change the constant to k)

so k(a^4)+k(b^4)+k(c^4)=k(a^4+b^4+c^4)?

[empty mason]

Okay good

[empty mason]

Yes, correct, but we only found an inequality for k(a^4).

[raw jetty]

why would it be any diffrent though?

[empty mason]

or k(b^4) or k(c^4). You need to add the 3 together. You left it the same though as for one only

[empty mason]

Look at the RHS here. It’s the same as k. It should be 3k

[raw jetty]

why would it be 3k?

ab+ac=a(b+c) no?

[empty mason]

Yeah ofc ik that

[empty mason]

Okay, so we got that a^3/(1-a^8) ≥ k(a^4). So as well b^3/(1-b^8) ≥ k(b^4) and c^3/(1-c^8) ≥ k(c^4). Adding the 3 inequalities together, we get that S ≥ k(a^4+b^4+c^4) where S is the wanted sum.

[raw jetty]

uhh i forgot we did some finnicky stuff back then

[raw jetty]

wait but didnt we define a^3/(1-a^8)>=k*a^4 instead of just k

[empty mason]

Oh wait, you’re right

I’m so sorry, mb

[empty mason]

Okay so then this is correct

[raw jetty]

alright then, thank you!!

.solved

[lone heart]

Channel closed

Closed by @raw jetty

Use .reopen if this was a mistake.

[empty mason]

No problem. So sorry for confusing you a lot. I’m horrible at algebra

[raw jetty]

me too

[empty mason]

At least we both can get better at it by grinding more problems (I seriously need to work on my proofs)

[raw jetty]

me too

[empty mason]

Also sorry for asking but do you do olympiads? Cause this is definitely a competition problem

[raw jetty]

yuh

why do you think i said oly nerds lmao

[empty mason]

Nice

[raw jetty]

carbonite also is an olympiad (a really good one)

[empty mason]

Oh yeah, you said that

[empty mason]

I can tell. I need to reach their level of algebra soon

[raw jetty]

guess their age

[empty mason]

They told me. I feel so old. 😭

[raw jetty]

IKRR

and i thought i was young

[empty mason]

We need an age limit on olympiads. Because honestly now I see that 9 year olds are taking olympiads like it’s nothing

[raw jetty]

yuh

like make amc10 actually for 9-10th graders??

you cant just let carbonite take part it ""because amc8 is boring""

ughh i envy him so much

[empty mason]

Exactly

[empty mason]

They let people who do well in amc8 take amc10. Just please wait until you’re old enough.

[raw jetty]

ikr

and why let top amc10 take part in aime

[empty mason]

Same, I asked if we could switch ages and he said no chance. And worst part is that organisations favour the young people because they have more potential

[raw jetty]

LMAO

can we switch is crazy doe

[empty mason]

Uh that’s how I made it that time. Thank goodness for that

[empty mason]

At least I tried

[raw jetty]

uhh

howd you do in aime

[empty mason]

Alright, like 10+

[raw jetty]

what the fuck does 10+ mean

[empty mason]

10 or above out of 15. I’m not revealing my actual score

[raw jetty]

cooked im falling behind

pretty please

[empty mason]

It’s okay, you have time young one

[empty mason]

I’m sure you will make IMO. I can foresee the future

[raw jetty]

a year flies by man 💔

[raw jetty]

thats my goal 🙏

[empty mason]

Me when time runs at the speed of light while I’m running at a snails speed.

[empty mason]

You better make it in. Then I can say that ik someone who made it into imo

[raw jetty]

im not telling you weather i made it in or not

gimme your full legal name first then maybe ill consider

[empty mason]

I’ll never know if you made it in

[empty mason]

Um no chance. I never reveal my name on here

[raw jetty]

me too

man i just realized my national mo is in like a week lmao

[empty mason]

The only thing I can say is that IMO is held in my country yet far away from where I live so I’ll never have the chance to go to the IMO, even as a volunteer

[raw jetty]

isnt you going to the imo subsided by your country

[empty mason]

Omg. You better start preparing. Best of luck on the national mo, I’m sure you do well. and btw don’t stress about it, I find that one does best by not stressing

[empty mason]

Yeah but it’s too far away for my parents to consider driving all the way there

[raw jetty]

oh your australian?

[empty mason]

Yes

[raw jetty]

but you participated in amc??

[empty mason]

This centre does outside competitions so I did the American amc as well, yes

Anyways I have to go now, but it was nice chatting with you!

[raw jetty]

you too!

[lone heart]

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[glossy kestrel]

.close

.reopen

😔

i fucked up

[proven leaf]

just open a new one—it's fine

[lone heart]

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[forest moth]

Is this accepted as a "showing"? Or do need to use an example? Thanks

[echo socket]

This is sufficient

[modern sedge]

You could add some words probably

[median oar]

i think that looks good

[modern sedge]

I'd guess that there is a lot of space because they expect you to write a bit more than 4 random matrices

at least describe what each matrix does

[median oar]

maybe put an arrow for the last matrix as well

but really the answer is to ask your teacher

[modern sedge]

oh the arrows do that

[forest moth]

Could be clearer tho yeah

[median oar]

we aren't the ones grading your test

[forest moth]

Yh I'll ask

[median oar]

criss crossing arrows are hard to understand

otherwise it's fine tbh

if i read this as a marker i'd be like

ok this guy knows what's up

[forest moth]

Thanks everyone for the quick response!

.close

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[pine wraith]

this is for an exam without a calculator so i calculated the arithmetic mean to be 6 and geometric mean as √35 so how would i find the difference

[toxic verge]

find sqrt 35 using differentiation

[pine wraith]

√36 is 6 but how would i approximate √35 to get an exact value from these answers

[pine wraith]

its a quick exam i wont have that much time

[toxic verge]

well then just see how many decimal places the subtraction goes till

the options are a decimal apart each

[pine wraith]

wdym ?

[toxic verge]

sqrt 35 will be something like 5.8

and 6- 5.8x

would be 0.08x

so which one is closest

[pine wraith]

why did u write x

isnt 6-5.8 = 0.2

[toxic verge]

x means "..."

since its not exactly 5.8 i think

Lmao w sqrt 35

[ocean seal]

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[toxic verge]

yeah its not exact

its more like 5.9 but the closest option is b

[pine wraith]

how did u get 0.08x

[toxic verge]

do you understand?

[toxic verge]

it's all approximation

it's closer to 0.1

but the options have 0.08

[pine wraith]

so why not 0.8

ik the ans os 0.08

[toxic verge]

0.8 is 8 times 0.1

0.08 is much closer

[pine wraith]

ohh ok sorry i confused 0.1 and 1

i got it now

thanks!

.close

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[magic drum]

how to slove this

[desert raptor]

relate the numbers

[tall shale]

solve it

[magic drum]

substitute them ?

[desert raptor]

You've been given a, b, c are in AP

so 2b = a + c

[magic drum]

yes

[desert raptor]

like that, relate p, q, and r

and ap, bq, and cr

[magic drum]

oh ok fine

thank you ,If any help can ping u ?

[desert raptor]

js ping the helpers role

@Helpers

[magic drum]

ok

[lone heart]

@magic drum Has your question been resolved?

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[stark flicker]

I have a set x & y position and a chain of lines defined by their angle off of the positive y direction and their distance from the endpoint of the previous line.

I also have variance values for the starting position, the angles, and the distances. I would like to generate a shape which approximates all the possible polygon configurations with the given variance values.

This is (relatively) easy to do with an open shape, but when the same method is a applied to a closed shape there are many configurations calculated which do not produce a closed shape.

I attached a sketch of an open shape and it's variance approximation, as well as a closed shape and (what I think) would be the end goal of the variance approximation

[stark flicker]

In my mind I feel like this is an inverse/forward kinematics problem, but I don't know enough to be sure

[modern sedge]

isnt this a problem for computer?

[stark flicker]

It is, but I can't write an algorithm to calculate it without the mathematical solution

[modern sedge]

Well

is the variance only for angles btw?

Or for distances as well?

For closed shapes, you could just generate the shape until the very last line

and for the very last line, accept only those that lead back to starting point

[chrome tiger]

So like, you have a sequence of angles and distances, and you can perturbate all angles by +-a, and all distances by +-delta, and the goal is to draw the region that will be covered by all possible options?

[charred summit]

So what I understand you want to generate an approximate polygon that represent the variance in x,y position, angles, and distances?

[stark flicker]

Correct

[chrome tiger]

alright

[modern sedge]

after this, you'd have to recursively delete the ones that didnt lead back to starting point smh

so i think that what we need is given some point, we need to find the area from which we couldve gotten there

that's the bit we will need for the recursion thing

oh

We could just go in 2 different directions

and then get the intersection of the 2 polygons

i think that should work

ill try to draw what i mean

[chrome tiger]

Wait so we're interested only in perturbations that lead to a closed shape?

[modern sedge]

open shape is easy was said

but yeah

[chrome tiger]

oh okay
I was thinking about some boundary calculation algorithm

[modern sedge]

so for one point, angle and distance the region would look like this if im not mistaken

section of circular ring

[chrome tiger]

So basically this problem is "finding all parameter values within a given region (in configuration space) that satisfy a system of 2 equations"
2 equations being (1) x coordinate of the end point = x coord of the starting point, (2) same for y

[modern sedge]

I think that we could just generate the region in the original direction first

then the region in reverse direction

and intersect them

ill have to draw this on paper, as it would took ages otherwise

cant find my phone to take a photo ;-;

okay

The red is our starting point

the red region is everywhere we can get from red

and there is one key thing to notice

if we reverse the direction, we can apply the same procedure to reverse it

i.e. the blue region is the set of all points, which could serve as a starting point to get to point blue

@stark flicker you here?

imma just continue, you can read it later

so now how can we use it

you got your set of directions and angles with variances

so you just start at the starting point and start generating the region as you normally would for open shapes

once you get to the end you stop

now the problem is that there is some extra region, from which we cant get to the endpoint

and we need to find and delete that region

so what we instead do is we find the region, from which we can get to the endpoint

[modern sedge]

and we do it using this

we reverse the list of directions and distances and also flip all the directions (e.g. add 180° to them)

then we start at the startpoint/endpoint and start making new region

this new region will be the set of all points, from which we can make it to the endpoint

and finally - we take the intersection of the 2 regions we got

so now every point in the region has guarantee that:

1. we can get to it from the starting point
2. we can get from it to the end point

[stark flicker]

Oooh, that is actually a very elegant solution

[modern sedge]

thanks :)

[stark flicker]

Thanks much, I was sitting here trying to code this in a vacuum for like 3 hours

[modern sedge]

np

what is this for btw?

[stark flicker]

I'm writing a mapping tool to generate roads and building/property edges from a list of azimuths and distances, and also to calculate single locations based on azimuth triangulation. Similar tools already exist, but none that do exactly what I need, and in particular there aren't any that give a good visualization of instrumentation error

[modern sedge]

oh that's pretty cool

good luck

[stark flicker]

Thanks :)

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[bitter pulsar]

i dont understand how to get from the RREF to this basis

[median oar]

What does the rref form say

Like what does that augmented matrix mean

[bitter pulsar]

that x_1 - x_3 - 2x_4 is 0 and x_2 - 3x_3 - 4x_4 is 0

[median oar]

Right

So in other words, x₁ = x₃ + 2x₄ and?

[bitter pulsar]

and x_2 is 3x_3 + 4x_4

[median oar]

Okay so if I write an arbitrary vector (x₁, x₂, x₃, x₄) but subject to these restraints

[median oar]

Then I can replace x₁ and x₂ with these

So in other words, the vectors look like (x₃ + 2x₄, 3x₃ + 4x₄, x₃, x₄)

Or, (1, 3, 1, 0)x₃ + (2, 4, 0, 1)x₄

Aha these can be the basis vectors

[bitter pulsar]

so its just how many of the free variables?

[median oar]

Huh?

[median oar]

What is?

[bitter pulsar]

these vectors

[median oar]

Okay I was being informal

[hallow vessel]

why is everyone awake

[median oar]

The set of vectors V, are ones that satisfy the leftmost augmented matrix

Incidentally by nature of Gaussian elimination, they also satisfy the augmented matrix on the right

Hence, V = {(x₁, x₂, x₃, x₄) ∈ ℝ⁴ | (x₁, x₂, x₃, x₄) = (1, 3, 1, 0)x₃ + (2, 4, 0, 1)x₄, for some x₃, x₄ ∈ ℝ}

[bitter pulsar]

i just dont understand exactly how we got there or why

[median oar]

Now you need only check that those 2 vectors (1, 3, 1, 0) and (2, 4, 0, 1) are linearly independent from each other

Where did you get lost

[median oar]

Since they definitely span V, if they are LI then it’s a basis of V by definition

[lavish cave]

ah, so yes it is a fair jump from the final RREF to that basis

[lavish cave]

if you accept that the basis has dimension 2 (4 - 2 = 2 by rank-nullity)

[median oar]

You don’t even need that theorem

[lavish cave]

then you could have two free variables u, v for the 1st and 2nd columns

[median oar]

An important thing is that basis is not unique

If you chose to set x₃ and x₄ in terms of x₁ and x₂ you will find a different basis

[bitter pulsar]

would it still be valid?

[median oar]

Yes

[bitter pulsar]

fucked up

[median oar]

It says to find a basis

It’s not fucked up

That’s just how it goes, bases are not unique

And it’s very useful that they aren’t unique

[lavish cave]

yeah to get that exact basis, they just set the first column to have coefficient 1 and the last column to have coefficient 0

so if a1 = 1 and a4 = 0, you get 1 * 1 + 0 * u + -1 * v + -2 * 0 = 0
1 - v = 0, or v = 1

so (1, anything, 1, 0)^t would satisfy

but then you also have to check the (0, 1, -3, -4) as well

so (1, x, 1, 0) dot (0, 1, -3, -4) = x - 3 = 0
x = 3

you must get (1, 3, 1, 0)

similarly repeat everything except now the first column has coefficient 2

[hallow vessel]

why r u not sleeping??

[lavish cave]

ok yeah I better sleep

[bitter pulsar]

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[surreal drum]

How can I fix the mistake because there ist (n+1I missing in the denominator but I need to have this solution?

[tacit arch]

what is "this solution"

[surreal drum]

I have to show that In+1(b) is equal to -(b+1)^(n+1)*exp(-b)+(n+1)!

[tight pier]

What's I?

Ah wait, first row

Assuming that's correct

[surreal drum]

Yes

[tight pier]

What if you muiltiplied b^(n+1) by (n+1)/(n+1)

Wait no that's dumb

I think you need to multply it by (n+1-k)/(n+1-k)

So that you can include the (n+1)-th into the sum

I mean you simply just included it just like that

But then again hmm

[tacit arch]

can you just show the whole problem and where you got I_n(b) from

[lone heart]

@surreal drum Has your question been resolved?

[lone heart]

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[glad terrace]

can someone help me with b.) and c.) please?

[tight pier]

For b) find the x coordinate by plugging in y=-1 and then plug both into the differentiated equation and for c) you need to show that dy/dx = 0 has no solutions.

Or rather the equation.

[glad terrace]

thank you man

I got it

but also just to learn

to prove there is a SP dy/dx = has a solution and to disprove it dy/dx can’t have a solution

[tight pier]

you mean dy/dx = 0

but yes

[glad terrace]

Okay tysm 😄

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[rocky trail]

Is this channel busy?

[tight pier]

Looks pretty unbusy

[remote heron]

no sir it is yours please ask away

[rocky trail]

Okay, thank you very much.

My question is about arithmetic, well I'm more looking for advice and suggestions.

I am an arithmetic student, and I want to know what advice or suggestions you can give me.

I study in this order:

1. Natural numbers

2. Ordinal numbers

3. Even/Odd numbers

4. Multiples and divisors

5. Prime and composite numbers

6. Fractions

7. Decimals

8. Integers

9. Rational numbers

Is this order okay, or is there anything else I should add?

[remote heron]

what are you looking for?

oh, about the order

IDK I think you could/should jump around as you are comfortable a lot of these topics are going to overlap and it really depends on what material about them you cover

[rocky trail]

I study arithmetic, basic mathematics :b

[remote heron]

its not really clear what youre lookin for I guess.

The collection of topics seems fine.

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@rocky trail Has your question been resolved?

[lone heart]

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[unborn surge]

how did they get 3/2(6b-2a)

because i got as far as 9b-3a

[tight pier]

You wanna write XZ in the form of λ(6b-2a) = λXY to show they are colinear. So they factored a 3/2.

3/2 • 6 = 9 and 3/2 • (-2) = -3

[unborn surge]

ohhhh

i get it now thx

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[unborn surge]

this is the answer

i got as far as 1/2(4a-b)

but i couldnt understand how to get AC

why is OC labelled 3/4 and 1/4

i thought it would be 3/7 and 4/7

[rancid scroll]

slight mistake

when ratio is 3:4 it means 3/7 not 3/4

so OA would be 3/7 and OC would be 4/7

[lone thicket]

This is not correct.

[rancid scroll]

why?

oh wait

i got it

[unborn surge]

im lost....

[rancid scroll]

mb bad just woke up brain not braining

[lone thicket]

If OA:AC = 3:4 , then it would be 3/7 and 4/7.

But we have OA:OC = 3:4 , where OC is the total length, so OA is 3/4 of OC.

[rancid scroll]

since there 2 ratios

make a ratio constant to keep it from getting messy

[rancid scroll]

what he did is correct

[unborn surge]

how do u know to split it into 4 tho 😭

[rancid scroll]

what he did was write the ratio as fractions

OA/OC=3/4

then he sent OC to other side

[unborn surge]

okay

got it

thx

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[rancid scroll]

yw

[livid notch]

Can someone help me? It's a surveying question. My teacher said the final sum is 88 hectares, but the maximum I could get was 77.

[lunar saddle]

!occupied

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[raw dove]

this is probably the simplest question on the planet but i'm just not sure what to do to start finding the equivalent exponential function (these are just optional homework questions)

[raw dove]

im most likely just overthinking it because I tend to do that and overlook the more simple ways of answering questions

[lethal belfry]

Is this graded

[raw dove]

nope!! just homework for practice

i can send the full page if needed??

[alpine sable]

Sure

[lethal belfry]

cool.
$y=5^x \implies log_5(y)=x$

yes?

[ocean seal]

wai

[vale wigeon]

\log btw

[raw dove]

sorry, just looked at the question again and remembered logarithms exist which is probably the whole point of the question, i'll try it on my own for a minute to see if i can do it

got it!! just used the base change formula, i was definitely just overthinking it for no reason

thanks for the help though!!🫶

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[radiant oasis]

can someone please check my work here?

[radiant oasis]

it says its wrong

@charred pond

no

shoot

sorry

meant to ping helpers

<@&286206848099549185>

[raw linden]

The major axis is x=3

[summer dirge]

I think the problem is that you've computed the foci as if the ellipse had a horizontal major axis

your ellipse has a vertical major axis though

so the foci are at (h, k +- c), not (h +- c, k)

after checking on Desmos, that does seem to be the error

[radiant oasis]

wait so its verticle not horizantle?

okay yeah got it then

ok thank you!!

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[summer dirge]

how does this proof look?

[summer dirge]

I'm feeling a little uneasy about it

[median oar]

i feel like something looks weird

||a + bi|| = a^2 + b^2?

oh that's a norm

shouldn't there be a sqrt

[pallid scarab]

and it's not good I think

[median oar]

also, lie groups dont have norms on them?

[pallid scarab]

I think they meant the modulus

[summer dirge]

oops, yes

but it won't change the fact that it's equal to 1

[summer dirge]

where do the issues... Lie?

[median oar]

i feel like i would've rather used 𝜑 to denote S^1

[summer dirge]

wdym

[median oar]

it looks like S^1 is 2 dimensional with the a and b

but really if you parametrise it in polar coords it's just 1 dimensional

[summer dirge]

I'm treating S^1 as a subset of C

[median oar]

what's U(1)?

[summer dirge]

unitary 1x1 matrices

[median oar]

are there only 2 of them?

oh it's zz* = 1

[summer dirge]

mhm!

[median oar]

yeah i mean you've convinced me that they're lie group isomorphic

although my proof for SO(2) and S^1 just sends (cos 𝜑, sin 𝜑) to (cos 𝜑, -sin 𝜑; sin 𝜑, cos 𝜑)

but that works too

[summer dirge]

all three?

[summer dirge]

I'm scared of angles

[median oar]

ya

[summer dirge]

I really ought to get more accustomed to that perspective though

[summer dirge]

okay, awesome

thank you so much for your time frosst

I really appreciate it

.solved

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[median oar]

this is great i must be learning something if i can understand your proofs

[summer dirge]

🤍

[median oar]

since when you were doing point set topology my head was dizzy

[summer dirge]

it's been a while since I did any raw pst

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[halcyon ore]

idk how to do this

[halcyon ore]

i think maybe i need to use the log rules

oh wait yea

i can do log(m-2/m)

log(m-2/m) = 2log(1/n)

okay stuck here

[solemn torrent]

wait cant you do now "(m-2/m) = log(1/n)"?

[mortal trellis]

brackets please

[solemn torrent]

like remove the log on the left and remove 1 log on the right

[mortal trellis]

m-2/m is not the same as (m-2)/m

[halcyon ore]

hi

[mortal trellis]

no

[halcyon ore]

sorry guys

i didnt notice

[solemn torrent]

ok im out dont wanna confuse anyone. 😭

[mortal trellis]

use another log law on the other side

[halcyon ore]

oh yeah hold up i know one

its the exponent one

log(m-2/m) = log((1/n)^2)

is this what you meant?

[mortal trellis]

yes

and now you can remove the logs

[halcyon ore]

now in this situation am i able to 'cancel out' the logs

ohhh

m-2/m = (1/n)^2

i need to write m in terms of n

hmm

is multiplying everything by m right or nah

[mortal trellis]

thats good

[halcyon ore]

m-2 = m((1/n)^2)

what can i do about the fraction and the exponent

hmm

[mortal trellis]

this is the same type of equation as m-2=17m would be

[halcyon ore]

huhh

ok im confused now

wdym?

[mortal trellis]

1/n^2 is just some term, might aswell be a 17

doesnt change anything for m

[halcyon ore]

ohh ok

then hmm

i want to subtract m from both sides

2 = m((1/n)^2) - m

oh wait

that might not be the right move

[mortal trellis]

it is

[mortal trellis]

its exactly the move you would do here

[halcyon ore]

ohhh yes

i see what you were trying to say now

now the issue is

if i divide (1/n)^2

then i will have m-m

which isnt good

[mortal trellis]

you have to factor out the m first

2=17m-m
2=(17-1)m

[halcyon ore]

2 = (((1/n)^2)-1)m

is this what you mean?

[mortal trellis]

yes

[halcyon ore]

now can i divide both sides by m?

or first do i subtract the 2

[mortal trellis]

here you would divide both sides by 17-1

[halcyon ore]

ohhh

so then

2/(((1/n)^2)-1) = m

[mortal trellis]

yes

[halcyon ore]

now i need to simplify the left side

hmm

oh wait

1/n^2

that what it can just become

the (1/n)^2

[mortal trellis]

yes

[halcyon ore]

2/(1/n^2-1)

is this still not fully simplified?

if its not now i need help

i cant think of ANYTHING else

[mortal trellis]

you can multiply top and bottom by n^2

or you can combine the fractions in the bottom and then simplify the double fraction you get

[halcyon ore]

i wanna do the multiply thing

it makes more sense

[halcyon ore]

when you say this do you mean like

it becomes

2n^2/1-1

or did i mess up

cuz nothing can be divided by 0

💔

[mortal trellis]

you messed up

[halcyon ore]

yea i understand

[mortal trellis]

you have to multiply the entire bottom by n^2

[halcyon ore]

ohhhh

[mortal trellis]

(1/n^2-1)*n^2

=1-n^2

[halcyon ore]

ahhhh okay

so rthen it becomes

2/(1-n^2)

[mortal trellis]

2n^2/(1-n^2)

[halcyon ore]

we multiply the top of the top fraction with n^2 too

i see

[mortal trellis]

yes

we multiply the entire fraction by n^2/n^2

aka we multiply it by 1

so we dont change its value

[halcyon ore]

ohh ok

so THIS is the fully ismplified version

and i just put an = m

[mortal trellis]

yes

[halcyon ore]

thanks a bunch

[mortal trellis]

yw

[halcyon ore]

oh wow

this was an excellence question

that means it had like a lot of points

tysm

(im not cheating this is a past paper i just mean to say im glad i understand it now)

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[wraith jolt]

Can someone help me to solve this?

a balloon is rising uniformly with a velocity of 10 fps, when a stone is dropped from it. The stone then reaches the ground in 3 sec, Find the height of the balloon when the stone hits the ground?

I’m getiing 114 ft

[vale wigeon]

fps?

[wraith jolt]

h = (-10 fps)(3s) + 1/2 (32.2 fps^2) (3s)^2

feet per second

[vale wigeon]

why not write ft/s

anyway where does this equation come from

it looks a bit weird

[viscid sage]

I feel like there should be more context

[vale wigeon]

like you need to find the initial height of the rock's freefall

[lavish cave]

it's a mechanics problem

[viscid sage]

O

[lavish cave]

I think you should be able to assume g = 10

[vale wigeon]

but you also need to account for the balloon rising while the rock is falling

[lavish cave]

gravitational acceleration = -10 m/s^2

[viscid sage]

Nvm then imma head out

[vale wigeon]

g = 32.2 ft/s^2 because burger units

[lavish cave]

oh SHIT right

yeah

any bets if that was the error they did

nvm

[wraith jolt]

this came from the kinematic equation

[vale wigeon]

can you write down the equation in full before you plugged any numbers into it

[undone ledge]

baloon go up , gravity pull baloon down

vector issue

you forgot to consider the opposite direction

you did not

i take it back

[wraith jolt]

aight lol

[vale wigeon]

ok right and s is displacement

[wraith jolt]

yes

[vale wigeon]

ok so this is correct for the rock's starting height.

but again,

the balloon still rises after letting go of the rock.

[wraith jolt]

right

[vale wigeon]

for 3 seconds, it keeps rising at 10 ft/s past that starting height.

[wraith jolt]

so 30 ft

[vale wigeon]

yes, add another 30 feet to it

[wraith jolt]

actually it’s 114.9 so it would be 145 ft if I add the 30 feet

Thanks!

.close

[lone heart]

Channel closed

Closed by @wraith jolt

Use .reopen if this was a mistake.

[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
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[lone plaza]

I need help caculate the lenght of this graph.

[mental finch]

Hej!

Har du tittat i sida 267?

[lone plaza]

Aa förstår formeln men fastnade sen på BC längden. Jag kan visa

Man ska bara använda L formeln. Men får inte till BC delen😭

https://tenor.com/view/sad-cat-kitty-depressed-guangdang-sad-meow-gif-2548323962849249220

[mental finch]

Så om jag förstår rätt så har du lyckats hitta vad f’(x) ska vara för AB och CD för att sedan använda det med L formeln?

[lone plaza]

Har ej gjort CD delen än eftersom jag fastnade på BC.

Men har försökt på flera sätt men tror att de är vinkeln som man ska räkna ut som förstör det. Har försökt att ex tänka Anta B(0.10) och C(3,10) och får begynnelsevillkor g(0)=10 g(3)=10 och xsym blir 1.5 och då g’(1.5)=0. Och anta g(x)=ax^2+bx+c

[mental finch]

Mhm

Hur gjorde du för AB? Den skulle hjälpa!

Det är så du kan få reda på vad BC måste vara

Ty, vi vet att att dem (parablerna) delar samma lutning vid punkten B

@lone plaza

[lone plaza]

Okej, hur ska man tänka då?

Kan ta fram hur jag gjorde på AB brb

[mental finch]

Kalla parabeln för AB, för f(x)

Dvs. y = f(x) beskriver parabeln AB

[mental finch]

Grymt!

[lone plaza]

Det är BC jag ej fattar😭

[mental finch]

Kalle nu parabeln för BC, för g(x)

[lone plaza]

Är du min matte lärare irl🤨

[mental finch]

Vad vet vi händer vid punkten B för bägge f(x) och g(x)?

[mental finch]

?

[lone plaza]

Skämt 😭

[mental finch]

Okej :3 lol

[lone plaza]

I’m completely lost😭 har problemet med vinkeln liksom.

[mental finch]

Mhm

Dem delar lutning där

Det ger dig ett extra villkor för g(x)

Du vet ni vad lutningen är i B

g’(x) är ju på formen ax + b

Du vet också nu att f’(8) = g’(8)

[lone plaza]

Visa mig🥹

Hur man löser plz

[mental finch]

Jag skriver ner det, ge mig några minuter!

[lone plaza]

https://tenor.com/view/das-war-ein-befehl-das-war-ein-befehl-cat-cat-austrian-painter-cat-painter-meme-german-cat-meme-gif-6662695980760100010

[mental finch]

Vill dock inte ge ut svaren direkt; så fundera gärna eller fråga, men jag kan ge en skiss på hur vi ska tänka

[lone plaza]

Är du min matte lärare reveal urselfffff😭👆

Okej ska försöka😔

Men kan du iallafall säger om de är rätt att man ska förskjuta grafen så punkten B blir (0,10)

Och C(3,10)

[mental finch]

Det är rätt!

Men tycker att du kan ha kvar så som det var innan

Dvs. Att g(8) =10

Och g(11) =10

[lone plaza]

Men då kan jag väll inte använda g(0)=10? För att få ut C. Tänkte så eftersom lättare ett få ut c variabeln?

[mental finch]

c konstanten kommer ju faktiskt inte bidra till något då vi endast behöver g’(x); men jag ifall det känns lättare gör så

Det kommer att bli samma ändå

Så det har du rätt i att göra

[lone plaza]

Men måste man inte veta C för att kunna räkna ut de andra variablerna. Förstår att vi deriverar försvinner c. Men om vi ska använda beygennelsevillkoren liksom.

[mental finch]

Tycker att din lösning blev bättre, kanske inte behöver skissa min lösningside då det blir onödigt komplicerat

[mental finch]

Mhm, jag tror det kan gå att undvika

Men gör så som du tänkte!

[lone plaza]

Men vi behöver dock ett begynnelse villkor för g’(x)

[mental finch]

Mhm, vi kan få reda på det mha av f’(x)

[lone plaza]

Jag tänkte g’(1,5)=0 eller om vi ej har förskjutitgrafen blir de g’(9,5)=0

[mental finch]

Det låter bra det

[lone plaza]

Så vilka begynnelsevillkor ska man använde för för den förskjutna g(0)=10, g(3)=10, g’(1,5)=0 eller icke förskjutna g(8)=10, g(11)=10, g’(9,5)=0

[mental finch]

För bägge så räcker dem på det sättet du tänkte lösa det!

Eftersom du har tre ekvationer och tre obekanta

[lone plaza]

Okej vilken ☠️

[mental finch]

Säg att du kör med den förskjutna

Veckla ut vad allt betyder, du har ju villkoren som du skrev

Det är ekvationer

Som du skrev är ju g(x) på formen ax^2 + bx + c, mhm?

[lone plaza]

Fick fram från derivata 3a=-b och vanliga med g(x) fick jag 9a+3b=0

Använt g(3) och g’(1.5)

[mental finch]

Mhm

Använda nu den sista!

[lone plaza]

Har redan använt alla 3 villkor

[mental finch]

Okej, ser du hur man löser ut a och b från det du skrev ovan?

[lone plaza]

Det tar bara ut varandra

När ja försöker använde de två ekvationer

[mental finch]

Ajdå

[lone plaza]

Bro wanna watch me suffer 😭

[mental finch]

Inte meningen!

Jag är ledsen om det känns så

[lone plaza]

Första gången jag gör en sån här uppgift därför jag är lite efterbliven

[mental finch]

Mhm det är okej

[lone plaza]

I’m not serious😭😭

[mental finch]

Vet man inte att symmetri linjen är kopplad till koefficienterna av andra gradaren?

Typ att b/2 är precis symmetri linjen

Vänta jo!

Det är -b/2

Som ju ska vara symmetri linjen

Vet du om detta också?

[lone plaza]

Använde xsym för att lutningen är 0 där

[mental finch]

Mhm men då vet vi vad b måste vara!

Dvs. Att -b/2 = 1.5

Ser du varför?

[lone plaza]

Nop, men det kanske blir fel i beräkningen då vi har förskjutit. För riktiga xsym på uppgift bilden kommer jue vara på 9,5

[mental finch]

Det okej, vi vill ju egentligen bara beräkna längden av parabeln

Då spelar det ju inte roll vart vi förskjuter den!

[lone plaza]

Okej men hur ska ja tänka

[alpine sable]

What's this language 🥲?

[lone plaza]

Bro spying on us😔

[mental finch]

Låt mig skissa

[lone plaza]

In my most vulnerable moment aswell😔

[tight pier]

Aslan is a good protector, jag vet

[lone plaza]

U guys prob making fun of meh🥲

[tight pier]

What is the math problem? Aslan are you handling it?

Also hi Aslan, long time no see

[alpine sable]

Heyy I'm trying to help not spy 😑

[lone plaza]

I’m trying learn this. It’s my first time with this kind of problem.

[tight pier]

Arc length?

Wait I think Aslan is doing a sketch

I dont wanna take his spot

[lone plaza]

Aslan is a pookie bear

[tight pier]

and in his eyes there is big hope in you!

[alpine sable]

Is this chemistry in maths?🤔

[lone plaza]

Nuh just need calculate the lenght of the graph. And I’m stuck on calculating the lenght on BC part.

[alpine sable]

I meant it as a joke

[lone plaza]

https://tenor.com/view/catstare-gif-8960785107294365444

[tight pier]

looks like a trigonometric function

[alpine sable]

How so?

[tight pier]

Actually, it's more a polynomial, now looking at it

[lone plaza]

I’m trying my best to learn a new concept instead getting violated 😔

[tight pier]

So you would need some sort of interpolation with points A, B, C and D

Or are you given the curve?

[alpine sable]

If this could be translated, i wouldn't be so lost

[mental finch]

Två sätt att tänka

[alpine sable]

This is correct

But how do u know that bc is quadratic?

[mental finch]

Du vet ju att g’(0) = -g’(3) av symmetri

[lone plaza]

We assumed on first part AB

[alpine sable]

Oh then it's plain and simple

[mental finch]

Också uh strunta i det jag skrev senare senare om -b/2

[tight pier]

Oh, it's probably piecewise quadratics

Anyway, nice to see you Aslan!

[mental finch]

Hi!!

[lone plaza]

Okej, men jag hade gjort så här. Hur ska man förutsätta ja försökte lösa ut a och be med ekvationerna 9a+3b=0 och 3a=-b men de tar bara ut varandra.

[mental finch]

Mhm, använda att g’(0) måste vara lika med -g’(3)

Ser du varför?

Det borde ge ett extra krav

[lone plaza]

Varför just -

[alpine sable]

Uhh guys it's not so tough

8,10 and 11,10 are roots of the quadratic equations

Ax^2+bx+c=0

[mental finch]

Okej, strunta i det

Det verkar inte hjälpa

Det är nog bäst om vi går tillbaka till min gamla strategi

[alpine sable]

Uhh well I think u guys got it so see ya

[mental finch]

Är det okej om vi gör det?

[lone plaza]

Ok

[mental finch]

Ditt extra krav på g(x) är nämligen f(x)

Titta på denna bilden

Håller du med om vad som pågår vid B?

[lone plaza]

Aa

[mental finch]

Mhm grymt, du vet ju dessutom vad f(x) är!

Så g’(8) = f’(8) ger dig en ekvation

[lone plaza]

f’(8) ska ja beräkna de med värdena jag fick på A och B i börja uppgiften

[mental finch]

Dock är detta för icke förskutnja g

[mental finch]

Japp!

[lone plaza]

Det blir -4,824

[mental finch]

Okej bra, då borde g’(8) vara lika med det, eller då g’(0) om du gillar den förskjutna

[lone plaza]

Okej om vi kör icke förskjutna. g’(8)=-4,824 g(8)=10 g(11)=10

[mental finch]

Mhm

[lone plaza]

Min fråga dock blir de inte problem med C? Isåfall men förstår vad du menar med lutningen nu iallafall.

[mental finch]

Du behöver ju egentligen inget med C att göra med senare när du använder L formeln, då du endast vill veta g’(x)

[lone plaza]

Okej ska prova

[mental finch]

Så egentligen är g’(8) = -4.824 och g’(9.5) = 0 tillräckligt

[lone plaza]

Kan man inte förresten (säger ej att vi behöver använde de nu) men skulle man också kunna tänka att g’(11)=4,824

[mental finch]

Dvs g’(8) och detta?

[lone plaza]

Alltså att de är samma lutning i c som B fast positivt (undrar bara generellt

[mental finch]

Ja precis

[mental finch]

Försökte säga detta förut, här

Men det verkade inte hjälpa för att lösa problemet

Ger ingen ny information eller krav på g

Men ja, förlåt att det tog sin tid

Hoppas detta har hjälpt

[lone plaza]

Iallafall har använt begynnelsevillkoren för g’(x) och kommit fram till 19a+b=0 och 16a+b=-4,824 tror du jag kommer behöva använda resten av g(x) begynnelse villkoren

[mental finch]

Inte enligt mig! Då du endast kommer att använda g’(x) i beräkningen av L

Men g’(x) har ju bara två obekanta

[lone plaza]

Okej ska försökte beräkna nu. Skriver när ja är klar

Fick 8.1m

Det är väll rimligt😭👆

För AB 22.1M

m”

[mental finch]

Ja det verkar rimligt

[lone plaza]

Okej bara CD kvar om du inte orkar behöver du inte hjälpa med de också

[mental finch]

Det är okej, den verkar ju lik AB så jag tror du klarar det galant!

[lone plaza]

Okej men om du orkar skulle du kunna rita en bild sen så ja kan på inspiration av hur ja ska rita tydligt sen när ja ska visa min lösningen för läraren. Ska börja lösa uppgiften nu dock.

Få”