help-42

[narrow hamlet]

but what did the answers work do, cuz i dont get it

[velvet osprey]

ok like

graphically

you're asked to prove that the graphs of y=f(x) and y=x^3 intersect somewhere between x=0 and x=2.

do you understand this

[narrow hamlet]

ohhh

yea

[velvet osprey]

they instead went for proving that y=f(x)-x^3 has an x-intercept somewhere between 0 and 2

[narrow hamlet]

and beacuase h(0) and h(2) from 1 to -4 crosses the x intercept

theres a value? that exists for c

its so funky and hard to visualise

[elder pawn]

then just follow the logic

[narrow hamlet]

im trying

[elder pawn]

by IVT h(c)=0 for some c

yes?

[narrow hamlet]

why are we defining h(x) in the first place im confused about, as in what the function describes

[elder pawn]

so we can use IVT on h

[narrow hamlet]

we have a fucntion f(x) that has a y value for each value between 0 and 2, and subtracting what f(c) is

but i dont see what the end result --> h(x) like shows, what does '1' or '-4' indicate

[elder pawn]

the existence of a root, thats it

[narrow hamlet]

huh

so why what does it show that we get 1 or -4

[elder pawn]

the important thing is this

[narrow hamlet]

like esdies the reason thats what the numbers give ofc

[elder pawn]

the claim is f and x^3 intersect

do u agree its equivalent to f-x^3 having a root

[narrow hamlet]

i dont get why

so not rly

[elder pawn]

f(x)=x^3 <=> f(x)-x^3=0

a solution existing for one eqn is equivalent to a solution existing for the other

[narrow hamlet]

so this 'and beacuase h(0) and h(2) from 1 to -4 crosses the x intercept
theres a value? that exists for c'

or rephrased

because f(c) crosses f(x) there must be a value c

[elder pawn]

more rigorously, 0 is between h(0) and h(2)

so by IVT h(c)=0 for some c in (0,2)

all g now?

[narrow hamlet]

ya ty

.close

[calm coral]

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[elder pawn]

np

[calm coral]

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[velvet osprey]

👀

[winter elbow]

👀

[deep violet]

Any quadratic polynomial can be written as k(x - alpha)(x - beta).
So k(x^2 - x(a+b) + ab) = (x^2 - x/4 - 1)k

[winter elbow]

Are these the roots of the quadratic function?

[deep violet]

No, a + b = 1/4 and ab = -1

[edgy leaf]

sums and products

[winter elbow]

mb

[deep violet]

The answer is THIS.

[kind quarry]

This works

That's one possible answer

[velvet osprey]

theirs is yours with k=4

you did it correctly, and in fact did a bit more than was asked by characterizing all quadratics with the given property

[deep violet]

Wot the heilll

[velvet osprey]

what's wrong?

[deep violet]

This.

[kind quarry]

Yeah no

[velvet osprey]

also might not want to spell it "heil"

[deep violet]

Okay, anyway the question asked to provide a quadratic polynomial, not the quadratic polynomial.

[velvet osprey]

that can have quite unsavory (if unintended) connotations

[velvet osprey]

yes and you are done

unless you have something to follow up with

[deep violet]

Okay.

Thanks y'all.

.close

[calm coral]

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[nocturne heron]

i want to show for $\Omega$, a quadrable domain, that $A(\overline{\Omega})=A(\mathring{\Omega})=A(\Omega)$

[potent lotus]

tm

[nocturne heron]

where A is the area

how could i do that?

[mortal orbit]

quadrable?

[nocturne heron]

quarrable en français

[mortal orbit]

ok bon du coup

[nocturne heron]

jcrois j’ai compris

genre att

[mortal orbit]

domain = ouvert?

ou juste "ensemble"?

[nocturne heron]

y’a juste marqué inclu dans R^2 lol

[mortal orbit]

et borné non?

au moins

[nocturne heron]

y’a pas marqué mais ça doit être sous entendu

bref, le truc A^- ça correspond à l’intérieur de omega et A^+ c’est l’adhérence de omega?

si oui jvois pk ducoup

[mortal orbit]

bon bah tu prends Omega quarrable

Alors on va prouver que son intérieur et sa fermeture sont quarrables

et de même aire

[nocturne heron]

ça doit pas être si difficile ça j’pense

[mortal orbit]

Il suffira de dire que

pour commencer, A-(fermeture) >= A-(Omega)

[nocturne heron]

hm oui

[mortal orbit]

et A+(Omega) >= A+(intérieur)

donc à méditer pourquoi

et ensuite

[nocturne heron]

ui jvois

[mortal orbit]

ensuite faudra prouver les autres inégalités

par exemple

attends excuse moi on a plein d'inégalités en plus si je ne me trompe

[nocturne heron]

😭

att au pire j’ai pensé à un truc

[mortal orbit]

A-(fermeture) >= A-(Omega) >= A-(ouverture)

A+(fermeture) >= A+(Omega) >= A+(ouverture)

je pense que toutes ces inégalités sont immédiates

ensuite:

[nocturne heron]

hm ui

[mortal orbit]

on va tenter de montrer que A-(fermeture) < A-(ouverture) + epsilon pour tout epsilon

pour ce faire, si tu considères un pavage contenu dans la fermeture

alors tu érodes un peu chaque pavé

et ça te fait un pavage contenu dans l'ouverture

avec autant de précision que tu veux

[nocturne heron]

oui

mais att 2s 😭 y’a équivalence entre le fait que Omega est quarrable et sa frontière est d’aire nulle ?

si oui on peut le faire + simplement

[mortal orbit]

euh bonne chance pour prouver ça j'y ai pas touché

[nocturne heron]

si oui, $\overline{\Omega}=\Omega \cup \partial \Omega$

[potent lotus]

tm

[nocturne heron]

et on utilise les propriétés de l’aire

[mortal orbit]

et ensuite que l'aire se comporte bien pour l'union disjointe

[nocturne heron]

ui voila

[mortal orbit]

il y a aussi des chances que la frontière ne soit pas quarrable

donc pour l'instant je m'avance pas trop

[nocturne heron]

si omega est quarrable ?

mais si omega est dans R^2, sa frontière sa sera une courbe

[mortal orbit]

mec j'ai fait tout un papier de recherche sur les fractales donc à un moment "la frontière sera une courbe" c'est pas vrai xdd

[nocturne heron]

😥😥😥😥😥

[mortal orbit]

t'as des frontières quarrables qui sont fractales

[nocturne heron]

ah ouais relou aussi

j’me disait l’aire d’une courbe c’est 0 et tout ça aurait peut être fait un truc 😥

[mortal orbit]

reste là dedans pour l'instant

[nocturne heron]

oki

[mortal orbit]

ça c'est peut être vrai sachant que A(frontière) = A(ferm) - A(intérieur)

mais pour le prouver avant de prouver ce qu'on fait là maintenant je pense que c'est pas une bonne idée

En tout cas Omega quarrable => A(frontière) = 0 (peut-être?)

mais la réciproque mmmh

[nocturne heron]

dans mon cours y’a marqué si la frontière est une courbe continue , de classe C^1 par morceaux alors omega est quarrable, la frontière aussi et son aire vaut 0

[mortal orbit]

ouais bah C1 par morceaux c'est important comme hypothèse 😭

[nocturne heron]

ah ui

bon bah ça tombe à l’o 😔

[mortal orbit]

allez tente de faire ça c'est pas si mal

[nocturne heron]

att faut que je m’y remette

[nocturne heron]

faut faire ça ?

[mortal orbit]

ouais je pense

c'est un truc classique

pour montrer que A = B avec A <= B évident

la technique classico c'est montrer A >= B - epsilon pour tout epsilon > 0

donc là la technique

comme A- c'est un sup

je prends un pavage de la fermeture

je peux toujours trouver un pavage de l'ouverture "aussi grand que le pavage" à précision epsilon

donc Aire(ce pavage de fermeture) < Aire(le pavage de l'ouverture trouvé) + epsilon

en passant au sup

Aire(ce pavage de fermeture) < sup(des pavages de l'ouverture)+epsilon = A-(ouverture)+epsilon

on passe au sup

A-(fermeture) <= A-(ouverture)+epsilon

[nocturne heron]

hm mouais mais c’est à dire genre mathématiquement ?😭

[mortal orbit]

bah ça mathématiquement

[nocturne heron]

mais enft jcomprends pas meme intuitivement

l’intérieur c’est plus petit que l’adhérence

[mortal orbit]

oui

donc si tu prends un pavage de l'adhérence

pour créer un pavage de l'intérieur

bah va falloir raccourcir les pavés

mais pas trop

pour respecter la précision

[nocturne heron]

ui mais si l’adhérence c’est + grand, pk les deux ont la même aire 😭

[mortal orbit]

parce que j'ai besoin de rétrécir mes pavés, mais tant que je les rétrécis, l'aire que je perds elle peut être aussi petite que je souhaite

[nocturne heron]

mais

[mortal orbit]

je vais te faire un dessin si tu l'as toujours pas

[nocturne heron]

att jvais dessiner sur mes notes mdr

A barre est plus grand que l’intérieur

[mortal orbit]

quand tu dessines comme ça oui

mais faut comprendre que la différence entre les deux c'est la frontière

qui est super riquiqui en aire

[nocturne heron]

bah elle a une aire nulle (si c’est une courbe c1)

[mortal orbit]

Si je prends un pavage de mon adhérence

bon bah chaque pavé c'est un rectangle fermé [a,b] x [c,d]

qui est complètement dans mon adhérence

mais qu'est-ce qu'on peut dire du rectangle ouvert ]a,b[ x ]c,d[ dans ce cas?

il est contenu dans quel ensemble?

[nocturne heron]

dans l’intérieur

fin c’est l’intérieur de l’adhérence ça

oula att

bah c’est la frontière ?

[mortal orbit]

?

[nocturne heron]

bah

l’intérieur de [a,b] c’est ]a,b[

[mortal orbit]

oui

[nocturne heron]

ça c’est l’intérieur de l’adhérence alors

ah mais c’est l’intérieur tout court mdr

[mortal orbit]

alors...

bon

je réfléchis de mon côté moi aussi mais c'est pas hyper évident non plus

[nocturne heron]

si A est fermé

[calm coral]

@nocturne heron Has your question been resolved?

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[remote mural]

why did we equate 4 with 2ab

[teal drift]

Because two complex numbers when both the real and imaginary parts of them are equal

[remote mural]

2ab is imaginary?

[teal drift]

In other words, the reason is the same for why you equated 3 with a² - b²

[teal drift]

The imaginary part of the right hand side, yep

[stoic oyster]

cause they typo'd

[teal drift]

Oh yeah I didn't even notice it

[remote mural]

how is 2ab imaginary tho what

[teal drift]

@remote mural

[remote mural]

what does this mean

[teal drift]

It should have been 2abi

[stoic oyster]

3rd line should be 3+4i = a^2 + 2abi - b^2

[remote mural]

ohh

makes sense

thanks guys

.close

[calm coral]

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[teal drift]

You're welcome 🤗

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[fickle flume]

hi it is me again

[fickle flume]

if i need to find all the roots, both purely real ones and complex ones

how do i go about doing that

for a polynomial of any degree

[velvet osprey]

if the degree is ≥5 and you need the exact roots and no rational ones can be guessed, then you are cooked.

[fickle flume]

okay in this case they seem to have spared me from being cooked then, it is of degree 4

[velvet osprey]

can you show yours

[fickle flume]

yes, and i want to say i did try to think about it before asking

i thought maybe i could substitue then use the quadratic equation

then divide it into two new polynomials

and use the quadratic equation again

z^4 + 2z^2 + 4

[velvet osprey]

quadratic formula not quadratic equation btw

[fickle flume]

sry that's what i meant

[velvet osprey]

this is your polynomial?

[fickle flume]

yes

[velvet osprey]

it is biquadratic

so indeed factoring t^2 + 2t + 4 and then substituting t = z^2 works.

[fickle flume]

i was gonna do z^2 = t

[velvet osprey]

yes you are on the right track

did you manage to factorize t^2 + 2t + 4

[fickle flume]

2t

but no not yet

let me try to do that

[velvet osprey]

yep typo sorry

[fickle flume]

it has to multiply to the coefficient

and add to the constant right

for the fast way to do the quadratic formula

[velvet osprey]

the fastest way here is completing the square actually.

[fickle flume]

oshit

[velvet osprey]

also if you're going to go for the quad formula then there is no question of "add to this" or "multiply to that" at all.

[fickle flume]

i forgot about that

that's when i add and subtract a number so it becomes a perfect square right?

[velvet osprey]

yes

[fickle flume]

so -3?

like i add 3 and -3

sry i mean like this: (t^2 +2t + 1) +3

[velvet osprey]

yes

[fickle flume]

i forgot how to continue with completing squares

sorry it has been a few months

i get (t+1)^2 + 3 = 0

now i treat this as a difference of squares??

(t+1)^2 - (sqrt(-3)^2) = 0

(t+1 - sqrt(-3))(t+1 + sqrt(-3)) = 0

[velvet osprey]

better to write those as sqrt(3)i not sqrt(-3)

so what you've got here now is that your original quartic is partially factorized as $$(z^2 + 1 - \sqrt{3} i)(z^2 - 1 - \sqrt{3} i)$$

[potent lotus]

Ann

[velvet osprey]

now you need to factor each of those brackets again

by finding the square roots of $-1 \pm \sqrt{3} i$ by whatever means

[potent lotus]

Ann

[fickle flume]

okay wow

there are 2 more roots

and you can write i instead of sqrt(-3) because sqrt(-1) * sqrt(3) = i * sqrt(3)

btw im still here im trying to factor them right now im just very slow

[velvet osprey]

i have to go now sorry

[fickle flume]

no problem thanks for the help anyway, i didnt know i had to actually go down this road to find imaginary roots

but now i do

.close

[calm coral]

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[rigid lintel]

can someone pleease just do this problem for me

[bronze adder]

you can rewrite the left as the chain rule

[rigid lintel]

i did that but i got it wrong

[bronze adder]

what did you get?

[potent lotus]

shawarma

[bronze adder]

this?

[rigid lintel]

uh i got y=sinx/x^3

lo

[fathom solstice]

dont forget about the +C

[bronze adder]

ah yeah okay the problem is you
forgot the constant

[rigid lintel]

ohh

so just the +C?

[bronze adder]

yeah

[rigid lintel]

oh ok thank you i was so confused lol

[bronze adder]

$yx^3 = \sin x + C$

[potent lotus]

shawarma

[bronze adder]

then divide by x^3

[proud halo]

yeah we find c by using sin to the power of x

[bronze adder]

what

[rigid lintel]

do you guys have any reccomendations for calc 2 like study tips or anythiung

[proud halo]

because y and x can =sin

[bronze adder]

honestly it's mostly just practice

[fathom shuttle]

<@&268886789983436800> can you stop this guy please

[bronze adder]

especially with integrals and stuff, the more you do the better

because it's about recognizing patterns

and DONT FORGET THE C

[calm coral]

@rigid lintel Has your question been resolved?

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[spice venture]

what ass law is this bro

"rate of separation is directly proportional to their separation"

as if we didnt know

[valid linden]

I think what it means is that larger the universe is faster is the rate of expansion

[rigid kettle]

maybe you're misunderstanding something

[spice venture]

yes probably

[rigid kettle]

its not something obvious

[spice venture]

ohhh

okay now help me solve please

[rigid kettle]

lets say milky way and andromeda are x light years away from each other.

Then they are separating at a rate of kx from each other.

and this k is the same no matter what x is

[spice venture]

oh ok

[valid linden]

Idk the value of 1 mega light year that's why I can't solve the part (a) but i can help u with part b

[spice venture]

yeah

part a im fine with

[valid linden]

okay so the velocity (v) = Hr where r is the separation

Do yk that we can write velocity as dx/dt?

[spice venture]

yeah

oh

dr/dt = Hr
dr/r = Hdt

integrate with limits

okay thanks man

[valid linden]

yupp

:)

[spice venture]

.close

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[vapid elm]

I have no clue how to start

Btw

[ancient grotto]

Translate?

[vapid elm]

Yeah I was about to say that

I) If 2^3a =729 the result of 2^-a is equal to 1/3
II) the result of the operation (numbers above) is equal to (numbers above)
III) if x^2 = 25^12 ; y^6 = 25^12 : w^7 =25^63. The value of the expression (XtimesYtimesW)12 is equal to 25^168

[desert nymph]

and then I imagine it's "only 1 is false", "only 2 is correct", "only 1+2 are correct", "only 1+3 are correct", "all 3 are false"?

[vapid elm]

Oh yeah I forgot about that

[desert nymph]

so for I, you can use a power law to split it up right

[vapid elm]

a) 1 false
b) 2 true
c) 1 and 2 true
d) 1 and 3 true
e) 1, 2 and 3 false

[desert nymph]

$x^{ab} = x ???$

[potent lotus]

PrettyPrincessKitty FS

[vapid elm]

Erm

I’m in 9th grade

[desert nymph]

but you must know about how to deal with powers right?

[vapid elm]

Yes

Like 2^3 right

[desert nymph]

yeah - and if you have e.g. 2^3 * 2^5

[vapid elm]

2^8

[desert nymph]

exactly, you do know the power laws 🙂

[vapid elm]

Nice but I still have no clue how to do that

[desert nymph]

so let's take 2^3a; ideally we want to get rid of the 3 somehow

[vapid elm]

Isn’t a 6?

Cause wait no

[vapid elm]

Ok

How come tho

[desert nymph]

so we know already that $x^{a+b} = x^a * x^b$

[potent lotus]

PrettyPrincessKitty FS

[vapid elm]

Mhm

[desert nymph]

what's the equivalent one for $x^{ab}$?

[potent lotus]

PrettyPrincessKitty FS

[vapid elm]

X^a * X^b

[desert nymph]

no-o, that makes x^a+b

[vapid elm]

What

What’s vno about

[desert nymph]

we don't have something that looks like $2^{3+a}$. we have something like $2^{3 * a}$

[potent lotus]

PrettyPrincessKitty FS

[vapid elm]

Oh

(X^a)^b

[desert nymph]

yeah great. so if we can put the 3 on the outside, and then take cube roots of both sides, we get 2^a = stuff

[vapid elm]

Ohhh

2^a root of 3

But how does the 3 get to the root

[desert nymph]

$(2^a)^3 = 729 => 2^a$ = cube root of 729

[potent lotus]

PrettyPrincessKitty FS

[desert nymph]

and of course $2^{-a}$ is just $1/(2^a)$

[vapid elm]

Oh ok cause we got rid of the 3

[potent lotus]

PrettyPrincessKitty FS

[vapid elm]

So where do u go from here

[desert nymph]

well you can work out the cube root of 729

[vapid elm]

Oh find the cube root

Yeah

Ok brb

It’s 9

So 1/2^9

[desert nymph]

yeah cool. if 2^a = 9, then 2^-a is 1/9

[vapid elm]

Oh

Oh yeah it’s the answer not the a

[desert nymph]

yeah - a is not gonna be a nice number but that's fine because we can leave it as a

[vapid elm]

lol

I have a project I need to do and I gotta learn how to do 10 questions and I’m gonna have to present how to do 1 of the 10 randomly chosen by the teacher and I also gotta have all 10 on a paper explained nicely so this is really gonna help

[desert nymph]

the other 2 are quite similar but obviously they use e.g. the x^(a+b) rule

[vapid elm]

These are way harder questions than I normally do

For the 2nd one can I do like (1,25 - 1,16) 10^-4-7

[desert nymph]

unfortunately not

[vapid elm]

so like 0,09 * 10-11

10^-11

[vapid elm]

😭

[desert nymph]

what you can do is take 10^-4 * stuff and make it into 10^-7 * other stuff

[vapid elm]

Huh

[desert nymph]

if you make the 10^-4 1000x smaller - to get 10^-7, and you make the other term 1000x bigger - you'll have the same number

but now it'll look like a * 10^-7 and now you can do your bracket stuff

[vapid elm]

0,00125*10^-7 - 1,16*times*10^-7

[desert nymph]

if you put \ in front of your * it'll not do the italics thing

like \*

[vapid elm]

Oh ok

[desert nymph]

or you can use ` backticks ` to get this

[vapid elm]

That’s cool

I didn’t know about the \

[vapid elm]

Yeah that I know

[desert nymph]

be careful - you've made both parts of the a * 10^-4 1000x smaller

so 0,00125*10^-7 is actually 1 million times smaller than 1,25 * 10^-4

[vapid elm]

But if I change 1 I have to change the other no?

Isn’t it scientific notation

Ohhh

I did it the wrong way

1250*/10^-7

[desert nymph]

$1.25 * 10^{-4} = 1.25 * (10^3 * 10^{-7}) = 1250 * 10^{-7}$

[potent lotus]

PrettyPrincessKitty FS

[vapid elm]

Yeah

[desert nymph]

and now you can do the subtraction because you can take a common factor of 10^-7 out

[vapid elm]

Mhm

I gotta write this on paper so gimme a few minutes

Ok I did it

1,24884*10^3

Is the answer for 2

So it’s also false

[vapid elm]

(X^2) *6 = (25^12)*6

@desert nymph

Does this work?

If ur busy lmk so I stop bothering you

For III)

[desert nymph]

sorry! I was away

[vapid elm]

Don’t need to say sorry Dw

[desert nymph]

note that if 1 and 2 are both false, then your only possible answers are "only 3 is true" OR "1, 2, and 3 are all false"

and only one of those is a valid answer 😉

[vapid elm]

Yeah ik but I have to be able to prove it cause imma be presenting it to my rescuer

Teacher*

Rescuer 😭

[desert nymph]

haha fair fair

so yeah for x^2 and y^6 and w^7, what we really want is to have them just like x = 25^a and y = 25^b

because then $xyw = 25^a25^b25^c = 25^{a+b+c}$

[potent lotus]

PrettyPrincessKitty FS

[vapid elm]

Uh

Yeah yeah

Mhm

[vapid elm]

But we need it to be ^12

[desert nymph]

yeah, but we just want to put the ^12 around the whole thing - which is fine

[vapid elm]

Yeah

[desert nymph]

first we need to get x y and w on their own

[vapid elm]

So just divide?

[desert nymph]

yup

[vapid elm]

Oh that’s easy

Lemme try to do on my own

25^224

So it’s false

[desert nymph]

I think I get ^204 but yeah either way it's false 🙂

[vapid elm]

Oh yeah I added wrong on my multiplication 😭💀

[desert nymph]

x=25^6, y=25^2, w=25^9, 6+2+9 = 17, 17*12 = 204

[vapid elm]

Yeah

[desert nymph]

but yeah otherwise spot on

[vapid elm]

I did that too

@desert nymph again, lmk if ur busy so i stop @ you

So like isn’t this just memorization

[desert nymph]

determine...in general...the something of 3^2009?

[sand raft]

unit place I think

@vapid elm do u see a pattern in the table for exponents of 3?

[desert nymph]

ahh, I see. yeah this is just asking you to be look at the table and see a pattern

[sand raft]

Yeh

Similarly for 2nd sub question

[vapid elm]

I see that it goes n*3 each time

Is that what I’m supposed to see?

I’m guessing not

Wait I see that if u add up the digits

[sand raft]

the question is asking u to find unit place yes?
See if u are able to observe something in unit place of the exponents

[vapid elm]

It’s a mutiple of 3

[vapid elm]

It’s asking me to determine 3^2009

[sand raft]

That’s the divisiblity test of 3 yes
Does it help with the question?

[sand raft]

What? It must be asking to determine the unit place of 3^2009

[vapid elm]

Erm

[ancient grotto]

It’s the units digit.

[sand raft]

What does “unidade” mean..???

[ancient grotto]

Unit.

[sand raft]

yeah

[desert nymph]

oh the last digit

[sand raft]

Yes

[vapid elm]

Oh yeah ur right

[sand raft]

Unit digit means the ones place or the last digit (unless there is decimal)

[sand raft]

Yeah

[vapid elm]

Portuguese isn’t my first language

[sand raft]

It’s ok

[ancient grotto]

(Roughly translated from Spanish lol)

[vapid elm]

I’ve lived in Canada most of my life but I went to Brazil like 4 years ago cause my grandparents are from here so we came back for some times

So I’m still learning

[sand raft]

too much exp w such questions lol
Easy to guess it can’t be asking the value of 3^2009

🙏🙏

I doubt even calci can work that out

[vapid elm]

lol

So I see the pattern at the end repeating

1397

[ancient grotto]

Yep.

[sand raft]

Yup

[ancient grotto]

So take 2009 mod 4.

[sand raft]

Yup

[vapid elm]

Mod?

Wdym by thst

[ancient grotto]

(Find the remainder when 2009 is divided by 4.)

[vapid elm]

What d9es that help with

[ancient grotto]

Okay, calculate the units digit of 3^5. Then 3^8.

[sand raft]

Basically it’s repeating every 4 times right
So u can place
1397 as positions
For power 1 - it’s 1
For power 2 - it’s 3
For power 3 - it’s 9
For power 4 - it’s 7
It’s a cycle and u gotta reach til 2009

[vapid elm]

The last digit?

[sand raft]

Yes

[vapid elm]

502 remainder is 1

[ancient grotto]

Therefore the position is?

[vapid elm]

So it repeats 502 times and the last repeat ends at 1st

[vapid elm]

1 or does it like go to 1 forward

Nvm I counted its 1

Is it

Oh no wait

[ancient grotto]

Forward.

[vapid elm]

The first one on the list is 0

So it’s 3

Cause the repetition starts at 3

Right

[ancient grotto]

Hm?

Yes.

[vapid elm]

Cause the first is when the power is 0

Idk if that counts for the repetition

[ancient grotto]

It worked, so just keep the method in mind.

[vapid elm]

Ok

0->1 1->2 2->3 3->4

Even though 0-4 goes through 5 numbers

It’s like 4 counts or whatever

01234

Kinda confuses me

[ancient grotto]

Eh.

You get used to it.

[vapid elm]

So if it starts at 0 u just add 1 to the number ur remainder got so it got 1 for me so I jump from 0 which is 1st in this and go to 1 which is 2nd in this

Would you say the sequence starts at 1 or 3

[vapid elm]

Oh

But if I got 1 remainder why do I have to jump to the 2nd number of the sequence

[ancient grotto]

Because you have 0, 4... corresponding to 3^0, 3^4...

[vapid elm]

So?

[ancient grotto]

If we have 3^1, what is our remainder?

Notice that 1 mod 4=1.

[vapid elm]

1, 3, 7, 9

So it would be 1 no?

Notice what?

[vapid elm]

Mod 4=1?

[ancient grotto]

Hm?

Here, let me show you.

[vapid elm]

Remainder is 4=1?

[ancient grotto]

Let’s take 3^3 instead.

[vapid elm]

Ok

[ancient grotto]

3^3=27, so the units digit is 7.

3 mod 4=3, since 3/4 has a remainder 3.

[vapid elm]

So wouldn’t the pattern start from 3, 9, 7, 1

Cause then it would make sense

[ancient grotto]

Now we have a sequence 1, 3, 9, 7.

[vapid elm]

Oh

[ancient grotto]

We start at 1, consider that position 0, then move 3 to the right of 1 to get our desired result.

[vapid elm]

So u do the power first then u see the result

[vapid elm]

Right

[vapid elm]

So starting at 1 we move 3 to the right ok

So if we get 0 remainder it’s 1

Cause 4=4

I see now

Thanks

How do I even write the explanation down on paper

I need to write how I got here cause imma present this in class

[ancient grotto]

Oh.

Just say that you noticed the pattern

[vapid elm]

I’ll tell you whatever I end up writing

We see that the sequence is 1, 3, 9, 7, so we divide 2009 by 4, the remainder is 1. That means it’s the position in the sequence where it ends up at. You have to move 1 digit to the right from where the sequence starts, so in our case it starts at 1: 1, 3, 9, 7. The answer is 3.

I put an arrow from 1 going to 3 to indicate the digit jumping

@ancient grotto

Sorry for the @ I just thought u might wanna see

[calm coral]

@vapid elm Has your question been resolved?

[calm coral]

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[calm coral]

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[wintry echo]

, rotate

[wintry echo]

It Is correct ?

[inner zealot]

,wolf integral cos^3(x)/sqrt(sin(x) dx

[potent lotus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by kookiemon

[frank jungle]

,w integral cos^3(x)/sqrt(sin(x)) dx

[potent lotus]

Results provided by WolframAlpha

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

Requested by dyxn06

[frank jungle]

okay fine be annoying

it looks alright to be

[cursive gulch]

yes

[wintry echo]

🙂

But was this the best method?

It was written to do it by substitution

[frank jungle]

which you did

[wintry echo]

SIN (X) = T

[calm coral]

@wintry echo Has your question been resolved?

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Send your question here to claim the channel.

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[fickle flume]

hi it is me again, i need more help with understanding absolute values

[fickle flume]

if i have |x-3| = 5

i have a number x-3 that is supposed to be 5 steps from 0 on the real number line

and there are two such numbers

because i can take steps in the negative direction or positive direction

and because of that

i can see that one x has to be 8, and the other has to be -2

but i dont get how i actually motivate that properly

how do I motivate that properly

[drifting seal]

didn’t you just do that

the distance from 0 is 5

[fickle flume]

i dont feel like my motivation is sufficient

[drifting seal]

|x-3| = 5 means x-3 = 5 or -(x-3) = 5

[fickle flume]

like i could write that in a bunch of words on paper

but there has to be a better way to mathematically motivate that

to get points on an exam

[drifting seal]

^

[fickle flume]

urhg i dont understand that

where does the - come from

[drifting seal]

definition of absolute value

|x| = x if x >= 0 and |x| = -x if x < 0

it’s defined piecewise

[fickle flume]

that definition hurts my brain and i dont understand why i find it so difficult to understand

im really having a hard time

[gusty stump]

There is a rule thats says:

|a| = b, means a = b or -a = b. You can multiply the second equation by (-1) and get a = -b

[drifting seal]

because if x >= 0 then just leave it, it’s already positive but if x < 0 you multiply by -1 to make it positive

[gusty stump]

Basically, if you want to take an absolute value of 5, you only write 5, but if you want absolute value of -7, you want to get 7, so you use the second rule which is -(-7), and that equals 7

[fickle flume]

okay some of these rules i didnt know about so

im happy to learn some new stuff

[fickle flume]

this seems slightly easier to understand

i shouldnt say understand

i dont understand this i just memorise it

[gusty stump]

I found that giving a concrete example is the way to go at times, an example that explains how the rule works

[gusty stump]

I don't recommend memorising things in math

[fickle flume]

i cant seem to do anything else

i dont think i understand anything within math at all

[gusty stump]

What level of math are you at?

[fickle flume]

truthfully im probably like a weak high school pre-cal student level, but im actually in uni and im asking questions about absolute values and complex numbers mostly because it is in my course for linear algebra

and i've been doing calculus as well where i got 66.7% of what i needed to pass on the test, but i didnt understand a thing i was doing

i just memorised

it is a very grim situation

i want to learn and i ask over and over but nothing seems to stick, i find a way that gets the right answer, memorise it as well as i can (i still forget though unless i keep practicing the steps), and then i move on

this is painful to read

|x| = x if x >= 0 and |x| = -x if x < 0

i can barely understand what this says

|a| = b, means a = b or a = -b is easy to memorise

so then maybe if my memory allows me to i will use this when appropriate

but i dont really understand it

[west shard]

you're just stripping the sign

[fickle flume]

i will probably use it inappropiately as well

[west shard]

so if you plot y = |x| on a graph

[fickle flume]

i do know what that graph looks like

[west shard]

note that the left hand side looks like y=-x

the right hand side looks like y=x

that's gives you it as a piecewise function

[fickle flume]

i just dont get how that turns into a bunch of other knowledge

[west shard]

now your question was |x-3|=5?

[fickle flume]

yes

[west shard]

how familiar are you with complex numbers?

[fickle flume]

only a little bit, i have been looking at them recently

[west shard]

ok

[fickle flume]

i know there are cartesian and polar forms for them

[west shard]

don't worry

imagine the number line

[fickle flume]

okay

[west shard]

draw a dot at x=3

then draw a circle radius 5

[fickle flume]

i will just do that on paper right now

[west shard]

with x=3 as the centre

[fickle flume]

super pretty circle

[west shard]

ok cool

[fickle flume]

ok i dont understand why we are drawing the circle

[west shard]

where does the circle cross the line

[fickle flume]

-2 and 8

x = -2 and x = 8

[west shard]

what were your solutions before?

[west shard]

^

[fickle flume]

but like does this count as motivating that answer

like i dont have to motivate the circle

[west shard]

are you trying to understand this for yourself or for your teacher?

[fickle flume]

for myself so i can start actually trying to understand math and not just memorise an algorithm to solve problems

[west shard]

ok

[fickle flume]

this is an example from an online pdf i found that is more complicated for me, i asked about help with it earlier but only got like half way

|x| = 2|x − 1| − |x + 1|

i was told that you can see the critical points straight up

like i dont even have to factorise or anything

[west shard]

remember that a circle is all the points a distance of 5 away from the centre

[fickle flume]

i dont understand that at all

[west shard]

ok let me think of a way to make this makes sense

[fickle flume]

also sorry for swapping away from the previous example

with the circle

[west shard]

in the meantime

the circle

[fickle flume]

are you trying to get me into thinking of the imaginary axis too?

[west shard]

no not at all

[fickle flume]

oh ok

[west shard]

although it might help in the future

[fickle flume]

i thought maybe that had something to do with compelx roots or something

like 3 + 5i or something

in this case

but maybe that's not a root since it is on a different axis

roots are on the x axis always right?

[west shard]

ok fine

in that case

do you know how to represent |x| = r?

on the complex plane?

[fickle flume]

no tbh i dont really know how to handle |x| still, but r is just the distance between origo and a complex number, like a point on a graph with an imaginary axis and a real axis?

like it is almost like a hypotenuse

[west shard]

(I'm also trying to figure out what you know and don't know)

[fickle flume]

that's okay with me thanks for asking

[west shard]

given a complex number z, |z| is the distance from the origin

[fickle flume]

i dont know anything i feel like i am genuinely so lost

|z| = r ?

right?

[west shard]

gives the set of points where it is a distance of r from the origin

[fickle flume]

idk what that means

sorry

i understand some of it maybe

[fickle flume]

i dont understand what yo umean by set of points

isnt it just a point?

[west shard]

no

okay let's go back to the 2d real world

can you plot y=2x

[fickle flume]

yeah i can do that

[west shard]

any point on this line satisfies y is twice of x

(you don't need to actually do it)

[fickle flume]

oh i did it anyway but yeah i can do that

i can do it for decimals to most of the time

if they are nice

like 0.25 or 0.5

maybe 0.1, 0.2

0.1 uses up a lot of the page

the decimals being the coefficient to x

[west shard]

ok that's fine

you understand how any point on this line satisfies y is x multiplied by m

(where m is your decimal)

[fickle flume]

sry this might be a language barrier thing but what do you mean by satisifies

[west shard]

it is true that

where are you from?

[fickle flume]

sweden

[west shard]

were you on CE server a few days ago?

[fickle flume]

CE server?

[west shard]

Chinese English

[fickle flume]

ooooh, that's interesting, i'm in one right now, but i haven't really chatted much in there at all i think

i checked and yeah i havent sent any messages there

but im in a chinese english server

[west shard]

no worries, I thought I chatted to a swede the other day

[fickle flume]

ok funny coincidence but it was not me

[fickle flume]

i dont understand this question

[west shard]

the graph y=mx

you've seen before?

[fickle flume]

i can graph y = mx

like given an m

[west shard]

I just want you to picture it

[fickle flume]

yes i can do that

[west shard]

the general shape

[fickle flume]

yep yep

linear

[west shard]

what happens if i increase m

[fickle flume]

and sometimes steep or less steep

big m steep

increasing steep

and low m steep, but decreasing

sorry

that is wrong

[west shard]

if m is above zero

[fickle flume]

low m

my bad

low m is almost compltely flat

non-negative low m

[west shard]

any point (x,y) where y=mx is on this line

otherwise it is not on this line

[fickle flume]

see this is like really basic stuff but im already slowing down trying to follow along with what you're explaining

it's not that i haven't seen it before i just have never understood it well

i agree there are an infintie number of points along the slope of y = mx

but i didnt know they could all be described by (x, y)

i dont really know what that part means

i understand a point has a y value and x value

[west shard]

Graph is y=x

is (1,1) on this line?

[fickle flume]

yeah

then (2, 2)

(3, 3)

for whole numbers

[west shard]

(3.141, 3.141)?

[fickle flume]

yeah i guess haha

that too

[west shard]

ok

[fickle flume]

and negative as well, like in the third quadrant

[west shard]

good

So all these points are on this line

[fickle flume]

yeah i agree with that

[west shard]

y=x is your rule for this line

[fickle flume]

it would be a different line if i changed the m value from 1 to something else

[west shard]

yes

[fickle flume]

is there a way to tie this back to |x| = 2|x − 1| − |x + 1|? sry if im being rude, i appreciate you taking the time to try to understand what i know, i just dont feel like know is the right word, i just memorised most of it

[fickle flume]

like when you asked me about this i have never really memorised this even if it is true

it is never really something i use when i solve problems

at least not actively

and frankly i dont think at all

something that would help me would be to hear what goes through your mind when you see a problem like this

|x| = 2|x − 1| − |x + 1|

for me i see absolute values and yeah i dont understand them very well yet

i understand now that they are a distance from to 0 on the number line

to some place else

on the number line in this case i think, but apparently (maybe) it can like crossover to the complex points as well

dont really get that very well

|x| = 2|x − 1| − |x + 1|

like more things pop up in my head but i cant use them because i dont understand them

there is a definition for |x|

that is somehow useful i think

but i dont understand how to apply it and i dont understand the definition

then apparently you can find critical points by just looking at |x| = 2|x − 1| − |x + 1|

like |1 − 1| = 0

so one critical point would be at x = 1

i dont understand how that becomes a critical point though

previously whenever i encountered critical points it has been with mostly fractions

like (x+1)/(x-1)

one critical point is -1, where it is defined

and another is 1, where it isn't

|x| = 2|x − 1| − |x + 1| this is a mess to me

apparently i can remove the absolute value sign by splitting something up into two cases

x = 2|x − 1| − |x + 1| or -x = 2|x − 1| − |x + 1|

i dont know if i did that right

but it doesnt feel like it helps me, because i still have two absolute value signs left

and from what i understand there is some kind of utility to leaving them there

i dont understand why it is useful to leave them there

can i just keep going like a mad man?

x + |x + 1| = 2|x − 1| or -x + |x + 1| = 2|x − 1|

x + |x + 1| = 2(x − 1) or -x + |x + 1| = -2(x − 1)

|x + 1| = 2(x − 1) - x or |x + 1| = -2(x − 1) + x

(x + 1) = 2(x − 1) - x or -(x + 1) = -2(x − 1) + x

0 = 2(x − 1) - x - (x + 1) or 0 = -2(x − 1) + x + (x + 1)

nah that doesnt make any sense

[west shard]

since this is linear

you can plot graphs

[fickle flume]

how do i plot this on paper

i dont even know where to begin

[west shard]

plug in at -1, 0 and 1

why?

because that's what your x's need to be to make the inside of one of the absolute values 0

[fickle flume]

i was told that, and it was supposed to make sense but i still dont understand why

like again, if im given a something like this (x^2 + 2x + 1)/(x-1)

[west shard]

can you agree that if you add linear terms, you get linear results?

[west shard]

that's a completely different matter

[fickle flume]

but why dont i have to factor first here

to determine the critical points

|x| = 2|x − 1| − |x + 1| i 100% agree that

if i put in x = 0 for |x|

that it returns a 0

and that for 2|x-1|, if i plug in 1, it returns a 0

and for -|x+1|, i plug in -1, it returns a 0

but that's individually

like if i plug in 0

for x

then i get 0 = 2 - 1

and that doesn't make any sense

if i plug in 1 i get

1 = 0 - 2

if i plug in -1 i get

-1 = -4 - 0

sry

[fickle flume]

this isnt right

i get 1 = 4 - 0

[west shard]

critical points are where the inside of |.| is zero

because that's where the edge is

[fickle flume]

but how can there be multiple different edges

[west shard]

because that's what your function is

[fickle flume]

|x| = 2|x − 1| − |x + 1|

but i dont get it like

if i plug in a critical point fo rx

for x*

the function doesnt return anything that makes sense

or

no i dont get it at all

is this even a function

[west shard]

$f(x) = 2|x-1| - |x+1| - |x| = 0$

[potent lotus]

Xetrov

[fickle flume]

i heard || is a function but

[west shard]

,w graph 2|x-1| - |x+1| - |x|

[potent lotus]

Results provided by WolframAlpha

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Requested by xettt

[fickle flume]

i dont understand how it looks like that at all

[west shard]

when everything is above 1, all of those absolute values take positive form

[fickle flume]

campus is closing in 15

i wish i understood more but im just so extremely lost

i dont understand this at all

|x| = 2|x − 1| − |x + 1|

[west shard]

maybe i can voice call at some point

text is annoying to explain things with

[fickle flume]

can i add you? i wont spam you a bunch

[fickle flume]

this is so weird to me

the middle part i assume is made out of the 2|x-1| and |x+1| bit

and the steeper half of the graph

is made out of the 2|x-1|

but i dont understand why that is or how it works

but yeah i really gotta get going unfortunately

campus used to be open past midnight but apparently homeless people and drug addicts would come here at nighttime

so they stopped keeping open during those times

ty for trying to help it was very much appreciated im gonna ask a lot of these questions to the student-lead tutors on monday-thursday next week

.close

[calm coral]

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Use .reopen if this was a mistake.

[calm coral]

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[hardy spindle]

!help

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[hardy spindle]

need help transforming 6 = a(x+1)^2 +8 into factorised form

which is f(x) = a(x-x1)(x-x2)

[bold bear]

What have you tried

[hardy spindle]

i did the (x+1)^2

which is

x^2 + 1

[bold bear]

That's the freshman's dream unfortunately

[sonic snow]

wait what 6-8 is -2 so ull have complex solutions

[bold bear]

$(x+1)^2=x^2+2x+1$

[potent lotus]

depression

[bold bear]

not necessarily bc a could be negative

[sonic snow]

oh nvm a can be -ve

yeah mb

[hardy spindle]

ok look

that form is

canonique

canonic

the one i just said

thats canonic

[hardy spindle]

f(x) = a(x-h)^2 + k

f(x) which is 6

a i dont know

x neither

h is -1

k is 8

[bold bear]

Sure

[hardy spindle]

but i dont think i have to know

x

[bold bear]

No x is a variable

[hardy spindle]

yeah but when u want to transform only

ok wait

so i gotta find x?

[bold bear]

If you're trying to factorise it then that gives you possible solutions for x

So sort of

[hardy spindle]

ok look

[bold bear]

So what have you tried to factorise it?

[hardy spindle]

the only info i have rn to find x thn

is my diagram

i mean graph

[sonic snow]

oh bruh show the grpah then

[hardy spindle]

nothing

[hardy spindle]

ight

[bold bear]

Okay

[hardy spindle]

ill show u guys

then youll know

[bold bear]

Okay

[sonic snow]

make sure to give any relevant diagrams to the Qs along with the Question next time

[bold bear]

So you've almost got it

[hardy spindle]

alr

[bold bear]

It looks like you've found that the equation is of the form $y = a(x+3)(x-1)$

[potent lotus]

depression

[hardy spindle]

yes exactly

[bold bear]

To solve for a you can just substitute in values for x and y

[hardy spindle]

what is x tho

[bold bear]

You substituted in a y value but not an x value

x is along the bottom

[hardy spindle]

but theres 2

[bold bear]

It's a function

y is a function of x

[hardy spindle]

ye but theres 2 x

[sonic snow]

the x1 an x2 are roots

the values of x for which the function is 0

[bold bear]

y is a function that takes any value for x in the reals

Choose your favorite x

And solve the equation $y(x_0) = y_0$