help-23

[uneven flax]

would the y" go like as in the drawing?

[fathom jewel]

x > 1 and x < -1

can you find a number x that is greater than 1 and smaller than -1

[uneven flax]

nope

[fathom jewel]

exactly

it's a contradiction

same here

[uneven flax]

i'll consult with the teacher after, i think the minus in the middle is not supposed to be there

but then again i think i would run into the same problem

since i would need a number greater than y and smaller than y

[fathom jewel]

that's not the issue, the issue is how the numbers are chosen

1 > y > 3 is just irrational in itself

as opposed 1 < y < 3 this would now make complete sense

[uneven flax]

In interval notation, this is:
[-7/3, -11/3)

[fathom jewel]

this also doesn't really make sense

[uneven flax]

i know

[fathom jewel]

it's like having (1,0] instead of [0,1)

[uneven flax]

but in theory that's what it would have to be expressed as, even though it's not valid

[fathom jewel]

wtf

[uneven flax]

im just as confused as you

[fathom jewel]

nah it's just wrong lmao

aint no way you can justify that

[uneven flax]

yeah, ill just write the reasoning behind why it doesnt make sense

thanks m8

.close

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[ember pelican]

Guys I need help for the shsat ELA section

[ember pelican]

Is there any study materials

Or a mentor?

I only have up to December 😭

[fleet condor]

this channel is for asking specific problems

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[clever hamlet]

How do I even start this trash?

to find the area of the orange fish

can i set up the integral

integrate (x+6)(x+4) from 1 to 5

so would that make the area of the orange fish 240.67?

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[bright pier]

can you help me solve this derivative? i’m not sure if the r is considered a constant or what

[austere cypress]

which are you taking the derivative

[bright pier]

i’m taking the derivate of the surface area of the cylinder with respect to the diameter of the cylinder.

[austere cypress]

if you change the radius you change the diameter so its not a constant

[bright pier]

but the radius is the radius of the sphere

help please @austere cypress 😔

[austere cypress]

its a cylinder inscribed in a sphere?

well, if you increase the sphere the cylinder will also increase

[bright pier]

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[subtle token]

Find the area of ​​the figure bounded by the given lines.

[subtle token]

how to understand which is the upper and which is the lower function

[hot hare]

yooo

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[dense wadi]

when a unit vector

[dense wadi]

or u here

isnt = to 1

what does it mean

[median vigil]

a unit vector has a magitude of 1 by definition

that formula will always produce a unit vector for any vector v, except if v = 0 in which case it will be undefined

[dense wadi]

here

i got 1/sqrt3

so if it always has a magnitude of 1 by def

how is it possible

or

am i wrong here

oh

[median vigil]

the magnitude of v is not sqrt(3)

[dense wadi]

if you add the 1/v * v together

[median vigil]

but the magnitude of that is 1, which is given by the pythagorean theorem, square root of the squares of the components

[dense wadi]

another example u = 1 if add them all together

[dense wadi]

here if i add them up its not 1, is what im questioning

so i dont determine magnitude

by adding them right

[median vigil]

the magnitude of a vector $\vb u = (u_1, u_2, u_3)$ is given by [ \norm{\vb u} = \sqrt{u_1^2 + u_2^2 + u_3^2} ]

[flat frigate]

cloud

[dense wadi]

ok that example

coincidentally

got 1 by adding them up so i got confused

ok ty

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[winged barn]

urgent help

[winged barn]

Solve (please)

I dont take this subject so I have no knowledge on how to solve this

There's been a debate

Stumped many people

(We dont know what the answer is)

[azure delta]

How can there be a debate? Can't you always tell which of two answers is better?

[winged barn]

No idea

Two answers seem logical

But they dont know what is what

B is a common answer

We just need to verify from a professional

[azure delta]

It's more than 2 hours.

If Edgar takes task 2 from Blake, then Blake can take task 1 from Carly (or task 3 from Dexter)

Task 2 to Edgar, then Task 3 to Blake, then task 1 to Dexter. Reduces it by 4

[winged barn]

So your answer is 4?

We have teachers and top students choosing B) as their answer

But 1 dux student chose 4

its either B or D

They both have logical reasonings

[azure delta]

I just explained why it's 4 though. What's the reasoning for 2?

[winged barn]

I'll ask

This is what they said
"Because the question simply says Edgar is replacing one activity
So one person
So the allocations shouldnt be changed"

And this is someones working

[azure delta]

"new allocation"

If someone isn't convinced that it's 4, that's just that person choosing to be wrong.

And there's nothing you can do about that.

[safe radish]

@winged barn Has your question been resolved?

[winged barn]

hm thanks though

We'll wait until the solutions gets published 👍

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[native tinsel]

nah cause that's crazy tho

[simple gazelle]

that wasn't math...

[native tinsel]

figured so

[simple gazelle]

that was just pornography

[native tinsel]

🙏🏻

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[native tinsel]

<@&286206848099549185>

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[native tinsel]

<@&286206848099549185>

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[pale star]

Have u tried drawing it out?

Do you need help with that

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[sacred crane]

The linear span of a vector space V is a subspace of V, namely the smallest one that contains x1 ... xn.

someone pls explain this to me why its the smallest subspace

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[fathom jewel]

Könntest du einen noch kleineren UVR vorschlagen, der die Vektoren x_1 bis x_n enthält?

[sacred crane]

nein

[fathom jewel]

Was waren die 3 Axiome für einen UVR?

[sacred crane]

meinst du diese skalarmultiplikation addition und o muss existieren ?

[fathom jewel]

ja genau

der Nullvektor mus enthalten sein
und die Abgeschlossenheit bzgl. Addition & Multiplikation

Jetzt nehmen wir mal an A = {x_1 bis x_n}

Das ist eine Teilmenge von V aber noch kein UVR

wir bräuchten den Nullvektor

und die Abgeschlossenheit bzgl. + und *

und das allein macht eben die lineare Hülle

[sacred crane]

was ist UVR

[fathom jewel]

Untervektorraum

abkürzung

die lineare Hülle enthält A, ist abgeschlossen bzgl + und * und enthält den Nullvektor

a_1x_1 + a_2x_2 + ... + a_nx_n

[sacred crane]

okay

[fathom jewel]

Jeder Untervektorraum von V

der die Vektoren A = {x_1 bis x_n} enthält

der hat doch wegen den Axiomen immer L(x_1 bis x_n) automatisch

weil du kannst ja die Vektoren {x_1 bis x_n} im UVR addieren und so alle Linearkombinationen kriegen

und weil also immer ein UVR der die Vektoren A enthält, somit auch L(x_1 bis x_n) enthält, macht das sozusagen L(x_1 bis x_n) den kleinsten UVR der die Vektoren {x_1 bis x_n} enthält

es ist der UVR der mindestens die Vektoren {x_1 bis x_n} enthält und die 3 Axiome erfüllt und nicht mehr

[sacred crane]

nichts verstanden😿

[fathom jewel]

Du hast eine Teilmenge von V von Vektoren x_1 bis x_n

ich habe diese Teilmenge A genannt

A = {x_1, ... ,x_n}

[sacred crane]

okay

[fathom jewel]

In einem Untervektorraum kann man Vektoren addieren und diese mit einem Skalar multiplizieren

Außerdem ist immer der Nullvektor enthalten

Jetzt kommts

Genau die Lineare Hülle erfüllt diese Anforderungen weil sie so definiert wurde

L(A) = alle Linearkombinationen der Vektoren x_1, ... ,x_n

Der Nullvektor ist drinnen, und die Abgeschlossenheit ist erfüllt

[sacred crane]

ja das verstehe ich

[fathom jewel]

Jeder andere UVR der auch A enthält muss die lineare Hülle L(A) auch enthalten

[sacred crane]

aber wie weisst du dass es der kleinste ist

[fathom jewel]

Weil jeder UVR der A enthält muss L(A) enthalten

[sacred crane]

was?

[fathom jewel]

Jeder UVR von V der {x_1 bis x_n} hat

[sacred crane]

wie beweist das, dass es der kleinste ist

[fathom jewel]

der hat doch automatisch auch die Linearkombinationen

weil du die ja addieren kannst

[fathom jewel]

Wahrscheinlich mit Widerspruch, man nimmt an es gibt einen kleineren

ich versuchs mal mit nem Bild

[sacred crane]

warum oder wie automatisch?

[fathom jewel]

Was kannst du in einem Untervektorraum machen?

[sacred crane]

addieren und skalarmultiplizieren?

[fathom jewel]

ja

siehst du?

[sacred crane]

ich sehe nicht

ich kann nicht sehen

ich verstehe jetzt

[fathom jewel]

nein

Die lineare Hülle enthält alle Vektoren von A und weils ein UVR ist auch alle Linearkombinationen und auch den 0 Vektor - nicht mehr

Jeder andere UVR der auch die Vektoren von A enthält muss automatisch L(A) enthalten, wegen der Addition und skalaren Multiplikation

[fathom jewel]

Angenommen du würdest einen kleinen UVR zeichnen in L(A)

dann würden ja Linearkombinationen fehlen

dann wäre die Addition und skalare Multiplikation nicht abgeschlossen wenn du ihn kleiner machst als die lineare Hülle

Ich hoffe es wird ersichtlicher

[sacred crane]

jetzt verstehe ich,

danke

[fathom jewel]

np kirsche

[sacred crane]

.solved

was habe ich falsch gemacht?

[fathom jewel]

du musst es neu eingeben ohne fehler

[sacred crane]

okay. aber ich habe jetzt andere Fragen

hast du lust?

[fathom jewel]

was für fragen

[sacred crane]

Austauschsatz von Steinitz

[fathom jewel]

ok

[sacred crane]

warum gilt es?

[fathom jewel]

es lohnt sich die beweise sich anzuschauen

[sacred crane]

ich habe es angeschaut

[fathom jewel]

und nichts gecheckt?

kannst du den Satz nochmal posten hier?

[sacred crane]

ich weiss nicht wie ich erklaren kann, aber es funktioniert nicht in meinem Kopf

w1, ... wm sind linear unabhangig bedeutet, dass sie basisvektoren sind oder?

[fathom jewel]

genau die bilden eine basis

[sacred crane]

wenn m 2 ware, können sie R2 erzeugen

[fathom jewel]

ja

ein

also nein

[sacred crane]

was

[fathom jewel]

was wenn V = R^3

dann wären das zwei 3-dimensionale Vektoren

[sacred crane]

was?

[fathom jewel]

(1,0,0) und (0,1,0)

[sacred crane]

aber sie konnen nur ein ebene aufspannen oder

[fathom jewel]

da steht nur dass w_1 bis w_m aus V linear unabhängig sind

[fathom jewel]

Ebene im Raum

Der satz würde für mich mehr Sinn machen wenn w_1 bis w_m eine Basis bilden würden

[sacred crane]

ja sie bilden eine Basis

aber was ist, wenn die lineare Hulle kleiner als die basis

[fathom jewel]

aber das setzt voraus das V m dimensionen hat was man nicht weiß

[sacred crane]

L((1, 0, 0), (2, 0, 0), (3, 0, 0))

[fathom jewel]

Ich finde die Formulierung macht mehr Sinn

[sacred crane]

was ist wenn die Lineare H[lle so wäre?

[sacred crane]

darf man es hier austauschen

[fathom jewel]

mit was

[fathom jewel]

mit dem?

[sacred crane]

ja wenn unsere lineare Hülle so wäre, durfen wir (1, 0 0), (0, 1, 0) austauschen

[fathom jewel]

nein

(0,1,0) kannst du nicht als LK darstellen von (1,0,0)

Die Idee ist dass du einen Vektor austauschen kann, der sich als LK der anderen darstellen lässt

[sacred crane]

die Lehrer hat es genutzt, um den Austauschsatz von Steinitz zu beweisen

[fathom jewel]

Angenommen den VR = R^2 und du hast die Basis B = {(1,0), (0,1)} und du möchtest (1,0) austauschen mit (2,1).

Dann muss du dafür sorgen, dass (2,1) als LK sich darstellen lässt mit den Vektoren von B, und genau dieses lambda von dem Vektor, den du austauschen möchtest, also hier (1,0), das darf nicht 0 sein.

(2,1) = 2 * (1,0) + 1 * (1,0)

Also bildet nach dem Austauschlemma die neue Menge B' = {(2,1), (0,1)} auch eine Basis von R^2

[sacred crane]

ja, aber es ist laut austauschlemma ?

[fathom jewel]

was

[sacred crane]

sind Austauschlemma und Austauschsatz von Steinitz nicht gleich?

oder sind sie gleich?

[fathom jewel]

jetzt wo du's erwähnst hmm

[sacred crane]

ich weiss nicht

[fathom jewel]

Was ich dazu sagen kann

Angenommen du hast L{ (1,0,0), (0,1,0), (0,0,1) } = R³
Jetzt haben wir die Vektoren (1,1,1) und (1,1,0) auch aus R³ und linear unabhängig.

Dann gilt L{ (1,0,0), (0,1,0), (0,0,1) } = L{ (1,1,1), (1,1,0), (1,0,0), (0,1,0), (0,0,1) } = R³

blöd gesagt, wenn die lineare Hülle den Vektorraum aufspannt

dann ist es egal welche anderen Vektoren du da hinzufügst, der wird immer noch den selben VR aufspannen

[sacred crane]

L{ (1,0,0), (0,1,0), (0,0,1) } = L{ (1,1,1), (1,1,0), (0,0,1) } ?

[fathom jewel]

So verstehe ich diesen Satz, aber den den ich kenne mit Steinitz hat mehr was mit Basen als mit der linearen Hülle zu tun

Was bedeutet bei euch das L?

[sacred crane]

aber was ist wenn L{ (1,0,0), (2, 0 ,0), (3, 0 ,0) } wäre

[fathom jewel]

was willst du damit anstellen?

[sacred crane]

Lineare Hülle

[fathom jewel]

ja dann ist dieses Lemma von euch trivial

[sacred crane]

wenn so wäre, ist es nicht unmöglich durch Vektoren (1,1,1) und (1,1,0) zu austauschen?

[fathom jewel]

du fügst einer linearen Hülle unnötig Vektoren hinzu

[fathom jewel]

Das kommt jetzt auf den VR an

L{ (1,0,0), (2, 0 ,0), (3, 0 ,0) } wäre im Raum die x-Achse

(1,1,1) und (1,1,0) liegen nicht auf der x-Achse somit kannst du es nicht anwenden

[fathom jewel]

w_1 ... w_m waren auch aus V

[fathom jewel]

ja genau

[sacred crane]

ach sooo

das habe ich nicht bemerkt

[fathom jewel]

schade

[sacred crane]

jetzt verstehe ich

[fathom jewel]

danke

[sacred crane]

du hast recht

[fathom jewel]

nein du

[sacred crane]

danke

was ist unterschied zwischen Dimension und Rang?

[fathom jewel]

Die Anzahl der Basisvektoren heißt Dimension des Vektorraums, zumindest so stehts bei mir... Bei Rang kann ich mir eher vorstellen, dass wenn man die Vektoren in eine Matrix packt und davon den Rang ausrechnet auch eine Aussage machen kann über die Dimension der Vektorraums

[sacred crane]

aber es klingt, als ob sie gleich wären

wenn ich es google, sehe ich nur Erklärungen zu Matrizen

[fathom jewel]

ja lol da steht es ja vor dir

wenn ihr das so definiert habt

[sacred crane]

sind sie gleich? ich glaube nicht

[fathom jewel]

https://de.wikipedia.org/wiki/Rang_(Lineare_Algebra)

[sacred crane]

darf mann lineare Hülle als ein Matrix schreiben?

[fathom jewel]

das macht keinen Sinn

[sacred crane]

tut mir leid

[fathom jewel]

Du kannst die Vektoren in eine Matrix packen und eben den Rang ausrechnen

[sacred crane]

Vektoren in der linearen Hülle?

[fathom jewel]

z.B L( (1,0) (2,0) ) dann die matrix ( (1, 2) (0, 0) ) und man kann sehen der rang der matrix is 1 also ist die dim(L) = 1

[sacred crane]

Ach so

Ich verstehe

[safe radish]

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[thin carbon]

do you need help cleo?

[forest lintel]

So for the harmonic series sum, starting from n = 1, if you sum the first term to the second one you get 3/2, and if you then sum that to the next one you get 11/6, if you keep doing this, the denominator will just be n!(n factorial), however the numerator becomes a bit trickier. Thinking about it algorithmically, you’re basically multiplying by “n” and then adding (n-1)! recursively to get your n’th term. I’ve seen that you can use Stirling operators of the first kind to do it, but I’d like to know if there is a notation for this kind of operation that I do not know about. Thank you in advance.

[forest lintel]

Yep lol

[thin carbon]

I am not equipped to help you with harmonic series

.close

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[azure delta]

some fractions won't introduce larger denominators

when you add 1/6, your denominator already has a 6 in it

But really, you're just looking for a closed formula for $\sum_{k=1}^n k!$?

[flat frigate]

Silkster

[safe radish]

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[forest lintel]

Is the question mark on purpose?

[azure delta]

I'm asking you if that's what you're looking for

[forest lintel]

Ahhhhh, not sure if that’s it

[cedar owl]

this is a fun one

[forest lintel]

What I’m trying to do is to find a closed form for the sum of “n” harmonic series terms, where im adding the first term to the second, and then adding the 3rd and so on, if I’m making the denominator the minimum common multiplier between the first and the second, 1 times 2, and then that with the third, 1 times 2 times 3, and keep doing that, the bottom term will be n!, my problem is finding a nice way to express the numerator of this sum

[azure delta]

I don't think it has a closed form

Also, you can see that the minimum common denominator will not be n!

When you get to 1/6, your denominator already has a 6 in it

[forest lintel]

That’s also true! But for standardization making it factorial looks nicer

It’s just so tantalizing that it has such a simple arithmetic sequence to it

Just multiply by n and then sum (n-1)!

I thought about expressing it as a sum using sigma notation inside a product using pi notation, and say that the 2 just swap to alter the expression each time but that seems a bit forced lol

[safe radish]

@forest lintel Has your question been resolved?

[safe radish]

@forest lintel Has your question been resolved?

[cedar owl]

i dont even know what the question is here lol

[forest lintel]

If you add the partial fractions of the harmonic series, and just multiply the denominators in order to get a common multiplier, I basically want a simplified form for the numerator given that I’m multiplying the denominators in order

And I don’t know if there’s a notation for that

If you start at 0, then multiply by 1, add (1-1)! you get 1, which is the numerator of the first term of the series

Then with that, multiply that 1 by 2 and add (2-1)! you get the numerator for adding the first two terms

So basically you can just multiply by n and add (n-1)! And you’ll get the next sum of terms

But that’s too algorithmic, I want to know if there’s a way to write that without giving instructions like my grandmother’s recipes

[short topaz]

doesn't exist unfortunately

ok well i looked online and apparently formulas do exist

but there's nothing that's like "the fractions is a/b where a is this nice function & b is this nice function"

[forest lintel]

Damn! Ok thanks!

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[tall pulsar]

How do I go about determining the bounds of my double integral? I tried to draw the domain that I'm integrating over and got that the bounds are 0 to 2 and 2y to 1 respectively when dA = dx dy w/ an integrand of sqrt(4-y^2). This was a really ugly integral and seemed off.

[fathom jewel]

can you show work?

[fathom jewel]

it should be from 0 to 2y

[tall pulsar]

oh

is the integrand correct initally correct

[fathom jewel]

yup

[tall pulsar]

alr thank you

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[fathom jewel]

https://www.desmos.com/3d/e25iuj6yqm

if it helps to visualize

also you could change order of integration to avoid that awful integrand

ah nvm i a stupid

the way you have it is perfect

lmao

[tall pulsar]

i was gonna say that its just a u-sub

[fathom jewel]

i mean yea both result the same anyway but yours is easier and that was my intent lol

[tall pulsar]

thank you tho

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[flat frigate]

Bean Man

[fathom yacht]

You only need to show they are opened (because going to the complement finishes the argument for closedness)

[flat frigate]

Bean Man

[dusk basin]

Like I get theres some point in C_2 arbitrarily close to some other point in C_1, but so what

[fathom yacht]

Assume C1 is not opened. Take a ball center in x in C1 and consider C1 union that ball

[dusk basin]

Take $C_1 \cup B_{\epsilon}(x)$

Is it connected?

[flat frigate]

Bean Man

Bean Man

[dusk basin]

Is this true??

Thats what I came up with

<@&286206848099549185>

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[tardy lark]

how did that happen?

[marsh walrus]

try factor by grouping

[upper mulch]

Factorize. $$r^3-r^2=r^2(r-1)$$ $$-r+1=-(r-1)$$

[flat frigate]

Good

[marsh walrus]

normally when you do this, you put the odd powers, and the even powers together

o

Good is smarter than me

[tardy lark]

it worked

[upper mulch]

[tardy lark]

this is very clever too

alright

thanks for your help

:)

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[mossy thunder]

please help

[lean otter]

Lol funny question

Any tries ?

The question is wrongly framed though

[mossy thunder]

well I did it, thanks anyways

Oh what?

[lean otter]

Unless the mention seat is perpendicular to the ground XD

[mossy thunder]

How

Ooopsss

[lean otter]

If he farts he should fly upward

[mossy thunder]

Seems accurate

[lean otter]

But the question says he's moving forward

[mossy thunder]

oh

[lean otter]

This is possible when the seat is perpendicular to the ground

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[void abyss]

Where did I go wrong?

[void abyss]

O ...

That's what I get for trying to do mental math

[void abyss]

Thanks!

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[tardy mango]

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[sour karma]

why are the second and third one not planes?

[sour karma]

is it because 2 has scalar multiples and 3 has a zero vector?

[jade flint]

What happens when you have a zero vector multiplying a free variable?

[sour karma]

the result is a zero vector?

[jade flint]

Yes

So

Any value the variable takes

The result will be?

[sour karma]

a zero vector?

[jade flint]

Yes

So basically that dimension is reduced to a zero vector

So basically a dimension is nulled

Each free variable adds a new dimension

Same thing with one vector being the scalar multiple of other

That basically means if we take the scalar away from the vec and multiply it with the free variable, the two variables are creating dimensions which lie on top of one another

[sour karma]

gotcha, im a little confused what u mean by the two variables are creating dimensions that lie on top of each other tho

[jade flint]

Like we have (1,2,3)a+(2,4,6)b

We can take scalar from 2 4 6

And write it as (1,2,3)a+(1,2,3)(2b) = (1,2,3)(a+2b)

By properties of scalar multiplication

U realise that the free variables are scalars right?

[sour karma]

yeah

gotcha thanks

i got it now

[jade flint]

Do u get the overlapping dimensions idea

We can basically treat a+2b as a single free variable c

[sour karma]

i think so yeah

[jade flint]

So only one dimension is actually at work here

Both free variables are scaling the same dimension

[sour karma]

ok yeah i think i got it now, no matter what values u have its the same as only having one of them

thanks

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[dense wadi]

for 15

[dense wadi]

if the orthogonal projection of u along a

is just 0

then what

[median vigil]

that would imply that u is orthogonal to a

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[dense wadi]

ok ty

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[craggy sedge]

go to bed and sleep

[dense wadi]

~~i just woke up ~~

[craggy sedge]

lol

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[quick crater]

transform everything into powers of 3

[rugged kettle]

And what bout n?

[hasty marsh]

Is this supposed to equal to something?

[rugged kettle]

6 i suppose

[quick crater]

if you did it correctly you can factor 3^(something*n) and cross

[hasty marsh]

Just do what skill issue said, make everything into 3 to the power of something

And then make it equal to 6

And solve for n

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[mortal moth]

pls help me find the red circled dimension, in terms of y. y is a given variable

[indigo valve]

you can equate left to right

Let that be x

So x + y = 66.25 + 6

[unkempt venture]

y+??? = 66.25+6

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[woven plaza]

find The equation for which a straight line is parallel to a straight line 4x- y + 1 = 0 and intersects with the y-axis at (0,-5).

[woven plaza]

how do i do this

[jolly moss]

y = mx + c

u have c already

as -5

and m is also given in the question

[proud dust]

yuh

what does it mean when 2 lines are parallel

[proud dust]

yes

but in terms of the equation, what similarities are there

[woven plaza]

idk

[proud dust]

think about it

in y = mx + b

what does m mean and what does b mean

[woven plaza]

b is intercept of the line

[proud dust]

yes

and m ?

[woven plaza]

slope

[proud dust]

nice okay

so how do both affect the line

take y = x for example

what will happen when you change the value of b to 1 or 2 or any number you want

[woven plaza]

they intercept

[proud dust]

yes they do but where

[woven plaza]

2 straight lines intercept?

[proud dust]

like for y = x, you know the line intersects the y-axis at x=0 and y = 0

this it what y=x looks like

you see how it goes through y=0 at x=0

now how would that change if you would have y = x + 1

[woven plaza]

im dumb dang

[proud dust]

nono you got this

[woven plaza]

can u link me video that teach these concepts

[proud dust]

uhh i'll try and find a video

but i'll also try and help

here just plug in x = 0

and see what y value you will get

[woven plaza]

in calc?

[proud dust]

calc as in calculator?

[woven plaza]

ye

[proud dust]

you don't really need a calculator to calcluate y = 0 + 1

[woven plaza]

1

[jolly moss]

yessir

[proud dust]

yes nice

so from that you can say that when x = 0 y = 1

so you have a point (0,1)

do you see it ?

the blue line is y= x + 1

do you see what changed ?

[woven plaza]

what does that mean

another line appeared

[proud dust]

this other line is another equation y = x + 1

blue : y = x + 1
red: y = x

these are 2 independent functions

but they have similarities

do you understand what a function is

like the realtionship y = x

[woven plaza]

idk

[jolly moss]

i think

you would learn better off from a teacher in real life or youtube maybe

its tough to explain in text

or understand

[woven plaza]

yeah

[proud dust]

yeah

[woven plaza]

I will close this thread

thanks for helping

[proud dust]

good luck!

[woven plaza]

.close

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[safe radish]

Please don't occupy multiple help channels.

[vagrant ice]

#help-1 message context

[tidal ermine]

Yes

can u help me

[vagrant ice]

you should have numerator being

a^4 + 4 - 4a + 2a^3 - 3a^2

so assume this is a perfect square

from the a^4 and the 4

[tidal ermine]

2a^3?

I got 2a^2

[vagrant ice]

we immediately get $(a^2 - ka - 2)^2$

[flat frigate]

south's secret twin brother

[vagrant ice]

yes should be 2a^3 instead

$\frac{2a^3}{4a^2} = \frac{2a}{4} = \cdots$

[flat frigate]

south's secret twin brother

[tidal ermine]

ohhh ok

[vagrant ice]

or (a^2 - ka + 2)^2 actually

[tidal ermine]

What is ka

[vagrant ice]

k is a constant

[tidal ermine]

ok

Wait a sec

I've done it upto here

Is that right

[vagrant ice]

go back

that's not the right form

[versed pendant]

You could not take the square root out like that

[tidal ermine]

This?

[vagrant ice]

no you need to go all the way back to (a^4 + 4 - 4a + 2a^3 - 3a^2) / 4a^2

[versed pendant]

Sqrt(a+b)≠sqrt(a)+sqrt(b)

[tidal ermine]

This

?

[versed pendant]

Yes

[tidal ermine]

ok

What's next

[versed pendant]

What is sqrt(4a^2)?

[tidal ermine]

2a

[versed pendant]

Yeah, take it out

[vagrant ice]

technically it's |2a|

[tidal ermine]

Is this right

[vagrant ice]

no, not the right method

[tidal ermine]

oh ok

[young nexus]

no

[versed pendant]

Oh yeah you are righter

[vagrant ice]

if you can figure out the square root of the numerator you're pretty much done

[tidal ermine]

How do I find the sqrt of 2a^2

[vagrant ice]

so okay so what I did was to assume the numerator equals $(pa^2 + qa + r)^2$ for some constants $p,q,r$

[flat frigate]

south's secret twin brother

[vagrant ice]

you need to find the square root of the entire numerator

[tidal ermine]

Ohhhh

So the entire thing

[vagrant ice]

yes

[tidal ermine]

Ok I wrote it down

Next?

[vagrant ice]

yes maybe it's easier to write it as $(pa^2 + qa + r)(pa^2 + qa + r)$

[flat frigate]

south's secret twin brother

[vagrant ice]

now we apply the method of comparing coefficients

there's only one way to make the a^4 term

$(pa^2)(pa^2)$

[flat frigate]

south's secret twin brother

[tidal ermine]

Yes I wrote it

The root is a^2

[vagrant ice]

yeah and we have a^4, so $(pa^2)(pa^2) = a^4$

[flat frigate]

south's secret twin brother
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tidal ermine]

For the term 4 it's 2

[vagrant ice]

cool

p = 1

we could have p = -1 but cause x^2 = (-x)^2 it doesn't matter

[tidal ermine]

Then how do I find the next 3 terms

[tidal ermine]

Yea

[vagrant ice]

similarly compare the constants, so we have r * r = 4

so now we have two choices, r = 2 or r = -2

[tidal ermine]

Alr

[vagrant ice]

it happens to be r = -2 btw for this question

[tidal ermine]

I don't understand I'm dumb

What's √4a

[vagrant ice]

that's not related

[vagrant ice]

basically the method of comparing coefficients is the same as expanding everything out, and then comparing it to the polynomial you have

https://www.youtube.com/watch?v=jc_QtKwrKtc

try watching this

[tidal ermine]

I know that formula

Should I write it in that form?

[vagrant ice]

what formula?

[tidal ermine]

(a-b)^3

[vagrant ice]

that's not related at all

[tidal ermine]

Then why does the thumbnail show it

Ok nvm I'll watch the vid

[vagrant ice]

the thumbnail does not show (a - b)^3

[tidal ermine]

okay

So should I trial and error it

[vagrant ice]

you don't need to if you follow this method

[tidal ermine]

Can I use the long division square root method instead

I cant understand what ur saying

[vagrant ice]

oh I see what you mean, okay so guessing factors of that polynomial

sure

you'll have to use synthetic division or long division twice

[tidal ermine]

Yeah

[vagrant ice]

so that you divide by two linear factors and get a quadratic

ok if that's more comfortable for you

[tidal ermine]

yes

Are you still her

I'm stuck here

[safe radish]

@tidal ermine Has your question been resolved?

[safe radish]

@tidal ermine Has your question been resolved?

[tidal ermine]

No it hasn't

[safe radish]

@tidal ermine Has your question been resolved?

[tidal ermine]

NOOOO IT HASNT

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[dense salmon]

Two ships leave from the same port. One ship travels on a bearing of 157° at 20 knots. The second ship travels on a bearing of 247° at 35 knots. (1 knot is a speed of 1 nautical mile per hour).

a. How far apart are the ships after 8 hours?

b. Calculate the bearing of the second ship from the first?

[dense salmon]

I solve question a already

@helper

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[analog marsh]

I need help to show that the series an=sin(n) doesn't converge:

1. n is in radians (and a natural number by definition of an infinite series).
2. I need to prove it by showing it isn't a cauchy sequence, you aren't allowed to use subsequences

[robust river]

Negate the definition of cauchy sequence and show that sin(n) satisfies the negated statement

[analog marsh]

I know, but how do i do it.

my idea was to choose epsilon=1 and show that for every N there are n,m>N
such that sin(n)>1/2 and sin(m)<-1/2 and that:
|sin(n)-sin(m)|>1

but I cant seem to find those m and n as they need to be natural

[robust river]

sin(n) > 1/2 if anf only if n happens to be in (pi/6 + 2 pi k, 5pi/6 + 2 pi k) for some integer k. So the goal is to show that we can choose proper n and k

[analog marsh]

if the distance between the end point is ≥ 1 then there is at least one natural number that satisfies this?

[robust river]

yes

(5pi/6 + 2pi k) - (pi/6 + 2 pi k) = 2pi/3 > 2

So we can choose an integer between them

[analog marsh]

and do the opposite for negative sine

[robust river]

yes, a similar argument will work

[analog marsh]

ok so I think that's it right?

[robust river]

yes

[analog marsh]

nice thanks🙏🙏🙏

is it possible to ask another question?

[robust river]

yes

[analog marsh]

let {an} be a positive infinite series that satisfies:
lim(a(n) * a(n+1)) = 1

show that if L≠0 is a limit of a subsequence of {an} then 1/L is also a limit of a subsequence of {an} (are those called partial limits in the US?)

[robust river]

I'm not from US either so not sure what those are called

Okay, let's denote $a_{n_k}$ the subsequence such that $\lim_{k\rightarrow \infty} a_{n_k} = L$. What can you say about the sequence $a_{n_k+1}$?

[flat frigate]

EQUENOS

[analog marsh]

wait you are right

lmao how did I not think about that

thanks

[safe radish]

@analog marsh Has your question been resolved?

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[lapis hazel]

could you pls help me to prove 2nd

[charred saffron]

what are a and b @lapis hazel

[lapis hazel]

b is the semi-minor axis of the ellipse and
a is the semi-major axis of the ellipse

<@&286206848099549185>

[frozen marlin]

!15m

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[lapis hazel]

sorry

[safe radish]

@lapis hazel Has your question been resolved?

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[hazy pewter]

What do you do here?

[ocean crypt]

what do you have to do?

[hazy pewter]

simplify

[ocean crypt]

It's already simplified

At least I think so

[cedar owl]

yeah 30 is already perfect

[hazy pewter]

oh so it was just a trick question

i was so confused 😂

.close

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[hazy pewter]

How is this right?

[hazy pewter]

We are JUST simplifying radicals

Is it right if i’m just supposed to simplify radicals?

The -6/12

[cold relic]

What you wrote can be simplified further

The sqrt(12)

[hazy pewter]

Even If i’m just simplifying the radical?

[cold relic]

I don’t know exactly what your teacher expects, but I would count simplifying sqrt(12) to 2*sqrt(3) as simplifying the radical, yes

[hazy pewter]

Ok, thank you. But can you explain how you get 12 sqrt 3

[cold relic]

Sure

$-6\sqrt{12}=-6\sqrt{4\cdot 3}=-6\sqrt{4}\sqrt{3}=-6\cdot 2\sqrt{3}=-12\sqrt{3}$

[flat frigate]

FirstNameLastName

[cold relic]

It’s what you wrote but simplified further

[hazy pewter]

So you would divide the 4 by 2?

to get -6x2

Ohhhhhh

I get it sorry

Thank you

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[worldly umbra]

Il y a des notions en terminal spe maths que je n arrive pas a comprendre

[sinful glen]

lesquelles?

[worldly umbra]

Enft c est sur les combinaisons et tout

J arrive pas a comprendre quelque chose

[sinful glen]

et quoi?

pose ta question au lieu de blablatter autour du sujet 😭

[worldly umbra]

Oui j avoue 🥲

Att je t envoie l exo que je comprend pas

[sinful glen]

ca marche

[worldly umbra]

Fais pas gaffe c etait une interro

[sinful glen]

tkt

[worldly umbra]

Ça s envoie

[sinful glen]

on a tous eu des interros horrible dans la vie

[worldly umbra]

Oui vrmt

Enft a la correction elle nous a dis d utiliser les combinaisons

Mais apres utiliser le factorielle

Mais je comprend pas quand faut l utiliser

Dans cette exo j ai compris mais dans un autre je serai pas capable

[sinful glen]

j'avoue que je comprends pas tes calculs

je ne comprends pas d'où vient le 210 ni le 6

[worldly umbra]

Non mais j avais vrmt pas compris 💀

Juste regarde l énoncé 🥲

[sinful glen]

ah ok ok

pas de soucis

[worldly umbra]

🥲

[sinful glen]

ok deja pour commencer tu sais ce que c les factoriels?

[worldly umbra]

Oui

[sinful glen]

parfait et tu sais quand on l'utilises?

[worldly umbra]

Non justement

[sinful glen]

d'accord

[worldly umbra]

Mis a part dans les formules bien sur

[sinful glen]

ok ok

les factoriels tu les utilises quand tu as un nombre limité de pièce pour constituer le résultat

[worldly umbra]

Mais justement c est pas deja fais avec les combinaisons?

Ou arrangements

On sait deja le nombre de possibilité

C est ça que j arrive pas trop a comprendre

[sinful glen]

alors oui

donne moi 2 secondes j'essaye de trouver le bon moyen pour expliquer

[worldly umbra]

Dacc pas de soucis

[sinful glen]

ok donc les combinaisons c'est toutes les possibilités d'arranger les pièces sans tenier compte de l'ordre. Tandis que pour les arrangements on tient compte de l'ordre

[worldly umbra]

Oui

Je suis d accord

[sinful glen]

Par exemple pour les arrangements , (A,B) et (B,A) sont 2 arrangements différents de l’ensemble {A,B}. Et pour les combinaisons (A,B) et (B,A) sont 2 combinaisons équivalentes de l’ensemble {A,B}.

Donc maintenant dans le contexte d'un mot de passe tu devrais deja te poser la question si l'ordre des lettres est important

[worldly umbra]

Oui

Dcp

[sinful glen]

tu peux en déduire qq chose déjà?

[worldly umbra]

C est un arrangements

[sinful glen]

exactement!

Donc est-ce que tu connais la formule pour les arrangements?

[worldly umbra]

Oui

Mais avec n et k

N!/(N-k)!

[sinful glen]

oui c'est exactement ce que je veux entendre 😄

[worldly umbra]

Mais enft justement

C est ça le probleme

Sur la correction la prof fais avec les combinaisons

Regarde

[sinful glen]

ok ecoute (j'aime pas ton prof)

mais c pas con ce qu'il fait

[worldly umbra]

🥲

Pk tu l aime pas?

Mais justement pk on fais pas un arrangements dcp

[sinful glen]

il utilise les combinaisons pour trouver tout les choix de caractéres possible

[worldly umbra]

Oui?

C est pas bon avec les arrangements?

[sinful glen]

parce que je connais personne qui fait ca de sa manière

[sinful glen]

si si normalement tu trouves le même resulat

[worldly umbra]

Justement non

C est ça le probleme

[sinful glen]

et ensuite après avoir trouver tout les choix de caractere possible il calcule toutes les differentes possibilités d'arranger les 6 caractères

[worldly umbra]

Mais pk on trouve pas pareil avec les arrangements

[sinful glen]

ok oui je vois pourquoi

ca fonctionnerait avec les arrangements si les chiffres et les caracteres speciaux étaient dans le meme ensemble

[worldly umbra]

Mais ils le sont pas?

[sinful glen]

non

[worldly umbra]

Oui dcp

Mais comment je fais pour pas me faire pieger comme ça

[sinful glen]

tu as une fois l'ensemble des chiffres et une fois l'ensemble des caracteres speciaux

[worldly umbra]

Oui

Mais la c etait un un piege mais comment je pour plus tomber dedans

[sinful glen]

de l'entrainement j'ai envie de dire

alors ce que je peux te conseiller pour ce genre de truc c'est d'essayer de visualiser l'Exercice.

donc tu regardes comment on veut construire le mdp par exemple

[worldly umbra]

Je vois

[sinful glen]

y'a pleins d'exo d'énumérations sur internet je suis sur que tu as bien moyen de t'entrainer

[worldly umbra]

Vasy merci

[sinful glen]

j'ai plus mes cours de terminale sinon je t'aurais passé mes exos 😅

[sinful glen]

de rien

[worldly umbra]

Tkt c est rien 😅

Merci pour l aide

Comment on fais pour dire qu on a plus besoin d aide 🥲

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[fleet zephyr]

How do I solve this to find u?

[fleet zephyr]

the context here is, I was doing some math to calculate the vector field of the reflection of light rays, reflected off of some surface f(x,y). So what I did was parametrize this vector field as Where R_{R} is the reflected light rays, and I_{NT} is the incidence ray

L is the parameter for this. What I am trying to do, is multiply R_{r} by some value u where for all values X and Y, when L = 1, div(R_{r}) = 0, meaning the vectors all converge at a single point

R_{R} is a vector that contains these components, <R_{x},R_{y},R_{z}> which are functions of x,y

And therefore, the divergence of R_{r} gives me this when I set it equal to 0

Essentially, u is just some scaling factor. But since this scaling factor depends on the each reflected vector rays, it is a continuous function of x and y. Therefore u = u(x,y)

R_{xy} is just R_{R} where R_{z} = 0

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✅

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[craggy pine]

whats wrong with this

[limber steeple]

Woah

Wat site is that

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[craggy pine]

https://texnique.xyz/

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[inner tapir]

this worked for me:
\oiint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A} = \frac{Q}{\varepsilon_0}

i guess it's more strict on fractions? i only tried once since it moves on immediately and i had to skip a million times to find the right one to test lmao

[craggy pine]

wait what that is so stupid if true

[inner tapir]

i'm always militant on braces when i tex 👀

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[twin dagger]

For an RSA cipher, doesn't the message m have to be coprime to p1*p2 in order for the cipher to work? The proof is based on euler's theorem, which only works for coprime numbers.

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[regal cypress]

Find the area of the shaded region bounded by the curve and the coordinate axes.

Hi ppl, I need help help here. so for this I have found that dy/dx is 8(3-2x)^-3 -1 and d2y/dx2 = 48(3-2x)^-4
I haven't yet been thought about Integrals yet
and I found that the coordinates of M to be (1/2, 0) after that I'm stuck

[finite igloo]

alright

have u tried setting up an integral for the area

[regal cypress]

nope, Idk anything abt intergrals

[thin bridge]

don't know how you're expected to do this without integrals

was there anything else provided?

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@regal cypress Has your question been resolved?

[limber steeple]

the purpose of finding M is just do u can figure out what ur bounds of integration are

which is just 0 to whatever the x coord of M is

then just integrate the function in that region

[limber steeple]

u havnt been taught integrals?

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[hardy rover]

i have a question

[hardy rover]

recently i while i was doing multipilcation and addittion i found that if 1+1=2 then why 1x1=2?

let me give you an example

2x4={2+2+2+2} right so 1x1={1+1=2} right then why?

btw i am an 8th grader

[pine horizon]

no

2x4 is 2 times 4

so 2+2+2+2

1x1 is 1 times 1

so just 1

[hardy rover]

why just why

[pine horizon]

2 times 4 means add 2, 4 times

1 times 1 means add 1, 1 time

[hardy rover]

sr to inteipt but the function of multipication is addition

2x4={2+2+2+2} so??

its 8

[pine horizon]

yes

[hardy rover]

its the same answer

[pine horizon]

you're interpreting 1x1 as 1+1

how?

why didnt you do the same with 2x4 -> 2+4

[hardy rover]

no no i dint mean that

i meant that 2x4 is 2 is added itself 4 times

so

so why 1x1=1?

[pine horizon]

you're misinterpreting it

2 is added to itself 4 times will result in 10

[hardy rover]

oh

[pine horizon]

more accurately its 2 added to 0 4 times so thats 8

so for 1x1 you add 1 to 0 1 time, thats 0+1 = 1

[twilit moat]

instead of "adding to itself" I think the term "lots of" makes a little more sense to me

[hardy rover]

-_- i ment that 2+2+2+2=8

[twilit moat]

so like 2*4 would be 4 lots of 2 or 4 2s = 2+2+2+2

[hardy rover]

yes yes!!

[pine horizon]

yes but the way you are understanding it is "2 added to itself 4 times" equal to 8, similarly "1 added to itself 1 time" is 1+1 = 2

which is wrong

[twilit moat]

instead it would be "1 lot of 1" or just 1

[hardy rover]

i even tried in calculator but the result was the same its 1

[hardy rover]

oh why is it wrong?

🤔🤔🤔🤔🤔

[twilit moat]

the point is for 1x1

[hardy rover]

yes

[twilit moat]

it's not "1 added to itself 1 time"

which is 1+1

it's more there is just 1 1

a singular 1

with 2x4 for example it's not "2 added to itself 4 times" it's more "4 lots of 2"

since it starts from 0 instead of 2

[mossy lotus]

According to you, 1x4 = {1+1+1+1} right?
How about 1x3? 1x3 = {1+1+1} ok?
And 1x2? 1x2 = {1+1}?
And now 1x1... Same logic makes it 1

[hardy rover]

sign i mean that the funtion of multiplication is addition so 2x4 = 8 or the fu ntion inside is 2+2+2+2=8 simmlary i meant that 1x1 = why 1

[twilit moat]

the function isn't addition

that's the point

[hardy rover]

ik ik but isnt it kinda smilar

[twilit moat]

I mean you can calculate it using it

but don't think about it this way.

It's more a 'multiple' of a number which means that's how many of that number there is

[hardy rover]

but this q was alway taunting me so i want to ask it -_-

[twilit moat]

this is the key point I think

which kind of addresses why 1x1=1

1 multiple of 1 is just 1

[twilit moat]

yeah that's fine

[hardy rover]

but isnt the funtion of multiplication is the same? like 'x' = '+' i can be wrong tho

[twilit moat]

nope

they are different operations

[hardy rover]

oh ok

[twilit moat]

now with something like 5x5 for example

[hardy rover]

but i tho they are the same

[twilit moat]

that's more like saying there are five 5s

[hardy rover]

25!

[twilit moat]

if that makes sense

if you want to calculate it you can imagine having 5+5+5+5+5, but just because calculating it employs addition doesn't mean that addition and multiplication are the same operation in this case.

[hardy rover]

5+5+5+5+5 = 25, 5x5 = 25

[twilit moat]

yes exactly

[hardy rover]

but after all in the end i think they are the same in my perspective idk about others

[twilit moat]

these two are the same yes

how about we apply it like this