help-33

[fervent rampart]

the zero vector is the vector (0,0,0) (or however many zeroes are appropriate)

[still temple]

I'm confused then

what does it even mean

to span r2

or r3

[fervent rampart]

every set of vectors in R^3 spans a particular subspace of R^3 (which is a subset of R^3 that is a vector space). if that set of vectors is linearly independent, then it is a basis for the space it spans. the number of vectors in a basis for a space is called its "dimension". if a set of vectors spans a 3-dimensional subpace of R^3, that subspace is R^3.

[still temple]

so then what is the criteria for it to span 3 dimensional space?

[fervent rampart]

for a set to span a 3-dimensional vector space, it must contain 3 linearly independent vectors

that is the definition of "3 dimensional"

[still temple]

even if it's 0?

[fervent rampart]

wdym 'if it's 0'?

[still temple]

if one of the vectors was 0

[spark otter]

then it's not linearly independent

since 1*0vector = 0vector

[still temple]

so then this isn't linearly independant

therefore ir doesn't span r3?

[spark otter]

because we're not looking to span R^3

[fervent rampart]

why do you think that's the case?

[still temple]

but it says it does

[spark otter]

we're looking to find a SUBSET of R^3

that is spanned

[still temple]

so rn it isn't spanned

[spark otter]

have you heard about Span(x1,....,xn)?

[still temple]

no

im new to this

[spark otter]

it's the smallest subset of the vector space that contains the vectors x1,....,xn

you're looking in this question for Span((1,0,0),(0,1,0))

[still temple]

just took a peek at the answers

it says xy plane

is that because we have values for x and y?

[spark otter]

well can you write the "xy plane" of R^3 in set notation?

perhaps you can see clearer

[marsh citrus]

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[marsh citrus]

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[grave storm]

please help

[marsh citrus]

Please don't occupy multiple help channels.

[worthy basin]

!show

[marsh citrus]

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[marsh citrus]

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[marble lichen]

Hello im stuck with this task (a) finding the angualer acceleration

[marble lichen]

need some help

[marsh citrus]

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@marble lichen Has your question been resolved?

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[plain crescent]

quick question

[plain crescent]

how did x-1 become x/e ?

[pliant cargo]

It's not x-1, there's a bracket problem I believe

Should be ln(x) - 1

And then 1 = ln(e)

[plain crescent]

oh

[pliant cargo]

So you get ln(x) - ln(e), which is ln(x/e) by one of the laws of logarithms

[plain crescent]

I SEEEE

Thank you so muchhh

.close

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[edgy sleet]

What/s Z/(0) in abstract algebra?

[tacit fjord]

quotient

[edgy sleet]

Z / (0) is Z why?

[tacit fjord]

Z is a ring.

(0) is an ideal

you should look up these terms.

[edgy sleet]

Yes, i know that quotient is mod

but how to mod 0?

[tacit fjord]

it is not mod.

You should look up what a ring quotient is.

[edgy sleet]

why do they call it mod then?

[tacit fjord]

its a casual way of referring to it

but you definitely need to formally understand what a ring quotient is

[kind solar]

Is Z a group or a ring? It doesn't really matter for the elements but I want to use terminology that's familiar to you

[tacit fjord]

yes there is a deeper reason why we call it 'mod'

but theres no use talking about this reason without the formal defn

[edgy sleet]

Ring / Ideal generated by 0 = Z / (0)

[kind solar]

The zero ring does not generate Z if I understand you correctly

[tacit fjord]

no thats not what they meant

the denom is the ideal generated by 0

[edgy sleet]

I have this in my notes but I just realized I don't really know what it means

I just intuitively think of it as a mod operator

[tacit fjord]

what level of study are you right now.

[edgy sleet]

UNiv

[tacit fjord]

have you done any group theory just checking

im guessing not

[edgy sleet]

Yup

Cosets

[tacit fjord]

did you do group quotients

[edgy sleet]

Yeah

[tacit fjord]

the ring quotient follows the exact same idea more or less

[tacit fjord]

what exactly is your confusion

[edgy sleet]

What is Z / (0)?

[tacit fjord]

R/S is also a set of cosets

{n + (0) : n in Z}

[edgy sleet]

Yes so Z / (0) is a set of cosets

[kind solar]

well anyway, if you have Z/{0}, then the elements of that set are all the cosets of {0} in Z with respect to the addition operation, so basically the {0} is telling you what kind of elements you're dealing with and the Z is telling you how you're modifying those elements to generate the quotient ring

So the elements that you have are
1 + {0} = {1}
2 + {0} = {2}
3 + {0} = {3}
... and so on

So Z/(0) is basically Z, except every integer is packed inside a set. There exists an isomorphism from Z to Z/(0) and it's not terribly difficult to find

[edgy sleet]

What would Z / (3) be?

And that makes sense I think

[kind solar]

The ideal generated by 3 is all the multiples of 3

[edgy sleet]

I didn't realize it was the same as groups

[tacit fjord]

You need to figure out what exactly that addition operator written there means.

and then what the underlying set of the ideal is

3 + I is a kind of lazy shorthand in a way

but it has a formal meaning

[edgy sleet]

By my estimates Z / (3) is {{..-3, 0, 3...}, {-4, -1, 2, ...} , {-5, -2, 1...}}

[kind solar]

So Z/(3) = Z/{..., -6, -3, 0, 3, 6, ...}

The "denominator" tells you what kind of sets you're dealing with and the "numerator" tells you how you're modifying those sets

So we have
0 + {..., -6, -3, 0, 3, 6, ...} = {..., -6, -3, 0, 3, 6, ...}
1 + {..., -6, -3, 0, 3, 6, ...} = {..., -5, -2, 1, 4, 7, ...}
2 + {..., -6, -3, 0, 3, 6, ...} = {..., -8, -1, 2, 5, 8, ...}
and those are in fact all the elements because after that it starts looping

So
Z/(3) = {{..., -6, -3, 0, 3, 6, ...}, {..., -5, -2, 1, 4, 7, ...}, {..., -8, -1, 2, 5, 8, ...}}

[kind solar]

exactly as you wrote here

well done

[edgy sleet]

R/I = {all cosets of R with addition}?

[kind solar]

all cosets of I in R with respect to addition

For computations, R is the more important set because it distinguishes the elements

But for understanding what the quotient looks like, I actually like to point my attention to what's in the "denominator" (the ideal)

and the first set is really just saying how you're modifying the denominator sets

Algebraists probably get really triggered when I call it a denominator 😂

[edgy sleet]

Z[x]/(3, x^2 + 1)

What is this one? I see two now

[kind solar]

I think that ideal might be a little difficult to visualize hmm

but I'm also not very fluent with algebra

[copper raven]

one other way of intuitively understanding quotients is that you add new equalities that didn't exist before in your ring

with Z/(3), well you start with Z but now 3=0

for Z[x]/(3, x^2+1), you have Z[x] but now 3=0 and x^2+1=0

[kind solar]

So the congruence class [x^2 + 1] in the quotient is the identity element?

I don't see why that is tbh

actually maybe I see why that is

because x^2 + 1 is contained in the ideal

and therefore it's contained in the 0+I coset

which is the neutral element

and by definiton also [x^2 + 1]

alright this is getting beyond my capabilities so I'll step out of the discussion now

[marsh citrus]

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[mint sonnet]

help?

[marsh citrus]

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[mint sonnet]

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[desert dirge]

thoughts?

[severe cypress]

Wouldn't I need to integrate the slope

Thats what I'm thinking

[desert dirge]

not necessarily no, you could just check the options if you wanted to

[severe cypress]

Wouldn't guess and check take too long

Realistically on a test

I mean it's calculator active so I guess I could put all of the equations into my calculator then plug in the values on the table

Didn't wanna have to resort to guess and checking and see if I could do it by showing work

ok wait

i cant even do it with a calculator

lol

[desert dirge]

3y+y/x

assuming y looks like g(x)e^{f(x)}

the derivative would be
g'(x)e^{f(x)}+f'(x)g(x)e^{f(x)}
=g'(x)e^{f(x)}+f'(x) y
and g'(x)=g(x)/x so g(x)=ax for some a

f'(x)=3 so f(x)=3x+c

y=ax e^{3x+c}

passes through (1,2)

2=ae^(3+c) a simple solution would be c=-3 a=2, thats the only one that fits your choices anyway

idk, i guess that kind of works

though it would just be easier to check the derivatives of the options

[silk sparrow]

It's a separable differential eqn so you can always say 1/y dy = 3 + 1/x dx, integrate to get ln|y| = 3x + ln|x| + c, solve for c and go from there

But yes

[silk sparrow]

This lol

[desert dirge]

ah didnt think of that lol

[severe cypress]

oh i lied

its not calculator activec

xd

[severe cypress]

how is it separable

how did u separate it

[silk sparrow]

1. Divide both sides by y, 2) job done XD

Probably should clarify that dy/dx = (3+1/x)y, even though they don't write the LHS in the question they told you (3+1/x)y is the gradient, i.e. y' i.e. dy/dx

[severe cypress]

i was just abt to say there wasnt another side

until u clarified

ty

@silk sparrow ok one last one, how tf would i do this because usually when i do this i integrate both sides to get one variable to go away and i can solve for one of them

but if i derive this time, they'll both still be there

[marsh citrus]

@severe cypress Has your question been resolved?

[silk sparrow]

About to sleep lmao but a) for a function to be differentiable it must be continuous on its domain (please note that while all differentiable functions are continuous not all continuous functions are differentiable, so use the rule carefully) so the values of f(x) must be equal at that point and b) the derivatives must also be equal at that point

Should be 2 equations 2 unknowns solve

Someone less sleep deprived can definitely take over if you try that and get stuck haha

[severe cypress]

ty i figured out with a system of equations

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[inland harness]

${\frac{x^3}{x^2-3x+2}}$

[elfin berry]

San

[inland harness]

How do u set up the partial fraction

[shadow nest]

long division first

[marsh citrus]

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[hazy lion]

Wanted some help on the hint

So clearly we can just define say $B_x (\cdot )$ as the bounded operator on $H$

[elfin berry]

jan Niku

[hazy lion]

I am unconvinced that this operator is linear though

by the given properties wont it be conjugate linear, and riesz theorem wont apply?

Riesz theorem requires a bounded linear functional

however it appears that for some fixed x in fact \begin{align*} B_x (ay+z) &= \overline a B_x (y) + B_x (z) \ &\neq aB_x (y) + B_x (z) \end{align*}

[elfin berry]

jan Niku

[glass silo]

Maybe the hint was supposed to fix the second coordinate?

Would make more sense for their linearity comment, and also that their later comment on B(x, y) = <y, z> is a bit

[hazy lion]

i think theyre just trying to match the form of the theorem that we know

[hazy lion]

yea im not sure, but this makes more sense huh

maybe ill try that

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[marsh citrus]

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[still temple]

anyone know why the 3 is improper? trying to understand these multi calc notes

[marsh citrus]

Please don't occupy multiple help channels.

@still temple Has your question been resolved?

[warm pendant]

Dude, the cylinder has radius 3 and height 4

Since you integrate a sqrt that attains the value 0 at y=3

And your little fraction

y^2/sqrt(9-y^2) is infinity at y=3

So you have a ‘improper integral’

Meaning change that integral limit 3 to a, and put a limit on the outside as a -> infinity to ensure the integral value exists

Sorry, the notes use b instead

It is how we treat integrals that approach infinity near the limits of integration

Gl

Sorry, meant a approaches 3 from below

This might help https://web.ma.utexas.edu/users/m408s/m408d/CurrentWeb/LM7-8-4.php#:~:text=An improper integral of type,on [a%2Cb].

See example 1

[marsh citrus]

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[cloud star]

im doing 7k^2 +4 = 564 and it says the answer is +,- 4 square root five and im just confused on how it got ther

[cloud star]

cause i keep getting k = -,+ square root 80😭

[cobalt sentinel]

Ok ur on the right track

U can simplify +-sqrt(80)

sqrt(80) = sqrt(16 * 5) = 4 * sqrt(5)

Lmk if ur confused

[cloud star]

im a bit confused yeah

[meager badger]

youre basically breaking up 80

[cloud star]

how do you get to the 16*5

[meager badger]

16 * 5 = 80

[cloud star]

oh yeah

but how do i know when to do that

like whats not letting me just stay at -,+ 80

square root 80**

[meager badger]

I mean technically this is the same thing

its just their answer is simplified

[cloud star]

oh alright. thank u!!

.close\

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[crude marlin]

Anyone know why my lines aren’t connecting?

[marsh citrus]

@crude marlin Has your question been resolved?

[crude marlin]

<@&286206848099549185>

[warm pendant]

You need to prove more info

Is this a parametrisiatipn of a circle?

Go into your graphing options and ensure the stepsize is small or automatic

provide*

the final result will not help, give us the equations and the settings you have too

[marsh citrus]

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[zealous glacier]

ive replaced y with x and x with y

but where do i go from there?

[pliant cargo]

Can probably just argue that x = 1 and x = - 1 result in the same y value

So the inverse isn't well defined

(actually true for any +/- a)

[tidal phoenix]

hsc?

[zealous glacier]

year 11 maths

[knotty frost]

Sir try to not give away the entire answer

[knotty frost]

But yes sir. For the inverse to be a function it has to pass the horizontal line test right

or else when you flip it around the line y=x (take the inverse) it fails the vertical line test and you wouldn't have a function

This is the main geometric idea behind the problem

Does that make sense?

[zealous glacier]

yeah

kinda

[knotty frost]

So you just need to show that it fails the horizontal line test. This would be true, for example, if the function was symmetric across the y - axis

[zealous glacier]

ok

[knotty frost]

What doesn't make sense?

[zealous glacier]

the line tests

do inverse relationshave to pass the horizontal line tests

[sacred idol]

yes

[zealous glacier]

for them to be functions

[knotty frost]

Ye

[zealous glacier]

oh ok

[knotty frost]

That's what i tried to verify up there

[knotty frost]

In this message

[sacred idol]

because there would be two answers for a single y

[knotty frost]

^ exactly

This is the more algebraic version of what i was trying to get with geometry

[zealous glacier]

oh ok

.close

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[knotty frost]

🔠

https://tenor.com/view/minion-gif-4256583569949039458

[zealous glacier]

https://tenor.com/view/erm-what-the-sigma-squidward-meme-tiktok-reaction-gif-574741893352575449

[marsh citrus]

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[woven plinth]

hello i have a question about power series representation

[woven plinth]

creating a power series to represent this function

this is what i tried but it doesn’t graph right on desmos

wait im stupid

that 3 outside the sigma should be a -3/2

.close

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[stark trail]

d

[hard hemlock]

.close

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[main flare]

Hello?

[main flare]

so..

do (tan x)/(x) is equal to 1 ??

[spare ether]

no

try on calc see for urself

[valid cape]

no merp?

[main flare]

merp

[main flare]

whuay 😭 ?

[valid cape]

no lol

[valid cape]

$\frac{a^c}{b^c}=\left(\frac{a}b\right)^c$

[main flare]

wait

[main flare]

so why its turning to 1 then?

ive asked this before

but

i just want to confirm but ended up confused again

[hard gull]

$$\lim_{x \to 0} \frac{\tan{x}}{x} = 1$$ @main flare

[elfin berry]

JustToPro

[hard gull]

$$\frac{\tan{x}}{x} \neq 1$$

[elfin berry]

JustToPro

[main flare]

if i change x to 0, then tan equals to 1 ??

[final fern]

tan^6x / x^6 can be rewritten as (tanx/x)^6

tanx/x approaching 0 is 0/0

thus you can apply lhopital

sec^2x/1 -> 1/cos^2x

1/cos^2x evaluated at x=0 is 1/1

[hard gull]

^^

if u havent studied lhopital u can also do smthing like this
$$\lim_{x \to 0} \frac{\tan{x}}{x} = \lim_{x \to 0} \frac{\frac{\sin{x}}{\cos{x}}}{x} =\lim_{x \to 0} \frac{\sin{x}} {\cos{x}\times x} =\lim_{x \to 0} \frac{\sin{x}}{x} \times \lim_{x \to 0} \frac{1}{\cos{x}} = 1 \times 1 = 1$$

[elfin berry]

JustToPro

[main flare]

so (sin 0)/(0) = 1 ???

[final fern]

that should be an identity you learn early in calc1

[hard gull]

yeah , thats a pretty known limit

[stoic slate]

Sin0/0 is undefined

[main flare]

so what is it then

[stoic slate]

You are talking about the limit

When x-> 0 of sinx/x

[main flare]

does it when only?

[marsh citrus]

@main flare Has your question been resolved?

[main flare]

.close

[marsh citrus]

Channel closed

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[main flare]

hello

?

.close

[marsh citrus]

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[drowsy totem]

how do i calculate the maginitude of those 2 arrows /vectors

[cyan parcel]

Unify them as 1 force pointing upward

[drowsy totem]

basically the i components wil cancel out

[cyan parcel]

Then since the unified force can keep the mass agaisnt gravity, u can get that force

[drowsy totem]

im guessing

[cyan parcel]

Then get the 2 vectors force from the unified forced

[drowsy totem]

ok

[marsh citrus]

@drowsy totem Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[jovial lily]

how do you find the position of a centroid bound by a curve, the curve is 16-x^2

[marsh citrus]

Please don't occupy multiple help channels.

[jovial lily]

.close

[marsh citrus]

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[marsh citrus]

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[mighty stratus]

im lost

[vagrant depot]

Do you know what completing the square is?

[mighty stratus]

i think its equal to (x-3/2)^2 - 4

but then what do i do

[vagrant depot]

How do you get 2^2

[mighty stratus]

1/2 times 3 squared

for complete square

man idk

[vagrant depot]

What's (x-a)^2

expansion

[mighty stratus]

so what do i need to do

[vagrant depot]

What is the expansion of (a-b)^2

[mighty stratus]

(a-b)(a-b)

[vagrant depot]

Expand it

Multiply it out

[mighty stratus]

a^2 +ab -ab +b

a^2 + b

[mighty stratus]

o im dumb

i meant that yeah yeah

[vagrant depot]

So basically, you want your equation in that form

with a constant outside the brackets

[mighty stratus]

uh

i think my dyslexia is acting up

so i make 2x - 3 + 1/4 into brackets?

[vagrant depot]

Forget about 1/4, try for x^2 - 3

[mighty stratus]

put x^2 - 3 in bracket?

[vagrant depot]

yes

how would you get x^2 - 3x in a form of (a-b)^2

[mighty stratus]

what

o

[vagrant depot]

mb

[mighty stratus]

(x-2)(x-1)

but ill get +2 on the side

[vagrant depot]

how'd u get rid of that?

[mighty stratus]

(x-2)(x-1) -2= -2

[vagrant depot]

Also, we need (a-b)^2

[mighty stratus]

wjat

[vagrant depot]

(x-2)(x-1) is wrong

[mighty stratus]

bruh

im so bad rn

[vagrant depot]

Try getting in terms of (x-a)(x-a)

[mighty stratus]

man

[vagrant depot]

x^2 - 2xb + b^2 = x^2 - 3x

x^2 = x^2

yes?

[mighty stratus]

idk

[vagrant depot]

wdym idk

x^2 = x^2

[mighty stratus]

yes

[vagrant depot]

thats a fact

[mighty stratus]

what was the bit before about

[vagrant depot]

so -2xb = -3x

[mighty stratus]

ok

[vagrant depot]

(x-a)(x-a) = x^2 - 2ax + a^2

[mighty stratus]

so -2b = -x

[vagrant depot]

-2b = -3

x cancels

[mighty stratus]

ok

[vagrant depot]

b = 3/2

[mighty stratus]

yeah

[vagrant depot]

(x-3/2)^2

expands to?

[mighty stratus]

(x-3/2)(x-3/2)

[vagrant depot]

further expansion

(a-b)^2 = a^2 -2ab + b^2

a = x
b = 3/2

[mighty stratus]

x^2 - 3x + 9/4

[vagrant depot]

How would you get rid of the 9/4?

[mighty stratus]

-9/4

[vagrant depot]

So whats x^2 - 3x

[mighty stratus]

x^2 - 3x + 9/4 - 9/4

[vagrant depot]

and what's x^2 - 3x + 9/4 (we just expanded)

[mighty stratus]

(x-3/2)(x-3/2)

[vagrant depot]

(x-3/2)^2

simpler form

[mighty stratus]

yuh

[vagrant depot]

So

x^2 - 3x = (x-3/2)^ - 9/4

yes?

[mighty stratus]

yes

[vagrant depot]

So you end up with x^2 - 3x + 1/4 = (x-3/2)^ - 9/4 + 1/4

Whats that simplify to

[mighty stratus]

what

[vagrant depot]

-9/4 + 1/4 =

[mighty stratus]

-8/4

[vagrant depot]

which is

[mighty stratus]

-2

[vagrant depot]

so you get?

[mighty stratus]

what has this got to do with the question

[vagrant depot]

This is called completing the square

[mighty stratus]

(x-3/2)^2 - 2

[vagrant depot]

= 0

[mighty stratus]

ye

[vagrant depot]

Now solve for x

[mighty stratus]

x = 3/2 +- root2

[vagrant depot]

yes

this answers ur question?

[mighty stratus]

yeahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhssssssssssssssssssssssss

tysm

.close

[marsh citrus]

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[still temple]

?

[terse dove]

i have to prove that 2^2n - 1 is a multiple of 3

the -1 isnt in the exponent

is this gonna be the solution

[vagrant depot]

what

[terse dove]

using mathematical indcution

[still temple]

do you have an idea about binomials?

nope

[terse dove]

yea binomials

degree 2

[still temple]

you want it by mathematical induction or binomial ?

[terse dove]

the question says to use mathematical indcution

induction

no clue? @still temple

[still temple]

i do

but it is way to lengthy

[terse dove]

i did it in like 3 steps

check the image

[still temple]

that is wrong

[terse dove]

skipped the writing part

from where

is it wrong

[still temple]

first you have to find for n=1

then assume for n=k

and then prove for n=k+1

[still temple]

❌

[terse dove]

exclty

i did the same

[still temple]

mathematical induction is used to prove statements

it can't be cut short

[still temple]

where is n=1 ?

[terse dove]

IT DOESNT MATTER

ik its gonna be true

just skipping to the difficult part ok?

[still temple]

is it your homework

[terse dove]

no

[still temple]

or are you learning ?

[terse dove]

i am practicing

[still temple]

1sec

min*

[terse dove]

lmao

sec ❌
min ✅

[wide grail]

You fucked up

[terse dove]

where

thats what i want to be told

[wide grail]

2^(2k+2)

You wrote as 2^2k * 2

Its 2^2k * 2²

[terse dove]

what

where

oh

i see

ima send ya again

a snap

check

[wide grail]

You forgot the 2k

One row it's 2k one row it's k

Otherwise its good

[terse dove]

row?

[wide grail]

Bro you dont have 16 rows

3rd row it's 2^2k*2²

Next row you wrote 2^K *2²

[terse dove]

YOU MEAN A LINE?

[wide grail]

You come to a linguistics class or a math help channel

[terse dove]

didnt get what u meant by rows

so

what next

[wide grail]

So thats it

[terse dove]

how do i tell that this is a multiple of 3

[wide grail]

You assumed 2^2k-1 is divisible by 3

And you're adding 3 to it

Which is a multiple of 3

If a and b are multiples of 3 then so is a+b

[terse dove]

no no no

[wide grail]

Yea yea yea

[terse dove]

the 2^2k -1

is being muliplied by 4

[wide grail]

Ok

Is 4*3 divisible by 3

[terse dove]

no

do you mean factor

[wide grail]

Wdym no

[terse dove]

yea yea its a factor

[wide grail]

So if 2²(2^2k-1) is a multiple of 3

And 3 is a multiple of 3

[terse dove]

yea thanks i got it now

[wide grail]

No

Np

[terse dove]

👍

@still temple

can you teach me the binomial method

ik that binomials are degree 2 polynomials

[still temple]

2^2n would be written as

4^n

4 can be written as 3+1

(3+1)^n - 1

when you expand it

you get it in the form

3k

which is divisible by 3

[terse dove]

k?

[still temple]

got it ?

[wide grail]

I think induction is much more intuitive and streamlined than this

[terse dove]

from where did the k come

we had n initislly

[still temple]

did you expand (3+1)^n ?

[wide grail]

Do you know about binomial expansion

[terse dove]

ik binomial expansion but the power here is a variable

i cant figure a way to handle that

[still temple]

however,mathematical induction is for proving very find formulas

[terse dove]

im used to expanding whole number powers

[wide grail]

Yeah but (3+1)^n is (n k)3^k * 1^(n-k)

Which is just (n k)3^k

For all k, this is divisible by 3 except k=0, when 3⁰ is 1

Which cancels with the -1

[terse dove]

WHERE DID K COME FROM

[wide grail]

Thats in the front

[terse dove]

i still have this question

[wide grail]

Binomial expansion is a sum

k is a number that goes from 0 to n

[still temple]

wait

k is a constant

[wide grail]

And you're adding up all of them

[still temple]

some constant

here bog and i use k for 2 different purposes

bog used it to represent the general term where as i used it as a constant

[terse dove]

then you reslve ur conflict first

lmao

[wide grail]

We're not in a conflict lol we know what we're saying

[still temple]

yes

bog you go on

i will leave

[terse dove]

bro can you expand

[wide grail]

No.no no

[terse dove]

and show me

[wide grail]

Let me.leave pls

U go

And explain

[terse dove]

you both go

[still temple]

okay

[wide grail]

Ok if you say so

[still temple]

bruh

[terse dove]

i hope so1 else comes

who is HELPFUL

lmao

bye

[still temple]

do you know hwo to expand (3+1)^ n?

[terse dove]

no

i do not

[still temple]

[wide grail]

Soz

[terse dove]

ik how to expand this

the upper one

but have no idea bout the 2nd one

how do i do that

[wide grail]

Thats kinda like saying you know 2+2=4 but dont get x+2=4

[still temple]

both are the same

[still temple]

instead of 5 you should replace the term with n

[terse dove]

n+1 power n?

[still temple]

what is that

[terse dove]

dont you guys have like

smth to write

[marsh citrus]

@terse dove Has your question been resolved?

[terse dove]

.close

[marsh citrus]

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[red skiff]

conceptual question :

[red skiff]

why doesn't the negative value of the dx itself create the negative area?

[onyx kindle]

u should’ve send this message first so it could be pinned

[terse dove]

use newtons laws of light

for hot medium

[red skiff]

my badf

.close

[marsh citrus]

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[marsh citrus]

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[glass silo]

,rcw

[naive rock]

am not how to solve this?

[naive rock]

this

[glass silo]

You’re happy with what you’re supposed to be finding, right?

(Remember that x, y and z all depend on the time t)

[naive rock]

ye but where does t go in the equation?

[glass silo]

Well, it implicitly is there

You won’t see explicitly how x, y and z depend on t, but you do know that they do somehow-

[glass silo]

(The word implicit should hopefully be a clue as to what to do!)

[naive rock]

so do i multiply dx/dt dy/dt dz/dt together?

[marsh citrus]

@naive rock Has your question been resolved?

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[still temple]

1 sec for me to take pic of work

i feel like i did something weong

wrong

oh i have an idea

do i do ... R3 - hR2

no -(h-1)

whats k then

huh

btw this is my help room lol

[red nimbus]

there is a unique solution if rang(A) = rang(A|b) = number of unknowns

[still temple]

ok idk that

that rang no bells

[red nimbus]

h a h a (you're just like me=

rang of a matrix describes basically the amount of linear independent vectors

to my understanding

[still temple]

ok so how do i get k

[red nimbus]

let me finsih my pizza farm girl

[still temple]

thats good

i want fries

[red nimbus]

@runic temple

That guy breeds fries

[runic temple]

the garlic breeds the fries

i have no control

[red nimbus]

so it happens against your will

[still temple]

ill take garlic bread too

my breath will keep ppl at a 5 mile radius

[red nimbus]

x^2+y^2=5^2 away

[still temple]

exactly lolol

[red nimbus]

actually

x^2+y^2>=5^2

[still temple]

ohhh

smartuy

smary

smarty

i think the answer is b

and its not

[red nimbus]

i forgot you sorry

lemme first check the system

[still temple]

thats fine lol

[red nimbus]

-1 - (-1(h-1)) = -1 + h-1 = h-2

last operation, or did i do something werong

@still temple

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

Huh

Oh

You’re right

[marsh citrus]

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[tulip fulcrum]

Is this correct? This is my first physics derivitive btw

[tulip fulcrum]

Btw I didn't pay attention to anything so any of the variables may be wrong

Like the 2/16 lul

[neat oak]

are u differentiating x(t)?

at t=2

[tulip fulcrum]

Trying to find the velocity at the point

[neat oak]

yeah its correct

[tulip fulcrum]

Thank you!

[marsh citrus]

@tulip fulcrum Has your question been resolved?

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[carmine bane]

I need help with creating a linear function

[red nimbus]

ok

[red nimbus]

my neck

[red nimbus]

,rccw

[carmine bane]

Number 3

[red nimbus]

ok!

So we have a function H(y) ...yearly house costs and y ...time in years

Linda pays $1,700 monthly

So Linda wants to find out the amount of years it will take her to pay that $30,000

are you here sara bear?

@carmine bane

[carmine bane]

Yes

[red nimbus]

If Linda pays monthly 1,700 then what would that be in a year?

[carmine bane]

20,400

[red nimbus]

What did you do?

[carmine bane]

1,700x12

[red nimbus]

very good!

So for one year we get 20,400

mathematically

H(1) = $20,400

[carmine bane]

Oooh

That makes more sense compared to what I did

[red nimbus]

It would make sense that in 0 years she has $0 paid

So c = 0

of y = mx + c

So you actually have y = mx left and you can calculate m by using h(1) = 20,400

Actually here of H(y) = my + c to H(y) = my

@carmine bane

Did I confuse you?

[carmine bane]

You didn’t

[red nimbus]

I utterly apologize

I will try better

[carmine bane]

It’s ok

[red nimbus]

I also agree that sleep isn't a thing

[carmine bane]

Im mainly just trying to complete my math so I can catch up on my writing and sociology class

Do you have time to help work through my other work sheet?

[marsh citrus]

@carmine bane Has your question been resolved?

[carmine bane]

.close

[marsh citrus]

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[dreamy hull]

Hello, can you help me to demonstrate this property please?

[marsh citrus]

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

[dreamy hull]

1

[stark trail]

Try proving it one direction at a time

Start by trying to prove Y subset X => Xbar intersect Y = emptyset

[dreamy hull]

sorry, I got the wrong exercise, it's this one:

[zenith python]

same idea

one direction at the time

[whole sleet]

You're missing context

[marsh citrus]

@dreamy hull Has your question been resolved?

[kind solar]

(=>) Assume there is some x in A intersection B. Since x is in A union B complement, we have x in B complement. Since x is in B, we have a contradiction

(<=) B complement is obviously a subset of A union B complement
Take x in A union B complement. If x is not in B complement then x is in A. x cannot be in B since if it were, we would have a contradiction. Therefore x is in B complement.

[dreamy hull]

I was able to justify the first conditional

[dreamy hull]

this is correct?

I mean what I did

[kind solar]

Well something is a little sus because no element can belong to the empty set

[dreamy hull]

hmm

[marsh citrus]

@dreamy hull Has your question been resolved?

[dreamy hull]

Isn't that what I want to demonstrate?

that this intersection is equal to void

[kind solar]

Well your argument probably makes sense in your head but it doesn't really make sense in mine

which is why it's good to use actual words in your argument

ok no I thought about it some more and now I also understand what you're trying to say

but you shouldn't say that x belongs to the empty set

because that just doesn't make any sense

you should instead say that you have found a contradiction and thus no such x can exist

and again, use words trust me it makes everyone's life so much easier

[dreamy hull]

this?

the other conditional I don't really know how to demonstrate it

<@&286206848099549185>

this is correct?

[marsh citrus]

@dreamy hull Has your question been resolved?

[marsh citrus]

@dreamy hull Has your question been resolved?

[marsh citrus]

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[magic heron]

After doing some long division on a quartic function, I got a quotient and remainder(red text), is there any issues with adding the the remainder like so and then proceeding to break down the cubic function to get the roots?

[fathom ridge]

You can simply factor an x out

and you have a simple quadratic

[elfin berry]

Sadie Carnot (η > 1)

[fathom ridge]

oh wait

i didnt understand what you meant

[magic heron]

yeah

I'm more asking about whether I can add the remainder

[fathom ridge]

you want the roots to a quartic?

[magic heron]

and then now solve the cubic

noo I started with a quartic and had one of it's factors

so I did some long division that I didn't show here

and got the quotient and remainder(red text)

[fathom ridge]

If you had one of it's factors

[magic heron]

so I'm asking if adding the way I did is legal

[fathom ridge]

then dividing by that factor shouldn't yield a remainder

[magic heron]

wait yeah wtf

I just realized

LMAO ty

wait yeah I'm tripping for even asking this question

I'll close this for now, but thanks again ahhaa

.close

[marsh citrus]

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[mild hearth]

.rotate

,rotate

[mild hearth]

I don’t get part b

What it means by “strictly increasing”

[marsh citrus]

@mild hearth Has your question been resolved?

[mild hearth]

<@&286206848099549185>

[marsh citrus]

@mild hearth Has your question been resolved?

[marsh citrus]

@mild hearth Has your question been resolved?

[vocal cape]

when the next y value is greater than the previous y value

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[hard gull]

how do u write equation of line l in normal form?

[hard gull]

i can write it in general form which would be
y - y1 = (y2-y1)/(x2-x1)(x-x1)

and i can also find angle by
m = tantheta
(y2 - y1)/(x2-x1) = tan theta

[storm dagger]

i think normal form is: ax + by + c = 0

just rearrange your equation into this form using algebra

[storm dagger]

ah right

[hard gull]

thats general form

and getting that is ez

[storm dagger]

just divide by sqrt(a^2+b^2)

[hard gull]

hm?

what about cos theta or sin theta?

do i make a triangle for that ? a is perp , b is base and sqrt(a^2+b^2) is hypt

[storm dagger]

yep, and the values of sin(theta) and cos(theta) will be the ratios of thisi triangle

[hard gull]

k

ty

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[solar fulcrum]

why is this part of the domain

[marsh citrus]

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[still temple]

.

hello I need help with this problem:

a girl has 4 skirts, 5 blouses, and 3 jackets
she picks one of each at random
how many diff ways can she dress?

[still temple]

my professor used permutations to solve it, but it didn't click for me why

try to think about it logically

she picks one of each, at random, so the order shouldn't matter

so we use combinations

does she have to wear at least one of each

she has to wear one of each, so yeeeees?

so

if you try to visualize it

how would you do that

I dunno how I would visualize it, but I know with logic it would be like
one of three jackets AND one of four skirts AND one of five blouses

the end would be multiplication

yes

what confuses me the one of smth part

well if she can wear a skirt and a blouse but no jacket then things have to change, there's more than 3*4*5 combinations

[still temple]

okay okay hear me out,
my answer was: 5C1 * 4C1 * 3C1=60
my professor's answer was: 5P1 * 4P1 * 3P1=60

we both got the same results, but the professor said we should use permutations (like she did) instead of using combinations (like I did)

is she right? if so why? or is she just gaslighting me ?

as long as order doesn't matter both should do the same

"The permutation is the number of different arrangement which can be made by picking r number of things from the available n things. The combination is the number of different groups of r objects each, which can be formed from the available n objects."

I'm pretty sure combinations should be used here though

so my professor is gaslighting me

thing is that combinations and permutations are different

there do exist 60 permutations and 60 combinations

so honestly whichever is less confusing for you

[still temple]

oh okay so she's right! bc we are picking one thing out of diff things, not creat diff groups of r objects from n objects

right?

technically but both are correct and both give the same answer

in this case it's more of how you define it

okay I think I understand it a bit now

Thanks a lott! >_<

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[still temple]

i have a doubt

why d

OH IS IT SPAN OF COLUMNS IN Rm?

[glass silo]

Well, more that you have a square matrix, and you're checking what the span of the columns are

Can you think of examples of square matrices?

[still temple]

no i cant think

jkjk like an identity matrix

but this m by n

no not square

[glass silo]

d says m = n

[still temple]

bro i cant read

i did read that but idk why my brain short circuited

[glass silo]

Happens

[still temple]

why not true

[glass silo]

Well, an identity matrix is an example of a square matrix, and it does happen that the span of the identity matrix's columns is R^n (the columns making up the standard basis vectors)

Can you think of another example?

[still temple]

maybe like one with not all pivots

[glass silo]

There you go

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[desert prairie]

I’m trying too see the pattern but I ain’t seeing it at all. When ever it comes to the ones where there more than 2 fractions I’m totally lost

[desert prairie]

For question 5b

[mystic minnow]

notice in your r = 2 row, you have -1/3
and in your r = 3 and 4 row, you have -1/3 + 2/3

Also in your r = 3,4 and 5 rows you have -1/4 -1/4 + 2/4.....

[desert prairie]

OH

I see it now

Thank you

Looks like I need to do more practice with these question

[desert prairie]

Are the “patterns” usually like this?

[mystic minnow]

in a telescoping series, yeah. It has some cancelation between terms.

so the series becomes the sum of the (usualy finite number) terms that don't cancel

[marsh citrus]

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[lucid vector]

For this

[lucid vector]

Is what I’ve done correct so far using lagrangian method

And if so what do i do after the final step because it’s not clear what lambda is?

I tried equating by putting 6y + 4x = 2x + 3y but didn’t give me right answer

[lucid vector]

Oh so here you wouldn’t use the lagrangian?

[granite trench]

It is

[lucid vector]

How though? It doesn’t write the function as f(x,y,lambda)

Like its correct and it makes sense

I just don’t wanna be penalised by my prof for not using their method

[granite trench]

Its ok

[lucid vector]

lol he does penalise us

[granite trench]

I think you have a strict professor

but I don't know

But you still have a solution from chatgpt plus, which I bought two months ago

But I still check that it also correctly resolves chatgpt plus

[lucid vector]

Yeah ik haha my friend dropped marks for using another method on coursework

Like a lot of marks

[granite trench]

Do you want the Lagrange method?

@lucid vector

[lucid vector]

Yes pls

I’m still trying and getting the wrong thing

[lucid vector]

Haha they skipped the part im failing on

Which is solving for x,y etc

Ill have to come back to this disgusting thing

Thanks

.close