help-10

[timid silo]

.close

[obtuse pebble]

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[dusk mauve]

how did they combine them here?

[civic zealot]

multiply the right fraction by x-6 on top and bottom

[dusk mauve]

is x + 5 multiply by x - 6 = x + 5?

[civic zealot]

yes, you multiply x + 5 by x - 6, they are not equal

[dusk mauve]

so the answer is x + 5?

@civic zealot

[civic zealot]

no?

[dusk mauve]

what is it then?

[civic zealot]

the answer to what?

[dusk mauve]

to multiplying the right fraction to x - 6

[civic zealot]

yes, multiply the top and bottom by x-6, you get what's in that red circle

[dusk mauve]

appreciate you man thank you

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[tardy kite]

The wording confuses me. I could use binomial series. But in terms of t/P I am not sure

[tardy kite]

My attempt.

@ruby fulcrum

[obtuse pebble]

@tardy kite Has your question been resolved?

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[tardy kite]

.reopen

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✅

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[meager hull]

Guys I fixed the image

[meager hull]

Ok anyways, finding second derivative through implicit differentiation but I messed up

I lost the 6xy smth that should be in my final alswer

Spoiler alert, 6xy smth isn't in my final answer atm

my xy stuff cancels out and i don't think it's supposed to 😭

[obtuse pebble]

@meager hull Has your question been resolved?

[meager hull]

<@&286206848099549185>

unless I'm overthinking it

and there really is no 6xy in the numerator of my final answer

[obtuse pebble]

@meager hull Has your question been resolved?

[dark stirrup]

Hint

Your second derivative may be easier if you do implicit differentiation of the first line you wrote. That way, you don't need to use quotient rule where stuff gets messy

[meager hull]

Ahh r

Alr

[obtuse pebble]

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[fallow needle]

How can I add (a/12) + 12 together in an equation

[timid silo]

12 + a/12 = (144+a)/12

if thats what u mean

[fallow needle]

Probably not, its a longer equation so should I post the whole equation?

[timid silo]

sure

[timid silo]

oh

ru trying to simplify?

[fallow needle]

Yessir

Need to find A

[timid silo]

also whats the thing in between a and 100

[fallow needle]

times

just think of it as 100a

[timid silo]

ok

[timid silo]

for the RHS you wanna get rid of the fraction on the bottom and top by multipling top and bottom by 12

do that first

[fallow needle]

100a=a+144

basically

?

[timid silo]

$\frac{100a}{150}=\frac{100a}{a+144}$

[warm shale]

😔😔😔😔😔😔😔

[timid silo]

you should end up with this

[fallow needle]

yessir thanks

So now I just solve like its a normal equation?

Oh?

[timid silo]

oops

that was wrong nvm

[fallow needle]

No worries

So now Ill just solve for a I guess?

[timid silo]

yeah

[fallow needle]

Thanks man

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[timid silo]

$\left(a+144\right)\left(\frac{100}{150}a\right)=100a$

[warm shale]

😔😔😔😔😔😔😔

[fallow needle]

Do I gotta factorise?

[timid silo]

uh yeah

[fallow needle]

I wanted to do cross multiplication

Guess thats wrong?

[timid silo]

wait

[obtuse pebble]

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[fallow needle]

.

[timid silo]

oh

[timid silo]

this turns into a quadratic

yeah u gotta factor i guess

[fallow needle]

Ohh

Kk ty man

Ill do that then

[timid silo]

$\left(a+144\right)\left(\frac{100}{150}a\right)=100a$

[warm shale]

😔😔😔😔😔😔😔

[timid silo]

so distribute

2/3(a) to a and 144

[fallow needle]

How can I distribute efficiently

[timid silo]

$2/3\cdot a^2+96a=100a$

[warm shale]

😔😔😔😔😔😔😔

[timid silo]

uhm

[fallow needle]

Yeah this is a quadratic definitely

[timid silo]

$2/3\cdot a^2-4a=0$

[warm shale]

😔😔😔😔😔😔😔

[timid silo]

oops its -4a

actually

[fallow needle]

Yea

[timid silo]

u can factor out a common factor

nice

[fallow needle]

a?

[timid silo]

$a\left(2/3\cdot a-4\right)=0$

[warm shale]

😔😔😔😔😔😔😔

[timid silo]

yep

now u have 2 factors

u can use zero product property

[fallow needle]

Much love man

U a GOAT

[timid silo]

thx

,w a(2/3*a-4)=0

[warm shale]

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[timid silo]

btw those r ur solutions

ok bye

[fallow needle]

Bye man

Thanks

[timid silo]

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[golden bone]

I have a question that I'm not sure how to word, it involves coordinates

[golden bone]

How do I transform coordinates so that the origin is at the center of one side of a square?

Let me try to make a visual example

[fallow needle]

So you mean mirror the coordinate?

Flip it?

[kind depot]

I’m fairly sure it’s just a translational transformation from the wording

[fallow needle]

Probably I guess

[kind depot]

No way to know for sure until we get a visual

[golden bone]

I am trying to programmatically place objects in a 2d plane, and I'm doing that based on their position in a 2d array, so [0, 0] would be at (0, 0) on the graph. I want to make it so that [0, 0] in the 2d array is placed half the width of the square over. If this even makes any sense 😅

So it would normally be like it is on the left, but I want it to be like it is on the right

[kind depot]

Is one of the corners already at the origin?

[golden bone]

I'm not sure what you mean

[kind depot]

If you have individual coordinates that correspond to locations, you would just need to translate each coordinate over half the width of the square

If that isn’t what’s happening then I don’t think I have the time to figure out what you’re doing unfortunately, I’m about to board a plane

[golden bone]

The problem I am facing is I want to be able to scale this square, and have the red point stay in the same place in the world space

I'm really bad at explaining

And normally if I scaled the square, the only point on the square that would stay in the same place would be in the corner. I want it to be halfway along one of the sides

[kind depot]

Are you trying to make a function that makes [0,0] = (W/2,0)?

[golden bone]

I think so, yeah

[kind depot]

That is a pretty interesting problem

I’m not sure if I have the time to figure this out

I feel like the easiest method would be to redraw the square such that the corners sit at (-W/2,0),(W/2,0),(-W/2,W), and (W/2,W)

If you have the exact width of the square then you can substitute it in for W, and you can add a multiplier c to each corner to scale it quickly

I’m assuming you’re coding it?

[golden bone]

I think I figured it out

I'm working with only odd widths, so [x, y] in the list would be at (x - (W/2 + 1), y)

I forgot to account for the scaling of the pieces

[kind depot]

Ahhh, alright

[golden bone]

Sorry for wasting your time 😅

[kind depot]

No worries

The time I have was made for spending

[golden bone]

I think I might be good for now, I'll be back if I can't figure out how to account for scaling lol

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[agile robin]

i dont understand top line

hows he getting 1/2 and -1/2

on right side

can someone explain

[nocturne minnow]

Partial fractions decomposition

[wild swallow]

is this a knockoff blackpendredpen

[agile robin]

LUL

[agile robin]

prob best i do it the other way, its easier anyway

ty

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[upbeat willow]

hi

this is a Green's Theorem problem

i dont understand how to approach this

<@&286206848099549185>

[timid silo]

Have you looked at the definition of the theorem?

@upbeat willow Can you identify L and M?

[upbeat willow]

yes

need help with the modulus thingies

to determine the limits for the double integral

[timid silo]

Are you familiar with desmos?

[upbeat willow]

hmmmm

yes but didn't think of it.

do I just break these down into 4 fragments?

and solve for line segments

[timid silo]

Worst case yes, sometimes you can identify a symmetry that would let you just multiply one quadrant by 2 or 4 to get the answer.

[upbeat willow]

hmm

alright

thank you very much, ill try this.

[timid silo]

Are you familiar with even/odd functions?

[upbeat willow]

yes

i've done line integrals for line segments from a coordinate to another coordinate with parameterization

will try that by taking these 4 coordinates of the rhombus

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[carmine surge]

i think i did everything here right but i’m just stumped with number 4

i think it’s simple but the answer won’t cross my mind

[warped agate]

congruency

RS=QU
QUT=RST
QTU=RTS

by AAS Congruency , RST =~ QUT

so by CPCT(Corresponding Parts of Congruent Triangles) , QTU=RTS

[carmine surge]

hmmm alright ill try it

u were wrong it was already given

[warped agate]

bruh

the reason for 4

u write as

Corresponding Parts of Congruent Triangles are Equal

CPCT

[carmine surge]

no it was given because we were given that information from the picture

thanks anyway

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[timid silo]

can someone explain why these two equations look so similar but the graphs go in opposite directions?

[nocturne minnow]

The slopes

It's perpendicular to each other, I believe

[timid silo]

oh ok thanks

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[wild swallow]

they definitely dont look perpendicular

[nocturne minnow]

Looks can be deceiving, but they are perpendicular

[wild swallow]

uh

no

they arent

,w {2, 3} . {2, -3}

[warm shale]

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[nocturne minnow]

Unless I did something wrong

[wild swallow]

their normal vectors arent perpendicular

one is a reflection of the other through the vertical axis

[zenith raft]

doesn't that show they aren't perpendicular lol

[serene dawn]

L

[nocturne minnow]

Oh wait

Wrong math

Neg recipical

Oops

[obtuse pebble]

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[swift root]

could i get some help with this?

[serene dawn]

Have you graphed it to help with intuition?

[swift root]

yes i have

[serene dawn]

So which equation is at "the bottom"

As in the lower bounding half

[swift root]

the z=x^2+y^2

because the the second equation starts at z=8

[serene dawn]

Yeah so subtract that from the other equation to get an equation that singularly described your solid

[swift root]

i did all of that and got 256/3pi

but answer is 64/3

pi

i converted to polar and everything

[serene dawn]

Can you show me please

Also are you allowed to integrate using volume of revolution?

Might be easier

[swift root]

ive never been shown how to do it

oh

thats calc 3

2

isnt it

[serene dawn]

I dont know the US calc system sorry

Can you show me your working

[frosty spoke]

set up the integral in whatever coordinate system you want to use first

and then you can always change coordinates later

[swift root]

i set up in xyz and converted into polar bc i dont feel like doing trig suvs

[frosty spoke]

this one lends very nicely to cylindrical coordinates I think

[swift root]

subs

i mean

cylindrical lol

[frosty spoke]

so what did you get for the integral then?

[serene dawn]

^

[frosty spoke]

okay

not the value

the actual integral

like whatever you set up

a number is meaningless

[swift root]

sec

[serene dawn]

I already asked for their working

[swift root]

boxed it

https://media.discordapp.net/attachments/1025579701621756056/1051355594214342757/IMG_0090.png?width=901&height=676

unless theres a typo in the key that is how it should be set up

[frosty spoke]

that doesn't seem right

[swift root]

feel free to check my work idrc

[frosty spoke]

volume is always a triple integral of 1

because each unit volume is literally counted equally to any other unit volume

(well, after you add in the Jacobian, it might be r)

[swift root]

forget everything else what boxed is what i concluded was the final integral

[frosty spoke]

but certainly not (8 - 2r^2)

yeah and what I'm saying is that's not correct

[swift root]

feel free to show me where im wrong

[frosty spoke]

I just told you

[swift root]

ur telling me but not proving it

[frosty spoke]

the volume is always a triple integral of 1 in Cartesian coordinates, and when you change to cylindrical, it should just be an integral of the Jacobian of the change of coordinates

prove what?

[swift root]

.close

[obtuse pebble]

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[barren vapor]

.open

how do I solve this

[nocturne sun]

I'll just give two examples,
f(2) means you look at the table for f and find 2 in the x row. This corresponds to a -3

f^{-1}(-3) means you do the same but look for -3 in the other f(x) row

so you get 2

also note that (fog)(a) = f(g(a))

[barren vapor]

so how do I calculate (f o g)(2)

[nocturne sun]

f(g(2))

[barren vapor]

yeah but then that's where I get confused

[nocturne sun]

you do it inside out

[barren vapor]

am I supposed to look for the x value of 2 in the g(x) section

[barren vapor]

I don't understand how to do it inside out tho

[nocturne sun]

think of what f and g are

or what functions are

they take some value x, do something, and spit an f(x) out

so you can draw (fog)(3) like this

the number to the left of the arrow is on the x row, and to the right is the f(x) or g(x) row

[barren vapor]

what's this topic of algebra called

I need to know so I can understand it better

[nocturne sun]

functions

[barren vapor]

yeah I know but what kind

like here I'm trying to find the values of a function from a table

I just answered my own question

fuck me bro 😐

[nocturne sun]

it's just the concept of functions in general

the first part I guess, if it's split into parts

[obtuse pebble]

@barren vapor Has your question been resolved?

[barren vapor]

eh

not really

but thanks for trying to help @nocturne sun

[spring sail]

composite functions

[obtuse pebble]

@barren vapor Has your question been resolved?

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[timid silo]

Is (2n+1)pi/4 and (4n+1)pi/4 same when it comes to writing trigonometric general solutions?

[timid silo]

if yes how?

[mighty geyser]

consider 3pi/4

[timid silo]

👍

[mighty geyser]

usually n is taken to be an integer

[timid silo]

right

yes

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[alpine panther]

where is the twerking bee

[obtuse pebble]

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[red hearth]

how can i solve this

[red hearth]

i found radius $arcsin(x^2)$

[warm shale]

TufanBob Kare Pantalon

[red hearth]

then i used disk method which lead me to this: $\int_0^1 \pi * \arcsin^2(x^2)$

[warm shale]

TufanBob Kare Pantalon

[red hearth]

and it didn't work

<@&286206848099549185>

[timid silo]

@red hearth

[red hearth]

yes?

@timid silo good, your's?

???

bro, why

...

bruh

[timid silo]

Hey bro i am reallyyyyyy sorry

Ts my friend typing

[red hearth]

okay np

[timid silo]

And being mad

I am really sorryyyyy

[red hearth]

okay, okay np

[timid silo]

It was a guilt for me 😞

[timid silo]

Was this your problem?

[red hearth]

yes

[timid silo]

Ok one sec

[red hearth]

<@&286206848099549185>

[pallid canyon]

Read the problem again, you are revolving around the x-axis

So the radius isn't arcsin(x^2)

[red hearth]

then what is the radius?

[pallid canyon]

Do you have any ideas?

[red hearth]

no

my brain stopped working

[pallid canyon]

lmao ok

Do you have a drawing of the graph?

[red hearth]

yes

[pallid canyon]

Okay

So since you want to use disk method, how do you think the disks would look on the graph?

[red hearth]

something like that

[pallid canyon]

You sure?

We are rotating around the x-axis

[red hearth]

x-axis??

it says y-axis

[pallid canyon]

read the problem carefully again

area is bounded by the y axis

[red hearth]

my english is not enough

...

bruh

i'm struggling with this question for like 2 hours

thanks stranger with anime pp!

[red hearth]

let me try this way

it didn't work

[pallid canyon]

oh btw if you use disk method you need to do some extra steps to get the volume

[red hearth]

i found pi

which steps

so i first find the radius

which is $\sqrt(sinx)$

[warm shale]

TufanBob Kare Pantalon

[red hearth]

then i used disk method $\int_0^(\pi/2) \pi \sin x$

[warm shale]

TufanBob Kare Pantalon

[pallid canyon]

this is the area you're rotating around the x axis

[red hearth]

yes

oh

it has a whole

[pallid canyon]

Yeah you need to take the volume of a cylinder minus the volume you got with disk method

[red hearth]

let me try

but wait

how can i get the volume of the cylinder?

[pallid canyon]

volume of cylinder = pi * radius^2 * height

[red hearth]

but which cylinder

[pallid canyon]

purple area rotated - red area rotated = yellow area rotated

[red hearth]

so i need to rotate y=1

get the area

then substract it from pi

am i right?

[pallid canyon]

Yeah

[red hearth]

okay let me try it

let's goooooooooo it worked

thank you so much @pallid canyon

[pallid canyon]

aye np :)

[red hearth]

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[obtuse pebble]

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[timid silo]

Confused with this question. Do i just list the set of the vertices $V_1$ and $V_2$? With them being:
[
V_1 = {v_1, v_2, ..., v_n}\text{ } (n \geq 1)
]
[
V_2 = {v_1, v_2, ..., v_m}\text{ } (m \geq 1)
]

[warm shale]

♡LexQa♡

[royal basin]

not like that

[zenith raft]

what are V_1 and V_2?

[timid silo]

just the set of vertices i suppose

[timid silo]

what's off?

[royal basin]

your V_i list the same vertices

[unreal spire]

9^x = x solve for x

[timid silo]

oh sorry

[
V_1 = {v_1, v_2, ..., v_n}\text{ } (n \geq 1)
]
[
V_2 = {u_1, u_2, ..., u_m}\text{ } (m \geq 1)
]

[warm shale]

♡LexQa♡

[royal basin]

@unreal spire open your own channel

[unreal spire]

i dont think i can

[loud tangle]

#❓how-to-get-help

[royal basin]

yes you can.

[unreal spire]

i got it thanks

[timid silo]

okay i think i got it, thanks

.close

[obtuse pebble]

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[ocean kestrel]

Are you allowed to solve the problem like this?

[royal basin]

do you have some doubts about a particular step?

there is nothing here that is wrong

[honest chasm]

i suck at math pls help

[royal basin]

@honest chasm open your own channel. read #❓how-to-get-help

[warped agate]

#❓how-to-get-help

[obtuse pebble]

@ocean kestrel Has your question been resolved?

[obtuse pebble]

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[vital frost]

can someone help me understand why x is sin n y cos

[timid silo]

There are.... a lot of stuff in that pic. You should probably specify what you mean

[drifting loom]

well those r components

[kind depot]

He’s asking why the horizontal vectors are represented as sin and the vertical ones are represented as cos when it’s usually the other way around

[timid silo]

The frame of reference is rotated and reflected i guess (from your usual form), try drawing triangles with those angles and write them into component form

[kind depot]

Basically the reason why it’s switched from the norm is due to how the angles are set up. Since sine is opposite over hypotenuse, it becomes the easiest way to represent the x direction with the given angles. Looking at T2, with T2 being the hypotenuse, the opposite side from the angle is the x direction, meaning the value of the side would be T2sin(theta)

[vital frost]

yh exactly

just posting again so i can look at pic n read ur comment

[kind depot]

It’s the same reason why the y direction is cos. If you look at T1, the adjacent side is the y direction, so the vertical component is T1cos(theta).

[vital frost]

u mind annotating on this

i strugggle with angles

[kind depot]

Give me a minute to grab my laptop

[vital frost]

thanks

[kind depot]

No problem

Force vectors can get confusing

[vital frost]

they can as always was told cos is x n sin y

as circles show

[kind depot]

Sometimes we need to break from the norm to solve problems

[vital frost]

i see

[kind depot]

Booting up laptop, then I’ll annotate the image

[vital frost]

thx

[kind depot]

I've annotated the steps for T2

[vital frost]

thanks as will be same for both

[kind depot]

I can't find where it saved to one second

[vital frost]

nw

[kind depot]

Sorry about the handwriting, I don't have a good way to draw on a laptop

[vital frost]

yh thats much clearer

[kind depot]

So in future problems, remember to look at the angle to figure out which direction is opposite and which direction is adjacent

[vital frost]

u familair with eignevalues/eigenvectors... simple question i have

[vital frost]

yh

[kind depot]

Somewhat familiar with those, although it's been a little bit since I've reviewed them

[vital frost]

do u kno why i divide by 2

obviously in the column vector.. is it to get that unity on rhs

ill share notes of this section

@kind depot thanks

[kind depot]

It does appear from the notes that dividing by 2, which is X3, would return the vector to unity

[vital frost]

but could i use 2.3232 also?

like how do i know what one to turn to unity

[kind depot]

It looks like you only need to turn X3 to unity

[vital frost]

ok

final question about earlier problem

why sintheta1 = x1 /l1

[kind depot]

l1 would be the hypotenuse

[vital frost]

as it looks like x1 is in x direction and l is y

oh yh its the solid line rather than dotted

right

thx

[kind depot]

No problem

[vital frost]

.closw

.close

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[timid silo]

Help me solve this geometry homework please

[obtuse pebble]

@timid silo Has your question been resolved?

[loud tangle]

u asked the same question twice

[obtuse pebble]

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[vital frost]

can this be done on a calculator.. does anyone know

[unreal musk]

Finding those roots? On what calculator?

[fierce lagoon]

Graph it

[unreal musk]

Some calculator models do, some don't...

[fierce lagoon]

On your TI-84

[vital frost]

casio fx 991ex

[fierce lagoon]

No idea

We do provide a calculator

[unreal musk]

I think it does, if it's the one I'm thinking of...

Send a picture of it please?

[fierce lagoon]

,w NSolve[2.5x^3 - 15.5x^2 + 22.4x == 5.4, x]

HA the bot's dead, I forgot

[unreal musk]

Yeaaaa, pain 😭

[unreal musk]

Try shift solve?

[vital frost]

cos i dont know how to do my hand for x^4

or 3 roots rather

[unreal musk]

(I had a calculator emulator of that back in the day, but my subscription ran out haha!)

[vital frost]

lol

[unreal musk]

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

[fierce lagoon]

Okay well you can always use desmos

[vital frost]

not in exam hall

prepping for exam

[fierce lagoon]

I see

[vital frost]

thanks

.close

[obtuse pebble]

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[cyan tinsel]

ik this sounds really dumb but how would I simplify 2sqrt5•5sqrt5

[cyan tinsel]

$$2sqrt5•5sqrt5$$

smh

[narrow fiber]

Use the fact that √a√b = √ab

[cyan tinsel]

so it’s root25?

10root25?

[narrow fiber]

Yes

[cyan tinsel]

says it’s incorrect rip

[narrow fiber]

(2)(√5)(5)(√5)
By rearranging
(2)(5)(√5)(√5)

10(√5)(√5)

What's (√5)(√5)?

[cyan tinsel]

5right?

[narrow fiber]

Yes

[cyan tinsel]

and then what do I do with that

10x5?

[narrow fiber]

= 10(5)

Yeah

[cyan tinsel]

oh it is

wow

cheers

[narrow fiber]

[cyan tinsel]

.close

[obtuse pebble]

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[spiral tulip]

does anyone know?

[spiral tulip]

i see

let me try to solve it

i have gotten it wrong

i dont understabd

is it 110?

[fierce lagoon]

What's the measure of arc IE

[spiral tulip]

i do not know

110?

[fierce lagoon]

Arc measures of a circle add up to 360

Yes, IE = 110

[spiral tulip]

yes i have gotten to that part

then divided by 2

[fierce lagoon]

How it says, specifically, "the measure of the angle formed by two tangents ... is half the difference of the measures of the intercepted arcs"

So you didn't do the difference

What is the difference between the arcs

[spiral tulip]

55?

[fierce lagoon]

Does 250 - 110 = 55?

It's asking for the difference

That's subtraction G

[spiral tulip]

i see

140

i have gotten 140

[fierce lagoon]

And then it says that the angle is half of that

So what's half of 140

[spiral tulip]

hm

70?

how did you kno what to do

@fierce lagoon

[fierce lagoon]

I read

[spiral tulip]

is that what thethoerem is called?

i will look into it

?

[obtuse pebble]

@spiral tulip Has your question been resolved?

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[storm heart]

the question is not necessarily hard

i just dont get how i will see that g(x) = 2/3

g'(2) = 2/3 *

yes

[high lily]

you could draw a tangent line (trying to be precise) in order to deduce the intended value

[storm heart]

yeah thats what i did

the thing is that i think its hard on the computere

so i get many different answers

i didnt know if there was a way to see it without drawing a line

[high lily]

not really other viable options

[storm heart]

okay then. thanks

.close

[obtuse pebble]

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[wicked knoll]

In this image above, the EPE is expected prediction error where X and Y are random variables. And they have a joint probability distribution Pr(X, Y). What does the Pr(dx, dy) mean and do in this integral above?

[obtuse pebble]

@grand sky Has your question been resolved?

[grand sky]

<@&286206848099549185>

[unreal musk]

Linear spaces being vector spaces yea? @grand sky

[grand sky]

yeah

for a) i worked through the axioms and I think it is but i'm not 100% sure

is it valid to say every element is in the form (a,a+nk) and go from there

[unreal musk]

From a quick look, I believe it is too!

Hmm the way I do it is to just go through the axioms one by one tbh

As you don't yet know if it's a vector space to have a basis for

[grand sky]

for b) how would i go about finding an additive inverse

[unreal musk]

But I believe many of them just follow from the real number properties?

[unreal musk]

Well, for, say, ar^{n-1}, wouldn't -ar^{n-1} do? It is a geometric sequence after all

[grand sky]

is it just showing if i make both components negative it's still in the set

yes that's what i was wondering, ty!

[unreal musk]

Hmm, but then would a linear combination of geometric sequences still form a geometric? That's what I'm thinking, hmm

[grand sky]

Same

[unreal musk]

Hmm, well 1 is a geometric sequence, and so is, say, 2^n, maybe that might form a counterexample...

Whether 1 + 2^n has a common ratio...

[grand sky]

you mean 1,1,1,1,1,1

[unreal musk]

(still thinking this live btw)

[grand sky]

right

[unreal musk]

Yea, first term and common ratio being 1

[grand sky]

it can't right if you just look at 3 consecutive terms

9/4 != 17/9

[unreal musk]

Ah yea, I was being slow and was gonna do it the "general" way, you could have just done that 😂

You right on that 😂 beat me to it!

[grand sky]

are you college level?

im just wondering when I would formally cover these things at college

[unreal musk]

Graduated ages ago 😂

[unreal musk]

I think by first year, you should get an intro to sequences and vector spaces pretty early on!

Personally believe the last one does form a vector space tho? (again by quick eyes)

Any linear combination of sequences that satisfy that, will also themselves satisfy that I believe

[grand sky]

yes and you can just make an additive inverse by negation again so i think it works out

[unreal musk]

Yep, that should be it! But of course, I'd say go through all the axioms and say why each are true!

[grand sky]

yep

[unreal musk]

Ahh, nice nice!

[grand sky]

for this one

can i say that as different basis vectors are linearly independent, if one is a subset of another they are not linearly independent - and so they must be the same basis vector

[unreal musk]

Hmm...

Not sure I see what you mean tbh

At least in the finite dimensional case, I'd just say that both being bases means they have the same size, and one being a subset of the other immediately forces them to be equal (else if not, you'd have something in one that you don't have in the other, contradicting the previous statement I made wrt them being the same size)

[grand sky]

hmm

i think i understand

[unreal musk]

Basically [ha!], if two finite sets have the same size and one is included in the other, then they must be equal, yea?

[grand sky]

OHH

very based

[unreal musk]

https://tenor.com/view/basedgod-gif-8096204

[unreal musk]

What you thinking for these?

(If you need me reply to this one or @ me)

[grand sky]

im thinking for r over r

it has to be 2 right

wait hmm

would it actually be 1

since one base takes you to every real

[unreal musk]

Yep, e.g. {1} forms a basis for the real numbers over the real numbers

[grand sky]

r over c would again be 1 right

waitttt

I'm not entirely sure

because you only need 1 basis for the reals

but is the dimension what's required to navigate your vector space

or what's required to navigate your field

[unreal musk]

Hint: ||is R a complex vector space?||

[grand sky]

no i dont think so

[unreal musk]

(tysm for teaching me how to do that btw )

[unreal musk]

Why not?

[grand sky]

wait

R is a subset of C

but R has no complex part

so there is no imaginary basis needed

so there's only the real basis?

[unreal musk]

Hmm, there's better wording for that...

Hint 2: ||can you find a vector space axiom that fails?||

[grand sky]

I'm not sure on that

can't you say if R was a complex vector space 'i' would be a basis but it isnt

[unreal musk]

Hmm, don't think that makes sense tbh

Anyways... take F = C, V = R

[grand sky]

this one

[unreal musk]

There we go! e.g. i is in C, 1 is in R, but i*1 = i is not in R

So you don't have scalar multiplication!

[grand sky]

ill try to do c over r on my own and I'll @ you once ive made a solid attempt

2 i think

both operations work

and 1 and 'i' work as bases

and 2 for C over C surely since 1,i are bases

and both operations work

@unreal musk

[unreal musk]

Yep, {1,i} is a basis for C over R (or 1 and i are "basis vectors" for C as a real vector space)

[grand sky]

polynomials I think for deg<=n it would be n+1

since linear combinations of x^n from 0 to n form every polynomial

so those can be the bases

[unreal musk]

Agree, facts 📠

[grand sky]

for greater than n

just >=n+2 i guess

[unreal musk]

Hmmm...

[unreal musk]

Well, that's technically true 😂

Can you find me a basis for that space? How many elements does it have?

[grand sky]

x^k s.t. k>=n

[unreal musk]

Yep, how many elements are in that set?

[grand sky]

k+1

[unreal musk]

Hmmm, not quite...

(And k is your "dummy variable" here)

You start from k=n... do you ever stop?

[grand sky]

no

so it's countably infinite?

[unreal musk]

Yep, it's infinite dimensional! (and clearly, that becomes the answer to the space of all polynomials...)

[grand sky]

for c) it should also be infinite right

since x(x-1)(x-a)(x-b)...(x-z) satisfies the conditions

and the polynomials with factor x(x-1) is also infinite

[unreal musk]

I think it is, personally, for both parts

[grand sky]

periodic functions can also be done with polynomials

since any polynomial composed of the trig functions is also periodic

so you can just use what we said early to say they're infinite (i think)

[unreal musk]

What do you mean by that sorry?

(This one I'm actually a little bit stumped on tbh )

[grand sky]

okay so

let's say we have a periodic function like sinx

it has period 2 pi

any polynomial composed of sinx

eg

Asin^2x+Bsinx+C

also has period 2 pi

hence

for any periodic function with period T

all polynomials of that function also have period T

[unreal musk]

That's true, I agree, but that doesn't say much about any periodic function, no?

[grand sky]

If it's infinite for one periodic function

its all good

[unreal musk]

Hmm, yeah, got you!

See where you're going with that!

[grand sky]

I don't really get part 5

what is a geometric vector

[grand sky]

luckily I don't have to prove anything rigorously so word explanations are fine

[unreal musk]

Think they just mean R^N there, you know? You know the normal usual vectors you think about in geometry

(Damn maths, I was gonna say "normal", then realised that also had another meaning in this context 😂 )

[grand sky]

omg michael penn is also doing linear spaces

on a livestream

[grand sky]

oh yeah like the norms of vectors

or normal like perpendicular

i think it would just be N dimensions then

[unreal musk]

Yea that 😂 whenever I have a conversation about maths, I always end up having to be careful

[unreal musk]

Yea pretty much, don't think there's a trick to that other than it literally just being the N dimensions

Probably put it there to have you fried 😂

[grand sky]

This is part of an online maths school for grade 11s

They're sadists

[unreal musk]

Hmmmm, part of me still feels a little bit off about this one tho

[grand sky]

any periodic function for T has an infinite family of polynomials of any degree that is also periodic with period T

[unreal musk]

Actually, no, I think it's fine, I'm thinking of Fourier series but those aren't finite either

[grand sky]

ohh I think I've heard of those

i think you've helped me through practically everything

this one isn't hard it's just a bunch of the same thing so i think ill be okay

i can show linear dependence by expressing one as a linear combination of another pair

so it's cool

[unreal musk]

Yea it's just that, set $\alpha_1 \mathbf{v_1} + \alpha_2 \mathbf{v_2} + \alpha_3 \mathbf{v_3} = \mathbf{0}$ and see whether the $\alpha_i$'s have to all be nonzero or not

[warm shale]

chartbit

[grand sky]

thank you so so much

lifesaver honestly

[unreal musk]

Hahah no worries, and thank you (and @timid silo) for teaching me that spoiler thing 💜

[timid silo]

It's nothing ahahha 💙

[obtuse pebble]

@grand sky Has your question been resolved?

[obtuse pebble]

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[runic rover]

This is Geometry

How do you find the x

[obtuse pebble]

@runic rover Has your question been resolved?

[obtuse pebble]

@runic rover Has your question been resolved?

[obtuse pebble]

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[frail shell]

how do I solve something like this?

[cedar lichen]

What does the notation (x, y) mean?

[frail shell]

horizontal and vertical

[cedar lichen]

But what does it mean in terms of a function?

For example, if y = x², then (2, 4) is on that graph. Why?

[frail shell]

mmm

idk

[cedar lichen]

Do you know how to graph functions?

[frail shell]

yeah

[cedar lichen]

How would you graph a function?

[frail shell]

just graph like xy

x first

y second

x horizontal

y vertical

[cedar lichen]

But what if you're not given points? How would you graph y = x²?

[frail shell]

its alright

im going to try to find a youtube video or something

[cedar lichen]

You don't know how to graph y = x²?

[frail shell]

I dont remember

[cedar lichen]

f(x) = y means (x, y) is on the graph

(2, 4) is on f(x) = x² because f(2) = 2² = 4

[frail shell]

oh

okay

btw

how is this incorrect?

.close

[obtuse pebble]

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[timid silo]

is it possible to make the x sqrt ln(x) = u ?

[haughty coyote]

Hint: 2 u-subs in a row

Both simple

[haughty coyote]

Try it

[timid silo]

wdym by in a row ?

[haughty coyote]

One after the other

[timid silo]

ok so I make this the u

[haughty coyote]

Yours probably works, but if I can, I'd rather spare myself computing the derivative of your expression

[timid silo]

is it possible to split it in 2 ?

1/ x

1/sqrt ln x

[haughty coyote]

Actually I'd expect sqrt(ln x) to work

Not what you said

[timid silo]

that implies I have to split it into 2 fractions

[haughty coyote]

It's already split into x and sqrt (ln x)

[timid silo]

ok

by doing u = sqrt ln x

i got integral 1/u du 2 sqrt ln x

the sqrt ln x I need to u substitute it too?

is this the second u sub ?

[haughty coyote]

That's the "complicated" u-sub actually

Which does it in one go

[timid silo]

ok

but it seems impossible from the looks of it

as there is another sqrt ln x

[haughty coyote]

That's literally 1/u du 2u

Which is just 2du

[timid silo]

du * u is du ?????????????

[haughty coyote]

1/u * du * u = u * 1/u * du = du

[timid silo]

du = 1/ 2x sqrt ln x

[haughty coyote]

So then

That's basically your integrand

[timid silo]

my integrand is

[haughty coyote]

It's clear then no ?

[timid silo]

being honest no, I am more confused now as I though we could not do these manipulations

2du just removes the integral sign ?

[haughty coyote]

No

Of course not

[timid silo]

ok

[haughty coyote]

But now it's the integral of 2du

Which is just 2(length of the u-interval)

[timid silo]

ok so integral of 1/x sqrt ln x

du is 1/ 2 x sqrt ln x

2 times that

removes the 2

[haughty coyote]

Don't forget dx when you write that

[timid silo]

this is even worse then the first one

[haughty coyote]

?

First what?

[timid silo]

hold on

2 du is this

integral of 2 du

it hasn't changed anything

this was the original integral

[haughty coyote]

Once you've reached that, the integral is trivial

The integral of 2 du is the same as the integral of 2dx. It's just a dummy variable

[timid silo]

ah ok

[haughty coyote]

That's why you can write

$2 = \int_0^\pi \sin n dn$

[warm shale]

mateo713

[haughty coyote]

People are going to scream because n is a terrible choice for an integration variable, but it's technically correct

[timid silo]

its supposed to give

2(e^2) - 2(e) + c ?

since integral of 2 is 2x

and there is no u

only a du

[haughty coyote]

1. why +c in a definite integral
2. is that really your u-interval ?

[haughty coyote]

Does u go from e to e^2 ?

[timid silo]

oh yeah my bad no c

[timid silo]

there is no u

only the integral

goes from e to e^2

[haughty coyote]

In terms of x

[timid silo]

yes since the integral goes from e to e^2

[haughty coyote]

Changing your variables changes your bounds

[timid silo]

did not know that

[haughty coyote]

It makes sense that it should

[timid silo]

im mostly familliar with u sub and primitives

sorry for my ignorance

[haughty coyote]

Otherwise would such sentences really have sense ?

[timid silo]

nope

[haughty coyote]

When you visualize the integral, it's the variable going from one bound to the other

Changing that variable changes the values

[timid silo]

ill have to check what value it changed to

I got it

Thanks

[haughty coyote]

Sqrt(2) - 1 ?

[timid silo]

2 sqrt 2 - 2

[haughty coyote]

Yeah right

[timid silo]

👍

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[obtuse pebble]

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[runic rover]

How do I find x in this question? This is geometry

[raven badger]

what are the properties of said figure? (isosceles trapezium)\

[runic rover]

The base angles are congruent

So angle B = angle C, angle A = angle D

[raven badger]

and what is angle B in terms of x?

what is angle C in terms of x?

[runic rover]

I don't know how to find x

[raven badger]

could you please transcribe? im having a hard time reading that

[runic rover]

It's asking to find x using angle B and angle C

B = 7x+16
C = (x2+x)

[raven badger]

so equate them as B=C right

simplify the quadratic

[runic rover]

Yea

[raven badger]

solve for x?

[runic rover]

Can you explain them using steps

[raven badger]

7x+16 = x2+x

[rancid ember]

x^2 -6x + 16

-16*

[raven badger]

x2 - 6x - 16 = 0

use the quadratic formula

[tranquil sonnet]

You can factor, (x-8)(x+2)

X is ofc positive so x=8

[runic rover]

I didn't learn the quadratic formula

[raven badger]

thats not necessarily true? the angles i.e 7x+16 and x2+x have to be positive. -2 yields 2 (unit irrelevant) and 8 yields 72. am i missing smth?

[tranquil sonnet]

oh yea, mb

[raven badger]

factorize by splitting the middle term

or u dont know how to solve quadratics anyway whatsoever?

in which case i recommend you to look up a video

[obtuse pebble]

@runic rover Has your question been resolved?

[obtuse pebble]

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[carmine surge]

can someone check my answers

and i need help with number 4

[cedar mantle]

Yeah other 5 are correct

For #4 could you draw out which angles they refer to on the graph