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[raven haven]

Some books denote vectors differently

some even use [3,2]

Don't worry too much

as long as its not {3,2}

this is wrong

[elfin depot]

Alright thx

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[crystal nebula]

i need help on # 32

[crystal nebula]

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[vernal thunder]

Need some help

[vernal thunder]

,rotate

[vernal thunder]

The answer is in the bottom right hand corner though I don’t know how to get to it

<@&286206848099549185>

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[solar monolith]

I have solved this question but I would like to know how I can visualize what the shape of I would be

[solar monolith]

All I know is it is some shape in R^3

[full frost]

it’s just like a circle in 3d

hollowed out

use a 3d visualizer if you wanna see it

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[wary stream]

Determine if it opens up or down, to figure out if it has a max or min, then find that max/min point

[full frost]

make it into vertex form

[wary stream]

Not needed

$\frac{-b}{2a}$ is the x value of the vertex

[ocean seal]

dldh06

[wary stream]

Then you can use that to find the y value

The coefficients from your equation

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[rancid pewter]

Hi! Need help with #5

This is my working solution, I’m particularly doubting d-f

The final answer should be 1 million something

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@rancid pewter Has your question been resolved?

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@rancid pewter Has your question been resolved?

[rancid pewter]

Still need help when anyone is around, thanks in advance.

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[cinder spindle]

am i on crack or are these answers wrong

the second line

how did they get $33^k$

[ocean seal]

chromium

[stray estuary]

they're definitely on crack

not you

i think thats a typo

LOL

[cinder spindle]

bruh

so what, the entire proof is fucked then

[stray estuary]

mm

[cinder spindle]

actually it doesn't really matter towards the end i don't think

[stray estuary]

it looks like the third line they fixed the typo

just that one line

[cinder spindle]

yea

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[rancid totem]

how do you graph this kind of peacewise function?

[rancid totem]

like

is it read as y is equal to -1 if x+2 is greater than 0?

[kindred anchor]

peacewise function 💀

[rancid totem]

peace-wise?

[kindred anchor]

It is 'piecewise'

[rancid totem]

oh

misread

[kindred anchor]

It is given that name because the function behaves differently on different pieces (intervals).

[rancid totem]

yeah

do i just put 3 lines at y=1,0,-1?

then not included on these points?

<@&286206848099549185>

[alpine sable]

Well it’s the same function

So the line is the same

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[rancid totem]

wait wdym

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[plain quartz]

You draw your lines till your point when the inequality doesn't hold, which is -2 with open circle and at -2 you have a black dot

Would be kinda like a step

[rancid totem]

ah

like this?

[plain quartz]

Yeah, but the lines go to inf

The dot in the middle should be black, others should look like they are circle

[rancid totem]

so domain would be (-inf,inf)

would range be included or no?

r=[-1,1]?

[plain quartz]

The function seems well defined for every point on x so yeah, imo should be -inf +inf for the domain

Range I think should be union of sets that have 1, 0 and -1

Since in [-1, 1] you have values that the function never takes

[rancid totem]

oh

so its

r=[-1,0,1]?

[plain quartz]

Think so, yeh

[rancid totem]

just a quickie before i close the channel

[plain quartz]

Ja

[rancid totem]

is the range and domain correct here?

since its a circle would it be correct to assume not included?

[plain quartz]

Yeah circle = not included / excluded, dot = included, seems correct to me

[rancid totem]

so the parentheses and brackets on the r and d are correct?

[plain quartz]

Given the circles, yeah

[rancid totem]

is a dot in a circle a thing?

or just included?

[plain quartz]

Tbf haven't seen dot in a circle in such things, just full circles and open ones

[rancid totem]

cause like two of the functions overlap on y2

one has greater than and the other has greater or equal

[plain quartz]

Oh ye, just saw that, in that case the point is included yeah

[rancid totem]

tysm!!!

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[vale beacon]

I don't really understand how to approach this... ^^; (deleted q)

[mortal trellis]

what if they were in R^2

[vale beacon]

Pardon? ^^;;

[mortal trellis]

it says they are in R^3

what if instead they were in R^2

[alpine sable]

what does "any pairs of them" mean here

[mortal trellis]

same question, just this thing changed

v1, v2 lin independent, v1, v3 independent and v2, v3 independent

"pairwise"

[vale beacon]

I'm not sure about that, either...

[vale wigeon]

do you know what linear independence means

and do you have a geometric picture in your mind for it

[vale beacon]

I vaguely understand linear independence, but I don't really have a visualisation.

[vale wigeon]

so first off, there's at least one thing you should try proving on your own, consulting the formal definition if necessary:

if a set of vectors contains the zero vector, then it is always linearly dependent.

what follows will thus concern sets of nonzero vectors only

for two vectors, linear independence is equivalent to them not being parallel

for three vectors, linear independence means not only that no two vectors are parallel, but also that the three vectors don't all lie in the same plane

(or to put it differently, no vector should lie in the plane spanned by the other two)

[vale beacon]

I see...

So, the answer to the question would be no then, right?

[vale wigeon]

indeed the answer would be no, but now it is up to you to produce a counterexample.

[vale beacon]

Alright. So I just need to come up with a set of vectors in which two lie on separate planes and one is in the same plane as one of the other two?

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[alpine sable]

what are you having trouble with

[ivory coyote]

starting

I don't know where to start

[alpine sable]

do you know how to deal with a system of linear equations in general?

[fluid ingot]

Multiply the 2nd equation with -1

Then add the 2 equations

Find the value of x

Substitute that value in any of the equations and find the value of y

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[alpine sable]

Hi, is does the height of the box have to be at a certain height or can I put it at any height?

[rocky grove]

It can be any height I think. It does not affect the actual distribution

[alpine sable]

Oh Ok Ty for the answer.

😊

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[delicate ice]

I need to find palindromic sums for n

How would you approach it?

[pliant cedar]

this is from project euler is it not

[delicate ice]

yup

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[delicate ice]

<@&286206848099549185>

[prime badge]

do you just need to count them?

a number n has 2^(n−1) sums

[cold lotus]

where did you find this

[prime badge]

find?

it didn't exist before you asked lol

well op asked

[cold lotus]

oh:)

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@delicate ice Has your question been resolved?

[delicate ice]

Yup, only count how many sums exist

[prime badge]

we write 6 as
(3)(0)(3)
(2)(2)(2)
(1)(4)(1)
and 6

4+2+1+1=8
(the middle number is not decomposed, 0 means no number)

if it was 7 it should still be 8

(3)(1)(3)
(2)(3)(2)
(1)(5)(1)
and 7

so the answer is 2^floor(n/2)

[delicate ice]

oh there is also a requirement for these sum

every sum needs to contain at least one 2

so i guess i can't just count how many sums exist

but i need to produce all these sums and verify them

[prime badge]

lol no clue

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[neat imp]

What does this "n" mean?

[prime badge]

(a,b)

[alpine sable]

its the unit vector

[prime badge]

uh

[alpine sable]

it is perpendicular to the plane that a and b are given in, it is what signifies the direction of the cross product vector

[prime badge]

okay it also says |n|=1 why does it says (a,b) as well

[neat imp]

oh I see, so it means "normal vector"

[prime badge]

it's the notation for this perpenficular?

[neat imp]

yeah I guess

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[static bloom]

.close

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[frigid zinc]

Anyone has ideas
A dice is thrown 7 times. What’s the probability of getting exactly one ‘4’,
two ‘5’ and at least
three ‘6’ at the same time?

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[rancid pewter]

<@&286206848099549185>

My working solution is above, thanks in advance.

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@alpine sable Has your question been resolved?

[icy sun]

Use a venn diagram for calculators with defective batteries and defective cases. The total amount of defective calculators is the the total % inside the two circles

Should help you visualize

Yeah

ah ok

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[drifting island]

I don't know how to do the first part of this question.

[drifting island]

<@&286206848099549185>

.close

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[alpine sable]

Can anyine help with C(ii)

[lofty heart]

can you send the table with the values

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@alpine sable Has your question been resolved?

[alpine sable]

Okay

I found an answer and according to the answerkey its correct, but IDK if the way is correct

[lofty heart]

if you got a correct answer through calculations then its probably a right method

[alpine sable]

Oh OK thx I just needed to know Tysm for jelping

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[distant thicket]

what do I use to solve : ((x-1)^2)

[vague coral]

wdym solving ?

differentiate ?

[distant thicket]

idk whats called in english

let me draw in paint

i know i have to use the things above

but idk which one

google translate translates "taps" but i think its very wrong

derived functions i think

im not sure

not good at math in eng

neither in my lng

[vague coral]

which language do u speak ?

[distant thicket]

romanian

[vague coral]

ah sadge, i dont speak romanian

[distant thicket]

:/

[vague coral]

if i use english term, will u understand ?

[distant thicket]

i hope so

[limpid turret]

$\frac{d}{dx}[(x-1)^2]$

[ocean seal]

SWR

[wind cloak]

$\dv{x}$

[ocean seal]

NEONPerseus

[distant thicket]

thats not what im looking for

[wind cloak]

You work too hard SWR

[distant thicket]

thank you

❤️

[wind cloak]

Can you give us an example of what you're meant to do?

[distant thicket]

.close

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[wind cloak]

Wut

[limpid turret]

I don't know what we did but he got his answer

[wind cloak]

Btw I meant you can do $\dv{x}$ instead of $\frac{d}{dx}$

[ocean seal]

NEONPerseus

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[limpid turret]

$\dv{x}[(x-1)^2]$

[ocean seal]

SWR

[limpid turret]

.close

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[royal plank]

I disagree.

[wind cloak]

Okay.

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[tall path]

Im having trouble setting upthis problem

Here is the graph for reference

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@tall path Has your question been resolved?

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@tall path Has your question been resolved?

[lone heart]

@tall path Has your question been resolved?

[eager ridge]

hi

I'm gonna help you today

[tall path]

hello!

[eager ridge]

So what are you stuck on?

I teach college

[tall path]

So at the moment I am trying to figure out what the outer radius when this graph is revolved around the y axis

although the values that Im given are throwing me off. I thought it would be 1 or 6 but I wasmistaken

[tall path]

<@&286206848099549185>

[eager ridge]

sorry

went afk

[tall path]

It's okay no worries!

[eager ridge]

Ok let me solve it and I will help you

[tall path]

Okay tyvm

[eager ridge]

Ok

[eager ridge]

So R (x) is vertical of 4,50

[tall path]

So the height of the rectangle would be 12/x^2 - 4?

.close

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[solid wren]

ive been stuck on this for a few mins, T-T

[meager roost]

Try putting that as an equation form

With a and b the two unknowns

Then play around a little bit

[solid wren]

Would the equation be x+y=-9 and x-y=-25?

hello?

[meager roost]

Yes!

[solid wren]

okk

how would I solve it?

[meager roost]

Well, find an expression for x with one of the equation, and plug it in, in your second one

[solid wren]

would it be x=-9-y+-25?

or x=-9-y-25

[meager roost]

More x=-9-y then change that into the other one such as -9-y-y=-25
Then solve

The goal is to reduce the number of unknowns

[solid wren]

2y=-9-25?

[meager roost]

You forgot a minus

And added one

But you've got the idea

[solid wren]

o

2y=-9--25?

im sorry im rlly confusd

[meager roost]

Ok, gimme a second

The equation is like those old balance. Basically, you can pretty much do everything, as long as it's the same on both side

[solid wren]

ohhhhh

thats a 9 right?

[meager roost]

Yes

[solid wren]

y would be 8

y=8

[meager roost]

Yep

You got this

[solid wren]

then i plug that in?

to my x?

so -9-8?

i got -17

so y=8 and x=-17

[lone heart]

@solid wren Has your question been resolved?

[solid wren]

I guess yea

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[vapid steppe]

yo

[vapid steppe]

how do i know if i take the positive or negative answer

[marsh rapids]

Poorly written question. Depends on the domain. Since f is assumed to be invertible, only one is true. So the answer is "the one that ends up in the domain of f", whichever that may be. You can't know if you don't know the domain. You may, however, assume the domain is R+ by default here and therefore take the positive root

[solid wren]

If you are referring about the symbols , I think it changes /switches places when it’s being multiplied by a negative and a negative since it turns to a positive .

[vapid steppe]

so like if it said x≥0 what would you do

[marsh rapids]

Take the positive root

[vapid steppe]

How do you know

[marsh rapids]

Think

[vapid steppe]

is it like the domain of original is range of the inverse and vice versa

[marsh rapids]

Yes

[vapid steppe]

so like since the original is x≥0 so the inverse has range y≥0

[marsh rapids]

And since the domain isn't given, you can't find the range you should have

[marsh rapids]

Yes

[vapid steppe]

and the only solution that fits that is the positive one

[vapid steppe]

do i keep the x≥-2?

for my final answer

[marsh rapids]

Yes

To define the domain

Be better than the lows your exercices allow themselves to reach

[vapid steppe]

ok ty

.close

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[olive oar]

How would one prove Tensor-Hom adjunction?

[olive oar]

the only proof i can imagine is taking flat, injective resolutions to bimodules

[pure folio]

Hi everyone, any recommendations for a guy who's trying to understand taylor's theorem? in particular, 'bout its proof and an intuitive perspective of it

Ok

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[vapid steppe]

can someone give me practice problems for inverse functions with domain restrictions

[olive oar]

$$f\left(x\right)=\frac{x^2-3}{11x+7}$$

[ocean seal]

ohNoiAmHere

[vapid steppe]

is there a domain restriction given

[olive oar]

$x\in R\ -{-\frac{7}{11}}$

[vapid steppe]

?

[ocean seal]

ohNoiAmHere

[vapid steppe]

what does that mean

[olive oar]

reals without -7/11

[vapid steppe]

oh so like x≠-7/11

[olive oar]

yea

[vapid steppe]

ok let me try

[royal plank]

you know, you can just do inverse function tasks without domain restrictions, all they do is make it simpler for you

[vapid steppe]

yea but i have a test tomorrow that i need to put in the restrictions

[royal plank]

and if you have an option (such as x=sqrt(y)) you can do both ways instead

[royal plank]

what kind of restriction?

[vapid steppe]

like it’ll give me something like (x-2)²,x≥2 and it’ll ask me to find the inverse with domain and range

[royal plank]

most likely one that just makes it easier to choose between the inverts?
like, x^2 can be sqrt(x) or -sqrt(x)

[vapid steppe]

don’t i have to use calculus to find the domain restriction for this one

[olive oar]

no?

[vapid steppe]

@olive oar

so like do i take plus or minus

[olive oar]

graph + then -

they should both make a function

[vapid steppe]

wait isn’t the original function not one to one

[olive oar]

yea its not

but individually theyre both +/-

it depends what part of the function you wanna invert

[vapid steppe]

so is this right

[olive oar]

well, thats not a function

think about the whole point of domain restrictions

its to choose one part of the function

so you can invert

sinply put

you choose the restrictions now

[vapid steppe]

so like x≥0

[olive oar]

sure

now youre gonna have to find the inverted functions domain

[vapid steppe]

the inside of the square root has a domain of R

[olive oar]

as long as the interval you choose is one to one you can invert

and both functions are defined

[vapid steppe]

ok

[lone heart]

@vapid steppe Has your question been resolved?

[vapid steppe]

do i need to add a domain restriction

,rotate

[vapid steppe]

do i need to add a domain restriction

.close

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[somber kiln]

So I know that a^3 can be (a)^3 but Idk about the other term

Should I use the formula (x-y)(x^2 +xy +y^2)?

[heady pollen]

what are you trying to calculate?

[tawny condor]

I guess factor

notice that 2sqrt(2) = sqrt(8)

[somber kiln]

Im supposed to factorize this

[wind cloak]

Sure you can (a^3) - (b√2)^3

[tawny condor]

bruh

ok

[somber kiln]

Oh ok then how can I subtract a-V2?

[wind cloak]

?

[somber kiln]

(a-V2)(a^2+V2*a+V2^2)
this would be the formula

[wind cloak]

Yes

[somber kiln]

but im not sure wht to do next

[wind cloak]

Well that's it lol

You've factorized it

[somber kiln]

I can't simplify it more?

Alr I understood

thanks

.close

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[last rampart]

can someone explain how this works

how does 1/(1-y_0) become the bottom

[wary stream]

Multiply everything by $(1 - y_0)$

[ocean seal]

dldh06

[last rampart]

nvm yea I figured itout

ty

[haughty axle]

i dont get how to do this

[lapis nacelle]

first identify the basic function in each case. For example in d, the basic function (where the x is) is x^2. Then, what transformations are acting on it?

-3/4 does something to that x^2. Then, that +10 also act

watch this https://youtu.be/sTCRB6hMsC4

[haughty axle]

i didnt really understand this lesson im doing but ill watch the video and lyk if i understand

[lapis nacelle]

@haughty axle you also must watch https://www.youtube.com/watch?v=NrmmR3-VxA8

[haughty axle]

thank you

ill watch that after i finish the first one

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[queen prism]

using the limit definition of the derivative. Not sure how to tackle this question. Just need some guidance

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@queen prism Has your question been resolved?

[queen prism]

<@&286206848099549185>

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@queen prism Has your question been resolved?

[queen prism]

.close

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[tender crystal]

for sample space calculation like this

[tender crystal]

for example get 0 from the 2 sample space, and 6 from the 7 sample space outta 2 from the 9 sample space

how do you calculate it?

I am not understaning how they did the math

Anyone?

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[alpine sable]

Can someone walk me through this please? I cant access the instructional videos n its due tn i finished the first question but I can’t wrap my head around this one

[pulsar aspen]

Equivalent: what is the probability of two independent coin toss are both heads

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@potent echo Has your question been resolved?

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[trim brook]

I have a general idea on how to do this question but I need to do it using my BA calculator. Can anyone help?

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@trim brook Has your question been resolved?

[alpine sable]

i would like to ask a proving question

if p and q are positive integers prove that root p+ root q is an irrational number

[north hemlock]

bruh get ur own channel

and that claim isn't even tru, let p = 4 and q = 25

2+5=7

7 is rational

[alpine sable]

im sorry i dont know how to use this channel

[north hemlock]

did u read #❓how-to-get-help

[alpine sable]

let me check

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[whole shell]

it says you got the right answer?

well also pro tip, you can check this with desmos, the fact that you cant see the blue line is because its correct.

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[ruby copper]

Can someone teach me how to find the x-intercept of 2(3)^x using log?

[coarse olive]

it doesn't have one

[ruby copper]

ok

lemme change that

what about finding the x-intercept of 2(3)^x - 3 using log

[coarse olive]

set y=0

ull get 0=2(3)^x - 3

like how u normally find x ints

[ruby copper]

so it'll go to like

[coarse olive]

3=2(3)^x

[ruby copper]

1.5 = 3^x

[coarse olive]

yea

[ruby copper]

alr

how do i use log to finish it

[coarse olive]

then u can log

log 3

[ruby copper]

so it'll go like

[coarse olive]

log base 3 i mean

[ruby copper]

log3 1.5 = x

?

[coarse olive]

yea

[ruby copper]

ok thanks

answer should be..

around 0.71

@coarse olive confirm pls

[coarse olive]

0.37

[ruby copper]

wait what

[coarse olive]

log_3(1.5) yeah?

[ruby copper]

how would i put that into the calculator

[coarse olive]

what calc?

scientific

[ruby copper]

yes

it has log

[coarse olive]

right

im not sure then

[ruby copper]

alr

[coarse olive]

i used a graphic one

[ruby copper]

ok

ty for help tho

[coarse olive]

no prob

[ruby copper]

.close

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[alpine sable]

Trying to prove by induction

Kinda getting lost

Fixed the six

Bur still a bit lost lol

[whole shell]

havent done induction in some time, but if you add (n+1)^2 to the formula, shouldnt it alll simply to the same formula but wwith n+1 instead of n

[alpine sable]

Im getting lost in the simplification

[smoky hedge]

Try replacing n with n+1 in the equation and then separating the formula for n from it

also you can take n+1 common and you'll get 2n^2 + 7n + 6

[whole shell]

@alpine sable couldnt you do polynomial long division

since you know (n+1) should be a factor

[alpine sable]

Lol i haven't done that in years

Yeah i see the common factor now

Solved

Thank you for the common factor

.close

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[wanton beacon]

Isn’t angle abc 128?

[wanton beacon]

Am I rigbt?

*right

[rich quiver]

138⁰

Check your calculations

[lone heart]

@wanton beacon Has your question been resolved?

[wanton beacon]

But aren’t opposite angles congruent

[rich quiver]

Which two triangles?

[wanton beacon]

No, it’s a kite

I’m pretty sure exactly one pair of opposite angles is congruent in all kites

@rich quiver Isn’t that true?

In that case, it wouldn’t make sense for it to be 138°

[rich quiver]

I know that

[wanton beacon]

Right

[rich quiver]

Can you show me your working

[wanton beacon]

Well since EDC is 64

The opposite angle must also be 64

Right?

[rich quiver]

Yep

[wanton beacon]

64 * 2

Is 128

[rich quiver]

Well the oppt angle is angle EBC

[wanton beacon]

Wait no but

I’m talking abt this

Like how the opposite angles r congruent

[rich quiver]

Δ ΑDE IS CONGRUENT TO ΔABE by Rhs

[rich quiver]

Is this you agree?

[wanton beacon]

Yes

[rich quiver]

Angle EDC = 64 NOT ANGLE ADC

[wanton beacon]

Right but

Opp angles r congruent

[wanton beacon]

Like

Plz explain

[rich quiver]

Can you wait for 10-15mins

I am busy right now

[wanton beacon]

Ok

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[torn isle]

Can anyone tell me how to solve this question if I make substitution x=e^t?

Like how I transform this Cauchy ruler equation into linear homogeneous equation with constant coefficient in t and solve it?

<@&286206848099549185>

[vapid shuttle]

,rotate

[vapid shuttle]

are you wanting to solve the linear using that substitution or make that substitution to transform it into a linear equation

and to confirm, you don't want to use the method that it gives as a solution for what reason?

[torn isle]

I want to solve this by making the substitution x=e^t

[vapid shuttle]

gotcha

so substitute x=e^t and then from there transform it into linear

with the linear how do you plan on solving

should it be seperable

[torn isle]

Nah, in the notes, it ask us to transform by making substitution

[vapid shuttle]

okay

if x=e^t

x^2=e^(2t)

[torn isle]

Like an alternative method to Cauchy Euler equation

[vapid shuttle]

,,e^{2t}y''+e^{t}y'+y=0

[ocean seal]

AustinU

[torn isle]

Yea, I did something like that, then I divide everything by e^2t

[vapid shuttle]

yes I think that would come next

[torn isle]

Then I don’t know how to continue 😅

[vapid shuttle]

,, y''+\frac{1}{e^{t}}y'+\frac{1}{e^{2t}}y=0

[ocean seal]

AustinU

[vapid shuttle]

well now we have a second order linear homogenous ODE

[torn isle]

Yea, but for this, normally we know that one of its solution is y=e^rx

So in this case, I sub x=e^t or I directly change y=e^rt?

[vapid shuttle]

one second

just thinking

if we substitute y=e^(rt)

then y' and y'' are easy to find yes

r^2e^(rt) and re^(rt)

so we can make all of our subs

and left with

,, r^{2}e^{rt}+\frac{1}{e^t}re^{rt}+\frac{1}{e^{2t}}e^{rt}

[ocean seal]

AustinU

[vapid shuttle]

which gives us the characteristic equation of the system

where e^rt is a factor of all terms

and is always positive

thus it can be divided out

leaving

[torn isle]

But this is not in the normal form that I got in the homogenous equation with constant coefficient

Normally we take e^rt out

[vapid shuttle]

,, r^2+\frac{1}{e^t}r+\frac{1}{e^{2t}}

[torn isle]

Then we will get the auxiliary equation

[ocean seal]

AustinU

[vapid shuttle]

ah you're right

sorry I am getting mixed up

we don't have constant coefficients

we have functions of t

my bad

one sec

[torn isle]

But in this case, the second term of e^t is cancelled

[vapid shuttle]

okay so I might not be the best help with this

but let me know if this is helpful

general method for 2nd order linear homogenous

which we have

is to find two independent y1 and y2

which are solutions

and then all solutions are

y=cy1+by2

[torn isle]

I search online, it’s using the relationship that t=Ln y

[torn isle]

Yea, this case I understood, the problem here is I can’t find the auxiliary equation

Let me try with the online method

[vapid shuttle]

sure go ahead

or continue to wait for someone else

sorry I can't help

good luck!

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@torn isle Has your question been resolved?

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[wheat isle]

Hi, i just want to make sure im correctly factorising

[wheat isle]

so I have 5x^2 + 14x - 3

so the first step is get the value of ac

and we get -15

so my understanding is that we need two numbers that are the product of -15 and the sum of -15

[pulsar aspen]

b^2-4ac

[wheat isle]

isnt that the discriminant

[pulsar aspen]

Yes

[wheat isle]

is that part of factorising

or just a different way of getting the final result

[pulsar aspen]

For a polynomial, factorization is the same thing as solving it

[wheat isle]

oh i see

does it work for every polynomial where you need to factorise

[gray isle]

sum of 14

[wheat isle]

not -15?

[gray isle]

sum to b

which is 14

[wheat isle]

oh alright

[gray isle]

not product of ac and sum of ac

[pulsar aspen]

It might not work for cubic polynomial or above, but you can use complex number for that

[wheat isle]

I just think it would be good for me to know because on tests it would likely want me to factorize in a specific way

[gray isle]

the zeroes and factors are linked
but that does NOT mean that factorisation and solving a polynomial are the same thing

[pulsar aspen]

But for all polynomial, if you solve it, you have factorized the polynomials too

It's just that as part of solving the polynomial, you sometimes get complex number as an intermediate value, even when the whole polynomial has real solution

[wheat isle]

so i need to find two numbers that make up the product and sum of 14 right

[gray isle]

product of ac= -15
sum of b = 14

[wheat isle]

ohh

so we need to find both ac and b

[gray isle]

no

you know the a,b,c already

a,b,c are the coefficients and constant of your quadratic

[wheat isle]

yes

[gray isle]

you want to find the pair of values that multiply to the value of ac, which here is -15
and sum to the value of b which here is 14

[wheat isle]

well that would be -1 x 15 and -1 + 15

[gray isle]

don't use x for multiplication

[pulsar aspen]

Use *

[gray isle]

but yes, your pair of values will be
-1, 15

just like before, use this information to split your 14x term

[wheat isle]

14x = -x + 15x

[gray isle]

yeh

[wheat isle]

glad i dont have to worry about that because complex numbers is a loooooooong way away for me

[gray isle]

and i hope you remember how to continue with grouping

[pulsar aspen]

True, but thankfully, if you encountered a complex number in quadratic polynomial, you know that you cannot factorize it to two real linear polynomials

[wheat isle]

so we have 5x^2 - x + 15x - 3 and then we group it from there right

[pulsar aspen]

Yes

[gray isle]

yeh

[wheat isle]

(5x^2 - x) + (15x - 3)

x(5x - 1) + 3(5x - 1)

is this right so far

[gray isle]

yeh

[wheat isle]

and then we factor out the other 5x - 1 so thatll leave us with (5x - 1)(x+3)

[gray isle]

its not "the other"

its factoring out (5x-1) from both terms

[wheat isle]

alright

also

lets say if i have something like

2x^14 - 512x^6

what the heck do i do to factorise that

[gray isle]

first thing to look for is whether all terms have common factor(s)

and if so factor that out first

[wheat isle]

well theres only 2 i think

or 2x

[gray isle]

well being able to identify any common factor,
although potentially inefficient, you can factor that out first
and repeat if needed

[wheat isle]

do we do anything with the powers

[gray isle]

knowing your exponent laws / definition of exponentiation helps with this initial step

do we do anything with the powers
yes

[wheat isle]

can we do something with

256 and 2

we can use 2 as common factor

[gray isle]

yes

[wheat isle]

2x(x^13 - 256x^5)

does that work

[gray isle]

no

why did the powers of x drop by one for no apparent reason

[wheat isle]

oh i meant to put 2x

[gray isle]

then its valid

determine whether the terms still have a common factor(s)

[wheat isle]

2x(x^13 - 256x^5)

[gray isle]

ideally you'd get the hcf/gcd, but any is fine

although potentially inefficient, you can factor that out first
and repeat if needed

[wheat isle]

i have a sneaking suspicion this will take awhile to factor

lets switch questions

how does factorising a non monic quadratic work with only 2 terms

4w^2 - 25

and would this technically still be considered a polynomial

[gray isle]

yes, the powers of your variable is a non-negative integer

[wheat isle]

oh wait we can just simplify these terms individually right

[gray isle]

i have a sneaking suspicion this will take awhile to factor
if you can only identify x one at a time yes

[wheat isle]

(2w)^2 - 5^2

[gray isle]

if you know your exponent laws, then no, it'll take only one line

[wheat isle]

can we do

(2w - 5)^2

[gray isle]

(2w)^2 - 5^2
you can express it like that yes,
but i wouldn't really call that simplification

can we do
(2w - 5)^2
no, freshman dream shennanigans

[wheat isle]

[gray isle]

expressing them as squares implies that you are aware of something related to a difference of two squares

and there is indeed a factorisation identity for that

what you just typed was not that

a^2 - b^2 = ?

[wheat isle]

i thought (2w)^2 - 5^2 was equal to (2w - 5)^2

[gray isle]

you thought wrong

[wheat isle]

because doesnt the exponent 2 apply to everything inside the brackets

ohhh wait

i know why its wrong

because (2w-5)^2 indicates that im talking about (2w-5)(2w-5)

and thats different from (2w)^2 - 5^2

[gray isle]

are you implying that
2^2 - 1^2 is (2-1)^2
that 4 - 1 is equal to 1?

[wheat isle]

no

[wheat isle]

i just realised my own mistake there^

[gray isle]

if you aren't familiar with the factorisation identity, you could approach it with the factorisation technique like before

[wheat isle]

so to factorise (2w)^2 - 5^2 should we do difference of squares

[gray isle]

yes

[wheat isle]

the formula for that is

(a-b)(a+b)

[gray isle]

yes, a^2 - b^2 = (a-b)(a+b)

[wheat isle]

and i get (2w-5)(2w+5)

[gray isle]

yes

good

[gray isle]

do you know exponent laws?

[wheat isle]

i think i might know a little bit of it but i dont know about everything

i do know some rules of it tho

[gray isle]

the one most applicable here would be
$$a^m \cdot a^n = , ?$$

[ocean seal]

ℝamonov

[wheat isle]

a^mn right

[gray isle]

no

[wheat isle]

i mean

oops

wait

wait

a^m+n

i didnt mean to put mn

[gray isle]

()

around the entire power

[wheat isle]

oh right

[gray isle]

it would otherwise without context be interpreted as $a^m + n$

[ocean seal]

ℝamonov

[wheat isle]

ohhh yeahh

that makes sense

[gray isle]

$a^m \cdot a^n = a^{m+n}$

[ocean seal]

ℝamonov

[gray isle]

can you try applying that to factor
$$p^2 + p^3$$

[ocean seal]

ℝamonov

[wheat isle]

yeah alright

wait

oh i think i know what you mean but its a little confusing because i havent applied it to factorise before

do we do

p^2(p+1)

[gray isle]

yeh

apply the same idea for what you had

[wheat isle]

does the exponent rule mean a^3 = aa^2

so if that is true would that work for 2xx

oh i just realised wouldnt we have to factorise 2x(x^13 - 256x^5) even further for this to work

do we also use exponent laws for this

and i dont know what the equation for part a means

is it saying that 12x^(2m+3) is multiplied by y^n

@gray isle

[gray isle]

yes

[lone heart]

@wheat isle Has your question been resolved?

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[alpine sable]

Hello

[alpine sable]

I need help

Evaluate

[sinful pollen]

If I understand correctly, you need to use column division

[alpine sable]

Bruhh

[alpine sable]

Noo

[lone heart]

@alpine sable Has your question been resolved?

[dawn void]

nnop

[long delta]

?

[long delta]

, rotate

[long delta]

, rotate

[long delta]

@dawn void, do you still need help with the above problem

If yes, let me know

Cuz

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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

Is anyone able to help me as to how to solve this question? I understand two sided limits though am a little lost on one sided limits

[lone heart]

@green crescent Has your question been resolved?

[green crescent]

I have to go though will check responses later on so feel free to close the channel is and when there is a good one.
Specifically in relation to this question is how do you know you take the lim x->2 and what do you do once you have limx->2((x^2-1)/(x-2)) to determine the vertical astemtope?

[lone heart]

@green crescent Has your question been resolved?

[lone heart]

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[ruby juniper]

"A relation R may be both symmetric and antisymmetric"
how can a relation be both symmetric and antisymmetric at the same time?

[ruby juniper]

is a reflexive relation both at the same time?

(can a reflexive relation be symmetric or is it never symmetric?)

[ionic trail]

A trivial relation like {(1,1)} is symmetric and antisymmetric.

[quasi scarab]

https://mikespivey.wordpress.com/2016/11/22/relations-that-are-symmetric-and-antisymmetric/

[ruby juniper]

oh ok ty

.close

[lone heart]

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[mighty falcon]

Hi! I've already asked this question in this server, but I've no response :( ... So I'm gonna try to be clearer. So I need to show that Lambda is a closed set using the Bolzano-Weierstrass theorem.

And we know Lambda is contained in the interval [0, |A|1] shown in the previous question

[mortal trellis]

can you translate the defs of P and Lambda?

[mighty falcon]

P = {vi >= 0 for all i such that v1.... vn = 1}

Lambda = {lambda, exits a v such that lalmbdavi <= (Av)i for all i}

So I should write a sequence lambda{n} ∈ [0, |A|1] for the theorem

with the theorem I know it exists c ∈ [0 , |A|1 ] and a subset ( lambda{nk} ),such that lim{k -> infty } lambda{n_k} = c

[mortal trellis]

well ok but you don't have to show anything for some random sequence

you only have to show something for those sequences of lambda_n which converge already

and then show that the limit is still in Lambda

you probably have to use bolzano on the corresponding vectors v_n

[mighty falcon]

so how can u assume it should be applied to the set P?

[mortal trellis]

just a feeling I have. cause I don't see anything else that makes sense to apply it to

[mighty falcon]

so we know v1+v2+...+ v_n = 1, but what's the point to use the theorem

[mortal trellis]

ok I shouldn't have used the letter n

[mighty falcon]

I feel so stupid, I feel this is obvious but 😭

you mean the vector?

[mortal trellis]

for each $\lambda_m$ you get a vector $v_m\in P$ which satisfies $(\lambda_m v_m)_i \leq (Av_m)_i$ for all $i$ and $(v_m)_1 + (v_m)_2 + \ldots + (v_m)_n=1$

[mighty falcon]

oooooh this way

[ocean seal]

Denascite

[lone heart]

@mighty falcon Has your question been resolved?

[mighty falcon]

and so I have to show that for each lambda_m, a vector v_m exists

[mortal trellis]

no

that is given by the definition of Lambda

[mighty falcon]

if I only have to show that the limit is still in Lambda as u mentioned
so maybe I should start with lambda_m v_mi <= (Av_m)i
=> lambda_m <= (Av_m)i/v_mi
=> lambda_m <= sum aij?

Hahahaha 🥲🥲

[alpine sable]

can i please get help

[mortal trellis]

so the v_m are in P

by bolzano weierstrass what do you know

[mighty falcon]

a bounded sequence has a convergent subsequence?

[mortal trellis]

yes

are the v_m a bounded sequence?

[mighty falcon]

y

in the v_ms are in [0,1]

[mortal trellis]

so P is in [0,1]^n which is bounded

yes

ok so what does bolzano give us now?

[mighty falcon]

then there exists c ∈ [0,1] and a subsequence (v_m)k such that lim{k-> inf} (v_m)k = c

i'm not sure of the application

[mortal trellis]

no

the v_m dont live in [0,1] so a limit of a subsequence of them can't live there either

[mighty falcon]

v_m are in P which is bounded
So w_mk is convergent ?

[mortal trellis]

to what

[mighty falcon]

I would say to 1, but it depends on how the w_m are ordered?

[mortal trellis]

no

it converges to some v

where does that v live

not in [0,1]

[mighty falcon]

ok we talk about the vector

[mighty falcon]

I think I miss something to really understand how to use the theorem

[mortal trellis]

P is closed

yes?

[mighty falcon]

P is closed bc it is in [0,1]^n which is bounded
I said a bounded sequence has a convergent subsequence.
What is the bounded sequence, w_m?

[mortal trellis]

the set (0,1)^n is also in [0,1]^n but is not closed

I only used [0,1]^n to quickly show that it is bounded

it does not show that P is closed

[mighty falcon]

so this is the conclusion of bolzano?

[mortal trellis]

no you dont need bolzano for that

do you know that subspaces are closed

[mighty falcon]

is it bc |w_m| =1?

[mortal trellis]

what is w_m and what do you mean with the absolute value here

[mighty falcon]

sry v_m

the norm of v_m, I'm not sure about the properties of a closed subspace

[mortal trellis]

which norm tho

[mighty falcon]

the 1-norm

[mortal trellis]

ok yes

do you know that the norm is a continuous function