help-0

[cinder sundial]

You’re biased then

[warped marlin]

lol

i found b2 > 4 fr

[buoyant saddle]

@cinder sundial what you say is gospel

[cinder sundial]

How

[warped marlin]

you gotta follow with me, step by step

since b2 = 2 ^ a2

lets first find a2

notice that we cant find a value for a2, since d cant be calculated

if a1 was fixed, we would be able to calculate d, thus a2

[flat abyss]

Oh wait please correct my grammar was the simultaneous question u from before

I can't remember if u got the solution or not

[cinder sundial]

I found that b_2 might be smaller than 4

[flat abyss]

Unless this was someone else

[cinder sundial]

Let me see

[fickle musk]

try gauss jordan elimination

[cinder sundial]

Why not do the Michael Jordan elimination

[fickle musk]

or cramers rule

since the values are kinda small

[flat abyss]

I solved it in my five minute walk to my next class

[fickle musk]

yea but that method is faster

[flat abyss]

And then my friend laughed at me for walking into smth

Ouch

[cinder sundial]

Let me check the answer for you, my friend

It is a perfect square, 100

[flat abyss]

O

Sweet

[cinder sundial]

Why you recalled it

[flat abyss]

Idk

I'm bored

On a train Rn

[cinder sundial]

Is x supposed to be the answer

[flat abyss]

Wha

U find the pronumerals

So y and z

And x

The value ahh thing

Uh

Anywaysv

[cinder sundial]

Hello Mickey!

[flat abyss]

Damn nice handwriting

[warped marlin]

since you are fixing d=1.5 why do you need inequality

you can just compute a2, no?

a2 = a3 - d = 4 - 1.5 = 2.5

and b2 = 2 ^ a2 = 2 ^ 2.5 > 4

see you're wrong

thats not how we do it 😛

[lone heart]

@cinder sundial Has your question been resolved?

[warped marlin]

you wanna follow with me how to actually do it?

since b2 = 2 ^ a2, we need to have a2

[cinder sundial]

For sure

[warped marlin]

but we cant, d cant be calculated

the best we can do is... find a range for a2

so find me that range

[cinder sundial]

a2 = (a1+a3)/2

Given a3=4

and 0<a1<2

[warped marlin]

yes, we get..

[cinder sundial]

2<a2<3

[warped marlin]

perfect, you rock

now just raise all sides to 2's power

[cinder sundial]

4<b2<8

[warped marlin]

🙂

[cinder sundial]

What if I do not want to apply the formula

[warped marlin]

you find the range of d first

the range of the common difference

[cinder sundial]

How could I find it

[warped marlin]

0 < a1 < 2

and a1 = a3 - 2d

a1 = 4 - 2d

[cinder sundial]

I see

Big brain

[warped marlin]

lol

0 < 4 - 2d < 2

what do you get

[cinder sundial]

8>d>4

[warped marlin]

nope

[cinder sundial]

It is supposed to be correct

[warped marlin]

noo

you fumbled somewhere

[warped marlin]

divide by -2, not multiply

[cinder sundial]

2>d>1

[warped marlin]

yes

now you can find the range of a2 using a3 and the range of d

[cinder sundial]

It seems like Im more of an idiot rather than the top mathematical in Mickey Mouse club

[warped marlin]

nah, dont say that

[cinder sundial]

You’re right

[warped marlin]

maybe you're just tired

or cant focus

[cinder sundial]

That is a bit tough

It would take a huge effort

To gauge the range

[warped marlin]

lol

cmon, gimme da range

[warped marlin]

ofc it should be the same

:p

[cinder sundial]

2d is between 4 and 2

Ohh

It is not the question

Yay

[warped marlin]

noice

[lone heart]

@cinder sundial Has your question been resolved?

[lone heart]

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[austere charm]

I'm doing 1a

I got y=1 (x-0)^2+1

How do I find the equation of the access of symmatty

[empty gulch]

-b/2a

[austere charm]

Never told thst

[empty gulch]

It’s usually something you remember

[austere charm]

Is there any other way

[lusty token]

find the vertex

x = x of vertex is AOS

-b/2a just does it faster

[austere charm]

Ik the vetext is 0, 1

[lusty token]

then x = 0 is your answer

[empty gulch]

Yup

[austere charm]

So does x=p in y=a(x-p)^2+q

[lusty token]

sure

[austere charm]

Wdym sure

[lusty token]

yes?

[austere charm]

Okay nvm

And to grind the domain it can be anything

Find*

Anything anything

So you can't find it

[lusty token]

why not

[austere charm]

And the range will be always greater or equall to 1 in this case

[austere charm]

Because it goes on forever

[lusty token]

so you know how far it goes

forever

[austere charm]

Okay yeah

And the range is

Equall or greater then

1 right

[lusty token]

are you writing in interval notation?

or inequality?

[austere charm]

Since q=1

[austere charm]

Idk man

Idk what that means

[lusty token]

seems right

actually how do you write all reals in inequality notation?

[raw vessel]

hmmm

[lusty token]

do u just write nothing and assume the domain of discourse is R?

[austere charm]

Idk

[austere charm]

Wbayys to find a random point to graph it I just insert soemthing right

Or how do I

[lusty token]

dw about what im asking

im just curious, irrelavent to your question

[austere charm]

Okay but how do I find a random intercept

So I can graph it

[austere charm]

<@&286206848099549185>

[lusty token]

just graph it from the vertex

why need random intercept

[austere charm]

So I can graph it

[lusty token]

why?

[austere charm]

So I can show my teacher that I did my work

When it asks to graph it in my textbook

[lusty token]

oh u actually need intercepts

[austere charm]

I just need some random point dude

Anything

That isn't the intercept

[lusty token]

nah it wants x and y intercepts right?

[austere charm]

Any point that isn't the vertex*

Any point

[lusty token]

????

u just said u needed intercept

but intercept isn't random

[austere charm]

It is

[lusty token]

in what sense?

[austere charm]

K whatever idk how to explain

I just need any point that isn't the vertex bro

Idc about anything else e

Preferably close

So would I take

Y=1(x-0)^2+1

And then

[lusty token]

p + 1 or - 1 is the easiest

[austere charm]

Yeah

Y=1(x-1)^2+q

So I need to solve for q correct?

[lusty token]

what problem r u even doing?

[austere charm]

The same one man

[austere charm]

Just tell me yes or no

[austere charm]

I'm doing 2a

Trying to sketch out y=x2 plus 1

I have Y=1(x-0)^2+1

I have everything besides the part it says "any intercepts"

How do I get an any intercepts

Do I just plug something random into it

But I'm missing y and x so idk how that works pls help

<@&286206848099549185>

I'm about to crash out bro

[austere charm]

So how do I do that then?mm

[raw vessel]

a

[austere charm]

Wdym

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[lone heart]

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[hollow drum]

hi, i do not understand why maxterms occur at 0,1,3,6. my own results are 0,3,4,6

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@hollow drum Has your question been resolved?

[hollow drum]

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[upbeat escarp]

so basicaly what i did first is put everything in the same base

[alpine sable]

ok

[gritty bramble]

the first question image u sent has different logs than the question you wrote in your solution

[upbeat escarp]

wait

whoops

[upbeat escarp]

i sent the wrong question

this is the one

[gritty bramble]

what do u have to solve for

[upbeat escarp]

x and y

i truly dont think it has solutions

[gritty bramble]

you cant get numeral solution from it

[gentle cloak]

you can get x in terms of y and vice versa

[gritty bramble]

but you can get the relation between x and y

[upbeat escarp]

yea

ty

the question is wrong

[gritty bramble]

or get x in terms of y and y in terms of x

[upbeat escarp]

yeye

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[lapis ingot]

$\displaystyle \sum_{i=1}^{n-r+1}i\cdot \binom{n-i}{r-1}=\binom{n+1}{r+1}$

[ocean seal]

Asagao 朝顔

[lapis ingot]

Having a hard time trying to prove this

tried induction on both n and r (didn't quite work out)

also tried expressing i as nCr(i, 1)

[lone heart]

@lapis ingot Has your question been resolved?

[lapis ingot]

<@&286206848099549185>

[somber crest]

what about a combinatorial proof? is that valid here?

alternatively maybe that one pascal-y equality helps with the induction

$\binom{n+1}{r+1}=\binom n{r+1}+\binom nr$

[ocean seal]

Flip

[somber crest]

you know, that one

so you induct on n and assume your claim holds for all k

maybe there's joy in that

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[proven maple]

I have a doubt regarding limits of multi-variabled functions. If two limits L1 and L2 are gotten from some paths C1 and C2, where L1 =/= L2, the limit L doesnt exist, but what if L1 exists but L2 is undefined, something like 0/0. Is that enough to say that the limit L doesnt exist?

[proven maple]

$\frac{x^3 - y^3}{x^2 + xy + y^2}, (x,y) \rightarrow (0, 0)$

[ocean seal]

Vali

[proven maple]

(as an example)

err thats probably a bad case, consider:

$\frac{x^2y + xy^2}{x^2 - y^2}, (x,y) \rightarrow (0,0)$

[ocean seal]

Vali

[proven maple]

for y = x + 1 as C1 we get 0, but for y = x as C2 we get 0/0

[somber crest]

if it converges from one direction but diverges from another, then it doesn't have a limit

[proven maple]

so then e.g y = x + 1 and y = x would be enough

cause in almost all examples in the book it seems like they always keep searching until they find converged values on L1 and L2 where they differ, but it seems easier to just find one and one diverging

[somber crest]

(x,x+1) doesn't equal (0,0) for any x

[proven maple]

not quite following how that relates

though i think that statement before answers the question

.close

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[light carbon]

Can any one help me with this question. I am not sure if I plugged in the values correctly

[light carbon]

Here is the work I have for it

Here is with values plugged in

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[astral creek]

can somebody help me out here? i’m completely stuck

[astral creek]

integration by parts

[alpine sable]

do you have to use integration by part? It is easier by inspection using chain rule

[astral creek]

not necessarily i guess but that’s what i’m learning right now, so i would like to know how to use that method. but if it is easier to do with chain rule then explaining that would also be helpful

[alpine sable]

if you differentiate $(3x^5-17)^{10}$ what do you get? Compare this with the integrand.

[ocean seal]

uop

[lament forge]

or, if you know about u-substitution: substitute u = 3x^5 - 17 and see what happens

this is roughly equivalent because u-sub is just the chain rule backwards, but it's potentially easier to spot

[alpine sable]

ah, totally forgot u-sub exists. It seems like nowadays I just see things instead of doing it "properly".

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@astral creek Has your question been resolved?

[silver plover]

@light carbon did u get the help?

[lone heart]

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[pale topaz]

Hey can someone help me with solving this statement if it is either true or false. I have watched a couple video's but I need a more systematic approach

[tacit arch]

Use the definition of big theta

[pale topaz]

theta = average bound

so I have n^2 >= n^2 right

so the statement is true?

hey @tight pier could you help me, if you have time, with some more examples

[tight pier]

wdym

[pale topaz]

I have some more examples but don't know exactly a right way to solve the statement

[tight pier]

hmm

post them maybe?

[pale topaz]

so I need to say if the statement is true or not. The only thing that comes up my mind is literally nothing

[tight pier]

,, \abs{(n+1)!} \leq C \abs{n!} \Leftrightarrow \abs{\frac{(n+1)!}{n!}} \leq C

We can divide by |n!|

[pale topaz]

I can't see it

[ocean seal]

bacc

[pale topaz]

What the hell 😭

[tight pier]

I applied the definition

[pale topaz]

this is way too advanced for me

[tight pier]

This just reduced to $\abs{n+1} \leq C$

[ocean seal]

bacc

[pale topaz]

so the statement is (n+1) <= n! what is the next step

[tight pier]

huh

[pale topaz]

am I doing it wrong

[tight pier]

The statement is this

[tight pier]

applying the definition means that for some n >= n_0 there is a constant C, so this holds

[pale topaz]

but why dividing by n!

this is first week computer science i am already struggling

omg

[tight pier]

So we simplify the expression

this is what came in mind

[pale topaz]

and how can I know if the statement is true or false

[tight pier]

Well I don't think it is because I don't think there is a constant that C bounds |n+1|

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[pale topaz]

I am too dumb for this I literally can't see it

😭

.close

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[tight pier]

draw it

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[dusky eagle]

can anyone explain why a(t) is 2/t instead of just 2

nvm

i asked gpt

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[fresh field]

I'm unsure how solids of revolution have been taught to you, but did it involve elemental discs?

Pi r² dr?

[subtle jasper]

https://youtu.be/OFNGpKGg9IQ
i think this guy can explain everything 10x better than i can

im pretty sure cylindrical shells and washer method refer to the same thing

just make sure you do it with respect to y

no i was wrong they are different

https://youtu.be/IOPw-54IFfI

i think this is the one

exact same question with different numbers

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@copper yoke Has your question been resolved?

[subtle jasper]

i think the bounds of integration would be -4 to 4

when you draw the graph they dont intersect at 0 iirc

np

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[strong blade]

what type of DE is this

[strong blade]

is it partial

[tacit arch]

It's not a differential equation. It's a total derivative

[strong blade]

oh so is there ways to solve this kind of equations.

[tacit arch]

Chain rule basically

[strong blade]

ok thanks

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[small galleon]

why do i need to sketch the derivative of g(x)

[small galleon]

i understand how to find intersections just not why

[gray isle]

you don't need to sketch the derivatives here

the functions are simple enough to solve this algebraically

but using a graphing calc would be a useful approach (if allowed)if your derivatives are more complicated

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[dusk viper]

Caculate HK

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[royal cloud]

If a · b = a · c, and a ✕ b = a ✕ c, does it follow that b = c?

[tardy stag]

I'm going to say probably not but you should try to see if it's true for 3 dimensions

if axb = axc then we know all three vectors are coplanar (right?)

[royal cloud]

yes

[sharp plaza]

i mean it depends on what a is i guess

[royal cloud]

wait

i thought its coplanar if a * (b x c) = 0

damn this question is tough

[tardy stag]

note that axb = a x (b + at) where t is any number

so there's a lot of freedom

[royal cloud]

mhm

[tardy stag]

what you can try doing is coming up with a and b

and then see if you can find an appropriate c

[royal cloud]

a=(1,0,0), b=(0,0,1), c=(1,0,1)

[tardy stag]

a . b ≠ a . c though

because the three vectors are coplanar you only need to use the first two components

[royal cloud]

ok i'm thinking of it a different way

a . b = a . c - > a . b - a . c = 0 - > a . (b - c) = 0

and a x b = a x c - > a x b - a x c = 0

a x (b - c) = 0

this means (b-c) has to be both parallel and perpendicular

to a

so that means b - c = 0

@tardy stag does that check out?

I think b = c

[tardy stag]

does axb - axc = ax(b-c)?

[royal cloud]

yes according to my textbook 😅

[tardy stag]

oh gaming

[royal cloud]

properties of cross product from my textbook

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[tardy stag]

this does assume the vectors are nonzero btw

[royal cloud]

yea

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[ocean seal]

ransik

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@alpine sable Has your question been resolved?

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@alpine sable Has your question been resolved?

[sly mantle]

in order to say that we also need to know u1,u2,u3 are independent. how can we know whether thats true?

yes the v's are LI but lets focus on the fact they form a base B of V

thats where this comes in

$[u_1]_B=(1,1,2,0)$ etc

[ocean seal]

RokettoJanpu

[sly mantle]

you should know that the u's are LI if and only if the [u]_B's are LI

and theres a very straightforward method for seeing if the [u]_B's are LI

youre directly using the def of LI, theres smth even simpler

in what im about to say we write column vectors

$v_1,\dots,v_n$ are LI if and only if $\m{v_1 & \dots & v_n}$ has a pivot in every column

[ocean seal]

RokettoJanpu

[sly mantle]

the matrix doesnt always look like that after elimination but yes in that case the pivots are a,b,..,n

actually i gave a bad definition

im kinda interchanging pivot and leading entry, i.e. the first nonzero element in each row after elimination

heres an example

,w rref[[1,2,3],[4,5,6]]

[ocean seal]

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[sly mantle]

the third column doesnt have a pivot, so the columns of the original matrix are LD

yes

another

,w rref[[1,2],[3,4],[5,6]]

[ocean seal]

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Requested by rokettojanpu

[sly mantle]

every column has a pivot, so the columns of the original are LI

i mean i AM doing gaussian by specifying "row reduce" or "rref"

and that becomes checking whether the matrix whose columns are [u]_B's has a pivot in each column

yeah

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[unborn burrow]

this would be yes right?

[unborn burrow]

but isn't arccos the inverse of cos

[sly stump]

Sorry

Ima delete

[unborn burrow]

this does contradict the definition of an inverse function right and I assume it's trying to imply that we must restrict the domain of cos

[obtuse knoll]

yeah

the domain of cos theta is all real values of theta but cos^-1 only has a range from 0 to pi

[lone heart]

@unborn burrow Has your question been resolved?

[south spade]

If you delve into complex numbers there is a solution for every angle, but in real numbers there's not

Because if you put smth like 10 in arccos(=cos^-1) it's not gonna work since it's not between 0 and π

[lone heart]

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[zealous folio]

Hey, my name is Aoi, can I ask for your help in solving this problem?

The quadratic equation 3x²-6x+9-0 has roots p and q. The value of 3p²q + 3pq² is

[zealous folio]

so what's the answer?

I'm really confused because I can't find the answer

[cinder tundra]

!noans

[lone heart]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[gray isle]

did you try applying the above? or somethig else?

[zealous folio]

yup

[gray isle]

can you show what you tried

[zealous folio]

Like this?

[alpine sable]

so basically

factorize the expression to be found

[gray isle]

get rid of those

=p
=q
at the end of those last two lines

[zealous folio]

ok, then what should i do to determine the result of 3p²q + 3pq²

just multiply as usual?

or what ?

[gray isle]

factorise it

and then you can use the known values of p+q and pq you've just obtained

[muted shadow]

You can take "pq "as a common out of the equation I guess

[alpine sable]

3pq

[muted shadow]

Yes

[zealous folio]

factorization?

like this?

3p²q + 3pq²

= 3pq(p+q)

[alpine sable]

yes

[muted shadow]

Exactly

[zealous folio]

then the next step is

3pq(p+q)

3.3(2)

9(2) / 9 x 2

18 ?

like this ?

[muted shadow]

Yes

[alpine sable]

yes

[muted shadow]

Why did you get for p+q and p. q?

[muted shadow]

Indeed correct

[zealous folio]

thx bro, thanks also to you guys for helping me

[muted shadow]

You're most welcome

Close the channel if you don't have any queries..

[zealous folio]

so, how do i delete this channel?

[muted shadow]

. close

[zealous folio]

.close

[lone heart]

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[lone heart]

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[strange fractal]

excuse me

[strange fractal]

i dont get what these symbols mean

[cinder tundra]

Which symbols?

[tidal summit]

I believe a_n, a_n+1, a_n+2... are terms in a sequence

[strange fractal]

what is A

[tidal summit]

its like the index of the term

mb

[strange fractal]

what does index mean

[tidal summit]

it denotes the term

[strange fractal]

is n just the number of terms

[tidal summit]

think of it like

a_1 = first term

a_2 = second term

a_3 = third term

a_n = nth term

[strange fractal]

how do i start this off

let n = 1...?

[tidal summit]

they gave you a definition of a_n+1

[strange fractal]

a_1+1 = 2a_1 - 1

[tidal summit]

a_1+1 is a_2 right

so find a_2 since you already know a_1

[strange fractal]

b ut idk a_1

[tidal summit]

read the question again

you do know a_1

[strange fractal]

oh

its 3

a_2 = 2(3) - 1

= 6-1

=5?

[tidal summit]

a_2 = 5

yes

and

5 = 2^2 + 1

so a_n = 2^n + 1 is true for n = 1, 2

so use induction to show that this is true for a_n+1

[strange fractal]

why is it also 2

oh

wai twhat i thought we only did n = 1

[tidal summit]

you found a_2 using a_n+1 = 2a_n-1

you used a_1 to find a_2

[strange fractal]

huh when did we do n = 2

we found a_2 but not n = 2

[tidal summit]

do you know what n means in a_n

a_n means nth term

the information a_1 = 3

and

a_n+1 = 2*a_n - 1

generates a sequence of numbers

a_1, a_2, a_3....

or

3,5,9,17 ...

in this sequence

the next term is generated by multiplying the previous term by 2 and subtracting 1

[strange fractal]

but i only subbed n = 1 into here so i only proved P(1) is true

[tidal summit]

the thing is

you also have to show that a_n = 2^n + 1

so you know that a_1 = 3

[strange fractal]

i sub in k next

[tidal summit]

which is also equal to 2^1 + 1

and

a_2 = 5

which is also equal 2^2 + 1

do you see now

[strange fractal]

they didnt do it like dat tho

[tidal summit]

https://tenor.com/view/facepalm-really-stressed-mad-angry-gif-16109475

I am going to say this again slowly

using the relation

a_n+1 = 2*a_n - 1

and that a_1 = 3

you were able to find that a_1 = 3

and

a_2 = 5

the question wants you to prove that a_n = 2^n + 1

[strange fractal]

how come they didnt write n=2 is also true

[lone heart]

@strange fractal Has your question been resolved?

[lone heart]

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[hidden gull]

Quick question, what does the x E R mean individually. As well as the colon

[paper cipher]

x belongs to real numbers

[coral flower]

x belongs to real numbers such that x is not equal to -2 or 1

[hidden gull]

Im guessing "belongs to" is the E, "real numbers" is the R, and "such that" is the colon?

or does separating them like that dont work

oh ight

[hidden gull]

last question, I noticed that you said -2 "or" 1, what if I want to say -2 "and" 1?

[coral flower]

i mean to say and , my bad :c

x cannot be equal to -2 and 1

[hidden gull]

oh ight

alr alr, thanks!

thats all

thank youu

.close

[lone heart]

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[jolly ridge]

6a and b

[jolly ridge]

A-alpha
B-Beta
C-gamma

i found ABC

idk how to find the top part

ABC is -3

[ivory fern]

For 6a, notice that it is the product of the roots of the new polynomial

The expression of 6b can be realised as a combination of those roots in a similar way

[jolly ridge]

wdym

[ivory fern]

$\dfrac{(\alpha+2)(\beta+2)(\gamma+2)}{\alpha\beta\gamma} = \dfrac{\alpha+2}{\alpha}\cdot\dfrac{\beta+2}{\beta}\cdot\dfrac{\gamma+2}{\gamma}$

[ocean seal]

d

[jolly ridge]

i get that

but

thats what i dont understand

where do i get

the answer from

[ivory fern]

You can use that information to obtain directly the solution

[jolly ridge]

how do i

get the top part

[ivory fern]

Essentially the same way you found $\alpha\beta\gamma$

[ocean seal]

d

[ivory fern]

Think about it

[jolly ridge]

i found that because there is a formula for it

theres no formula for the top part

[ivory fern]

And what does that formula say?

[jolly ridge]

its

-(d/a)

[ivory fern]

So that means that you can find in that way the product of the three roots of any polynomial

[jolly ridge]

how tho

the formulas are limited

[ivory fern]

But it's the same formula

Just applied to a different polynomial

[jolly ridge]

its not

[ivory fern]

The name of the roots may change

[jolly ridge]

could you write it out for me?

[ivory fern]

The important part is that if you multiply the three roots of some polynomial, you obtain the same number as if you divide the last number by the first one with a minus sign

[ivory fern]

No, you have to do the problem by yourself

[jolly ridge]

i ltierally cant

i dont know how to do the top part

[ivory fern]

I'm literally telling you what you have to do

[ivory fern]

Read this

[ivory fern]

With this

[jolly ridge]

man

im tweaking

woult the top part be the same as $(alpha+2)^3$

[ocean seal]

ripico

[jolly ridge]

because you could considerthem all as the same

[ivory fern]

Forget about the top part and the bottom part

[jolly ridge]

then what do i focus on

[ivory fern]

Precisely on this formula

[jolly ridge]

i understandthat formula

[ivory fern]

You were given two polynomials

The original one, which was x^3 + 3x^2 - 1

[jolly ridge]

yes

[ivory fern]

Let's call it f(x)

And the second one, y^3 - 3y^2 - 9y + 3

Let's call it g(y)

[jolly ridge]

ok

[ivory fern]

So if you apply the formula for f(x) you obtain:
product of roots = -d/a

In this case, the roots are alpha, beta and gamma

So the product of roots is αβγ

[jolly ridge]

yeah

[ivory fern]

Therefore, the formula becomes:
αβγ = -d/a

[jolly ridge]

yes

but i applied -d/a to g(y) not f(x)

[ivory fern]

In the other hand, for this polynomial, d is the last number, -1, and a is the first number, 1

[jolly ridge]

yeah

so its -1

[ivory fern]

With a minus sign

It becomes 1

So you have αβγ = 1

And this is the result of applying the formula to f(x)

Now you can apply the formula to g(y)

Who are the roots of g(y)?

[jolly ridge]

$(α+2)/α$ aswelll as replacing α for β and α for γ

[ivory fern]

Yes

And what happens if you multiply the three?

[jolly ridge]

the roots?

[ivory fern]

Remember that we are applying the formula

product of roots = -d/a

[jolly ridge]

so it becomes 3

?

[ivory fern]

So in the left hand side we have to multiply the roots

[ivory fern]

No, we are in the left hand side
That comes later

[jolly ridge]

left hand side of what

[ivory fern]

Of this

[jolly ridge]

ooh

[ivory fern]

So what is the product

[jolly ridge]

so u times the product of roots by 3 and that is -3d/a

[ivory fern]

No

You don't multiply by three

You multiply the three roots

[jolly ridge]

oh

so u multiply it by alpha beta gamma

[ivory fern]

No, those are the roots of f(x)

[jolly ridge]

oh

so the other roots

u multiply what by those roots

[ivory fern]

Since we are applying the formula tog g(y), everything must come from g(y)

The roots

d

a

[jolly ridge]

huh

-d/a of gy is -3

[ivory fern]

Yes

[jolly ridge]

sowhat

[ivory fern]

So product of roots = -3

[jolly ridge]

but why

[ivory fern]

Because the formula says so

[jolly ridge]

but how is

(a+2)(b+2)(c+2) = abc

[ivory fern]

Where do you obtain that from?

[jolly ridge]

because

its the top over bottom

[ivory fern]

Forget the top and the bottom

f(x) has roots α, β, γ, therefore its product of roots is αβγ

[jolly ridge]

yes

i get that

[ivory fern]

The new polynomial z^3 - z has roots 1, -1 and 0, therefore its product of roots is 1·(-1)·0 = 0

[jolly ridge]

huh

where did you get the new polynomial from

[ivory fern]

The polynomial g(y) has roots (alpha+2)/alpha, (beta+2)/beta and (gamma+2)/gamma, what is its product of roots?

[ivory fern]

I invented it, it is an unrelated example

[jolly ridge]

if u multiply them all it turnes into g(y)

if u multiply all the roots

[ivory fern]

No

There is no y

In the roots

[jolly ridge]

i mean

it turns into

a

[ivory fern]

It must be something like alpha·beta·gamma

[jolly ridge]

thats for the first one which has equations alpha beta gamma

[ivory fern]

Yes

So instead of that

[jolly ridge]

for the roots of g(y) if u multiply them all u get

a

((a+2)(b+2)(c+2))/abc

[ivory fern]

Yes

Precisely that

And because of the formula

(product of roots = -d/a)

[jolly ridge]

so its 3?

[ivory fern]

Yes

[jolly ridge]

-d/a would be 3/1

but its not 3

the answer is -3

[ivory fern]

Or -3

[jolly ridge]

where do u get -3 from i dont get it

[ivory fern]

Because d is 3 right?

[jolly ridge]

ys

yes

[ivory fern]

And a is 1

[jolly ridge]

Oh

[ivory fern]

So the formula says -d/a

You have to change sign

[jolly ridge]

omg

i am

so stupid

[ivory fern]

No, it was not really obvious

Especially if you never saw this kind of solution

But now do something similar to 6b

[jolly ridge]

it would just be the negative reciprical

for 6b

no?

[ivory fern]

Why negative?

But yes, it's the sum of the reciprocals

[jolly ridge]

wdym

sum of reciporcials

it would

be

-c/d

[ivory fern]

The roots of g(y) are (alpha+2)/alpha et cetera, here you have alpha/(alpha+2) + ...

[ivory fern]

Yes probably
Maybe c/d

I don't know the formula exactly

[jolly ridge]

thank u bro

but

i dont understand

why you found

abc

[jolly ridge]

?

[ivory fern]

?

[ivory fern]

You mean this?

[jolly ridge]

yes

[ivory fern]

Just to clarify what I was doing

[jolly ridge]

oh

i have a list

of the formulas

[ivory fern]

Ohh I didn't know the third one in the red box, that one is actually very interesting

[ivory fern]

You could actually hava a formula for almost any expression of the roots

[jolly ridge]

yea

but

i still dont get

why we did

the -d/a

according to the formulas

[ivory fern]

It was the last formula of the blue box

[jolly ridge]

yes

but

why

its not abc

its (a+2)(b+2)(c+2) over abc

[ivory fern]

Oh

It's because the alpha, beta and gamma are not literally the alpha, beta and gamma

They have to be replaced by the actual roots

Substituted

For f(x) the roots are actually alpha beta and gamma but for g(y) you have to substitute them with (alpha+2)/alpha et cetera

Just like the -d/a is different for f(x) and g(y)

For f(x) it was -1 and for g(y) it was 3

[jolly ridge]

but

[ivory fern]

Likewise, the alpha beta gamma is different for each of them

[jolly ridge]

why is it

yes but

why is it

1d/a

-d/a

ifnits not the formula

[ivory fern]

It is

[jolly ridge]

OHH
I

[ivory fern]

Here it says so

[jolly ridge]

I GET IT I GET IT

so the roots of g(y)

are just

abc

but

written differently

[ivory fern]

Yes!

[jolly ridge]

OHHH OKAY

tysm bro

like

its not that difficult

its just

a mindgame

[ivory fern]

Yeah

And if nobody ever tells you it can be hard to see it

[jolly ridge]

tysm broo

[lone heart]

@jolly ridge Has your question been resolved?

[lone heart]

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Please don't occupy multiple help channels.

[proud lodge]

example 36

[lone heart]

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as this action is irreversible, and this channel may abruptly lock.

[tidal summit]

ya you gonna have to go to a diff channel

and post it there

go to help 18

[lone heart]

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[nova hound]

I spent 8 months without touching math and forgot the basics, anyone could help with this and some others questions?

[gray isle]

apply formula for area of a triangle

[nova hound]

but how i am supossed know the perimeters without the left side measurements?

[gray isle]

you don't need them

[nova hound]

just this?

[gray isle]

if you really wanted, you can apply pythagoras a few times,
but completely unnecessary for what's being asked

[nova hound]

oh

[random dirge]

how would you even apply pythagoras here

[gray isle]

on the right triangles, with the given info about the sides

[lone heart]

@nova hound Has your question been resolved?

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[slate barn]

I know how solve this problem but i dont rlly understand what this question is asking me to do

[slate barn]

Like what does 2 parameter family mean

[ocean whale]

,rotate

[slate barn]

And also the part that says y” + y = 0

Its says find the solut of the second order but this derivative is only first order?

[lone heart]

@slate barn Has your question been resolved?

[lone heart]

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[ashen pawn]

hey guys i need help with some math homework

[buoyant saddle]

[ashen pawn]

find (fog)(x) and it domain
f(x)=x+1,g(x)=3x^2-2x-1

[wooden bolt]

(fog)(x) = f(g(x))

[ashen pawn]

there two boxes one says simplify and the other says find the domain type in interval notation

[wooden bolt]

do this first:

(fog)(x) = f(g(x))
= f(3x²-2x-1)
then find the domain

[lone heart]

@ashen pawn Has your question been resolved?

[lone heart]

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[main cape]

isn’t this a inverse of a square root

[jagged cobalt]

that would be half a parabola

no?

[main cape]

idk

because is that a function

[tacit arch]

,w plot y=log(6-x) for -5<x<5

[main cape]

that’s the equation?

[tacit arch]

Nope, I can't figure it out with logs

But it could be something else

By the picture and definition of inverse function, you should be able to tell if the inverse is a function

[main cape]

yeah

[tacit arch]

Do you know the definition of inverse

[main cape]

that’s what i’m having through ke with

yeah it’s the oppisiste of the function i know we have to graph it again usuing the defintion

but when i do i still forgetting what’s the equation

it looks so much of a square root inverse but it’s wrong i think

[tacit arch]

You don't need the equation

Just graph the inverse

[main cape]

oh yeah

so it’s a function right?

that’s all they asking for

[tacit arch]

Yes

[main cape]

thanks

.close

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[worn hinge]

Hii!

Suppose you have the following variable.
F23 = 22:00
G23 = 04:00

And you have the variable
B2 = Time In
C2 = Time Out

I want to make a function I that calculates the total number of hours that is not in the range F23-G23

Example:
i(B2, C2) = ???
i(00:00, 06:00) = 2
i(06:00, 12:00) = 6
i(12:00, 18:00) = 6
i(18:00, 00:00) = 2
i(18:00, 02:00) = 4

[exotic canopy]

isn't it just 24 hours - (F23 - G23)

[worn hinge]

No?

[tacit arch]

Your question is not well defined

F23 and G23 are independent of B2 and C2

[worn hinge]

Yes.

Basically. I want to calculate the regular hours of a person

https://docs.google.com/spreadsheets/d/1T1gL5-F4tqKrDwzaynGGAeuKl-SyRRkidkXFFLUzayo/edit?usp=sharing

Here's the link

[tacit arch]

No one's gonna open a spreadsheet to help you

[tacit arch]

Do you mean working hours not between F23 and G23?

Instead of total number of hours

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@worn hinge Has your question been resolved?

[worn hinge]

Yes

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[green osprey]

for htis question

[green osprey]

I used similar triangles

nad I did 5/(y+2) = 3/2

and got y=4/3

is that correct?

I noticed that the topic angle of the triangle was the same as the top angle of the second triangle

[livid sage]

yes both your equation and your answer are correct

[green osprey]

tysm

.close

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[snow needle]

I forgot how to do thjs

[proven leaf]

For a general quadratic $y=a(x-k)^2+h$

The vertex is found at $(k,h)$.

The equation of the axis of symmetry is given by $x=k$

The parabola opens up if $a>0$, and opens down if $a<0$. If $a=0$ then you don't have a parabola lol.

The y-intercept is found when $x=0$, or in this case, $y=ak^2+h$, because we plugged in 0 for $x$.

[ocean seal]

PajamaMamaLlama

[azure needle]

Yeah

[proven leaf]

forget to square the k whoops

[azure needle]

😅

[snow needle]

thank you soo much!

do u flip the sign of k for vertex?

???

???

???

??

???

???

???

???

???

???

???

???

???

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[rose sigil]

???

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[fallow plinth]

vertex is coordinate of min point

so cts and then find coordinate of min point

[jagged cobalt]

(or max)

[forest marsh]

(If a >0 <=> vertex is min)
(If a <0 <=> vertex is max )

[snow needle]

How do I complete the cts

[tacit arch]

,tex .cts

[ocean seal]

pizzanator

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[ocean seal]

RiverineDolphin

Hello everyone. I am learning about eigenvectors and eigenvalues in linear algebra, and I recently received some feedback from a quiz that I'm trying to understand. I have the following matrix A, where:

\begin{matrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
\end{matrix}

Now, I know how to find the eigenvalues and eigenvectors of A without any problems. They are:

\begin{align}
    \mathrm{Det(A-I\lambda)} &= \begin{vmatrix}
        \cos\theta -\lambda & -\sin\theta \\
        \sin\theta & \cos\theta - \lambda
    
    \end{vmatrix} \\
    &= (\cos\theta-\lambda)^2 + \sin^2\theta = 0 \\
    &\implies \lambda_1 = \cos\theta - i\sin\theta \\
    &\implies \lambda_2 = \cos\theta + i\sin\theta
\end{align}

And their corresponding eigenvectors:

\begin{align}
    v_1 &= \bmatrix{i \\ 1} \\
    v_2 &= \bmatrix{-i \\ 1}
\end{align}
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.51 \begin{matrix}
                   
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

LaTeX Font Info:    Trying to load font information for OT1+lmr on input line 5
1.
(/usr/local/texlive/2023/texmf-dist/tex/latex/lm/ot1lmr.fd```

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[wary root]

i need help solving this

[wary root]

2tan^2(120deg) + 3sin^2(150deg) - cos^2(180deg)

do i have to draw each individual angle and the solve like that?

[tacit arch]

Those are special angles so you should know the trig value of each of them

No drawing necessary

[wary root]

wdym by "special angles"

i know the Pythagorean identities, but idk what to do with them if i have to use them here

[surreal crystal]

just gonna rip this off the internet

the unit circle

[wary root]

i have to have that memorized?

[surreal crystal]

yes

i did, at least