help-33

[lime heron]

Oh

i dont see how it can be a CONSTANT

EITHER

IF THERES' A VARIABLE

Btw i have the answer

k = 2

but i dont see HOW it came about

[desert dirge]

in the expansion somewhere youre going to have a term that has a 1/x^7 somewhere in it

so when you distribute the x^7 outside you get 1, a constant term is thus born

[lime heron]

TWait what-

So If I have a term 1/x^7 how would you distribute it outside

[desert dirge]

you dont
you have x^7 [ expansion]

[lime heron]

Srry i am confused

[desert dirge]

you distribute the x^7, not terms in the expansion

[lime heron]

Ohh

and how would distributing get you 1?

[desert dirge]

because there will be a term with x^(-7) somewhere in the expansion

and it cancels

[lime heron]

Which would make it 1

?

Oh-

The whole thing cancels?

[desert dirge]

just the x's

leaving a constant term

[lime heron]

I fear I do not know how to get the xs to cancel

[desert dirge]

if you try compare this to the bottom of the image you may kind of get what i was saying

just keep in mind its not the same n

[lime heron]

I must be honest Im so lost-

[desert dirge]

yeah im not explaining it too well, i recommend watching a quick video on the binomial expansion, even if you dont quite get it, it can be useful in many situations outside this one

no loss

[lime heron]

Since the constant term is 168, how would I figure out what position the constant term is in?

[desert dirge]

by finding my n

[lime heron]

my n?

[desert dirge]

this n

[lime heron]

OH

[lime heron]

the n in nCr?

[desert dirge]

the r

[lime heron]

Oh yea

n = 7

in nCr

[desert dirge]

yeah

[lime heron]

7Cr =

what would it equal?

not 168 because thats not possible

[desert dirge]

wed have to find r first

[lime heron]

how?

[desert dirge]

ie my n

[lime heron]

i dont know how t ofind the r-

[desert dirge]

you can solve this equation

just remember (a^b)^c=a^(bc)
and that if a^b=a^c then b=c

[lime heron]

Uhhh

[lime heron]

there's vriables in the exponents-

[desert dirge]

yeah?

[lime heron]

x^(28-4n)*x^(-3n)

= x^-7

?

[desert dirge]

yup

[lime heron]

so i can do

WHAT

HOW DO I ALGEBRAICALLY SOLVE

IN THE EXPONENTS

[desert dirge]

a^b * a^c=a^{b+c}

[lime heron]

28-4n -3n = -7

?

[desert dirge]

exactly

[lime heron]

r = 5

OMGGGG

IM A GENIUS

UR A GENIUS

OMG

[lime heron]

ik its right bc in the book

it also says r = 5

However, I'm wondering

[x^4]^(7-n) * [x^{-3}]^(n)=x^(-7) where did this come from?

[desert dirge]

if you look at the general expression in the sum
replace n with 7
and k with n, thats all i did

and i wanted the product to be equal to x^(-7)

probably should have used something other than n

like r

[lime heron]

Hmm let me see

[desert dirge]

only looking at the x's and nothing else

[lime heron]

and whata bout k?

what is k?

in that general term

[desert dirge]

k is being indexed over
k=0,1,2,3,...,n

[lime heron]

oh wait

its so weird

cuz the variables

mean different things

[desert dirge]

just replace k with r if it makes it easier to think about

[lime heron]

but you cant use r

well now we can but before

we didnt know what r is

What would a and b be?

[desert dirge]

(x^4)/2 and k/(x^3)

[desert dirge]

im speaking generally

r is indexed, we just wanted to know which r to look at specifically

[lime heron]

Ohh

[lime heron]

Yk Im not that smart

So i think that's about as far as i understand

Thank you so much though

Figuring out r helped so much

[desert dirge]

no worries

[lime heron]

Tysm again

ur an amazing helper :)

.close

[marsh citrus]

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[marsh citrus]

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[queen swan]

I really seriously need help on this problem about the law of sines if anyone’s available to help

[queen swan]

I’m not even sure we’re to begin I understand law of sines but I don’t understand bearings at all

[glass silo]

Presumably their use of bearings is the angle you have to turn clockwise from facing north

Does the fact that B is [10mi] east of A at least make it easy to sketch that part out?

[queen swan]

I really just don’t know how I’m supposed to sketch it to start the problem, I feel like I could try better if I knew how to start it off

[glass silo]

Well, alright, it's kinda difficult to explain in words

But, draw some point A, and, for example, a line going straight up (i.e. north)

[marsh citrus]

@queen swan Has your question been resolved?

[marsh citrus]

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[tacit gyro]

am i correct?

<@&286206848099549185>

[primal nest]

It’s only been 3 minutes😭

[marsh citrus]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[storm ember]

what do you think?

[tacit gyro]

ya but i saw you needed help i think cuz u were typing so i wanted to be respectful instead of taking ur time for my ez problem

[tacit gyro]

id assume the inverse is just flipped

[storm ember]

over what

[tacit gyro]

the origin

[primal nest]

No it’s fine thank you tho you’re so nice!😊

[storm ember]

that's ambiguous

[tacit gyro]

well i also think the inverse is calculated by flipping the sign of multiple points on a line or parabolic function

[storm ember]

you flip graphs over lines in this context

[tacit gyro]

such as 4, -6 becoming -4, 6???

[tacit gyro]

ya

[storm ember]

but graphically for inverse functions they would be flipped over the y = x line

[lone heart]

The inverse isn't flipped over the origin, it's over y = x

[tacit gyro]

got it

so my answer is correct

[storm ember]

yes

[tacit gyro]

woot woot tysm

,close

.close

[marsh citrus]

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[lone heart]

And yes this was correct

You swap x and y for the inverse

[marsh citrus]

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[old girder]

This what I got so far

[dreamy pollen]

Find f°g(x) then find g°f(x) , then equate both of them to find x

[old girder]

Wdym equate

[wise stag]

For f o g(x) you have to put g(x) inside of f(x) so you would have
(X+2)^2

[wise stag]

Idk how u got gx^2 + 2

[old girder]

I see

[dreamy pollen]

f°g(x) = g°f(x)
For an example if

g°f(x) =3x+1
So 
2x = 3x+1```

[old girder]

And then does this get squared

Right?

(X+2) (X+2)

Distribute

Or nah

[wise stag]

Yeah

And do it the other way around as well

And then set them equal to each other

[old girder]

So for the first one

X^2+4x+4

Right

[dreamy pollen]

Yep

[wise stag]

Yeah

Good

Why is my message not sending

There u go

[old girder]

Okay so how do I solve the second

I’m stuck at

X^2+2

[wise stag]

Yeah that’s all you’d have

Then set it equal to the first one

And solve for x

[dreamy pollen]

@old girder you got one as x²+4x+4 and another as x²+2
Now equate then in this way to find value of x

[wise stag]

Yeah

[old girder]

So now I solve

,rotate

[old girder]

Is this right

[wise stag]

Yeah

[old girder]

I don’t see how

There can be a answer

[wise stag]

Solve for x

[old girder]

If I plug in 2

[wise stag]

Don’t plug in anything

[old girder]

4+2=4+8+4

Wdym

[wise stag]

Dawg u not supposed to plug in anything

U just solve through

Like normal equation

Subtract

And divide

[old girder]

I’m lost gimmie 5 mins

[wise stag]

Okay u got this

[dreamy pollen]

Do you know how to solve equations consisting of variables like x,y etc ?

Like

Subtracting 5 both side 
x+5-5=2-5
x=-3```

[wise stag]

How are u getting that

Like

The black box

With stuff inside

[dreamy pollen]

Enclose your text inside ```
Start and end both

[wise stag]

‘’’ hi yoooo waaauoo ‘’’

[old girder]

So basically

I gotta make it one side

[dreamy pollen]

Not that ```

[dreamy pollen]

Yep x one side, constant another side

[old girder]

So I subtract 2

From both sides?

[wise stag]

My phone doesn’t have that 😿

[wise stag]

Yes

[dreamy pollen]

[old girder]

So is that the answer

[wise stag]

No

[dreamy pollen]

You can remove x² also

[wise stag]

Yes

[old girder]

Ahhh

So

4x+2

[dreamy pollen]

You need the final answer as
x= some number

[old girder]

X=4x+2

[dreamy pollen]

Noo

😅

[old girder]

X^2=4x+2?

[dreamy pollen]

x² = x²+4x+2
Now if you subtract x² both side
0=4x+2
Now subtract 2 from both side

[old girder]

-2=4x

Order

[wise stag]

Yess

[dreamy pollen]

Yes now you want to keep just x

How can you do that?

[old girder]

Divide by 4?

[dreamy pollen]

Yep

Do it

[old girder]

-1/2=X

[dreamy pollen]

That's it

[dreamy pollen]

I use gboard

[wise stag]

Bet bet thanks

[fervent rampart]

many phone keyboards include ` as a press-and-hold alternative of '

[old girder]

Question

When I go F(g(x))

And I do F(cx-5)

[wise stag]

``` YOURE MY KING ``‘

[old girder]

For F do I just put -3(cx+5)

Or C-3(CX+5)

[wise stag]

Are u asking for f o g?

[old girder]

what

[old girder]

Look what I asked

So when I’m doing F(CX+5)

And F=CX-3

[wise stag]

Because you’re going to be putting g inside of f so it’ll be

c(cx+5) -3

[old girder]

Got it

So only X leaving

[wise stag]

Yeah

Replace for x

[old girder]

And then from here do I multiply C to the inside parenthesis

[wise stag]

Yeah sure you can

[old girder]

So (c^2 X+5)-3

[wise stag]

C5

[old girder]

What about the CX

[wise stag]

U didn’t distribute the c to the 5

The first part inside parenthesis is right but the 5 part is missing the c

[old girder]

So C squared X +C5

[wise stag]

Yeah

Then the -3

[old girder]

Would they be in parenthesis

[wise stag]

Doesn’t matter u can remove the parenthesis once u distribute

[old girder]

So after

Do I get rid of the -3

By combining like terms

So turning +5 into +2

[dreamy pollen]

There are no combining terms

Are there?

[wise stag]

Woah what

[old girder]

I

What do I do from here

[wise stag]

It’s c^2 x+ 5c -3

[dreamy pollen]

a(b+c) is ab+ac

[wise stag]

Yeah

I gotta stop giving the answer

[old girder]

What happend to the X

[dreamy pollen]

They missed accidentally

[wise stag]

That’s on me

Mb

[old girder]

Is that all for F.g

[dreamy pollen]

Yep

Now g°f

[old girder]

Roger that pip squeak

[dreamy pollen]

Then f°g = g°f like you did for last question

[wise stag]

☠️

[old girder]

Boom is this right

[dreamy pollen]

Cool

Now equate

[old girder]

Did I do something wrong

Or is it 8c=8

[dreamy pollen]

Yep

That's correct

You can subtract x both side

[old girder]

So the answer would be 1 right

[dreamy pollen]

Yeah

[old girder]

You the man thubdwr

Thunder*

[dreamy pollen]

Haha hope you understood the concept

[old girder]

Yes sir

[dreamy pollen]

!done

[marsh citrus]

If you are done with this channel, please mark your problem as solved by typing .close

[old girder]

,rotate

[old girder]

@dreamy pollen so for this

X≠0

Because 0X+2-2=0

And we can’t have 1/0

There fore it’s any real number beside 0

[dreamy pollen]

Yep

Perfect

[old girder]

Don’t would be D

Because A

Is basically including 0

[dreamy pollen]

Yeah D looks good

[old girder]

Okay got one more bear with me

[dreamy pollen]

Sure

[old girder]

Is this like the one we did earlier

[dreamy pollen]

Yep

Same

Find both then equate

[marsh citrus]

@old girder Has your question been resolved?

[old girder]

Got it boss man

[old girder]

I feel like I’m wrong

[dreamy pollen]

Yep

It was f°g(x) right?

[old girder]

There’s 2 different ones

[old girder]

There both correct?

[dreamy pollen]

One is 1/x³ +1 correct

[old girder]

And what about the top

[dreamy pollen]

Another one will be (1/x +1)³

[marsh citrus]

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[old girder]

.reopen

[marsh citrus]

✅

[old girder]

Is that the top one

Why parenthesis for that one

Is this correct

[dreamy pollen]

Because you need to replace x in f(x) with g(x) value

So f(x) = x³
g(x) = 1/x + 1
So if you replace x in f(x) by (1/x +1) it will become
(1/x + 1)³

[old girder]

@dreamy pollen

[dreamy pollen]

@old girder

[old girder]

So is this correct

And I just equal them out

[dreamy pollen]

Yeah

Simplify (1/x +1)³

[old girder]

1/X=x right

[dreamy pollen]

(a+b)³ = a³+b³+3a²b+3ab²

[old girder]

Or 1x

[old girder]

^

[old girder]

?

[dreamy pollen]

What is f°g(x) value?

[old girder]

U said simplify this

So we start with 1 divided by X right

[dreamy pollen]

Yes

No

[old girder]

Then what

1/X+1

[dreamy pollen]

You need to use this formula

[old girder]

1/X=A

And B=1

[dreamy pollen]

Yes

[dreamy pollen]

(1/x)³ = 1/x³ not 1³/x

(1/x +1)³ = (1/x)³ + (1)³ + 3 * (1/x)² * 1 + 3 * 1/x * 1²

1/x³ + 1 + 3/x² +3/x

[old girder]

Is there a simplier way

[dreamy pollen]

Nope

[dreamy pollen]

Equate this with 1/x³ + 1 and find x

[old girder]

@dreamy pollen

[dreamy pollen]

Not -3

3+3x=0

Subtracting 3 both sides

[old girder]

-1

[dreamy pollen]

Yes

[old girder]

Got it

.close

[marsh citrus]

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[lone thicket]

hey

[lone thicket]

any help

just how to start

[sand fable]

consider how you can get a u-substitution to work here

[lone thicket]

yeah thats what im looking for. what s the U and du

[sand fable]

from inspecting the integrand, what would you say a natural choice for u might be?

[lone thicket]

either U = csc or U = cot

[marsh citrus]

@lone thicket Has your question been resolved?

[marsh citrus]

@lone thicket Has your question been resolved?

[marsh citrus]

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[wet tendon]

how do I solve this?

i know there are multiple methods

should I use the graph or turn it into cosin or what ;-;;;

is there a way to solve it without having to memorise graphs?

[small vector]

it's usually a nice first step to turn things into sin and cos

[wet tendon]

i would then get (1/(cos -pi/2)) -2

and cosine of negative pi / 2 = 0

which would get me (1/0) -2

but where do I go from there?

[marsh citrus]

@wet tendon Has your question been resolved?

[wet tendon]

<@&286206848099549185>

;-;;

[wet tendon]

pretty pleaaaase :3

[still temple]

Er.....

Im gonna just yolo but.

i wrote wrong.

$sec \frac{\pi x}{2} = \frac{1}{cos \frac{\pi x}{2}}$ No?

[elfin berry]

Sukiyaki

[still temple]

Then just pluck in -1 , it got me -0.999

to the entire equation.

@wet tendon

[wet tendon]

first of all i truly do appreciate your time ;-;;

but sir that isnt an answer choice ;;-;;;;;

[still temple]

b?

hehe

[wet tendon]

the answer is a 😔 i just dont know why

[still temple]

LOL

[wet tendon]

THANK YOU SM THO

[still temple]

goodluck

[wet tendon]

THANK YOU 🙇‍♀️

[slate yarrow]

it is a

[marsh citrus]

@wet tendon Has your question been resolved?

[loud idol]

Let's rewrite this as 1/cos(pi*x/2)
Notice that x-> - 1- means that it's something very small aka like - 1.0...1 Meaning our cosine will be approaching a -0
1/(-0) is just -infinity so +2(-1) does not affect it, and it is A)

Another way to rewrite this is as x = 1+E (usually epsilon is used) where epsilon approaches -0 this will give you a very clear 1/(-0)

[marsh citrus]

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[red nimbus]

$\textbf{Exercise.}\$
Let $a,b,c > 0$ with $a+b+c=1.\$
Show that
[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq 3+2\cdot \frac{a^3+b^3+c^3}{abc} ]

[elfin berry]

bacc

[red nimbus]

I would like someone to check for my proof please

[elfin berry]

bacc

[red nimbus]

.close

[marsh citrus]

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[marsh citrus]

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[alpine bone]

$\mathbb{F}$ is just a stand-in for either $\mathbb{R}$ or $\mathbb{C}$

[elfin berry]

Alvaro Neto

[alpine bone]

How would you go about doing that? As in, what is the reasoning process?

I know that $U \oplus W$ means direct sum, and all that must be shown is that the only way to express the additive identity 0 is by having all elements be 0

[elfin berry]

Alvaro Neto

[stuck pollen]

There’s a few ways to think about it, but I would find the basis vectors of F, then try to construct the basis vectors for U. And then whatever is left is your basis for W.

[alpine bone]

I don't know about basis vectors yet

that's the next chapter

This is chapter 1 of 'Linear Algebra Done Right'

[stuck pollen]

Hmm, well if it’ll be a little less rigorous

[alpine bone]

I have a solution that I found online, but all it shows is a working solution, not how to reach it

[stuck pollen]

Basically, try and find a vectorspace W such that adding it to U gives you 4 independent elements.

[alpine bone]

Not at all trying to be facetious, I really, really appreciate the help! I'm going to try think about it a bit more, given what you said... but it sounds to me like only a restatement of the question.

[stuck pollen]

Yeah, well. Pretty much. A vector space is quite literally defined by its basis vectors, so it’s kinda hard to explain it with any rigour. Someone else may be better at finding the words than me, sorry.

[alpine bone]

I appreciate it. I'm giving it a crack, then show how far I've got

So given what you said, here is how I went about it: trial and error from the 0 vector as a starting point

does this look right? @stuck pollen

[stuck pollen]

Yeah, that’s it. But you just have to show that the instersewction of U and W is trivial.

[alpine bone]

Okay. trying that now

[sweet birch]

uh hello guys

i would like some help with this problem

[alpine bone]

There are other available help channels

look under MATH HELP (AVAILABLE)

[sweet birch]

oh ok thanks

[alpine bone]

Here's my attempt

[marsh citrus]

@alpine bone Has your question been resolved?

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[native tapir]

hi

[native tapir]

how does this work

[lethal bridge]

This is my favourite algebra class

Looks like they just solved for H+

[still temple]

Yes, they solved for H+. If you don’t understand, You probably need to learn how equations with "log_10" work.

[native tapir]

can u pls help me

[lethal bridge]

Water you confused about

[native tapir]

nvm i gyat it

its kinda easy

[marsh citrus]

@native tapir Has your question been resolved?

[marsh citrus]

@native tapir Has your question been resolved?

[marsh citrus]

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[shut pilot]

without actual calculating cubes find value of (-12)^3+(7)^3+(5)^3

[shut pilot]

i know the identity a^3+b^3+c^3-3abc but i do not know how will i put that here

???

[proud basin]

does the question want you to use that?

[shut pilot]

it's not given

to use it

but i think i have too

cuz ch-polynomial class-9

[proud basin]

maybe you could use 7^3+5^3-(5+7)^3

[shut pilot]

how will this came

what you did?

[proud basin]

well you can notice that you can get the minus out from (-12)^3

[shut pilot]

???

hmmm

k after that

now no identity will work what would we do?

[proud basin]

expand (5+7)^3

after that there shouldnt be any cubes and be left with squares

[shut pilot]

the meaning of the question is this only we cannot do that

we have to a identity i think

let me check

yeah that's correct

thx

.close

[marsh citrus]

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[night estuary]

When finding lim x->0, how is x/x = 1

[night estuary]

I get its same number over same number but I dont get why we dont just say like indeterminate

[whole thorn]

i physically blew raspberry

[static quarry]

when calculating lim x->0 you're looking at the behavior for x near zero, not actually equal to zero

[night estuary]

wait

im dumb

omg

im sorry

its 6 am lmfao

ty

.close

[marsh citrus]

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[static quarry]

drink coffee

[lethal bridge]

bungoose

[static quarry]

☕

[lethal bridge]

🪿

[marsh citrus]

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[lethal steppe]

Hi when i try to solve this function it comes zero over zero can someone solve it and send a picture with the steps if you guys don't mind..

[lethal steppe]

but the answer should be 4

[mellow crag]

simplify (x^2-4)/(x-2)

[lethal steppe]

i do it (x-2)(x+2)

?

over x-2 right?

[mellow crag]

yeah

[lethal steppe]

thank you so much

.close

[marsh citrus]

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[lethal steppe]

hello what about this one?

.reopen

[marsh citrus]

✅

[mellow crag]

again, same thing

if you can simplify the fraction first, do so

[lethal steppe]

what do we simplify?

[mellow crag]

(x-5)/(x^2-25)

[lethal steppe]

i tried to simplify x^2-25 but didnt work

[red nimbus]

(a+b)(a-b) = a²-b²

[lethal steppe]

hm

can you explain more?

[red nimbus]

Write the 25 as a square

[lethal steppe]

ok?

then

[red nimbus]

Use the formula

Then you will see

[lethal steppe]

5² = 25

https://tenor.com/vEsu.gif

[red nimbus]

Yep

x² - 5²

Use the formula now

To write it in linear factors

[lethal steppe]

(x+5)(x - 5)

am i cooked?

[red nimbus]

No you cooked

[lethal steppe]

NAH

no way

☠️

[red nimbus]

Now

[lethal steppe]

i take the x-5 and the x-5

[red nimbus]

(x-5)/[(x-5)(x+5)]

[lethal steppe]

but it would still be 0 over 0

☠️

[red nimbus]

🐐

[red nimbus]

No

You then have left

[lethal steppe]

OHHHHHHHHHHHHHHHHHH

broo

Yeahhh

[red nimbus]

1/(x+5)

[lethal steppe]

yeahh

10

thank you bro

🐐

1/10

[red nimbus]

Yea

[marsh citrus]

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[brave spire]

yo people

[brave spire]

$$\int_{0}^{1} \sqrt{\frac{2-x^4}{1-x^4}} dx$$

[thorny bison]

Ello

[brave spire]

working on this big boi

shit missed a square root

[elfin berry]

rak³en

[brave spire]

yes

[thorny bison]

it looked better without a square root

[brave spire]

so if I first split this

I get

[brave spire]

frfr

[thorny bison]

√1/1-x⁴ + 1

[brave spire]

yes

then split 1/(1-x^4) to get

$$\int_{0}^{1} \sqrt{1 + \frac{1}{2(1-x^2)} + \frac{1}{2(1+x^2)}} dx$$

[elfin berry]

rak³en

[thorny bison]

Hmm that square root is the main prob other wise splitting was perfect how u gonna deal with it

[brave spire]

now if I put $x = \tan \theta$, $dx = \sec^2 \theta d \theta$

[brave spire]

I aint lol

[elfin berry]

rak³en

[brave spire]

lag

1 sec

yes back

so

I get

$$\int_{0}^{\frac{\pi}{4}} \sec^2 \theta \sqrt{3-\sec^2 \theta - \sin^2 \theta} d \theta $$

[elfin berry]

rak³en

[brave spire]

finally

so could someone verify this?

because WA just told me I messed up somewhere

the numerics of these two integrals do not match

pls ping me if you do end up checking this out 🙂

[thorny bison]

How did you get that

Im getting something worse a little too bad by trig substitution

[brave spire]

wat

why

1/1+tan^2 gets yeeted to cos^2/2

1/2(1-tan^2) gets yeeted to sin^2/2(cos^2 - sin^2)

adding everything gives u that by taking LCM

[marsh citrus]

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[marsh citrus]

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[sharp sedge]

does equal roots mean a single root?

or 2 roots?

[desert dirge]

one root

[sharp sedge]

but why is it roots then and not root

[desert dirge]

repeated root, equal roots

[split comet]

For quadratic equations there are 2 roots and it's said in the question that it has equal roots meaning the two solutions would be the same

[sharp sedge]

oh ok

thanks

wait

so in this case

the discriminant is = 0

right?

[split comet]

You're absolutely correct

[sharp sedge]

k

thanks

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[sharp sedge]

when it says something has 2 equal roots what does that exactly mean?

[split comet]

Didn't I just tell you?

[sharp sedge]

thats equal roots

[gilded cedar]

crosses axis at a single point

[split comet]

They'll have two roots which will be equal

Discriminant = 0

[sharp sedge]

equal roots and 2 equal roots the same then

[sharp sedge]

ah ok thanks

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[clear prism]

Hi, not sure to what extent I can ask this but I'm just confused on the M[x,y] part of this theory on right/left inverses

[clear prism]

I'm quite slow on linear algebra theorems and logic so I may be missing obvious stuff, but this whole line I don't really get

and what is concluded after

It's a vector x matrix multiplication right? and suddenly it turns into standard vectors?

[devout mauve]

it starts with matrix matrix multiplication

but if you look at the columns of the second matrix and the columns of the product, you'll see that matrix matrix multiplication is basically just matrix vector multiplication repeated several times for each column

matrix * first column of second matrix = first column of result

matrix * second column of second matrix = second column of result

and so on

so thats whats happening here

M_R has two columns which are called x and y

so the product M*M_R has the columns Mx and My

but you want that this product equals the matrix with e1 as first column and e2 as the second column

so Mx=e1 and My=e2

@clear prism

[marsh citrus]

@clear prism Has your question been resolved?

[clear prism]

right so that would be 2x - 3x for the first column with Mx?

Or you're saying whole matrix * first column

I'm seeing it as row (2 -3 0) * x = 2x - 3x but I don't see that as M*vector x really. But I generally confuse myself

But I do think the [e1, e2] is just the logic that M*MR is = I and I (in R2?) is [e1, e2]

[clear prism]

Ohh wait, is it hypothetically x with n entries and then matrix multiplication with those entries not with x lmao
Shown quickly as M*x (matrix multiplication with a matrix)

x and y must have n entries for this to be defined, right?

[devout mauve]

x and y are the columns of M_R

they each have 3 entries, yes

[clear prism]

I think I saw them as a normal x first even though I knew they shouldnt have been

[clear prism]

Thanks helpful

[devout mauve]

yw

[clear prism]

If I have another question on this lecture topic, should I make a new question?

[marsh citrus]

@clear prism Has your question been resolved?

[marsh citrus]

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[fallow tusk]

is this not 2i sqrt 3 i sqrt 3

[fallow tusk]

i forgot if u hvae to do something when the sqrts next to eachother

or do u multiply them

cuz theyre next to each other

so -2 sqrt 3

[limber hearth]

Idk what to think of this since its an heresy to me to put a negative under a radical symbol

[fallow tusk]

ok well rn my answer is -2√3

[smoky plover]

Ever seen quadratic formula?

[fallow tusk]

what the quadratic formula gotta do with this man

[limber hearth]

Yes but we don't write negative under square root, we do different kind of writing

[smoky plover]

Almost, you forgot to multiply the sqrt(3) together

[fallow tusk]

wdym

multiply -2 and sqrt3 ?

[smoky plover]

Yeah it is kinda cursed

[fallow tusk]

its not 3^2 its like √3

if thats what u mean

[smoky plover]

Sqrt(3) and sqrt(3)

[limber hearth]

Well if you have sqrt(-12)*sqrt(-3) its sqrt(-3(-12))

[smoky plover]

because in sqrt function multiplicativity holds for positives

[smoky plover]

Yup this is what I was getting at

[smoky plover]

I’ll just assume you want 2i * sqrt(3) * i * sqrt(3)

[limber hearth]

It leads to same result

[smoky plover]

No

[smoky plover]

6 and -6

[limber hearth]

Ih no

[limber hearth]

Ye

Missed the i while reading mb

[smoky plover]

Anyways @fallow tusk you did 2i * i=-2, now you need to do sqrt(3) * sqrt(3)

[limber hearth]

I let u do it, u seemed more at ease with this writing

[marsh citrus]

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[humble lichen]

Looking for some help understanding why an answer is incorrect

[humble lichen]

this is the question, I repeated the euler method here

I think I'm misunderstanding what it's asking of me or something

oh wait

I see it

nvm lol .close

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[molten crown]

I have done this problem and I had a question, should I include f(x)=√x -1, also if I should I do +- √x or just √x

[marsh citrus]

@molten crown Has your question been resolved?

[molten crown]

<@&286206848099549185>

[patent sundial]

Actually...

Whenever we take the square root of something, we get 2 values

positive and negative

But we know that for a function to be a function

We need to have exactly one value of y

So, the qeustion asks to restrict the domain such that there is only one value of y

Like just write that only positive x can work.... Or something like that...

[molten crown]

ok

[patent sundial]

That just restricts the function to have that negative value as well...

[molten crown]

ok so i would do y=√x -1

also should i write f-1(x) = √x -1

[patent sundial]

Yeah just say:
f-1(x) = √x -1
and also specify the domain of this function such that there is exactly one value possible

[molten crown]

alright, so i could write that x >= -1

[patent sundial]

not really...

Actually they are asking to restrict the domain of f(x)

not f-1(x)

They are asking that restrict the domain of f(x) such that f-1(x) is a function

[molten crown]

what would the restriction to the domain be?

[patent sundial]

If you say that the domain of f(x) can only be positives, then f-1(x) will be a function too...

Just say the domain of f(x) is x belong to [0,inf)

got it?

[molten crown]

yea thank you

[patent sundial]

no problem

[marsh citrus]

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[marsh citrus]

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[harsh viper]

Please help, Our teacher gave us this as an assignment telling us to understand this step by step process thingy.

[harsh viper]

I just need an explanation for every step. I don't get anything. For the second part I can understand that it's by dividing 7 and -9 but how come there's still a 7 and a -9 outside of the parenthesis?

then the next... I really don't get it please help 😭

[golden prairie]

youre basically completing the square with both the x^2 and y^2

you factor out the 7 and the 9 because you want to have the coefficient in front of the x^2 and y^2 be 1 when completing the square

do you get that part

[harsh viper]

Wait let me take a moment to comprehend

no I dont get it

😭

[harsh viper]

what step are you talking abt?

[golden prairie]

did you learn completing the square as a technique when you learned how to solve quadratics

[harsh viper]

I probably forgot can you give an example?

[harsh viper]

Oh

[golden prairie]

these are called perfect square trinomials

look at x^2+10x+25

you can recognize that as a perfect square trinomial because if you divide the middle coefficient by 2 and square it you get the 3rd term

10/2 = 5

5^2 = 25

etc

[harsh viper]

No nobody has taught me about perfect square trinomials just binomials

[harsh viper]

OH WAIT YEAH

yeah I remember now

😭

[golden prairie]

yes so thats what youre trying to do with both the x and y

https://cdn.discordapp.com/attachments/907022703230345237/1284105977271947314/image.png?ex=66e56c62&is=66e41ae2&hm=d4b082390ec2063cc7113e869459fe1781150c0742ae522f5cb5e3f12a40f502&

lets go back here

you factor out the 7 and 9 because when doing this you want to have no coefficient in front of the x^2 and y^2

do you get the 2nd step

[harsh viper]

No... when will I do the perfect square trinomial part?

[golden prairie]

youre setting it up now

[harsh viper]

I get it but idk when to do this part

oh

[golden prairie]

ya its a process

see in these examples its just x^2

and not 7x^2

[harsh viper]

yeah

[golden prairie]

or 9y^2

thats why youre factoring those out

[harsh viper]

but what happened to -28? howd it turn into -4x?

[golden prairie]

you factored out the 7 from both the x^2 and -28x

because its in the parenthesis

[harsh viper]

OHhh

Ohhhhhh thanks

[golden prairie]

ya the distributive property

[harsh viper]

now what about the +4?

why tf is it there 😭

[golden prairie]

in the 3rd step

you add the final term of the perfect square trinomial

[harsh viper]

Ion get it...

[golden prairie]

hmm

basically youre trying to manipulate what you have into a perfect square trinomial

so thats why you add the 4 to the left

you add a 4

because its the square of half the number in front of the x

its also why you add the 1

[harsh viper]

I don't get it sorry...

[golden prairie]

thats ok

what do you remember about completing the square

like do you still remember the process on how to do it?

or just vaguely

[harsh viper]

no

[golden prairie]

ok

[harsh viper]

nope

is it adding and multiplying?

2 things to get the x^2 and x?

is that it?

[golden prairie]

what do you mena

mean

[harsh viper]

nvm thats not it

😭

[golden prairie]

https://youtu.be/f6yhfmW41wI

would it be possible for you to watch the first 2 minutes of this?

[harsh viper]

sure

OH I WAS RIGHT

ggrrr

[harsh viper]

.

):<

[golden prairie]

oh with the adding and multiplying

[harsh viper]

YEAH! ADDING AND MULTIPLYING 2 THINGS!

[golden prairie]

yeah technically you can think about it like that

but with these you can just kinda memorize the pattern

anyways back to the problem sorry

https://cdn.discordapp.com/attachments/907022703230345237/1284105977271947314/image.png?ex=66e56c62&is=66e41ae2&hm=d4b082390ec2063cc7113e869459fe1781150c0742ae522f5cb5e3f12a40f502&

[harsh viper]

its fine lol

[golden prairie]

to find what number you want to add in each parenthesis

just take the coefficient in front of the x or y

and divide it by 2

and then square it

and boom

so for the x you have -4

-4/2 = -2

(-2)^2=4

[harsh viper]

yeh but what about the y

[golden prairie]

same process

[harsh viper]

-2/1?

[golden prairie]

-2/2

[harsh viper]

Oh

Ohh omg im so dumb

😭😭😭😭😭😭

[golden prairie]

(-1)^2=1

so you add 1

[harsh viper]

😭 I WAS DIVIDING IT BY 1

NOT 2 THATS WHY I WAS CONFUSED

[golden prairie]

but you cant just add numbers to one side of the equation you have to add it to both

[harsh viper]

Thank you ❤️

[golden prairie]

which is why you add 7(4)

[harsh viper]

i get this part dw

[golden prairie]

ok cool

so then you just factor your perfect square trinomials and divide by 63

[harsh viper]

this is factored right? like simplifying the parenthesis so it becomes this

[golden prairie]

yes

[harsh viper]

how do you turn this part

into this

[golden prairie]

adding and multiplying

[harsh viper]

Oh ok

[golden prairie]

what two numbers multiply to get 4 and add to get -4

[harsh viper]

I have never factored trinomials before thats why I got confused why it looked like this

[golden prairie]

wait isnt this algebra 2

[harsh viper]

Yeah but they only taught us binomials

[golden prairie]

wtf thats weird

[harsh viper]

ikr

then out of nowhere they expect us to know this in precal

[golden prairie]

most schools teach factoring trinomials in algebra 1 way before this stuff

[harsh viper]

🙄

[golden prairie]

i learned a little in 8th grade

and then had geometry in 9th

and algebra 2 in 10th where we learned the bulk of factoring and then stuff like hyperbolas

[harsh viper]

thats interesting, ig its because we live in different countries?

[golden prairie]

im in the US

wbu

[harsh viper]

Oh im in the ph

[golden prairie]

philippines?

[harsh viper]

Yess

how about in science what were your lessons?

[golden prairie]

earth sciences i think in 8th?

and then biology in 9th

chem in 10th

AP chem in 11th

and AP physics in 12th

[harsh viper]

oh like it's not mixed? just one topic?

[golden prairie]

yeah one topic per year pretty much

[harsh viper]

For us it's mixed, earth science, chem, bioligy, physics per grade lvl

basically our lessons in 10th were the same as 8th n up just a little advance

and practically a summary lol

[harsh viper]

Anyways thank you so much! (´▽`ʃ♡ƪ)

[golden prairie]

ofc do you think you have it from here

[harsh viper]

hm? sorry I didn't understand your question, I haven't been sleeping well from too much cramming. I don't even remember eating meals well 😭

[golden prairie]

like do you understand it well

[harsh viper]

Yeah I guess, I can understand things pretty well even if I'm half asleep hehe

[golden prairie]

ok gl in your class

[harsh viper]

Thank you! Have a nice day ^^

[golden prairie]

you too

[marsh citrus]

@harsh viper Has your question been resolved?

[indigo coral]

2+2

[marsh citrus]

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[hazy dragon]

x_i != x_j, for every 1<=i,j<=n? or does it not matter

[hazy dragon]

var der monde

[marsh citrus]

@hazy dragon Has your question been resolved?

[marsh citrus]

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[civic drum]

how do you approach this?

[civic drum]

<@&286206848099549185>

[marsh citrus]

@civic drum Has your question been resolved?

[civic drum]

.close

[marsh citrus]

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Closed by @civic drum

Use .reopen if this was a mistake.

[civic drum]

.reopen

[gilded cedar]

dy/dx ko 1/2 slope put krke differential

[marsh citrus]

✅

[civic drum]

so no need of knowing of curve?

[gilded cedar]

directly kaise btaoge kaunsa h

[civic drum]

ye bhi h

and koi aur approach nhi hai?

[gilded cedar]

maine to aise hi kiya tha baki pta nhi

[civic drum]

accha okk

and ek aur

yaha pe critical point kyu nikala?

[gilded cedar]

ye aisa symmetric aayega

to area xf(x) hoga

to usko hi maximize kiya

P(x)= area = xf(x)

[civic drum]

accha ok bhai, thanks

[gilded cedar]

np

[civic drum]

.close

[marsh citrus]

Channel closed

Closed by @civic drum

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[edgy kestrel]

In ABC triangle C=90⁰, CD is an altitude and BC=2BD. Find AD/AB

[split pike]

you mean ad divided by ab or both?

[edgy kestrel]

Ad divided by ab

[split pike]

oh ok

so

it seems like this could apply to the pythagorean and euclidean theorem

since its a right triangle

now, what we can do here is imagine these heights and lines are numbers

[edgy kestrel]

Oh I just remembered the Euclidean theorem

[split pike]

i imagined BD is 4 and BC is 8

[edgy kestrel]

Alright

[split pike]

makes cd the square root of 48

now, remember that h (height) squared is 48

you need any more help?