help-0

[gray isle]

wdym

[shrewd whale]

the common factor

[gray isle]

don't overthink this

[shrewd whale]

2pi?

[gray isle]

no

[shrewd whale]

I am overthinking

[gray isle]

and i'm telling you not to do that

[shrewd whale]

ok is it 2?

[gray isle]

yes.

[shrewd whale]

but what about the pi^2, and the h^2?

[gray isle]

not much you can do to simplify

[shrewd whale]

they are in a sqrt, can't they be taken out?

[gray isle]

not cleanly

[shrewd whale]

why?

[gray isle]

any attempt to do so legitimately would make your expression look more complicated

[shrewd whale]

wouldn't it just put them outside the sqrt?

[languid bolt]

because there is + 8 pi A

[shrewd whale]

wouldn't that just leave the pi inside ?

[gray isle]

the terms under the sqrt don't have any common perfect square factors

[shrewd whale]

oh all of them

ok

I get it

thx

I gtg

.close

[lone heart]

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[gray isle]

i suppose since you're supposedly dealing with stuff like length and area, you'd take the positive case

[lone heart]

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[low cliff]

how is (tan-1) = (1-tan)

[rugged sun]

Factor -1 from tan -1

[lean depot]

When you square a number, it doesn’t matter if it’s negative or not. (-x)^2 = x^2.

Similarily (1-tan)^2 = (tan - 1)^2

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[orchid citrus]

Hey, can anyone help me understand how i end up with (A,B)=(-1,4)? understand the rest but not how i get (A,B)=(-1,4)

[marsh rapids]

Just the system at the end ?

[orchid citrus]

yes

[marsh rapids]

2 = -2A so A = -1.
Then 3 = A + B becomes 3 = B - 1 so B = 4

[orchid citrus]

ahhh, thanks 🙂

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[worn fox]

Don't use up a help channel to shitpost pls

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[bronze nacelle]

need help on this question

[alpine sable]

well it has zeros at plus or minus 1

and it has a y-intercept at -2

[bronze nacelle]

ok

[alpine sable]

so y=a(x-1)(x+1) which is equal to y=a(x^2-1)

[warm perch]

and it is quite narrow so it is A

it will be wider if the coefficient is 1

[bronze nacelle]

so its a?

[warm perch]

ye

and u can test the functions on different x values

like sub it in

and see if it matches the output or not

[lone heart]

@bronze nacelle Has your question been resolved?

[lunar dawn]

why is it not B

[lone heart]

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[stoic adder]

1.4.2

[lone heart]

@stoic adder Has your question been resolved?

[stoic adder]

<@&286206848099549185>

damn nobody wanna solve this one

[tired echo]

can you show your work so far ?

@stoic adder

[stoic adder]

i got nothing

i don’t know where to start

[tired echo]

is this your first time solving linear systems of equations?

[stoic adder]

i’ve never seen k1 k2 k3 stuff

[tired echo]

do you know the methods for solving systems of equations like this ?

[stoic adder]

do i solve the system of equations first?

i can do that

[tired echo]

i mean you didnt post anything else wdym?

oh above

[stoic adder]

oh yeah my bad

[tired echo]

can you say specifically which one ?

1.4.2?

[stoic adder]

yes

[fluid basin]

could you restate what the problem is

[stoic adder]

it says there

[stoic adder]

1.4.2

[tired echo]

so this

[clever folio]

You need to "solve" k1(eqn1)+k2(eqn2)=eqn3 (for k1,k2)

[fluid basin]

ah, that is helpful

[stoic adder]

it refers back to 1.17

sry for confusion

[fluid basin]

oh,. i ee what it's asking you to do

[clever folio]

Well one equation you would need to be true is 6k_1+14k_2=36

[stoic adder]

okay

first of all how do k1 k2 k3 variables world

*workl

are they the same as normal variables

[clever folio]

There is no k_3

They're just variables in this case

[stoic adder]

okay then i have a different question later

but not now

[clever folio]

You could try equating coefficients of a,b,c maybe to produce more equations?

I don't really like the wording of this problem.

[stoic adder]

its just words to me

i wasnt taught any of this btw

im fresh outa geometry

[clever folio]

So, like you want k_1(a+b+c)+k_2(a+2b+3c)=2a+5b+8c

[clever folio]

You have a book right?

[stoic adder]

yes

[clever folio]

You should read it.

[stoic adder]

but becasue of the wording it basically explains nothing

i dont understand it

also there is simply no explanation for this kind of problem

[clever folio]

I usually ask my professor for explanations when that happens

[stoic adder]

this is supposed to be no help but thats not happening

ill get a zero along with everyone lese

[clever folio]

What I'm getting at here is maybe try equating the coefficients of a,b,c

[stoic adder]

i dont know what that means

[fluid basin]

it's relatively easy to find a combination by inspection

try to find some way to combine the first two equations so as to get the third

like add twice the first to the second, or subtract the first from twice the second, etc

[stoic adder]

2a +3b +4c = 20?

[fluid basin]

like you want some m + 3n = 8 but also m + 2n = 5

where m is what you multiply the first equation by and n what you multiply the second by

[stoic adder]

are we trying to solve the system of equations rn

[fluid basin]

no, you're trying to figure out how the third equation is linearly dependent on the first two

[stoic adder]

im lost

[clever folio]

Where are you getting 9 and 6 from?

[fluid basin]

oh, sorry, typo

[clever folio]

I think romeofoxtrot is talking about equating coefficients like I am?

[fluid basin]

yes, i am

[clever folio]

Here

[fluid basin]

that is what you need to do

yes that

[clever folio]

You need k_1+k_2=1 for ex because after you expand this

The coefficients of a need to add to 1

[stoic adder]

how do i solve this

[clever folio]

Also the coefficients of b need to satisfy k_1+2k_2=5

[fluid basin]

expand the left side out and group , then set coefficients equal to one another

[tired echo]

oh you can write everything in terms of c

it even had it on the page you sent before i did it

[clever folio]

I don't see immediately how it helps to do that lol

Ig we're technically trying to show the third equation is a linear combo of the other two or something.

But it seems like we want the coefficients to work out a certain way.

So having 1 equation vs 3 seems less helpful

[stoic adder]

at this point its ok if u solve it and dont explain how

[clever folio]

I'm not going to do that

[stoic adder]

ok fair

[clever folio]

Take this

[clever folio]

Do this

[stoic adder]

how

[clever folio]

Romeofoxtrot explained how

I'm assuming part of the explanation is unclear to you yeah?

[stoic adder]

wdym expand

[clever folio]

Expand means to multiply

Like distribute

A(b+c)=Ab+Ac for ex

Group is a synonym for collecting like terms. For ex 3h+2y+3h+3y=6h+5y

You "collect" all the h's and y's together by adding their coefficients.

[stoic adder]

ok

[clever folio]

For constants it's a little trickier I guess. If you have $k_1 x + k_2 x$ you can group using distributive property as $k_1 x + k_2 x= (k_1 + k_2 )x$

[ocean seal]

DootDooter

[stoic adder]

what next?

@clever folio

[clever folio]

Equate coefficients

a is multiplied by one thing on the lhs but another thing on the rhs

Same for b, same for c

[stoic adder]

lhs?

[tired echo]

left hand side

[stoic adder]

ah

[tired echo]

did you group all the a, b, and c terms ?

[clever folio]

You have (some stuff)a on the lhs and (some other stuff)a on the rhs, you want (some stuff)=(some other stuff)

Same for b and c

[tired echo]

you end up with another 3 system of equations with k1 and k2

[stoic adder]

yeah idk how you got there

[tired echo]

ill throw you a bone so this can hopefully elucidate

i factored an a out of each term with an a, and likewise for all the other variables

the next step you end up with something
a(...) + b(...)+c(...) = 0

[clever folio]

For ex if you had (X+Y)a+(T+U)b=3a+4b one way this can happen is if X+Y=3 and T+U=4

[clever folio]

Explaining this

[stoic adder]

this helps with what

sorry im completely lost

[clever folio]

Equating coefficients

[tired echo]

factor out a, b, and c from everything there first

[stoic adder]

here?

[tired echo]

yes

[stoic adder]

bruh i cant even do this part

[tired echo]

a(k1+k2- 2) …

All the terms are being multiplied by a

[stoic adder]

$a=(5b-k1b-2k_2(b)-4k_2(c))/(k_1+k_2-2)$

idk what im doing

[ocean seal]

byjos

[clever folio]

Well repeatedly tell us you don't know what you're doing is kind of unproductive right?

[stoic adder]

i dunno

[clever folio]

It sounds unproductive to me.

You need to ask more specific questions.

[tired echo]

Do you know how to factor out terms/distributive property?

[stoic adder]

yes

[tired echo]

That’s all you’re doing to that second equation

Factor out an a with everything with an a

[stoic adder]

which one is the second one

[tired echo]

2nd one here

[stoic adder]

theres no parentheses

[tired echo]

Yea

You have to make them yourseld this time

Factor OUT

Not distribute

[stoic adder]

ok so im doing the opposite

[tired echo]

Yes

[stoic adder]

$k_1(a+b)+k_2(a+2b+4c)-2a-8c=0$

[clever folio]

That's not an equation

But also you want (some stuff)a+(some more stuff)b+(even more stuff)c=0 in this case

[ocean seal]

byjos

[stoic adder]

did i not do that

[tired echo]

No

[clever folio]

So rather than factor the k1,k2 you want to be factoring the a,b,c

[stoic adder]

OH

[tired echo]

Theres a k1 c missing

If u factor out k1 and k2 you get back to the start

[stoic adder]

k1 c where

[clever folio]

k_1 (a+b+c)

[tired echo]

From what you expanded out

[clever folio]

Not k_1(a+b)

[stoic adder]

i only see k2 with c

[clever folio]

You are copying josemom2's equations wrong.

[tired echo]

The original thing youre working with is k1(a+b+c)+k2(a+2b+3c)=2a+5b+8c

[stoic adder]

ok

so what am i trying to find again

[clever folio]

The way josemom2 was doing it was to move everything to the lhs

[tired echo]

a(something in terms of k) + b(something in terms of k) + c(something in terms of k) = 0

[stoic adder]

so the answer will look like that

[clever folio]

No

But it will get you closer to the answer

We're doing this to find more equations in k_1 and k_2 that we can solve to get the answer

[stoic adder]

$ak_1+k_1b+k_1c+ak_2+2k_2b+3k_2c=2a+5b+8c$

[ocean seal]

byjos

[tired echo]

Yes

You can even solve from here

K1 + k2 = 2

[stoic adder]

$a(k_1+k_2-2)+b(k_1+2k_2-5)+c(k_1+3k_2-8)=0$

[ocean seal]

byjos

[tired echo]

Yuss

There you go

[stoic adder]

im doin math

ok

any mistakes tho

[tired echo]

Now treat each ‘inside’ as an equation to set to 0

Nope

[stoic adder]

hold on lemme write this down

[tired echo]

You can think of the right as 0k1+0k2+0

[stoic adder]

ok

[stoic adder]

so do i treat the parentheses as a system of equations all equal to 0

[tired echo]

Exactly

[alpine sable]

Byjos the learning app

[stoic adder]

$k_1=-1, k_2=3$

[ocean seal]

byjos

[stoic adder]

can i swear

[clever folio]

I doubt anybody cares if you swear lmao

[stoic adder]

LETS FUCKING GOOOO

ok im done

[worn fox]

omg did you just swear??

[stoic adder]

😳😳😳

.close

[lone heart]

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[alpine sable]

just finished some combinatorics problems, if anyone could have a quick look, would be greatly appreciated

[alpine sable]

A committee with 8 members two from each department, Math, Chemistry, Biochem and CS has to be formed. The chairperson must come from Maths and the Secretary from CS. Mathematics, Chem, Biochem and CS have 6,7,7,5 eligible committee members. How many committees can be formed?

If Papadakis is a member of a group of 10 men and 15 women, how many committee rosters can be made to include him in their roster if the committee must have 2 men and 4 female members of this department?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

.close

[lone heart]

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[lone heart]

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[terse cosmos]

i have to get the left side same to the right side

[last ether]

Rewrite log(40) and then factor

[terse cosmos]

I didnt get it

[lone heart]

@terse cosmos Has your question been resolved?

[terse cosmos]

<@&286206848099549185>

[iron mulch]

Are you aware of the property logs have when they are multiplied by something on the outside?

[terse cosmos]

Yes

<@&286206848099549185>

Could someone help

[last ether]

You only get to ping them once

Like I said

Rewrite log(40)

Recall that log(ab) = log(a) + log(b)

[terse cosmos]

Ok i know this

Re write it for log4 + log 10

?

[last ether]

You could, I guess

[terse cosmos]

Ok

Wait

@last ether

[last ether]

Gimmie a sec

[terse cosmos]

Tyt

[last ether]

https://www.desmos.com/calculator/pbcwhwtvr2

i have steps in here

@terse cosmos

[terse cosmos]

Thank you very much

[lone heart]

@terse cosmos Has your question been resolved?

[lone heart]

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[lone heart]

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[nocturne sluice]

I need help regarding limit and continuity for preparing my mid exam. In the sample question, can anyone explain me the number 2?

[nocturne sluice]

I want the answer of 2 (i, ii, iii)

mention me if someone help

[real gazelle]

what do you think the answers should be

@nocturne sluice

ii is probably the easiest

what do you think lim x->-2 f(x) should be

[nocturne sluice]

Yes sir! teach me

[real gazelle]

do you have any guesses

[nocturne sluice]

1?

[real gazelle]

yup

that's correct

[nocturne sluice]

Idk why :3

I just randomly chose 1

[real gazelle]

well what was your reasoning

for 1

[nocturne sluice]

I chose 1 because it was on the y anix -.-

I’m so bad at calculus sorry

Can you tell me why 1 is correct?

[real gazelle]

yes that is right

it is on the y axis

[nocturne sluice]

What? It was correct?

[real gazelle]

do you see how when the x value approaches -2

[nocturne sluice]

How! Why?

[real gazelle]

the graph approaches 1 on the y axis?

[nocturne sluice]

It would help me a lot if you start from beginning

2.i

[real gazelle]

I think ii is easier

we should start with ii

[nocturne sluice]

Ok

[real gazelle]

it will make more sense

[nocturne sluice]

I solved this kind of maths

|x|

[real gazelle]

okay yes

what is $$\lim_{x\to -2^+} f(x)$$ here

[ocean seal]

Eric Tao (he/him)

[nocturne sluice]

Umm let me think for a minute

For the question I solved?

[real gazelle]

for the question you posted

[nocturne sluice]

Or the one I gave first?

[real gazelle]

b

[nocturne sluice]

Ok

Well it is on -2 in x axis but it can’t be 2 so it’s 1

Am I right or wrong?

[real gazelle]

what do you mean by it can't be 2

[nocturne sluice]

Nevermind

I messed up

With another math

[real gazelle]

there are no calculations you have to do

everything you need you can find by looking at the graph

[nocturne sluice]

Okay tell me

[real gazelle]

for part b i, ii, iii

[nocturne sluice]

Then I’ll try i

You solve the ii for me

I’ll try i

[real gazelle]

do you know what the $$-2^+$$ means?

[ocean seal]

Eric Tao (he/him)

[nocturne sluice]

Yes

[real gazelle]

what does it mean

[edgy wind]

Approaching from the right

[real gazelle]

yup

so as x approaches -2 from the right

we see that the y value approaches y=1

right?

[nocturne sluice]

What does the + means?

[real gazelle]

from the right

[nocturne sluice]

Beside -2

Oh

Thanks

What after then?

I’m writing all down

[real gazelle]

okay can you figure out what $$\lim_{x\to -2^-} f(x)$$ is then?

[ocean seal]

Eric Tao (he/him)

[nocturne sluice]

Well it’s -2 from 0

And 1 from x axis

I still don’t know why its 1

Do I need to follow the graph carefully?

[real gazelle]

what do you mean by -2 from 0

[nocturne sluice]

From right

-2 from right

[real gazelle]

well

what does the - mean

[nocturne sluice]

Can you give me the answer? So that I can figure out? 😢

[real gazelle]

next to the -2

sorry I can't give out any answers

[nocturne sluice]

It’s not for my exam tho the question is from 2019

[real gazelle]

I still can't

[nocturne sluice]

Oh

Okay

[real gazelle]

what does the - mean to the right of the -2 though?

[nocturne sluice]

Then I’ll just follow your steps

-2 means sufficient close but less than 2

[real gazelle]

do you mean "-2^- means sufficiently close but less than -2"

[nocturne sluice]

Yes

[real gazelle]

okay

good so that means it approaches from which side

the left or right

[nocturne sluice]

Right side?

[real gazelle]

which side is less than -2?

[nocturne sluice]

Left side

[real gazelle]

yes

so now what is this equal to
$$\lim_{x\to -2^-} f(x)$$

[ocean seal]

Eric Tao (he/him)

[nocturne sluice]

So it approaches from right side right?

[real gazelle]

no

if the - means less than then it means approaching from the left side

[nocturne sluice]

Can you tell me why I’m wrong?

[real gazelle]

the + means approaching from the right side

[nocturne sluice]

Oh

Got it

Got it

[real gazelle]

okay so what is this limit equal to

[nocturne sluice]

I still don’t know but I’m learning little by little

I understood what - and + means

[real gazelle]

okay good

but when x approaches -2 from the left

what does the y value approach?

[nocturne sluice]

1

But it has no connection with the -2 right?

That’s why I’ve been saying it’s approaching from the right side

[real gazelle]

are you having trouble visualizing what from the left side and from the right side mean?

[nocturne sluice]

Yes

I mean, let me clarify

[nocturne sluice]

You’re saying -2 is approaching from left side

[real gazelle]

-2 doesn't move, -2 is just a number

x is the one that moves

towards -2

[nocturne sluice]

@edgy wind here he said right side

[real gazelle]

yes

[nocturne sluice]

Ahh i see i see

[real gazelle]

that has a + on it

this is "approaching from the left side"

[nocturne sluice]

So -2 is just a number

[real gazelle]

$$\lim_{x\to -2^-} f(x)$$

[ocean seal]

Eric Tao (he/him)

[real gazelle]

yes

[nocturne sluice]

Where x moves to -2

[real gazelle]

yes, from the left side

[nocturne sluice]

Let me tell what I understood

[real gazelle]

this is "approaching from the right side" or $$\lim_{x\to -2^+} f(x)$$

[ocean seal]

Eric Tao (he/him)

[nocturne sluice]

Yes

Understood

But why the answer is 1?

[real gazelle]

do you see how the y value is going up to 1

[nocturne sluice]

Yes

Let me give a try to i

The answer is -3

Right or wrong?

[real gazelle]

partially correct

which side does it approach -3 from?

left or right

[nocturne sluice]

Umm

Left

[real gazelle]

yup

what does it approach from the right side?

[nocturne sluice]

Didn’t understand what you said

[real gazelle]

okay well

if it approaches -3 from the left side

what number does it approach from the right side

[nocturne sluice]

I don’t know 😐

[real gazelle]

well what was your reasoning for -3

can you draw out where the -3 came from

[real gazelle]

okay perfect that's good

but there's more of the graph to the right of x=1, correct?

[nocturne sluice]

Yes

[real gazelle]

so what is the limit as x=1 approaching from the right of x=1?

[nocturne sluice]

-1?

[real gazelle]

how did you get that

[nocturne sluice]

Because the line starting from -1

From the very left

[real gazelle]

can you draw where on the graph you mean

[nocturne sluice]

From (-2,-1)

[real gazelle]

but there is no point (-2, -1) on the graph...

[nocturne sluice]

Sorry

(-2,1)

[real gazelle]

we want to approach from the right of x=1 though

-2 is not to the right of x=1

also there is no -1 there

[nocturne sluice]

I corrected here

[real gazelle]

okay let me put this simply

we want to find the limit as x approaches 1 from the right

so only the stuff to the right of x=1 matters

can you draw the graph and circle the relevant part so that I know we're on the same page

[nocturne sluice]

Why right?

[real gazelle]

because you already found the limit from the left

so now you have to consider the limit from the right

[nocturne sluice]

Ohh

[lone heart]

@nocturne sluice Has your question been resolved?

[lone heart]

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[steady pasture]

Hello! Can i have help with this?

[steady pasture]

I don’t know how to solve it

[fervent timber]

what don't you understand

[steady pasture]

I just don’t know how to start haha

[fervent timber]

do you know what the picture means

[steady pasture]

Yeah it’s a parabola

[fervent timber]

do you know what it means to be a "solution"

[steady pasture]

Sorta

Not rly though

I kinda forgot how to do parabolas, i haven’t touched this subject since algebra 1

[fervent timber]

well in this context I think they are talking about the equation f(x) = 0
so basically what value can you plug into the graph that will get you 0

[steady pasture]

1?

well no

none of those options

[alpine sable]

the question in itself is a bit wrong

I mean there will be complex solutions

it should say "real" solutions

[alpine sable]

u still have the no solution option

[steady pasture]

no solution then?

[alpine sable]

ya

[median oar]

Assuming they haven’t learned about complex numbers it makes sense for it not to say real solutions

It might just confuse them more than help them

[steady pasture]

i have i’m just reviewing alg1 rn

[median oar]

Well if you have then there are 2 solutions

[alpine sable]

not necessary there are a max of 2 solutions over C

oh wait

y

2 solutions

[steady pasture]

wait soo

[median oar]

The only way it has 1 solution is if it touches the x axis at the vertex

[steady pasture]

if i made the parabola into an equation and solved complex solutions from there, would it be easier to solve from there??

[median oar]

All other parabolas will have either 2 real or 2 imaginary solutions

[alpine sable]

then saying no solutions is fine

[median oar]

Yeah

[steady pasture]

oh okay cool

[median oar]

Depends on the context of the question

[steady pasture]

yeah i think i’ll just like make it into an equation cos there’s a lot of problems like this on the review

thank you!! 😅

[alpine sable]

if u actually want to get the roots u can figure out stuff like -b=2a and get the equatiom for the parabola and get the roots

[median oar]

If “the question hasn’t learned” about complex numbers then I’d say assume they don’t exist

[steady pasture]

Oh okay got it

[median oar]

Like maybe you’ve learned about it but the course hasn’t

[steady pasture]

true

[median oar]

So for someone doing just the course they won’t know

[steady pasture]

yep

anyways thanks for the help!

means a lot

.close

[lone heart]

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[languid bolt]

you cant just cancel the i^3 like that

it's addition

not multiplication

[terse frigate]

oh yea... totally forgot about that. I've just gotten into the habit of doing so after working with indices

alright

so then how do i approach this question

[languid bolt]

what is i^2

[terse frigate]

-1

[languid bolt]

ok so

i^3 = i^2 * i

right?

[terse frigate]

right

[languid bolt]

so i^3 = -i

[terse frigate]

yes

[languid bolt]

simplify the numerator now

[terse frigate]

-1

[languid bolt]

yes

now do the same for the denominator

[terse frigate]

oh shi

i haven't even taken complex numbers yet

im an IBDP student so it's taught in DP2 in my school

[languid bolt]

same

[terse frigate]

lol

thanks tho

this act prep is exhausting

[languid bolt]

np

good luck

[terse frigate]

thanks

.close

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[alpine sable]

Here is a DE i was practicing on and I got ln25/ln2 but wasnt sure if what i did was legit. here is the problem:

Say a new puppy is born, and for the first 6 months of its life it gains weight at a rate proportional to its current weight at that time.
Propose also that this double its weight in two months.
How long will it take the puppy to quintuple his weight?

[alpine sable]

im pretty sure its from calc 1 or something but im tripping

[last ether]

Alright well

We can restrict our answer choice to be between 0 and 6 months

And then set up an equation

$g(x) = A_0(2)^{0.5t}, t\in[0, 6]$

[ocean seal]

Umbraleviathan

[last ether]

Reasonably, we wanna find when the puppy reached 4A_0

So we can set g(x) = 4A_0

$$4A_0 = A_0(2)^{0.5t}$$
$$4 = 2^{0.5t}$$
$$2^2 = 2^{0.5t}$$
$$2 = 0.5t$$
$$4 = t$$

[ocean seal]

Umbraleviathan

[last ether]

Idk how you got log2

Were you doing continuous growth @alpine sable ?

like somehow $A_0e^t$?

[ocean seal]

Umbraleviathan

[alpine sable]

no just differential equations

but isnt quintuple 5 times?

[last ether]

I thought it said quadruple

Uh

[alpine sable]

lol

no i get the point tho

lol

[last ether]

I mean I still wouldn't go into the derivative realms

[alpine sable]

k yea thanks

[last ether]

And yeah you'll get log2

In that case

But spare yourself the pain of going into calculus

[alpine sable]

lol

.close

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[drifting coyote]

is this true or false?

[raven kraken]

True

That upside down T symbol implies orthogonality which that the two vectors are 90 degrees apart

[drifting coyote]

my linear algebra slides say that two vectors are orthogonal if they are non zero and their dot product is 0

wouldn't this mean that the zero vector cannot be orthogonal to the zero vector?

so that would mean a subspace W_1 cannot be orthogonal to the W_2

[jagged raptor]

the zero vector is orthogonal to every vector

[drifting coyote]

does the definition say that?

[jagged raptor]

nope its not part of that definition

[drifting coyote]

so is this wrong?

[jagged raptor]

that definition only talks about nonzero vecotrs

its correct

[drifting coyote]

then how could we know about zero vectors?

[jagged raptor]

notice it says "if they are non-zero and the dot product is zero" not "if and only if"

[drifting coyote]

right

[jagged raptor]

you cant from this definition

[drifting coyote]

(non zero) and (dot product = 0) => orthogonal

right ^

[jagged raptor]

correct

[drifting coyote]

then if we have a zero vector we cannot conclude if it is orthogonal

[jagged raptor]

from this definition alone thats correct

[drifting coyote]

the issue is that this is the only defintion of orthogonal that my professor provided me with

[jagged raptor]

ohh

[drifting coyote]

these were the only two definitions i was bascially given

[jagged raptor]

yeah those say nothing about the orthogonality of the zero vector
if we instead define orthogonality as the dot product being 0, then the zero vector is orthogonal to all other vectors of the same dimension (which is what we normally do)

[drifting coyote]

right

[drifting coyote]

i got my question marked wrong on my homework assignemnt cause i answers this false

from those definitions

[jagged raptor]

i guess it comes down to can we even answer that question if we dont have a notion of "orthogonal" with zero vectors (from the given definitions)

[drifting coyote]

right exactly

[jagged raptor]

can you bring it up with your professor

[drifting coyote]

yea this is my email

@jagged raptor what you think

[jagged raptor]

i think its better if we say that "it would make it impossible to say/determine if the subspace is orthogonal to another subspace"

its certainly not impossible for them to be orthogonal, you just cant say yes or no given those definitions

_ _

the definitions also dont back this up (and generally with other definitions where we say the zero vector is orthogonal to all others this isnt true)

[drifting coyote]

i changed the middle paragraph to this:

[jagged raptor]

that looks good except you put "impossible to say" twice

just make sure you include this bc its the most important part

[drifting coyote]

also i noticed one more thing:
(non zero) and (dot product is zero) => orthogonal
contrapositive is:
not orthogonal => zero or dot product is not zero

does this contrapositive help

[drifting coyote]

lol thanks

[jagged raptor]

uhh i dont think so

btw in the future youre much more likely to get good advice in #linear-algebra

[drifting coyote]

yea cause we cant go from zero => not orthogonal

oh rly okay

also are you in undergrad?

thanks a lot for your help

[jagged raptor]

im going into my first year undergrad this fall

[drifting coyote]

oh wow nice

[jagged raptor]

yep np

[drifting coyote]

i just finished my first year at uni

[jagged raptor]

oh nice so 1 year above me

[drifting coyote]

yep

[jagged raptor]

wait

[drifting coyote]

no way!

wtf

[jagged raptor]

how do you know them LOL

[drifting coyote]

go to #discussion

.close

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[raven kraken]

Oh lol

I did this for a project

hold on

first of all put this into a matrix

and then row reduce it

@acoustic plover

Do you have matlab by chance? well I could do it for you here

Create a matrice this specific matrice takes place in the form

[1 0 0 -1 | 475]
[0 1 0 1| 150]
[0 0 1 -1 |900]

My aologies it's
[1 0 0 1 | 475]
[0 1 0 -1| 150]
[0 0 1 1 |900]

[lone heart]

@acoustic plover Has your question been resolved?

[raven kraken]

Off of this matrix we can also construct ouir own arbitrary nX1 vector'

such that

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✅

[lone heart]

@acoustic plover Has your question been resolved?

[raven kraken]

Yeah you are correct

My project we had a problem similar to this about traffics

and we had to code it using matlab

or it's like the math version of c++

anyways, so essentially yes you do need to solve the system of equations

and the way you do that is you create an arbitrary 4th nx1 vector

and that makes a mxn matrix that you can solve for

The other way you can do this is solving via transitivity

because if x1Rx4 and x4Rx2 and x4Rx3

then that means x1Rx2Rx3

so essentially solve for x4 in the first and then move downward by the row

@acoustic plover I can do the work

related to

well lets just do simple

and create that arbitrary 4th row of the vector so we can reduced echelon form it

Oh you did?

oh shit I didn't notice lemme look

so I'm noticing something that culd also fit in the system

add up R1 + R2 + R3

then you get x1 + x2 +x3 +x4 = 3050

hold on a second lemme work this lol

ALRIGHT I SOLVED IT @acoustic plover

so essentially minimizing

x1 means adding the following row x1 - x4 = 0

not finished yet

from the matrice you gave me

it is assumed now that x4 is a free variable

we can set x4 to whatever value we want

also if x1 is 0 475 btw

x4 = 475

https://i.imgur.com/pgKnMMA.png

According to the information provided here

if you set x1 to 0

but just to make sure I'm gonna graph this

OH THAT was n't eh full equations

okay but first of all though

you should graph it and find the system of equaiton for the minimum

Oh okay

well now that I see the full system of equations

yes that works out just fine

row reduced I got this

yeah that is funcitonally the same

Okay so I just

tracked the eigen value of these variables

x2 has approximately no weight upon the equation

but I agree with your model of making x1 0

I think you'll be fine with it

yeah the math is correct

I got the same thing

remember though that x4 is the real free variable here

so you can't exactly just adjust

x1

so instead of setting x1 = 0

thee line of logic basically says that you have to set x4 to 950

and then x1 will natrually be 0

and if that's true that means x1's relationship to x4 is going to be x1 <= 950-x4

therefore x1+x4-950 <= 0

[lone heart]

@acoustic plover Has your question been resolved?

[lone heart]

@acoustic plover Has your question been resolved?

[nocturne sluice]

Can anyone teach me how to solve question number 2?

[gleaming current]

bruh I am doing the same thing

[nocturne sluice]

Really?

[gleaming current]

yeah

[nocturne sluice]

Can you help me?

[gleaming current]

I am in the same situation as you

🙂

[nocturne sluice]

Damn

[gleaming current]

I think you just gotta solve it hard bro

At least that's what I found

@nocturne sluice f(x) = y ⇔ f− 1(y) = x

[nocturne sluice]

I tried man

[gleaming current]

oh

[nocturne sluice]

Sorry

[gleaming current]

kk

we move out

bro

@thick sleet

Go to one of the empty channels

and then post your question

[lone heart]

@acoustic plover Has your question been resolved?

[lone heart]

@acoustic plover Has your question been resolved?

[elfin snow]

@acoustic plover try asking in #linear-algebra

[lone heart]

@acoustic plover Has your question been resolved?

[robust geode]

@nocturne sluice @gleaming current I can teach you guys but yall will need to go onto DMs so that nobody's bothered in help channels

[nocturne sluice]

I understood the maths

I followed a YouTube video of Brian McLogan

[gleaming current]

Well, I would like to but we certainly at least have a 6-8 hour time diffrence

thanks anyway

[robust geode]

Nah it's okay lol

We can do it now if it's okay

I'm EU btw

[lone heart]

@acoustic plover Has your question been resolved?

[alpine sable]

Lmfao

Still not resolved?

[mortal trellis]

sushi it would help if you clarified what you were asking, what was answered and what you still want to know instead of only answering no to the bot. I doubt anyone wants to read through the whole channel

[alpine sable]

I don’t think sushi used any reactions here lol

It’s only other people

[mortal trellis]

you can go through the list of reactions btw. sushi also reacted. otherwise the channel would be closed by now

[tacit arch]

Yea mods letting channels live for so long (more than 24 hours now) is partly why I quit being a helper. It's ridiculous

[mortal trellis]

how does that have anything to do with being a helper. just ignore a channel if it stays open that long

[tacit arch]

That's exactly why.

Seeing the same channel not go anywhere is annoying

More often than not, they've abandoned the channel

So coming and helping is useless

[mortal trellis]

and that is different when you are not a helper how?

[tacit arch]

Yea cuz I don't get pings or help as much

Whereas before, being a helper, I wanted every channel worked on

[real gazelle]

Lol

if sushi doesn't respond in like a couple hours, I'll close it

[alpine sable]

can all helpers close channels? what other perks am I missing out on lmao. I'm just scared to get the role because of the pings

[slow hound]

same lmfao

[fervent timber]

doesn’t it close if the channel is abandoned?
if they’re constantly answering “no” to the bot, I don’t think they’ve abandoned it

[real gazelle]

yeah all helpers can close channels

[real gazelle]

you can turn off role pings

[alpine sable]

true

[real gazelle]

also I don't care if they keep reacting no, if they don't respond or clarify or smth in the next few hours, I'm closing the channel

[fervent timber]

@acoustic plover

hi

[tacit arch]

Not for over a day sometimes. Mods get mad if you close channels too much. One time I got my helper role taken away because of it

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[zinc cove]

Completely unrelated but thanks for helping me yesterday -- I didn't realize what you were talking about until I looked at the problem again before bed

[zinc cove]

.close

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[tacit arch]

Because you didn't ask your question clearly. Can't help when nobody knows what you're stuck on

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[modern wren]

assuming we are only looking for diophantine solutions. how could i prove these obvious statements?

[lone heart]

@modern wren Has your question been resolved?

[molten pivot]

Mod

Take the second equation mod 3

[modern wren]

can you elaborate for me? im a layman not a student of mathematics

i understand what mod 3 is, im just wondering what applying mod 3 directly implies 😄

@molten pivot sorry to ping you ^

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[meager vapor]

Determine f –1(inverse)(x) if f of x is equal to the quantity x plus 2 end quantity over the quantity x minus 3 end quantity where x ≠ 3.

[weary wyvern]

bruh

$f(x) = \frac{x+2}{x-3}, x\neq 3$

[ocean seal]

giannis_money

[meager vapor]

Would that be the inverse function though?

Wait

No

What would be the inverse of that?

[weary wyvern]

$f^{-1}(f(x)) = f^{-1}\left(\frac{x+2}{x-3}\right) = x$

[ocean seal]

giannis_money

[weary wyvern]

so now

$f^{-1}\left(\frac{x+2}{x-3}\right) = x$

[ocean seal]

giannis_money

[weary wyvern]

Let $y = \frac{x+2}{x-3}$

[ocean seal]

giannis_money

[meager vapor]

See unfortunately that's not one of the answers available

[weary wyvern]

$f^{-1}(y) = x$

[ocean seal]

giannis_money

[weary wyvern]

now, you find x in terms of y

[meager vapor]

What do you think of f-1(x)=x+2/x-3?

[weary wyvern]

i think it could be the right answer but i'm not sure

how did you get that

[meager vapor]

I put it into the calculator

[weary wyvern]

i've checked your answer, it's wrong

[paper ridge]

I think it's clearer to say $x=f(f^{-1}(x))=\frac{f^{-1}(x)+2}{f^{-1}(x)-3}$, then algebraically isolate for $f^{-1}$

[ocean seal]

Tymelord14

[zinc cove]

guys

[paper ridge]

yea

[zinc cove]

f(inverse)(x)=(3x-2)/(x-1)

[weary wyvern]

indeed

[weary wyvern]

good job. but that doesn't help the op

[zinc cove]

Ik

[weary wyvern]

acc nvm it's wrong

[zinc cove]

f(inv)(f(x)) = x, so what you need to do is... excuse me

it's correct

[meager vapor]

What is?

[zinc cove]

the inverse of f (assuming f is a one-one function) maps outputs back to inputs, so what you need to do is solve your original equation for x

this can be done by multiplying both sides of your original equation by (x-3)

distributing

factoring out the x

[molten pivot]

This is a lot simpler than yall are making it

[zinc cove]

and then yeah

[weary wyvern]

no it isn't

[molten pivot]

Yes

Convert the improper fraction into a mixed number

[zinc cove]

it's literally just this

sorry for blur

[molten pivot]

(x+2)/(x-3) = 1 + 5/(x-3)

[meager vapor]

3x + 2, Mada?

[weary wyvern]

didn't know you liked potatos so much

[zinc cove]

excuse me dude, what's with the sass

anyways, yes @meager vapor

[meager vapor]

Alright so it was what I got out of the calculator

Thank you Mada

[zinc cove]

np lol

[meager vapor]

.close

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[zinc cove]

"Akctuallly yr wrong" lol

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[zinc cove]

I made a typo on my original post. My bad but you could be a bit nicer (case in point above):

[zinc cove]

.close

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[proud spear]

so uhh how do I work the bot

[zinc cove]

it should open automatically when you type something I think

hm

.close

weird, I suggest going to a different un-occupied channel in the meantime to ask

whyy

.close

ahhh

.reopen

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✅

[zinc cove]

.close

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[zinc cove]

.reopen

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✅

[zinc cove]

https://tenor.com/view/michael-scott-smh-stressed-gif-20493459

[paper ridge]

just close it and give it some time, it should reset eventually (hopefully)

[zinc cove]

alright lol

.close

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[alpine sable]

What are the diagonal lines supposed to represent?

It's neither the domain nor range nor the image of f, so what are they

[proven cedar]

you see the first one

f

[alpine sable]

huh?

[proven cedar]

its a function that take a real number and give a real number

R --> R

so theres diagonal lines everywhere

the second one

is a function that take a positive real number

and give a real number

[alpine sable]

Yeah, but what are the diagonal lines?

I understand what they represent, but how would you describe their set?

[proven cedar]

in R --> R

the first R correspond to the x axis

the second one to the f(x) axis

its just this

its the intersection of where its defined , and where it give values in

[alpine sable]

No it's not

If that were true, then the second one would not be distinct from the third one

[proven cedar]

Ok gimme 1s

[alpine sable]

But anyway I think I got it

Actually I think I got it myself, it's just ${ (x,y) \mid x \in \text{dom}, f, y \in \text{range},f }$

[ocean seal]

n/c

[proven cedar]

yeah

thats what my paint was pointing

[alpine sable]

but the way they did it made me think it's supposed to be something else, as if it was just the dom/range/image...

Anyway..

.close

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[outer zodiac]

The question is "model 6/(x-9)"

[outer zodiac]

An (unhelpful) example

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[outer zodiac]

No.

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[fickle nacelle]

Don’t know how to do #2

[fallen verge]

,rotate

[fallen verge]

Use pythag

[sage violet]

yeah

[fickle nacelle]

Wont I get g(t) then?

I need g(f)

[fallen verge]

f(t) is one of the legs, the other leg is 90ft

[fickle nacelle]

Yes and by using Pythagorean I get g(t)

I need g(f) in terms of distance f

I am really confused sorry if I am writing something dumb

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@fickle nacelle Has your question been resolved?

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