help-38

[limber minnow]

Because we have that "2x is twelve more then 4y"

[barren hearth]

Wait

If 2x is 12 more

Then why are you adding

To 4y

[limber minnow]

Hmmm...

What if I said "Four is one more than 3"

Does that sentence make sense?

[barren hearth]

Yeah

[limber minnow]

And both 4 - 1 = 3 and 4 = 3 + 1 are true

But 4 + 1 = 3 and 4 = 3 - 1 aren't

So "2x is 12 more than 4y" would translate to 2x - 12 = 4y or 2x = 4y + 12

[barren hearth]

To me it would be 2x + 12 = 4y

Explain how if it says 2x IS 12 MORE your SUBTRACTING 12 from your value instead of SUBTRACTING 12
from the other side to make it 12 less

Instead your making it 12 more

[limber minnow]

Hmmmmmm

Well lets pretend that x = 10 and y = 2, so "2x is 12 more than 4y" would turn into "20 is 12 more then 8"

Is that sentence true?

[barren hearth]

How did it turn into 20 and 8

[limber minnow]

It's because 2x is 12 more than 4y, we need to remove 12 from 2x in order to make it equal to 4y

Ah sorry, I should have been a bit more explicit

[barren hearth]

So because it’s 12 more

You have to make it 12 less

To make it equal?

[limber minnow]

x = 10, y = 2 gives 2x = 20, 4y = 8

[barren hearth]

Huh

[limber minnow]

Yeah, like I said before "4 is 1 more than 3" is a true statement and can be turned into 4 - 1 = 3 or 4 = 3 + 1

Make the bigger side 12 less, or the smaller side 12 more in order to make them equal

[barren hearth]

So your skipping a step basically

So instead of doing

For example

2x + 12 = y

You do 2x = y - 12

But isn’t it already even

[limber minnow]

Not really?
How about this, let's introduce a new variable A and set it to A = 2x

So now "A is 12 more than 4y"

Does that mean that A is bigger than 4y?

[barren hearth]

We don’t know

[limber minnow]

Yeah I'm pretty sure we do, because that sentence gave us that A is "more" than 4y

[barren hearth]

Oh

Yeah if

Ig

[limber minnow]

I'm actually gonna drop the A, not as useful as I first thought

But yeah, so since the sentence gives us that 2x is "more" than 4y, we know that 2x > 4y, right?

So if we want an equality with 2x on the left side and 4y on the right, we would need to either remove the difference from 2x or add the difference to 4y

Does that kinda make sense?

[barren hearth]

Yeah

But how do we know the difference between them is 12

[limber minnow]

Because of the sentence "2x is 12 more than 4y"

[barren hearth]

Ah

So 2x = 12+4y

Ok so then what from there

Cause we can’t solve x and y at the same time

[limber minnow]

Right! Well that's where the equation x + y = 180 comes in

[barren hearth]

Still, we have x and y together

So that would mean x = -y+180

Which we don’t want

[limber minnow]

Yup!

So, do you think you can use x = -y + 180 to solve 4x = 2y + 12 ?

[trim jolt]

@barren hearth Has your question been resolved?

[barren hearth]

Maybe

It would be hard

[limber minnow]

Hmm, well do you have an idea on how you would do it?

[barren hearth]

No

I guess that’s the only way

[limber minnow]

Well using x = -y + 180, do you think you could get rid of the x in 4x = 2y + 12 ?

So that you can solve for y?

[barren hearth]

Yeah cause x would turn into y

And you can solve for y

[barren hearth]

Yeah

[limber minnow]

Alright, so what do you get?

[barren hearth]

hm

One sec

I got y=1

X=-4

that wouldnt make sense

I think once I got x=2y-6 I did something wrong

Is x=2y-6 wrong

[limber minnow]

Yeah there's a couple of issues

First, you're using 2x = 4y - 12, when it should be 2x = 4y + 12

[barren hearth]

uh

It was 2x+12>4y

[limber minnow]

And second, 4y = 4y doesn't give you y = 1

[barren hearth]

What is it

y=0?

oh yeah it is

[limber minnow]

It doesn't actually give you anything

[barren hearth]

so

its nothing

[limber minnow]

Yeah

So you did have the right idea though with replacing x

[barren hearth]

So why is it 2x=4y+12

I moved 12 to the other side

I had to do that by subtraction

2x+12>4y

move 12 to make it even

you have to subtract 12

wait ig I get iut

it

Its 12 more so before setting it up I'd move it

Because if its 12 more and I move it to the other side it'd make both sides equal

24>12 move 12 from 24 makes it 12=24

WAIT WHAT

wait nvm bad example

Idk how it works

Cause if your moving 12

doesnt that make it

24 to 12 to 12 to 24

[limber minnow]

Okay, do you remember what the sentence was that got us the equation?

[barren hearth]

Yeah

Twice the angle of 6 is 12 more than 4 times the angle of 7

or something like that

so it makes it 2x is 12 MORE than 4y

[limber minnow]

"Twice the measure of <6 is twelve more than four times the measure of <7" Yeah

So, 2x is 12 more than 4y

[barren hearth]

yep

[limber minnow]

If you add 12 to 4y, you get 2x

[barren hearth]

so if you move 6 to the other side

wouldnt it be equal

cause if you added 6 the one side loses 6 while the other gets 6 so its a 12 difference between the two

if you added 12 to the other side it'd take away 12 and give 12 which makes it unbalanced by 12

[limber minnow]

You might be thinking about it too hard

[barren hearth]

Well my teachers say that a lot

Well think about it

If you have 12 more than somebody and you have them all 12 of something you have 0 and they gain 12

so its a difference of 24

so then the one side would have 12 more since you went down and the difference is 0 to 12

ugh im definitely making it too hard

i just dont get it when you put it in simple terms

its like

not

iudk

idk

[limber minnow]

Okay, sorry my meds are wearing off and that makes me really scatterbrained so I'm having difficulty finding a place to start

So sorry if I wind up typing for 10 minutes just to send 5 words ^^;

But anyway, I think I might have an idea where you're getting confused?

[barren hearth]

lol ur good

[limber minnow]

How about this, if I have 12 more apples than someone else, then I get rid of 12 apples, wouldn't we have the same number of apples?

Rather than give the 12 apples to the other person, I just toss them (subtract them)

[barren hearth]

show in numbers

no you cant say 12 to 36

btw

ok go

[limber minnow]

If I have 23 apples and you have 11 apples, I have 12 more apples than you

[barren hearth]

Ok give me twelve now I have more

lolol

[limber minnow]

If I threw away 12 of my apples I would have 11 apples left, giving me the same number as you

[barren hearth]

threw away??

Wth

You cant throw away numbers

[limber minnow]

That's sorta what subtracting is though?

It doesn't matter how I got rid of those 12 apples, I now have 12 less apples than I did before

[barren hearth]

Yeah but the other number gets subtracted by it

Its impossible to use it in numbers

if you have 12 more

and give someone 12

now you have 12 less

its simple

if you give 6

you got the same

24 apples to 12

give 6 and you have 18 and they have 18

boom

[limber minnow]

Okay, so how about 2x - 6 = 4y + 6 ?

[barren hearth]

ok do +6

2x=4y+12

[limber minnow]

So this only really applies if you already have an equality, like if I had that a = b then I can't say a + 2 = b but I can say a + 2 = b + 2

But we don't have an equality yet, we're trying to make one

[barren hearth]

ok but the original wasn't set up like that so you can't do that for this example. Original had 2x +12 you cant just subtract 6 out of it, you have to do the whole 12.

ok just go back to the original

this is getting too off track

[limber minnow]

Okay, so we have that 2x is 12 more than 4y

[barren hearth]

can I do this

wait no I cant

ughhh

[limber minnow]

Let's say that y = 10, what would that make 4y equal?

[barren hearth]

40

[limber minnow]

And what is 12 more than 40?

[barren hearth]

52

[limber minnow]

So if 2x is 12 more than 4y, 2x = 52 right?

[barren hearth]

no

not rlly

cause its 12 more so its not =

unless you already added the 12

[limber minnow]

Yeah, but we got 52 by adding the 12 to 40

[barren hearth]

Im just gonna try to erase the thoughts in my brain that say removing 12 and adding 12 to another side is a 24 difference

so im just gonna accept the fact adding 12 from taking 12 away is 12 ig

[limber minnow]

Yeah, the difference should only be 12

[barren hearth]

Your making it equal

So

If its 12 more

then

[limber minnow]

We're not adding 12 to one side and removing it from the other, we're taking away the difference so that they are equal

[barren hearth]

So you just get rid of the 12

How

[limber minnow]

Subtraction!

[barren hearth]

you cant subtract from anything

no like terms

[limber minnow]

You can when you're constructing an equation

We're not modifying an equation, we're building one

[barren hearth]

wait wait wait

so your telling me

I can just like

if something is 24 more than something

I can just remove the number

like

Just throw it out the window

to solve your problems

that cant be legal

[limber minnow]

Not really to solve the problem, but to build the equation that we can use to solve the problem

Or rather, we're trying to translate an english statement into a math equation

[barren hearth]

ok so its just

2x=4y

[limber minnow]

So doing like a + 24 = b and throwing away that 24 to get a = b? Super not allowed

[barren hearth]

Nah you cant do that

Only for making equations

[limber minnow]

But if I have that b is 24 more than a? Then I can definitely say that a + 24 = b

[limber minnow]

Sorry no

[barren hearth]

I thought you just said you can remove numbers when your making equations

so you can throw away the 12

Or are you still adding it to the other side to make it a 24 difference

ughh

i keep with this 24 thing

i just cant think of adding a number to something and making it just 12

thats it

just 12

no removing it from your side as difference

Okay lets go by that

so its 2x = 4y+12

to make it 12 more for the y side ig

[limber minnow]

Sure

[barren hearth]

x=2y+6

[limber minnow]

Yup, and x = 180 - y

[barren hearth]

2y+6=4y+12

[limber minnow]

This is just going to give you y=y again, which doesn't actually mean anything

Try replacing with x = 180 - y instead

[barren hearth]

How am I supposed to remember this on a test

x+y=180

ok ill do y=-3 for y

nvm

im lost

[limber minnow]

Okay so, we have that x = 2y + 6 and x = 180 - y right?

Do you think we could say that 2y + 6 = 180 - y ?

Since both of those things equal x, and thus equal each other

[barren hearth]

How did you get x=180-y

isnt it x+y=180

Which is equivelant to x=180-y

so its the same>

?

[limber minnow]

Yeah, we were talking about that earlier

[barren hearth]

k go on

[limber minnow]

So do you think you could solve for y with this equation?

[barren hearth]

hmm yeah

could we also do x+y=180

and plug in x

[limber minnow]

Sure!

[barren hearth]

to make like terms

[limber minnow]

Could you do that then?

[barren hearth]

sure

is it 64.66

i did it in my head

wait no

no

no

58?

y=58

[limber minnow]

🎊

[barren hearth]

ok so 122

[limber minnow]

Yup!

[barren hearth]

I just realised my teacher purpousely left out this question

I wrote it on my agenda

im in honors classes she said its too hard for her honors people

so she removed it from the book

it was in her ap book or something

[limber minnow]

Huh

[barren hearth]

advanced placement

sometimes college courses

Not saying this is a college question

But

She said none of her students could figure it out

so she removed it from the books

my bad

but at least ik how to solve it lol

im gonna try to do it again but

all at once

cause it was a lot of steps

[limber minnow]

Okay, good luck!

[barren hearth]

So the measure of 6 is 122 or 58

how do I know which is which

it doesnt say which is the supplementary angle

wait now I have to substitute x and y

to find the angle of 6 and 7

[limber minnow]

x was the measure of 6

[barren hearth]

idk

im gonna plug it in

Cause it was 2x

so would it be 244 lol

that would be if it was 360!

So I guess I already solved it for 180

Yeah I got it right

ofc the next questions are easy

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[drifting storm]

Guys, i’m stuck

[burnt mulch]

Quadratic formula/complete the square

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[drifting storm]

Ohhh okok

[drifting storm]

Thankuuu

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[weak vigil]

Like d f(x)/dx is asking, if I change the value of x just a little bit, how many times with the will of f(x) change compared to that small change in x.

[weak vigil]

What is d f(g(x)/ dg(x) saying like this?

in general what does derivative of f(x) mean wrt anything other than x

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[weak vigil]

.reopeb

.reopen

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✅

[weak vigil]

Like d f(x)/dx is asking, if I change the value of x just a little bit, how many times with the will of f(x) change compared to that small change in x.

What is d f(g(x)/ dg(x) saying like this?

[nova spire]

well

if I change the value of g(x) just a little bit, how many times will f(g(x)) change compared to that small change in g(x)

[weak vigil]

that is what I thought

But then what is this saying d f(g(x))/dx

[nova spire]

if I change the value of x just a little bit, how many times will f(g(x)) change compared to that small change in x

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[lunar galleon]

A ball is thrown vertically upwards with an initial velocity of 20ms-1. The ball returns to its original position after 5 seconds.

[lunar galleon]

how do I find how high it went

[undone sierra]

ever heard of the kinematic equation?

[lunar galleon]

s = ut + 1/2at²

that's one I use right

[undone sierra]

well its the same

so sure

[lunar galleon]

.close

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[brittle root]

hi

[brittle root]

what is wrong with this solution

[wraith hinge]

i think it is the mutliplication

a+bi)(a-bi) = a^2-b^2

[brittle root]

(a)^2 - (bi)^2 = a^2 - (b^2)(i^2) = a^2 + b^2

but it does not change anything

its still not zero

[wraith hinge]

yeah that is correct

[brittle root]

and also is this correct? any shorter and easier solution?

[brittle root]

pls help me with this one too

[fresh jewel]

Complex no.?

[brittle root]

yes

[brittle root]

.

[brittle root]

.

<@&286206848099549185>

[fresh jewel]

Use the information that they are on 1 line

You know the complex plane right?

[brittle root]

yes

thats why they have the main form of a+bi

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.close

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[short cave]

A hand of 13 cards is taken out of a well-shuffled pack of 52 cards. How many different hands of 13 cards consist of at least THREE aces and at least THREE kings?

[short cave]

i know how to do this by just adding cases

but i just want to know why (4c3)x(4c3)x(46c7) doesnt work

what is that overcounting ?

[short cave]

ie choosing 3 aces and 3 kings from the 4 aces and kings, then choosing the rest

[nova spire]

so you're allowing the last ace/king to be chosen after the others have already been chosen

thus creating order

for example choosing Aces of spade/heart/club and THEN choosing ace of diamonds in the last 46 cards can also be done by choosing Aces of spade/heart/diamonds and THEN choosing ace of clubs in the last 46 cards

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[ripe valley]

yo

[ripe valley]

for a fully positive function on some set, is it necessary that the sum is greater than the integral ?

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[steel spruce]

Can someone help me understand what the question means realisticly, i am struggling to comprehend what this discrete maths assignment is really asking

[steel spruce]

like I dont understand how to go about finding the complement

[violet gust]

g should be such that suppose
g mod p = a_1
g^2 mod p = a_2
...
g^(p-1) mod p = a_p-1
Then a1, a2, ... are all distinct and {a1, a2, ... ,a_p-1} forms the set {1, 2, ..., p-1} when you sort them in order

say, g = 2 and p = 5, then
2mod5 = 2; 4mod5 = 4; 8mod5 = 3; 16mod5 = 1. So the set becomes {2, 4, 3, 1} which is equivalent to {1, 2, 3, 4}. This makes 2 a companion of 5

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[steel spruce]

Thank you, this logic makes a lot of sense now

[steel spruce]

I understand it was random pick of g = 2, in my case, p = 7, but how can I find g

[violet gust]

You can play around with some random p,g pairs and youd see some properties about companion numbers

e.g can 2 be a companion to 4 ever? what about 3?

(btw 2 was not a random pick. If you can find the pattern, youd see why)

[steel spruce]

But just to clarify what you understand and what I understand, g and a will never be more than 6 in the case of question A because of the range,

[violet gust]

yes

thats what we put as a limit within the definition

[steel spruce]

3 will be a companion but 2 will never be

[steel spruce]

is it something to do with primes

[violet gust]

something like that yes

you can check if 4 is companion to 7

[steel spruce]

in order for my refernce is
4 2 1 4 2 1, so the companions result in a pattern?

In case of original question,
1 has no companion, 1 1 1 pattern?
2 has a 2 4 1 pattern?
3 is a companion
4 has a 4 2 1 pattern?
5 is a companion
6 is a 6 1 pattern?
Thus, even has a pattern, odd has a companion

<@&286206848099549185> I am very confused on what this meant to do?

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.close

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[shut flame]

T, S in End(V) are linear projections
Find a Necessary and sufficient condition on the Kernels and Images in order for ST to be a linear projection.

Can anyone please help me with this question I have been trying to solve it but failed

[sleek canopy]

what is the definition of linear projection ?

[shut flame]

a linear transition T is a projection iff T=T^2

[sleek canopy]

ok

$(ST)^2:=STST=ST\implies ST(ST-I)=0 \iff Im(ST-I)\subset ker(ST)$

[solid kiln]

everg

[sleek canopy]

a sufficient condition could be instead

$ST=TS$

[solid kiln]

everg

[shut flame]

the question wants a condition on kerS kerT, ImS ImT not for ST

[sleek canopy]

oh ok

[shut flame]

ye i saw this

[sleek canopy]

maybe by consider also $ST(ST-I)=0= (ST-I)ST$

[solid kiln]

everg

[sleek canopy]

also $STST=ST\implies S(TS-I)T=0$ 😂

[solid kiln]

everg

[shut flame]

ye i don't think this will help us 😂

[sleek canopy]

who knows xD

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[sleek canopy]

mhh.. interesting

[shut flame]

what do you think of this?

[sleek canopy]

sorry i cannot follow it ... the several steps don't seem trivial to me... I would need more time

[shut flame]

same here lol

if u find anything please tell me

[sleek canopy]

sure

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[shut flame]

.close

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[wraith hinge]

How does 6/sqrt 3 turn into 2 sqrt 3

[ionic pendant]

multiply by (sqrt 3)/(sqrt 3)

[wraith hinge]

Ohhh right

Thanks!

.close

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[marsh forum]

So I'm trying to prove or disprove this statement

[marsh forum]

so I was thinking

$Ax_1=b;Ax_2=b \implies A(x_1-x_2)=0$

[solid kiln]

nobody

[marsh forum]

that's it?

[tepid prism]

is A a matrix

[marsh forum]

I know

[tepid prism]

and are x, vectors

its a question

[marsh forum]

I get that

yes

[tepid prism]

matrices are linear operators so yeah

[marsh forum]

so A(x_1-x_2)=0 implies either A is a null matrix

or (x_1-x_2) is the 0 vector

[tepid prism]

why? its just the kernel of the matrix, if the matrix doesnt have full rank its not always the case?

[marsh forum]

What's a kernel

[tepid prism]

or is my linear algebra memory bad here?

[ionic pendant]

well note that we can use linearity to distribute the matrix multiplication

[marsh forum]

I've only had 2 LA lectures until now, excuse my ignorance

I'm just self studying LTs to gain a head start

[tepid prism]

this matrix here is not the zero matrix

and its not the 0 vector either

[ionic pendant]

i.e. A(x + y) = Ax + Ay

[tepid prism]

this statement is wrong, shown by my example

its only true, when the matrix has full rank

[marsh forum]

oops

Sorry

[tepid prism]

np

[ionic pendant]

think about how we can use this to show A(x1 - x2) = 0

[tepid prism]

yeah and use that matrices are linear operators

[marsh forum]

so Ax_1=Ax_2

so Ax_1-Ax_2=0

[tepid prism]

how would you write down this proof?

[marsh forum]

uh, as x_1 and x_2 are both solutions

$Ax_1=Ax_2$

[solid kiln]

nobody

[marsh forum]

so $Ax_1-Ax_2=0$

[solid kiln]

nobody

[marsh forum]

$A(x_1-x_2)=0$

[solid kiln]

nobody

[tepid prism]

yeah nice, now you might also see that this works for every linear operator

because thats all you had to use

good job

[marsh forum]

Cool

thanks

also congrats on getting helpful?

[tepid prism]

I got it some month ago, but I dont know when

can only mods give this rank, or how does it work?

[marsh forum]

It's auto-assigned

you can also access #「helpers-lounge」 now!

[tepid prism]

very cool, thanks!

[marsh forum]

I think this is true

if A is an arbitrary matrix

[tepid prism]

yeah you can use a very simple example of a matrix here to see this

[marsh forum]

so let A be an arbitrary $n \cross m$ matrix

[solid kiln]

nobody

[marsh forum]

I have to prove this

[tepid prism]

is A a fixed matrix

or is x fixed

[marsh forum]

I mean we can't take an example of A

$\begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1m} \
a_{21} & a_{22} & \cdots & a_{2m} \
\vdots & \vdots & \ddots & \vdots \
a_{n1} & a_{n2} & \cdots & a_{nm}
\end{bmatrix} \begin{bmatrix}
v_1 \
v_2 \
\vdots \
v_n
\end{bmatrix}$

[solid kiln]

nobody

[marsh forum]

wait

[tepid prism]

my question is just, what are we given, we are given a linear combination of vectors in R^n, in a specific basis, in a generell basis?

is the vector x fixed? is A fixed?

[marsh forum]

oh wait

R^n

oops

[marsh forum]

this is the OG questio

[tepid prism]

so we are given a linear combination and want to find a Matrix A and a vector x, such that the linear combination is equal to Ax

if the statement is true, if it is wrong, we want to find a linear combination that cant be expressed as a Matrix vector product

[marsh forum]

I mean given an arbitrary basis and A, we can span the entirety of R^n

that;s the question, right

[tepid prism]

we are not given A

we are given a linear combination of vectors in R^n

we want to find a vector x and a matrix A, such that the linear combination of vectors is equal to R^n

[marsh forum]

What are we given

I don't follow

[tepid prism]

we are given a linear combination of vectors in R^n

in a given basis

we want to show that

"Every Linear Combination of vectors in R^n can be written in the form Ax"

so we take any Linear Combination and need to show we can write it as Ax

[marsh forum]

Hmm

so I'm given $x_i \forall i \in {1 \leq i \leq n}$

[solid kiln]

nobody

[tepid prism]

maybe recall the definition of a linear combination

[tepid prism]

close, right idea

or maybe im wrong

what are you x_i here?

[marsh forum]

Oh wait

there's a trivial counter example

let all $x_i$ be scalar multiples of one another

[solid kiln]

nobody

[marsh forum]

to any linear combination WON'T span $\R^n$

[solid kiln]

nobody

[tepid prism]

what are your x_i?

[marsh forum]

say $\zeta_{i} x_i$ where $x_i \in \R^n; i \in {i|(1 \leq i \leq n) \land i \in \N}; \zeta_i \in \R$

[solid kiln]

nobody

[tepid prism]

so basically just one basis vector scaled?

now you need to add them all together and thats your linear combination

$\sum_{i=1}^n \zeta_i x_i$

[solid kiln]

tobi

[marsh forum]

Well, this DOESN'T span $\R^n$

[solid kiln]

nobody

[tepid prism]

yeah it just spans a subspace I agree with that

but this doesnt disproof the statement

[marsh forum]

Why doesn't it

[tepid prism]

x_i are your basis vectors

and now your subspace just lifes on a 1D line in R^n

however I can still find matrix and a vector such that $\sum_{i=1}^n \zeta_i x_i = Ax$

[solid kiln]

tobi

[tepid prism]

you would know need to show that you cant express your vector, as Ax

you can always express your vector in R^n using the standard orthonormal basis, would you agree with me on that

we can do the proof later, maybe focus on intuition first

this is just an example, but given this basis of R^2, we can show that any linear combination of this basis can be expressed in terms of the standard basis of R^2

[marsh forum]

yes

I think the issue is I don't fully understand what;s given and what we have to prove

[tepid prism]

you are given a set of linear independent vector, lets say ${b_1, ... , b_n}$ that span the whole $\mathbb{R}^n$

now given a linear combination

$\sum_{i=1}^n \zeta_i b_i$

show that

$\sum_{i=1}^n \zeta_i b_i = Ax$ for a Matrix A and a vector x, in other words, find A and x

[solid kiln]

tobi

[tepid prism]

where zeti_i are scalars

thats how I understand the question

[marsh forum]

oh

That shouldn't be too hard

[tepid prism]

because to be given a linear combination does only make sense if we are given a basis, so lets take an arbitrary basis

[marsh forum]

What to I assume A's order is

n cross n

or m cross n

[tepid prism]

n cross n, dimensions need to match

[marsh forum]

true

oops

[tepid prism]

maybe recall what a matrix vector product looks like in its summation notation

[marsh forum]

\begin{bmatrix}
\zeta_{11} & 0 & \cdots & 0 \
0 & \zeta_{22} & \cdots & 0 \
\vdots & \vdots & \ddots & \vdots \
0 & 0 & \cdots & \zeta_{nn}
\end{bmatrix} \begin{bmatrix} b_1\b_2 \ \vdots \ b_n \end{bmatrix}$

This spans $\R^n$

[solid kiln]

nobody

[tepid prism]

I don't really like this notation here

[marsh forum]

wdym

[tepid prism]

what are your b_i

[marsh forum]

Any arbitrary linearly independent basis vectors

[tepid prism]

then what is this?

[marsh forum]

[solid kiln]

nobody
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[marsh forum]

there we go

[tepid prism]

is b_1,...,b_n a vector?

[marsh forum]

yes

[tepid prism]

so b_1, ...,b_n are scalars?

and in what basis is your vector

[marsh forum]

R^n

[tepid prism]

usally this [] notation is only used for the standard basis meaning (1,0,0..,0) and so on

[tepid prism]

this basis?

[marsh forum]

yeah

or any set of perpendicular vectors

[tepid prism]

you are given a set of linear independent vector, lets say ${b_1, ... , b_n}$ that span the whole $\mathbb{R}^n$

now given a linear combination

$\sum_{i=1}^n \zeta_i b_i$

show that

$\sum_{i=1}^n \zeta_i b_i = Ax$ for a Matrix A and a vector x, in other words, find A and x

[solid kiln]

tobi

[tepid prism]

b_1, .., b_n are vectors here

I mean you could say that

$\sum_{i=1}^n \zeta_i b_i = \sum_{i=1}^n c_i e_i$

[solid kiln]

tobi

[marsh forum]

eh, I could say that $b_i = \sum_{j=1}^{i} \omega_j e_i$

[solid kiln]

nobody

[marsh forum]

I suppose

[tepid prism]

you could say that $b_j = \sum_{i=1}^n c_i e_i$ for all $1 \leq j \leq n$

[solid kiln]

tobi

[marsh forum]

yup

[tepid prism]

or in other words this

$\sum_{i=1}^n \zeta_i b_i = \sum_{i=1}^n c_i e_i$

[solid kiln]

tobi

[tepid prism]

given zeta and b

you can solve a linear system of equations to get c_i

[marsh forum]

yup

So that prove it, right

[tepid prism]

what vector and what matrix would you choose here?

at the end

[marsh forum]

vector $e_i

hmm

let me think

is this supposed to be an easy question

[tepid prism]

yes

its already written

[marsh forum]

I think I first perform some row operations?

[tepid prism]

we are using the change of basis here! this a very important statement about vector spaces, I would suggest checking out the proof of this or trying to proof this yourself.

[marsh forum]

I actually don't think my book has even covered basis or vector spaces yet

[tepid prism]

what does the last vector look like in this [] notation?

[marsh forum]

yup, chnage of basis is chapter 9

vector spaces is chapter 6

I'm in 2

Let me think though

[tepid prism]

okay then I guess this question was supposed to be way easier, I think the book wanted you to proof this with the standard basis, not any basis

[marsh forum]

I'l still try for any basis

Because I've already done it for the standard basis

[tepid prism]

you need this then, which is intuitive, but this would need to be proven as well

[marsh forum]

I mean that's not hard to prove

Let me see if a soln is given

[tepid prism]

yeah you are right

e_i are basis vectors of R^n so thia always works

for any given vector

and b_j is just another given vector of R^n

[marsh forum]

This is what my book says

[tepid prism]

sorry I have to sleep now

good luck

[marsh forum]

Thanks

gn

ohh

I think I got a proof

$\begin{bmatrix}
\alpha_{11} & \alpha_{12} & \cdots & \alpha_{1n} \
\alpha_{21} & \alpha_{22} & \cdots & \alpha_{2n} \
\vdots & \vdots & \ddots & \vdots \
\alpha_{n1} & \alpha_{n2} & \cdots & \alpha_{nn}
\end{bmatrix} \begin{bmatrix} e_1\e_2 \ \vdots \ e_n \end{bmatrix} $

[solid kiln]

nobody

[marsh forum]

So assuming consitncy, we can express each vector in terms of the basis vectors

is this fine?

[trim jolt]

@marsh forum Has your question been resolved?

[marsh forum]

<@&286206848099549185>

[marsh forum]

Is this fine?

.close

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[wraith hinge]

is the solution to this question valid? tan(90-x) = 3 and tan(y) = 2/3 and since tan(x) is increasing on (0,pi/2) U (pi/2, pi), we have that 90-x > y

[wraith hinge]

it looks correct to me and it probably is correct but just need clarification cuz i have brain damage 😭

[burnt mulch]

Is fine

[wraith hinge]

yooo thanks 💅

[wraith hinge]

the last time i posted this shit we wuz all yapping like crazy

for one line

in here too

.close

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[trim jolt]

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

I don't understand the first step. Is long division method used here?

[pale juniper]

Um i have a slightly different method

I'll write it

[ashen quiver]

Sure

[split chasm]

i don't know what they're doing there

[vestal token]

Which klass bro

[ashen quiver]

12th klass

[split chasm]

but long division is usually what you'd start with

[vestal token]

Dayum bro

[pale juniper]

This is what I did

I first split it

Hope u got it

[split chasm]

Please try to avoid giving out fully worked solutions like that.

[pale juniper]

Oh yeah I'm sorry

[split chasm]

Instead guide them through step by step and have them try do the work themselves.

[ashen quiver]

Yeah I'm Trying to understand

[pale juniper]

Will do from next time

[ashen quiver]

So, I divide 1-x² by x-2x²?

[split chasm]

that'd be how i'd approach it

[pale juniper]

Did u understand

[naive otter]

How about 1/4 ×( (1-4x²) + 3 )/(...)

[ashen quiver]

I'm trying the long division method still

[pale juniper]

Ok

[ashen quiver]

Ok I got it, thanks y'all.

.close

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[marsh forum]

!status

[trim jolt]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@frank inlet Has your question been resolved?

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[lethal bone]

With a model like this, how do I know if the line MD is passing through the plane, going behind it, or in front of it?

I really struggle with seeing 2D drawings in a 3D perspective so any tips on figuring this out would be lovely

[trim jolt]

@lethal bone Has your question been resolved?

[lethal bone]

<@&286206848099549185>

[cinder sentinel]

Yes?

[lethal bone]

question above

[keen lark]

which syllabus does the question belong too

[lethal bone]

wdym

[keen lark]

like in which class do u learn this

[lethal bone]

geometry

@frail heron you helped me before, ru still available?

[trim jolt]

@lethal bone Has your question been resolved?

[lethal bone]

what do the dot lines mean?

[lethal bone]

tysm

.close

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[jovial anchor]

Hi

[rotund grove]

hi

[jovial anchor]

I need to find the contact point with the X axis for each formula

I'm kind'a new at this

how can I do that?

[rotund grove]

each formula?

[jovial anchor]

yes

[rotund grove]

wdym

each question?

[wraith hinge]

on the x axis, the y value is 0, so set y = 0 and solve for x

[trim jolt]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[jovial anchor]

y is always 0...

[wraith hinge]

no

[jovial anchor]

This is not homework

[wraith hinge]

set y = 0

[jovial anchor]

It's a course I'm taking for an exam

[jovial anchor]

set?

how do u set y

[wraith hinge]

Yes

those are functions and you want to solve for x where y = 0

[jovial anchor]

okay

[rotund grove]

you see, the point of contant will be on the x axis and therfore the y coordinate will be 0

so the coordinates are of the form (x,0)

put y = 0 in the equation

0 = 3-x

x = 3

[jovial anchor]

so instead of y = mx + b

[wraith hinge]

no sols

[jovial anchor]

we have 0 = 3 - x

right?

[rotund grove]

i mean you just put the values

[rotund grove]

that's the general representation of a line

[jovial anchor]

so then it's pretty simple x = 3

[rotund grove]

yes

the main part was knowing that y = 0 at the x-axis

[jovial anchor]

so then this would be correct

[jovial anchor]

oh ok

[charred atlas]

shit

[rotund grove]

no, (3,0)

[wraith hinge]

no

[charred atlas]

whats kind of lang is this

[rotund grove]

-3 ain't 3

[jovial anchor]

oh right 💀

there we go

okay, onto the next one

uh...

we have 1/3x + 6

wait, how do we know that y = 0 in all of them?

[rotund grove]

1/3x + 6 = 0

again set y = 0

yeah bro!

[jovial anchor]

but how do we know y is always 0

[jovial anchor]

so just set it on 0 by default? why's that

we get 0 = 1/3x + 6

[rotund grove]

yeah

because they're asking you to tell the point of intersection of the x-axis and the line, in which case the ordinate is always 0.

[jovial anchor]

okay? so

In the second one, I got to 6 = 0.3x

I don't think that's right

okay so then 6 * 3 (because 1/3) is 18

and since it switched sides we get -18

so x = -18?

I think I got it right, I didn't get these two perfect tho

I didn't understand em

[pure totem]

hi

[jovial anchor]

I'm also supposed to do this

In translation: I need to find the cotact rate points of the purple line with the axes (y and x)

y=2x+6 (green)
y=−4x+12 (purple)

how can I do that? what does that even mean

.close

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[hidden lance]

How do I prove that this equation equals arctan(1/2)?

[sleek canopy]

maybe taylor expansion could be useful here =

?

[hidden lance]

Oh I didn't try that

But what would be the next step after the taylor expansion though?

[wraith hinge]

find the radius of convergence

[hidden lance]

ok

[wraith hinge]

help fr

how do i find B

found A

[hidden lance]

go to an available help channel

#help-48

[wraith hinge]

okk

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[vast light]

This is trigonometry with analytic geometry, the topic is similar triangles. I'm so confused on how to even start on this any help is appreciated

im just now starting trig and already struggling 💀

[wind cloak]

ratios of corresponding sides of similar triangles are equal

[charred atlas]

So if the ratio here

Its 51:1

[wraith hinge]

[charred atlas]

Find the length of C
$$51.1,1=56,1$

When you figure it out 2 length, C and B

You can get a lil creative xd

[vast light]

oh ok so the ideal would be to figure out one length to figure out the other?

[charred atlas]

Pythagorean theorem or sth

[vast light]

yea

[charred atlas]

Yea

[vast light]

ok thank you one sec

[charred atlas]

Bc its cool xd

Or just stick with the ratio

[vast light]

@charred atlas b=51 is a leg right that wouldnt be the hyp?

[civic forum]

hyp is the longest side of a right triangle and its the opposite of the 90 degrees angle

[vast light]

ok thank you

[charred atlas]

C =56.1

And

Nvm he solve ur problem

[vast light]

no im still trying to figure out how u got the C

[civic forum]

''ratios of corresponding sides of similar triangles are equal''

so since b is 51

and the ratio for b is 1

ah its a bit rough to explain

do you know how ratios work?

[vast light]

not really sorry my teacher for this class doesnt really explain anything but the main basics such as finding the X through different problems

do u know a youtube i could watch?

[civic forum]

i dont really watch anything besides calculus videos but i can try my best to explain ratios

lets imagine there is a number k

[vast light]

i found a video

[civic forum]

lets not imagine there is a number k

oh wait

actually i found it

you see the sides of the triangle?

[civic forum]

on the left

the hypotnuse is 1.1
the leg is 1
the height is 0.47 right?

to define the ratio we can give a little letter k next to it, so on the right triangle its gonna be
$c=1.1k
b=1k
a=0.47*k$

bruh

[civic forum]

this k is gonna be our ratio constant and it will help us solve all the problems

for example on question 13 it says b=51 right?

and we made b equal to 1*k

so that means 1*k=51

and so k=51

now to find c and a we need to just put 51 instead of k

you basically do the same thing for all the rest of the question

@vast light

[solid kiln]

keb abs

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[hidden lance]

How do I calculate the Taylor's remainder of this equation?

[hidden lance]

Apparently I did it wrong but i'm not sure where i went wrong

0.9 < x < 1.1 and c = 1

.close

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[lofty elm]

This is a proof that I want to analyze before googling the answer, does my reasoning here seem valid?

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[tawny kettle]

i think it makes sense. the last two paragraphs seem weirdly worded, but i think it's correct if you mean $n^2=2 \rightarrow n=\sqrt{2}$ causes a contradiction, then $n^2=2\cdot2k$, so $\frac{n^2}{2k}=2$ when k is a perfect square $k<n^2$ as you defined.

[solid kiln]

fish

[lofty elm]

Thanks!

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[dry holly]

is there any way to solve a system of 2 quadratic inequalities algebraically

[dry holly]

without graphing

[nova spire]

you can solve inequality per inequality. after that you take the intersection to find the real numbers that verify both inequalities

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[dry holly]

that works with a system of 2 variable quadratic inequalities?

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[nova spire]

no matter how many equalities you want, and doesn't have to be quadratic

when you want to solve "f(x) > ... AND g(x) > ... AND ..." as an example

you solve one at a time

for example you start with "f(x) > ..."

gets you to a set of solutions, call it $S_f$

[solid kiln]

rafilou2003

[nova spire]

then continue one by one

get $S_g$

etc...

[solid kiln]

rafilou2003

[nova spire]

and then the set of solutions for ".... AND ... AND..." will just be $S_f \cap S_g \cap ...$

[solid kiln]

rafilou2003

[dry holly]

i see so like

i just graph both and the union of both or however many inequalities there are is the set of answers

but my question was like is there a way to solve it without using any graphs whatsoever

wait so if i had something like

y > (x-4)(x+3)

and y > (x-5)(x+2)

for the first one (-3, 4) is a set of solutions

for the second one (-2, 5) is a set

so for both of them as a system (-2, 4) is the set of solutions?

oh shoot i get it now i was tripping

thanks tho i appreciate it

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[manic jackal]

Let ( d : M \times M \to \mathbb{R} ) be a function such that:

1. ( d(x, y) = 0 ) if and only if ( x = y ) \
2. ( d(x, z) \leq d(x, y) + d(y, z) )

Show that ( d ) is a metric.

[manic jackal]

I firstly tried d(x, x) <= d(x,y) + d(x, y) = 2d(x,y) implies 0 <= d(x,y)

[solid kiln]

Halex

[marble wharf]

pay close attention to the order of x and y

you dont know d(x,y)=d(y,x)

[manic jackal]

yep, if I had that from 2. the triangle inequality is followed, I also tried to use 2. to show the symmetry property

[nova spire]

I'm not even sure myself this is possible without symmetry or an additional property

[marble wharf]

=0 is possible

[nova spire]

to prove

[marble wharf]

but I am really not sure about symmetry

[nova spire]

not even sure you can prove that

[marble wharf]

you can

no wait, mistake on my part

hmm, maybe not

[nova spire]

d(a,b) = 1, d(b,a) = -1?

am I missing something

[marble wharf]

you mean M={a,b}?

[nova spire]

yeah

[marble wharf]

dont have paper on hand, cant check if it satisfies all option for the triangle inequality

[nova spire]

well it's either stay + forth or back + forth combo

so -1 <= 0 - 1, 1 <= 0 + 1 or 0 <= 1-1

game over

[nova spire]

your question is impossible to prove because it's false

[nova spire]

the following over M = {a,b} disproves it

you at least need symmetry to work

[manic jackal]

How is d(a,b) = 1 and d(b,a) = -1?

[nova spire]

I just defined it that way

[manic jackal]

it should be symmetric for it to be a metric

[nova spire]

if you add a third condition for symmetry

[nova spire]

then this reasoning would have worked

where does this problem come from btw?

[manic jackal]

was thinking the symmetry had to be proven from the properties of this function

[manic jackal]

from my lectures

the main textbook is elon lima's

[nova spire]

real analysis?

If you have a picture/screenshot/link or anything that shows the problem I'll take it

[manic jackal]

"curso de analise" the author is Brazilian

[manic jackal]

It's one of the exercises proposed in my class notes

[nova spire]

damn

then do show your teacher the counterexample I gave you

M = {a,b}

d(a,a) = d(b,b) = 0

d(a,b) = 1, d(b,a) = -1

[manic jackal]

does this example satisfies the 2. property?

[nova spire]

this part

you can go through d(a,a) + d(a,b) and d(b,b) + d(b,a) and similars

which either give you -1 <= 0-1 or 1 <= 0+1 or swap the order

so the "stay + forth" cases are taken care of

and you're left with d(a,b) + d(b,a) and d(b,a) + d(a,b) cases, which are both 0 <= 1 - 1

[manic jackal]

d(a, b) <= d(a,a) + d(b,a)

[nova spire]

no that's not true

you don't have symmetry

to go from point a to b with one step in between, you can't go from b to a

you either station at a and then go to b

or go to b directly and then station at b

so you can either bound d(a, b) <= d(a,a) + d(a,b)

or d(a, b) <= d(a,b) + d(b,b)

but there's no "going from b to a" as in d(b,a)

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[marsh forum]

So I'm trying to prove $|a+b| \leq |a|+|b|$, $a,b \in\R$