help-28

[noble pelican]

Vector Space

You imagined correctly

[surreal hawk]

Would it be a good idea to first write a basis for the kernel and of the pre-image?

[nova basin]

it would be a good start to find all the information you can get about F from its kernel

and the 2 given values

[noble pelican]

Ok, I'll review my notes and see what I can come up with, hold on

[nova basin]

You should be able to find the images of each basis vector of R^5 directly. I found them in the order ||2->5->1->3->4||

[noble pelican]

Ok, I got a lead, I should probably reverse engineer the linear system given to me, to find the matrix associated with F

Although in this case the matrix that I'll find through Ker(F) it's gonna be a Gaussian reduced matrix, can I still consider it associated with F?

[noble pelican]

The standard Basis of R^5 should be B = { 10000, 01000, 00100, 00010, 00001} yeah?
But how do I find the images, I don't have F yet

[nova basin]

the kernel gives you information

[noble pelican]

Oh ok, then back to work

[nova basin]

you can write F(x1, x2, x3, x4, x5) = x1P1 + x2P2 + x3P3 + x4P4 + x5P5 where the P_i are polynomials

the kernel gives you info on the P_i

[noble pelican]

I thought you had some advanced mindblowing Algebra theorem to find them just with a look

[nova basin]

with the 2 values given, you have just enough to find F

[full forum]

@noble pelican Has your question been resolved?

[noble pelican]

I'm missing something, a connection
I tried reverse engineering the matrix, and found the T(b1), T(b2) .... T(b5) but they do not look right
(Image 1)
So then I went and found the space of the soution of Sol(Ker(F), 0) but I have no idea what to do with that information:
Sol(KerF, 0) = [ a(1, 0, -1, 1, 1, 0) + b(0, 1, 0, 0, 0) + c(0, 0, 0, 0, 1)]

[nova basin]

have you find F before trying to find its matrix ?

[nova basin]

because you don't need to think of it as a matrix to find it here. Arguably the abstraction can even make it harder

[nova basin]

ok lemme check

I don't think P5 = 0

notice that P1 = -P5

similarly P3 = -P4

that's how you find all the polynomials: you use this to simplify the 2 given values

[nova basin]

but yes, P2 = 0

[noble pelican]

I wrote it wrong

I wrote x1 + x3 = 0 instead of x1 + x5 = 0*

[nova basin]

I didn't look at the matrix actually

[nova basin]

I just looked at the polynomials you wrote below

[noble pelican]

Let me try again with this new values

Hints that T(b2) = 0 keeps appearing in my calculations, but I do not understand why its implied

[nova basin]

look at the kernel

is (0, 1, 0, 0, 0) in Ker F ?

[nova basin]

don't know how you can end up with some t² and t^3 if you're only combining linear functions

[noble pelican]

I don't see it, which means that T( 0, 1, 0, 0, 0) != 0?

[nova basin]

it verifies both equalities

therefore P2 = 0

since there is no constraint on x2, it means that x2 always verifies the constraints, so if x1, x3, x4, x5 do as well, then it's in the kernel. Therefore F(0, 1, 0, 0, 0) = 0

[noble pelican]

Okay, let's back up, I need to review the basics here, first of all:

1. x = (x1, x2, x3, x4, x5) is a vector of COOORDINATES of the vector v that belongs to V
2. Ax = 0 has solutions Sol(A, 0) which are a bunch of vector of Coordinates, they do not make the basis of Ker(F) yet.
3. Through the use of the basis of V + Sol(A,0) (vector of coordinates) we can find the vectors to form basis of KerF.

Is what I stated right?

[nova basin]

1. x is vector of R^5, not a polynomial

2. Fx = 0 is equivalent to x in ker F, that's more than the basis

3. don't know what you mean

[noble pelican]

In linear Algebra, when we talk about the ker(F), don't we get it's System of linear equations through a matrix?

[nova basin]

Ker F is a vector space. You describe it however you want

it can be through a system of equations, like what they've done here

[noble pelican]

okay isn't the next step to solve that system of linear equations?

[nova basin]

that's not the goal here

[noble pelican]

Okay, I know you already told them to me, but can you explain me how did you solve this exercise, step by step, with a logic explanation behind each one?
I really wanna understand your point of view but it feels like I was taught one way to do stuff that's useless in this context, although related.

[nova basin]

is it fine to take P2 = 0 as a starting point or do I need to reexplain that ?

[noble pelican]

Let me see if I understood correctly, in the System of linear equations (SLE) only x1, x3, x4 and x5 are involved, they describe kerF(?), and since there is no x2, it has no constraints(?) which in turns means it verifies this constrains( I don't understand what are the constraints in these context)

So P2 = 0

(?)

[nova basin]

(0, 1, 0, 0, 0) verifies the equations that describe Ker F right ?

[noble pelican]

Yeah, the constraints of the SLE

Since if we apply it, there is no problem

[nova basin]

alright, so P2 = 0

[noble pelican]

Yes

[nova basin]

then we had F(0, 1, 0, 0, 1) = 1 + t + t²
Can you find P5 from that ?

[noble pelican]

A clarification, with P2 you mean T(01000)?

[nova basin]

I'm still sticking to that

[noble pelican]

Yeah just subtract T(01000) from it which leaves it with the same value

[nova basin]

yes

so P5 = 1 + t + t²

but we had P1 = P5 didn't we ?

because F(1, 0, 0, 0, -1) = 0

that means P1 - P5 = 0

so P1 = P5

right ?

[noble pelican]

Uhm

[nova basin]

I didn't actually do the computations so I guessed wrong

mental math is not as reliable as written math

[noble pelican]

I agree with that thought

[nova basin]

so is that good ?

[nova basin]

because here x1 + x5 = 0 so we're in the kernel

[noble pelican]

x1 + x5 you mean.

[nova basin]

yes

[noble pelican]

Sorry if I' a bit slow to catch on but these are all new logic steps to me

But I understand

P1 = P5

👌

[nova basin]

then you need some practice. Can you use some similar reasoning to find P3 and P4 ?

[noble pelican]

Yeah, I'll try thanks

[nova basin]

how is it going ?

[noble pelican]

Good, I found P3

P3 = -1 + t -t^2 -t^3 = P4 right?

[nova basin]

yes

[noble pelican]

it took me a few minutes to remember they gave those two images.
Also I rewrote every step to find P2, P5 and P1

Okay, I'll go ahead and write F (I think I can do that right?)

[nova basin]

I sure hope you can add a few polynomials

[noble pelican]

It's a bit long though:
F(x1, x2, x3, x4, x5) =
x1+x5-x3-x4 + (x1+x5+x3+x4)t +(x1+x5-x3-x4)t^2 - (x3+ x4)t^3

[nova basin]

it is a bit long yes

P1 = P5 = 1+t+t²
P2 = 0
P3 = P4 = -1 + t - t² - t^3

[nova basin]

agreed. With that you should be able to find the matrix that represents F

[noble pelican]

Question, do you know what the linear extension principle is?

[nova basin]

I can take a guess, but I know nothing of that name

is it that a linear application is uniquely determined by the image of a basis of the input space ?

[noble pelican]

It allowed the professor's solution to unite both the vectors that form the basis of Ker(F) (found through the SLE) and the vectors the he has gives us to find a basis of R^5, and basically helped him find every Pi.

By just combining these vectors

He also used it as a reason to prove that this linear map was unique (which was a request of the exercise).

[noble pelican]

Nvm then, thank you a lot for your help!

I find your solution cleaner btw

[nova basin]

sounds like completing a basis

where you have Ker F = Span({(1, 0, 0, 0, -1), (0, 1, 0, 0, 0), (0, 0, 1, -1, 0)}) and you add 2 vectors to get a basis that's fitted to F

and then rewrite it to get it in terms of the standard basis

[noble pelican]

how do you rewrite it?

[nova basin]

some expand then factor similar to this

I suppose

no

you find the linear combination that gives each basis vector and find its image by F

that's how I'd do it

which feels like a rather long and tedious process

[noble pelican]

You mean you can add ANY two vectors and it will still form a basis of F?
Aren't there any rules or restrictions to this, seems a bit OP.

[nova basin]

of course not

2 vectors that make a free family

so you have a free family of 5 vectors for a 5D vector space, so it's a basis

[noble pelican]

What's a free family?

[nova basin]

synonym to linearly independent

[noble pelican]

Oh ok, so I guess he somehow verified that together they were linearly independent

[nova basin]

probably

[noble pelican]

Well then I think that will be all for today, thank you for your time and efforts, if you ever come to Italy, I'd like to offer you a pizza and Coca-Cola 🙂

[nova basin]

questions 2 and 3 are fine ?

[nova basin]

actually, the reason he did that was because it directly answers Q2

[noble pelican]

oh yeah

but wait, first he needs to prove that those two vectors alone are a subvector space U

[noble pelican]

Actually no

But I thought that keeping you here for 2 hours was enough

[nova basin]

it's actually been 2h30 yeah

but I do other stuff in the meantime

[noble pelican]

depression

[nova basin]

I was more interested by Q3 actually

the more interesting stuff is the more complicated questions

[noble pelican]

Okay then give me 10 min. and I'll be back, I need to process this last 2h30min

[noble pelican]

I can relate but on a different context

actually I'll be back in 20 min

[nova basin]

brb, probably > 30min

[full forum]

@noble pelican Has your question been resolved?

[nova basin]

still here ?

[noble pelican]

yeah

They are linearly independent

But I can't find a feasible way to prove they form a subvector space of R^5

[nova basin]

that's regular vector space theory

for any S subset of V, span(S) is a subspace of V

also since it's a linearly independent family of 5 vectors in a 5D vector space, it is a basis

[noble pelican]

Yes S has to be a subset, but it also needs to respect the conditions for it to be a vector space, i.e. we have to verify that any linear combination of its element has to stay in S

[nova basin]

that's what the span operator is for

[noble pelican]

guess what

the professor never explained us what a span operator is, I can guess that it represents the linear combinations of the vectors involved in the span operation

[nova basin]

given a subset S of V, span(S) is
the smallest subspace of V that contains S
the set of all linear combinations of elements in S

$Span(S) = {\sum_{s \in S} a_s s \setminus (a_s)_{s \in S} \in \mathbb{R}^{(S)}}$

[glossy valve]

themateo713

[nova basin]

i.e. all finite linear combinations of elements of S

[noble pelican]

So to explain to my prof. that the subset of {v1, v2} is a subvector space I just have to tell him Span(v1, v2)?
I don't think that's gonna fly, on what basis I can confidently say that U it's a vector space, therefore a subvector space?

[nova basin]

{v1, v2} isn't a vector space

Span(v1, v2) is, by definition, a vector space

[noble pelican]

Span( (0,1,0,0,1) (0,0,1,0,1) ) by definition is a vector space then

[nova basin]

yes

compactly it's {(0, a, b, 0, a+b) \ a, b in R}

[noble pelican]

U = {(0,1,0,0,1)(0,0,1,0,1)}
Span(U) is a vector space
But U isn't or at least I haven't proven it yet

Oh no, don't tell me the answer is just writing out:
The subspace vector is Span((0,1,0,0,1)(0,0,1,0,1))

[nova basin]

that's a way of writing it

it depends on how you want to express it

that's why a lot of math questions are "prove this" rather than "find that" because they want you to figure out how to write it a certain way

[noble pelican]

It feels like cheating, then again in the solutions the professor never stated why those two vectors form a basis

[nova basin]

they don't

it's the 2 vectors + a basis a Ker F. That makes a basis

if the 2 vectors are chosen correctly

[noble pelican]

A subspace vector U found is the one that has as basis (0,1,0,0,1)(0,0,1,0,1)

its what he wrote

[nova basin]

as a supplementary to Ker F ?

[noble pelican]

"Il sottospazio U cercato è quello che ha per base (0,1,0,0,1), (0,0,1,0,1)"

[nova basin]

gl with that handwriting lol

[noble pelican]

doesn't state it

[nova basin]

sottospazio ?

[noble pelican]

No kidding, he mastered the writing of a doctor

His notes are incomprehensible

[noble pelican]

subspace / subvectorspace

[nova basin]

is that his entire answer ?

[noble pelican]

yes 😦

[nova basin]

damn

rip

the least you could do is prove it is one

[noble pelican]

all of his answer are disjointed and minimal (As stated in my original call for help, I mean question)

[nova basin]

a better way would be to explain where it comes from

rip

[noble pelican]

Yep, and why

if you want we can move to the next question sigh

[nova basin]

btw here it's that you know they're not in Ker F because you're given their image. And it's quite obvious that they're linearly independent

then it should be linearly independent with the kernel's basis. I think that might be provable in the general case, but I don't want to say something I can't give a proof for

[noble pelican]

yep, not in KerF, checks out

[nova basin]

think that's true

If they were linearly dependent, then you could express 1 as a function of the others. Applying F to that, we find that their images are linearly dependent, which is wrong

[noble pelican]

They are linearly independent, if you solve the System of Linear Equation associated with those 5 vectors that form the basis of R^5 (set to 0), you'll get only the solution (0, 0, 0, 0, 0) for their xis.

[nova basin]

actually it's only wrong in this case

but I think it should be true

[noble pelican]

Which means that the only way for you to have a linear combination of those elements to result in 0, is if all of their factors are 0 ( the definition of linear independence of vectors )

So anway

Question 3)
Find a basis for V/Im(F)

[nova basin]

I just realized that I'm familiar with quotient groups but not quotient VS lol

that's what I thought though

If F + G = E is a direct sum, then E/F ~ G

which is very nice

the question is equivalent to finding a supplementary to Im F and finding a basis for it

[noble pelican]

not sure what you mean

[nova basin]

if you find G then it's much easier

not sure that's the simplest way though

Im F = {a+(a-2b)t + at² - bt^3 \ a, b in R}

[noble pelican]

what's G?

[nova basin]

for any vectors spaces E, F, G with F and G subspaces of E

[noble pelican]

oh okay

[nova basin]

if P1 = ax^3 + bx² + cx + d and P2 = ex^3 + fx² + gx + h are equivalent, then
a-e = -v, b-f = u, c-g = u-2v, h-d = u for some u and v
Good luck with that, whichever way you want to do it !
For now I'm going to eat lol

[noble pelican]

don't worry about it!

Thanks for all the help

Though it's a bit discouraging how much I studied(3+ months) vs. how much I apparently lack

But I'll keep pushing through, like I've been doing all this time

have a nice lunch/dinner?

my parents are fed up with me because I can't pass this damn exam...
I don't know what to tell them anymore 😦

oh well

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[delicate gust]

For this problem, when I did usub I changed the bounds but I think the answer key doesn’t. Why is that?

[delicate gust]

In what cases do you not have to change bounds

Is it since u just sub sinx back layer

Later *

[hot herald]

,rotate

[hot herald]

if you're using u-sub for definite integrals, ideally you'd update bounds as you go
(although technically not necessary if you have the proper notation)

[delicate gust]

Wdym not necessary with proper notation

[hot herald]

well after changing your variable, the bounds are implied to be for that variable unless stated otherwise

[delicate gust]

wait huh

so which bounds are right for the problem, the [1,0] or [0,pi/2]

[hot herald]

after changing variable to u, [1,0]

[delicate gust]

oh so im correct so far?

wait when i sub sinx back in then what happens

[hot herald]

the whole idea of update the bounds as you go is that you don't have to sub back your sin(x) at all

[delicate gust]

ohhhhh okkk i see

thats where i was confused

[delicate gust]

ah

[hot herald]

that should be a u btw

[delicate gust]

tyty

[hot herald]

if you wanted to keep them as 0 and $\frac{\pi}{2}$ you'd need to explicitly write something like
$$\int_{x=0}^{x=\frac{\pi}{2}} \dots$$

[delicate gust]

oh i could also just sub sinx in and then keep the bounds, orrevt?

[glossy valve]

ℝamonov

[delicate gust]

correct*

[hot herald]

if you explicitly indicate the bounds like above

you can if you really want but
its a bit inefficient (many go this route and do it improperly from just finishing indefinite integrals)

there have been cases where the bounds temporarily disappear completely only to come back later which is also bad

[delicate gust]

gotcha

thank you

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[stiff wagon]

Would I just substitute the values of x in to solve to get the slope?

[full forum]

Please don't occupy multiple help channels.

[light sonnet]

You have another channel open, close that channel

[stiff wagon]

Rather close that one. The other one was never gotten to

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[misty raft]

how do you do this?
I4x-1I = 2x

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@misty raft Has your question been resolved?

[misty raft]

fuck

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[languid stone]

You still need help?

[languid stone]

Fack.

.close

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[misty raft]

@languid stone nah

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[tawny sierra]

How do i solve a system like this, I tried everything i could think of.

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[tawny sierra]

<@&286206848099549185>

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@tawny sierra Has your question been resolved?

[robust slate]

Assuming you want integer solutions, I’ll give a few hints.

1. ||(a+b)^2 = a^2 + 2ab + b^2||
2. ||Consider the expansion of (a+b+c)^2||

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[torn jolt]

These are both the same problems but I have no clue which one is correct. I’ve been doing this for like a full day haha

[torn jolt]

Basically on the bottom one I did a log rule to xtan(x) and the entire answer changed

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@torn jolt Has your question been resolved?

[torn jolt]

<@&286206848099549185>

[torn jolt]

<@&286206848099549185> I am still unsure

[torn jolt]

looks good to me

wait

yeah

Which one looks good? haha

both are the same problem but idk which one is correct

I feel like the bottom one is since I did the log rule of xtan(x) and took it apart.

[torn jolt]

it is the same thing either way

since you can divide everything by 2xtan(x) in the first one and get the same answer as the second

ohh!

[torn jolt]

my god. thank you so much

you don’t understand how frustrated i was. i felt like it could be the same but didn’t even think about dividing the tan

thank you

no problem ❤️

[torn jolt]

what I am unsure about is the dy/dx part since it seems like you already took the derivative of ln(y)

So I think that should disappear

hmm

well I was supposed to solve using logarithmic differentiation

and in all the example problems the 1/y was just cancelled out and then the dy/dx was kept

It might be something I forgot about gimme a sec

sure !

nevermind you did it right

https://en.wikipedia.org/wiki/Logarithmic_differentiation

Good job

yayaya

thank you sm 💜

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[strange stag]

pls help

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[strange stag]

<@&286206848099549185> pls help

Could you guys help me to tell how can I simply it further?

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[halcyon wave]

Find the distance from the point A(-2,3) to the line y=1/2x+1. Round your answer to the nearest tenth.

[halcyon wave]

the answers 2.3 units however i don't know how they got that

when my teacher went over it it was like a different answer

like an ordered pair

[lilac sky]

Hi

[halcyon wave]

hey

[lilac sky]

I think your teacher identified the point on the line which you use along with the first to then find the distance between

[halcyon wave]

what kind of answers can you use to answer that one specific question

i always see numbers with square roots but i never saw him go over that

and im hella confused

[lilac sky]

Ok, lets think about how to answer this sort of question from scratch

To there are several ways this can be done, but thinking about the most intuitive

[halcyon wave]

i found an example on the textbook

[lilac sky]

Ok

So they are always using a line perpendicular to the one you are given

Do you know why

[halcyon wave]

so theres an intersection point?

[lilac sky]

Well even if we use a non perpendicular line, provided it is not parallel, there will always be an intersection

[halcyon wave]

well idk

[lilac sky]

With these questions, it is implied that we are looking for the shortest distance

Other non perpendicular lines still represent straight paths to the line given, but that path will be longer

[halcyon wave]

huh so why are they always using a line perpendicular

[lilac sky]

Bc that line represents the path to the line given which is shortest

Imagine you are walking to a road

You want to get there with as little walking as possible

Will you walk straight towards the road or at some diagaonal

[halcyon wave]

straight

so would the line

go straight

wait

so it would

intersect the other point

which point would it intersect

[lilac sky]

Your walking path will always have the red point as that is where you are starting from

But we need to choose the gradient which will minimize the distance

So we can either use some general point on the blue line find the distance to it, then use differentiation to minimize that distance, or we can think geometrically and say that the perpendicular line will have the smallest distance and get that number directly

Since the first way is really used to prove that the perpendicular line gives the shortest distance, we can just use that fact and do the second way

Is that all ok?

[halcyon wave]

how would u solve it

its just all this explaining kinda hurts my head

especially in school

[lilac sky]

Well as your example does it

Sorry about that

You're using desmos above, right

[halcyon wave]

yes

[lilac sky]

Ok, so can you enter an equation in for me with the slider m

Enter this

y-3=m(x+2)

[halcyon wave]

\

[lilac sky]

Along with the red point and blue line above you had

So have all three in the same graph

[lilac sky]

Nice

Now move m around

See how the distance from the black point to the red line changes (the green line is the path)

[halcyon wave]

kinda

[lilac sky]

Now calculate the gradient perpendicular to 1/2

What would that be

[halcyon wave]

h u h

i have no clue how to

[lilac sky]

Ok there is a formula

The gradients of two perpendicular lines m1 and m2 satisfy the equation

[halcyon wave]

whats a gradient

[lilac sky]

,,m_1 m_2=-1

[glossy valve]

Social Capital Gainer

[lilac sky]

The slope of the line

[halcyon wave]

ok

[lilac sky]

So the gradient of the red line above is 1/2

So of we make m1=1/2

What would m2 be?

[halcyon wave]

so are we trying to find hte alternate equation for y

cause in the example

[lilac sky]

Yes, we wanna chose a specific value for m in the above graph

[halcyon wave]

theres y=-x+3 and y=x-1

[lilac sky]

Yes

See how in the first line in red text it says the productof the slopes of perpendicular lines is -1

That's what we are about to do

[halcyon wave]

ok so y=1/2x+1
m=1/2
3=2(1/2)+b
4=b
y=1/2x+4?

[lilac sky]

But using this formula

So set m1 to 1/2 and solve for m2

[halcyon wave]

how do i solve for m2

[lilac sky]

So did you put 1/2 into m1

[halcyon wave]

1/2m2=-1

would it be -2

[lilac sky]

Yes, well done

That's the gradient of the perpendicular line

Put -2 as m in your desmos graph fpr the green line

[lilac sky]

There it is

See how the distance from the black dot to the red line is thr shortest it can be with this gradient

[halcyon wave]

and then you can estimate?

2 units down and about 1/3 of the 3rd square so 2.3

[lilac sky]

Ypu don't estimate

[halcyon wave]

ok

[lilac sky]

Find the point of intersection with those lines

[halcyon wave]

how am i supposed to

[lilac sky]

So the two equations are

y=1/2x+1

And

y-3=-2(x+2)

[halcyon wave]

wait where did u get the other stuff

before m1m2=-1

[lilac sky]

You mean this equation?

[halcyon wave]

yea

[lilac sky]

So what I did there was

Using the form of the straight line through a point (a,b):

,,y-b=m(x-a)

[glossy valve]

Social Capital Gainer

[lilac sky]

Have you seen a straight line written like that before

[halcyon wave]

no

[lilac sky]

Ok

Have you seen this general form before

,,y=mx+c

[glossy valve]

Social Capital Gainer

[lilac sky]

Perhaps with different letters

[halcyon wave]

ye

[lilac sky]

Ok, what is c in that equation?

What does it represent

[halcyon wave]

its where u start on the graph

[lilac sky]

Yeah

[lilac sky]

Now in this line, we can make it look like the one you have seen, but it will be more useful to us to use this one then make it look like yours

So to use this

The straight line passing through the point (a,b) with slope m has the equation

,,y-b=m(x-a)

[glossy valve]

Social Capital Gainer

[lilac sky]

So in our case, the point it passes through is (-2,3)

And the slope is -2

So can you put those numbers in for me

a is -2, b is 3 and m is -2

[halcyon wave]

y-3=-2(x-2)

[lilac sky]

Yeah

Thats how i got it

[halcyon wave]

but wouldny x or y be -2 or -3

[lilac sky]

Well x and y are the variables, but (a,b) is a point which the line definitely passes through
This means that x and y will be -2 and 3 at some point, but the equation still works because it gives 0=0 and that is true

3-3=-2(-2+2)

0=0

Basically, an ordered pair (x,y) is on the line if we can put x and y into the equation and the equation is true

So for example

If you put (0,-1) into the equation

-1-3=-2(0+2)

-4=-4

Since the left side is the same as the right side, the equation works

But if we try (1,1)

1-3=-2(1+2)

-2=-6

This is wrong

So (1,1) is not on the line

But (0,-1) is on the line

For the same reason (-2,3) is on the line

Is that ok?

[halcyon wave]

this hurts my brain

[lilac sky]

Oh, ok, all you need to know is that you can make this line as you did above, you don't need to know why it works

,,y-3=-2(x+2)

[glossy valve]

Social Capital Gainer

[halcyon wave]

what do you do after?

[lilac sky]

So now lets make it look like the line yoi are familiar with

So add 3 to both sides

,,y=-2(x+2)+3

[glossy valve]

Social Capital Gainer

[lilac sky]

Then expand the brackets

,,y=-2x-4+3

[glossy valve]

Social Capital Gainer

[lilac sky]

So what is the equation of the line?

[halcyon wave]

y=-2x-1

[lilac sky]

Yeah, well done

[spiral grail]

I can explain why point slope works if u want

[halcyon wave]

what

[lilac sky]

We already passed that

[spiral grail]

Your equation was in point slope and u made it into slope intercept

[lilac sky]

Anyway

Now we have two equations

y=1/2x+1

And the one you just rearranged

y=-2x-1

So we set y=y

And find x

So we do

y=y

1/2x+1=-2x-1

Can you find x for me by solving this?

[halcyon wave]

2 1/2x=-2

[lilac sky]

Yep and then

[halcyon wave]

uhh

[lilac sky]

Turn your mixed number into a top heavy fraction

[halcyon wave]

idk

x=-2/2.5?

[lilac sky]

We will get a neater answer if we use

,,2\frac{1}{2}=\frac{5}{2}

[glossy valve]

Social Capital Gainer

[halcyon wave]

wha

[lilac sky]

Well 2.5 is 5/2 right

[halcyon wave]

ye but u have it on the other sidew

oh wait

ok

what do u do if u wanna put it into -2

[lilac sky]

So we have

5/2x=-2

Times both sides by 2

5x=-4

Then divide both sides by 5

[halcyon wave]

x=-4/5

[lilac sky]

Yeah

So that is the x coordinate of the point of intersection

Now to get the y coordinate

Choose either of the two straight lines, doesn't matter which

Maybe y=-2x-1

As the slope is not a fraction, but it doesn't matter

[halcyon wave]

uhh

[lilac sky]

And put x into it

[halcyon wave]

y=-2x-1

[lilac sky]

The x you just found

[halcyon wave]

what

ok

y= 1 3/5-1

[lilac sky]

Yeah, it will be easier if you work with top heavy fractions. Mixed numbers like you are using are not often useful in these contexts

But actually you can carry on from here pretty easily

[halcyon wave]

alr

[lilac sky]

So what is

1 3/5 take away 1

[halcyon wave]

3/5

[lilac sky]

Yeah

Specifically y=3/5

So that is the y coordinate

Now we have the point of intersection

(-4/5 , 3/5)

Now just find the distance from (-2,3) to that point

Using the distance formula, or however you like to do it

[lilac sky]

By the way, this ordered pair is probably what your prof found

And stopped there

[lilac sky]

So how would you find that distance

[halcyon wave]

distance formula

[lilac sky]

Oke

[halcyon wave]

ok ty

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[lilac sky]

No problem at all

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[torn jolt]

I need to find the nth term of this geometric sequence, but the common ratio isn't constant

[wild sleet]

you can't simplify a factorial

[torn jolt]

could you elborate

[rain star]

From the pattern I'm noticing, wouldnt you increase the number you multiply the previous answer by by one each time? 1x2=2, 2x3=6, 6x4=24, 24x5=120, etc

[wild sleet]

yeah that's the pattern i see

[torn jolt]

yeah but like geometric sequences have a common ration that is constant

[wild sleet]

you can;t do anything instead of doing the multiplication

[torn jolt]

im not sure how to solve one that isnt constant

is there any formula for a geometric equation that doesant have a constant common ratio???????????

[full forum]

@torn jolt Has your question been resolved?

[fast peak]

well then it's not a geometric sequence anymore

every sequence has a nonconstant ratio

if there really was a simple trick like this we could easily solve every sequence. sadly we can't

[torn jolt]

sorry im so dumb

this is actually a factorial sequence

not a geometric one

i was confused because i just dont remember this stuf

thanks for the help everyone

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[runic bloom]

is my work right? (so far)

u = cosx btw

it ends up as irrational stuff so I’m not sure

sorry for the bad handwriting btw lol

[dusk quest]

looks fine

[cloud prairie]

one mistake

first line

you put everything on the left-hand side right

it should be -1 instead of 1

[dusk quest]

ye

[cloud prairie]

so at the end equation should have -3 not -5

[dusk quest]

yup

[cloud prairie]

$4u^2 -2u -3$

[glossy valve]

kjmkty

[dusk quest]

yes

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[runic bloom]

.reopen

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✅

[runic bloom]

wait lemme read it lmao

[runic bloom]

oh whoops

so I fixed the -1 but this seems wrong.. can’t arccos those values

[woven snow]

,w 4x^2 +2x -3=0

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Deep.#0001

[woven snow]

,calc -sqrt(13)/4 -(1/4)

[glossy valve]

Result:

-1.151387818866

[woven snow]

,calc sqrt(13)/4 -(1/4)

[glossy valve]

Result:

0.651387818866

[woven snow]

Check your calculations again

[runic bloom]

yeaa I messed up my calc lmao

okay so 0.65 is the only possible one right?

soo 0.86?

[woven snow]

Yea

,w arccos(0.651387818866)

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Deep.#0001

[runic bloom]

ty!

[woven snow]

Seems fine

Np

[runic bloom]

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[slender stump]

My task is to find x and y in the equation, using substitution:

the two equations:
3y + x = -2
y - x = 10

x can be found by:
x = -2 -3y

then i calculate y by:
y - 1(-2 -3y) = 10
y + 2 + 3y = 10
4y = 8
y = 2

and at last find x:
x = -2 -3(2)
x = -8

is this the correct way?

[torn jolt]

balls

idk i got -2 for x

on the first baby

😉

but i got snail scent so who knows

it might be all seeing

[hot herald]

y + 2 + 3y = 10
4y = 12
incorrect manipulation in that step

[slender stump]

thx brudas

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[torn jolt]

== 2x^2 - 3x + 5 - ( 10/(x+3) )

so y = 2x^2 - 3x + 5 is an asymptote ;; right?

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[torn jolt]

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[atomic blade]

2x^2-3x+5 are not asymptotes

In fact that's not even the right terminology

Asymptotes are where a function will approach such as that the distance between the function and the asymptote will essentially become 0

I'm assuming you're trying to find the vertical asymptote

You've done the long division

But note the remainder: you can't cancel x+3 further

Set the denominator (of the remainder or leftover factors after division) equal to 0 to find your vertical asymptotes

[torn jolt]

how about horizontal asymptote

how would you find that?

[atomic blade]

A horizontal asymptote only exists if the leading term of both the numerator and the denominator are the same power

Are they the same power?

@torn jolt

bruh

[full forum]

@torn jolt Has your question been resolved?

[hybrid geyser]

wait I thought if it’s less than or equal it’ll exist?

pls I have an exam soon 😭

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[simple totem]

umbral is right, but sloped asymptotes exist if the top polynomial is one degree higher than the bottom

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[torn jolt]

I got to K = (10-C)/4
I also got C=-4K+10
I think that I am supposed to use sys. of equations?
Graphing it gave me the points (2,2)

[modern sentinel]

at x=0,y=-5

[torn jolt]

Inputting those values into the variables gives -5 = K+C

[modern sentinel]

ok

[torn jolt]

I dont understand what to do with that equation

[dawn rose]

put C = -5-K

then put it on first equation

[dawn rose]

i mean this

[torn jolt]

Input it on the Original Y=k*2^x+c ?

[dawn rose]

no

[dawn rose]

this one

just place -5-k instead of C

2 variables , 2 equation

[torn jolt]

K=15 is what I ended up with.

plugging that into the C= equation gave

10 = 15*4+c
10=60+c
C=50```

That was the correct manor to do that equation in?

I suppose it is. Thank you guys.

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[torn jolt]

how does he find the LCD

[torn jolt]

how do i find the LCD here

3x2

6 is a multiple of 3

if it were not then the LCD would be 3* that number

Nice u r Muslim

i am yes

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[modern bobcat]

hi! i have an assignment based on shapes that i need some help with... the assignment is to recreate an image of a sailboat and theres one particular line i dont know how to recreate - the first image is the assignment requirements and the second image is mine. i am unsure of how to create the line that makes the bottom of the sailboat - could anyone help me with that? thanks in advance!!!

[scenic reef]

To me it looks like a parabola, so you could start with a parabola in turning point form

And then sub in one of the end points to find the dilation

[modern bobcat]

thank you so much

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[torn jolt]

Hey

[torn jolt]

Looking for help

What I did was

not sure if this is the right way to go

[light sonnet]

Not quite

[torn jolt]

how would you go about it

[light sonnet]

When you rationalize, you want to flip the sign separating the terms

So instead of $2 \sqrt{2} -1$, you need to use $2 \sqrt{2} + 1$

[glossy valve]

dldh06

[torn jolt]

ah

i see

that is smart

[light sonnet]

That's what rationalizing is, nothing smart about it

[torn jolt]

oh

lol

thanks for your help

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[candid zinc]

Teach me how to do piecewise function please 🥺

[light sonnet]

YouTube can do this

Just look up videos on this

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[torn jolt]

oof im struggling with this trig identity stuff....how did they get cos^2theta = 1+cos2theta/2 ?

[fast peak]

well it's just a well known identity

[torn jolt]

which one is it? half angle?

[fast peak]

sure or double angle. essentially equivalent up to rearranging

[torn jolt]

i see one that says cos(theta/2) = +- sqrt(1+costheta/2)

[fast peak]

yeah that one after squaring and rearranging a bit. or wikipedia also has a list of power reduction identities with this one

[torn jolt]

its so hard to remember all of these damn identities

=c

[fast peak]

yeah tbh I just look them up on wikipedia all the time. good idea to know which ones exist tho

and well hopefully in an exam you can use a cheat sheet

[torn jolt]

we are allowed formuala sheets for the exams

im not sure about the quizzes though

they didnt say

i will ask and see

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[tame tree]

how do I find the inverse of the given function. The Ln is throwing me off

[robust slate]

You can raise both sides to the power of e

[tame tree]

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[royal glade]

how would I go about x = 1 / y + 3

[royal glade]

working with inverse functions in algebra

[torn jolt]

trying to solve for y?

[royal glade]

right, but the steps, like whats first when dealing with a fraction

[torn jolt]

first bring the 3 to the other side

[royal glade]

x + 3 = 1 / y?

i mean - 3

x - 3 = 1 / y

[torn jolt]

yes

and then reciprocal

[royal glade]

sorry, noob... how do i do that

[torn jolt]

just do 1/z on both sides

1/(...) both terms

BUT assume that x≠3

[royal glade]

yeah, x cant equal -3, right?

[torn jolt]

it can

it cant 3

[royal glade]

what do you mean 1 / z on both sides

like, multiply 1/y on both sides?

[torn jolt]

like transform x - 3 to 1/(x - 3)

and 1/y to 1/(1/y)

[royal glade]

then what?

[torn jolt]

do you know what 1/(1/y) is?

[royal glade]

no

[torn jolt]

its just y

[royal glade]

why?

no pun intended lol

so the ones just cancel out?

[torn jolt]

good question, not sure if there is a good satisfying answer

do you know how 1/y = y^-1?

[royal glade]

yeah

what would 1 / 1/y be in that notation?

[torn jolt]

and if we have 1/(1/y) its (y^-1)^-1

and do you know (x^a)^b = x^(a*b)?

[royal glade]

yeah

[torn jolt]

so if we do that, we get -1 * -1 in the exponent

[royal glade]

ah, which is 1

good explanation

[torn jolt]

oh another way to see it

you can multiply the numerator and denominator by y

1 / (1/y) = (y*1) / (y*(1/y))

[royal glade]

right

so if y = 1

it would work out to 1 anyway

y*1/y

[torn jolt]

1 / (1/y) = (y*1) / (y*(1/y))
= y/1 = y

[royal glade]

sick! thanks bro! thats some deep math (for me at least)

[torn jolt]

well going back to the original problem

we had
x - 3 = 1/y

and then doing the 1/(...) trick we get
1/(x - 3) = y

[royal glade]

ok, that makes sense now that I understand it

thanks again m8!

[torn jolt]

btw another way would be to first multiply by y:

x - 3 = 1/y
(x - 3)*y = 1
and then divide x - 3
y = 1/(x-3)

[royal glade]

ooohhhh! thats super!

thats a little easier right?

makes perfect sense

[torn jolt]

yes depends on what you find more intuitive (it just requires more steps)

[royal glade]

one more question if you have time

this isnt one to one because its a parabola right?

and to restrict the domain to make it one to ine, i'd just basically cut it in half?

[torn jolt]

wait i need to think about this

have you looked at the graph of the function?

[torn jolt]

does the question want you to have the domain only contain positive values or does it just have to contain positives?

[royal glade]

i think it has to contain positive values

[torn jolt]

ofc [-9/2, infinity) would do the job

[royal glade]

nice, that was right!

is that where the lowest point is?

[torn jolt]

yes, at -9/2 there is the global minimum

[royal glade]

so how to do you if its not (infinity, -9/2]?

[torn jolt]

what exactly?

[royal glade]

like, we measure the domain from right to left always right?

i mean left to right

lol, i meant you do you know**

[torn jolt]

oh you mean how we notate intervals?

[royal glade]

kind of

like, what was your process of solving that?

[torn jolt]

we always put the smaller number left and larger number right

[royal glade]

ooohhh, ok

[torn jolt]

(-infinity, -9/2] would also have been a one to one domain

[royal glade]

so find the global minimum and notate the rest of the graph (- or + infinity

[royal glade]

right, that makes sense

[torn jolt]

well just look at the vertex form

the -9/2 is the x coordinate of the vertex

we can directly read it from x - -9/2

[royal glade]

right. so from here, how do I find the inverse functon?

[torn jolt]

look at the vertex form and set it equal to y and solve for x

[royal glade]

x^2 + 9x + 81/4?

its wrong, but after factoring, thats what i got

i had (x + 9/2)(x+9/2)

x^2 and 9/2 + 9/2 = 18 / 2 or 9

[torn jolt]

im not sure how you got there

[royal glade]

lol me niether

[torn jolt]

you take the vertex form, set it equal = y

then add 81/4 both sides

take square root both sides

[royal glade]

then subtract 9/2?

[torn jolt]

yes

[royal glade]

so it'd be

sqrt(y+81/4)+9/2

= x

[torn jolt]

ahem, subtract

[royal glade]

oh right

[torn jolt]

also, strictly speaking you would need to put ±sqrt there

but, now think about what the domain for x is

[royal glade]

cause now its a parabola

right>

?

[torn jolt]

what is parabola?

[royal glade]

nvm i got it

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[slow gulch]

Ayon

[slow gulch]

Ayo

someone help

with math

[wide sundial]

[slow gulch]

So fried right now

[hybrid geyser]

y=2x-3, substitute into the 2 given eqn then find x and then y

[wide sundial]

Ok it’s just some simultaneous equations

[hybrid geyser]

ure here too HAHA

[slow gulch]

Aight

[slow gulch]

Just posting my questions every server i get

Dont mind me

[slow gulch]

Aight

Another question

[hybrid geyser]

HAHAH maximum efficiency

[wide sundial]

Which part are you stuck at

[slow gulch]

The whole thinf

[hybrid geyser]

a=4/3 x (lgx) /(r^2) where r is the common ratio
Substitute into formula for sum of geometric sequence to find r

[wide sundial]

It’s also simultaneous equation

Can you see they are simultaneous equations?

And can you see what formulas would relate them together

[slow gulch]

Wait let me try the methods both of you suggested

Thx for the help guys

Afk

[hybrid geyser]

okii

atb!

[slow gulch]

Ayeee aight got it

.close

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[stiff ermine]

Can someone help explain this math problem and what i am not getting about it?

[atomic blade]

Send the original

[stiff ermine]

What do you mean?

[atomic blade]

Like the original problem

You just showed me work

[stiff ermine]

Oh ok srry i just joined this discord

[atomic blade]

That's fine

[stiff ermine]

113

I am working with scientific notation

[atomic blade]

Hm okay

Rearrange the factors so that there are no negative powers

Combine like terms

[hybrid geyser]

wha-

114 qns

[hybrid geyser]

can also first convert the 0 power to 1 and expand the rest,, then can simplify the terms

[stiff ermine]

I dont know why it came out sideways

[atomic blade]

Again, get rid of negative powers

[hybrid geyser]

how to rotate

ik there’s a way but I forgot

[hybrid geyser]

hihii

[stiff ermine]

.clos3

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[hybrid geyser]

HAHA the timing omg

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[slow gulch]

help

[slow gulch]

<@&286206848099549185>

[prisma tusk]

r u his son fr

[slow gulch]

LMAO dont mind that name

Made this dumb name 5 years ago

[atomic blade]

Seems like it decomposes

From whatever that is

[hybrid geyser]

why chem question

[atomic blade]

I don't know what the OSO3H group is

[hybrid geyser]

isn’t this for maths HAHA

[slow gulch]

Lmao bro

I just post it everywhere

[hybrid geyser]

HAHAHA

from the looks of it

shd be hydrolysis and condensation

for 1 and 2

[slow gulch]

BRUH that’s right

Lmao wait

[hybrid geyser]

phew HAHA

did u understand why

I can explain if u don’t see it

[slow gulch]

Lol nvm I see it now

Thx

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[hybrid geyser]

oh okii thats gd

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[torn jolt]

This is what my teacher wrote on the board but I don’t think I’ve ever learned what is or how to find LCD. I watched a few videos but couldn’t learn it. Can someone explain how he got 3x please

[torn jolt]

here is another example shared by my teacher

he was teaching Solving Rational Equations, but I dont understand this step of what is or how to find the LCD

or share a video if you have one

[silk flame]

Lowest common denominator is the lowest number you can make the bottom number to make both fractions bottom number the same. In the example, your teacher saw that 6 is 2x3 therefore they could make the lcd 6. For your question, there is not rlly a way to only modify one fraction, you need to change them both to make the denominator the same. What could you multiply 3 by and what could you multiply x by to make them the same

[torn jolt]

Okay I get it now

[silk flame]

What is the lcd?

[torn jolt]

Oh

Wait the teacher put 3x as lcd is that incorrect

Or multiply one by 3 and the other by x and the third by 3x

To get the common denominator of 3x

[silk flame]

The common denominator is 3x, ignore the one after the equals sign for now

[torn jolt]

So for the first one would u only multiple the numerator by x since the bottom already has 3

Making it x^2 over 3x

[silk flame]

Perfect

[torn jolt]

Ok great

Thank u

[silk flame]

And the other one 27/3x

[torn jolt]

Let me share what the teacher wrote

Ok