help-23

[obsidian oracle]

k = log_5(n)

so T(n) = T(5^k), not T(k)

[crude star]

im so confused rn

where did I say that T(n) = T(k)

[obsidian oracle]

"assume that T(k) = ..."

[crude star]

oh the hypothesis

[obsidian oracle]

and then "T(k+1) = ..."

[crude star]

so what should it have been

that T(k) = 5^k ?

for hypothesis

[obsidian oracle]

no

T(5^k) = ...

and T(5^(k+1)) = ...

[crude star]

right so that means my base case is wrong too right

oh wait no im dumb 5^0 = 1

[obsidian oracle]

base case : k = 0

would have cause less confusion to write it in terms of k

[crude star]

ok so for indcuctive step im suppose to prove that

T(5^(k + 1)) = (k+1)^log5(7)a + ((k+1)log5(7))/6 − 1/6

[obsidian oracle]

exactly

[crude star]

uh im stuck

[obsidian oracle]

wait I didn't even notice

[crude star]

oh what

[obsidian oracle]

on one line you write 7^log(n) and on the next you write n^log(7)

xd

[crude star]

its the same no?

[obsidian oracle]

oh maybe

[crude star]

but idk what to do after this

[obsidian oracle]

even if it's the same, keep it written as 7^log(n)

[crude star]

oh alright

thing is

im confused now

so when I write it as that

then instead of 7k^(log_5(7)) I get 7 * 7^(log_5(k))

but i graphed

its not the same

so idk

nvm im dumb I thought 7 * 7^(log_5(k)) = 49^(log_5(k))

smh

idk what to do now

how do i manipulate that

[safe radish]

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[wind abyss]

Hi

So i have evrything here

How do i attain the y intercept?

[pulsar pecan]

obtain

[wind abyss]

like cna i just do e^-1+4

oh

lol

i bad english

[pulsar pecan]

Yes

[wind abyss]

perfect ok

thnxs

uh

i think +4 is not need

i got 8.37037037 when i added 4

[pulsar pecan]

try calculating it again

[wind abyss]

ohk

so e^-1

will be 1/2.7?

oh i got it

lol

.close thnxs

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[grim badge]

how do i get the transformation matrix for a reflection across line y=2x

[grim badge]

please help me im very desperate

i understand that the matrix is [cos2x sin2x}

[sin2x -cos2x]

sorry i meant to write theta instead of x

2=tan(theta)

idk how to go from tan(theta) to cos(2xtheta) or sin(2xtheta)

thats my issue

please ask if u want me to elaborate

ill repeat it:

let q be the angle line makes with positive x-axis

how do i go from tan(q) to sin(2q) or cos(2q)

<@&286206848099549185>

[glacial snow]

https://discord.com/channels/268882317391429632/1168153444083847178

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[visual depot]

uhhh

[visual depot]

my graohic calcukator says there is no y intercept

like am i ttippin or sum

cause i either get a complex solution or 1

for x=0

[proven ore]

i think u should do power firstly

[safe radish]

@visual depot Has your question been resolved?

[visual depot]

wdym

[worn zenith]

my says there is

I think it's because you replaced the ^2 thing

[visual depot]

huh

[worn zenith]

and now half of graph is gone

[visual depot]

woah

ok

lemme try reset my calc

[worn zenith]

This is how it should look like

wait

I think it really is a problem with calculator and you did nothing wrong

[rotund glacier]

there is a y intercept

your calc doesnt show it for some reason

it only considers the positive part

you want this

blue line

[safe radish]

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[icy frigate]

Can someone explain to me how did get -2 in the inequality

[tall bough]

,w |x|<2

[flat frigate]

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Requested by fractalogist

[tall bough]

Just replace x with x-1

[icy frigate]

Oh that’s how absolute value works?

Because |-2 | will give 2

But greater than -2 or 2 will break the inequality

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[sturdy sapphire]

Exercise: "Write down the elements of this set"

I don't know where to start because there are two things I don't understand. Firs: What is the meaning of the "i" next to the 3 above intersection. Second: This is a set with infinitive elements, right ?

[quiet juniper]

the i is the index seen under the union

[sturdy sapphire]

I see. And how do I translate that

Does it mean "3 * 2"

Or "32"

[cold relic]

6

For i=2

It’s 3*i

[cold relic]

It looks to me like this is infinite, yes

[sturdy sapphire]

If it is infinitve, how can I solve it ? How can I find the elements of this thing ?

[cold relic]

How to note them down or how to find them?

For finding them, look at the set and see what elements it contains

Then use what you know about intersections and unions to determine what’s part of the set and what isn’t

[sturdy sapphire]

They have all numbers from 6 upwards in common

[cold relic]

Yes

[sturdy sapphire]

Ok

But then I have to increase i

I is now 3

I have to do the same for all "i"s

[cold relic]

Now think about why that doesn’t matter

Why you DONT need to do that

[sturdy sapphire]

Ok, let me think about that. It will take a while, I am a little slow on these things...

I can assume that the elements of "i = 3" are already inside "i = 2"

[cold relic]

No

[sturdy sapphire]

😦

[cold relic]

How does the intersection work as you increase i?

[sturdy sapphire]

Let me think for longer...

You mean this would be wrong ?

[cold relic]

Oh whoops

Yeah I made a mistake

[cold relic]

You were right, they are already part of that so it stays the same

Sorry, my bad

[sturdy sapphire]

Ok nice

Thx

.close

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[tough canopy]

Hello, can someone help me find a simpler solution for..

[tough canopy]

.close

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[junior wagon]

If a function is holomorphic, why can we conclude that its line integral is 0, given that it is (piecewise) smooth, simple, and closed?

[peak estuary]

cauchys integral theorem

you can look up a proof? not sure what you want to hear

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[junior wagon]

Ok. I found a proof using Greens theorem.

Just wanted to be sure why it is true.

ty

.close

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[shy canyon]

anyoen can help me with these couple of geometry problems they simple but i cant really understand it

[lean thorn]

take two known points (any two points), $(x_1, y_1)$ and $(x_2, y_2)$, and find the slope $m = \frac{y_2 - y_1}{x_2 - x_1}$. Then use the point slope formula $y - y_a = m(x - x_a)$, where $(x_a, y_a)$ is any point on the line.

[flat frigate]

MellowDramaLlama

[lean otter]

I think you can just do y=mx+b

[shy canyon]

yea but then they wnat like the final equation

[lean otter]

M is y2-y1/x2-x1 (pick any random points)

[shy canyon]

like on the graph

[lean otter]

Get m

[shy canyon]

Ok

[lean otter]

Did u find m yet?

[safe radish]

@shy canyon Has your question been resolved?

[shy canyon]

Nah im not really understanding it and i gotta submit is by a ceratin time ill just ask my teacher tmr

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[obtuse plover]

What have u tried

$\frac x {-8} = -20$

[flat frigate]

Stephen

[tall bough]

bruh

you said u got it

[obtuse plover]

If I gave u the problem

[tall bough]

liar

[flat frigate]

Stephen

[obtuse plover]

How would u do it

@lean otter

[bold ferry]

you’re trying to get x by itself

so what operation can undo division?

[tall bough]

https://www.youtube.com/watch?v=crJI4iZ_DbI*

just watch the vid

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[burnt nest]

It's true?

[burnt nest]

Anyone 👀

.close

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[lean otter]

Hello

[fossil shell]

hi

[lean otter]

(-inf, c] U (a, +inf)

[frigid spruce]

?

[lean otter]

Is the Union of this (a, c]?

[spice grove]

No.

That's intersection.

[lean otter]

I'm not sure, is it (-inf, c]

This is how the teacher has written it down

[flat frigate]

Anne S

[lean otter]

No

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[torn shore]

I think it matters. If my low quality MS paint drawing is the be believed, when a<c, all real numbers are in the union. When c<a, all real numbers are included except on the interval (c,a]

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[exotic jewel]

is velocity = Atcos(t)-Atsin(t) in this case?

[safe radish]

@exotic jewel Has your question been resolved?

[exotic jewel]

is it okay?

<@&286206848099549185>

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@exotic jewel Has your question been resolved?

[exotic jewel]

<@&286206848099549185>

[exotic jewel]

.close

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[lean otter]

This is not a basis for R^4 because...

it cant be written as a linear combination?

[raven heart]

wdym it ?

[lean otter]

to check linear independence vs independence, u have to solve for Ci. but i think u can skip that when it comes to finding basis

[lean otter]

any vector in V

like we have R^4 and only 3 vectors

[raven heart]

you also need linear independence to have a basis tbf

[raven heart]

yea exactly

[lean otter]

yes ik

[raven heart]

all bases of R^4 have 4 vectors

[lean otter]

can u show that like

can u draw a diagram lol

[raven heart]

wym?

idk what diagram I'd do

[lean otter]

idk the words arent clicking for me

i need a cartoon :((

[untold sky]

drawing a 4 dimensional space is slightly difficult on 2d paper

try constructing some points in R^4 that your vector set can’t reach. you may want to scale some vectors for that.

[lean otter]

idk like this

so we're missing a vector

ok yeah

i get it now lol

thanks

.close

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[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

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@lean otter Has your question been resolved?

[marsh walrus]

do we take mx=sqrt(x)

i think this is the approach

to be honest im still not 100% what the question is asking @lean otter

but you could solve this for a further relation

if you solve for x, i think you get the x coordinate

just creating a relation gets you the y

actually, yea

its not so complicated

take mx=sqrt(x)

this gives you x=1/m^2

substitute y/m = x = 1/m^2 provides y=1/m

then $P(m) = \qty( \frac{1}{m^2}, \frac{1}{m} )$

[flat frigate]

jan Niku

[marsh walrus]

alternatively, assume that division by x in your solution process is not okay because x=0. This leads to the second solution

I'm just gonna give what i got as the answer since you been waiting and i guess we wont be able to meet up

https://www.desmos.com/calculator/8xo9eawrpo

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[solar needle]

i learn in class A = X lambda X^-1 but honestly my teacher did a horrible job explaining it

[solar needle]

what is the point of it

i know similar matricies is like A = B C B^-1

is it related to that

also how do you know if a matrix is non diagonizeable

[placid oak]

I'm not familiar with what lambda refers to, but you can tell that a matrix is nondiagonalisable if it has fewer linearly independent eigenvectors than its dimension

the entries in a diagonal matrix are exactly the eigenvalues of the transformation

because you're writing the transformation in an eigenbasis

[solar needle]

sorry this is my fault, could i show you how it was used in context

maybe that is better?

[placid oak]

if you don't have enough LI eigevectors, you can't find such a basis, so it's not diagonalisable

[placid oak]

yeah sure

[solar needle]

a basis for the matrix?

[placid oak]

a basis for the space the matrix acts on

such that the eigenvectors are basis vectors

[solar needle]

basis is a spanning set with linearly independent vectors right

[placid oak]

yes

but here I mean basis in the sense of your choice of coordinates for a space

[solar needle]

can i ask in the context of eigen vectors, what does algebriac and geometric mulplicity mean

sorry i am on crack

[placid oak]

algebraic multiplicity is the number of times the eigenvector apears as a root of the characteristic polynomial

the geometric multiplicity is the dimension of the kernel of A - lambda*I (or equivalently, the nullity of A- lambda*I)

where lambda is your eigenvalue

[solar needle]

you wrote a textbook??????

[placid oak]

they're mostly old notes

[lean otter]

Sheeesh

[placid oak]

oh, the linear algebra section 29.8 might be relevant to what you're doing though

[lean otter]

I don't understand what you guys saying 101 what topic is that?

[placid oak]

linear algebra

[solar needle]

yeah idk im really struggling with linear algebra

[lean otter]

Ohh

[solar needle]

a lot of vocabulary

it confuses me

[placid oak]

I'd try focus on a more geometric understanding

proofs are a lot simpler when you can think about how a matrix is transforming space

plus it also helps for multivariable calculus and matrix valued functions, etc.

[lean otter]

I don't understand anything i am still in quadratic equation lol Good luck In algebra linear

[solar needle]

yeah I have not taken multivariable calc

i find it very difficult to conceptually understand linear algebra

[lean otter]

You can do it

[placid oak]

https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

[solar needle]

im not sure what to do about that, this is my worst class and i need to severely improve my grade

[placid oak]

this gives a very good geometric overview of linear algebra

[lean otter]

What your grade right now?

[solar needle]

uhhhhh

[placid oak]

I'd recommend looking through it at least once

and my notes are written to supplement it with extra details and exercises

[solar needle]

im extremely determined to get an A on my next midterm

[lean otter]

You are definitely smart bro

[lean otter]

You should get A++++++ WHAHAHAHAH

[solar needle]

.close

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[eternal sparrow]

Hi there

[eternal sparrow]

I have a few questions about maths

[safe radish]

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[eternal sparrow]

<@&286206848099549185>

These two

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[toxic fjord]

,tex $\theta = \frac{\pi}{3}$

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[oak nest]

Can someone explain B

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@oak nest Has your question been resolved?

[hasty wagon]

since the length and width are known
we just have to make sure that the x is within 170 long and 70wide

[oak nest]

How about range? What does rhe explanation mean

So we need to find domain forst since the variable is x

[hasty wagon]

yes

[hasty wagon]

since we know function P is increasing

so we find starting point and ending point

plug into P to get the range

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@oak nest Has your question been resolved?

[oak nest]

Oh amkes sense

Thanks

Would it be teh same if we had the range

[hasty wagon]

what do you mean?

[oak nest]

So would we also plug in the range to get the domain

If we only know the range

[hasty wagon]

probably yea

not sure, because it's an "if"

can't tell without further info

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[wise schooner]

Can someone help find my mistake

The first row of the rref is supposed to be 1 0 -3 | -1 but I can’t find my mistake

[haughty turtle]

that makes more sense thanks

Im going to make X1 and all that be other values so it is easier to understand

give me a second

are you trying to find what x1 and x2 and x3 equal?

[wise schooner]

ye

[haughty turtle]

ok

[wise schooner]

the bottom two rows of my rref are correct but the first one is wrong

[haughty turtle]

I dont know what a rref is

Ill try to understand tho

[delicate sphinx]

Found it, at this step, when you did R2 + R3
R2 = [0 -2 -10 | 2 ]
R3 = [2 -2 -16 | 0 ]
Doing that you get [2 -4 -26 | 2]
What you meant to do is R2 - R3
So it's
R2 = [0 -2 -10 | 2 ]
-R3 = [-2 2 16 | 0 ]
To get [-2 0 6 | 2]

[wise schooner]

thanks

,rotate

[delicate sphinx]

,rotate

[wise schooner]

could you also help me find the mistake in here

[delicate sphinx]

You made it harder on yourself when you multiplied R2 by 3 when you could have done R1/3

But when you did R2 - R1, not sure where you got 2 from

Because R2 = [3 6 3 | -15], R1 = [3 3 6 | -6]
So R2 - R1
R2 = [3 6 3 | -15]
-R1 = [3 3 6 | -6]
You should have gotten [0 3 -3 | -9]

[wise schooner]

yeah that's it thanks lol I'm horrible with mental math

I always mess up

so used to calculators but we're not allowed them on the exam :/

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[opaque girder]

hlep

[opaque girder]

Im stuck on what the angle part even means

[safe radish]

@opaque girder Has your question been resolved?

[safe radish]

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[lean otter]

I need help

[plucky elk]

do you know mean value theorem

[lean otter]

Yes

We consider f a function differential in [0,1] and positive strictly on ]0,1[ and f(0)=0

Show that

[plucky elk]

apply that to f(x) on [0, 1]

[lean otter]

This is a better transition

[lean otter]

What is the function ?

[plucky elk]

f(x)

try that first

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

fx only ?

i dont get it my bad

[safe radish]

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[modest holly]

help?

[lean otter]

well what does "end behaviour" mean to you

[modest holly]

function that ends at x axis

[lean otter]

no

end behaviour is what the function is converging to as it approaches infinity or minus infinity

[modest holly]

yea

im assuming its c

[lean otter]

yes

[modest holly]

help?

[safe radish]

@modest holly Has your question been resolved?

[safe radish]

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[void fox]

[Calculus] I'm working on my Implicit Functions homework, and am looking for someone to point out what went wrong in my work

(Cengage) (webassign.net)

[void fox]

Oh, the negative one

is that it?

[chrome wave]

try with -x+5y/-5x+y

[hard crest]

all the other steps seem right to me

[void fox]

That worked on the homework, I'm gonna look back on the work to see if I can work that out

Ah, I see now, thank you!

had these for previous answers on attempt 1 and 2, and the original screenshot shows attempt 3

Keeping track of the negative signs is important

Thank you!

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[scarlet wagon]

Since (1, 2) and (2, 2) are in the relation it should be transitive if (1, 2) is in the relation

[rough storm]

aRb, bRc -> aRc

also

(1,1)->(1,2)

what s hould be true if this is transitive

[scarlet wagon]

(1, 2)

[worldly breach]

Hi

[potent bay]

yes, it should be

[rough storm]

mhm

[scarlet wagon]

oh okj

[worldly breach]

@scarlet wagon

[scarlet wagon]

i guess this website im using is wrong then lol

[potent bay]

it worked with others?

[scarlet wagon]

not sure

[potent bay]

try it

maybe it needs a specific input syntax

[scarlet wagon]

i think it understood the input correctly

[final halo]

yeah the code is flawed

it thinks its not transitive if one of the off diagonals in the top left square is 0 when the other is 1

thinks that isnt transitive for that reason, but is happy with this

[scarlet wagon]

oh okay

thanks for confirming

.close

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[normal bay]

Hello! How would I start this one?

[rough storm]

,r

,rotate

[normal bay]

16 yeah, sorry

[versed folio]

16 ?

[normal bay]

Yup

[rough storm]

so first, I will focus on 3/5. we want to 0=x-3/5

multiply both sides by 5

$0=5x-3$

[flat frigate]

Moosey

[versed folio]

Oh okay i looking for "answer" xd

[rough storm]

this has integral coefficients, so we are done with the 3/5 part

[normal bay]

Oh! Is this using long/synthetic division?

[rough storm]

I am focusing on each part of this polynomial, you know how you can write a polynomial as a product of its roots yeah?

[normal bay]

Yes

But unsure how to relate that with the problem

[rough storm]

I am focusing on each of the roots, basically finding each part of the product, then you can multiply them out

so far we have (5x-3)()()

[normal bay]

Oooh

[rough storm]

the 7 as a root should be trivial yeah

[normal bay]

So the others would be (x-7) and (x+4)?

[rough storm]

mhm

[normal bay]

Alright

[rough storm]

then multiply out

[normal bay]

Ok

[rough storm]

(5x-3)(x-7)(x+4)

[normal bay]

Alright, thank you! I multiples it out and got the correct answer

Thanks so much

.close

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[wanton willow]

Question is on Pythagorus Triple

[wanton willow]

Find K is the goal

2D is original question

[weary osprey]

is k^2 on both sides like that?

[wanton willow]

I think so

[weary osprey]

this is straight up 1=2

[wanton willow]

How to write 2D?

Its a triple ✅

We just need to find K

[weary osprey]

(k+2)^2= k^2+8^2 ?

[wanton willow]

Lemme try

[weary osprey]

i see your mistake

(a+b)^2 is not equals to a^2 + b^2

[wanton willow]

Thank u

Yep got the answer

I forgot to put brackets

Thanks again

.close

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[normal crown]

hello

[normal crown]

i need help

idk how to graph that

[last heath]

Do you know how absolute value graph works?

[normal crown]

kinda

just not that one specifically

cuz the fraction

makes it confusing

like

my guess is

[last heath]

what about transformation rules?

[normal crown]

ye

im good with that

[last heath]

then use it

it's faster to graph it

[normal crown]

i wrote them down

i just need directions

cuz

okay

my guess is i just go down 7

on the y axis

and then

i go up 1 and right 3

idk if thats right though

[last heath]

,tex .transformation rules

[flat frigate]

Spooky Akira 👻

[last heath]

does this help better?

[normal crown]

yyeye

[last heath]

you can like

[normal crown]

thank you

i got it now

thank you

[last heath]

okay

you're welcome

[normal crown]

.close

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[quiet bobcat]

im stuck on how to proceed with integrating this i dont want the answer cause i feel stupid i just want a nudge in the right direction on the integration

[quiet bobcat]

.close

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[tight marsh]

10 and 11

i need urgent help

[iron flare]

do you know what slope-intercept form is?

[tight marsh]

y= mx + b

but im confused on this

[iron flare]

where have you become confused in 10. ?

[tight marsh]

i dont know how to convert it into slope intercept form

[iron flare]

a first step you can always do if you're confused is to make one side 'y ='

[tight marsh]

wdym

[iron flare]

we see on the left side of the equation " 12x - 18y "

but slope intercept form means one side should just be " y "

[tight marsh]

alr

[iron flare]

so you need to rearrange " 12x - 8y = 90 "
into " y = [something] "

[tight marsh]

alright

so whats the first step on that

[iron flare]

on the left hand side we see an x term, 12x, and a y term, -8y

we want to isolate the y term so that only the y term is on the left side

[tight marsh]

what if the equation also asks to write it with x intercept

[iron flare]

the only difference is that instead of trying to get " y = mx + b" you want " x = my + b "

so the first step will be different: write one side as x = , typically starting by isolating the x term

[tight marsh]

so get rid of the y term instead?

[iron flare]

yes

in 10. do you know how to get rid of the x term so you have the y alone?

[tight marsh]

12x - 12x?

and 12x - 90?

[iron flare]

close

you take away 12x from each side

[tight marsh]

wait so then

90 - 12x

[iron flare]

yes

an easy way to think of it is when you are moving terms that are + or -

to move it to the other side, you just swap the sign

so 12x 'disappears' from the left hand side, but 'moves' to the right hand side as -12x

• swaps to -

[tight marsh]

so now were left with -18y = 90 -12x

[iron flare]

thats correct

do you know what you do next, so that we are left with y = on the left hand side?

[tight marsh]

uh

divide -18y next?

[iron flare]

if we divide by -18y you will get 1 = on the left hand side

because -18y / -18y = 1

[tight marsh]

so were left with just y

on that side

now 90 - 12x / -18?

[iron flare]

oh you mean -18, yes thats right

yes, so y = 90 - 12x / -18

the question asks you to simplify fracitons however so you will need to simplify the fractions on the right hand side

[tight marsh]

so that means

90 / -18

and 12x / -18

[iron flare]

correct

[tight marsh]

wait so do we divide -12x / -18 like how we normally do it

[tight marsh]

-12x

[iron flare]

you want to write it as ' mx '

which just means you can leave the fraction multiplying x, since -12x / -18 is the same as (-12/-18)x

you just need to simplify (-12/-18)

[tight marsh]

so -2/3?

or 2/3

since two negatives when dividing are positives

[iron flare]

correct, so 2/3 is the right value for m

now similarly you can simplify the other fraction, 90 / -18

[tight marsh]

-5

[iron flare]

that's right

so, what is your final answer to the question?

[tight marsh]

y = 2/3x -5

this is easier since i have a sheet of paper in hand

usually i dont and thats probably why i can hardly understand

[iron flare]

looks right to me, though be careful when typing 2/3x as it can be seen as you meant x is on the bottom of the fraction

[tight marsh]

ye ik

so now my problem is

how do we convert slope intercept or point slope into standard?

for 11

[iron flare]

do you know what the standard form of an equation is?

or at least, one with just 'x' and 'y' terms as well as other numbers

[tight marsh]

well i kinda forgot

but

i know how to identify standard forms with numbers in them

i just forgot the formula

[iron flare]

the standard form you're looking for is

"ax + by = c"

[tight marsh]

wait so what does a stand for

[iron flare]

in this case, a is a number multiplying x

for example

5x + 3y = 10 is a standard form equation

here, a = 5, b = 3, c = 10

a b and c are numbers, x and y are the algebra variables

[tight marsh]

alright

[iron flare]

so with y = 3/2 x - 11 , do you know what you would do to write this in standard form?

[tight marsh]

nope

well

maybe i do like with the last one

but i need to learn the steps

[iron flare]

it is quite similar to what you did for 10

you are just moving things around to put it in the right form

except, this time it is even easier

you will not need to divide to make something just 'y =' or 'x ='

your goal is to just move all x and ys to on eside, and simplify it

and the lone numbers to the toher side

[tight marsh]

so how do we do that

[iron flare]

let me show you as an example

14x + 5y + 16 = 3x + 2

you should think 'i want one side to just be ax + by'

so move all the x and ys to one side

we see the 3x is alone so we should move it to the other x

[tight marsh]

this doesnt look familiar

[iron flare]

its an equation i made up

[tight marsh]

oh

[iron flare]

just to show you how to move it into standard form

[tight marsh]

but thats not slope intercept

[iron flare]

it can be

it is just very rearranged

say, for example, y = 5x + 14

you could rearrange this by splitting up the 5x into y = 3x + 2x + 14

and then maybe swap the side of the 2x, y = 2x = 3x + 14

that doesnt look like slope intercept but you can change it back by doing the reverse

so it is

does that make sense?

[tight marsh]

so you broke apart the slope

[iron flare]

yes as an example for how some equations that dont look like they can be slop eintercepts in fact can

in fact, any equation that consits of just x and y's (that arent quadratic, if you have come across that) can be a slope intercept

that aside, it does not really matter in this case, we ar ejust making a equation standard, not necessarily a slope intercept one

[tight marsh]

i havent come accross splitting the slope yet

but

nice

[iron flare]

oh it isnt something you like, do

[tight marsh]

in the class we dont

[iron flare]

well, let me give a different example then

3x + 4 = 5y

to make standard, we want x and y terms to be on the same side

so we move the 5y

3x - 5y + 4 = 0

but we also want the other numbers that arent x or y terms to be on the opposite side to x and y

so, we move the 4

[tight marsh]

isnt 4 the lone number

[iron flare]

3x - 5y = -4

yes

[tight marsh]

or y intercept

should it be on the right of =?

[iron flare]

Yes

i just did it one step at a time

so first i moved 5y, then i moved 4

[tight marsh]

oh

[iron flare]

the final answer is 3x - 5y = -4

[tight marsh]

could we do it with familar slope intercepts?

[iron flare]

alright, but i think you should be expected to do this in any equation

but lets look at the answer we got last time in 10.

[tight marsh]

i dont think its in the test tho

[iron flare]

y = 2/3x -5

we see here we have on each side [ y term = x term + lone number ]

but we want [ x term + y term = lone number ]

so we move 2/3 x

y - 2/3 x = -5

and we are done

that is now in standard form

[tight marsh]

alright

wait isnt the lone number the y intercept

[iron flare]

in this case yes, im just being very general and calling it a lone number because

[tight marsh]

oh

[iron flare]

a standard form is not quite the same as an equation for a graph that you are used to

[tight marsh]

so we already have the x term and lone numbe

number

we need to find out whats with y?

[iron flare]

what do you mean?

[tight marsh]

y doesnt have a number next to it

like what a standard equation should have

[iron flare]

it actually does

we just dont need to write it

y = 1y

but we know 1*y is just y by itself

so we just write y

'y' is in the form ay

[tight marsh]

oh alr

what if the answer has like a number besides one

and in the question were supposed to find out

[iron flare]

can you give me an example, im not entirely sure what you m ean

[tight marsh]

like what if we had an answer thats like 5y for the y term

how do we find out in the first place that its a 5y

[iron flare]

im still not 100% understanding you, i am sorry
what do you mean by 'in the first place'?

[tight marsh]

nvm

sorry

so wait

the answer is 1y + 3/2x = 11?

[iron flare]

yeah but dont write the 1

[tight marsh]

alright

so now how do we do it with point slope to standard

for the other one

[iron flare]

oh i suppose it is in that form, i didnt even notice lol

y+21=-4(x-2)

[tight marsh]

what

[iron flare]

that is the original equation

i think

have i typed it wrong?

for 11 b)

[tight marsh]

alright

well it doesnt matter

since this is the review

i just want to know in general how to convert it

[iron flare]

the general idea is to write 'ax + by = c'

you do this through rearranging the equation

[tight marsh]

alright

[iron flare]

other than memorizing what the different forms are i think some practice with rearranging equations should help you with your test

[tight marsh]

yeah wait

[iron flare]

as that is what a lot of it boils down to

[tight marsh]

so is it possible you could write down all types of combination with combining forms?

you dont have to

[iron flare]

all types of combination? do you mean the strategies for rearranging an equation

[tight marsh]

well

yes

[iron flare]

i think so, but ill just need a moment

[tight marsh]

wait so how long will it take

[iron flare]

a few minutes?

[tight marsh]

cuz u dont have to do it if it takes awhile

alright

i really appreciate it man

[iron flare]

im still not 100% as to what you're asking for so sorry if this is too much detail

as a reminder the core idea is that, since both sides are equal, what you do to one side you do to the other

i.e 4x = 5y will stay true if you multiply both sides by two: 8x = 10y
however if you multiply one side by two and the other by three, for example 4x * 2 and 5y * 3, you get 8x = 15y which is not the same, and may no longer be true

this is the logic we apply when we move terms by adding and subtracting them

consider 4x + 5y = 3

if we want to get rid of 5y from the left side, we would want to do - 5y, so that we have 4x + 5y - 5y = just 4x. this also msut happen to the other side though as we must do everything to both sides, so 4x+5y-5y = 3 -5y

this leads to the trick:

when getting rid of (by moving) a term, you move it to the other side and flip its sign
4x = 3-5y is done by this

similarly if we have negative terms this is done the exact same way:

if we want to move the -5y back, we can just move it and change the - to a +
like 4x+5y=3, as we started

with multiplying and dividing you seemed to already ahve the hang of it: multiplying and dividng affects the entire of both sides

these are all the basics of how to add, subtract, multiply and divide in your equations

[tight marsh]

yo no way

listen i dont know how ill ever pay you back

[iron flare]

in 11 b. there is also the notation -4(x-2), which is just the same as "-4 multiplied by (x-2)"

[tight marsh]

yeah ik

i think thats enough

[iron flare]

i think you should also learn to group like terms

[tight marsh]

thank you so much

[tight marsh]

will do

[iron flare]

but if you are tired and want to stop since we've been through a lot that's alright

[tight marsh]

yes

[iron flare]

👍 hope you understand everything a little more than you started

[tight marsh]

once again i dont know how ill ever repay you back

cya

.close

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[fresh birch]

Does my graph look correct for 5b?

Or is this right?

[safe radish]

@fresh birch Has your question been resolved?

[fresh birch]

Could someone plz help???

<@&286206848099549185>

[safe radish]

@fresh birch Has your question been resolved?

[safe radish]

@fresh birch Has your question been resolved?

[fresh birch]

Anyone???.?.?.

[safe radish]

@fresh birch Has your question been resolved?

[safe radish]

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[terse bane]

Trying to answer q24, but it looks like it will require the solution of q21 which i just proved

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[safe radish]

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[lean otter]

Just curious but where do I ask about language/logic questions?

[lean otter]

I have this following question

If someone does have an idea here that'd be great too

[buoyant shadow]

someone recently said #foundations

the description fits

[lean otter]

Okay thanks

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[oblique garden]

can anyone provide me the proof?

[oblique garden]

sigh

the image isn't loading my data is shit rn

A adj(A)=adj(A)A= I |A|

[dull sequoia]

$A^{-1}=\frac{1}{det(A)}adj(A)$

have you seen this before

[flat frigate]

Frosst

[oblique garden]

I have but I don't know the proof

idk how we got that so can you explain

[peak estuary]

think about the entries you get if you multiply out A*adj(A) and what you get if you expand various rows and columns of A

[oblique garden]

is there any nice proof or do I just have to prove it manually

[peak estuary]

the only proof I know is doing it "manually"

[oblique garden]

okay I'll try to see

ok so

detA = A adjA I saw this somewhere but how is a numerical value equal to a matrix?

since detA is a number and A adjA is matrix multiplication

[peak estuary]

well its not

where have you seen it

[dull sequoia]

nowhere does it say that

[dull sequoia]

how did you get to the first line

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[lean otter]

So uh, really, what is the reasoning behind this axiom?

Why does this need to be declared?

[dry scarab]

Just to tell you obvious thing

[obtuse jackal]

we're taking inspiration from things like Z, R, or even Z/nZ

in practice, 0 is never 1
Might as well not forbid properties that only fail in fictional instances

[viscid ore]

0^-1 times 0 = 0 = 1

to avoid this ig

[obtuse jackal]

Well if 1 = 0, that's fine

it's very incompatible with 0*a = 0 though, since if 0 = 1, you have 0*a = a

so it would break distributivity

hence the only ring with 0=1 is the singleton/trivial ring

[lean otter]

okay yeah this makes sense now thank you

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[obtuse jackal]

reminder: 0*a = (1-1) *a = a - a = 0

hence it relies on distributivity

[lean otter]

ah so 0*a = 0 itself would not be an axiom

[obtuse jackal]

well it breaks one way or another

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[proper moat]

hi

[proper moat]

,tex $4x^{21} - 3x^{12} + 2x^2 + x = (x^3-x)Q(x) + R(x)$

[flat frigate]

JoeTheLazy1

[proper moat]

what is R(2)

[quasi bison]

anything else known about Q and R?

[proper moat]

nope

[quasi bison]

then there is not enough info

are you 110% sure there is nothing else?

[proper moat]

why? it's just normal divison

[quasi bison]

!xty

sddgggldfkj

so this is a division of polynomials.

[lean otter]

uh i have a sneaking suspicion R(x) is remainder and Q(x) the quotient of the polynomial on the left being divided by (x^3-x)

[quasi bison]

Q is the quotient and R is the remainder, yes?

[proper moat]

yes

there is an absoulte trash way

and a good way

[quasi bison]

ok you should have said that when i asked lol

otherwise we could not know

anyway yeah there is a better way i think

find R(-1), R(0) and R(1) first.

[proper moat]

well that is the hard way

the teacher solved it in a number theory sort of way

he said x^3 - x ≡ 0

so 4x^21 + 3x^12 + 2x^2 + x ≡ 4(x^3)^7 + 3(x^3)^4 + x^2 + x

and the rest

[proper moat]

I want to know the reason behind this

[quasi bison]

oh i see what hes doing

right ok so

do you know modular arithmetic?

[proper moat]

yes

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[quiet bobcat]

hi just wanted to check if my answer was right

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@quiet bobcat Has your question been resolved?

[quiet bobcat]

<@&286206848099549185>

hello?

[junior smelt]

Seems right to me

[quiet bobcat]

okay thankl you xxx

wait is the final function odd

i tried it and it gave me even but it has to be odd

]because the ft is odd

[junior smelt]

\catthink actually think the power should be negative, so the final line should be $\frac{\sqrt{\pi}}{\pi} e^{-\frac{t^2 \sigma^2 }{2} }$

[junior smelt]

wait how do you mean?

[flat frigate]

@junior smelt

[junior smelt]

The final function ends up being even by the looks of it, as is the original function you're working with

[quiet bobcat]

okay thank you xxxx

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[noble knoll]

hello

[plucky elk]

what's your question?

[noble knoll]

my exponential is both negative and irrational

there has to be another step?

[plucky elk]

do you know the domain of the logarithm function?

and why is irrational bad?

[snow matrix]

I don't think there is a nother step to this

[noble knoll]

yes?

[plucky elk]

what is it

[noble knoll]

so a has to be greater than 0

[plucky elk]

so you shouldn't be getting a negative value for a then

[noble knoll]

yes but i dont see what should i do then?

[snow matrix]

so you want to solve for a?

[noble knoll]

yes

[plucky elk]

you should be getting a positive value for $8^{-2/3}$

[flat frigate]

riemann

[snow matrix]

use rational exponential rules

[noble knoll]

oop i see one sec

[noble knoll]

so?

[snow matrix]

yeah, it can still be simplified

if there are no exponents, it is in its simplest form

unless stated in the problem that it needs to be in a certain form (i.e.: decimal, by a factor of 4, etc.)

[noble knoll]

okay okay thank you very much

got it