help-4

[runic scroll]

and y = 0 at x = 0 means it should cross (0, 0) - the origin.

[west rapids]

oh mb

also im drawing on microsoft paint so yeah

irl i can do much better

[runic scroll]

you could have drawn on paper.

but either way, so long as you know what's happening here, all good.

[west rapids]

yeah but ion have a phone near me rn so i cant send a pic

[west rapids]

yup thanks alot for the help

and sorry for wasting your time

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[twin crow]

I am substituting x = u and y = uv. The Jacobian here should be u. Since the absolute value of the Jacobian should be taken, I am confused as to why they multiply by x (which is equal to u) in the first integral when they should multiply by -x.

[pine prairie]

Huh?

What’s the problem

Going from dy to dv means you need to do dv/dy dy = dv

[twin crow]

What I don't get is that x seems to be going to from negative infinity to 0 in the first double integral, so given the absolute value of the jacobian must be taken, they should multiply by (-x) instead of x.

[pine prairie]

dv/dy is 1/x so it’s dy = xdv

It’s not absolute value of the jacobian

[midnight pier]

very random but i wanna say kudos to you i dont even understand this shit 😭

[pine prairie]

It’s the determinant of the jacobian

[twin crow]

as you can see here, they always use the absolute value of the determinant (Jacobian).

[jade ivy]

I'm not sure I see what the issue is, they have the absolute value on the last line

They are just doing it via a single variable change of variables instead of using the jacobian

[twin crow]

yes, but multiplying by -x immediately after substitution results in a different expression

[jade ivy]

Honestly, I never really look at the limits of integration when I do this step so I haven't tried what you are proposing before. I know the single variable substitution is a special case that doesn't need the absolute value, but I would need to work it out with a pencil and paper to see how the two perspectives would end up agreeing with each other

[vale dock]

@twin crow Has your question been resolved?

[jade ivy]

You'll have to write it all out, or at least I'm not going to. But you need to be a little bit more careful here since between your first image and in the theorem for the second, the order of integration changes. It's not so easy to just glance at it and see what is going on.

[twin crow]

I did write it out and in the end I got a different expression.

the bounds remain the same but in the first term, I have negative |x| instead of |x| if I subtitute x = u and y = uv.

nvm I think it was because I forgot to put the bounds in increasing order after taking the absolute value.

https://math.stackexchange.com/questions/856654/why-absolute-values-of-jacobians-in-change-of-variables-for-multiple-integrals-b

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[hardy coral]

Just expand it?

[red cobalt]

how so

[hardy coral]

Expand (x-1)(x+3) into something in parentheses, and then expand that with (x-5)

[red cobalt]

ohhhh

i get it

thx bro

i really appreciate it

much lovee

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[golden ember]

For each $x \in A$ we have that there exists an open neighborhood $U$ such that the union of these $x$ is open and gives us the interior of A.

For each $x \in \overline{A}$we have that $X - $Clo$(A)$ = int$(X-A)$

Since the interior (X-A) is open, then its compliment clo(A) is closed

[stark wedge]

\in btw not /in

$x \in \overline{A}$

[rocky lotus]

Ann

[golden ember]

okay is see, I have to figure out where those are, my native latex interperator uses /

[rocky lotus]

RealTek

[golden ember]

Anyways this is my attempt at this. If I am down the right path please let me know 🙂

[vale dock]

@golden ember Has your question been resolved?

[woeful trench]

"...the union of these x is open" not quite sure what you mean by that

[golden ember]

The union of hte nhbds sorry

[woeful trench]

You've only mentioned one neighbourhood so far (namely U), and haven't said much that defines it

(did you want x in A, or?)

[golden ember]

I was just thinking that for the interior of A we could have multiple elements that each has their own open nhbd, and the union of those is open, which would be the interior.

and I could define each U to be a b(x,e)

ah where b(x,e) subset A

then int(A) = union_ x in int(a) b(x,e)

Okay thanks ill re-do this tommorow.

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[tardy widget]

i'm having trouble figuring out where to start here, so far i've assumed not p and q

[tardy widget]

i'm thinking i can prove p and q separately and then use ->i

just not sure how to go about it

[hardy coral]

You need to prove that's a tautology?

[tardy widget]

just the validity of it

using deduction rules

[hardy coral]

Do you have a list of rules?

[hardy coral]

Ok I'm not quite familiar with this, do you have a previous question to show so I can see how you've done it?

[tardy widget]

which part aren't you familiar with?

[hardy coral]

The notation mainly

[tardy widget]

okay, let me know if that example isn't enough then

[hardy coral]

Ok I can follow this

Let's see.. you don't have a premise, so you can just start with the assumption ~p or q

[tardy widget]

yes, i was going to assume that, then prove p->q and then use ->i to prove the whole thing

[hardy coral]

Yes exactly

[tardy widget]

but proving p->q is where i'm stuck lol

[hardy coral]

You can split like you said, prove with p and then with q

Or rather with ~p

Take ~p as your assumption (that would be the second step)

[gleaming kindle]

https://i.gyazo.com/7609b88c044d9603114f7a0fc7fd1cc2.png

are you allowed this rule?

[hardy coral]

I don't see it in there so I assume not, but good call

[tardy widget]

i think he's brought it up but doesn't want it being used

that would be the nicest way

[tardy widget]

wait, could I use LEM if it's two different variables? cause this is up there but only with -phi or phi

mm no i think he specifies when he wants LEM being used

[gleaming kindle]

no

[gleaming kindle]

this is called disjunctive syllogism. are you sure he explicitly forbade using it?

[hardy coral]

That's odd, I would've expected another rule like [(~P, ...) => P->Q]

[tardy widget]

he has stated he only wants us using the rules in the above diagram

that's what we'll be given on tests

[hardy coral]

For the other part, with the assumption q, you have a rule that gives you the conclusion

You see it, right?

[tardy widget]

honestly no

i only see it with the assumption p

probably being blind

[hardy coral]

I might be stupid

What exactly is this notation? Having all of these as true like a conjunction?

[gleaming kindle]

are you referring to the alt def of implies?

[tardy widget]

so assuming phi, prove psi with alternate rules

[hardy coral]

Oh that makes more sense

[tardy widget]

then you can conclude phi -> psi

idk i've been thinking about this one for like 3 days to no avail

[hardy coral]

It does feel like none of the rules are helpful at all

[tardy widget]

wait

[gleaming kindle]

i have smth but its very ugly

[tardy widget]

anything will help lol

[gleaming kindle]

assume p to setup ->i then assume !q to setup PBC

[tardy widget]

i could kind of see it

could you do the same with p and -p instead?

[gleaming kindle]

theres lots of ways to do it including disjunctive syllo which wouldve made this infinitely faster

but if the prof insists on a limited rule set then this is as direct as i can be

[tardy widget]

wait what would the PBC do

[gleaming kindle]

heres the whole setup

[tardy widget]

.. i just found the solution in his notes

looking at his notes on PBC

[gleaming kindle]

ok i wont type it out

the gist is nested setups for ->i

[tardy widget]

yeah that was kind of what i was thinking

this is his solution, one of the rules i actually don't know how to use which is probbly why

[gleaming kindle]

at some point we must prove q so i assume !q to setup PBC

[tardy widget]

i don't know how the upside down T thing works

for elimination

[hardy coral]

Contradiction implies anything

Steps 4 and 5 are these:

[gleaming kindle]

aka principle of explosion 💥

[hardy coral]

But the phi in the first part is p, and the phi in the second part is q

[tardy widget]

so i can do WHATEVER?

that's crazy

[hardy coral]

Well... kinda, but you need to combine afterward

[tardy widget]

why do we assume q again?

ohh wait

[gleaming kindle]

for \/e

btw you can prove this using explosion https://i.gyazo.com/7609b88c044d9603114f7a0fc7fd1cc2.png

[hardy coral]

So step 7 is this, where [phi or psy] is the premise, phi is ~p, psy is q (and chi (?) is also q)

Step 6 is just assuming q to set it up for that rule

[tardy widget]

yes i see

one issue, thi question is slightly different because -p or q is the premise

whereas in mine, it's an assumption

[hardy coral]

Yes you need an additional layer

[hardy coral]

The premise is nothing, the conclusion is [step 1] -> [step 8] from ->i

[tardy widget]

did i box this right 😭

nvm, i think i got it

thank you both for your help, it's really appreciated

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[hardy coral]

I think you always need a first line of "premise", even if it's empty

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[glad pagoda]

I am studying college algebra right now and have been curious about ordinal numbers. What exactly are ordinal numbers? Are ordinal numbers part of the real numbers? Will I ever see it in college algebra?

[frank salmon]

U meant complex numbers?

[drifting hornet]

probably not in algebra, ordinal numbers are related more to set theory

[glad pagoda]

I mean ordinal numbers

[drifting hornet]

if you have a set theory subject, you might see it in there

[glad pagoda]

From my understanding they are like 1st 2nd 3rd…

[drifting hornet]

maybe in some proofs class as well

[drifting hornet]

yes, thats how it starts

then there are infinite ordinals and thats when it gets more complex

[glad pagoda]

Are they apart of the real numbers?

[drifting hornet]

no, not really

[glad pagoda]

So unless it is explicitly stated, is it assumed the numbers are part of the reals?

[drifting hornet]

and they are not really that significant from an algebraic perspective afaik. You can do some basic algebra on them, but they are mainly significant from the perspective of set theory

[drifting hornet]

ordinal numbers are not part of the reals

nor are reals part of the ordinal numbers

ordinal numbers and reals are just something completely different

more different than e.g. natural numbers and complex numbers

[glad pagoda]

I mean, if I see “9” I can assume it is a real number unless its explicitly states

[drifting hornet]

Yeah, sure

this isnt usually an issue though. It's usually clear from the context whether you talk about reals or ordinals

[glad pagoda]

How I started to think about this was when I thought about how days work.

When someone says “in 2 days” it often doesnt mean 48 hours. They usually mean the day after tomorrow

Would you consider this an ordinal number?

[drifting hornet]

its best not to think about ordinals as things we use in everyday life

they are abstract mathematical objects

and it will make it easier for you to take them as such, because once you introduce infinite ordinal numbers, it will no longer make any sense to mention them in everyday use

[glad pagoda]

So what do you do when you are working with order

Like 5 places after 3

[drifting hornet]

3 + 5 = 8

naturals

i dont see any reason why one should use ordinals here

irl, we only ever talk about finite orders, so naturals are sufficient

[glad pagoda]

When are ordinals uses?

[drifting hornet]

what makes ordinals special is that they can order infinite sets as well

infinite ordinals is what makes ordinals interesting

only finite ordinals are really uninteresting. In fact, they are same as naturals

they are not used to tackle any word problems from real life. They are used only in pure and abstract math

it might be possible that they are used in some super fancy physics or other science, but i know nothing about that

[glad pagoda]

So even in problems with position you can just use natural numbers

[drifting hornet]

yes

[glad pagoda]

So natural numbers can represent amount or position?

Like 8 apples or the 3rd place

[drifting hornet]

yeah, sure, why not

only when you get to infinite stuff youll need cardinals for amoutn and ordinals for order

but for finite stuff, naturals work just fine

[glad pagoda]

Alright, its just something I have been thinking about

I guess I felt like the real numbers only represented quantity

[drifting hornet]

well its really up to you what they represent. They give you some structure with the addition and multiplication that you can work with. And for example 3rd place + 5 places = 8th place, so the structure seems to be the same as the one on naturals and reals, so why not just use them? Why invent something which would behave the same way

we dont need naturals for quantity, naturals for order, reals for length, reals for area, reals for quantity, ... because one naturals and one reals just work for all those applications and it wouldnt make any sense to just copy and paste them

[glad pagoda]

I see

I think I am getting to ahead of myself

[drifting hornet]

You can wait for uni to learn about ordinals and cardinals or if you wish to, pick up a set theory book and u can learn about them rn. Once you do, you'll understand how they differ from reals and naturals and what makes them interesting enough to be a "number system" on their own

[glad pagoda]

For sure, yeah I am studying for meteorology

Thank you!

[drifting hornet]

Np

[glad pagoda]

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[haughty roost]

i was writing some game and i want to know where i should find something similiar to the concept of what im making

[zealous pendant]

Sure, what is the concept

[haughty roost]

those are weight distributions of the grid if the blue square randomly moved in any direction after 71 steps

i want to know where i can learn more about this concept and if there is any way i can improve it into a continous rather than "9 possibilities after 1 iteration"

[ebon glade]

well this is called a random walk

[haughty roost]

i see, very self discriptive

[ebon glade]

there are continuous versions of that

there is also brownian motion/wiener processes which I think are also continuous versions

but I dont really know a lot about them

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[heavy sigil]

Can anyone explain to me how similar triangles work

[wraith heart]

https://www.cuemath.com/geometry/similar-triangles/

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@heavy sigil Has your question been resolved?

[honest stone]

still waiting on the answer to your hypothesis bro

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[sweet sinew]

could someone explain how the answer is -oo

[steady charm]

it looks like $\frac{-n^2}{n^{3/2}}$ for large $n$

[rocky lotus]

κλαοδ ☁ (cloud)

[sweet sinew]

isnt this the same thing

as the function inside the limit

[steady charm]

no because $\sqrt{x^2} = \abs {x}$

[rocky lotus]

κλαοδ ☁ (cloud)

[sweet sinew]

so in this case, my teacher said to add a negative infront of the sqrt sign

[steady charm]

that `fixes it' because $-\abs x = x$ for $x < 0$

[rocky lotus]

κλαοδ ☁ (cloud)

[sweet sinew]

but how do we know when to add the - infront

[steady charm]

if the expression inside is negative

[vale dock]

@sweet sinew Has your question been resolved?

[grim hemlock]

whenever limit is infinite i’d advise u to divide by higher power of x

and then js put infinity instead of x

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[wintry creek]

Can anyone help me with this geometrical problem-

Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium. (Use midpoint theorem)

[wintry creek]

Please ping me if anyone answers

[vale dock]

@wintry creek Has your question been resolved?

[vale dock]

@wintry creek Has your question been resolved?

[hazy pivot]

Well draw a diagram to start with

[cloud coral]

after drawing the diagram, you may want to connect some midpoints

[vale dock]

@wintry creek Has your question been resolved?

[wintry creek]

Ok then

[vale dock]

@wintry creek Has your question been resolved?

[wintry creek]

<@&286206848099549185>

[cunning sky]

hi

[wintry creek]

Help me

[cunning sky]

draw a diagram and connect some midpopints

haev you drawn it or

[jovial edge]

send it here bro

[cunning sky]

have*

[wintry creek]

I can't draw

I am not able to understand the question

[jovial edge]

then why you said ok

[wintry creek]

I thought I could do it

[flint phoenix]

.......

[cunning sky]

fr

[cunning sky]

when u ask a euclidean geometry question post ur drawing here next time

was in the middle of a pf match

anways you want to use midpoint theorem like stated

[wintry creek]

yup

[cunning sky]

use thales

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

[wintry creek]

my teacher only wants midpoint

not bpt

[cunning sky]

how do you understand the midpoint theorem

[cunning sky]

aa1 and bb1 intersect at c

prove cde and cab is similar

[wintry creek]

I have not yet read similarity

[wintry creek]

It is a theorem

[cunning sky]

only solution i can think of tbh

imma leave this image for any helpers to come back k?

[wintry creek]

Can u use congruency

[wintry creek]

fr

[jovial edge]

what do we have here?

Are we trying to prove a theorem...

[wintry creek]

We want to prove a question

using midpoint theorem

refer to the pinned msg

[jovial edge]

what's midpoint theorem lol, I'm not familiar with that name...

Gimme a min

sorry my native language is not english

[wintry creek]

https://en.wikipedia.org/wiki/Midpoint_theorem_(triangle)

Check it

The paragrph written is enough

after the link

[wide fossil]

wt we have to do prove it?

[wide fossil]

i can

[wide fossil]

see lets say a trapezium ABCD and draw its diagonal lets say diagonal AC has midpoint N and BD has M now join MN and we know midpoint theorem says that midpoint of two sides of triangle are always parallel to the third side i.e is base taht means MN parallel to DC and since dc was parallel to AB that implies AB parallel MN parallel DC @wintry creek

[woeful trench]

,rccw

[wintry creek]

Ok thanks

.close

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[humble flicker]

does someone mind confirming my answers and working out

[normal hollow]

Solution 1 seems alright

[humble flicker]

what abt solution w

2

[normal hollow]

Correct as well

[humble flicker]

so why is that we can replace dx with d(…)

[normal hollow]

Since u subbed u=1+4x³
du=12x² dx
U can write dx=du/12x²

[humble flicker]

sorry i meant for solution 2

[normal hollow]

And u=1+4x³
So dx=d(1+4x³)/12x²

[normal hollow]

I think the sub was used for both

[humble flicker]

i’m doing the working following this

and for solution 2 they’ve done the same

which is the second last line

i just don’t understand how they can replace the dx with d(ax + b)

[normal hollow]

They didnt repplace dx with d(ax+b) ,they replaced dx with d(ax+b)/a

[normal hollow]

Js like they have written y=f(x)
So ir was possible to replace y in dy with f(x)

[humble flicker]

but how do u know what x is

[normal hollow]

Lemme write it down and show u

[humble flicker]

okay thanks

[normal hollow]

Also they determined what dx is not what x is

[humble flicker]

isn’t it meant to be dy/f’(x)

[normal hollow]

Oops yea

Lemme give u a refined one

[humble flicker]

okay thanks

cus i wanna put that into my notes

[normal hollow]

What u have been given is ur note already
But u r prolly js missing this that
The y in dy is the function of x i.e. (ax+b)^n

[humble flicker]

and there’s a rule that allows u to move the a to the outside right

cus it’s a constant

[normal hollow]

Yes

[humble flicker]

so the stuff in blue

the adx

like is that confusing

[normal hollow]

Did u use f(x) for (ax+b) or (ax+b)^n

[humble flicker]

i’m pretty sure ax + b

[normal hollow]

It seems alright
There is not supposed to be any adx if u take this approach
The a dx would be required if u shifted it to dy

[humble flicker]

personally it’s confusing me

but i understand ur method

so i’ll just remove it

but ig i sorta understand the blue as well

cus the logic is you’ve ‘introduced’ an adx so u place the fraction outside to ‘cancel’ it but in reality there’s actually not an adx present

[normal hollow]

The cancelling out would work when u converted d(ax+b) to adx

[normal hollow]

Yes there is no adx in this approach

There is a possibility that whoever mentioned this mixed it up with another approach

[humble flicker]

yes

[humble flicker]

perhaps

im simply just following along

it’s easier now that i understand it

[normal hollow]

Whatever blue is referring to is taking u back to the original question

Also the tick marked portion =>
U can't write it in this way

The presentation is wrong
Cuz with respect to y everything within should be a function of y(but to make u understand, i went along)

[humble flicker]

so how should it be then

just straight from d(ax + b) to adx?

[normal hollow]

Either that(that's the best thing to do)
Or u could express the (ax+b)^n=y^n
And then shift it back to (ax+b)^n in next step
Hassle

[humble flicker]

ight

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[long parrot]

Are addition rules with sig figs based off the place values or decimal points
like 7 000 000 + 10 or 1200 + 40
do you round them too cus they dont have any decimals

[steady charm]

you would take the least significant digit of each number and round it to whichever position is larger

[long parrot]

so 1200 + 40 the least sig digit of 1200 is the hundreds place and 40 is the tens place so it would be 1240 -> 1200?

[steady charm]

yes

[long parrot]

ok thanks i think i get it

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[runic sedge]

Help please, I have no idea where to go from here

[runic sedge]

This is for competition math, so it’s not like a typical geometry problem

[wild linden]

If you exchange the triangle and the square on the left you preserve the area and it might be easier to decompose, into a trapezoid and a triangle

But I've not tried yet just glanced at it

[wintry raptor]

1. Put square DEIJ with side 18 so
   D = 0,0 E = 18,0 I = 18,18 J = 0,18
2. Let C be the point on DJ with Coords C = 0,t for some 0<=t<=18
   Since ABCJ is a square and JC is one side of it, the side length of that square is 18-t then
   B = (-(18-t) t) = (t - 18, t)
3. Let H be the point on EI with coords H=(18,s) for some 0<=s<=18
   Since EFGH is a square with side s, we get
   G = 18 + s, s
   Now compute the area of the quadrilateral BDGI with vertices in hpthat order
   B = (t-18, t) D = 0,0 G = 18 + s,s) I = 18,18
   Use the shoelace formula
   Area = 1/2 | (t-18) * 0+0s+(18+s)18+18t -(t0+0 (18+s)+s18+18*(t-18)) | = 1/2 | 0+0+18(18+s)+18t-(0+0+18s+18t-324) | =1/2 | 324 + 18s+18t-18s-18t+324 = 1/2 (648) =324

[runic sedge]

Thank you so much @wintry raptor

[open niche]

another way is to just complete the rectangle around this and substract the tringales

everything cancels out except 324

@runic sedge

[runic sedge]

@open niche thanks for the other solution

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[silk magnet]

right now i figured out a1 which equals 13.5

i wanna figure out theta

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@silk magnet Has your question been resolved?

[haughty jackal]

You wanna use the fact that the tangent drawn to a circle at any point is perpendicular to the radius at that point.

[silk magnet]

i did that

i drew a line between the centres of the two circles as well

now what do i do

[haughty jackal]

Use some trig

[silk magnet]

i got the angle as 68.9* how would i go about finding the whole length of the belt?

[haughty jackal]

Divide it into components

[haughty jackal]

Like this

[silk magnet]

the length* of the arc would be theta (rad) * R

for each of them

and then the sides of the triangles i can figure them all out with pythagoras

add them up

[haughty jackal]

Not here

Theta doesn't subtend a sector on the circle

Try using the fact that all the angles of a quadrilateral add up to 2π rad

[silk magnet]

would theta/2 be the angle for each edge of the triangle beside the middle

[haughty jackal]

This angle?

If so, yes

[silk magnet]

Yea

alr thank you

[haughty jackal]

Np

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[silk magnet]

how would i solve for the arc length here?

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[distant galleon]

They told you tbf

[distant galleon]

^

They central angle of each sector is this orange angle

Which is 2pi-this blue angle

[distant galleon]

Which you can find using this ^

[marble arch]

Also we can use similarity among triangles to find the ratio between a1 and a2 to find a1

Using the fact that a1 + a2 = L

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[novel zealot]

\textbf{Find all values of real number $h$ for which each equation has a solution or show no such $h$ exists.}
\\
\textbf{(a) $\begin{cases}
x_1 + 3x_2 -x_3 = h + 2 \
2x_1 + x_2 - x_3 = h \
-3x_1 + x_2 + x_3 = h + 1
\end{cases}$}
\\\
$\begin{pmatrix}
1 & 3 & -1 & h+2 \
2 & 1 & -1 & h \
-3 & 1 & 1 & h+1
\end{pmatrix}$

is this just a row reduction exercise lmao

[rocky lotus]

toast

[novel zealot]

or is there a better way of doing this?

cuz row reducing is so annoying

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[daring harness]

Is this .NET answer correct or way off base: [System.Math]::Cos([System.Math]::Tau/360*90.0); 6.12323399573677E-17

[gritty sinew]

it's essentially 0, ...E-17

so floating point error really

[daring harness]

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[lucid nimbus]

im stuck as to how to find F2

from the equation

As F1 cancels out

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[cloud coral]

Wait isnt that how you solve linear systems

basically you need to make sure force in x direction and force in y direction are 0

Another way is to take advantage of the fact F1 F2 are perpendicular

Then you resolve in F1 and F2 directions

[lucid nimbus]

so from where I am

do I rearrange for f2

basically would I solve the equation

I made for f2

[lucid nimbus]

so we are using the elimination method?

[wintry oxide]

Have you studied linear algebra ?

[lucid nimbus]

yh

ig

why

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[ashen prawn]

by the component breaking components i got this and thus this

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[novel zealot]

am i just stupid or is part b legit a carbon copy of part a?

[novel zealot]

if the rows of A are linearly independent, then the columns of A span F^n?

then we're asking if the rows of AB are linearly independent, arent we just asking if columns of AB span f^n as well?

[keen tundra]

No this is not the same thing

The rows of A being independent means that
x-->xA is injective

[novel zealot]

wait just in general, if A has all linearly independent rows, then its column space has to span f^n?

might be dumb af

[keen tundra]

Ok this is true

[novel zealot]

yeah

[keen tundra]

But this is a theorem

If the rows of A are independent then rkA >= n so dim(col(A)) >= n

[novel zealot]

i mean could we say something like

A has linearely independent rows <=> col A spans f^n <=> AB spans f^n (by part a) <=> AB has linearly independent rows)

cheese but whatever

[keen tundra]

Its cheese

[novel zealot]

Lol

but works right?

i mean ill try to prove it without part a

[keen tundra]

Especially if you haven't learned the theorem where
rkA = dim(colA)=dim(R(A))

[keen tundra]

.

[novel zealot]

uhhh

yeah we havent learned this yet

ok but ty

[keen tundra]

This can help

[novel zealot]

hmm

ill look into it now

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[zenith lantern]

How do I determine if the following sequence converges or diverges? Can I get a hint

[leaden kraken]

are you allowed to use a calculator to find e^pi/pi^e?

[zenith lantern]

I'm not sure. But what if I could or couldn't?

[leaden kraken]

if you could, then you would be able to resolve the question: for any real number a, if a is nonnegative and less than 1, then as n goes to infinity, a^n goes to 0

[zenith lantern]

Right

[leaden kraken]

and if it’s equal to 1 or larger than 1, then you can find the limit relatively easily in those cases as well

[zenith lantern]

If it's larger than 1 the limit is infinity right?

[leaden kraken]

Yes

[zenith lantern]

What if I couldn't use a calculator

[leaden kraken]

Then you can still use this method, but now you have to find out whether pi^e/e^pi is less than 1, equal to 1, or greater than 1 by hand

[zenith lantern]

How would I do that

Or rather, is there a way that's not time consuming to do it

Because I doubt we'd have to do some time consuming calculation on an exam

[leaden kraken]

There’s a clever way, but I only know it because I’ve seen it before

[keen tundra]

you compare e^pi and pi^e by taking their ln

e ln(pi) ? pi ln(e)
ln(pi)/pi ? ln(e) / e

and ln(x)/x is decreasing for x >= e

[leaden kraken]

note that $e^\pi = e^{\frac {\pi e}{e}}$ and $\pi^e = \pi^{\frac {\pi e}{e}}$. Consider the function that sends $x$ to $x^{\frac 1x}$

[rocky lotus]

soup_norm

[leaden kraken]

this is proven by differentiation, right?

[keen tundra]

yes

[leaden kraken]

ohh

similar to the one I know about then, but this one is a little bit more natural

[zenith lantern]

What do the question marks mean here

[keen tundra]

I don't know which one is bigger

so I put a ? for the relation between them

[zenith lantern]

I'm still not sure how to do the comparison

@keen tundra

[keen tundra]

You want to comapre e^pi and pi^e
you take ln on both of them and since ln is increasing the result of the comparison won't change

then ln(e^pi) = pi ln(e)
ln(pi^e) = e ln(pi)

[zenith lantern]

yeah

[keen tundra]

now we also divide by e * pi so we need to compare ln(pi)/pi and ln(e)/e

and ln(x)/x is a decreasing function

when x>=e

[zenith lantern]

Divide each of these by e*pi?

[keen tundra]

yeah

[zenith lantern]

So since ln(x)/x is decreasing when x >= e, and since pi > e, we would have ln(e)/e > ln(pi)/pi ?

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[rapid violet]

I'm working on this one. I have proven that 0 is a lower bound for the sequence [moreover, that the entire sequence is bounded by (0,1)], and that the sequence is monotone and strictly decreasing, so it should converge downwards. I am having trouble showing that it converges to 0 specifically though. I also haven't been able to find an explicit form, and it would be great to avoid that if I can while still showing convergence to 0.

[rapid violet]

Also, the sequence should eventually become irrational no matter the starting value, but I'm not sure how to prove that or use it

[worldly zinc]

Shouldn't it be enough to show that the sequence is strictly decreasing and the lower limit is 0?

[rapid violet]

No. 0 can be a lower bound but not the lower limit

like -1 is a lower bound of the function x^2

[ruby sleet]

if you show the sequence is decreasing, then monotone convergence theorem ->... Now use the fact that the limit of x_n (call it l) is also the limit of x_(n+1)....

[rapid violet]

hmm

[rapid violet]

ok, I'm going to try setting up a contradiction by supposing the limit a>0 and seeing what that does to the epsilon definitions for convergence of x_n and x_n+1

[ruby sleet]

well you can do that, but it's not necessary

just take this equality

and make n-> infinity...

[rapid violet]

I'm not sure I can use limits for this

[ruby sleet]

why not?

if x_n converges to l

what does the LHS converge to

what does the RHS converge to

[rapid violet]

well it's a sequence so I wasn't aware there was really a LHS and RHS for this convergence

[ruby sleet]

this is an equality between sequences

call (L_n) the sequence of left hand sides (L_n = x_(n+1))

(R_n) the sequence of right hand sides

they're both convergent sequences (why?)

and they're equal

so their limits are...

[rapid violet]

I suppose they would both be subsequences of the monotone and decreasing (x_n) which I don't have the explicit form for, so they should both converge to the same limit

[ruby sleet]

the left hand side is a subsequence of x_n yes

[rapid violet]

ok, I did the proof by contradiction, and I found that if both converge to an a>0, then 1-sqrt(1-x_n) < a < x_n, so that should be enough to show it converges to 0, since both need to converge downwards?

[ruby sleet]

but the RHS is f(x_n) where f is continuous

so it converges to f(l)

[ruby sleet]

how is 1-sqrt(1-x_n) < a

and for which n

[rapid violet]

wait it should be epsilon + that is less than a

[ruby sleet]

Makes less sense

[rapid violet]

well, I've got to go to class now. you are probably a little frustrated with me but thank you so much for helping. There are many new pathways for me to think about that I wasn't before

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[low flame]

hello everyone

ok so we are finding alpha

heres what i tried

i found the equation of the two lines which is

y=1/2 x +3
y= 3x-7

[woeful trench]

If I asked you to find the angle that the line y = 3x - 7 makes with the x axis, would you be able to do that?

[low flame]

sorry i only use gradient yes i realise

and then im thinking

i move the x axis up and y axis on B to FORM RIGHT ANGLE TRIANGLE

one second allow me to draw

i thought this was one way of doing it

[woeful trench]

That is one way you could do it, sure, there is a slightly better way though

If you give me one second as well

[low flame]

here is my second way

i have an idea but im not sure how to execute the working

[woeful trench]

Almost that, that's really close to what I wanted(!)

[low flame]

oh

[woeful trench]

But basically the orange and purple angles are "Really Easy™" to find here, those ones come with a tiny bit of work from the gradient

Then from that, alpha is easy to find too, from [purple] - [orange]

[low flame]

ok i see

is that parallel to x axis

the orange line

[woeful trench]

The yellow line is, yep (sorry, poor choice of colour there )

[low flame]

OK YES thank you charbit i see how this works

i make triangles on the paper

should i be doing this

tanx = 3

[woeful trench]

Yep that's it, same with the other one

[low flame]

thanks mate and thanks for letting me figure it out

have a good day

[woeful trench]

You too, have a wonderful one

[low flame]

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[devout skiff]

Hello, i had a pretty simple question on math syntax for a physics problem

[devout skiff]

so we haven't learned about limits yet, i have a physics test tomorrow on optics. I'll have to talk about how when an object is far away the image gets closer to the focal point and prove a certain identity from that

I wanted to know if it was coherent to write "\lim_{x_A \to \infty} \frac{1}{x_A} = 0"

\lim_{x_A \to \infty} \frac{1}{x_A} = 0

[distant latch]

In optics, particularly with lenses or mirrors, you’re likely dealing with the lens formula (for lenses) or the mirror formula (for mirrors), which relates object distance ($ u $), image distance ($ v $), and focal length ($ f $):
$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$
Your statement about an object being "far away" and the image getting "closer to the focal point" suggests you’re working with a situation where the object distance ($ u $) becomes very large. Let’s explore this physically and then tie it to the expression you provided.
Physical Intuition (Without Limits)
When an object is placed very far from a lens or mirror (think of a star or a distant tree), the rays from that object are nearly parallel when they reach the lens or mirror. For a converging lens or a concave mirror, these parallel rays focus at the focal point. This means the image distance ($ v $) approaches the focal length ($ f $) as the object distance ($ u $) gets very large.
Let’s use the lens formula to show this without formally using limits:
$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$
Rearrange to solve for the image distance $ v $:
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$$
$$\frac{1}{v} = \frac{u - f}{u f}$$
$$v = \frac{u f}{u - f}$$
Now, if the object is very far away, the object distance $ u $ becomes extremely large. When $ u $ is very large compared to $ f $, the denominator $ u - f \approx u $ (since $ f $ is typically small, like the focal length of a lens, e.g., 10 cm, while $ u $ might be meters or more). So:
$$v \approx \frac{u f}{u} = f$$
This shows that as the object distance $ u $ gets very large, the image distance $ v $ approaches the focal length $ f $. This is exactly what you described: the image gets closer to the focal point.

[rocky lotus]

LilyGranger2048

[distant latch]

In the lens formula, let’s identify $ x_A $ with the object distance $ u $. So, $ x_A = u $, and the term $ \frac{1}{x_A} = \frac{1}{u} $. Your expression suggests that as the object distance $ u $ becomes very large (approaching infinity), the term $ \frac{1}{u} $ becomes very small, effectively zero. Let’s see how this fits into the lens formula:
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$$
If $ \frac{1}{u} $ is very small (because $ u $ is very large), then:
$$\frac{1}{v} \approx \frac{1}{f}$$
$$v \approx f$$
This matches our earlier conclusion: when the object is far away ($ u $ is large), $ \frac{1}{u} $ is negligible, so the image distance $ v $ is approximately equal to the focal length $ f $. Thus, the idea behind $ \lim_{x_A \to \infty} \frac{1}{x_A} = 0 $ (with $ x_A = u $) is directly applicable—it explains why the image forms at the focal point.
However, since you haven’t learned limits, using the limit notation might not be expected or necessary for your test. Instead, you can express this idea in words or with simple algebra:

In words: “When the object is very far from the lens, the object distance becomes so large that the term $ \frac{1}{u} $ is extremely small, almost zero. This makes the lens formula simplify to $ \frac{1}{v} \approx \frac{1}{f} $, so the image distance $ v $ is approximately equal to the focal length $ f $, meaning the image forms at the focal point.”
In algebra: Show the lens formula and explain that as $ u $ gets very large, $ \frac{1}{u} $ becomes negligible, leading to $ v \approx f $.

[rocky lotus]

LilyGranger2048

[distant latch]

sorry for long explanation hope it helps

[devout skiff]

Yes absolutely, it helps

[devout skiff]

however i just need to know if this is syntaxically correct

like if i can write this and not lose points or something

cuz 1 + 1 = 2 is understandable, you're not gonna write 1 + + 1 = 2

thansk for the explanation btw, i read it

i'm assuming it is a proper way to write it then

[distant latch]

Yes, the expression $ \lim_{x_A \to \infty} \frac{1}{x_A} = 0 $ is syntactically correct in mathematical terms. It follows standard limit notation:

$ \lim $ indicates a limit.
$ x_A \to \infty $ means the variable $ x_A $ approaches infinity.
$ \frac{1}{x_A} $ is the function being evaluated.
$ = 0 $ states the result.

yep

it is

[devout skiff]

okay thank you very much 🙏

have a blessed day

[distant latch]

same to u best of luck for yr test

[fleet burrow]

this was a 🔥 read

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[devout skiff]

w read indeed, genuinely

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[unkempt bane]

Hello! ^.^ I am stuck here... and do not know how to continue... 🤔

[unkempt bane]

Are you able to help? 🙈

[ivory dragon]

im stuck on this question could i please get help solving this

[sage glacier]

@ivory dragon go to a channel that hasn't been claimed yet

[unkempt bane]

oh we must have claimed it almost same moment

[vale dock]

@unkempt bane Has your question been resolved?

[woeful trench]

Just to try and take a bit from what you've done, you got up to the point where you had the two equations 2x - 5y = 1 and 3x - 2y = 4? (and then did some extra work, but imma ignore that for a second )

[unkempt bane]

👍

[woeful trench]

Cool, are you happy in general with solving linear simultaneous equations?

[unkempt bane]

eh, depends. I dont usually have issues but this one is first for me. (first issue)

[woeful trench]

Well, you know there's substitution (where you rearrange one of the equations to make one variable the subject, and put that into the other equation),

and there's elimination (where e.g. you multiply the equations such that you have one of the variables having the same coefficient, at which point you can happily subtract/add the equations to eliminate that variable) right? (if those don't sound familiar, let me know )

[unkempt bane]

the first one sounds familiar, not the second 👀

elimination where you multiply equations does not sound familiar.

[woeful trench]

Alright, do you wanna do the first method then? If anything I can run by how the second one works after we're done, it might be familiar, or it might be something you haven't done, in which case, it's something to at haha

[unkempt bane]

Will try! ^.^ thank you, I d love to hear the second one too please (:

I ll come back as soon as I try the first one!

[cedar moss]

just solve it?

[woeful trench]

Of course, ping me when you're done, or if you get stuck

[woeful trench]

And how is that helpful when OP's already discussing how to do it

[vale dock]

@unkempt bane Has your question been resolved?

[unkempt bane]

I am legally suffering

xD

I am trying the first method.... but I feel like I am overthinking so much I make more mistakes and make things worse xD

possibly the second method is easier?

[unkempt bane]

I dont think its even worth seeing the pic xD 🙈

[woeful trench]

No no, you've done quite decently so far, for someone who tried to do it alone
A suggestion that makes things a bit easier is that from here, you could have multiplied everything by 5, and it would be really similar to what you've done in places, then it's just a matter of a bit more rearranging

Doing so should get you 15x - 4x + 2 = 20, then you can of course move that to one side and all

[unkempt bane]

thats so true.....

amma try it away

[woeful trench]

Perfect, ping me when you've made some progress, or if you need me again

[unkempt bane]

aye aye!

[unkempt bane]

I got some scary looking numbers there... 🙈

[woeful trench]

They do be scary looking, but that doesn't automatically mean they're wrong just give me a moment

[woeful trench]

Oh wait a moment, I didn't notice something earlier
I think the equations we should have should be 3x - 2y = 4, and instead 2x - 5y = -1 rather (the exponent got us to 1 + 2x = 5y, and of course we also could have rearranged that to have made y the subject from there too, which would have given us y = (1 + 2x)/5, which we could work with too )

[unkempt bane]

OH

[woeful trench]

Yea that will give us some more "normal" numbers

[unkempt bane]

https://tenor.com/view/daeth-funi-periflight-dies-from-gif-22212023

[woeful trench]

If only I had spotted that sooner, soorrrryyyyy

[unkempt bane]

NOOOOOO

not ur fault at all cmooooon

xDDDD

[heady imp]

Yh

[unkempt bane]

I am ashamed x'D 3h on this.. my second day today xD

[heady imp]

Trust me it's worth ut

It*

[unkempt bane]

I shall put this to use x'D

[woeful trench]

Awwwww sometimes it's the small things, it took me so long for me to get simultaneous equations when I was doing them, then when I understood them, I was like "really, I struggled with that!? "

That's growth for you

[unkempt bane]

https://tenor.com/view/feels-right-in-feels-gif-13802802

thank you so much ❤️

[heady imp]

Hahaha

[woeful trench]

Anyways, we can do it once we take y = (1 + 2x)/5 and put that into 3x - 2y = 4, we just do a lot of the same work, and we are happy

A much more natural number, if you'll excuse the pun

[unkempt bane]

x=2 ... is this real? x'D

[rocky kettle]

hi

im here again haha

[unkempt bane]

hello again! ^.^

[rocky kettle]

is the question pinned or is it another one

[unkempt bane]

its pinned (:

there is a mistake tho 2x-5y=-1

[woeful trench]

Yep that's it

Of course you can then find y

[unkempt bane]

https://tenor.com/view/otter-shock-otters-wow-wtf-gif-24606115

y=1

thank.you.so.much

[woeful trench]

Let's see if we were right

[rocky kettle]

oh i get here and its solved everything is good xD

[rocky kettle]

you were

[woeful trench]

,w 125^x/25^y = 625, 2(4^x) = 2^y

What shush Wolfie

[unkempt bane]

😳

[woeful trench]

I thought so

Did you still want to hear the other method btw?

[unkempt bane]

yes if possible ^.^

[woeful trench]

Ya sure

So if I take the equations
\begin{gather}
2x - 5y = -1 \
3x - 2y = 4
\end{gather}
we can, for example, multiply the first equation by 3, and the second equation by 2, and that would give us
\begin{gather}
6x - 15y = -3 \
6x - 4y = 8
\end{gather}
at which point, we could e.g. subtract the third equation from the fourth, that would give us
[
11y = 11
]
and from there we get $y = 1$, as we found, then we can of course find $x$ by putting that into one of the original equations \catthink

[rocky lotus]

@woeful trench

[woeful trench]

You could instead have multiplied the first equation (1) by 2, and the second (2) by 5, then y would have the same coefficient and you could cancel that out instead too, but that's another way you could go about it

It's only really useful for linear equations, but it might be a bit less work, but of course if you disagree, you can ignore it and just pretend you're just chilling to see it

[unkempt bane]

I ll defo use it!!!

looks far more simple in my eyes. defo worth knowing

How do you decide on which numbers you d multiply those by?

is it because 2x and 3x have those numbers?

[rocky kettle]

any of those

either x or y

[woeful trench]

Yea the idea is that you want to have one of the variables, in the case I did the x, to have the same coefficient, so I got them to both have the common coefficient 6 (the signs can differ of course, but instead of subtracting you'd add to eliminate)

In some cases you may only need to multiply one equation, or might not need to multiply any equations, but whatever means you can add or subtract and one of the variables cancel out like that

[unkempt bane]

I really feel like answers would differ if I did multiply only 1 D':

[unkempt bane]

oooooooooooooooooooooooooooh this is awesome

thank you very much, just tried it :3

to keep practice I ll keep trying in next tasks! ^.^

[woeful trench]

In this case they would, it's equation dependent, I would try and make up an example, but I'm terrible with doing those

[woeful trench]

Awwww, I'm happy

[woeful trench]

Yep of course! Bear in mind that I've kind of tl;dr'ed how you'd do it, that's like a short example, there are some people who can explain it better than I can, but of course if you ever want more help, you're welcome to come here and have someone explain it more for you, or link you to a video that does a better job

[unkempt bane]

It was very simple to understand! ^.^ If I struggle I ll check some vids. but for now I defo got the gist of it! ❤️

If I hella struggle u ll be seeing me again haha xD (Roy knows)

thanky you 🙏 I shall close this now then! ^.^

.close

[vale dock]

Channel closed

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[rapid violet]

the closest I have been able to get to it is (x_n + y_n)^2/4 > 3x_n*y_n/4

[stark wedge]

eh hold up tho.

(a) is literally just AM-GM innit?

[rapid violet]

AM-GM?

[stark wedge]

arithmetic mean ≥ geometric mean inequality

[rapid violet]

yes

[stark wedge]

is that... not kosher somehow?

barred from usage?

[hardy coral]

Maybe the trick is that x and y must be different, but that's pretty easy to see

[rapid violet]

well... yeah, the point is to prove the inequality for two different numbers

[stark wedge]

i will say this, intentionally informal but formalizable:

if an equality ever occurs, then it propagates backwards along both sequences

but it's false for n=1 so it couldn't have happened ever

[hardy coral]

Yea

@rapid violet any doubt about this?

[rapid violet]

so, this would be like a reverse-induction proof? assume the geometric and arithmetic means are equal, then use that to show it would imply the starting values are equal?

[hardy coral]

Sure

[light saddle]

you can do it by "regular induction" though

[hardy coral]

Both ways are pretty easy

[rapid violet]

awesome! got it done! Thank you!

[rocky kettle]

!done

[vale dock]

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

@rapid violet Has your question been resolved?

[vale dock]

Channel closed

Closed due to timeout

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Send your question here to claim the channel.

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

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

[wintry cape]

Looking at number 14
So my teacher explained like a lot of different factoring methods and I struggling to stay on track
Can someone walk me through these problems?

[wintry cape]

So I’m looking over my notes that’s a 4 term polynomial right so I would spilt it into 2 groups?

[red tulip]

are you supposed to find the roots

[marble arch]

What are you supposed to do here?

Factor it right?

[red tulip]

you can notice that the x^3 coefficient is 4 times the x^5 coefficient

[wintry cape]

Yeah

[red tulip]

and same with the -32 and 8x^2

[wintry cape]

Factor completely

[marble arch]

Well look for similarities, like he said

See how taking something of 2 pairs of them common would give a similar pair on both of them

Like x² - 3x + 2, here if you break it into x² - 2x - x + 2

And take x common from first two and -1 common from second two, you get x(x-2)-1(x-2)

See how x-2 is common in both of them, you have to look for something similar there too

Like see how x⁵ and x³ are exactly two powers different, and so are x² and -32

[wintry cape]

I’m looking at these steps cus I’m pretty sure it’s a 4 term polynomial

[marble arch]

Yes yes that

That is correct

[wintry cape]

Am I done

I get confused with the difference of two squares and sum/diff of squares etc etc

But can I continue factoring or

[marble arch]

No no, this is fine

You could potentially break (x² - 4) into (x + 2)(x - 2) though

[wintry cape]

I think that’s the final answer because I could continue factoring the (x^2-4)

[wintry cape]

Yep

[marble arch]

Yeah that's fine

Also, (x³ + 8) could be broken into (a³ + b³)

[wintry cape]

Ok well that was easier then I thought but now this one 😭

[marble arch]

IDK how much strict they are other there

[marble arch]

Well, take the only x ones one side and the ones with the y on the other

[wintry cape]

Idk they want it fully factored

[marble arch]

Ahhhh, so you might have to do that, especially if they've taught the formula of a³ + b³

[wintry cape]

So I would assume i would keep on going till there’s nothing else I can do

[wintry cape]

I was probably shown this but I’m a bit confused

Bc it seems unfamiliar

[marble arch]

Well you can keep it like this and ask your teacher if you're supposed to that

[wintry cape]

Alright

[wintry cape]

Number 16 looks a lot more difficult

[marble arch]

Well send it

You got 15?

[wintry cape]

I mean 15*

Number 15

Can I combine like terms?

[marble arch]

Yeah well take the ones with only x on one pair, and the ones which have y on the other

[wintry cape]

Is it another 4 term polynomial?

[marble arch]

Well, you have to factor them out

So combine the ones with only x and the ones with y in them

So x⁸ + 27x² - x⁶y² - 27y²

[wintry cape]

Ah ok

[marble arch]

Now see how there's a difference of x⁶ in both (x⁸ + 27x²) and -(x⁶y² + 27y²)

[marble arch]

Yes that's correct

Now take (x⁶ + 27) common from both of them

[wintry cape]

Now what can I do can I do a sum of cubes with the x^6 +27

?

[marble arch]

Yes that's correct, now you could break (x² - y²) into (x + y)(x - y), and (x⁶ + 27) into (a³ + b³), but confirm with your teacher first

[wintry cape]

is that what you mean?

[marble arch]

Ahhh well if you have them then you probably should, also in the previous question

[marble arch]

Yes

[wintry cape]

Alright I will

Looking back at number 14 for a second
Does x^3 +8 turn into x^3 + 2^3

[marble arch]

Yes, it's x³ + 2³

[wintry cape]

But js the a and b are x and 2

Right

[marble arch]

Yes

[marble arch]

Wait a is 2

8 is a³

So a² would be 4, and -ab would be -2x

[wintry cape]

A is x
B is 2

No?

[marble arch]

Well okay, yes

So it's (a + b)(a² + b² - ab)

Now b² would be 4, not 64

[wintry cape]

Oh okay

[marble arch]

And -ab would be -2x

[marble arch]

Also correct the -8x

[wintry cape]

All good?

[marble arch]

Yes now it's fine, you could also combine the (x + 2)s to a whole square

[wintry cape]

Like that

?

[marble arch]

Yes, no more simplication is valid now

It's finished

[wintry cape]

Alright thank you for being so helpful so far

Going back here

X^6 +27 is also a sum of cubes

[marble arch]

Yeah here take x² = a and 3 = b

[wintry cape]

How does that looo

Look

[marble arch]

Yes that's correct

[wintry cape]

Amazing okay

17- my first thought is to deal with the ^ but i don’t think so

Especially since I’m factoring

[marble arch]

Well they've basically laid it out for you

[wintry cape]

Oh wait

[marble arch]

All you have to do is take common from both of them

[wintry cape]

😭

What about the ^3 and ^2

[marble arch]

Well include them, (2x + 1)² is common in both of them, and so is x⁵

So take both of them common

[wintry cape]

Like that

Or should I ^5 both numbers from the start

[marble arch]

Well, you could take x⁵ out

Or wait

It should be (2x + 1)², because that's what common in both of them

And now take x⁵ out

[wintry cape]

Wdym

[marble arch]

Well if I gave you, let's say (x⁷ + x⁵) and ask you to simplify it, what would you do?

[wintry cape]

x^4(x^3 + x) ??

[marble arch]

You can take one more x out

How about x⁵(x² + 1)?

[wintry cape]

Yeah

[marble arch]

Okay, so do the same with x⁶ - 6x⁵

Take x⁵ common

[marble arch]

Yes that's correct

[wintry cape]

Can’t do anything else?

[marble arch]

Whoops, forgot the whole square at (2x + 1)

[marble arch]

Yes, as far as I can see

[wintry cape]

Oops

Okay

The next one looks similar let me try to start myself

[marble arch]

Alright mate

[marble arch]

Yes, just look to take common whatever you can

[marble arch]

Ahhh, you missed (x - 5), you've taken (x - 5)³ common and there was one (x - 5)⁴, so it'd still have one (x - 5) left behind

[wintry cape]

So ^4 not ^3

[marble arch]

No no, see if you take x³ common from (x⁴ - x³), you'd be left with (x - 1), so there'd be one x still inside

Similarly, there'd be one (x - 5) still inside

[wintry cape]

So (x^2 - 5)^3

[marble arch]

No, okay see, try to expand it back

You get 2x(x - 5)³ - 4x²(x - 5)³

But it was 2x(x - 5)⁴

Alright I'll tell you this time

It's supposed to be (x - 5)³(2x(x - 5) - 4x²)

Now if you expand it, it becomes the question

[wintry cape]

I’m lost give me a second to try to understand

[marble arch]

Alright, let me know if you need help in understanding it

[wintry cape]

The question started at ^4

[marble arch]

Yes

[wintry cape]

Not ^3

[marble arch]

Yes, but we'd get 3 if we expand your answer

[wintry cape]

How did we just lose the ^4 and go to a ^3 not changing anything else in that part

[marble arch]

Okay do you know how to expand

Like if I give you x²(x³ + 1) and ask you to expand it, what would you give

[wintry cape]

X^5 +x^2

[marble arch]

Yes

[marble arch]

Okay so, expand your answer, what do you get?

[marble arch]

Omg no

[wintry cape]

😭 im sorry I sorta knew that’s wrong

But wasn’t sure what u really wanted me to do

[marble arch]

Okay see

[marble arch]

Like I wanted you to expand x²(x³ + 1), now I want you to expand this

Basically send 2x back in (1 - 2x), and multiply both terms inside with (x - 5)³

Tell me what you get

[wintry cape]

So start at this?

[marble arch]

Okay that's nice, you got the first step

Now open the bracket, multiply the terms inside with (x - 5)³, don't break them down or anything, just show that they multiply, like they've done in the question

[wintry cape]

Like that?

[marble arch]

No no, keep (x - 5)³ as it, just multiply it to (2x - 4x²)

Like 2x(x - 5)³ - 4x²(x - 5)³ looks right to you?

[wintry cape]

Ah ok

[wintry cape]

I don’t think so

[marble arch]

Well, take (x - 5)³ common

What do you get

[wintry cape]

So foil right

[marble arch]

I didn't understand, what?

[marble arch]

Ahhhh unfortunately no

[wintry cape]

Had a feeling so what did you mean

[marble arch]

Okay, take (x - 5)³ common here

[wintry cape]

Well this is what the question is right

Oh wait you made both ^3

[marble arch]

Nope, it's (x - 5)⁴ in the first one

[marble arch]

Yeah, that's what your answer gives if you expand it back

[wintry cape]

Ok I see

How does that help factor

I understand how like ^3 ^3 will get you back to the original answer

[marble arch]

Okay I'll ask you a question

If I had x⁴ - x³, how would you simplify it?

[wintry cape]

X^3 (x-1)

[marble arch]

Okay

Now if I had (x - 5)⁴ - (x - 5)³, then would there be any difference in the format?

[wintry cape]

Unsure

[marble arch]

Alright, let x - 5 be y

Now it's y⁴ - y³, how would you simplify it

[wintry cape]

Y^3(y-1)

[marble arch]

Alright, now substitute (x - 5) back to y

[wintry cape]

X-5^3(x-5 -1)

[marble arch]

Bravo

[marble arch]

Also it's (x - 5)³, just confirming

So see, there's an (x - 5) still inside

Then only would it be (x - 5)⁴ - (x - 5)³

[wintry cape]

X-5 is left still

[marble arch]

Yes, inside the bracket, (x - 5) would once be left inside