help-17

[compact eagle]

how do we know/figure out that epsilon/6 is equal or bigger than delta

[cobalt crypt]

we don't

[compact eagle]

what if epsilon/6 was smaller

[cobalt crypt]

we're choosing delta

delta isn't a fixed value

we can just pick some delta <= epsilon/6

its never a problem

[compact eagle]

so we just WANT delta to be equal or less than epsilon/6

cuz we want a smaller delta?

[cobalt crypt]

yes

[compact eagle]

we want the delta to be as small as possible

[cobalt crypt]

not really

just small enough

[compact eagle]

small enough to...

[cobalt crypt]

meet your enemy's challenge

[compact eagle]

cuz our enemy wont like it if we have a big delta

[cobalt crypt]

yeah

[compact eagle]

cuz its cheaty

[cobalt crypt]

if you say that |f(x) - L| < epsilon when |x-a| < 0.1 for example

then it will still work when |x-a| < 0.01

because you're even closer to a

so if you find one delta that works

then any smaller delta will also work

[compact eagle]

so lemmeget this straight

rn we know that |x+1|<epsilon/6

this is a fact

we also know that |x+1|<delta

[cobalt crypt]

no that was a goal

[compact eagle]

oh

but |x+1|<delta is definitely a fact

[cobalt crypt]

yes

[compact eagle]

what does epsilon/6 really tell us then

[cobalt crypt]

the original goal is |(x+1)(x-3)| < epsilon

we converted that to 6|x+1| < epsilon by making delta <= 2

now this is the same as asking for |x+1| < epsilon/6

so our goal is |x+1| < epsilon/6

[compact eagle]

so epsilon/6 is really only assuming that delta is equal or smaller than 2

[cobalt crypt]

yes

we need that condition

[compact eagle]

so what can we do with this assumption

rn assuming delta <=2, |x+1|< e/6

what is our goal

rn

[cobalt crypt]

|x+1| < epsilon/6

[compact eagle]

so our goal is to prove that true?

[cobalt crypt]

yes

[compact eagle]

how can we prove that true

[cobalt crypt]

known: delta <= 2
|x+1| < delta
|x+1| < 2
|x-3| < 6

goal: |x+1| < epsilon/6

we select delta <= epsilon/6

[compact eagle]

well how do we know delta is <= 2

we just said that it is

[cobalt crypt]

yes

so thats known

its our choice

[compact eagle]

okay

[cobalt crypt]

delta is in our control, its perfectly fine to declare that

[compact eagle]

but say this

say epsilon is freakishly massive

like 10000

how would delta only be less than or equal to 2 in our function

[cobalt crypt]

we just pick it to be

theres no requirement on delta other than for |(x+1)(x-3)| < epsilon

so delta can be <= 2

[compact eagle]

but like

on a graph how can that be true?

visualized

with such a massive epsilon wouldnt the delta also be massive

[cobalt crypt]

no

you just choose values of x within 2 of -1

,w graph (x+1)(x-3) for -3 < x < 1

[twin meteor]

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[cobalt crypt]

these values are all within 10000 distance of 0

if your enemy gives you a large value of epsilon, you should be happy

because some tiny value of delta will probably just work

[compact eagle]

so delta is, in our terms, 2. from a which is -1, so we can pick any x value between -3 and 1

[cobalt crypt]

yes

[compact eagle]

the epsilon is the distance from the limit of the function at -1

[cobalt crypt]

yes

so if you want your function to be < 10000 distance away from the limit

well

thats not a very strict requirement is it

10000 is huge

just stay close to -1, and you're g

[compact eagle]

ok right so actually a massive epsilon is a good thing for a delta cuz that means it covers a larger amount of the functioin

the line of the graph

if it was a tiny epsilon

that means a very small portion or even no part of the function is captured?

or actually

u cant have no part of the function

[cobalt crypt]

well if you're trying to do a limit proof, you need to be able to produce a delta

otherwise you won't be able to complete the proof

[compact eagle]

okay

lets continue

we want to prove |x+1|<e/6

[cobalt crypt]

so select delta <= epsilon/6

doesn't matter which

[compact eagle]

and why cant delta be larger than e/6

is it cuz our enemy is like

fuck you

thats not dfair

[cobalt crypt]

yes

if delta > epsilon/6, we can't guarantee that we're safe from our enemy

[compact eagle]

but if the delta is massive, wont that mean more space is covered

on the graph

[cobalt crypt]

|x+1||x-3| could exceed epsilon if we're not careful

that would be bad

because you don't want to stray too far

epsilon is the limit to how far you can move around

[compact eagle]

oh we want to stay in the confines of the epsilon space

[cobalt crypt]

yes

if you move outside of that your enemy zaps you

[compact eagle]

ok

so knowing |x+1|<e/6, and delta <= e/6,

[cobalt crypt]

delta is the number such that |x+1| < delta

so |x+1| < delta <= epsilon/6

which achieves the goal

[compact eagle]

|x+1|<delta<=e/6

right

so delta has gotta be bigger than |x+1| whilst also being smaller or equal to e/6

[cobalt crypt]

i mean

thats kinda the wrong way to think about it

|x+1| has to be smaller than delta

delta is a constant that we pick

x is the thing thats varying

[compact eagle]

alr

ok we

gotta hurry

i got 3 homeworks due

fuck

alright shit

whats next

[cobalt crypt]

we're done basically

to summarise

recall that we needed delta <= 2

and also delta <= epsilon/6

so one such choice of delta that satisfies these requirements is delta = min{2, epsilon/6}

the lesser of the two numbers

[compact eagle]

delta works for both <=2 in satisfying the proof as well as e/6

and delta given whatever the epsilon value truly is is going to be at most 2

[cobalt crypt]

yes

[compact eagle]

or at most e/6

is that right

[cobalt crypt]

yes

exactly

this is what your thing did

but using delta <= 1 instead of delta <= 2

[compact eagle]

so there can be multiple

answers?

[cobalt crypt]

yes

like i said earlier

you just need a delta that works

[compact eagle]

so picking a smaller number generally = better?

[cobalt crypt]

theres no single "answer" for delta

not really

1 is just an easy number to work with

but as we just saw, 2 also works just as well

[burnt solstice]

what is the definition of delta

[craggy crypt]

wow, I can't believe you're pre-university and learning straight up analysis

that's insane

[compact eagle]

what is analysis

is this what analysis is

im in calculus 1

idk if this is like

super easy or what

[craggy crypt]

well, it's a topic taught in analysis courses usually

[compact eagle]

for calculus people

im 1 chapteri nto calculus

it hasnt even been a full month

since we started

[craggy crypt]

it's a field of math, and its introductory principles are often taught in analysis/real analysis courses

and I'm surprised to see you being made to learn it in calc 1

because it's a hard topic

[compact eagle]

so to summarize this whole process we wanna prove that |fx-l|<epsilon and |x-a|<delta, i assume that delta is going to be 2 so i find the boundaries of |x-a| given that delta value and use the values given by thiose boundaries to find the maximum value possible for |f(x)-l|<epsilon to be true given my delta. knowing the maximum possible value given delta=2, that maximum possible value must still be less than epsilon for my proof to be true. i used algebra to find the epsilon value that confirms the proof so now i know that our delta is either going to be at most 2 or at moste/6

ok im glad people see this as a hard topic

cause

im ngl

im about to break my monitor

[craggy crypt]

but also I hate how they don't seem to use quantifiers

quantifiers make the definition of a limit way easier to understand

[compact eagle]

especially becausr the textbook did NOT

explain this at all

it literally just threw it at me

the only 'explanation' given was a definition or two and the examples

[craggy crypt]

lmaoooo

[compact eagle]

with shitty expkanations

[cobalt crypt]

sometimes in life, you have to develop your own intuition

a textbook cant handhold you through everything

[craggy crypt]

to be fair, most analysis courses go into a lot more and deeper with limits and continuity, but still, it's impressive to see it being in your calc 1 course

[craggy crypt]

ah, let me see if I can find a good explanation

[burnt solstice]

can somebody ans me

[bitter pilot]

ans

[craggy crypt]

I really like this definition from taylor's book on 1v analysis

specifically, 3.1.2

it's a little too general

but take $d_{something}(x, y)$ to mean $|x-y|$

if it's continuous at a point x, the limit exists at x

[craggy crypt]

(not exactly exactly true, but pedants, pls allow me to be a little lax for pedagogical purposes...)

[twin meteor]

00100000

[vocal sleet]

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[digital aurora]

??

[digital aurora]

Happy Holi everyone 🎉🥳

[daring bramble]

Can I ask questions regarding Statics?

[jagged cargo]

idk what holi is, but happy holi to you too

do you have a question?

[digital aurora]

Its the festival of colours , representing victory of good over evil

[daring bramble]

How does one treat this UDL extending on angle member?

[jagged cargo]

!occupied

[vocal sleet]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[daring bramble]

oh sorry, Im on the wrong chat

[jagged cargo]

i see. but this is not the place for casual conversations, if you don't have questions we can help with, then please close this channel

[digital aurora]

I actually have a question, just give me a minute

[vocal sleet]

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[tough falcon]

hi im confused with solving this graphically for acd. when looking for solution(s), i know that it is where two functions intersect. B worked but acd gives me ---=0. does this mean i could just use the x-int as my solutions, as solutions sometimes mean the x-int?

[dull bear]

How do you mean "acd gives me ---=0"? Can you show what you plotted?

[tough falcon]

oh questions A, C, D

so like as an example

when algebraically solved question A, it was x=11 in the end

[paper depot]

that sounds kinda sus

11 is not a solution of equation a). you messed up for sure. show your work

[tough falcon]

3x+5x(x-2)=x+4

3x+5x^2-10x=x+4

??

oh i messed up the next line

i think

idk why i got 3x^2

[paper depot]

so far so good based on what you've typed

[paper depot]

ah.

[tough falcon]

ok ill cont

[paper depot]

still don't know what ---=0 is supposed to mean tho

[tough falcon]

3x+5x^2-10x=x+4

[tough falcon]

--- was just any random equation

??why did i do smth

[dull bear]

(If you're doing it graphically, then as you mentioned, are you not looking for where the two functions intersect? Hence why I asked for what you graphed, so we could see what you did)

e.g. for a), you want to plot y = 3x/(x - 2) + 5x and y = (x + 4) / (x - 2) and see where those two intersect

[tough falcon]

yeah but i think im allowed to rearrange the equation a lil bit algebraically and then draw two functions

yeah

[paper depot]

i mean if you rearrange it into the form f(x)=0 then yeah it becomes looking for the x ints

instead of intersections between two curves

[tough falcon]

ok so if it becomes f(x)=0, look for the x int instead of the intersection between two functions, as there is only one function.

[tough falcon]

wait here

i got (5x+2)(x-2)=0

when solved algebraically, i state x=2 or -2/5

ignore verifying for now)

but when graphing, am i allowed to move one of the expressions to the other side of the = sign

so 5x+2=-x+2, graphing y= 5x+2 and y=x+2 and look for their intersections

[dull bear]

You could either plot the two separately and see where they meet, or rearrange and plot one function and find the points it intercepts the x axis

(in some cases, one may be easier than the other)

[tough falcon]

yeah

oh wait

i cant do what i just mentioned since that is a property of multiplication right-

or could i

[dull bear]

Not in the way I suspect you're doing, no

Hence my careful wording

[tough falcon]

ok...im confused w graphing...

so for c, in teh end it goes like x^2+2x-15=0

(x+3)(x-5)=0

x=-3 or 5

[dull bear]

(I'm taking your word for that for now)

[tough falcon]

now idk what to do w graphingg

is makng to quadratic vertexform the only option

[dull bear]

Do you have to plot those by hand?

[tough falcon]

kind of

(x+1)^2-16

if not then what shoudl i be doing

[paper pelican]

you have to graph the original equation and find the point it meets

[paper pelican]

not this

[dull bear]

Well, you could, though that does seem a little bit "circular reasoning" to me, in that you find the solutions in the process of figuring how to plot, but then using the plot to find those solutions

[paper pelican]

btw have you learned extraneous solutions

[tough falcon]

its a reciprocal function

[tough falcon]

yesh

[paper pelican]

yeah it is

are you sure you have to graph it by hand?

[tough falcon]

yeah

[paper pelican]

what is it?

[tough falcon]

what is..?

[paper pelican]

What is an extraneous solution

Or look back at your algebraic answer bc your pretty close

[tough falcon]

extraneous root is smth u solved it algebraically but gets rejected as it is pt of the restriction

[paper pelican]

Yes when it returns an answer thats not valid when plugged into the original equation

So going back to your answer, do you have any extraneous solutions?

[hot pine]

So computationally dense

Exactly why I hate high school

Btw you have to do substitution back though

Graphically is not hard too for a just let (X+4)/(x-2)=1-6/(x-2) I think

Probably hard to draw in a way but doable

If you try to graph without some factorization it would be really hard, some of them especially right or left hand side can be simplified significantly if you do a little manipulation on the side I think

6x/(x-3)=(6x-18+18)/(x-3)=[6(x-3)+18]/(x-3)

Something like these

And you will always have some coordination to begin with

Btw I am not really good at calculation so go with my sol without check might lead you into trouble 🫣🫣🫣

[vocal sleet]

@tough falcon Has your question been resolved?

[tough falcon]

thanks so much but i dont think i am understanding what u wrote

[hot pine]

Why?

[tough falcon]

which is ma faute fs..

[hot pine]

?

It’s just basic algebra though

[tough falcon]

OH MY

nvmnvm

yeah i got the same stuff when algebraically solved

i just dont know how to graph so...

[hot pine]

Just factor

The method I mentioned

Once you simplify it can be drawn

[tough falcon]

so for D, before figuring out the solution, it says (3x+1)(x-1)=0

thats my factored answer

idk what to do w it though..

[hot pine]

You have to factor this way for RHS

X/(x-1)=1+(1/(x-1))

This is much better than with two X in a fraction right?

Of course if you just represent them in a root formula you can also write Lagrange interpolation which can help you draw somehow maybe

[tough falcon]

???im sorry im bad at reading fractions via computer

[hot pine]

Using computer? Then what’s the problem here

[tough falcon]

ik what an interpolation is but whats an lagrage interpolation

[hot pine]

Just graph and get the answer

[tough falcon]

i think the curriculum and the terms my school uses are quite different..

[hot pine]

If it’s just using computer to draw just get the answer

I don’t see any potential problem you can mess up thing though

It’s only a bit troubling when you do by hand

[tough falcon]

yeah im supposed to be drawing it w hand..

[hot pine]

Then just do what I said well if you want computer to help you then just use computer

[tough falcon]

ok...

[hot pine]

X/(x-1)=1+(1/(x-1))

This isn’t hard to understand right?

[tough falcon]

yeah

huh why we get that though..

[hot pine]

This isn’t that hard to draw right?

[tough falcon]

yaeh

[hot pine]

Why would you draw anything like X/(x-1)

This is harder to draw

And you can maybe factor out constant to cancel for other side

[tough falcon]

its just VA at 1 and HA at 1 and table of values

oh

?

sorry i just dont think my head is workig rn

[hot pine]

So 4-1 and then the left side is easier too

And simplify the LHS

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

Again you can further simply

Same trick

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

[tough falcon]

trying to process those fractions

lemme write down the fractions i cant read them 😦

[hot pine]

With our earlier manipulation

RHS is +1/(x-1) right since have have already factored out an one and moved it to left

This means 1/x(2+1/(x-1))+3=1/(x-1)

so many (X-1) annoying

How about substitution

Easy

Now right

[tough falcon]

mm

[hot pine]

I mean this is pretty much very easy to draw

It doesn’t matter you just love these term for grouping

Or just proceed with separate comparisons

It would work

If I may have done some algebraic mistake since I am lying on my bed so I didn’t use any paper or pen

Just used my phone

So do a double check to understand the concept

Actually all these questions you could use this trick

It’s always easy to play with constant plus a curve

Rather than two nonlinear equations

I am so tired can’t open my eyes and can’t even read properly i am gonna sleep please do some revision then you’ll be fine

[hot pine]

You really need to try though your inability for now to read fractions and also unable to sense these manipulation all pointing to lack of exercises, and I feel you can do much better since at least you’re listening

Math is different than other study, it has been a game in its nature, without playing the game you never get leveled up

[tough falcon]

ill try to work on it by writing them down

wait so what you did is:
X/(x-1)=1+(1/(x-1))
(2x-1)/x(x-1)=1/x(1+x/(x-1))+3
(2x-1)/x(x-1)=1/(x-1)
1/x(2+1/(x-1))+3=1/(x-1)

i copied pasted stuff u wrote but erm..

[hot pine]

Yes but do double check please

I am too sleepy to make my brain functional, but this trick is essential the way for reducing complexity

[tough falcon]

same im also still kinda confused w what the trick is but.. thanks

[hot pine]

Just simplification

Basically

[tough falcon]

wait did u find this strategy from like a video or...

i would like to revisit but im afriad that this chat will go up and i cant see it then so..

[hot pine]

Yes you often just assume to do it but I guess yes there might be video like factorization techniques basically search this up

Cross method, or simply try to complete the formulation by addition or subtraction to make it more favored in algebraic closed form

[tough falcon]

alralr

[hot pine]

It’s a simple concept that YOU DO NOT DRAW this directly it’s daunting

[tough falcon]

lol i think my tteacher wants us to so..

ill see how i do on my test tmr

[hot pine]

Yes they want you to draw it by reducing complexity first

Mathematics is about to reduce a complicated problem to simpler one

[tough falcon]

yeah but once i reduced it there's nothing left so

[hot pine]

These equations even professors wouldn’t draw them directly

[hot pine]

You only need to reduce it to a point where comparison or drawing is easy

[tough falcon]

yeah

[hot pine]

In fact in mathematics, you’ll learn Jacobian transformation, it’s to reduce complicated equations to simpler one

So you can simply proceed

[tough falcon]

...year

uh one last question

[hot pine]

Yes

[tough falcon]

when we're simplifying rational equations, we multiply the lcd to all terms

[hot pine]

That’s so mechanical forget it

Do it just exactly how it fits best

[tough falcon]

there, would the lcd and cancling out denominators become a POD

[hot pine]

Well if you prefer mechanical way it can be or you simply multiply the entire denominator on both sides

In some case you can use like conjugate too

[tough falcon]

yes but when we're cancling those denominators out, would those denominators be POD?

[hot pine]

POD?

[tough falcon]

point of discontinuity

hole

[hot pine]

That one is actually interesting

Have you learn how to do long division

[tough falcon]

not in school but ik what it is

[tough falcon]

so would that be a no..?

[hot pine]

If you want the most elegant way to figure out the discontinuity

Then you can do long division directly on a fraction

[tough falcon]

uh what

[hot pine]

Yes sometimes if you cancel things out you potentially can miss some discontinuity

Long division

Can preserve the discontinuity

[tough falcon]

so cancling out (simplifying) always have pod right?

[hot pine]

No, canceling can lead to loss of point of discontinuity

That’s why it’s actually an interesting question

[tough falcon]

??ok..

well

[hot pine]

But why you want discontinuity?

[tough falcon]

bc that is ont he test

anywasy

tysm for helping

[hot pine]

Btw you know how to do long division right?

On fraction

[tough falcon]

no haha...

[hot pine]

I’ll show you

Let me get up

[tough falcon]

ok can i get back to u in like 10 min if thats fine

im so soryr

ill be rly quick

[hot pine]

It’s fine I will use b as example

[tough falcon]

ok

[hot pine]

Here you go

How you do it

Haha even I forgot to use this method when I was writing the answer for you earlier

I just realized I made many mistakes earlier 🥲

[tough falcon]

uh

hm makes sense

ok well tysm

[hot pine]

Practice more, I mean you’ll actually know the most abstract concepts learned isn’t what you got from high school but elementary

Counting
Long division
Commutative
Distribution..
These are highly abstract

You’re welcome

[tough falcon]

screen shotted(?)

havea good onee

[vocal sleet]

@tough falcon Has your question been resolved?

[vocal sleet]

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[polar wren]

help me here

[vocal sleet]

Please don't occupy multiple help channels.

[manic flax]

where are you stuck, or do you not know how to approach the problem?

[vocal sleet]

@polar wren Has your question been resolved?

[polar wren]

dont know how to approach

[vestal heart]

evaluate alpha first

x/(x+1)-->1 for limx-->inf

||your second bracket then becomes 1/e-1||

[polar wren]

e/1-e * ( 1-e/e)^ infinity

[vestal heart]

||times together gives alpha=1||

[polar wren]

but second bracket is raised to power x?

[vestal heart]

no?

all is raised to power of x

this thing becomes 1/e - 1

=(1-e)/e

top and bottom cancels giving this is 1

raised to inf is still one

so alpha is 1

then your answer is actually 0

@polar wren do you get now

calculator answer if you dont believe me

bye

[polar wren]

thank u

.close

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[white mesa]

Prove that every positive integer $n$ has a unique factorial expansion, meaning there exist unique integers $k > 0$ and $f_i$ such that $0 \leq f_i \leq i$ for each $i = 0,1,\dots,k-1$, with $f_k > 0$, and

$n = 1! f_1 + 2! f_2 + \dots + k! f_k$

Given the factorial expansion $(f_1, f_2, f_3, \dots, f_k)$ of

$N = -8! + 16! - 24! + 32! - 40! + 48! - \dots + 2000! - 2008! + 2016!$,

determine the value of

$f_1 - f_2 + f_3 - f_4 + \dots + (-1)^{k-1} f_k.$

[twin meteor]

WD_Nessuno

[white mesa]

It should be about divisibility and factorisation but have no idea on how to prove this

[stone stream]

you first need to analyze the factorial expansion of N

[white mesa]

so

i can write n as

[stone stream]

Each factorial term represents a unique contribution to the expansion. The coefficients $f_1$ are uniquely determined by how each factorial contributes to the total sum.

[twin meteor]

donut

[white mesa]

mhmm

so like

[stone stream]

oh sorry it should be $f_i$

[twin meteor]

donut

[white mesa]

i have $n=f_1 \times m_1!+ r_1$

and same for $r_1$

[twin meteor]

WD_Nessuno

[stone stream]

hmm so when you observing the alternating pattern in sign, we see that each factorial contributes with a coefficient of 1(or -1, and depending on the sign.

[white mesa]

ok so every couple of numbers is $(n+8)! -n!$

[twin meteor]

WD_Nessuno

[stone stream]

that thing implies to $f_1 - f_2 + f_3 - ... + (-1)^{k-1} f_k$

[twin meteor]

donut

[stone stream]

and it's equal 1

so the answer is 1

[white mesa]

but it says to me it's 504

[stone stream]

because each coefficient contributes to a telescoping sum that simplifies to 1.

[stone stream]

hmm idk why?

[twin meteor]

WD_Nessuno

[white mesa]

i dont get why it should even be one

[stone stream]

here is the solution from chatgpt i just searched

[white mesa]

ChatGPT always get things wrong why even bother asking him

[white mesa]

increasing*

why would it even be 1

[stone stream]

idk you can lookin ur book already

[white mesa]

the first part of unique i got that

[white mesa]

theres the answer but how would that help me

[stone stream]

there isn't an explanation?

[white mesa]

nope

[stone stream]

😩

[vocal sleet]

@white mesa Has your question been resolved?

[vocal sleet]

@white mesa Has your question been resolved?

[manic flax]

💀 1 raised to infinity is actually an indeterminate formhttps://www.youtube.com/watch?v=Ea23eufsN8E

[grim oasis]

!nogpt

[vocal sleet]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[grim oasis]

why even help someone if youre not gonna bother to understand it yourself

@white mesa have you tried to put any simpler differences of factorials in factorial expansion form?

for example, what would 6! - 3! be in this factorial expansion?

[grim oasis]

once you know what this looks like youre just summing over them

[vocal sleet]

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[keen tide]

help

[keen tide]

i dont know how to get the answer for this

[runic frost]

hi guys

[keen tide]

this the answer in the ms but idk how this the answer

[runic frost]

im a complete beginner on math

[obtuse oxide]

What is your answer for the earlier parts?

[keen tide]

for which question?

because i believe this question can stand on its own

even tho its a sub question

[oak magnet]

Would be wierd if it start by hence

May we see?

[obtuse oxide]

This says "hence", so it assumes u have done stuff already you can use

[obtuse oxide]

Have you done part iii yet?

[keen tide]

ye

its just sub in

wait i just realised the answer for ii) is the same as iv)

but idk how that relates

@oak magnet @obtuse oxide

[obtuse oxide]

sorry im on a train rn had really bad signal for a bit

[keen tide]

ok

[obtuse oxide]

what is your answer for iii?

[keen tide]

4 for both

[obtuse oxide]

so if y is the same for both

you can substitute $y=|2x-5|$ i think

[twin meteor]

AMysteriousStranger

[obtuse oxide]

so u get $9|2x-5| = 80x - 16x^2$

[twin meteor]

AMysteriousStranger

[obtuse oxide]

um

idk what to do next tho

ig you can sub in your values of x from part ii?

[keen tide]

ig

still confusing tho

[obtuse oxide]

so at the points where y = 4, this is true, which you proved in iii

and because you have the values of x where they intersect

[keen tide]

so after that?

[obtuse oxide]

you can state when the line $y=|2x-5|$ (which is in red here) is less than the other line

i think

[twin meteor]

AMysteriousStranger

[keen tide]

i think i get it

[vocal sleet]

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[vital scroll]

Hiii could someone please tell me if I did this right and my method is correct?

[knotty zodiac]

yeah it is correct

[vital scroll]

Okay thank you

Also are there any things that need to be remembered while doing these sort of questions?

Like if I'm doing it in an exam are there tips or anything yall can give me?

[vocal sleet]

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[tepid hound]

@slim kettle

[tepid hound]

So like this ?

[slim kettle]

Yes sir

[tepid hound]

c = -a

b+d=a²

a(d-b) = -24

bd = 55

I writed;

b+d = a²

d-b = 24/a

Did you mean that?

[slim kettle]

Yes

Then add em

[tepid hound]

Ok now 2d = a²+24/a

So d = 1/2 ( a² + 24/a)

Now ?

[slim kettle]

Well u could substitute it for d in second and fourth eqn

Sorry for the lengthy method i cant think of anything else

[tepid hound]

Ok

Wait are you sure its correct?

Because a² = b + d

@slim kettle

Maybe bd = 55

I can the Couples (11,5) (5,11) (-11,-5) and (-5,-11)?

<@&286206848099549185>

[slim kettle]

Yeah

Hello

@tepid hound

Now you can eliminate the third and fourth pair because of a^2 = b+d

Because a is real therefore b+d should be non negative

then you can get value of a=+-4

You can assume a case where d>b or b>d

Then put values accordingly

Then you can arrive at a solution

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[naive mirage]

edexcel further pure one a level maths, vectors

it's just c i want to understand how to do

thanks guys

<@&286206848099549185>

[drifting whale]

For a you should find the normal vector using the points A B C

Do you know the equation for a plane?

[naive mirage]

r.n = d

also hi

i know how to do a and b

pretty sure

for a you do b - a and c - a and cross product for the normal vector to the plane and then sub in one of the points to get d

[drifting whale]

For CDE im fairly certain you need to use the properties of the dot product

[naive mirage]

and for two its just the formula for volume of a tetrahedron

[drifting whale]

Yea

[naive mirage]

what are the properties of the dot product

[drifting whale]

So basically A*B = ||A||||B||Cos(theta)

[naive mirage]

yeah a.b = moda mod b cos theta

[drifting whale]

Length of a time length of b times cosine theta

Yea

So you can rearrange that and take inverse cosine

But remember we are looking for the smaller angle

[naive mirage]

i know that but its just what to put into that formula

[drifting whale]

So depending on the dot product we can see if it is obtuse or not

You have to solve for vectors

Using the points

[naive mirage]

how

[drifting whale]

It goes from C to D to E

Try drawing it out on paper

[naive mirage]

how can i draw out a 3d coordinate system on paper

[drifting whale]

Bro just draw it

[naive mirage]

btw thanks for helping

[drifting whale]

Not the 3d

But draw it

[naive mirage]

what do you mean just draw it hahaha

[drifting whale]

Just draw an angle

And label the points

[naive mirage]

an angle has been drawn

[drifting whale]

I am not at my desk rn so I cant draw it

Ok do you see how you have two vectors

Use that for the dot product

[naive mirage]

ok

[drifting whale]

Okay

Do you get it now

[naive mirage]

uhhhhh

kinda

its ok thank you though

i have to go now anyway

byeeeee

thanks for the helo

broski

[drifting whale]

You’re welcome

[oak magnet]

<@&268886789983436800>

Stealth one

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@naive mirage Has your question been resolved?

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[merry prawn]

hi

I have this problem and its solution. But I am struggling in finding the valid n
I'll write my reasoning...

[merry prawn]

i see F is not defined for n<0
the partials i get to compute the curl are the same and = nxyr^(n-2)
this means the partials are not defined when n<2
So i have 2 cases:

• n<2 F is not well defined
• n>=2 F is well defined and the region is simply connected

I think this is different than what the solution is showing.

i think this part is wrong.

[vestal heart]

Its not. Lim x goes to inf. And 1 to inf is 1.

[vocal sleet]

@merry prawn Has your question been resolved?

[merry prawn]

help

[vocal sleet]

@merry prawn Has your question been resolved?

[obsidian stream]

I have no idea what part of Case 1 you think is wrong

[merry prawn]

n>=1 is not continuously differentiable

cause the partials are not defined for n<2

[obsidian stream]

You'll have to show some work to justify that. It's clear the only problem is near the origin and intuition suggests that checking n = 1, since that's the easiest to do in your head, you get functions like x^2/sqrt(x^2+y^2) and the limit exists. I don't see anything at a glance that would make any other value of n particularly sacred that would cause a sudden failure.

[merry prawn]

i wrote above, the partials are nxyr^(n-2)

so if n<2 then r will be on denominator, and when r is zero the partials are undefined

so if partials are undefined, it is not a simply connected region if n<2

[obsidian stream]

Why are you writing the partials half substituted like that? x and y are still functions of r.

[manic flax]

bruh did you even watch the video it literally explains why ur wrong

[merry prawn]

was simpler and shorter , and the poblem itself wrote the function with r like that.

but i did the calculation with x and y, not r. I agree i find it weird but since they write like that i thought it made sense for most people... dont know

[obsidian stream]

So do you still have a problem with that mixed partial or is this resolved?

[merry prawn]

yes

my problem is not the mixed partial, my problem is the n

i have a feeling you are not reading

[obsidian stream]

Well, there is an easy way to leave no doubt about that.

[merry prawn]

<@&286206848099549185> can i get help?

[merry prawn]

https://tenor.com/view/puss-in-boots-cat-cute-kitty-begging-gif-11965101

[vocal sleet]

@merry prawn Has your question been resolved?

[austere cliff]

can i please get help with percentages

any one please😭😭

breh

[robust cape]

I can help

[austere cliff]

yes please

pleae help me

[robust cape]

ok

[austere cliff]

basically my teacher in set 3 wants me to find precentages of a number without using calculators

[robust cape]

m

[austere cliff]

but i dont know how i dont understand anything

[robust cape]

Claim a channel

[austere cliff]

what?

[robust cape]

Wait

[austere cliff]

what does that mean?

[robust cape]

Nvm

[austere cliff]

ok

[robust cape]

Give me one of ur questions

[austere cliff]

so i need to find 15 percent of 67 with noi calculator

nvm not 15

19 percentage of 67

[robust cape]

alr

turn 19% into a fraction

U know how to do that?

[austere cliff]

how do i do that?

no😭

i didnt learn how to do that yet

they showed another way but i couldnt understand anything

[robust cape]

What percentage is equal to 1

[austere cliff]

100 [ercentage

[robust cape]

So put 19 over 100

[vocal sleet]

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[robust cape]

19% is also equal to 19/100

[austere cliff]

ok

[robust cape]

Then put 67 over 1

[vocal sleet]

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[tiny oracle]

If I have some function representing a transformation say f:R^2->R^2 is it standard to write f(X) where X is some set of points in R^2 to mean the set of {f(x):x\in X}?

[hybrid flicker]

yes

with f:A->B, f(X) with X subset of A is standard notation

"image of X by f"

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@tiny oracle Has your question been resolved?

[tiny oracle]

thanks

[vocal sleet]

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[magic quartz]

Can somebody dumb this down/explain this to me, pls? Specifically the corresponding stuff because I don't rlly understand how to corresponding stuff works

[steady plover]

Corresponding y value is the output of the function

Eg: for f(0) draw a vertical line through 0 and see where it intersects the graph, the y coordinate is your corresponding value

[magic quartz]

(this is going to sound really silly) if i draw a verticle line thru 0, how would i know what my y is

[steady plover]

Look at the intersection

[magic quartz]

OH

[steady plover]

Between the red line and the black graph

[magic quartz]

ohhhh

[steady plover]

[magic quartz]

thxx :3

.close

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[zenith dagger]

.reopen

Points $X$ and $Y$ lie on sides $AB$ and $BC$ of triangle $ABC$, respectively. Ray $\overrightarrow{XY}$ is extended to point $Z$ such that $A, C$, and $Z$ are collinear, in that order. If triangle$ ABZ$ is isosceles and triangle $CYZ$ is equilateral, then the possible values of $\angle ZXB$ lie in the interval $I = (a^o, b^o)$, such that $0 \le a, b \le 360$ and such that $a$ is as large as possible and $b$ is as small as possible. Find $a + b$.

[twin meteor]

lj

[zenith dagger]

$help$

[twin meteor]

lj

[vocal sleet]

@zenith dagger Has your question been resolved?

[vocal sleet]

@zenith dagger Has your question been resolved?

[vocal sleet]

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[granite anchor]

Im doing a physics project where I drop a weight on rubber bands (like a bungee jump). Stupidly I did even weight first then odd which means that when I got to my max even weight the rubber lost its strength so to say. Now I have a shift in my data.

I said to my teacher can I draw two curves one with even and then one with odds weights but he said no.

How could I mathematically adjust the data to work without faking my results?

[hard atlas]

repeat the experiment?

[granite anchor]

yeah thats what I said to him

but it isn't an important project

its like a practice one for smth to come

[heavy yoke]

it seems like you will just have a certain amount of variance in the data. it seems to me that it's better that the effect of the rubber getting less elastic is at least distributed throughout the data rather than concentrated on one end

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[lilac sluice]

The letters of the word PRINTER are arranged in a row. Find the probability that there are at least three letters between N and T

[lilac sluice]

How should I approach this question?

[paper depot]

!nosols

[vocal sleet]

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

[lilac sluice]

it’s not 240

[paper depot]

and don't blurt answers out either

[green veldt]

Mb

[paper depot]

there can physically only be 3, 4 or 5 letters between N and T

[lilac sluice]

right

[paper depot]

N___T__
_N___T_
__N___T
N____T_
_N____T
N_____T

& the same but with the N and T swapped

[lilac sluice]

oh yea

so 12 ways

[paper depot]

yes

out of 7*6 in fact

[lilac sluice]

wdym

12 ways out of 42?

[paper depot]

yes

cause you want the probability

[lilac sluice]

42 what tho

[paper depot]

and the total number of ways to allocate the N and T is 7*6

[lilac sluice]

could you explain how?

[paper depot]

to count the total number of possibilities, the N can go in 7 places and the T can then go in 6

[lilac sluice]

ohhhh

oh i get it now

but that doesn’t include T then N right?

[paper depot]

wym

[lilac sluice]

like yk how u said N can go in 7 places then T can go into 6?

what about T can go in 7 palces and N can go in 6

does the latter not matter when we do 12/42?

[paper depot]

it's an alternative way to count the same thing

[lilac sluice]

oh so regardless the final answer will be 12/42?

[paper depot]

yes

[lilac sluice]

okay

[paper depot]

well, simplify that, but yes

[lilac sluice]

mmm i was hoping you’d do another way that’s similar to my textbook solutions

i’ll show u hold on

i’m just a little confused with how they do it

i get the 5C4 part cus there’s 4 letters in between N and T

same with 5C5

but idk abt the rest

[vocal sleet]

@lilac sluice Has your question been resolved?

[lilac sluice]

<@&286206848099549185>

[kind light]

im p sure 5c4 is just tbe ammount of ways to choose 4 letters from the 5 excluding N and T

[lilac sluice]

yes

what about the rest tho

[kind light]

oh whoops misread that

4! is the stuff for between N and T, 2! is for N and T itself, and 2!/2! is uhh im not sure

[lilac sluice]

hmm

would it be the R’s?

like there’s 4! factorials of rearranging the letters between N and T right

so 2!/2! where the numerator is the amount of ways u can rearrange the R’s and the denominator cus that’s just the number of repetitions?

[vocal sleet]

@lilac sluice Has your question been resolved?

[lilac sluice]

.close

[vocal sleet]

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[vocal sleet]

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[vital scroll]

help please so im watching a video on bearings and this is how the question was done in the video, but when i do the question in my workbook and draw both bearings i keep getting 347 for my answer and im not sure why ive redrawn it a few times

[paper depot]

how are you getting 347

[vital scroll]

ill redo it and show u

So I've got this

But I can't work out how to measure the small angle-

[paper depot]

wdym measure

are you trying to do it with a protractor??????

and still i don't see how 347 comes into the picture at all.

[vital scroll]

Yes

[vital scroll]

So then because I'm struggling to measure the small angle

I do the big one instead

So up to 180 and then i do it from there until the line for p

And I get 167

And then I add 180 to 167 and get 347 for the big angle

[paper depot]

bad!

are you 120% sure that your drawing is accurate to within 1°??

[vital scroll]

No

[vital scroll]

Oh-

[paper depot]

then your protractor measurements cannot be reliable

[vital scroll]

My maths teacher taught me to do it with a Protractor

[paper depot]

you don't need one here...

[vital scroll]

Oh..

[paper depot]

all you need is this

the north lines are parallel by definition

[vital scroll]

Wait the question was 149 not 147 but I get it so if I do 180-149 the small angle is 31 which means the big angle is 329

[vital scroll]

I understand now thank you

[paper depot]

yes

[vital scroll]

I still need to use a protractor do draw bearings though right?

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[median nacelle]

Hi guys im really strugging with this problem here, because I'm not sure how to maximise one function and minimise another, we have to solve our problems using solver from excel and im just not getting the right solution

[median nacelle]

<@&286206848099549185>

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[west vigil]

I dont really know how to do this, learnt it about two weeks ago and cant seem to get it into my head

[brittle topaz]

do it step by step

for each operation do a new column in your table

[proud terrace]

fill the table : $P | Q | P \vee Q | \neg(P \vee Q)$ with their verity statements

[twin meteor]

tm

[proud terrace]

idk how to make a table in latex

[true swallow]

(•)(•)

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@west vigil Has your question been resolved?

[west vigil]

idk how to unfortunately

j was hoping i could get help on 1.a

then mimic it for the 1.b

i would appreciate it as a stepping stone yk?

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[fleet lark]

Stuck on question D the graph one

[paper bridge]

You can eliminate option D quite obviously

The horizontal asymptote is y=-3

and vertical is x=5

so uh

you should be able to narrow it down

[willow pike]

one with an asymptote at x=5
and y = -3

yeah

it's not hard to narrow down given the previous questions

[paper bridge]

@fleet lark bro it's a dead giveaway, don't react with that

just look at each graph

[fleet lark]

some of the graphs have the same asymptotes

[paper bridge]

yes

certain asymptotes

there is one option with both x=5 and y=-3

[willow pike]

do some have the same PAIR of asymptotes?

[paper bridge]

basically, choose the one with positive vertical asymptote and negaitve horizontal asymptote

[fleet lark]

dawg im mad confused

A

[paper bridge]

No

[fleet lark]

B

[paper bridge]

positive vertical asymptote

negative horizontal asymptote

[fleet lark]

C

[fleet lark]

so then its C?

[paper bridge]

congrats

[fleet lark]

[fleet lark]

thanks

how do i close this?

[sullen condor]

.close

[fleet lark]

thanks

.close

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[polar wren]

Why the variance is wrong here

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[ivory basin]

what does this mean ?

[ivory basin]

is it 3 log base 3 statement 8 ?

[paper depot]

"statement" is a strange word to be using there

do you know what $\log_3(8)$ means?

[twin meteor]

Ann

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[karmic crane]

I'm still not sure why $a=\frac{n}{\gcd(n,m)}$ is the smallest $a\ge1$ such that $n|ma$

[twin meteor]

Luca M

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[sudden cloud]

Hey could anyone help me w the n=k+2

[paper bridge]

,rccw

what the

?!

,rotate

[sudden cloud]

Ok

Lemme rotate for u

[paper bridge]

thank you

bot is not working bro

[sudden cloud]

Np

[paper bridge]

I think you can use the a^(2n+1)+b^(2n+1) factorization

but lets see

hmm

$87^2500^k+87^k500^2+587Q\mod 587$

[twin meteor]

vengeance

[paper bridge]

so then we just analyze $87^2500^2(500^{k-2}+87^{k-2})\mod 587$

[twin meteor]

vengeance

[paper bridge]

which obviously, according to our rule, is divisible by 587, thus proven

[twin meteor]

Operation timed out.

Mar 16, 2025, 1:20 AM

[sudden cloud]

Huh

[paper bridge]

try to look through it

[sudden cloud]

But u cant do tis no?

[paper bridge]

i expanded the $(87^2+87^k)(500^2+500^k)$ here

[twin meteor]

vengeance

[paper bridge]

we have it from the assumption

[sudden cloud]

Oh I see

[paper bridge]

see how last term will expand

[paper bridge]

wtf

[sudden cloud]

💀😭

[paper bridge]

now the bot decides to work

😭

[sudden cloud]

Wait

But

From the last term is it not 87k x 500k

Not 87^k + 500^k

U cant use the assumption

[paper bridge]

oh oop

there is probably some bs

[sudden cloud]

😭😭

[paper bridge]

mesus now

[paper bridge]

try

[sudden cloud]

Uhm