help-23

[nova creek]

Exactly

The function is less than zero between -2 and 2, and since we are allowing the function to equal 0, we include -2 and 2 in our answer

So how do we write this with an inequality?

Or with interval notation, whichever you prefer

[timber atlas]

-2 is less than or equal to x is less than or equal to ?

[nova creek]

less than or equal to what?

[timber atlas]

X

[nova creek]

The second part. You didn't write the last number

[nova creek]

Yes

That's the answer

[timber atlas]

Ok I understood thanks!

.close

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[visual forge]

help

[visual forge]

bro

[worthy hemlock]

Someone else has the channel, open your own

[worthy hemlock]

You're given the ratios, you just need to plug in the proper ones

A hint is the numerator and denominator should cancel out

Until you get the desired unit

[visual forge]

i dont get it

[worthy hemlock]

You want to cancel miles out, from the first fraction, so miles in numerator cancels out miles in denominator

If the miles in currently in the numerator, where should miles be to cancel out, numerator or denominator?

[visual forge]

numerator

[worthy hemlock]

Where should miles be to cancel out?

It's already in the numerator

[visual forge]

oh

denominator

[worthy hemlock]

So then what fraction has miles in the denominator?

[visual forge]

5280/1 mi

[worthy hemlock]

So then do that

[safe radish]

@visual forge Has your question been resolved?

[visual forge]

oh

what about

for the other ones

same thing?

[worthy hemlock]

Yes

[visual forge]

ok

.close

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[untold wadi]

Apparently I'm using the Jordan Closed Curve Theorem, which we haven't learned about but I'm given a hint

[untold wadi]

not sure what it all means

Here's the diagram

[swift flame]

i hate math

and it is not fun

[untold wadi]

Math is

Having lots of love-hate relationships

[swift flame]

ik

[untold wadi]

I just don't like geometry but taking it for the challenge

[swift flame]

i like fortnite

[worthy hemlock]

Why are you interrupting someone's channel with pointless comments?

[swift flame]

not pointless

[worthy hemlock]

If you can't help, then don't say anything

[swift flame]

k

[untold wadi]

Anyways, I'm not sure how to imagine what it's asking

Never seen this theorem before and the professor never explained anything

[final loom]

Have you read the wiki article on Jordan Curve Theorem?

[untold wadi]

Yes, but I'm still confused

[final loom]

For your question, it's only asked to show whether you can figure a continuous path from A to B in fig. 3.73

[untold wadi]

Like a maze?

[final loom]

mhmm(was the first thing that came to mind)

now, by Jordan curve theorem, if any of these two points lie in the interior region of any of the two loops, your arc must intersect the loop

[untold wadi]

Okay so it's a yes that there is a continuous path from A to B

[final loom]

thinks

I keep getting lost in the maze tho lmao

[untold wadi]

I kinda drew a light line lol

[final loom]

but if you did find one such path, notice how you assumed that point A kinda lies within this bounded loop that you drew your line in

[final loom]

when in actuality, A and B can be lying between the outer walls of two seperate loops

[untold wadi]

Oh I see

Must every arc joining A and B intersect curve C? That's the question from the figure. What does this mean?

[final loom]

why'd you stop the light line you drew from A to B when it was (kind of) reaching the end of one such loop

[untold wadi]

Oh so that's why?

[final loom]

what the question means is, pick C.. now figure if either of A and B are inside the loop C and the other is outside it

only then, must your arc always intersect C

[untold wadi]

How do you know if it's inside?

[final loom]

(@_@;) trace C

the same maze thingy

[untold wadi]

Trace C to any of A or B?

[final loom]

from what it looks like to me, C can merely just be the outermost boundary of the maze

[final loom]

if your diagram is topologically equivalent to this one

your task is also to trace all of C, and the one other loop present in there, and figure which among these two does A and B lie in

[untold wadi]

Okay

So when I trace all of C, I'm not getting anywhere

[final loom]

show mie +_+

[untold wadi]

I lied

I found a path to B

[final loom]

path to B? whaa

okay, here's the trick

ifffff

A or B, lie inside the loop C,

[untold wadi]

I lied again xD I didn't oml these lines are too tiny

[final loom]

by Jordan Curve Theorem, you can pick any point outside C, and the continuous arc joining this point to A or B must intersect C 👀

[untold wadi]

Okay, so if I start at the "entrance of this maze"

That could be one point that's outside of C?

[final loom]

mhmm

[untold wadi]

Okay okay

This time it's real

I found a path from that point to B

But I did find a path from B to A, so does this mean that both are in?

[final loom]

[untold wadi]

[final loom]

If you found a path from entrance to A, B, then both are outside C no?

[untold wadi]

Oh

Ohh

And if I didn't, then it would be inside of C?

Or if one had a path from that point to just B, then B is outside and A is inside

[final loom]

(@_@;) but you did find a path from B to A earlier

[untold wadi]

I meant it in a hypothetical way

[final loom]

*mhmmm

[untold wadi]

lol

Okay

So does this mean that in the diagram of my problem, both A and B would be outside of the figures

[final loom]

(ヘ･_･)ヘ┳━┳

not really

you didn't figure if they're going to be inside the other curve

but oh well, if you can justify that both the loops in Fig. 3.73 are closed, then same argument applies -> pick any point outside the maze, A, B can be connected to this pt. without intersecting any interior loop

[untold wadi]

That would be the argument?

[final loom]

what'd you think (@_@;)

[untold wadi]

🥺

Well there was a line that went from the point (imma say D, the one outside C) connected to B and then to A

[final loom]

mhmm

now you also need to show that both your "topologically equivalent" curves from the maze are indeed closed

[untold wadi]

What do you mean by closed?

[final loom]

oiiii

JCT is for "plane simple closed curve"

[untold wadi]

Oh

😅

Okay, ignore that question then

[untold wadi]

Going back to this, how would you know that

[final loom]

give me 2 mins, I'll have to prolly draw it for you (ヘ･_･)ヘ┳━┳

[untold wadi]

Okay

[final loom]

here's how :o

lmfao

basically, for a closed loop, you start with any one point, walk around the loop, and end up at the same point

[untold wadi]

Oh, you're going based on the exact line

I was actually going inside xD

[final loom]

ALSO

Did you figure the "two closed loops" in fig. 3.73 that the question is talking about?

[untold wadi]

I didn't

[final loom]

one is C and the other is

there

that sketchy loop disguised in there

notice that both C and this green loop are closed

hence, you pick any point outside of C and Greenie, now if there is a continuous path connecting "D" to "A", then by JCT, A must not be lying inside C, same argument for greenie

also, since B and A can be connected without intersecting any curve either, B also must not be lying inside C

[untold wadi]

How did you know that A is not there

Sorry for replying late, I had to run to the store 😅

Ohh

Nvm!

I think I understand now

[final loom]

Gg

[untold wadi]

Thank you Ansh

Always helping 🥺

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[mossy lake]

Given f(x) is a 3rd degree polynomial, with three distinct real x-intersects, where two of them are x=2 and x, =0.5(-3+square root 5). Determine the equation for f(x) if it passes through (3,19)

[safe radish]

Please don't occupy multiple help channels.

[mossy lake]

Please help me

[meager chasm]

close this channel

@mossy lake

[mossy lake]

how

[meager chasm]

.close

[mossy lake]

.close

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[lilac trout]

how can i attempt this ?

[meager chasm]

2b?

[lilac trout]

help with all 3 pls

[meager chasm]

we can't help with 2a

it's not a question per se

but rather a prerequisite to this question

[lilac trout]

yea makes sense

anyway for 2b

woudl you bring x over or no

[meager chasm]

first thing to notice

in order for $(a - x)^{\frac{1}{2}}$ to be well-defined, $a - x > 0$, ie $x < a$

[flat frigate]

Chromium

[meager chasm]

correct?

[lilac trout]

yes

as you cant square root a - number

[meager chasm]

yea

[meager chasm]

so everything we do onwards will be under this prerequisite

agreed?

[lilac trout]

yea

[meager chasm]

first of all do you know how to solve square root equations

[lilac trout]

no

[meager chasm]

do you know how to solve quadratics, then?

[lilac trout]

yes

[broken yew]

The 2nd bottom equation is a quadratic.

[meager chasm]

square root equations are basically quadratics

$x = (x - a)^{\frac{1}{2}}$

[flat frigate]

Chromium

[meager chasm]

so $x^2 = x - a$

[flat frigate]

Chromium

[meager chasm]

right?

[lilac trout]

yep

[meager chasm]

$x^2 - x + a = 0$

[lilac trout]

so then bring it to the other side and solve for x ?

[flat frigate]

Chromium

[meager chasm]

yea

do you know determinant?

[lilac trout]

no

[meager chasm]

😦

thought you know quadratics?

do you know the quadratic formula, at least..?

[lilac trout]

yes

we havent learnt determinant

[meager chasm]

do you know the prerequisite for a quadratic equation to have real roots, then?

[lilac trout]

and thats further maths i dont do that :(

[meager chasm]

it's an important concept

shame your course cut it out

[meager chasm]

anyway

[lilac trout]

discriminant has to be a positive

[meager chasm]

yea

[lilac trout]

either a or c has to be a negative number

[meager chasm]

do you know the determinant of this equation

$ax^2 + bx + c = 0 \land b^2 - 4ac \geq 0 \implies$ solutions has real roots

[flat frigate]

Chromium

[meager chasm]

this is from the quadratic formula

[zinc crown]

Not true. Consider: x^2 + 4x + 4

[meager chasm]

yea that works

[lilac trout]

true

[lilac trout]

yes

[meager chasm]

you see how this is true right

[lilac trout]

yep

[zinc crown]

So what is ur question

[meager chasm]

so here, the determinant is $1 - 4a$

[flat frigate]

Chromium

[meager chasm]

and it has real roots when $1 - 4a \geq 0$

right?

[flat frigate]

Chromium

[lilac trout]

so would a just be =1/4 ?

[meager chasm]

wdym =

[lilac trout]

a= 1/4

[meager chasm]

$a \leq \frac{1}{4}$

[zinc crown]

That would be when there would only be one solution to the quadratic

[flat frigate]

Chromium

[meager chasm]

not equal

[zinc crown]

That is, it’s a perfect square

[meager chasm]

this is the range of a

[lilac trout]

since it says its one real solution

[zinc crown]

then yeah a would be 1/4

For one solution

[meager chasm]

finally, our goal is to solve for range of x

[lilac trout]

we dont need to find x for this question

it just says find a which has one real solution

[meager chasm]

yea oof

[lilac trout]

so a would be a= 1/4

and for the second bit

a < 1/4 ?

[meager chasm]

why are you always saying =?

$\leq$ or $\geq$ will do

[flat frigate]

Chromium

[lilac trout]

isnt one real solution the same as =

[meager chasm]

<=, >= if you dont know tex

[meager chasm]

just write this lol

[lilac trout]

aight

[meager chasm]

oh shit didn't see your full question

yea at $a = \frac{1}{4}$ there is only one solution

[flat frigate]

Chromium

[lilac trout]

bruh

lmao

then for part c?

[meager chasm]

for the first one try to convert it to a quadratic

for the second one throw the square root to one side and everything to the other

since from there you can square it

[lilac trout]

it says Use an appropriate substitution

[meager chasm]

.

yea

[lilac trout]

so substitute ( 1/4 - x )^1/2

in every x ?

[meager chasm]

hm?

i'd use $\sqrt{x} = y$ instead

it becomes a quadratic in y

[flat frigate]

Chromium

[lilac trout]

ok

i have gotten myself confused

@meager chasm wait does does mean to use another letter to solve it ?

[meager chasm]

if you solve for y

you can solve for x

[lilac trout]

so it would be y^2 +3y-1/2 = 0?

find y

[meager chasm]

yea

[lilac trout]

then squart root to find x?

[meager chasm]

find y

the solutions for x are y^2

[lilac trout]

yea thats what i meant

[lilac trout]

.close

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[lilac trout]

.reopen

[safe radish]

✅

[lilac trout]

how could i do this

[zinc crown]

Well you want to make it a quadratic

So what would the substation be, noting that sqrt^2 = the thing itself

[lilac trout]

its just the square root idk how to deal with it

[zinc crown]

how about starting by letting u = sqrt(x+2)

[lilac trout]

ok

would you substitute x as u aswell ?

but it would be x = u^2 -2 ?

[safe radish]

@lilac trout Has your question been resolved?

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[lean otter]

need help

[lean otter]

confused on the whole concept

honestly

please halp

[plucky elk]

read this and try
https://tutorial.math.lamar.edu/classes/calci/tangents_rates.aspx

[lean otter]

ight bet

.close

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[hard smelt]

Would anyone mind helping

[worthy hemlock]

Distance formula

Do you know it?

[hard smelt]

yeah

[worthy hemlock]

So get the coordinates of points A and B, use the distance formula

[hard smelt]

ive done that

but need to check the answer

feels wrong

and its a decimal

[worthy hemlock]

Check with a calculator

Decimals are a thing

You're not going to get whole numbers every time

[hard smelt]

i mean

could you still check my working out

[worthy hemlock]

If you got a decimal, it's right

[hard smelt]

also

should i round it

to 2dp?

if i do get a decimal

or

[worthy hemlock]

Or you could keep it exact

You can leave the radical

,w sqrt{(3--2)^2 + (2--1)^2}

[flat frigate]

Results provided by WolframAlpha

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

Requested by dldh06#2659

[worthy hemlock]

If you apply the formula properly, you get that

There's nothing confusing about it

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@hard smelt Has your question been resolved?

[hard smelt]

.close

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[shy barn]

for f(x) = x+5, what is the value when x = 2?
for f(x) = x², what is the value when x =2?
Are they the same?

[cosmic grove]

think about limits on the left and on the right but yeah that's the idea

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@wraith sable Has your question been resolved?

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[wild copper]

Expanding all those expressions will be error prone.

There is a simpler way to solve it.

[woven kiln]

find as x tends to 2, is it the same value ?

(ii) if a = x and b =y
and if a = y and b = x,
is a*b = b*a
are they the same
(iii) is a*(b*c) = (a*b)*c

i think that’s what it’s asking

not fully sure

yeah, it doesn’t look associative

so (iii) you can just say it’s not associative

simple as that

[steady hinge]

.close

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[waxen chasm]

Here’s my work and this is the question: Let A and B be nXn matrices and assume that the product AB is invertible. Prove that both A and B are also invertible.

[waxen chasm]

Can someone tell me if this is right please?

And just to double check since the wording is kind of confusing online.. if A is invertible, then there’s a matrix it can be multiplied by to get the identity matrix? And if that’s possible, then we denote A^-1 = AB?

[light shoal]

The first part is OK. For the second part, why is it OK to say "assuming A is invertible"? That's what you are trying to prove. Instead, just do what you did in the first part: AB(C) = I, therefore A(BC) = I, therefore A is invertible and BC is its inverse.

[weak dirge]

Yeah idk honestly

[waxen chasm]

Thank you I see what I did

Lollll me too tho

[weak dirge]

You can use the same argument that $AB=I \implies \exists A^{-1},B^{-1}$

[flat frigate]

KurtDee

[waxen chasm]

What language is this

Hahaha I have no clue what the Flipped E means

[weak dirge]

there exists

[waxen chasm]

But thank you I’ll try

[weak dirge]

honestly it's a bit abusive

but yeah like

I'm not sure if that's saying the same thing or not

Your part 1 and part 2

[waxen chasm]

But someone said I can’t use that argument

Cause that’s assuming

[weak dirge]

No no no

[waxen chasm]

I can?

[weak dirge]

You basically do the same thing twice

[waxen chasm]

Really? But I’ve shown the inverse of A and B

In 2 separate equations

[weak dirge]

I'm not sure I understand what you mean

Like by strict definition of an inverse, A is invertible if there exists B such that $AB=BA=I_n$

[flat frigate]

KurtDee

[weak dirge]

So what exactly are you saying here

[waxen chasm]

What does that even mean tho? All these equal signs in algebra confuse me. So A times B is the same as BA which gives an identity matrix? Pretty sure my interpretation is wrong

[weak dirge]

What exactly is the question?

Like if you're given CAB=I=ABC that

[waxen chasm]

This is it: Let A and B be nXn matrices and assume that the product AB is invertible. Prove that both A and B are also invertible.

[weak dirge]

Okay

[waxen chasm]

I just asked my prof and he said denote C by AB = In

But I’m not sure why part 2 is wrong

[weak dirge]

part 2 line 2 assumes A is invertible

[waxen chasm]

Better?

True I’ll have to remove that

[weak dirge]

that's sufficient, but can you prove that CAB = ABC

[waxen chasm]

Crap do I have to? 😅😅😅

[weak dirge]

It's actually really easy

But yes

[waxen chasm]

Ok let me try

Thanks

[weak dirge]

C = (AB)^-1

so it's pretty much given

once you have that you pretty much have the definition of inverse

which is $MN=NM=I$

[flat frigate]

KurtDee

[weak dirge]

for some matrix M,N

[waxen chasm]

Omg I’m dumb

I’m not able to prove it

I can’t wrap my head around inverses

[weak dirge]

Actually I'm wrong sorry you dont need to prove it

but it's pretty straight forward

ABC=AB(AB)^-1=I=(AB)^-1AB=CAB

by definition of inverse

[waxen chasm]

But where’d the C disappear to in the first part on the RHS?

Thank you by the way

Appreciate it

[weak dirge]

$C=(AB)^{-1}$

[flat frigate]

KurtDee

[waxen chasm]

Thank you

I set it up as CAB = ABC like you mentioned but I can’t get it alone like you did

I think I have my brackets in the wrong places

[weak dirge]

I mean I don't think you really need it

[waxen chasm]

True

[weak dirge]

now that I look again

[waxen chasm]

But I really wanna understand inverses

Cause all my prof says is AB = BA = In without explaining

Like i thought it wasn’t commutative

[weak dirge]

That's the definition

of an inverse

[waxen chasm]

Why are we switching the order

[weak dirge]

It's just a property of it

[waxen chasm]

So I don’t have to worry about it too much? Just know?

[weak dirge]

No like it's something that's important

Like $AA^{-1} = A^{-1}A = I_n$

[flat frigate]

KurtDee

[waxen chasm]

Ohhhh

[weak dirge]

It could be multiplied from the left or right

[waxen chasm]

thank you!!!

So it’s showing B is the inverse of A?

[weak dirge]

that's still the "inverse" function if you want to call it that

Yeah

[waxen chasm]

And that you get the identity of you multiply

[weak dirge]

yeah pretty much

[waxen chasm]

Thanks godly math hero

[weak dirge]

you could do it like that

So $CAB = I$ BECAUSE $C=(AB)^{-1}$

[flat frigate]

KurtDee

[waxen chasm]

What if I wanted to say A is the inverse of B? Can they be worded differently or is this a whole different statement?

Thank you

[weak dirge]

well in square matricies, they're the same thing

A inverse B or B inverse A

[waxen chasm]

If they’re the same why do I have to prove both?

I can just prove one 😎

[weak dirge]

You're not saying A is the inverse of B, you're saying that (AB) having an inverse implies A,B have an inverse

[waxen chasm]

Can I say (CA)^-1 = B
And B^-1 = (CA)?

Thanks

Wish u were my prof

Lol

[weak dirge]

Sooooo you can say that after showing CAB = I believe

Because that's showing it has an inverse

[waxen chasm]

Ok

Thank you so much

[weak dirge]

Hope it helped

[waxen chasm]

It really did, thank you 🙏

[weak dirge]

https://math.stackexchange.com/a/470160

Here's a link showing that $AB=I \implies BA=I$

[flat frigate]

KurtDee

[weak dirge]

idk if you might need this or not

[waxen chasm]

This part helped cause it’s basically what you said

I just gotta study more but this really helped me

Thanks so much

Hope you have a good day

[safe radish]

@waxen chasm Has your question been resolved?

[safe radish]

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[ruby sequoia]

can anyone help me simplify this? with w being a 7th root of unity

[ruby sequoia]

yeah i tried using that but it wasnt very fruitful

[final loom]

np, I'm just wondering about a useful polynomial transformation that could help

[ruby sequoia]

i mean you might get something useful out of it x)

[final loom]

oh right

[ruby sequoia]

<@&286206848099549185> anyone?

[broken yew]

Checking there isn't a typo

like 1/(1+w^2) on the first would make slightly more sense

but if it isnt, then nvm.

[ruby sequoia]

hm no i dont think so

[broken yew]

I would consider attempting to draw this

it looks totally possible

[ruby sequoia]

huh but how would you do that?

oh yeah w isnt equal to 1 i forgot to mention

[broken yew]

i just mean a sketch

the elementary arithmetic operations have geometric meaning in the argand plane

[ruby sequoia]

oh even the rational functions?

[broken yew]

???

x

/

^

[ruby sequoia]

im stupid

[final loom]

oof, it really is a pain 🤦‍♂️ mental math with complex no.s smh

[ruby sequoia]

🙃 yeah that must be pretty painful

[final loom]

$\frac{-w^3 - (w^6 + 1)}{(w^2+1)(w^4 + 1)} + \frac{w^5}{1+w}$

[final loom]

yep, totally no idea what that last term is about ..

can you check the q for me rq?

[ruby sequoia]

i mean i feel like even without it it's still pretty nasty

yep it's correct

[final loom]

hmm (ヘ･_･)ヘ┳━┳ hands down.

no mental math when sleepy lol

[ruby sequoia]

yeah i went through the same stuff as you did but it was to no avail...

[lean otter]

all i know about that is that

$\frac{1}{w^{2}+1}=\frac{w^{4}-w^{6}}{w^{3}-w^{7}}$

[flat frigate]

aspwil

[lean otter]

at least i think its true

i dont think its help full

[ruby sequoia]

dunno

[lean otter]

but do with it what you will?

[ruby sequoia]

itll probably just make things more complicated but sure

[lean otter]

desmos says its zero

[ruby sequoia]

probably just comes from euclidean division

[lean otter]

in fact it seems to be true for

$\frac{x^{4+a}-x^{6+a}}{x^{3+a}-x^{7+a}}=\frac{x}{x^{2}+1}$

[flat frigate]

aspwil

[lean otter]

exactly why i have absoultly no idea

in fact

[ruby sequoia]

well all you did was multiply the top and bottom by some polynomials lol

[lean otter]

oh shit your right

lmao

[ruby sequoia]

lmao

[lean otter]

lmao there goes one my oldest unsolved math things lol

[ruby sequoia]

what

you mean youve never seen this is true..?

[lean otter]

well it was just an old image from my phone from like 7th grade

[ruby sequoia]

lmfao

[lean otter]

i never really revisited it other then "huh neat"

just noticed it on desmos one day

[ruby sequoia]

thats just hilarious

[lean otter]

more specificly i noticed that for

$f\left(x\right)=\ \frac{1-x^{2}}{1-x^{4}} \g\left(x\right)=\frac{x^{6}-x^{4}}{x^{3}-x^{7}} \ g\left(x\right)\ =\ -x\cdot f\left(x\right) \ f\left(x\right)\ =\ x\cdot g\left(x\right)+1$

[flat frigate]

aspwil

[ruby sequoia]

bro are you sober

[lean otter]

yes?

[ruby sequoia]

well then

[lean otter]

hold on wait im probs distracting from the help

[ruby sequoia]

its fine at this point

this question will probably just die

[lean otter]

yeah i just found that the 2 equations f(x) and g(x) have this weird simple relation to eachother

given by equation 3 and 4

[ruby sequoia]

i mean its just a matter of simplification...

[lean otter]

yeah but i never really dove into it lol, remeber this is just an old image that been sitting in my phone unsolved for like 7 years

cause i never revisited it lol

[ruby sequoia]

mhm ig thats plausible

but also quite funny at the same time

[lean otter]

yeah

the other intresting part was that g(x) relates to itself by proxy

[ruby sequoia]

that doesnt sound right

[lean otter]

like g(x) = -x * (x * g(x) + 1)

[ruby sequoia]

hmm if it's that involved then it's possible

[lean otter]

yeah i was like "woah thats cool"

[ruby sequoia]

yeah i think i had moments like those too lmao

little me playing with simple formulas thinking im hot shit

[lean otter]

or g(x) = -x^2 * g(x) - x

[ruby sequoia]

but look at me now 😔

[lean otter]

yeah lol

[lean otter]

as it turns out the first 2 equatte to really simple eqatuions when simplified

sorry i got to go now though fun chat though.

[ruby sequoia]

it sure was

👋

[safe radish]

@ruby sequoia Has your question been resolved?

[ruby sequoia]

.close

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[snow robin]

whatsup?

[timber atlas]

I can’t get how to make the equation

Of b

[snow robin]

the first sentence means

A : area

A is proportional to x^2

[timber atlas]

Ok

[snow robin]

which means that A = c x^2

with c being some constant

3

6

20004238528934

any constant

[timber atlas]

Ok

[snow robin]

if A is 108

and x is 3

what is c

[timber atlas]

12

[snow robin]

exactly

so A or in the way the question writes it y

y = 12x^2

that's the equation

[timber atlas]

I didn’t get this part

[snow robin]

the first sentence

says

A or y is in direct proportion to x^2

this means that y relates to x

[timber atlas]

Yep

[snow robin]

in a way that it goes up with x^2 as x goes up

[timber atlas]

Ok

[snow robin]

but that doesn't mean that y = x^2

because there could be some constant

that relates the two

[timber atlas]

Yes

[snow robin]

some number that turns the proportion to an equation

and you give that a symbol

i called it c

and so

y = c x^2

[timber atlas]

Oh ok

[snow robin]

and they give you some numbers to calculate c

[timber atlas]

Yep

[snow robin]

so that you can find the equation

which is y = 12 x^2

[timber atlas]

I get it now

[snow robin]

🙂

[timber atlas]

Thanks

For your help

👍

.close

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[urban gale]

Im a little stuck

[urban gale]

I've multiplied the exponents to the log and got 14log(-9x)+12log(-20x+17)-4log(6x)

[dense galleon]

If you want to condense it to one logarithm, you should move everything inside the log

Also 14log(-9x)+12log(-20x+17) is invalid because the power is only on the x

[urban gale]

so i dont multiply the powers of the x to log?

[mystic tundra]

It may seem long but I think what you should do is log [((-9x^2)^7)*((-20x^2 + 17)^6)]/(6x)^4

[urban gale]

is it simplified? it looks kinda long

[plucky elk]

Use algebra to simplify it further

[urban gale]

ahh.. i got it

thanks everyone

.close

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[lost sorrel]

Determine the derivatives F(x)=x^3 using First Principles

[lost sorrel]

I know how to set up the equation but idk how to cancel the H in the bottom

[pulsar condor]

expand (x+h)^3

and simplify

[lost sorrel]

i factored by cubing

[sleek light]

Isnt it 3x²

Where s the H

[lost sorrel]

yes, but u have to use first principles

[worthy hemlock]

Definition of the derivative

This thing

[sleek light]

Ahh i see

[pulsar condor]

ok, well the h should cancel instantly

(x+h-x)(stuff)/h

[lost sorrel]

how do u get the (x+h-x) ?

ohh... nvm

i got it

lol

ty ty

[safe radish]

@lost sorrel Has your question been resolved?

[safe radish]

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[silver sigil]

Set A has 5 elements and set B has 4 elements. The elements of set S are all of the possible numbers of elements that could be in A∪B. Find the sum of the elements in set S.

[silver sigil]

Would this be the sum of 1 through 9?

[sleek light]

It cant be 1 2 3 i think

[silver sigil]

why not?

it never said the digits had to be distinct

[sleek light]

How can 4 and 9 elements can add and become one?

[silver sigil]

if all the elements in set A are one

and all the elemnts in set B are one

wouldnt that make the union just 1

so it would be one element long

[sleek light]

Hmm

But a set can have {1,1,1,1} ?

[silver sigil]

hmm

im not sure if a set has to distinct values

[nova creek]

It does

[silver sigil]

oh

then is it 35

sum of 5 thru 9

[nova creek]

{1, 1, 1, 1} = {1}

[silver sigil]

oh

i see

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[vivid veldt]

I'm looking for a neat function that generates some sufficiently shuffled-looking permutation on a list A by mapping the index.
p: {0..N} -> {0..N}, p bijective and "shuffled looking"

some offset+modulo come to mind but thats just a constant offset, not shuffled.
Then there are the sattolo permutation shuffle, but thats an algorithm, not a neat fuction.

I feel like I'm overlooking some simple auto bijection for a finite set.

Thanks for the help

[safe radish]

@vivid veldt Has your question been resolved?

[safe radish]

@vivid veldt Has your question been resolved?

[fair dagger]

@vivid veldt Can you specifiy more closely what is a "neat function" to you?

[vivid veldt]

Any sort of expression.

[fair dagger]

You can write algorithms as expressions if you try hard enough.

I would just use use some pseudo random number generation thing

and turn it into a shuffle thing.

[vivid veldt]

I just want to avoid actually shuffling the list, I just want to map the index to the shuffled index, preferrably using a simple expression, instead of an iterative algo like sattolo

[fair dagger]

You want to map a single index?

If I understand correctly you want to map single offest deterministically in a way that is more efficient than figuring out all permutations?

Is this a practical or a theoretical question?

[vivid veldt]

Practical

[fair dagger]

Is N very large? (say a million or more?)

[vivid veldt]

Small-ish, a few k

[fair dagger]

Hm.

[vivid veldt]

I could totaly to a deterministic shuffle on the list, but I find that very un-beautiful 😄

[fair dagger]

How about combining a couple simple shuffles together to make it unpredictable?

[vivid veldt]

Are there any that can be expressed simply?

Hmm, multipkication and modulo could work, but that would need to use a prime field

[fair dagger]

Value_{n+1} = DeterministicShuffleN(Value_n + RandomOffsetN)

repeat until satisfied

[vivid veldt]

Well, that's what shuffling looks like, but how do I do that without materializing the list

[fair dagger]

Aha

so memory is a concern as well

hm

[vivid veldt]

an example would be
shuffleindex(x) := (x + 1337)mod N

But thats not shuffled, just rotated

[fair dagger]

I think we need a second operation to modulo sum

that commutes very badly with sum.

[vivid veldt]

I just feel dumb not being able to come up with a simple bijective function on a finite set thats not super simple.

Oooh, i forgot about https://en.wikipedia.org/wiki/Perfect_hash_function

[fair dagger]

Mutiplication should work, right?

I don't thinky ou need a prime modulo.

[vivid veldt]

Hmm

[fair dagger]

Just multiply and sum

Wait no, multiplication is not 1-1 without prime.

[vivid veldt]

But without an inverse it wouldn't be bijective

[fair dagger]

Okay I was no longer in the bijective case.

yeah...

[vivid veldt]

😄

[fair dagger]

But you can do it anyway?

[vivid veldt]

I mean, I could just use a prime sized list

[fair dagger]

Select a prime subset of elements

and shuffle it with rotation.

If you do this often enough it will move essentially everything

[vivid veldt]

If you are curious, this all came about because I spoilered myself trying to reverse the word selection in wordle.

They just modulo on the day since the start of the game, meaning the words are in there in-order

I was thinking to implement this for a topic specific wordle and was looking for a way to blind myself.

Well, thanks for the brainstorm @fair dagger !

.close

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[fair dagger]

You are welcome.

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[lean otter]

I'm stuck on this

:V

[strange helm]

Think about the number difference between them

That should get you a pattern with "n"

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

@strange helm

so like it goes +4 +8 + 12

which I can like make a recurrence formula of an = an-1 + 4n?

I think I'm not sure

[strange helm]

Something like that, be back in like 10

[lean otter]

k

[strange helm]

So you are on the right path. You can associate the figure's # as part of a series - for your "n"

[lean otter]

how?

I can't just do 100 without knowing 99 and I cant know 99 without 98

and it goes on

[strange helm]

You can, you just need to find out the more precise function for the relationship "4n" is part of it as you have identified

"an = an-1 + 4n" wouldn't really work for Figure 0, no?

[lean otter]

no

it wouldn't

so like a condition where n>2?

[strange helm]

Not sure how to hint at.... "what mathematically gives you '1' when it is effected by '0'? Multiplication, addition, exponent, etc..."

[lean otter]

additon

or subtraction

[strange helm]

Not exactly..... n+0 = n, not 1

[lean otter]

hmm

im confused

what?

[strange helm]

Have you learned exponents?

[lean otter]

yes

ofc

[strange helm]

What happens when you raise something to the zero?

[lean otter]

it becomes 1

ooohhh

so how would I apply that to my formula?

[strange helm]

Yeppers

So, since our figure 0 has 1, then it is likely that the exponent value will our Figure number

[lean otter]

hmm ok

an = an-1 + 4^n?

wait no

an = an-1 + 4n

idk

bruh

[strange helm]

I need to calculate it too, one sec

[lean otter]

k

[strange helm]

But it shouldn't have a(n-1)

[lean otter]

hm alright

@strange helm

I have no idea

I took a while to think but I didn't even know where I was going

it goes 1, 5, 13, 25

with difference of 4n

[strange helm]

Ok, figured it a bit more. It does increase by a multiple of 4, but it doesn't work exactly that way with the sequence. 2 does

[lean otter]

wdym?

figure 2 does?

[strange helm]

the number 2

[lean otter]

what number 2

like the number "2"?

[strange helm]

instead of "4n", it relates to "2n"

[lean otter]

how so

?

[strange helm]

The pattern relies on the figure number, and since the pattern formula uses the current figure number and the figure number after it, 4 is too high. The exponent stuff from earlier is wrong too.
Example for Figure 1: this is the box_count_figure_0 + 2(Figure_number * Figure_number_plus_1) => 1 + 2(1*2) = 1+4 = 5

[lean otter]

not really understanding

[strange helm]

Since either figure_number or figure_number_plus_1 will always be even (some math proof about consecutive numbers having one odd one even), your multiplicative factor will always be even (another math proof), so 4 is just a little too high

[lean otter]

[strange helm]

Sure. Looking at Figure 2: 1 + 2(2*3) = 1+2*6 = 13

Guess and Check method also works

[lean otter]

an = a0 + n(n3)

is that the formula

[strange helm]

an = 1 + 2 ( n * n+1)

Except for a0

Actually it works for a0 too

[lean otter]

can u use syntax to help me visulize

[strange helm]

Try that formula on figure 3, since 1 and 2 are above

n = figure number

[lean otter]

an = 1 + 2 ( n * n+1)

$$an = 1 + 2 ( n * n+1)$$

[flat frigate]

痛苦

[lean otter]

$$an = 1 + 2 ( n ^2+1)$$

[strange helm]

Op! Sorry

[flat frigate]

痛苦

[lean otter]

this?

[strange helm]

$an = 1 + 2 ( n * (n+1))$

[flat frigate]

Buxy

[lean otter]

whys it so complicated

I feel like it wouldn't be so complex

[strange helm]

Deceptively simple math never is.

$$$but if you plug 100 into $1 + 2 ( n * (n+1))$, you should be good$$$

[flat frigate]

Buxy
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lean otter]

$$$but if you plug 100 into $1 + 2 ( n * (n+1)), you should be good$$$

[flat frigate]

痛苦
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lean otter]

$$but if you plug 100 into $1 + 2 ( n * (n+1))$, you should be good$$

[flat frigate]

痛苦
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lean otter]

bruh what

[strange helm]

$1 + 2 ( n * (n+1))$

[flat frigate]

Buxy

[strange helm]

Use ^

For the requested n = 100

[lean otter]

20201

@strange helm

?

[strange helm]

That is what I got

[lean otter]

wew thanks for helping but I didn't really get it

[strange helm]

Finding sequences is a lot of.... guess and check sadly

And there is, like in most math, a million ways to find the answer

[lean otter]

ok

[strange helm]

Sorry I couldn't be more helpful, but

[lean otter]

.close

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[lean otter]

quick question I feel like its super easy and im over thinking it

[wanton blade]

Solve n square plus n minus 156 = 0

[lean otter]

why

it forms an arithmetic sequence

@wanton blade

[wanton blade]

12 is the answer

[ember bough]

kermit you are not bring helpful at all

[lean otter]

true

I want to know why

[ember bough]

alright

the number of passengers on the train after n stops is 1+2+3+...+n

[lean otter]

yep

and it adds to 78

[ember bough]

i wonder if there's an identity you can use to rewrite that expression

[lean otter]

prob is

[ember bough]

do you know the identity?

[lean otter]

nop

[lean otter]

where do I go from there

[ember bough]

1+2+3+...+n = n(n+1)/2

[lean otter]

ok

[ember bough]

(do you want proof for that?)

[lean otter]

no

[ember bough]

oh okay

[lean otter]

I have that formula writen down

[ember bough]

then you have n(n+1)/2 = 78

[lean otter]

so $$\frac{n(n+1)}{2} = 78$$

[flat frigate]

痛苦

[ember bough]

multiply both sides and then expand

[lean otter]

and then solve the quadratic?

[ember bough]

yep!

[wanton blade]

So u see why it’s not linear

I already told u the equation

Lol

[ember bough]

that's not helpful!

[wanton blade]

Because u didn’t understand my provision

[ember bough]

i did, but that person wont understand your equation and why it's there

so please dont just give answers

[wanton blade]

You don’t give solutions here

[lean otter]

all I needed to know was the explicit formula n(n+1)/2

[wanton blade]

U give hints

Then they ask

And u tell

U did that here RYS

After he asked

I would have eventually told him that

[ember bough]

i'm not giving solutions. i'm guiding them (where did you see me giving the answer?)

[ancient escarp]

this is perhaps the least helpful thing you could possibly say

as a helper, you should know and have read the rules

you should know how to help, hence the role name, and know you're not supposed to give away answers

[wanton blade]

Because I didn’t complete explaining all that @ancient escarp

[ancient escarp]

that sentence makes no sense, and let me make this clear, i don't want you to type another one for me

[wanton blade]

@ember bough interrupted with his thoughts

[ancient escarp]

just don't give away answers, and if someone tells you not to, say "oh my bad", delete your answer, and move on

[wanton blade]

When I give answers I’m usually checking with the person if I’m correct or not

[ancient escarp]

.

[ember bough]

thanks disappointing son, i bet your father would be proud of you

[wanton blade]

So I was expecting a “yes this correct and now tell me how u came up w that”

[ancient escarp]

.

[lean otter]

my lesson doesn't explain stuff well like the information is easy but the questions are super hard I already looked back at it but still dont understand

[frigid geyser]

we don't give away answers, period

[lean otter]

so a,b,c make a AP so we can like say its a, a + r, a + 2r

[wanton blade]

Alright I get it.

[lean otter]

and a, c, b make a GP which we can say a, ar, ar^2

[ember bough]

geometric series with negative common ratio means...?

[lean otter]

but like I can going nowhere with it

[lean otter]

it converges?

or it gets smaller

[wanton blade]

Answers never help either however. I get it tho.

[wanton blade]

If a person is really trying to learn how

Bad for someone here cheating on tests tho: I get it

[lean otter]

this is my practice problems

[ember bough]

uh no
not necessarily
you can try a geometric progression with common ratio -2 and see where it goes.

[lean otter]

it goes positive then negative

like it switches back and forth

-2, 4, -8, 16

[ember bough]

yep!
that's the behavior you're looking for here, right?
the second term is larger than the first term

[lean otter]

yes

oh I understand

wait no

an arithmetic sequence doesn't always increase with a ratio

it could decrease

4,2,0,-2

[ember bough]

actually that's just info on geometric sequences as i think that might be useful. you can get the answer with just algebra.

[lean otter]

?

what

@ember bough

[ember bough]

as you've said before (common difference = x; common ratio=r):
$b = a+x = ar^2$\
$c = a+2x = ar$

[flat frigate]

RYC for mod⁴ (meg)

[plucky elk]

Plug in the definition of arithmetic and geometric sequence

[ember bough]

focus on $a+x=ar^2$ and $a+2x=ar$

[flat frigate]

RYC for mod⁴ (meg)

[lean otter]

um

systems of equations?

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[lean otter]

could I get some help on how to start this, I'm a little confused

[lean otter]

Because I don't think I can assume the if statement, because that's just SAA right there

[queen parcel]

I’m not super confident I know this stuff super well, but I’d start with a diagram consisting of 2 triangles - ABC and DEF

And labeling the information you’re given

[lean otter]

okay, I see that much. I'll see where I can go from there, thanks!

.close

[safe radish]

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[safe radish]

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• 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.

[gleaming yacht]

m^2 - n^2 = 10k
m - n = k
Which one is correct?

1. -2n = 9k
2. 2k = -9n
3. 2k = 9n
4. 2n = 9k
   I know it's so easy but I do not know why I got stuck in it.
   k = m-n = m+n / 10
   Found the answer lol