help-23

[simple galleon]

if f(x) is not equal to g(x) then what does 'f(x)=4x^2+6x equals to g(x)' mean?

anyway i dont need rigor that much, sorry if that was unclear

im just doing this for the fun

[humble egret]

f(x) equals to g(x) over some constant that gives fx , sorry but isnt that it 😭 ??

what are u confusing this for

[simple galleon]

again, counterexample is translation

[upbeat swan]

https://tomrocksmaths.com/wp-content/uploads/2021/05/parabolae-similarity.pdf

[simple galleon]

bois??

[upbeat swan]

What?

[humble egret]

ohhhhhh

lol i didnt see this

[simple galleon]

read the first paragraph

[upbeat swan]

I dont get it

[simple galleon]

what does 'all the bois' mean

[upbeat swan]

Lmao

Idk what that means

I mean

Its probably a slang for boys

Like

[simple galleon]

so a teacher who is desperately trying to sound hip

i see

[upbeat swan]

Nah Tom is actually a cool guy

He isnt pretending

[simple galleon]

what the hell does 'all the bois' mean then

ive never heard anyone use 'bois' like this

[upbeat swan]

He's just too cool

Too swag

[quiet juniper]

Man is stuck 10 years behind in meme years

https://tenor.com/view/yeahyeahyeah-yeah-oh-frog-unicycle-gif-12211532

[safe radish]

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[vernal tree]

I am unsure of what to do as that matrix can't be multiplied

[split moss]

You wrote A as rows and used the formula for columns

[raven heart]

A should have the vectors of the subspace you're projecting on in column form

so transpose your A

@vernal tree

[safe radish]

@vernal tree Has your question been resolved?

[vernal tree]

Huh?

[vernal tree]

Oh shit

Sorry

but then the determinant is 0?

so there's no inverse?

[split moss]

?

The subspace spanned by (1,0,0) and (0,0,1) is just the xz-plane, since every vector in it has the form (x,0,z).

So the orthogonal projection of a general vector $(x,y,z)$ onto that plane is simply
$(x,y,z)\mapsto (x,0,z),$

[flat frigate]

Paul Dirac

[split moss]

because you keep the x and z components and remove the y component.

wait . . .

[vernal tree]

see in the class I am taking it was explained as just using A(AA^T)^-1 A^T

[flat frigate]

Paul Dirac

[split moss]

Finally . . .

[split moss]

?

This is the orthogonal projection matrix

You can also dont use formula

Are you just saying that they gave a formula in class and you are trying to use that?

[vernal tree]

I figured it out#

I had it backwards

A^T(AA^T)A

[vernal tree]

yes

he didn't explain what it does

or what it is

[split moss]

Did you understand my procedure?

[vernal tree]

just this formula

[vernal tree]

yes

[split moss]

Then I would focus on the geometry first, because the formula makes more sense once you know what the projection is doing

[vernal tree]

I understand that it is projecting the x and z vectors

[vernal tree]

what's the geometry?

[split moss]

it is projecting onto the plane spanned by the x and z directions, not “projecting the x and z vectors

[split moss]

The geometry is: the span of (1,0,0) and (0,0,1) is the xz plane

[flat frigate]

Paul Dirac

[vernal tree]

So it's projecting something in 3D to a 2D plane along XZ?

[split moss]

Just a sec...

[flat frigate]

Paul Dirac

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[obtuse kettle]

for part b, how would u know when u should expand the brackets first before integrating and when to dive straight into using by parts or other strategies?
in the second pic, i initially did by parts but it looked a bit too complicated for just 4 marks

[opaque fern]

if your denominator is a single term, usually just expanding it works well

also the whole point of integration by parts is to make your new integral simpler than it originally was

your new integral is simply

ya know

worse

[obtuse kettle]

yeah 😭

[opaque fern]

so that wasnt a good idea

[obtuse kettle]

whenever i see a product or something where i can turn it into a product, i just jump into doing by parts

[opaque fern]

i mean try to make sense of it right

usually ibp is good if differentiating a term makes it vanish or reduce its order

[obtuse kettle]

ahh i see

[opaque fern]

like

x to 1

x^2 to 2x

[obtuse kettle]

gotcha

thankss!

.close

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[opaque fern]

ofc

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[fierce nimbus]

<@&268886789983436800>

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[fierce nimbus]

theres like 2 more channels with this too

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[twin prawn]

hello

[twin prawn]

is this correct?

im extremely unsure about the middle graph

[plucky elk]

Well no because your graph isn't a function at all

[twin prawn]

it fails vertical line test?

[covert rain]

in certain cases, a fixed x corresponds to several y's

[twin prawn]

in limits i assume

[twin prawn]

whats this have to do with it? sory for askng

[covert rain]

for example at x = -3 does the function you drew return only one value?

[twin prawn]

oh crap it fails the vertical line test

[covert rain]

yes

[plucky elk]

Show the entire question. Looks like you're supposed to graph a different function for each limit option

[twin prawn]

no

this is like all in one

[twin prawn]

the bullet points are properties of the function f(x)

[covert rain]

why is there a vertical asymptote at + infinity at -2^-?

It looks like you need to delete some unnecessary parts

[twin prawn]

the question or the graph?

oh woops

ill restart it, i got the asymptotes right at last

least

[covert rain]

@twin prawn did you manage to solve it?

[twin prawn]

im stuck :/

[covert rain]

why?

[twin prawn]

i did lim -2 from the left first and lim 2 from the right

that seems intuitive

[covert rain]

mark the conditions you have now and see which ones you are missing

[covert rain]

try to finish the vertical asymptotes

[twin prawn]

but i dont know how to do the vertical asymptotes tho

the -2^+ seems like an imposibility

[covert rain]

why?

[twin prawn]

i cant think of how to draw it

[covert rain]

however you can also delete the horizontal asymptote at y=0

[twin prawn]

wdym delete

like ignore it?

[covert rain]

mark

also the last point

you already have those conditions on the graph

[twin prawn]

yeah, it shows na 0 is the ha

[covert rain]

whats the problem?

[plucky elk]

Then show the entire question

[twin prawn]

this is from a friend and i think theyre asleep

i decided to do it as well for review for finas

finals

[plucky elk]

I think this works now

[twin prawn]

i see

[plucky elk]

,w plot -x/((x+2)^2(x-2)) for -3<x<3

[flat frigate]

Results provided by WolframAlpha

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

Requested by rie.mann

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[dry pivot]

Particle P travels in a straight line from A to B, the velocity of P at time tS after leaving a is denoted by V ms^-1 where V =0.04t^3 +ct^2 +kt. P takes 5s to travel from A to B and it reaches B with speed 10 ms^-1, the distance AB is 25m
i) Find values of the constants c and k
ii) Show that the acceleration of P is a minimum when t =2.5

I checked the marking scheme and for some reason they omitted a constant of integration when going integrating velocity to obtain displacement

My guess is that since c and k are already constants, c +k + C(constant of integration) is just another another so that's why they omitted it but that logic doesn't really sit right with me and I am very unconfident, could someone please explain why they didn't add a constant of integration?

[grand basin]

what is the integral of f(x) from A to B equal to, in general?

[dry pivot]

Wdym? Integral of 0.04t^3 +ct^2 +kt?

Integral of velocity is displacement

[warm warren]

general

[grand basin]

no, any function in general

[dry pivot]

Area under the curve?

[grand basin]

thats the usage of a definite integral

i am asking about the formula

[dry pivot]

Idk

[grand basin]

ok ok, you know that f(x) represents the derivative of F(x), yes?

[dry pivot]

Nop

Ohh, yea

[grand basin]

if 2x = f(x)
then F(x) = ?

[dry pivot]

I am used to writing it in a slightly different way 🤷‍♂️

[grand basin]

writing does not matter, it's the idea

[grand basin]

?

[dry pivot]

x^2 +c?

[grand basin]

very good

we call F(x) an antiderivative of f(x)

the same way we call f(x) a derivative of F(x)

now question

how do you express the integral of a f(x), from A to B

using F(x)?

[dry pivot]

integral(A to B)?

[grand basin]

yes, integral A to B of a function f(x)

do you know what that equals

in terms of the antiderivative F(x)

we have $\int^{b}_a f(x) dx = F(b) - F(a)$

[flat frigate]

Jun

[grand basin]

are you familiar with this?

[dry pivot]

Yup, ranges

Definite integration

[grand basin]

great

[dry pivot]

Upper limit - lower limit

[grand basin]

now you may ask me "Jun, where is the constant of integration?????"

easy.

you add a constant of integration to both sides, it simply disappears!

[dry pivot]

Nah it goes away with definite integration

[dry pivot]

😨

What

[grand basin]

we have $\int^{b}_a f(x) dx = (F(b)+C) - (F(a)+C)$

[flat frigate]

Jun

[dry pivot]

I see I see

[grand basin]

this why you hear people say "definite integrals means no constant"

its not that it doesnt exist

it DOES exist

but it just cancels out

now back to your problem

were you asked to "Give an antiderivative" or "calculate a definite integral"?

"giving an antiderivative" = there are many solutions, we denote all of them as "+c"

"calculating a definite integral" = using any given antiderivative and substituting the bounds, making the constant cancel out, so we end up with a clear value

i hope that clears out the difference

[dry pivot]

Right

Is my question one with definite integration tho? Would the bounds be 0 and 5?

And since theres no constant I guess we can just ignore the lower limit 0?

[grand basin]

of course!

you have bounds, A and B

that makes it a definite integral

[dry pivot]

I seeeee, thank you. I will re-solve just give me a moment

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[safe radish]

Please don't occupy multiple help channels.

[round egret]

<@&268886789983436800>

[radiant isle]

dont advertise your server here

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[winter whale]

damn

hello mods

long time no fight

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[real imp]

construct a triangle given two sides are equal to 6cm, 5cm with an enclosed angle of 30 degrees

[split kayak]

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[errant scarab]

4 6 7

[real imp]

1

[split kayak]

Do you know anything about trigonometry?

[real imp]

yea

@split kayak

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[dark widget]

How to solve this question? I tried with method of multipliers but still not getting the answer.

[safe radish]

@dark widget Has your question been resolved?

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@dark widget Has your question been resolved?

[tardy mango]

why do I feel like this question was typed incorrectly

[plucky elk]

Where's the PDE

Is PDE partial differential equation?

[dark widget]

I have tried this one, and i believe that certainly the question has some error.

[spark sentinel]

Where is $\frac{\partial}{\partial x}$?

[flat frigate]

YeetusDeletus5

[dark widget]

p= del z/del x, q= del z/del y

[plucky elk]

Amazing context

[safe radish]

@dark widget Has your question been resolved?

[safe radish]

@dark widget Has your question been resolved?

[dark widget]

<@&286206848099549185>

[hollow swift]

?

[wraith ember]

looks lowkeyuinely incorrect qn to me

i solved a similar qn and it was x+y instead of x^2 + y^2

Due to the non-linear term on the right, the solution is typically expressed as: Φ(u,v) = 0 Where u and v are the constants of integration C_1 and C_2 obtained from the auxiliary equations....

I dont have much knowledge about this tho

I might be incorrect too

[safe radish]

@dark widget Has your question been resolved?

[upper forge]

ts unsolvable 😭

[tawdry hamlet]

tried to do it, it was impossible. ask prof or whoever u got this question from about it

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@dark widget Has your question been resolved?

[dark widget]

.close

[tardy mango]

.close

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[median sleet]

this is for part a

[spark sentinel]

Lower bound for second integral is 0?

[median sleet]

should be no?

[spark sentinel]

The first integral’s lower bound is also 0

You might recount area here

[median sleet]

wait which second integral are you referring to here

[spark sentinel]

After the plus

[median sleet]

are you saying it shouldnt be 0?

[spark sentinel]

I think it would recount area otherwise, no?

[median sleet]

well im also only spreading the slices all up until the red line with the outter integral

thats the goal

<@&286206848099549185>

[median sleet]

.

<@&286206848099549185>

[golden jolt]

Ok, so when you set up your inner integrals, you seem to be doing it correctly, but when you do the outer ones you are including an extra variable term when it should just be evaluated from two different points.
The inner integrals take care of the equations while the outer ones just set the bounds essentially

Plus when its an area integral like this the f(x,y) should just be 1, I believe. Im not sure if the problems exactly saying that it should be f(x,y) in the integrand though

[median sleet]

yeah true they should be constant

so the inner integral slices the region with slices with respect to one variable which then finds each slice's area

and the outer one is just supposed to sum them up

my issue is that i was treating the outer integral the same as the inner one

[golden jolt]

yes that sounds right

[median sleet]

thinking that because its an integral it must also be summing up rectangles

but turns out that thats not necessarily the case and it just happens so the operation of summing all those slices is a sum that resembles the riemann sum

which can be rewritten as an integral

if im not mistaken

i mean i lowkey dont understand the bounds tho

like i get the intuition visually but not the evaluation

[golden jolt]

its kind of like a regular integral with its bounds. area integrals are kind of a fancy way of writing normal, single integrals. for example, the inner integral, along with the fact that the f(x,y) term is 1 just creates the function in which youre finding the area, then you integrate across the bounds as you would any single integral

i like to think of it as calculating area between two curves. when you do that, you have to subtract the bottom curve from the top curve. in area integrals if you have a function like x = y^2, you could solve for y and set up a double integral as shown in this image. the bounds of the inner integral would essentially create the difference between those functions, then the outer integral would just be the bounds themselves.

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[safe radish]

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[median sleet]

.reopen

wtv

[safe radish]

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[lean otter]

can anyone teach me how to solve normal parabola?

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[upbeat swan]

do you have a specific question in mind?

[lean otter]

I havent even started but my private teacher told me I should start today, but dont know where and how

I have some example exercises tho

[thin bridge]

show a question you need help with

[lean otter]

I mean the question is if someone can teach me from start to finish, or at least show some resources so I can start somewhere

[thin bridge]

khan academy

people are unlikely to teach you from scratch

but they will guide you along if you put the effort in

[lean otter]

they dont have this topic

[thin bridge]

ideally post specific questions or
note you want explained more clearly

[lean otter]

I see

[thin bridge]

search quadratics

[lean otter]

found some videos thank you!

its unfortunately the hardest topic of my finals so mastering this would free up alot of time

if you have more resources please send it

[thin bridge]

organic chem tutor on yt

[safe radish]

@upbeat swan Has your question been resolved?

[upbeat swan]

yes

youtube is great

especially organic chemistry tutor

just watch those videos and do the exercises and come back if you are stuck somewhere

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[echo gazelle]

excuse me

[echo gazelle]

how do i do dis?

im getting kind of confused

[fathom jewel]

What's confusing

[echo gazelle]

wat do i do now

how do i find the x and y values

[fathom jewel]

Well you have two identical conditions

[echo gazelle]

yerr

[fathom jewel]

So it's just one condition actually

2x+2y=2

which represents a line

[echo gazelle]

x + y = 1

[fathom jewel]

yes

[echo gazelle]

oemgeeeeeeeeeeeeeee

so 1,0 0,1?

[fathom jewel]

No

[echo gazelle]

oh..

[fathom jewel]

it's a set of candidates along the line

withtin that square region

y=1-x with 0=<x=<1

for example x=1/2 is also valid

[echo gazelle]

what... so theres like 100000 solutions it just needs to be on the line???

like 0.49 and 0.51

[covert rain]

I think you need to parameterize the curve

[echo gazelle]

wurt u mean

[covert rain]

like x=t and y=1-t 0<=t<=1

if I remember correctly

[fathom jewel]

No it just means you have the same local minimum along that line

[outer river]

these are only candidates for the absolute minimum

[echo gazelle]

o

[fathom jewel]

Now'd you have to check the boundaries too

[echo gazelle]

so u can use one and it'll be fine?

1,1?

[fathom jewel]

https://www.desmos.com/3d/ypyutdrziy

[fathom jewel]

yes, since it's asking for the absolute minimum, not where it lies

[echo gazelle]

ahh

[fathom jewel]

also since you have a region your function is constrained on, you need to check the boundary too

x=0, x=1, y=0 and y=1

[echo gazelle]

ok thank u

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[lavish magnet]

I have a small question about constant speed and constant velocity that I'm hoping someone here can clear up for me.

So if a car is moving at 30km/h North and then after 5 minutes it makes a turn when it makes the turn the velocity is changing but the speed is constant right?

My current definition of constant speed and constant velocity is:
Constant speed = Speed that stays the same during the change of motion, direction is irreveant because its a scalar.
Constant Velocity = Same speed and direction throughout the change in motion.

Please correct me if I have made any mistakes in my statements or observations.

[upper forge]

yeah speed= magnitude
velocity = magnitude + direction

[plush spruce]

Yes, speed is a scalar and velocity is a vector

[upper forge]

needs to remain same

[lavish magnet]

constant means pretty much that right?

[upper forge]

yeah

[lavish magnet]

That the speed/velocity should stay the same during the motion

[plush spruce]

Yeah, what you said is correct

[lavish magnet]

Okay thank you so much guys

.close

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[lavish magnet]

.reopen

[safe radish]

✅ Original question: #help-23 message

[lavish magnet]

In this video https://www.youtube.com/watch?v=pJphJVezld4&list=PLEs_zKbA7sG-GK4Xorc-UWYruQDqVMDYT&index=2 at timestamp 9:00 to calculate the gradient of the displacement time graph he did (5-20)/10-0, I'm confused as to why he did that, I know he used the y2-y1/x2-x1 formula to calculate the gradient but i did the same thing and I got (20-5)/10-0. He gets a negative value as the result and I get a positive value as my result, can someone help me find where I'm making a mistake?

[calm cloak]

make sure the x2,y2 corresponds to the same point on the graph which in this case maybe either of (0,20) or (10,5)

[lavish magnet]

here is the graph in question, im still struggling to see how my values are incorrect

[calm cloak]

when t=0, notice where the graph actually is

the graph is not the dotted line, if that helps clear the confusion

[lavish magnet]

ohhhh i see it now

when x is 10 y is 20, when x is 0 y is 5

final value of x is 10 which corresponds to the final value of y

i see where i messed up

thanks for ur help

.close

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[calm cloak]

what

bruh

[upper forge]

😭

[calm cloak]

when x is 10, y is 5

[upper forge]

shi ts is wrongerer (not a word though but you get the idea)

[lavish magnet]

wait what

ohh

bruh wtf did i just say before 🤦‍♂️

yeah yeah ur right idk what i was yapping

[calm cloak]

lol its okay

[lavish magnet]

i got it the other way around

[calm cloak]

yup

[lavish magnet]

when x is 10 y is 5, when x is 0 y is 20
y2 = 5, x2 is 10, y1 = 20, x1 = 0.

y2-y1/x2-x1 is hence 5-20/10-0 thus gives a negative gradient

okay double checked it this time so i dont look like an idiot

[calm cloak]

yeah

[safe radish]

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[graceful lichen]

Is there another way rather than considering many cases?

[brave wolf]

4x <= 5y is almost equivalent to x <= y for such low numbers

there is a very small number of exceptions

[graceful lichen]

Yes I considered that

[brave wolf]

have you found the exceptions then?

[graceful lichen]

In mind I calculate some numbers will satisfy this condition but some will not. I didn't find then

I'll check tho

Rn am eating

[brave wolf]

alr, lemme know once youre done

the basic idea is that once you find the exceptions, you can basically solve the question for the relation x <= y (which is much easier) and then deal witha those exceptions separately

[tardy mango]

<@&268886789983436800>

[safe radish]

@graceful lichen Has your question been resolved?

[graceful lichen]

@left gyro

Oops

@brave wolf

[brave wolf]

oh well

[brave wolf]

since youve already written it all out, you could just count it

4x <= 5y
is also true at (5,4) btw

otherwise its true where x <= y

[graceful lichen]

5<4 then how?

[brave wolf]

yeah, (5,4) is the exception

[graceful lichen]

Yes exception

[brave wolf]

these should be the ones where 4x <= 5y holds

so you can just count them now and get m

[graceful lichen]

16

Oh wow nice method bro

[brave wolf]

sounds about right

yea, 16

[graceful lichen]

Now n

[brave wolf]

its even possible to do it without writing it all out btw, if you can identify the exception (5,4), it remains to count the number of pairs where x <= y.
x <= y iff x = y (which is 5) or x < y.
There are 25 total pairs, 20 of which are such that x =/= y. Half of them will be x < y, so there are 10 with x < y.

5 + 10 is 15 x <= y pairs
Now add the exception (5,4) and you get 15 + 1 = 16

but in this case, its only 25 pairs, so you can afford to list it all out

[brave wolf]

for n, you need to make the relation symmetric

[brave wolf]

symmetric relations are symmetric around the diagonal

[graceful lichen]

xRy => yRx

^right?

[brave wolf]

yep

if you think about it, it basically means that its symmetric about this diagonal

[graceful lichen]

Row 1 and Column 1 gives symmetric relariknz

Oh yes

About the diagonal

[brave wolf]

so how many red dots do you need to add to make it symmetric

[graceful lichen]

26 is answer then

[brave wolf]

not quite

i think you forgot about the (5,4) dot which is already there (its in the relation, because 5*4 <= 4*5)

[graceful lichen]

11 dots?

[brave wolf]

hmm, i only count 9

[graceful lichen]

Yes got it 9

[brave wolf]

the total is then 25

retrospectively, there was an easier way to solve it

instead of focusing on m and n alone, you could realize that to make R symmetric, you'd need to add all the remaining pairs

and so m + n is just gonna be the total number of pairs in A x A

which is 25

[graceful lichen]

Yes, That will be all elements in A x A. Then Which remaining pairs are you talking about about?

[brave wolf]

by remaining pairs, i meant all the pairs that arent in R but are inA x A

[graceful lichen]

So we remove those pairs from A xA that are not satisfied by the pairs in R

[brave wolf]

i mean that m is the pairs that are in R
n is the pairs that you need to add. Which is all the remaining pairs in A x A, which are not yet in R.

Either way, the total is just the pairs in A x A

[graceful lichen]

Yes because R is subset or equal to A x A aswell

So R will contain all elements of AxA

[brave wolf]

yeah and R surely contains all pairs with x <= y, so youll have to ad all the remaining pairs

[brave wolf]

after you try to make it symmetric, yes

before that, it only contained 16

[graceful lichen]

Yes symmetric with the given condition 4x<=5y

So we can conclude that our condition is 4x<= 5y(initial), it says n is from A x A and that R to be proved symmetric. For this we can add all the pairs which do not satisfy the initial condition after adding set R is ready to be called symmetric

Therefore 25 elements of A x A

@brave wolf

[brave wolf]

there are 25 elements of A x A because there are 5 elements in A, so you do 5^2

i think youre confusing a couple of things here

R contains all the pairs (x, y) which satisfy 4x <= 5y (there are 16 of those, but you dont need to know that)

m counts the number of pairs in R (so m = 16)

then n counts all the pairs that we'd need to add to make R symmetric

but since R contains all the pairs with x <= y as a subset, we will need to add all the remaining pairs with x > y to make it symmetric. So to make R symmetric, we'd need to add all of A x A - R (i.e. all the pairs, that arent yet in R)

but then n + m is just |R| + |AxA - R| = |AxA| = 25

and we dont even need to know what |R| itself is

but this approach is a bit complicated, so its also okay if you just list it all out as you did

[graceful lichen]

but i gotta adapt this conceptual method to rise my thinking

[brave wolf]

ill try to explain it with pics then

[graceful lichen]

thx you

[brave wolf]

so R is the set of all pairs with 4x <= 5y

notice that a subset of this are the pairs with x <= y, do you understand why?

if we just imagine drawing it (you dont actually have to draw it, just the mental picture suffices), this would be a subset of R

and the size of it would be m

but then to make it symmetric, youd certainly need to add all this. Otherwise it cannot be symmetric

and in total (so m + n), it makes all of the pairs

and how many pairs are there?

5 x 5 = 25

so the answer is 25

[graceful lichen]

Got it

Thx

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[lofty viper]

Could someone help me with this?
(BC=12√3 also, not 12/3, the translator just sucks)

[zenith maple]

Try to find all the angles and use trigonometry

[lofty viper]

Ty!!

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[rich dune]

Whys the integrating factor here x4+5?

Why not 4x3 over that

Nvm im dum

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[rare haven]

if a function is not differentiable in any of its points does it lack a derivate or does it have a derivate but its domain/the original function's differentiable domain is the empty set?

[plush spruce]

Well, if the domain is the empty set there is no application of the function, i mean, there are no inputs

[visual kraken]

Yeah, you could say it has a derivative that isn't valid anywhere, but that's essentially the same as just not having a derivative

[minor goblet]

id say its more correct to say the derivative doesn’t exist but depending on the context theres no reason to not say the second

[errant finch]

it's more useful to say the set of all points at which the function is differentiable is empty

functions on an empty set are not very interesting objects

[rare haven]

ty

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[pearl field]

Why can a loan of 10 thousand dollars with 20% interest for 46 months turn into 43 million 887 thousand dollars?

[vague phoenix]

exponential

[stoic dune]

Is that 20% per month? That's 791% interest per year, which is how we usually express interest.

And yeah, 791% interest can make anything happen

You likely did not mean to use 20% per month

[normal moss]

Well, you start with $10,000. Next month you have $12,000. And then you get 20% on this new balance, so + $2,400 = $14,400. And then 20% on 14.4k, and so on. Notice, each month you get interest on the previous interest, not the original amount. Which ends up skyrocketing fast

But yeah, 20% interest per month is ludicrous.

[quiet juniper]

I dont wanna borrow from this mafia

[normal moss]

I would be happy to be the owner of such an organization, lending out these loans

[pearl field]

Yea drug dealers be like

[pearl field]

But how? It’s just 10 grand

And it’s not 43 mil?!!!

[stoic dune]

Again, 791% interest is insane

30% is considered a "loan shark rate"

[normal moss]

I don't understand, did you read my message?

You begin with $10,000. You end up with approx. $43,887,144 after 46 months

[clever tinsel]

,w 10000(1.2)^46

[flat frigate]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by yukarin_167

[clever tinsel]

that high a monthly interest works wonders.

[normal moss]

I personally wouldn't even stop at 20%

Never seen before ∞% interest

[clever tinsel]

now if a bank would offer this interest on FD accounts...
alright, sorry, sidetracking.

[pearl field]

Don’t be passive aggressive

[clever tinsel]

so do you still have any further questions?

[safe radish]

@pearl field Has your question been resolved?

[pearl field]

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[pearl field]

can anyone tell me the rules to know if a nubmer can be divided by :

2
3
4
5
9?

[safe radish]

Please don't occupy multiple help channels.

[quiet plume]

That's like three wildly different questions within 5 minutes lol

[clever tinsel]

respectively:

• last digit is even
• sum of all digits individually is a multiple of 3
• last two digits is a multiple of 4
• ends in 5 or 0
• sum of all digits individually is a multiple of 9

[twilit spindle]

2: last digit even
3: sum of digits divisible by 3
4: sum of last two digits divisbile by 4
5: last digit 0 or 5
9: sum of digits divisible by 9

[pearl field]

also 8

[twilit spindle]

sniped 😔

[pearl field]

i need also 8

[pearl field]

actually 2 is if the last number is 2, 4, 6, 8 or 0

[clever tinsel]

last three digits divisible by 8.

[twilit spindle]

in general, write (a_0a_1a_2\dots=a_0\cdot 10^0+a_1\cdot 10^1+\dots) then reduce (\bmod D)

[flat frigate]

ΠαϳαμαΜαμαΛλαμα

[pearl field]

but 6?

[clever tinsel]

and what about 0?

6 is even number + rule for 3.

[pearl field]

okay.

.clos

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[ancient kernel]

pls hint for this problem, topic is pigeonhoe principle

[twilit moat]

I'm guessing you should look at the number of arrangements of 2x2 squares and then the number of possible arrangements of colours in those squares

[ancient kernel]

theres 81 ways to color a 2x2 square

and 54 2x2 squares in total

so at least 2 of them have the same cooloring but idk how thats useful

[twilit moat]

I think you can also maybe discuss the relationship between squares as squares share boundaries

[ancient kernel]

hmm, could you elaborate a bit more i don't understand

[small rampart]

Look at the strips of 4

[ancient kernel]

yes

what about them? two of them have teh same color

[small rampart]

Each strip must have at least 2 squares of the same color

[ancient kernel]

yes

[small rampart]

Define a strip to be red if it has at least 2 rep squares,... etc

[ancient kernel]

oh

[small rampart]

What can you say about the number of red, blue, yellow strips (by pigeonhole)?

[ancient kernel]

oh i see it

[small rampart]

There must be at least how many strips of the same color?

[ancient kernel]

i think i got it

thanks for the help! really appreciate it.

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[safe radish]

Please don't occupy multiple help channels.

[main mural]

<@&268886789983436800>

[lean otter]

lol

mrbeast

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[open wedge]

WHO SNIPED MODS

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[fathom jewel]

. @hazy vault

[hazy vault]

#help-12 message is this how you link message?

[fathom jewel]

that's better

Do you understand at least the issue I pointed out

[hazy vault]

yes I do understand. it was why I was skeptical with the proof in the first place

[fathom jewel]

Ok now can you at least use double integrals?

[hazy vault]

I can use it personally but the lecturer doesn't want us to use it

[fathom jewel]

Ok

[hazy vault]

Is there any other method other than double integrals?

[fathom jewel]

I am trying to think of how we can justify the limits

when you use IBP

I mean

but ok wait

we dont know if lim a and lim b exist that's the whole problem

[hazy vault]

ah yeah it's kinda cooked

[hazy vault]

then doesn't it depend on F_X(a)? fubini oesn't come across that problem

[fathom jewel]

well you have 0 * oo

[hazy vault]

I can't think of anything other than F_X(a)-1=-P(X>a) and it's<=E(X^n)/a^n in the discrete case

but that's the discrete case

and we're trying to prove it in the continuous case by using limit

[hazy vault]

ahh

[fathom jewel]

oops I mean greater or equal in the last inequality

[fathom jewel]

But yeah this should justify it rigorously why it has to go to 0, since we also know IE(X^n) exists

[hazy vault]

ahh tysm

[fathom jewel]

You should have encountered the squeeze theorem I suppose

[hazy vault]

yeah I have thankfully

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[plush spruce]

Is this again about induction?

[spiral saddle]

ye

[plush spruce]

What have you tried and where are you stuck?

[spiral saddle]

say for example that

the side is of length 2^n

and the other side is 2^n - 1

then we need to prove that 2^n / L is a integer

and that (2^n-1) / L is an integer

[plush spruce]

Knowing this is induction, have you tested the base case?

[spiral saddle]

n = 1 ?

yeah, take n = 1 then we have 2x2 = 4 squares you remove 1 we have 3

[plush spruce]

okay, in the test case, it perfectly fits one piece in L right?

[spiral saddle]

ye

[plush spruce]

okay, have you tried the inductive step?

[spiral saddle]

its hard

[plush spruce]

yeah

i think of creating divisions on the board

mmm

i think i got it

[safe radish]

@spiral saddle Has your question been resolved?

[plush spruce]

This is what I thinked

You subdivide this in 4, each of the 4 subdivisions has one corner without the piece, you take the corners that are colliding in this 4 subdivisions, and you could introduce a L piece, the unique thing I'm missing is one square

[spiral saddle]

can you make a drawing with a down to earth example

[plush spruce]

btw, the exercise is tricky, I am spanish,and I don't know if:

"Probar que es posible cubrirlo usando fichas en forma de L"

Refers to, prove that you can to cover the board leaving one edge without a squar, or if it means, that you have to prove you can cover the little square.

I don't know If I'm explaining myself

I'm pretty sure that you cannot cover just a square using Ls without putting one above the other

Let me draw it for a 4x4

I'm still drawing xD

Something like this

The red L is to clarify that is the last L you need

[spiral saddle]

you are spanish?

[plush spruce]

Yes

[spiral saddle]

the third one

[plush spruce]

So, for a 4x4 it's possible to fill a board leaving that corner empty

That's what I think of the exercise, I don't think it's possible to fill just one corner with Ls

So I think what is asking is to prove that it's possible to cover a board leaving one corner empty using Ls

[spiral saddle]

yes of course

[plush spruce]

So that drawing is the main idea,that you can use Ls in the subquadrants, rotate them and use more Ls

[spiral saddle]

help

[plush spruce]

You already had that idea?

[spiral saddle]

visually is simple but with math is hard

[plush spruce]

You may already knew that and only ask for the math

[spiral saddle]

correct?

[spiral saddle]

well there must be a pattern somehow

[plush spruce]

yeah

prove this mathematically is going to be another thing xDD

[spiral saddle]

any ideas?

i mean, as long as we can do it like you did

divide the square into subsquares of 2x2

the solution will always be possible

[spiral saddle]

yes something like this, but with math

[plush spruce]

well, you divide it into subsquares 2^n to prove it for 2^{n+1}

I'd first try to setup the board mathematically

I mean

I'd say that the board is something like T = { 1,.., 2^n } x {1, ... , 2^n}

and the elements are the "tiles"

are you following_

this is kind of hard tbh😅

[plush spruce]

With this, the Ls occupies 3 tiles of T, this Ls are disjoint from each other, and the union of the Ls has to be T \ { c } being c the square we don't have

[split kayak]

You dont really even need induction

Like, by showing that the "n+1" grid is just a collection of 4 "n" grids.

[spiral saddle]

the exercise is on the induction section

[split kayak]

mb

[plush spruce]

finally a professional came to the help xD

I'm also a bit stuck

[spiral saddle]

trinome?

[plush spruce]

I came to the help but got outplayed :/

[plush spruce]

oupss, I meaned Ls

[quiet plume]

It's part of a section they're doing on induction but the issue is that the problems they've chosen to put in there are almost easier to frame with without induction kek

[spiral saddle]

can you help

[quiet plume]

You've done the base case haven't you?

[spiral saddle]

i got stuck with the base case

[quiet plume]

How?

n=1. You have a 2x2 grid. Remove one corner. What are you left with?

[spiral saddle]

3 squares

[quiet plume]

Can you cover three such squares with a three-piece L?

[spiral saddle]

with am L

[spiral saddle]

ye

[quiet plume]

So your base case is done..

For the inductive step, you have a 2^(k+1) by 2^(k+1) grid. This gives you four 2^n by 2^n grids. One of the corners is removed from one of those 4 grids, so we can deal with this one with the induction hypothesis.
What about the other three? We would like to place an L somewhere such that what you're left with is three 2^n by 2^n grids with one corner missing.

You can draw it to try things out.

[spiral saddle]

idk how to say it formally

[quiet plume]

If you don't have an intuitive understanding of how you can do this by drawing it first, then how do you think you'll write something that is formal.

Again formal doesn't mean a syntactic logic proof. It mostly means convincing. Demonstrate that you're able to do this for smaller square. Argue why this process can also be done in the general case.

[spiral saddle]

is like what @plush spruce said, but formally is hard

[spiral saddle]

how?

[quiet plume]

Argue why this process can also be done in the general case.

[spiral saddle]

is hard

[quiet plume]

• Why was this specific L chosen in particular as opposed to another L you could have chosen?
• Why does this simplify to 4 smaller cases of 2x2 squares missing a corner?
• How does this relate to 4 smaller cases for a general grid like the one you have to deal with?
• If you have a 2^(n+1) by 2^(n+1) square can you do the same? Is there a "middle" 2x2 square you can place an L in in the same manner?
• Would this also reduce to 4 smaller cases that the induction hypothesis can deal with?

If you answer those questions you'll have your proof.

[plush spruce]

Now that you zoomed in I'm embarrassed about my artistic habilities xD

[spiral saddle]

• i am not sure
• i am not sure
• because its the same except here we have n = 2
• maybe
• not sure

[quiet plume]

Then you have to think about it and sketch things until you can answer those questions.
If you're not convinced yourself because someone has done the work for you, how are you meant to write down a proof to convince your professor?

[late thorn]

you can prove via induction that any L shape that consist of 3 squares with sides 2^n can be filled with small L triminos

[spiral saddle]

• because otherwise is impossible to make it work
• not sure
• because is the same except we have here n = 2
• if its n = k + 1 we need to do something similar notice that in that case we have one 2x2 square and 3 of 3x2 recrangles, then we can intercake the Ls put them one upside fown obe upside up and fill the entire 3x2 blocks
  -unsure

@quiet plume

[quiet plume]

because otherwise it's impossible to make it work
Alright, but what about choosing this specific L made it so you get 3 other 2x2 squares missing a corner?
If the initial missing corner was different, how would you change the L accordingly for instance?
In the 2^(k+1) by 2^(k+1) grid, do you have a middle 2x2 square? Can you place an L like in the 4x4 case? Does this result in the same thing as in the 4x4 case, namely 3 other smaller squares with a missing corner?

Again answering all those questions with certainty will give you the proof.

[spiral saddle]

i already explained the k + 1 case

[quiet plume]

You talked about 3 3x2 rectangles.

Draw it again. Try for an 8x8 square.

Try to do the same as in the 4x4 grid.

The goal is to break it up into smaller squares that you know you can deal with using the induction hypothesis.

You don't need to cover the whole 8x8 grid missing a square, but just to see if you can use the same argument as for the 4x4 case.
If so you've got two good examples that seem to generalize to each n, saying that by placing some L, you can reduce the problem to smaller cases. This is exactly what you want to do for induction.

[spiral saddle]

@quiet plume

[quiet plume]

This is 5 by 5.

It doesn't have much to do with the problem.

[spiral saddle]

@quiet plume

[quiet plume]

And again, you don't need to cover the whole thing.

The important part is for you to see that you end up with four 4x4 grids, each of which is missing one corner.

And that in general, if you do this with a 2^(n+1) by 2^(n+1) grid, you end up with four 2^n by 2^n grids , each of which is missing one corner.

[spiral saddle]

it's still impressive on it's own, no?

[spiral saddle]

why

[quiet plume]

Because you're concerned with square grids that have powers of 2 as sides.

The argument you're currently working on doesn't apply to a 5x5 grid because there is no middle 2x2 square.

[spiral saddle]

how to formalize

[quiet plume]

Well now that you've seen it work for 2 cases, think about the questions I sent above. Why does choosing to place the L in this specific way give 4 smaller grids that all miss a corner?

[spiral saddle]

idk how to explain it you just rotate the grids to form L with the empty spots

@quiet plume

[quiet plume]

It's again important that you think of those examples not as exhibiting how you can cover the 4x4 or 8x8 grids with Ls explicitly.
They should exhibit that from such a grid, the placement of an L strategically in the middle 2x2 square ensures you get 4 smaller square grids missing a corner.

This is how you prove things using induction.

[safe radish]

@spiral saddle Has your question been resolved?

[spiral saddle]

ok

but idk how to describe it with math

[safe radish]

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[spiral saddle]

.close

[safe radish]

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[calm prawn]

im not sure if you have to completely describe it with math alone, im pretty sure it is to get you to do it in that inductive process (start with base case --> assume true for n = k --> show true for n = k+1 --> if true for n = 1, then it is also true for n = 2,3,4 ... )

[primal gazelle]

this is a very nice one its quite well known

[safe radish]

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[hallow agate]

What is 2a+ 2a

[scenic phoenix]

4a

[hallow agate]

Ok

Thank u

What about finding x

[scenic phoenix]

from what? there is no x

also yoiu would need an equation

[hallow agate]

Like

For example

2x +5x is 8 what is x

[scenic phoenix]

we get 7x=8 correct?

[hallow agate]

I don’t understand

[timid snow]

You can combine like terms, since we have 2 and 5 times x that is equivalent to 7 times x =7x

[hallow agate]

Oooo thank u

[timid snow]

No problem!

[timid snow]

How would you solve this for x from here?

[scenic phoenix]

we can divide both sides by 7 to get $x=\frac{8}{7}$

[flat frigate]

pinkishnova cgt queen

[hallow agate]

Thank u

how do u solve x for angles

[timid snow]

do u have a specific question u are doing? like is it a triangle and if so what kind

[hallow agate]

Like a alternate extriror angle

[timid snow]

ok of what shape?

if u have a picture that would be helpful too

[hallow agate]

Like a two parallel lines inside the lines there are two angles that both equal 65

And there is one big angle

That

Is 180

[timid snow]

Is it like this?

[hallow agate]

Yes

Yes

[timid snow]

Ok

which angle did u want to know? the 115° one?

[hallow agate]

No how to solve x

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[summer coral]

don't make helpers draw it for you, please provide your question so its easier to help

[timid snow]

to solve for x we need to know what x is representing first

[summer coral]

or draw it yourself so we dont have to keep guessing

[hallow agate]

Ok

Good point

Wait

One minute

I will draw

[timid snow]

ok 👍

[hallow agate]

Oh no

I draw it wrong

One sec

[timid snow]

ok no problem

[hallow agate]

I think x is 180

Must be

[timid snow]

angles around a point add to 360°, so x is 360-180-65=115°

[hallow agate]

Oooh

Oh my god

[timid snow]

also, 180° is half a circle

so the other half must also be 180°, meaning 65+x=180
x=180-65=115°

[hallow agate]

No a reflex angle is half a circle

Ok

[timid snow]

how many degrees is a full circle ?

[hallow agate]

360

[timid snow]

yes

so half a circle is half of 360, so 360/2?

[hallow agate]

Ok

I one

Sec

Half is 180

Oooo

I understand what you’re saying

[timid snow]

a reflex angle is between 180° and 360°

[timid snow]

great

[hallow agate]

Ok

Thanks for help

[timid snow]

no problem

[hallow agate]

Bye

[timid snow]

bye!

[hallow agate]

Talk next time

[safe radish]

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[merry grove]

question is : The number 916238457 is an example of a nine digit number which contains each of the digits 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. How many such numbers are there?

why cant i do case by case basis where:

1. i have 2 together 1 seperate first make sure the 6 is behind the 12345 therefore 5x(7x3C2x6x3P3) because u pick 3 out of 2 and put them into either 7 slots and then u have the final where u can pick the last number in 6 and then u order them
2. 789 are altogehet 5x7 x3P3
3. 789 are seperate 5x7C3x3P3

altogehter cases is 5040 and the actual answer is
2520
any combinatoric masters that could possibly debug my solution ik there is a faster solution but i couldnt think of it cuz i gaslighted that my solution is correct

[dull sequoia]

Here’s how I think about it

? 1 ? 2 ? 3 ? 4 ? 5

Here’s where the 6 can go

So there’s 5 slots for the 6

After you put the 6 somewhere, there are now 6 numbers, and the next number can go in any of the 7 slots (7 slots because now the numbers can go after 5, but previously 6 couldn’t go after 5)

After this there’s now 7 numbers and 8 slots for the number 8

Similarly again, after this there’s now 8 numbers and 9 slots for the number 9

,calc 578*9

[flat frigate]

Result:

2520

[safe radish]

@merry grove Has your question been resolved?

[summer coral]

<@&268886789983436800>

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[safe radish]

Please don't occupy multiple help channels.

[clever tinsel]

please do not open multiple channels with the same question.

[misty ice]

I am sorry

[dusky totem]

Can you open multiple with different questions?

[solar hazel]

no

[dusky totem]

Oh

[clever tinsel]

the official stance (as in via #❓how-to-get-help) is no. but if you are not sure just ask the mods.

[safe radish]

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[trail gust]

I have completed parts a) and b), but I am unsure of how to approach part c).

[trail gust]

From the last two parts, I know that the general solution to the PDE, excluding the boundary conditions (which I frankly do not understand the purpose of for this problem), is u(x, y) = f(y + 4x) + g(x - y).

The PDE can also be factored as (u_x - 4u_y)(u_x + u_y) = 0, from which one can solve the two first order equations separately.

The characteristic curves I found from solving u_x - 4u_y = 0 were of the form 4x + y = C, and the ones I found from solving v_x + v_y = f(y + 4x) were of the form y - x = C.

I suspect that part c) has something to do with these curves, but I cannot see what.

[safe radish]

@trail gust Has your question been resolved?

[opaque fern]

I mean those are initial conditions

Not boundary conditions

Yes you don't need to actually solve for the exact equation using those, but their existence establishes that information is propagating outward from the x-axis as y increases.

[trail gust]

Oh, I misspoke.

Thank you for the correction.

[opaque fern]

Well yhe characteristic curves are gonna be y = x and y = -4x if you plug in those initial conditions

Do you know of d'Alembert's principle?

[trail gust]

I do know of it, yes.

I.. am afraid I do not understand how it will help me for part c) though.

[opaque fern]

Well like

If you pick some arbitrary point (x,y) in the plane, the value of your solution at (x,y) is determined entirely by the initial conditions on the segment of the x-axis between where its charactristics intersect the x-axis.

[trail gust]

Then is it the case that the only points on the plane that are affected by a change in value at (0, 0) are the points on the curves that pass through the origin?

Specifically, the ones on the curves 4x + y = 0 and y - x = 0?

[opaque fern]

I think that depends entirely on what values you change

[trail gust]

Hm, I'm afraid I do not follow what you mean.

[opaque fern]

Well like

I agree if we are talking about u(x,0)

Changing the initial 'velcoity' aka u_y(x,0) would not agree with your statement

Because the function lives inside the integral, so the integral will 'catch' any disturbances at x = 0 as long as 0 falls within those integration bounds

[trail gust]

Ah, I see.

[opaque fern]

Maybe i misinterpreted your question sorry

[trail gust]

Well, it is certainly a bit ambiguous.

This is a bit troubling. I think I will skip this part for now, and perhaps come back later on.

[opaque fern]

Ah understandable

[trail gust]

I do not want to spend too much time hashing out something that may not be extremely important anyways, so I will put this on the shelf for now.

[opaque fern]

Seems fair enough

[trail gust]

There is a lot of other review questions I need to get to as well, after all.

In any case, I appreciate your help as always, Lex.

[opaque fern]

You bring a lot of interesting questions here

[trail gust]

Thank you for your time and effort. It means a lot to me.

[opaque fern]

Of course! Anytime 💜

[trail gust]

In that case, I will close this channel for now.

But expect me to ask plenty more in the coming days, for my exam is next Tuesday.

.close

[safe radish]

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[opaque fern]

And expect me to be there to help you with them :D

[safe radish]

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[upper forge]

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[upper forge]

1

[safe radish]

@upper forge Has your question been resolved?

[upper forge]

<@&286206848099549185>

[dusky totem]

Hint: for the first equation first exponentiate both sides then spam log(a)+log(b) = log(ab)

[astral cliff]

Hey this is irrelevant but what's the best ai for maths doubt ??

My guy ?? Sorry for ping but it's urgent @dusky totem

[dusky totem]

Idk I have chatgpt plus

I don’t use it a lot for math tho

[upper forge]

ook lemme try

[warm warren]

!noai

[safe radish]

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).

[dusky totem]

No but AI can be seriously big brain quite a few times

[safe radish]

@upper forge Has your question been resolved?

[clever tinsel]

for a learner who cannot verify their output I'd still recommend against it.

[upper forge]

.close

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[novel iron]

Can someone please tell me why the following inference rule is logcally sound?

[novel iron]

The ansswer sheet is the following:

oh

no Ij ust realized

the logical conclusion can be true, even if the premise is false

F -> T yields true

oh nvm

.close

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[slate tapir]

hlo i need help

[slate tapir]

guys

in hcf and lc

lcm

in theri word problems

their

how to know which is hcf and lcm

[round egret]

hcf refers to higest common factor while lcm refers to lowest common multiple

[slate tapir]

ye ik that

but in word problems

it is complete diff sory

story

[round egret]

try to look for keywords maybe