help-26

[noble laurel]

he is my master 🙈

[ionic oar]

@inner dove

[inner dove]

I didn't understand

do you mean x=tanx?

i am asking for first step🙈

@ionic oar

@noble laurel

[inner dove]

Bob the solver

[ivory sorrel]

First re-write this as $\int_0^{\infty} ln(\frac{x^2+1}{x})\frac{dx}{1+x^2}$

[noble laurel]

\left( \right)

[thorny flame]

ƒ(Why am. I here)=I don't Know

[ionic oar]

yeah

[inner dove]

ln 1/x dx

[ivory sorrel]

uh, I'd suggest $u=arctan(x)$

[thorny flame]

ƒ(Why am. I here)=I don't Know

[inner dove]

Now?

ln 1/tanx dx

u=arctanx
Du= 1. Dx/(1+x^2)

u=tan^-1x
x=tanu
X^2=tan^2u

[ionic oar]

[ 2\int_0^{\pi/2} \ln \sec x \dd{x} - \int_0^{\pi/2} \ln \tan x \dd{x} ]

[thorny flame]

Bob l'éponge

[ionic oar]

right you got it

[noble laurel]

It’s 3:25 am

[inner dove]

3:55PM

[noble laurel]

, ti

[thorny flame]

The current time for austinu is 03:25 AM (PDT) on Sun, 23/06/2024.

[noble laurel]

@inner dove

[inner dove]

You sikly silly

Go n sleep

[noble laurel]

You silly silly

We’re you chanting my name?

[inner dove]

I like uour name

[noble laurel]

[inner dove]

How you made that

Ultra pro coder

[noble laurel]

The cobra is

[inner dove]

Hahaha

Traiotor neon🫣

So you have nitro

[topaz sinew]

@inner dove Has your question been resolved?

[inner dove]

I got πlog2

[topaz sinew]

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[tulip verge]

This is a problem I came up with but it's generic enough that I imagine someone has thought of it before. I think I solved it but I wanted to know if anyone recognizes it and could point me to some resource that elaborates on it.

[tulip verge]

You have a ring of N keys, and you're trying to open a door with a very special lock. At any one time the door needs one specific key that you have to open it, but you don't know which one. And every time you try a key that doesn't work, the lock re-arranges itself according to some algorithm, and may now require a different key.
Is there a sequence of keys that you can try that guarantees you will be able to open the door, no matter the algorithm and the initial key requirement?

[torpid sparrow]

Re-arranged with a pattern or randomly?

[tulip verge]

Yeah I forgot to mention, with a specific pattern.

Since there are finitely many patterns or algorithms for how the required keys change, theoretically you could just keep track of all of the combinations and rule them out one-by-one. I'm just wondering if there's some convenient writeup for it.

[torpid sparrow]

How do you know they are finite?

[tulip verge]

Well because every algorithm can be boiled down to what keys the requirement shifts after every incorrect try. There are n*(n-1) key required-incorrect key tried pairs, and n possible shifts for each which grants n^(n(n-1)) algorithms. And then n possible first key requirements for a final n^(n(n-1)+1) initial states.

[torpid sparrow]

According to what I read, the exercise’s doesnt specify there not time involving. For example, the locker could have a timer that decides the next key based on how long it took to decide for you to choose the key.

Maybe the question should have the restrictions well defined, or maybe I am not reading it correctly

In any case, if you know all the algorithms the only way to guarantee it would be finding the algorithm, and for that you should try them all

[tulip verge]

Have you ever seen a question similar to this before?

nvm, ty anyways

.close

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[mystic siren]

explain how he could prove this, for case a > 1

[mystic siren]

so you have to prove this of course

for a > 1 this is obvious so this is understandable

so now he needs to prove this

which is this ofc

and using the bernouli inequality

he just has to make this holds true

but he DIDNT

how do i prove this?

[pseudo jetty]

wdym he DIDNT

[mystic siren]

he just assumed it's correct

[pseudo jetty]

well intuitively like

epsilon is some positive number

and you can just choose n to be ginormous

and obviously itll surpass a

[mystic siren]

this is bs

[pseudo jetty]

[mystic siren]

it could be proven better

[pseudo jetty]

well thats why i said intuitively ...

if you want the proof

a > 1 so a - 1 > 0

[mystic siren]

thats einstein level inution

[pseudo jetty]

then you want εn > a - 1 > 0

this is true by the archimedean property of the reals

[mystic siren]

which is for every real number thare's a larger natural number right?

[pseudo jetty]

well that is true...

[mystic siren]

this should've been in the proof

[pseudo jetty]

but its also for every pair of positive real numbers x, y, theres a natural n such that nx > y

here, you're applying it to ε and a - 1

[pseudo jetty]

i mean

its in some sense "obvious"

[mystic siren]

this was the final step right?

[pseudo jetty]

sure

[mystic siren]

no definetely not obvious in the 3rd calc 1 lecture

[pseudo jetty]

its not obvious that you can scale a small number really large by multiplying it by a ginormous number?

[mystic siren]

he didnt even show the proof for the archimidean property i think

[pseudo jetty]

its not meant to be formal reasoning

like

[mystic siren]

it's not obvious if you dont tell it in the proof

[pseudo jetty]

just from daily life

surely you've multiplied numbers before

[mystic siren]

this is a regirous alegbric proof

[pseudo jetty]

well

[mystic siren]

im just saying

[pseudo jetty]

not every step needs to be filled in if there are parts that are "obviously true"

but okay sure if you want a more complete proof

go fill in the gaps

[mystic siren]

this was not

[mystic siren]

yes

ty mate

.close

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[pseudo jetty]

👍

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[still goblet]

Can someone explain me why N> θ^2/r

[topaz sinew]

@still goblet Has your question been resolved?

[shadow salmon]

it isnt a conclusion

but rather a premise

so for n > θ^2/r, then...

[still goblet]

Yes but why specially θ^2/r

[shadow salmon]

To find the concluding bound i would assume

Theta is fixed, while r is some abritrary postive number

so we can find n large enough such that n > θ^2/r

[still goblet]

How theta is fixed though

[shadow salmon]

or in other words, that is why we probably used something like θ^2/r, since we have a nice bound to then conclude the limit

[shadow salmon]

Well what is it?

[still goblet]

An angle

[shadow salmon]

So some fixed number

if it wasnt, we would have to fix it anyways

[still goblet]

Yes but it can be any number

[shadow salmon]

Yeah any number, aka fixed

just different wording

[still goblet]

“While r is some arbitrary positive number “

It’s the same ?

[shadow salmon]

yeah

i thought it sounded better because saying fixed twice in the same sentence doesnt roll off as nicely imo

[topaz sinew]

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[still goblet]

Yes

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[dawn abyss]

is this the correct inverse of the function y=(x+7)/(x-5) -- (-7-5x)/(1-x) =y ?

[shadow salmon]

I would recommend writing down the process of arriving at it

and explain where youre troubled

[dawn abyss]

i multiplied both sides by x-5 and got xy-5x=y+7

i subtracted 7 and xy on both sides and got -7-5x=y-xy

factored the right hand side and got y(1-x) and divided that on both sides and got

(-7-5x)/(1-x) = y

[shadow salmon]

You multiplied by x-5?

I guess you have a hidden step, that is rewriting

y=(x+7)/(x-5) as x=(y+7)/(y-5)

[thick lily]

yeah that's right, I'd probably multiply the numerator and denominator by -1

[dawn abyss]

what?

[thick lily]

if you know linear algebra there is a trick to doing it fast

[dawn abyss]

im in alg 2

[dawn abyss]

but then i would have to multiply y by -1, no?

[shadow salmon]

How did the RHS suddenly become y+7? instead of x+7?

It's confusing, since you said you mulitplied by x-5

but that should be y-5 if you keep the consistency

[dawn abyss]

sorr thats what i meant

y-5 on both sides

[shadow salmon]

I see

It's good practice to be precise when writing this stuff down, that way; you dont have to be unsure if you've been correct or not

So in this case I can see that you've done everything right and as Mero pointed out it's correct!

[dawn abyss]

but when i put it into mathway its smth else

[shadow salmon]

Something else? As in written in different way I suppose?

It might be the same

[thick lily]

multiplying numerator and denominator by the same number is the same as multiplying by 1 so doesn't change anything

[dawn abyss]

they multplied both sides by -1

[shadow salmon]

as Mero pointed out, you can take away a factor of -1

[dawn abyss]

but u have to multiply y by -1

then, no?

[thick lily]

nope, here's an example for you

$$y=\frac{3}{2}$$
$$y=1*\frac{3}{2}$$
$$\frac{-1}{-1} * \frac{3}{2}$$
$$y=\frac{-3}{-2}$$

[thorny flame]

Mero

[thick lily]

1 is special in that multiplying anything by 1 doesn't change it, so we're really writing 1 in a fancy way as -1/-1

[dawn abyss]

Ohhh

i see

[thick lily]

if you really wanted you could multiply both sides by -1/-1 and then just simplify -1/-1 down to 1 on the side that has y if that helps

but probably that is more confusing adding in extra stuff yup

[dawn abyss]

thx either way

.close

[topaz sinew]

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[thick lily]

yup you're welcome

[topaz sinew]

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[crisp glacier]

.reopen

[crisp glacier]

how do you prove the existance of the NPC

i had a proof but i forgot and cant recreate it

[cursive patrol]

if only one of those points was labeled P

[crisp glacier]

which obne?

[cursive patrol]

ok that stands for nine point circle

[crisp glacier]

yeah

lol

[cursive patrol]

do you know how your proof started?

there's a lot of ways of going about this

[crisp glacier]

cyclic stuff

[topaz sinew]

@crisp glacier Has your question been resolved?

[crisp glacier]

.close

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[neon iron]

hiii guys can anyone help me with quadratic equations

[half edge]

Send ur question

[neon iron]

a tiled path surrounds an 8-by-12- metre swimming pool the path has an width of x metres on all sides the are of the path is 4x² + 40x m²

A
show that this is correct

😭

help

[neon iron]

.

@half edge

[topaz sinew]

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.

[neon iron]

it was lil bro

[topaz sinew]

@neon iron Has your question been resolved?

[topaz sinew]

@fast quiver Has your question been resolved?

[swift nebula]

the 4x^2 stands for the edges, which are each x^2 and there are 4 edges, so 4x^2

And the 40x for the total length (2x8+2x12) multiplied with the width x

[topaz sinew]

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[winged depot]

hello

[winged depot]

i'm interested in whether, when solving matrices using the gauss jordan method, the units must be strictly in the diagonal at the end or not?

talking about this one

[prisma mesa]

In this case, yes

[winged depot]

in which case they dont have be diagonal?

[prisma mesa]

if the matrix was non-singular, then it could happen

or if it wasnt square

ill give example in a moment...

[winged depot]

thanks

[prisma mesa]

this can be reduced row echelon too

(to be exact, its RREF form of this matrixú

Generally, in RREF, all pivot points must be 1 and everything in the columns with pivot points must be 0

[winged depot]

oh

[prisma mesa]

and they should be also ordered

[winged depot]

hmmmm

idk i just cant get this matrix correct

[prisma mesa]

Basically this + row echelon

what's your current progress?

you can either send photo or type the matrix simply like this:

1 2 3
2 4 0
3 6 3

[winged depot]

ok

i will send a pic

Here you go

now i get stuck going further

[prisma mesa]

,rccw

[prisma mesa]

okay now you need to make 0 from that -3

How can that be done without changing the first column?

[winged depot]

hmmmm

multiply 2nd row woth 3

[prisma mesa]

it's like standard gaussian reduction atm

[prisma mesa]

and add it to the third, yep

[winged depot]

oh

i mean i someow know how to solve them when someone gives me a tip

but alone its quite impossible

[prisma mesa]

it's just like an algorithm

[winged depot]

do yo have some shortcuts to help me understand it easier

[prisma mesa]

firstly you always start with everything under the 1, 1 entry of the matrix

and you turn it to 0

now what do you have to multiply the first row by, such that when you add those 2 you get 0?

[winged depot]

do you see that replacement of 1st and 3rd row

[prisma mesa]

a * 2 + 1 = 0

[winged depot]

i didnt get it by myself acutally

[prisma mesa]

Yes I see, that easens it a bit

[winged depot]

so how do i recognize those stuff

[prisma mesa]

generally, its nice to have 1 in the "main" row. Main in the sense that you will use it to turn everything else in the column to 0

so that swap easens the things a bit

because then, you'll have to be adding multiples of 1st row to row 2 and 3

and multiples of 1 are much easier to calculate

so its good to have 1 in that place

now this

so you need to get rid of those 2

you do that by adding -1 * first row and -2 * first row

respectively

to the 2nd and 3rd row

[winged depot]

oh

ok so ive done some work

and came to this

thats by multiplying 2nd row with 3 then adding it to 3rd

[prisma mesa]

now you need to get all 1's on the diagonal

you do that by scaling the non-1 entries

(and their rows)

there is just one in this case

[winged depot]

Can i like multiply 2nd row with 2 then add it to 1st to get rid of 4, then again multiply 2nd with -3 then again add it to 1st to get rid od 1(now 3)

[prisma mesa]

not really

[winged depot]

oh wait no

I still have -2 down

[prisma mesa]

the trick is, you need to do it column by column

[winged depot]

Ohhhhh

[winged depot]

ok imma write that down

[prisma mesa]

start with the third column

the -7 is the only non-zero entry that can stay there, everything else needs to go to 0

so you will need to find some a, such that a * -7 + (-2) = 0

to get rid of the -2 first

Or maybe it would be better to scale down the third row by -7 first

[winged depot]

why dont i leave -7 till end, then i just divide it by (-7)

and i get 1 there?

or thats wrong

[prisma mesa]

because you will have harder time working the third column

[winged depot]

oh ok then

[prisma mesa]

its not wrong, but it makes stuff computationally harder

[winged depot]

ok

just a sec

ok so now i try to get rid od -7

well i dont know how to do it

🥲

[prisma mesa]

You just need to make it 1

so multiply by -1/7

the third row

[winged depot]

ok

this is what i get

[prisma mesa]

mhm

now you need to get rid of those 2

[winged depot]

multiply 2nd by 2 then add to 1st row?

to get rid of 4

[prisma mesa]

sure that can work

and then for the -2, you can use the third row

[winged depot]

yes

This is the result

for now

[prisma mesa]

yes

[winged depot]

so get rid of -2 in 2nd row by multiplying last one with 2 and simply adding it back to 2nd

[prisma mesa]

correct

[winged depot]

Now i get this

[prisma mesa]

nice

[winged depot]

multiply 2nd by -3 then add it to 1st?

[prisma mesa]

and we are done with third column

[prisma mesa]

perfect, we now move to the 2nd column as you proposed

In the "gaussian" part of the reduction, we do it column by column from left to right, and then with the "jordan" part, we do it again column by column, but this time from right to left

[winged depot]

Ohhhh

See nobody told me this

And final answer it is

[prisma mesa]

mhm

[winged depot]

why mhm🥲

[prisma mesa]

,w solve 2x+5y+3z=2, x+2y+2z=1,x+y+z=1

okay thats weird

[winged depot]

The last one is x+y+4z

not z

[prisma mesa]

ah

,w 2x+5y+3z=2, x+2y+2z=1,x+y+4z=11

[winged depot]

oh f*ck the scripture says its 12, -5, 1

as solutions

x+2y+2z=4

not 1

[prisma mesa]

,w 2x+5y+3z = 2, x + 2y + 2z = 4, x + y + 4z = 11

[thorny flame]

Results provided by WolframAlpha

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Requested by methisalwaysright

[prisma mesa]

i see

[winged depot]

what did we do wrong

[winged depot]

I substracted 1 and 3 row*

in the beginning

[prisma mesa]

I think the mistake happened here

[prisma mesa]

do you mean swapped?

[prisma mesa]

somehow after multiplying first row by -1 and adding it to the 2nd, you got -2 in the last column

but 4 + (-1) * 2 = 2

[winged depot]

Yes

Sorry

[winged depot]

💀

[winged depot]

isnt it 4*(-1) + 2

-4+2=-2

[inner wren]

yes

[prisma mesa]

im talking about the 4th column

the completely last one

[winged depot]

o sh*t

you are right

wait let me get it right now

Give me a min

[thorny flame]

MæthIsAlwaysRight

[winged depot]

Laat one is 2 2 -2

Not 2 2 2

[prisma mesa]

oh right

[thorny flame]

MæthIsAlwaysRight

[winged depot]

Yes

Thats it

and not 3, its -3

Just a sec

Like this

[prisma mesa]

it should be 3, no?

[winged depot]

bro...

i think i should take a break from studying

it is 3 sorry

i think my brain is just fried from 5hr of doing math

i mean i appreciate your help

a lot

i will surely take a break but as soon as i finish this one 😂

[prisma mesa]

well fuck

[winged depot]

what?

[prisma mesa]

,w x + y + 4z = 2, x + 2y + 2z = 4, 2x + 5y + 3z = 2

[thorny flame]

Results provided by WolframAlpha

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

Requested by methisalwaysright

[prisma mesa]

this is already incorrect

lmao

this is terrible

[winged depot]

na bro

wait

a shit.....

you know what i did wrong

[prisma mesa]

what?

[winged depot]

there are 2 matrix, both are the same

except one ends with +z, other with +4z

and i mixed them up

[prisma mesa]

LOL

[winged depot]

hold on a sec

[prisma mesa]

x + y + 4z = 11 it was?

if so then it's not too bad

it can be fixed quite easily

luckily

[winged depot]

you see

[prisma mesa]

yes

so x + y + 4z = 11

[winged depot]

i was looking at wrong one whole time..

[prisma mesa]

this is the right one, right?

[atomic stream]

why is there 2 images?

[thorny flame]

MæthIsAlwaysRight

[winged depot]

bc i was looking at one while solving other

and messed up

[thorny flame]

MæthIsAlwaysRight

MæthIsAlwaysRight

MæthIsAlwaysRight

MæthIsAlwaysRight

[winged depot]

Yes

[prisma mesa]

12, -5, 1

[winged depot]

Thats is correct

[prisma mesa]

you've encountered enough pain today so there it is done for you lol

ill just get a shower rq, brb

[winged depot]

Thanks bro

i really appreciate this

i think im gonna take a shower too

a cold one

[prisma mesa]

okay back

[prisma mesa]

np

You can try the 2nd system for yourself

remember, column by column from left to right

once you do the gaussian part, column by column, from right to left

[topaz sinew]

@winged depot Has your question been resolved?

[winged depot]

yes

i will

appreciate your help

wish you all the best

.close

[topaz sinew]

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[prisma mesa]

You too :) Good luck

[topaz sinew]

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[narrow cradle]

(PHYSICS) Hey guys, my physics teacher has taught me that in a current network if two points are connected by a conducting wire with no resistsncr the potential is the same between those two points which i understand, but he gave a transformation which I don't. Can anyone help?

[lime perch]

,rotate

[lime perch]

you can adjust the wires to make it look like that

[narrow cradle]

How exactly?

I am unable to visualize the wires moving

One second, trying to do that.

@sage siren Thanks a lot!!

@lime perch You too

,cloae

,close

.close

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[thorny flame]

Couldn't find an attached image in the last 10 messages.

Jun 23, 2024, 8:13 PM

[inner wren]

just one less. whatever.

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[crude acorn]

hello

[crude acorn]

so here I have an excercise along with its solution

Where it says the series converges according to the p series test in which p>1

but doesn’t the polynomial denominator affect this?

also, if it doesn’t affect the criteria for convergence then why take out the constant 2^2 at all

[shrewd cave]

yapping

[inner wren]

yap - talk at length in an irritating manner.
Source - Oxford Languages via Google

[crude acorn]

💀

wdym

[inner wren]

i think he is allergic to large long sentence

[crude acorn]

yeah no I get that but like what about my question lol

I tried to put it simple but I think it required some explanation at least

[worthy flax]

You can still use regular p series test

If you combine with the comparison test

[crude acorn]

so why did they take out the constant

wasn’t really necessary was it

[worthy flax]

I mean like...

$\frac{1}{(n + \frac 12)^2}$ and $\frac{1}{n^2}$ are basically the same at large $n$

[cinder sequoia]

not necessary but just to make the comparison to a p-series easier

[thorny flame]

Pseudonium

[worthy flax]

So one converges iff the other does

[crude acorn]

ok awesome

thanks!

.solved

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[thick lily]

yeah I don't like this either, it seems better to just say it has the odd terms with the even terms of the p series taken out at p=2,

[crude acorn]

wdym

.reopen

[topaz sinew]

✅

[thick lily]

so like the denominator has (2n+1) in it, if you did the same series with (2n) and add it to it, you'd have all the even and odd terms

sum 1/(2n+1)^2 + 1/(2n)^2 = sum 1/n^2

[crude acorn]

.solved

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[stuck hatch]

how do I evaluate this derivative ?

[gilded edge]

Sleep bro

[stuck hatch]

given that N and P are:

and m < 1

[topaz sinew]

@stuck hatch Has your question been resolved?

[keen raptor]

first get rid of the 1 cause it's constant

and move the negative sign outside the derivative

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[neon iron]

How do I solve this question

[neon iron]

I am confused because I got 16 as my final answer

An explanation would be appreciated

[lime perch]

What is your calculation

[neon iron]

my final calculation was 16

So first

Here ill explain gimme a sec

bro this is the thing

I DONT KNow

how

here

I did this

500/4 blue material
1500/8 white material

5 extra blue material and 5.5 extra white material

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[hearty storm]

f(x) = 1/x and im given (f(x)-f(a))/x-a. i need help simplyfing it

[hearty storm]

by just plugging in you get ((1/x)-(1/a))/x-a

but idk where to go from there

the answer is given as -1/(ax) and i just want to understand how to get there

[jaunty dawn]

Try to bring 1/x-1/a to a common denominator.

[hearty storm]

how would i do that?

[jaunty dawn]

expand 1/x with a and 1/a with x

[hearty storm]

can i do that? if i multiply 1/x with a wouldnt i also have to do the same to 1/a?

[jaunty dawn]

you not multiplying with it

your expanding it

1/x=a/ax

1/a=x/ax

[hearty storm]

ok so that would give ((a/ax) - (x/ax))/x-a

[jaunty dawn]

Yes

[hearty storm]

what would be the next step to make it look like -1/(ax)

[jaunty dawn]

Well

How can you rewrite a/ax-x/ax? (substracting fractions with same denominators)

[hearty storm]

a-x/ax

[jaunty dawn]

Now, in general, what is (a/b)/c equal to?

[hearty storm]

i genuinely have no clue

[jaunty dawn]

Okay, maybe thats the step, thats missing for you to get to the result.

(a/b)/c = a/(bc)

Therefore ((a-x)/ax)/(x-a) = (a-x)/(ax*(x-a))

And now just cancel (a-x) [ =-(x-a) ] with (x-a)

youll get -1/ax

[hearty storm]

a-x is equivalent to -(x-a)?

[jaunty dawn]

Yes 🙂

-(x-a) = -x - (-a) = -x + a = a + -x = a - x

[hearty storm]

hahaha

okie

thats cool

thank you very much

[jaunty dawn]

🙂

[hearty storm]

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[vestal hatch]

Need help completing this, have incorrect entries

[vestal hatch]

Need help completing this, have incorrect entries..

Here is my graph

[latent agate]

You forgot the square root

[vestal hatch]

For which ones?

[latent agate]

That is why you are getting an incorrect answer

[latent agate]

In the graph, there has to be a square root on top

[vestal hatch]

I only need help with A. and B. by the way

Ah okay

[latent agate]

like $sqrt(5x^3+9x^2)$

[thorny flame]

MoonCaller27

[latent agate]

Let me know if that fixes the issue you are having.

[vestal hatch]

well

it changes the graph to this.

the local max y

would be

it says 2.078

ah okay

yeah I think this is right

[latent agate]

Should be, yeah

[vestal hatch]

I think i hjust have to figure out the sedcondary poitns of increase and decrease

any recommendations

[latent agate]

Hmm

Look closely specifically at the parts that are increasing

[latent agate]

For example, one portion where it is increasing is the (-1.8, -1.2) interval

[vestal hatch]

Imm still doing something wrong I dont get it 😭

v

and Idk what the yellow checkmark means I've never seen that before

i feel like its saying its correct but partically correcrt

@latent agate

[latent agate]

Let's see

We're looking for the interval for which the interval is increasing

$(-1.8, -1.2)$ is technically part of the correct answer

However, From (-1.2, 0), the interval is decreasing, so the answer (-1.8, inf) would not be correct.

[topaz sinew]

@vestal hatch Has your question been resolved?

[vestal hatch]

ah

so i got this far now

but its still showing those two as yellow

[latent agate]

MM makes sense

Wait sorry that is my mistake, the interval is decreasing from (-1.2,0)

The other portion of the graph that is increasing is (0, inf) because it scales upwards

$(-1.8,-1.2) U (0,inf)$

[thorny flame]

MoonCaller27

[wispy vortex]

My open math is always fun

[vestal hatch]

mmm

but how could that be

ohhhhhhh

[latent agate]

Ok so

[vestal hatch]

i see

[latent agate]

yes

[vestal hatch]

ok

👍

[latent agate]

nice

[vestal hatch]

yes 🐱

IM having to find variables of average rate of change

@wispy vortex

I start college in the fall

so im doing college algebra prepartion

[wispy vortex]

It’s decreasing on the interval (-1.2,0)

[vestal hatch]

yeah i see that

I solvd it out

now im doing this one

[wispy vortex]

It’s also increasing from (0,inf)

[wispy vortex]

Kk

[vestal hatch]

but I think i can figure tihs one out myself.

[wispy vortex]

👍

[vestal hatch]

thx!

/close

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[wispy vortex]

Gl

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[vestal hatch]

thx

[vestal hatch]

.close

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[distant linden]

guys how did

(1-cosx)(1+cosx) give sin^2x

[humble spoke]

,tex
$
\sin^2(x)+\cos^2(x)=1\
\sin^2(x)=1-\cos^2(x)\
\sin^2(x)=(1-\cos(x))(1+\cos(x))
$

[thorny flame]

lILi

[distant linden]

ohhh p

[brave oyster]

blud i thought u got it

[humble spoke]

if you need further proof, rewrite as the definitions of sin and cos with o, a, and h (with h written as sqrt{o^2+a^2}), and solve algebraically, you'll end up finding that sin^2x+cos^2x=1

[distant linden]

I only got the first part

😭

[brave oyster]

;-;

[distant linden]

v_V

[distant linden]

Ok ty

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[upper prawn]

Hi guys, im doing a maths assignment and i have had to create a position function using calculus.
Im pretty sure its right but when i put it into desmos with a launch angle of 0 degrees launches at an angle not flat like i would expect

[upper prawn]

I assume its because i have split it into x and y but how do i explain why this is happening

[raven sparrow]

Is this meant to mimic free fall? If so why is there acceleration in both x and y components?

[upper prawn]

its meant to represent a object being launched at 0 degrees from a height

i am eventually going to consider drag so assume i need acceleration in the x?

[raven sparrow]

Yeah but for now you can assume it's zero, maybe add it back later.

What's for sure is that it won't accelerate at the same rate due to drag

Can I see how you've formulated it as well? Like your full formula?

[upper prawn]

Ok but without it it still goes up

This is probably mostly wrong im just getting started

[raven sparrow]

That's ok. What about on Desmos? Can you link it or show me your whole expression for r?

[upper prawn]

i just removed the a in x for that one

[raven sparrow]

Ok. So the issue is Desmos does not see this as a parametric equation.

Try and write it explicitly like $r(t) = (first component, second component)$

[thorny flame]

Azyrashacorki

[upper prawn]

It doesnt see it as i and j?

[raven sparrow]

No, it just adds both things, so you're just drawing a random quadratic with some constant terms as it is.

Once you've written it down with a parameter t, it should ask for a range for t, which you can choose depending on your context.

[upper prawn]

like that?

[raven sparrow]

Yeah. That should work. Note that with the acceleration in both components, it will look sort of oblique, because the 'drag' brings it back onto the x axis as fast as "gravity" is

[upper prawn]

should the t's be x's? because im getting a error

[raven sparrow]

Oh wait you have to put it in one big parenthesis

And remove the (t) at the beginning. I forgot Desmos is a bit finnicky

[raven sparrow]

Like just r instead of r(t)

[raven sparrow]

Yeah that's the acceleration thing. If you want to keep it, maybe write $a_x$ in the x component and $a_y$ in the y component and you can control both separately.

[thorny flame]

Azyrashacorki

[raven sparrow]

The "no drag" case is with a_x = 0

[upper prawn]

ok thanks

without a it looks like that, is that the path of the projectile?

[raven sparrow]

Yeah it looks pretty good. You can extend the interval for t under the formula for r

Play with the angle a bit it should match, same thing with initial height and stuff

[upper prawn]

yeah thankyou, what does the x and y axis represent now?

[raven sparrow]

Axes are position, and it draws over t, so time

[upper prawn]

obviously the y is the height but what is the x, is that time? and howcome the t interval doesnt match up with it

Thanks for all your help

appreciate it alot

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[round jungle]

how come it goes from 3x-1 to 3x+1

[noble laurel]

Have you tried doing the derivative

[round jungle]

hmmmmm

let met ry

me try

i found it

thanks

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[wary tulip]

15 people are to be assigned to teams of various sizes. each person must be assigned to exactly 3 teams, all of different size (so e.g., assigning a person to a size 4 team, a size 2 team, and another size 2 team is not allowed)
there are:
5 teams of size 1
4 teams of size 2
3 teams of size 3
2 teams of size 4
3 teams of size 5
there is no order for players within teams, and teams of the same size are to be treated as indistinguishable. so e.g. if the size 3 teams are named t1, t2, t3, then
assigning p1 p2 p3 to t1 and p10 p11 p12 to t2
is the same as
assigning p10 p11 p12 to t1 and p1 p2 p3 to t2

how many ways are there to assign teams?

[wary tulip]

:)

i did not have any very good ideas

[neon iron]

layla

[wary tulip]

pun pun

one thing to note is every person must be assigned to exactly one team of size 5

which seems like it should simplify the problem a lot because we can recast it as just assigning each person to 2 teams and ignoring the teams of size 5 for now BUT ITS STILL A MESS

[neon iron]

is this combinatorics

[wary tulip]

yes

[neon iron]

oh ok

[topaz sinew]

@wary tulip Has your question been resolved?

[unreal kite]

Maybe you can look at it as distributing placements of the elements in the multiset ${p_1,p_1,p_1,p_2,p_2,p_2,...,p_{15},p_{15},p_{15}}$ into 'posts' for different team sizes. Like there are 5 posts for teams of size 1, 8 for teams of size 2, 9 for teams of size 3, 8 for teams of size 4 and 15 for teams of size 5. \
Then, you notice that you cannot assign the same person to 2 teams of equal size, so each person $p_1,p_2,...,p_{15}$ has to be part of a team of 5, since there are as many places for members of teams of size 5 as there are persons.\
Then the problem reduces to finding distributions of assignments in ${p_1,p_1,p_2,p_2,...,p_{15},p_{15}}$ into 5 posts for teams of size 1, 8 for teams of size 2, 9 for teams of size 3 and 8 for teams of size 4.

[thorny flame]

Crystopher

[wary tulip]

well yes that’s what i was talking about here

it doesn’t seem to be very nice after that though

like you could fix 5 people to the size 1 teams and then take cases on where to put the last 10

but then… many more cases?

or maybe it’s better to fix the size 3 teams

but either way soooo many cases

need something more industrial

[unreal kite]

Something almost straightforward is using multinomials, like $\binom{45}{5,8,9,8,15}$ but then we have the problem that it also counts combinations where the same person may be assigned to several teams of the same size.

[thorny flame]

Crystopher

[topaz sinew]

@wary tulip Has your question been resolved?

[wary tulip]

yea it is kind of like that but i think those numbers should be further partitioned into team sizes (so e.g. 5,5,5 instead of 15)

then the only issues is players being assigned same size teams

hmmmm maybe some inclusion exclusion bs

🥵

this is tough

eepy time for me

[topaz sinew]

@wary tulip Has your question been resolved?

[shadow salmon]

Always start small with these kinds of problems

Reduce it to a simple enough case such that you get some feel for it and build up from there

[topaz sinew]

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[ember lynx]

the solution is there but I don't know how they got x-4, I got x-8 because I calculated the whole equation x2 to get the x^2/2 to just x^2. Is that unnessacary?

[prime flame]

They just take the derivative of f(x)

[ember lynx]

how do they do that

[prime flame]

Do you know how to take the derivative of polynomial

[winter temple]

You cannot just multiply by 2
That’s a whole different function

[ember lynx]

oh

[prime flame]

If you multiply by 2 then it is 2f(x)

It is different

[ember lynx]

sorry I'm trying to translate derivative of polynomial to german but I don't know what that is

can you explain it, what is done`?

[prime flame]

derivative is f'(x)

[ember lynx]

ah, yeah I know that

[prime flame]

well, if $f(x) = x^n$ then $\frac{d}{dx}x^n = n(x^{n-1})$

[thorny flame]

NKH

[ember lynx]

but I thought I had to remove the division to do it with x^2, otherwise how do you do it with x^2/2?

[winter temple]

why did you multiply it by 2 in the first place?

[winter temple]

Yea

[ember lynx]

that's why I multiplied by 2

[winter temple]

uhh no i mean you just take the derivative of x^2/2-4x+4

[ember lynx]

how do you do this f'(x) with x^2/2?

[prime flame]

rewrite it like this: $\frac{d}{dx}\frac{1}{2}x^2$

[thorny flame]

NKH

[prime flame]

1/2 is just a constant

[ember lynx]

yes but how does it change

[prime flame]

You can bring it out of derivative: $\frac{1}{2}\frac{d}{dx}x^2$

[thorny flame]

NKH

[prime flame]

And then differentiate normally

[ember lynx]

sorry, I don't understand it like this but for example with f'(x) 4 dissapears, -4x turns into just -4, the x falls off and what do I do with the x^2/2??

[winter temple]

How would you take the derivative of 4x^2 ?

[ember lynx]

you would have 4x

wait

no

one sec

[prime flame]

I am confused. What are you trying to do, though?

[ember lynx]

you would have 3*4x, right?

so 12x

[winter temple]

Nope it’s 4*(2x)

Anyway the thing here is you can take out the constant when doing derivatives

[prime flame]

Yeah, constant doesn't have anothing to do with derivative

You don't need to simplify or anything like that

[ember lynx]

oh right sorry, my bad

but what do I do with the x^2/2?

how do you take out the constant

[winter temple]

What is the constant here?

[ember lynx]

I don't know what the constant is in german so I don't know what it actually is

is it 1/2?

[winter temple]

That’s right

[prime flame]

Yeah

[ember lynx]

okay so you take it out and have 1/2 * x^2? or

[prime flame]

$\frac{x^2}{2}$ = $\frac{1}{2}x^2$

[thorny flame]

NKH

[ember lynx]

ohhhh and then you put 1/2 * 2 and it's just 1, so you get left with x-4?

[prime flame]

Yeah

[ember lynx]

okay and then you put 3 as x into the original term?

to get the point where the slope has the value 1

[prime flame]

You know that f'(x) give you slope of tangent line at x, right?

[ember lynx]

uhh

it's the slope yes

[prime flame]

Then if you have f'(x) = x-4

and you need slope = 1

It will give you the equation: x-4 = 1

and solve for x with give you x at which the slope of tangent line is equal to 1

[ember lynx]

but that isn't the same as in the solution

[ember lynx]

the solution has somehow -3,5

[prime flame]

Now that you talk about it

x = 5 is the solution of the equation

how do they get x = 3?

[ember lynx]

no wait

x is 3

but how is y -3,5

wait ur right

[prime flame]

I think there is a mistake

[ember lynx]

yeah

that's insane, that's from an exam lol

[prime flame]

To get y, you just plug x = 3 to f(x)

[ember lynx]

into the original f(x) not the f'(x) right?

[prime flame]

Yeah

[ember lynx]

okay

[prime flame]

plug x = 3 to f'(x) will give you slope at x = 3

Not y coordinate

[ember lynx]

oh right

oops

okay, thank you for helping

that was weird with the mistake, it's an official solution from our teacher

[prime flame]

You should ask your teacher if there is a mistake

[ember lynx]

yeahhh we have an exam today so it's a bit too late for that 😅

but yea, thank you, goodbye! :)

.close

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[gusty delta]

i need help cuz idk what to do next

i think i was going the right way for the first bit

[hollow plover]

, rcw

[topaz sinew]

@gusty delta Has your question been resolved?

[pliant tulip]

try finding the area of the yellow (its a right triangle)

actually (area of blue) = 225-2(area of yellow)

[gusty delta]

thx

.close

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[idle osprey]

yo

any 12th grader

[potent flicker]

9th, but I imagine I can help you in maths

[keen raptor]

you should help them figure out how to open a help channel

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[boreal mulch]

if im solving a differential equation

[boreal mulch]

and i get x(2t-1)/(2t+1) = (2t+1)/2 - ln(2t+1)+C

do i have to expand RHS and merge the 1/2 with the C

[potent flicker]

you want t=f(x)?

and is this a fixed solution or

[boreal mulch]

its given when x=3 t=1

(2t+1)/2 - ln(2t+1)+C is obtained after integrating (2t-1)/(2t+1

[potent flicker]

ok

ok ok ok

get c first

[boreal mulch]

do i have to merge 1/2 with C

or keep it there

[potent flicker]

No i mean

you put x and t values into the equation first

so u can get c

[boreal mulch]

why don't i have to merge 1/2 with C

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@boreal mulch Has your question been resolved?

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@boreal mulch Has your question been resolved?

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[nova path]

An electron is in the fundamental state of a simple harmonic potential, which minimizes the Heisenberg uncertainty relation, and has a kinetic energy of 1 eV. Determine the uncertainty in the measurement of the electron's position. how can i solve this???

[ionic oar]

fundamental state of a simple harmonic potential

does this jargon just mean the ground state

[ashen nest]

Sounds like it lmfao

I would've just used heisenbergs formula

And solved for ∆x

[nova path]

how do i get the ∆p?

without ∆v (p = mv)

[lime perch]

$\Delta x\Delta p\geq \frac{\hbar}{2}$

[thorny flame]

Flappie

[ionic oar]

kinetic energy I guess

[ K = \f{p^2}{2m} ]

[iron blaze]

1/2mv^2 >= h/2 ye

[thorny flame]

Bob l'éponge

[iron blaze]

wait nvm

[ionic oar]

Kind of confused though

[ \Delta K = \f{p}{m}\Delta p ]

[ashen nest]

What values do we have to work with

[thorny flame]

Bob l'éponge

[lime perch]

are you sure that this is everything youre given?

[iron blaze]

so solve for mv (p)

[nova path]

unfortenately yes

[iron blaze]

and let △x △p = h/2

then find △x

mass of an electron = 9.109 * 10^-31 kg

reduced planck's constant = 1.055 * 10^-34 J s

@nova path ?

[nova path]

sorry

[nova path]

for mv?

[iron blaze]

mb v

idk why i said mv

solve for v

then find mv

1/2mv^2 = 1 eV

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@nova path Has your question been resolved?

[nova path]

so i assume that delta v is equal to v? @iron blaze

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@nova path Has your question been resolved?

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[noble crest]

what i need to do: turn these problems into a matrix equation

context:
this is for a puzzle game i play

its somewhat hard to explain what i need to exactly do

for a more detailed explanation, this is the guide https://ktane.timwi.de/HTML/ReGret-B Filtering.html

but in short

i need to do 3 operations to a hexadecimal number in about 2 minutes

and these are just a few of them but

[brittle belfry]

what game is this

[noble crest]

keep talking and nobody explodes

[brittle belfry]

oh

[noble crest]

so for example

for the image with the "1"

i would need to split the number into pairs
C7 64 66

convert it to decimal

199 100 102

do the same thing to the bottom number

[brittle belfry]

excuse me but is this linear algebra?

[noble crest]

no

229 166 236

then i need to take the multiplicative average of them both

and then convert it back to hexadecimal

the reason im asking here

in simple terms uhh

someone else is really good at doing the puzzle

and apparently you need matrices to do it faster

[brittle belfry]

uhm...

[noble crest]

here are all the possible functions

[brittle belfry]

i'm just a high schooler btw...

[noble crest]

• : additive average
  x : multiplicative average
  / : inverse average
  OR : convert hexadecimal to binary instead and OR operation
  XOR: XOR operation
  AND: AND operation

so for example

the solution to the first one was

this is what it looks like int he calculator i use

the problem with this is that its too slow

and also i am supposed to delete everything and type it all manually every time

to give you a general idea of how fast im supposed to do it and why i need to know this

[long stirrup]

use python or something like that

[noble crest]

i

coding is banned

[long stirrup]

ah

[noble crest]

well actualkly

its not but

i would have to type out the entire program each time

[long stirrup]

got it

[noble crest]

and also i dont know how to code

[noble crest]

Opponent's time: 1:30
Tepel(fella who knows the matrix): 1:00

and 30 seconds is a really, really big time for what i am doing

[long stirrup]

https://tio.run/##y/r/PzU5I18hIy1FwUZPQdVKQUtfwcjUTFkvM69MwdAsKdnczMTMDMRKNU00S03@/x8A