help-38

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@wet delta Has your question been resolved?

[wet delta]

s0 9x?

[swift sable]

Misread sorry

[wet delta]

<@&286206848099549185> anyways help me out

[forest egret]

ok we had x^2+b^2=(3x)^2

solve for b first

[wet delta]

huh

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[vague thistle]

Kinda confused what R^{nxn} is here, is it the set of real-valued matrices? Then how does it take two vectors from R^n as input?

[stoic iron]

yeah this feels like it should be R^n x R^n

[vague thistle]

Yeah I thought so, thanks for confirming

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[stoic iron]

can also try check the reference

[umbral dragon]

.reopen

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✅

[umbral dragon]

I’ll just leave this open until I’ve checked

Cause I’m uncertain and it’s better to confirm

Yeah pretty sure it should be 2n. After all, for n=1 we have R^2, not R

@vague thistle

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[daring dragon]

I'm not sure how to do the rest.

[kindred pier]

Can you calculate $\arcsin{\sqrt{2}}$?

[solid kiln]

SWR

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[wet delta]

in a right triangle, the long side is three times the length of the shorter short side. Find the length of all three sides when the area is 24.0 cm2 https://media.discordapp.net/attachments/1172135383753039922/1172135545892245524/IMG_0284.jpeg?ex=655f374b&is=654cc24b&hm=80dbc62cca58e6633f0f182ba81686e060aae5661dcca19609b0e0199121e3f6&=&width=1366&height=1024

[wet delta]

@forest egret

[wispy sparrow]

!status

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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

[remote thistle]

I would start this by drawing a right triangle and setting the short side equal to x. The other side is three times as long, so it'll be 3x. Try plugging into the formula for area from there

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[zinc ginkgo]

!msgdel

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[wraith hinge]

Hello Everyone

[wraith hinge]

This question makes no sense

so when you take a of n aka a(n) it is 3/1 right?

if you plug anything into that it will always stay 3 which will never be zero

so how does it converge?

the website says it converges

i selected diverges but it was wrong

[viscid flower]

im not sure what you mean it will always stay three

you know you can just pull the 3 out of the sum, right

so $3 \sum_{k=0}^\infty (-0.8)^k$

[solid kiln]

jan Niku

[viscid flower]

so its really just a question of whether that sum converges, the 3 doesnt play a role in this

either the infinite sum is finite, and the answer is 3 times something finite

or its infinite and 3 doesnt change it to finite

[wraith hinge]

does a(n) have a role to play at all?

because I watched a video and they said take a(n) where you include everything besides the alternating part

this question is very different

[viscid flower]

so the geometric sum as i know is like

$\frac{1}{1-x} = \sum _{k=0}^\infty x^k$ given $|x| < 1$

[solid kiln]

jan Niku

[viscid flower]

im not sure what you mean a(n)

[viscid flower]

can you explain what a(n) is?

[wraith hinge]

a(n) is like everything besides the part that makes it alternative

like lets say there is something like ((-1)^n)*3

we remove the -1^n

because it isn't part of a(n) because it makes it alternate

[viscid flower]

i dont think you should remove it

unless you are looking at the alternating series test?

but theres no real reason to use that here

[wraith hinge]

aaaahhhh

so its in the wrong hw lol

[viscid flower]

what i mean is, its a geometric series, you should recognize it because you dont need any crazy tests

the condition for a geometric series converging is straightforward

everything raised to the k has to have magnitude less than 1

if it does, it converges, and you can write down the total

[wraith hinge]

oooo

so i can write it like 1/1+0.8

[viscid flower]

yes

,calc 1/1.8

[solid kiln]

Result:

0.55555555555556

[viscid flower]

5/9

[wraith hinge]

and since its less than one it converges

[viscid flower]

yes

|-0.8| = 0.8 < 1

[viscid flower]

dont forget the 3

theres still a 3 floating around

[wraith hinge]

so * 3

[viscid flower]

to me its easier to take it out though

yea

so 15/9

[wraith hinge]

now its bigger than 1

[viscid flower]

the total is bigger than 1, thats fine

the important part is that the thing thats being raised to the k has magnitude under 1

sometimes this is written like

[wraith hinge]

so the ratio should be less than one

something like that

[viscid flower]

$\frac{a}{1-r} = \sum_{k=0}^\infty a r^k$

[solid kiln]

jan Niku

[viscid flower]

r for ratio

only r matters for convergence

given a is just some number obviously

a is not raised to the k

[wraith hinge]

alright these two things are not the same

a/1-r

and 1/1-x

[viscid flower]

the two sums?

no theyre the same

[wraith hinge]

not the same

oh

[viscid flower]

here

start here

multiply each side of = by a

you get the same result

swap x for r

[wraith hinge]

okay so the thing getting raised by k should have the magnitude less than one and that makes it converges

[viscid flower]

yes

[wraith hinge]

and to find what it converges too

[wraith hinge]

you use this

[viscid flower]

yup

you will learn to love these once you recognize them

they are the easiest series of all

and you can get an answer

[wraith hinge]

I hope I will love them I was having a hard week with physics and neglected my calc 2

so today

i went from studying sequences

all the way to root/ratio test

in one day lol

i have test on monday

it lowkey doesn't feel bad though

[viscid flower]

you can definitely pick up which test to use pretty quickly

if you see a lot of examples

geometric series is a special case and its like a gift

[wraith hinge]

telescoping too right?

[viscid flower]

thats alright too

im more partial to geometric

lol

you know calculus?

[wraith hinge]

geometric is nice cuz u can just do it right away

[viscid flower]

i mean

obviously

this also gives you the series for log(1-x)

if you integrate naively

so its a good one to know

you can sub x -> -x to get the one for log(1+x)

its just nice

anyways good luck on your exam

[wraith hinge]

thanks I appreciate the help and clarification its rare someone can explain to me via text whats happening

have a good one 😁

.close

[viscid flower]

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[placid valve]

hello

[placid valve]

I need help understanding how to get the angle when converting to polar form

I understand how to get the cos and sin angles but how do we arrive at the conclusion that the angle is 2pi/3

[thorn trout]

angle = arctan(1/root3)

because it’s in quadrant 2

you get 2pi/3

[placid valve]

ah I have to use the unit circle?

I completely forgot how to use it

[thorn trout]

oh lol

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[gentle sleet]

how do u do this question?

if T is left handed then det A_T < 0. and if det A_T < 0 then isnt det A_T^(k) < 0 for all k in Z aswell?

@whole coral what did I do wrong

i dont get it

[whole coral]

Think about the matrix of A_T^k in terms of the matrix of A_T

[gentle sleet]

i dont get it

so we know that if T is left handed so is T^-1 which means A_T^-1 < 0

so is A_T^k = A_(T^-1)^(-k)

[whole coral]

Well there's at least one trivial counterexample for this one here: think about k=0

[whole coral]

That is true, but say you have $T$ as it is, you can represent that by the matrix $A_T$. How would you represent $T^2$, using only that $A_T$?

[solid kiln]

@whole coral

[gentle sleet]

A_T^2

[whole coral]

As in $A_{T^2}$, or $(A_{T})^2$?

[solid kiln]

@whole coral

[gentle sleet]

A_(T^2)

[whole coral]

Well... that is technically true, but the answer I was looking for was the latter one

[gentle sleet]

wait what

[whole coral]

Applying T twice is equivalent to multiplying by A_T twice, no?

[gentle sleet]

ohhhh

so det of A_T^k = det(A_T)^k right ?

[whole coral]

Yep yep

[gentle sleet]

so if T is left handed then det(A_T) < 0. and to find a value k s.t det(A_T) > 0. we gotta make the power an even number

so k has to be an even number

[whole coral]

That works nicely

[gentle sleet]

tysm

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[plush canyon]

Call a positive integer $N$ a 7-10 double if the digits of the base-$7$ representation of $N$ form a base-$10$ number that is twice $N$. For example, $51$ is a 7-10 double because its base-$7$ representation is $102$. What is the largest 7-10 double, given that it is 3 digits?

[solid kiln]

Aurora

[plush canyon]

can someone explain the last paragraph of this solution

source: 2001 AIME I p8

cant wrap my head around this sol for some reason

[tough iron]

Do you get why they picked a=6

[naive dirge]

if a was more than 6, it would go into the next digit i think?

[tough iron]

Exactlt

[plush canyon]

yeah

[tough iron]

Its like picking a 9 base ten

So then you have an equation 12=4b+c and you want your b to be as big as possible

[plush canyon]

like if i pick 815 base 10 it could be = 1630 b7 obv this isnt true but like what if?

[tough iron]

Its like asking to pick 2 digis so that 9\ _\ _ is largest

[tough iron]

Thats a 10-7 double

Not a 7-10

No wait

[plush canyon]

ohh true

[tough iron]

Lol

[plush canyon]

oh nvm

we jumped down the hole together

anyways

[naive dirge]

not too bad of an aime problem

[plush canyon]

how do you know it's not >315?

[tough iron]

I guess because it has to have 3 digits base 7 too

?

[raw magnet]

I remember doing AIME in hs

I got like an 7 or 8 I forgot

but it was def harder than this

[tough iron]

Thats the only reason why anything above a 600 base 7 doesnt work

[plush canyon]

yeah but it could be like 80-->160

160 is base 7

[tough iron]

Yeah but thats not the largest one

[plush canyon]

or 800-->1600

how do you know its not smt like that

[tough iron]

My only guess looking at the solution is that they assume its a 3 digit number in both base

[naive dirge]

aime answers are 000 - 999

so if cannot be 4 digits

[raw magnet]

^^^

[plush canyon]

ik

but the answer can be 3 digits

[naive dirge]

yes

we are looking for the largest one

[plush canyon]

in my 800-->1600 example, the answer is 800

[raw magnet]

800 in base 7??

💀

[plush canyon]

800 in b10

[naive dirge]

the answer is in b10

[plush canyon]

to 1600 in b7

so the answer would be 800

oh wait

i see i was doing the prob a diff way

[naive dirge]

what

oh

[plush canyon]

but for the first line right "let the number be in base 7"

i let the # be in base 10

and still got 100a+10b+c=98a+14b+2c

what went wrong here

[naive dirge]

wdym

we just rewrote it in their respective bases with relation to a b and c

[plush canyon]

the number is let to be in b7

[tough iron]

Its the same thing lol

[naive dirge]

so let’s say the digits of our number is a b and c

[plush canyon]

ik, i rewrote it the same way but in base 10

[naive dirge]

in base 10

moo

ok*

[plush canyon]

so why can't a b or c be 9 then

[naive dirge]

9 doesn’t exist in base 7?

[tough iron]

Its still 2* the base 7 representation no matter if you first defined abc as a base 7 or base 10 representation

[plush canyon]

hm

[tough iron]

If you did it your way, then you would halve the abc base 10 so you'd get 50a+5b+½c = 49a + 7b + c aka a = 2b + ½c

[plush canyon]

im thinking

[naive dirge]

ok

[tough iron]

Which is just 2a = 4b+c halved

Its completely the same reasoning

[naive dirge]

pretty interesting problem tbh

[plush canyon]

im just too stupid for aime lmao

[naive dirge]

better then a bunch of the amc10 probs

[naive dirge]

what’s was ur amc score?

[plush canyon]

115.5 this yr

amc 10

[naive dirge]

so you’ll get in

[plush canyon]

yeah

im trying jmo

[naive dirge]

gl

i still gotta take b

[plush canyon]

ty im still thinking lol

prolly wont make jmo this yr but hopefully next yr

i feel like i kinda understand but 50% dont

ohhhhh

i get it now

the equation only cares for 3 digits

even if it was base 10

[naive dirge]

yep

[plush canyon]

once i get to 4 digits the equation breaks

[naive dirge]

yep

[plush canyon]

plus it cant be 4 digits

cuz its 2 abc = abc in base 7

ok ty!

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[raw magnet]

what

[placid valve]

?

[raw magnet]

explain your reasoning

bc you kinda messed up

big time

[placid valve]

Used this formula

Is this formula wrong?

[raw magnet]

$\int _{x^5}^{x^7}\left(2t-1\right)^3dx=\int _{x^5}^0\left(2t-1\right)^3dx+:\int _0^{x^7}\left(2t-1\right)^3dx$

@placid valve

[solid kiln]

nosqldb

[raw magnet]

yes?

$= -\int _0^{x^5}:\left(2t-1\right)^3dx+:\int _0^{x^7}\left(2t-1\right)^3dx$

yes?

[solid kiln]

nosqldb

[raw magnet]

then just use fundamental thm with some chain rule

and a lil common sense

and ur chilling

@placid valve .close when ur done here

[placid valve]

is this formula wrong tho?

[wraith hinge]

it isn't

[raw magnet]

it's not incorrect

but I would prefer you expand it out

and use the actual FTC

that way you won't make a mistake in the application of the formula

[raw magnet]

wrong

[raw magnet]

pls use the method I gave you

[raw magnet]

@placid valve

[placid valve]

I understand but im trying to see why what I did is wrong

[raw magnet]

let g(x) = x^5

[raw magnet]

you will understand with my method

f(x) = $\int _0^x:\left(2t-1\right)^3dx$

[solid kiln]

nosqldb

[raw magnet]

the derivative of that is just the inside function

$\frac{df}{dx} = (2x-1)^3$

[solid kiln]

nosqldb

[wraith hinge]

b(x) isn't (2x⁷ - 1). it's x⁷

[raw magnet]

$f'(g(x)) * g'(x) = (2x^7 - 1)^3 * (7x^6)$

[solid kiln]

nosqldb

[raw magnet]

@placid valve you should see ur mistake now

[placid valve]

do you not have to use the chain rule?

[raw magnet]

you are incorrectly using chain rule

[placid valve]

Is this right

@raw magnet

[raw magnet]

omg

yay

!!!

.close

[placid valve]

its right?

.close

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[proven violet]

$u=\frac{x^2}{y}, v=\frac{y^2}{x}$

[solid kiln]

george clooney real account

[proven violet]

how can I find $\frac{\partial x}{\partial u},\frac{\partial x}{\partial v},\frac{\partial y}{\partial u},\frac{\partial y}{\partial v}$?

sorry i mean dx/du

oops lemme rewrite

[solid kiln]

george clooney real account

[proven violet]

^^^

[raw magnet]

holy shit

[opaque linden]

u can write x as a function of u and then do partial differentiation

[brisk solar]

Find partial u partial x and then flip it lol

[raw magnet]

^^

[proven violet]

but i can't really find x as a fdunction of u

[opaque linden]

why not

[proven violet]

because taking square root

isn't inverse function

[raw magnet]

good work clooney

[proven violet]

like how do i know if to take positive or negative branch yk?

i can try take reciprocal i guess

but is that generally true?

ok i googled it

it is

nvm

.close

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[crude ravine]

I received a list of math exercises, I managed to do them, except for this one where I can't modify the equation, to that it looks like something solvable by ordinary equations

[eager vault]

Take y=vx

[raw magnet]

nvm

ignore

[crude ravine]

oh

[brisk solar]

omg

🧠

[eager vault]

Haha lol

Nice one guys

[crude ravine]

thanks guyss

[eager vault]

Wtf im laughing at

Im a r**tard

[crude ravine]

.close

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[crude ravine]

Here I am again, with the same question... I have to put the answer in y(x) but i cant pass that part

[edgy cedar]

Can’t read this

[crude ravine]

oh

1min

[edgy cedar]

Can you post original question

[crude ravine]

its here

i got arcsin(y/x) = ln|x| + C

and i know the answer is

but i dont know how to get to that

[edgy cedar]

Why would it be that

the two comes out of nowhere

[crude ravine]

the answer on the list is that

[edgy cedar]

One sec I’ll solve it

[crude ravine]

and on symbolab is too

Here is my try more clean

[edgy cedar]

It should be arcsin(y/2x)

[edgy cedar]

Oh that’s meant to say asin

One sec

[crude ravine]

oh

[edgy cedar]

Looks like you got the right thing into doing the integral

[crude ravine]

now it makes sense

ty

[edgy cedar]

Np

[crude ravine]

.close

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[edgy cedar]

Which one?

Part a?

Okay what did you get for the inverse of a

Okay so isn’t it just the case of plugging that back into f

What are you stuck with exactly

So we’re looking at $$\log_2 {\left( \frac{\frac{100}{1+2^{-x}}{100 - \frac{100}{1+2^{-x}} \right)}$$

@amber python did u just steal my emote

[solid kiln]

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

[edgy cedar]

I give up

[amber python]

[\log_2 \left( \frac{\frac{100}{1+2^{-x}}}{100 - \frac{100}{1+2^{-x}}} \right)]

[edgy cedar]

Thank you lord snow

WTH

It’s animated

I need that

[edgy cedar]

From here can you simplify the denominator

No just make a common denominator

And add fractions

[solid kiln]

Snopes

[edgy cedar]

No how did you get this

[edgy cedar]

Do this first and see

Well first of all, it’s 1+2^-x

And $2^{-x} \cdot 100 \neq -200^{-x}$

[solid kiln]

$Pure$

[edgy cedar]

?

100 - 100/(1+2^(-x))
= 100 - (2^x • 100)/(2^x +1)
= [2^x • 100 +100 - 2^x • 100]/(1+2^x)
=100/(1+2^x)

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[loud wasp]

Q5

[loud wasp]

I think I'm doing it wrong

Formula I'm using

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[empty iris]

While doing partial fraction decomp would the s not disaphere from e^s? Im really struggling with understanding why it is sitll there

[empty iris]

@zinc ginkgo The channel timed out, but why would the s stay? I understand that it would equal the top but how would it work mathamatically?

[edgy cedar]

Wdym why the s stay

[empty iris]

so if u do the partial fraction decomd

decomp

one sec writing it out to send

why would the s stay in the answer as when you do the math it gets a seperate value

unless im missing something in my decomp

[zinc ginkgo]

your starting equation is wrong

[empty iris]

oh?

one sec let me look at all the cases

[zinc ginkgo]

https://math.stackexchange.com/questions/1047549/partial-fraction-decomposition-with-exponent-in-numerator

[edgy cedar]

You have to factor it out and do partial fraction separately

[empty iris]

oh.... thats interesting

so I take out the e^(-2s-2) and then multiply it through my partial fraction decomp?

[edgy cedar]

Here’s how I’d do it. $$ e^{-2s-2} \cdot \frac{1}{s^2 + 5s +6} + \frac{2s+5}{s^2+5s+6}$$

[solid kiln]

$Pure$

[edgy cedar]

Then partial fraction decomp the 1/… and (2s+5)/…

[empty iris]

hmm ok I see how that would work

any reason that we cannot use e inside the partial fraction tho

like mathamatically?

[zinc ginkgo]

polynomials don't multiply to exponential functions

[empty iris]

ah

ok well thank you

ill try to redo it

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[wraith hinge]

hi guys

[wraith hinge]

I dont think the f(n) makes sense

ill show yall my work

How is it 4^3 power¿

[nimble stone]

denominator of a(n+1) is 4^{3n+5} not 3n+3

3(n+1)+2=3n+5

then when you cancel, you get 4^3

[wraith hinge]

Thanks man

[nimble stone]

nw

[wraith hinge]

.close

.close

[nimble stone]

how odd

[wraith hinge]

bruh waht

[nimble stone]

lol dw about it

[wraith hinge]

well ig its gonna stay mine than

ill jsut chat if i have more questions here lol

unless it closes

[nimble stone]

👍

[rugged latch]

.close

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[rugged latch]

Oops

.reopen

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✅

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[quaint apex]

python related problem

[mystic veldt]

wtf

check out #old-network

[quaint apex]

bro

its all math at the end of the day'

[lethal anvil]

what's the problem?

[quaint apex]

my formatting is off

look at my actual output

compared to expected

not including commas too

not sure how to approach this

[lethal anvil]

ah ok

well for the commas there's a few ways

what version of python are you running

[quaint apex]

split(",")?

im using 3.11.5

[lethal anvil]

no you can just do f'{value:_}' and do a format string

[quaint apex]

huh

[lethal anvil]

so we're you're printing {counters[i]:_}

try that

[quaint apex]

woah

what!

[lethal anvil]

cool right?

[quaint apex]

how

did u knwo this!

[lethal anvil]

I had to use it recently at work lol

now

for your other problem of alignment...

[quaint apex]

thats sick

[lethal anvil]

give me a second to think about that

I'm not sure off of the top of my head

Oh wait!

actually

hmm

[quaint apex]

actually i guess i dont need to format it to have the same spacing

[lethal anvil]

hold on

[quaint apex]

i just passed the code interpreter by adding that small bit

thats incredible

[lethal anvil]

great!

sounds like you're good to go then, easy fix :O)

[quaint apex]

but hold on

how the heck

did u find that feature

[lethal anvil]

I googled it. I just googled "python commas numbers format"

and found either a SO forum or an article, I can't remember

[quaint apex]

that is awesome

thank you!

.close

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[lethal anvil]

no problem!

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[near urchin]

somebody please help me solve for julia🔥🔥

[mystic veldt]

,rccw

[mystic veldt]

!status

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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[near urchin]

1

[mystic veldt]

well you know that total salary of ur julia is her base salaray + 15 percent of her sales

first question says that her total sales is x

it what should the equation be

according

to yo

[near urchin]

20,000 +15%(x)

??

[mystic veldt]

yes

is equal to y

[near urchin]

how do i state domain and range for this

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[raw magnet]

hmm

[raw magnet]

I wonder if austin can give me some guidance to this problem

you CANNOT tell me the solution

"An 8-by-8 “lattice” is represented above. A knight arrives at a1 at time T=0, and wishes to travel to the upper-right corner (h8). This lattice respresents a 3-dimensional landscape; the numbers at each “point” represent their altitudes.

But there’s a big problem here, which is that this grid was built on a swamp, and the lattice points are prone to sinking. Whenever the knight arrives on a lattice point of altitude A, that point and all others with the same altitude start sinking at the rate of 1 unit per n minutes, where n is the number of lattice points of altitude A. Furthermore, the lattice point diametrically opposite the one the knight is on rises at the rate of 1 unit per n minutes. This sinking (and rising) continues only while the knight remains stationary, which may be for as much or as little time as the knight chooses. No other points move or begin to move during this time. (Altitudes can become negative. In the event that the “opposite” point is at the same initial altitude, it neither rises nor sinks, and is ignored for the purpose of calculating n.)

The knight is only allowed to make 3-dimensional jumps that are permutations of (0,±1,±2). Jumps take 0 time to complete. In the spirit of a knight’s “tour”, your goal is to find the path to the upper-right corner that takes as much time as possible. "

I just want guidance

preferably from @knotty locust

[knotty locust]

Hmm I'm not sure how to do this without thinking about it quite a bit which I don't exactly have time for right now, sorry

good luck though

[raw magnet]

ty Austin

@stoic iron I'm sorry to ping, but would you be able to provide some guidance

[stoic iron]

no thanks, and please don't ping me

[raw magnet]

.close

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[dense panther]

Physics just need help on 4 and 5. Please help fix mistakes.

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[dense panther]

<@&286206848099549185>

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[supple parcel]

||x−x′|−|y−y′|| ≤|x−y|+|x′−y′| does anyone know how to proof this

[supple parcel]

Ah mid 💀

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[ripe valley]

Hello

[ripe valley]

I have come back with a stronger (but I asume tougher version) of the problem I posted yesterday

Find $f(x)$ such that $f(x) = \frac{x}{2}$ if $x$ is even, and $f(x) = 2x$ if $x$ is odd

[solid kiln]

ItzKraken

[ripe valley]

so far, no attempts, no ideas. Looking for a few hints and some guidance on how to start

[haughty oasis]

only needs to be defined on the integers?

[ripe valley]

no counting numbers

i.e. 1 to infinity

[haughty oasis]

you want to adjust the coefficient based on the parity, so find the midpoint of 1/2 and 2 and note that a(-1)^n equals a for even n and -a for odd n

[ripe valley]

Hmm

[haughty oasis]

then choose a to be the distance from the midpoint

[ripe valley]

so...$\frac{3}{4} \cdot (-1)^x = f(x)$?

[solid kiln]

ItzKraken

[ripe valley]

wait this isnt N to N

this is N to R

I need N to N

let me reread what u said

[haughty oasis]

yes, this is N to N

[ripe valley]

huh? x=1 doesnt give N, it gives R

[haughty oasis]

because that’s not the function

you want to add what you have to the midpoint of 1/2 and 2 and then multiply by x

[ripe valley]

1 sec

[haughty oasis]

f(x) = (coeff dependent on x) * x

[ripe valley]

midpoint of 1/2 and 2 is 5/4 right?

[haughty oasis]

yes

[ripe valley]

so $(\frac{3}{4} \cdot (-1)^x + \frac{5]{4}) \cdot x = f(x)$

the crap

[haughty oasis]

you have 5] instead of 5}

[ripe valley]

oh 🤦‍♂️ thanks I couldnt spot it

$\left(\frac{3}{4} \cdot (-1)^x + \frac{5}{4}\right) \cdot x = f(x)$

[haughty oasis]

almost correct but just check what happens for even/odd x

[solid kiln]

ItzKraken

[ripe valley]

lemme see

so f(1) should be 2, f(2) should be 1, f(3) should be 6..nope not the case

f(1) = 1/2

[haughty oasis]

the coefficient in particular

[ripe valley]

f(2) = 2

wait these are twice for even and half for odd

[haughty oasis]

the coefficient is 1/2 for odd x and 2 for even x, which is the opposite of what you want

[ripe valley]

so I multiply 5/4 with (-1)^x? instead of 3/4?

[haughty oasis]

so change when 3/4 gets subtracted/added

[haughty oasis]

nope

[ripe valley]

change how ...Do I change x to x+1?

[haughty oasis]

this will give you values other than 1/2 or 2

[haughty oasis]

exactly

[ripe valley]

let me see

[haughty oasis]

or equivalently add a - in front of the 3/4

[solid kiln]

ItzKraken

[ripe valley]

yeah same thing

nice

the next one is even more trivial

[solid kiln]

ItzKraken

[ripe valley]

this might be some trig thing

[haughty oasis]

ax + b such that a is 1 iff x is even and b is 1 iff x is even

(and 0 otherwise)

[ripe valley]

huh?

shouldnt it be the otherway around

[haughty oasis]

yes

my bad

so you just need an even/odd detector

[ripe valley]

I was looking for something like the above instead of a piecewise function

[ripe valley]

(-1)^n?

[haughty oasis]

which is 0 when even and 1 when odd

so you can modify that

[ripe valley]

hmm let me think

so I need a function

that takes (-1)^n

and gives 180 for even n and 90 for odd n

right?

[haughty oasis]

uhh

you want to use trig?

don’t need

[ripe valley]

well it feels simpler imo

[haughty oasis]

okay sure

[ripe valley]

what other ideas do you have in mind..?

[haughty oasis]

let me think about a trig solution for a second

[ripe valley]

okay

[haughty oasis]

yes okay, it might be simpler with trig

sin(n*pi/2) is 0 for even n and plus or minus 1 for odd n

so you can square this to get exactly what you want

[ripe valley]

a is the input to our function right?

[haughty oasis]

yep, I’m using n sorry

[ripe valley]

its fine

so

$f(x) = \sin^2 \left( x \cdot \frac{\pi}{2} \right) \cdot (x+1)$

[haughty oasis]

No

[ripe valley]

1 minute

[haughty oasis]

looks good

oh wait

forgot to square it

[ripe valley]

oh right

[solid kiln]

ItzKraken

[haughty oasis]

you could’ve also used (1-(-1)^n) / 2 instead of the sine

that was my original idea

[ripe valley]

oh rightttt

dayum that goes better with the first function we found

Well tyvm!

.close

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[frail owl]

How to find largest x-digit multiple of a number y, without rigorous multiplication and trial and error ?

[frail owl]

So let's take a number 120. How do we find the largest 3 digit multiple of it without checking it with each number until we reach a 4 digit number.

[shrewd ridge]

trial and error

120 is fairly easy, because 1200 is a multiple, and it's close to 3 digit

it would be harder with like 249

still easy, like 390 maybe

oh, 88

no that's still one step fom 880 to 968

it has to be like, largest 4-digit

oh

you just divide it lol

largest 4 digit is 120×floor(9999 / 120)

is dividing rigorous multiplication?@frail owl

i assume no

[frail owl]

Is trail and error the only solution ?

[shrewd ridge]

no, long division

[frail owl]

Can you tell me how did you came up with this equation ?

My mathematics is pretty weak.

[shrewd ridge]

no

you divide the largest number it could be, so 999 in this case

subtract the remainder from 999

[frail owl]

Understood.

[shrewd ridge]

but that just uses trial and error anyway

[frail owl]

No it isn't trial and error exactly tbh.

[shrewd ridge]

because long division does

[frail owl]

It's better.

That is so amazing.

I really wonder how you came up with that equation @shrewd ridge

n * (pow(10,n)-1 // n)

[shrewd ridge]

this problem is division

it's not coming up, it's recognizing that division is this problem

[frail owl]

It's floor division. What's the problem ?

[shrewd ridge]

B//A means "find the largest multiple of A under B, but instead of the multiple itself give me n that i can multiply by A to get the multiple"

[frail owl]

That clarifies everything.

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[true coral]

given pyramid ABCD and a sphere inscribed inside ABCD, with centre I, what can we say about its relation to the triangles ABC,BCD, ACD, ABD

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[true coral]

<@&286206848099549185>

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[true coral]

<@&286206848099549185>

[frigid kayak]

yh

[true coral]

Yeah

[true coral]

<@&286206848099549185>

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[wraith hinge]

Can someone please explain why x-h is right?

[gaunt harness]

wdym "x-h"

[wraith hinge]

(x-h)^2 or (x-h)^3

[gaunt harness]

bro what

you mean like definition of derivative??

[wraith hinge]

Just transforming graphs

Moving them around.

[orchid wagon]

give more context

[wraith hinge]

Ok. So let's say I want to graph (x+2)^2, why does it shift to the left?

[tulip violet]

Because the general form is (x - h)^2 and so (x + 2)^2 = (x - (-2))^2 hence the shift left

[wraith hinge]

Oh, so if it were positive, it would (x-2) since there isn't two negatives to cancel?

[tulip violet]

Yes

[wraith hinge]

Thank you

I appreciate it

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[twilit kraken]

As far as i know, the latter.

I think so.

[verbal sedge]

I feel like they probably are related if they have the same name

I’m not sure exactly how tho but it’s a good question

Yea I don’t really see the relation either but it might be like a geometric interpretation of 1/cos(x) visually looks similar to a secant line

That’s just a guess tho idk but I agree that it seems weird for it be a coincidence

Yea I searching it up and don’t see anything abt it so maybe it is a coincidence but it is very strange

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[uneven parcel]

what are my local max and mins

im so confused

[unique oriole]

Are you learning calc?

[uneven parcel]

no

shes just asking u

us

to identify

them

thats all

[unique oriole]

The local max and min is between the point in between where the slope is 0 at it's highest and lowest respectively

[uneven parcel]

highest point is (0,6)

is that the max?

[unique oriole]

In the desmos one?

[uneven parcel]

ya

[unique oriole]

Yes that is a local max

[uneven parcel]

then the local min

is the one at -54

?

for y

[unique oriole]

there is one more

[uneven parcel]

2??

wait is there like a vertex

local max and min

or no

[unique oriole]

Yes

so the 2 local mins are....

[uneven parcel]

the two at the bottom

y at -15 and -54?

[unique oriole]

Yup

[uneven parcel]

so ur saying theres 2 min

[unique oriole]

Yeah

Cause it’s a local min

[uneven parcel]

yea

okay si (0,6)

is the only local max

[unique oriole]

Yup

[uneven parcel]

kkkkk

i see

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[chilly basin]

how do u do this?

[inland orbit]

,rotate

[chilly basin]

HELP

hello?

[river willow]

expand and simplify the left

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[neon vortex]

Is this correct i also need help on #4?

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[ornate light]

Hey guys, I need help with this integral. Thanks!

[tough iron]

its easy to integrate x/(1+x^2)^2

and its easy to differentiate lnx

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[ornate light]

ok, i got this

what do i do from here?

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[spring sorrel]

try partial fracture decomposition, on the Integral

sry, partial fraction expansion, or just partial fraction

but you could have also started different, by substitution with u = ln(x) , or u = x^2. Both work when done properly

hope that helped

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[ornate light]

alright, thx

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[dim kite]

How would i do b and c?

I tried setting the total force to 0

There are 4 forces

The boom gravitation force, the rod gravitational force, the tension and the pivot force

And then i try finding the pivot force but its wrong

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[lucid raven]

How can I write this series so that it contains z^k instead of (i + z)^k ?

[marble wharf]

well you can multiply it out with the binomial theorem and then rearrange a bit. but it will probably look like shit

what do you want to achieve?

[lucid raven]

i also thought about the binomial theorem, but I was hoping there would be something less ugly...

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[lone mauve]

i'm not sure how to use REF/RREF to get scalars with this question

it says You can use this row reduction tool to help with the calculations. but I'm not following.. anybody know?

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[stoic iron]

If you call the scalars a,b,c,d the equation at the bottom is a linear system of 4 equations

I hope by now you know how to convert a linear system into a matrix in order to row reduce

[lone mauve]

So do I convert this 4x4 matrix into a 4x5 augmented matrix?

and try to solve for 0 on the fifth column?

[stoic iron]

Augmenting it to a 4x5 would be pointless

Your last column would be all zeros

[lone mauve]

yeah

isn't that the goal here? to get all rows equal to 0?

via scalars

[stoic iron]

Sure but no need to write the zero column all the time

Just write it at the end

[lone mauve]

and I cannot use 0 for scalars

w + x -2y + z = 0 3w + 2z = 0 -2w -3y -z = 0 -w -3x + 3y + 2z = 0

[stoic iron]

The point is if the scalars all being 0 is the ONLY solution, then the vectors are independent

[lone mauve]

and the vectors are linearly independent

so i should do that? 0,0,0,0 is the final answer?

[stoic iron]

Solve the system

[lone mauve]

which method would you recommend for checking linear independence?

[stoic iron]

To determine if the only solutions is the scalars being all zero

[stoic iron]

The exact same method

If you get non zero solutions they are dependent

[lone mauve]

but based on this, i already know the vectors are independent. .. so the only solution for scalars has to be 0?

[stoic iron]

How do you already know?

[lone mauve]

i used a calculator to check lol

https://www.emathhelp.net/en/calculators/linear-algebra/linear-independence-calculator/

[stoic iron]

Well you shouldn't really have done that

[lone mauve]

hmmmm

I got it correct, but I will need to review this

[subtle scarab]

Can someone help me with this task pls:
Determine all integers n for which real numbers x,y exist in such a way that the following equations are satisfied:
x+y=n
x³+y³=n²

[stoic iron]

Yeah you kind of missed the whole point of the question by plugging it into a calc

[stoic iron]

!help

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[bitter sleet]

I am studying old work for my AP calc test and I am looking at the awnser to this question to try and work backwards. We are using the derivitives of trig functions to solve these questions and I understand the section of the question highlighted in green (cscx -> -cscxcotx) but do not understand how the yellow one on the left turned into the yellow one on the right.

[bitter sleet]

here is the original problem

[tough iron]

cotangent is cos/sin

also

the csc*1 part is not yellow's part

its greens part of the product rule

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[wraith hinge]

Can someone please tell me if I rewrote this formula correctly?

[wraith hinge]

So that it would become this?

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[echo jasper]

i feel like i get everything

until the part there took ln

why take ln instead of log

it confuse me

[charred trail]

it doesnt really matter what base is taken as long as its consistent

ln is more common thats why its taken i would imagine

[echo jasper]

O

like

if i used log

it would be the same =0

[charred trail]

yes

as long as its the same on both sides of the equation

[echo jasper]

ok

oh man

did something wrongg i think

ohhh

nvm